Q&A Visualizer

Question:

Balance the equation for the following reaction taking place in aqueous acid solution: Cr_2O^2-_7+ I_2 \rightarrow Cr^3++ IO^-_3

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Solution:

The equation in this problem involves both an oxidation and a reduction reaction. It can be balanced by using the following rules: (1) Separate the net reaction into its two major components, the oxidation process (the loss of electrons) and the reduction process (the gain of electrons). For each of these reactions, balance the charges by adding H^+, if the reaction is occurring in an acidic medium, or OH^- in a basic medium. (2) Balance the oxygens by addition of H_2O. (3) Balance hydrogen atoms by addition of H. (4) Combine the two half reactions, so that all charges from electron transfer cancel out. These rules are applied in the following example. The net reaction is Cr_2O^2-_7+ I_2 \rightarrow Cr^3+ + IO^-_3 The oxidation reaction is I^0_2 + 2lO^-_3+ 10e^- The I atom went from an oxidation number of O in I_2 to + 5 in IO^-_3 , because O always has a - 2 charge. You begin with I_2, therefore, 2 moles of IO^-_3 must be produced and 10 electrons are lost, 5 from each I atom. Recall, the next step is to balance the charges. The right side has a total of 12 negative charges. Add 12 H^+'s to obtain I_2 + 2lO^-_3+ 10e^- +12H^+ To balance the oxygen atoms, add 6H_2O to the left side, since there are 6 O's on the right, thus, I_2 + 6H_2O \rightarrow2lO^-_3 + 10e^- + 12H^+. Hydrogens are already balanced. There are 12 on each side. Proceed to the reduction reaction: Cr_2O^2-_7+ 6e^- \rightarrow 2Cr^3+ Cr began with an oxidation state of + 6 and went to + 3. since 2Cr^3+ are produced, and you began with Cr_2O^2-_7 , a total of 6 electrons are added to the left. Balancing charges: the left side has 8 negative charges and right side has 6 positive charges. If you add 14 H^+ to left, they balance. Both sides now have a net + 3 charge. The equation can now be written. Cr_2O^2-_7+ 6e^- + 14H^+ \rightarrow 2Cr^3+. To balance oxygen atoms, add 7H_2O `s to right. You obtain Cr_2O^2-_7+ 6e^- + 14H^+ \rightarrow 2Cr^3++ 7H_2O. The hydrogens are also balanced, 14 on each side. The oxidation reaction becomes I_2 + 6H_2O + 2lO^-_3 + 10e^- + 12H^+ The reduction reaction is Cr_2O^2-_7+ 6e^- + 14H^+ \rightarrow 2Cr^3+ + 7H_2O. Combine these two in such a manner that the number of electrons used in the oxidation reaction is equal to the number used in the reduction. To do this, note that the oxidation reaction has 10e^- and the reduction 6e^-. Both are a multiple of 30. Multiply the oxidation reaction by 3, and the reduction reaction by 5, obtaining oxidation:3I_2 + 18H_2O \rightarrow 6IO^-_3+ 30e^- + 36H^+ reduction:5Cr_2O^2-_7 + 30e^- + 70H^+ \rightarrow 10Cr^3+ + 35H_2O Add these two half-reactions together. 3I_2 + 18H_2O \rightarrow 6IO^-_3+ 30e^- + 36H^+ +5Cr_2O^2-_7 + 30e^- + 70H^+ \rightarrow 10Cr^3+ + 35H_2O 3I_2 + 18H_2O + 5Cr_2O^2-_7 + 30e^- + 70H^+ \rightarrow 10Cr^3+ + 35H_2O + 30e^- + 36H^+ Simplifying, you obtain: 3I_2 + 5Cr_2O^2-_7 + 34H^+ \rightarrow 6IO^-_3 + 10cr^3+ + 17H_2O This is the balanced equation.

Question:

A 25.00-ml sample of sea water is titrated with 2.50 M AgN0_3 solution, using K_2CrO_4, as an indicator, until the presence of excess Ag^+ is shown by the reddish color of Ag_2CrO_4 at the end point. The density of the sea water is 1.028 g/cc. Find the chlorinity and salinity of the sea water if 53.50 ml of AgNO_3 solution were used to complete the precipitation of silver halides.

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Solution:

To solve this problem use the following equation: M_1V_1= M_21V_2, where M = molarity and V = volume. This equation will determine the number of moles of Cl^- consumed in the titration. The equation of the reaction is AgNO_3 + Cl^- (aq) \rightarrow AgCl \downarrow + NO^-_3 (aq) The problem states that 0.0535 liter of AgNO_3 solution at 2.5 M concentration reacts with an unknown Cl^- concen-tration. What actually needs to be known is the total weight of Cl^- in 25.00 ml of sea water. (2.5 moles / liter) (0.0535 liter) = 0.13 moles AgNO_3 However, according to the stoichiometry of the titration reaction, 0.13 moles of Cl^- also is consumed. Therefore, the weight of Cl^- in the sea water is (0.13 moles Cl^-) (35.5 g/mole) = 4.62 g. The total weight of sea water is (25 ml) (1.028 g/ml or cc ) = 25.7 g sea water. Thus, the percentage of Cl^- in sea water is [(4.62 g Cl^-) / (25.7 g sea water)] × 100 = 18.0 % Cl^- The chlorinity of sea water is 18 %. The salinity is determined from the following equation % salinity = 1.805 (Cl %) + 0.03 %. Thus, % salinity= (1.805) (18 %) + 0.03 % = 32.52 %.

Question:

A sample of the poisonous compound nicotine extracted fromcigarette smoke was found to contain 74.0 % by weight ofcarbon (C, atomic weight = 12.0 g/mole), 8.65 % by weightof hydrogen (H, atomic weight = 1.01 g/mole), and 17.3 % by weight of nitrogen (N, atomic weight =14.0 g/mole). What is the empirical formula of nicotine?

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Solution:

The empirical formula states the relative ratio of the atoms of the variouselements in any substance; the actual number of atoms (molecular formula) is not implied. Problems of this sort are solved by assuming a sampleof some convenient mass and then calculating the number of molesof each atom in this sample. Once the number of moles has been calculatedfor each component, these numbers are divided by the greatest commonfactor in order to obtain the proportions in which the various componentsappear in the molecule. There is no error in assuming a sampleof definite mass, since the size of a sample of molecules does not affectthe composition of the molecules. Assume a sample weighing 100 g. Then the masses of C, H, and N in this sampleare: massC = 74.0 % × 100 g = 74.0 g massH = 8.65 % × 100 g = 8.65 g mass N = 17.3 % × 100 g = 17.3 g We convert the mass of each element to the corresponding number of molesby dividing by the atomic mass of that element. Thus we obtain the followingnumber of moles of each element: molesC = (74.0 g) / (12.0 g/mole) = 6.2 moles molesH = (8.65 g) / (1.01 g/mole) = 8.6 moles molesN = (17.3 g) / (14.0 g/mole) = 1.2 moles The greatest common factor is 1.2 moles. Dividing the number of moles of each element by 1.2 moles, we obtain (molesC) / (1.2moles) = (6.2moles) / (1.2moles) \allequal 5. (molesH) / (1.2moles) = (8.6moles) / (1.2moles) \allequal 7. (molesN) / (1.2moles) = (1.2moles) / (1.2moles) \allequal 1. Hence, C, H, and N appear in the proportions 5, 7, and 1, respectively. The empirical formula of nicotine is thus C_5H_7N.

Question:

Theglomerularfiltration rate is controlled primarily by alteration of theglomerularcapillary pressure. What are the two factors which produce these changes?

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Solution:

Theglomerularfiltration rate is defined as the volume of blood filtered by theglomerulusper minute. One of the factors expected to influence theglomerularfil-tration rate is the arterial blood pressure. A decrease in arterial pressure would decrease the filtration rate by low-ering glomerularcapillary pressure. An increase in arterial pressure would serve to increase theglomerularfiltration rate. However, GFR stays relatively constant de-spite changes in mean arterial pressure. This is due to an intrinsic ability of the kidneys toautoregulate. The direct factor influencing filtration is the diameter of the affer-ent arterioles. A decrease in the diameter of the afferent arterioles lowers blood flow to theglomerulus, loweringglomerularcapillary pressure and filtration rate. Con-versely, an increase in the diameter of the afferent arteri-oles increases the volume of blood flow to theglomerulus, increasingglomerularcapillary pressure and filtration rate. The main stimulus that controls afferent arteriolar size is the mean arterial pressure (MAP) . When the MAP falls, the afferent arteriole dilates to maintain GFR. The converse occurs when the MAP increases.

Question:

John Dalton found water to be 12.5% hydrogen by weight. Calculate the atomic weight of oxygen if Dalton assumed water contained two hydrogen atoms for every three oxygen atoms. Assume 1 H atom weighs 1amu.

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Solution:

This problem can be solved if one sets up a ratio. If it is assumed water is 12.5% hydrogen by weight, then water is 100-12.5 or 87.5% oxygen by weight. Dalton assumed that there are two hydrogen atoms for every three oxygen atoms. Therefore, 12.5% / 18.5% = 1 / 7 = (2 H'S) / (3 O'S). It is given that 2 H atoms weight 2amu. Thus, 1 / 7 = 2amu/ (3 × atomic weight of O). Solving for the atomic weight of O, atomic weight of O = (2amu× 7) / 3 = 4.67amu.

Question:

To what tension must a brass wire of cross-sectional area 10^\rule{1em}{1pt}2 cm^2 be subjected so that the speed of longitudinal and transverse waves along it may be the same? Young's modulus for brass is 9.1 x 10^11 dynes\textbulletcm^\rule{1em}{1pt}2. Is this situation physically realizable?

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Solution:

The speed of transverse waves in the wire is \surdS/\mu, and of longitudinal waves \surd(Y/\rho), where Y is the Young's modulus of brass, S is the tension in the wire, \rho is the density of brass, and \mu is the mass per unit length of the wire. In this problem we require that \surd(S/\mu) = \surd(Y/\rho). \therefore S = (\mu/\rho) Y = (mass per unit length/ mass per unit volume) Y = AY where A is the cross-sectional area of the wire. Therefore S = 10 ^-2 cm^2 × 9.1 × 10^11 dynes\textbulletcm^\rule{1em}{1pt}2 = 9.1 × 10^9 dynes Since the formula for Young's modulus is Y = (S/A)(\Deltal/l), then S = AY implies\Deltal/l =1. In other words, the wire must stretch by an amount equal to its length l. The elastic limit would have been passed long before this point. The situation is therefore physically unrealizable and longitudinal waves will always travel faster than transverse ones in the wire.

Question:

Radioactive material spontaneously disintegrates at a rate that depends on the amount of material that remains. That is, a given fraction of . the remaining material disintegrates each day. Write a FORTRAN program to simulate this system, using the modified Euler's method, from time t = 0 to time t = t_f if the amount of material remaining at time t is N(t); N(0) = N^f_0; and P is the percentage (per unit of time) of the remaining material that decays. Compare the modified Euler approximations with the exact values.

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Solution:

If Ṅ is the rate of disintegration the behavior of the system can be diagrammed as in Fig. 1. The amount of material that remains is a result of past disintegra-tion, and the present decay rate is a negative percentage of the amount of remaining material. Stating this mathematically: Ṅ = (-P/100)\textasteriskcenteredN(t)(1) The exact solution of equation (1) is: N(t) = N_0e^-Pt/100(2) Note that, as t \ding{217} \infty , N(t) \ding{217} 0 as expected since the "goal" of this system is complete disintegration of the material. The program uses the variable AMOUNT for N(t). DATA T/0.0/ COMMON P READ,N,TFIN,AOCUR,P,NO PRINT,T,NO REALN = N DT = TFIN/REALN AMOUNT = NO DO 10 I = 1,N T = T + ET CALL MEULER (T,AMOUNT ,ACCUR, DT) EXACT = N0\textasteriskcenteredEXP(-P\textasteriskcenteredT/100.) ERROR = ABS ( (EXACT-AMOUNT)/EXACT) \textasteriskcentered100. PRINT, T, AMOUNT, EXACT, ERROR 10CONTINUE STOP END FUNCTION G(W) COMMON P G = (-P/100)\textasteriskcenteredW RETURN END

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Question:

Acyclindercontaining gas at 27\textdegreeC is divided into two parts of equal volume, each of 100 cm^3, and at equal pressure, by a piston of cross-sectional area 15 cm^2 . The gas is one part is raised in temperature to 100\textdegreeC; the other volume is maintained at the original temperature. The piston and walls are perfect insulators. How far will the piston move during the change in temperature?

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Solution:

The heating of one side of the cylinder increases the pressure of the gas in that portion. If the piston were fixed, the volumes on the two sides would stay equal and there would be a pressure difference across the piston. Since the piston is movable, it alters its position until there is no pressure difference between its two sides: the hotter gas expands and thus drops in pressure, and the cooler gas is compressed and thus increases in pressure. When equilibrium has been reached, the two pressures are equal at p_0; the cooler gas now occupies a volume smaller by an amountdVand the hotter gas a volume greater by a corresponding amountdV. The ideal gas equations for both compartments are p_0(V -dV) =nRT(1) p_0(V +dV) =nRT' where n is the number of moles in each compartment, T and T are the temperatures. Dividing (1) by (2), we get (V -dV)/(V +dV) = T/T' VT' -dVT' = VT +dVT V(T' - T) =dV(T + T') or(dV/v) = (T' - T) / (T' + T) Therefore, the change in the volume of each compartment is dV= [{(373 - 300)\textdegreeC}/ {(373 + 300)\textdegreeC}] × 100 cm^3 = 10.85 cm^3. The piston has an area of 15 cm^2. Hence it moves a distance of 10.85 cm^3/15 cm^2 = 0.723 cm.

Question:

A space probe explodes in flight into three equal por-tions. One portion continues along the original line of flight. The other two go off in directions each inclined at 60\textdegree to the original path. The energy released in the explosion is twice as great as the kinetic energy possess-ed by the probe at the time of the explosion. Determine the kinetic energy of each fragment immediately after the explosion.

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Solution:

Take the direction in which the probe is moving immediately prior to the explosion as the positive x-direction and the point at which the explosion takes place as the origin of coordinates. Let M be the mass of the probe and let the quantities applicable to the probe and its fragments be given subscripts as in the diagram. Momentum must be conserved in the x-direction, the y-direction, and the z-direction independently. It follows that V_1^\ding{217}, V_2^\ding{217}, V_3^\ding{217}, and V^\ding{217} must be coplanar and that MV = (1/3) MV1+ (1/3) MV_2 cos 60\textdegree+ (1/3) MV3cos 60\textdegreeand (1/3) MV2sin 60\textdegree= (1/3) MV_3 sin 60\textdegree. From the second of these equation V_3 = V_2, and thus the first equation becomes MV = (1/3) MV_1 + (1/3) MV_2 cos 60\textdegree+ (1/3) MV_2 cos 60\textdegree = (1/3) MV_1 + (2/3) MV_2 cos 60\textdegree = (1/3) MV_1 + (2/3) MV_2(1/2) = (1/3) MV_1 + (1/3) MV2 ButE = (1/2) MV^2 = (M^2V^2)/(2M)orMV = \surd2ME. Similarly, E_1 = (1/2) [(1/3) M]V_1^2 = (M^2V2_1) / [(3)(2M)] = [(3 M^2V2_1) / (2M)](1/9)or (1/3) MV1= \surd[(2/3) ME_1] and(1/3) MV2= \surd[(2/3) ME_2] The first equation thus becomes \surd(2ME) = \surd[(2/3) ME_1] + \surd[(2/3) ME_2] or \surd(3E) = \surd(E_1) + \surd(E_2). \therefore3E = E_1 + E_2 + 2\surd(E_1E_2). The original kinetic energy of the probe plus the energy released by the explosion must equal the sum of the kinetic energies of the fragments, since no energy can be lost in the process. (That is, we assume an elastic "collision" occurs.) Hence E + 2E = 3E = E_1 + E_2 + E_3 = E_1 + 2E_2. \thereforeE1+ 2E_2 = E_1 + E2+ 2\surd(E_1E_3)orE_2 = 2\surd(E_1E_2). \thereforeE2_2 = 4E_1E_2orE_2 = 4E_1. Thus, 3E = E_1 + 2E_2 = E1+ 8E_1orE_1 = (1/3) E. \therefore E_2 = (4/3) E. Thus the fragment that continues in the line of flight has one-third of the original kinetic energy. The other fragments each have four-thirds of the original kinetic energy. The sum of these kinetic energies is three times the original kinetic energy, as required by the conserva-tion principle.

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Question:

Determine the weights of CO_2 and H_2 O produced on burning 104 g. of C_2 H_2. Molecular weights are CO_2 = 44 and H_2 O = 18. The equation for the reaction is 2C_2 H_2 + 5O_2 \rightarrow 4CO_2 + 2H_2 O

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Solution:

Two methods can be used to solve this problem. One is called the mole method and the other the proportion method. Mole method :According to the equation, 2molesof C_2 H_2 react with 5 moles of O_2 to produce 4 moles of CO_2 and 2 moles of H_2 O. Thus, in this problem, the first thing one has to determine is how many moles of C_2 H_2 are contained in 104 g. The molecular weight of C_2 H_2 is 26, by dividing the amount of C_2 H_2 present by the molecular weight of the com-pound, the number of moles present is found. number of moles of C_2 H_2 = [(number of grams of C_2 H_2) / (molecular weight of C_2 H_2)] number of moles of C_2 H_2 present = [(104 g) / (26 g/mole)] = 4 moles. In the equation 2 moles of C_2 H_2 are burned to form moles of CO_2 and 2 moles of H_2 O. Here there are 4 moles of C_2 H_2 present, which is twice the amount in the empirical equation. Therefore, twice as much CO_2 and H_2 O will be formed. 8 moles of CO_2 and 4 moles of H_2 O will be formed. 2 × (2C_2 H_2 + 5O_2 \rightarrow 4CO_2 + 2H_2 O) = 4C_2 H_2 + 10O_2 \rightarrow 8CO_2 + 4H_2 O To find the weight of 8 moles of CO_2, the molecular weight of CO_2 is multiplied by 8. weight of CO_2 = number of moles present × molecular weight of CO_2. weight of CO_2 = 8 moles × 44 g/mole = 352 g. The same procedure can be followed for H_2 O. weight of H_2 O = number of moles present × molecular weight of H_2 O weight of H_2 O = 4 moles × 18 g/mole = 72 g. Proportion method: An alternate method of solution to this problem is the proportion method in which molecular weights (multiplied by the proper coefficients) are placed below the formula in the equation and the amounts of substances (given and unknown) are placed above. Here one has for CO_2: 104 gX 2 C_2 H_2 + 5O_2 \rightarrow 4 CO_2 + 2 H_2 O 2 × 26 g4 × 44 g This becomes (104 g) / (52 g) = (X/176 g) Solving for X one has X = [(104 g) (176 g)] / [52 g] = 352 gr gr CO_2 CO_2 A similar method can be applied to the H_2 O 104 gX 2 C_2 H_2 +5O_2 \rightarrow4CO_2 +2 H_2 O 2 × 26 g2 × 18 g (104 g) / (52 g) = (X/36 g) X = [(104 g) (36 g)] / [52 g] = 72 g H_2 O.

Question:

A bicycle and its rider together weigh 200 lb. If the cyclist free-wheels down a slope of 1 in 100, he has a constant speed of 10 mph, and if he free-wheels down a slope of 1 in 40, he has a constant speed of 20 mph. Suppose that he free-wheels on the level while holding on to the back of a moving truck. Find the power ex-pended by the truck in maintaining his speed at 15 mph. Assume that air resistance varies as the square of his speed, while frictional forces remain constant at all times.

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Solution:

Let the frictional force be F^\ding{217} and the force of air resistance F'^\ding{217} with magnitude kv^2, where k is a constant and v is the speed of the bicycle. On a slope, the forces acting on the bicycle and rider are the weight W^\ding{217} acting downward, which can be resolved into components parallel to and perpendicular to the slope, the normal force exerted by the slope N^\ding{217}, and the forces of friction and air resistance act-ing up the slope opposing the motion. The forces per-pendicular to the slope are equal and opposite and are of no further interest. Since the bicycle is moving with constant speed, the forces parallel to the slope must also cancel out. Hence W sin \texttheta = F + F' = F + kv^2. For the two cases given, values can be inserted. Thus 200lb × (1/100) = F + k × 10^2 mi^2 \bullet hr^-2and 200lb × (1/40) = F + k × 20^2 mi^2 \bullet hr^-2. \therefore2 lb = F + 100k mi2\bullet hr^-2and 5 lb = F + 400k mi2\bullet hr^-2. \therefore300k mi^2 \bullet hr^-2 = 3 lbor k = 1/100 lb \bullet(mph)^-2 and F = 1 lb. For the case of the bicycle traveling on a level surface, a force P^\ding{217} must be supplied to overcome the forces of friction and air resistance and keep the bicycle moving with constant speed. Since there is no acceleration, P^\ding{217} must just balance F^\ding{217} and F'^\ding{217} , or P = F + kv^2 = 1lb + (1/100) lb \bullet (mph)^-2 × 225 (mph)^2 = 3.25 lb. The rate of working (the mechanical power) is P^\ding{217} \textbullet v^\ding{217} and v = 15 mph = 22 ft \textbullet s^-1. \therefore P × v = 3.25 lb × 22 ft \textbullet s^-1. But 1 hp = 550 ft \textbullet lb \textbullet s^-1 Therefore: The rate of working = [(3.25 × 22 ft \textbullet lb \textbullet s^-1)/(550 ft \textbullet lb \textbullet s^-1/hp)] = [(3.25 × 22)/550] hp = 0.13 hp.

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Question:

Calculate the velocity of a transverse pulse in a string under a tension of 20 1b if the string weighs 0.003 1b/ft.

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Solution:

Consider the section of string shown in figure A, which is under the tension T. This segment can be interpreted as being part of the pulse being trans-mitted along the string. The net force acting in the y direction is F_y = T sin \textthetax + dx\rule{1em}{1pt} T sin \texttheta_x Since \texttheta_x and \texttheta_x _+ _dx are small, we have approximately, sin \texttheta_x \approx \texttheta_x \approx tan \texttheta_x sin \textthetax + dx\approx \textthetax + dx\approx tan \textthetax + dx Therefore, F_y = T(tan \textthetax + dx\rule{1em}{1pt} tan \texttheta_x) buttan \textthetax + dx= (\partialy/\partialx)x + dx tan \texttheta_x = (\partialy/\partialx)_x(See fig. B) \partialy and \partialx have the same meaning as dy and dx, they are small increments of y and of x, respectively. The \partial symbol indicates that more than one variable is under consideration. Hence, F_y = T [(\partialy/\partialx)x + dx\rule{1em}{1pt}(\partialy/\partialx)_x](1) Now, by Taylor's theorem, given a continuous and differentiable function G(x), G(x + dx) = G(x) + (\partialG/\partialx) dx + ... G(x + dx) \rule{1em}{1pt} G(x) \approx (\partialG/\partialx) dx(2) Higher order terms have been neglected because dx is so small that terms involving (dx)^2 or higher must ne-cessarily be negligible. If G(x) = (\partialy/\partialx), then we have from (2) (\partialy/\partialx)x + dx- (\partialy/\partialx)_x = (\partial/\partialx) (\partialy/\partialx) dx = [(\partial^2 y)/(\partialx^2)] dx Hence, (1) become F_y = T [(\partial^2 y)/(\partialx^2)] dx Denote the mass per unit length of the string by \mu (i.e. \mu = dm/dx). The mass of the segment dx is then m = \mudx. By Newton's second law, the vertical accelera-tion of the string segment, a_y, is F_y = ma_y T [(\partial^2y)/(\partial x^2)] dx = (\mudx) [(\partial^2 y)/(\partial t^2)] [(\partial^2 y)/(\partial x^2)] = (\mu/T) [(\partial^2 y)/(\partial t^2)](3) Equation (3) is of the form of the wave equation [(\partial^2 y)/(\partial x^2)] = (1/v^2) [(\partial^2 y)/(\partial t^2)] where y(x, t) is the displacement of the wave at position x and at time t, and v is the velocity of the wave. For the transverse wave, then v = \surd(T/\mu) Returning to our original problem, T = 20 1b, \mu = (0.003/32) {(1b/ft)/(ft/sec^2)}, v = \surd[(20 1b × 32 ft/sec^2)/(0.003 1b/ft)] = 461 (ft/sec).

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Question:

What are the rules of Assembler Language? Explain them in detail with an illustrative example.

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Solution:

A programming language is defined by a set of rules. Almost every commercial computer has its own parti-cular assembly language. The basic unit of an assembly language program is a line of code. The general rules for writing a mnemonic assembler language for the basic computer are given below. Rules of the language - each line of an assembly lan-guage program is arranged in three fields. The fields spec-ify the following information . address . pseudo-instruction . In the label field a symbolic address consists of one, two, or three, but not more than four alphanumeric charac-ters. The first character must be a letter; the next two may be letters or numerals. A symbolic address in the la-bel field is terminated by a comma so it will be recognized by the assembler as a label. The instruction field in an assembly language program specifies one of the following items. A memory-reference instruction occupies two or three symbols separatedby spaces. The first must be a three-letter symbol defining an MRI operation.The second is a symbolic address. The third symbol, which mayor may not be present, is the letter I, depending on direct or indirect addressinstruction situation. A non-MRI is defined as an instruction that does not have an address part. A non-MRI is recognized in the in-struction field of the program by one three-letter symbol. A pseudo-instruction is not a machine instruction but rather an instruction to the assembler giving information about some phase of the translation . The following is an illustration of the symbols that may be placed in the instruction field of program. CMAnon MRI ADDOPRdirect address MRI ANDPTR Iindirect address MRI ORGpseudo instruction A memory reference instruction, such as ADD, must be followed by a symbolic address. The letter I may or may not be present. A symbolic address in the instruction field specifies the memory location of an operand . This location must be defined somewhere in the program by appearing again as a label in the first column; it is absolutely necessary that each symbolic address that is mentioned in the instruction field must occur again in the label field. The third field in a program is reserved for comments. It must be preceeded by a slash for the assembler to re-cognize the beginning of a comment field. Example 1 ORG 100 /ORIGIN OF PROGRAM IN LOCATION LDA SUB /LOAD SUBTRAHEND TO AC (MRI) CMA /COMPLEMENT AC. (NON MRI) INC /INCREMENT AC. (NON MRI) ADD MIN /ADD MINUEND TO AC (MRI) STA DIFF /STORE DIFF (MRI) HLT /END OF INSTRUCTIONS MIN, DEC 50 /MINUED. MIN IS LABEL SUB, DEC -10 /SUBTRAHEND DIFF, HEX 0 /DIFFERENCE STORED HERE END /END OF SYMBOLIC PROGRAM

Question:

What are some factors that interfere with our obtaining a completeand unbiased picture of life in the past from a study ofthe fossil strata?

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/Users/wenhuchen/Documents/Crawler/Biology/F28-0731.htm

Solution:

There are five major rock strata in the earth's crust, each of which is subdividedinto lesser strata. Theoretically, because the strata were formedby the chronological deposition of sediment, the charac-teristics of eachancient era or period should become evident upon examining the fossilcomposition of each stratum and substratum. Also, one should be ableto estimate the duration of each era or period from the thickness of eachlayer and the known rate of sediment deposition. Knowledge of the fossil strata has been useful in the past and is still beingused today in those cases when radioactive dating is not applicable. In relative dating, a time or era can be assigned to a fossil or ar-tifact relativeto other fossils or artifacts of known age in otherstata. But this viewis oversimplified. The relative length of each ancient epoch cannot, in reality, be determined in great accuracy since the rate of sedi-mentary depositionvaried at different times and in different places. While the layers ofsedimentary rock should occur in the sequence of their deposition, with thenewer, later strata on top of the older, earlier ones, geologic disturbancesat some point of history may have changed the order and relationshipof the layers. It is known that certain sections of theearth's crusthave undergone tremendousfoldingsandsplittings, and old layers maynow cover new ones. In addition, in some regions, the strata formed previouslyhave emerged from the sea due to the continuous addition of newlayers; the top layers were then very easily carried away by running wateror other natural disturbances, so that re-latively recent strata were depositeddirectly on top of very ancient ones. Such factors make it difficultfor us to obtain a complete, accurate picture of the ancient ages in chronological order.

Question:

If a disk is rotating at 1200 rev/min, what torque is required to stop it in 3.0 min?

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/Users/wenhuchen/Documents/Crawler/Physics/D04-0184.htm

Solution:

If the disk is to decelerate from 1200 rev/min to 0 rev/min uniformly, then the angular acceleration (\alpha) will be constant. Hence \alpha = constant. But \alpha = (d\omega)/(dt) where \omega is the angular velocity of rotation. Therefore, \alpha = (d\omega)/(dt) ord\omega=\alphadt ^\omega\int_\omega0d\omegat=t\int_t_=0\alphadt \int \omega - \omega_0 =\alphat(1) where \omega = \omega at t = t and \omega_0 is the initial angular velo-city of rotation. \omega_0 = 1200 rev/min. Since 1 rev/min = (1/60) rev/sec \omega_0 = 20 rev /sec. But 1 rev = 2\pi radians and \omega_0 = 40\pirad/sec. t = 3.0 min = 180 sec Substituting in (1) 0 - 40\pirad/sec = \alpha(180 sec) \alpha = [(-40\pi)/(180)]rad/sec^2. This is the acceleration which must be applied to the disk if it is to come to rest in the required time. Because the disk is rotating about a fixed axis, the torque L is the product of the angular acceleration of the disk and the moment of inertia of the disk about the axis of rotation. L =I\alpha = (38 slug-ft^2)[-{(40\pi)/(180)}rad/sec^2] = -26 lb-ft. Hence, a torque of -26 lb-ft must act on the disk in order to bring it to rest in 3 minutes from a velocity of 1200 rev/min. The negative sign is consistent with a retarding torque.

Question:

For a cell reaction of the type T + R^+2 \leftrightarrows R + T^+2 , calculate the value that E\textdegree must have so that the equilibrium concentration of T^+2 is a thousand times the R^+2 concentration.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E16-0602.htm

Solution:

In this question, we are asked to obtain the standard cell potential (E\textdegree), not the actual cell potential (E). Since there is no net electron flow at equilibrium, E = 0. There exists an expression which relates E\textdegree , E and concentrations. This expression, called the Nernst equation, states E = E\textdegree - (.059/n) log Q, at a temperature of 25\textdegreec where n = number of electrons trans-ferred and Q = mass action expression. From the chemical reaction, you see T goes from an oxidation state of 0 to +2 . It lost 2 electrons. Thus, R^+2 gained 2 electrons. Therefore, the number of electrons transferred, n, equals 2. From the same chemical reaction, Q becomes [T^+2] / [R^+2] . You are asked to determine [T^+2] to be 1000 times [R^+2], which means Q = 1000. You already determined that E = 0. To find E\textdegree , substitute these values in the Nernst equation. You obtain 0 = E\textdegree - (0.059 / 2) In 1000. Solving, E\textdegree = .088 volt.

Question:

Give a brief summary of various types of operating systems.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G01-0022.htm

Solution:

Batch: Runs a collection of jobs sequentially with little or no interaction with the user. Good for large jobs such as automatic generation of monthly reports. Interactive : Composed of many short transactions where the result of the next transaction depends on the current one. The user is informed of the result (normal completion, error) at the end of each transaction and takes the appropriate steps accord-ingly. Time-shared : Uses multiprogramming and CPU scheduling where each job (or process) is given a small time quantum to allow interactive use of the computer by multiple users. Usually, a command processor is provided (Job Control Language-JCL, Shell, Digital Command Language- DCL) for interpreting the commands entered from the terminals or consoles . Real-Time : The system must respond (a program must complete the execution ) within a predetermined (fixed) amount of time for correct function . Often used for dedicated (non time-shared) applications with priorities assigned to jobs as required. Distributed : Distributes the computational load among several CPUs (processors) that do not share common clock or memory. Processors communicate with each other using high-speed buses within a small area (aroom or cabinet) in tightly-coupled sys-tems. The communication takes place via telephone lines or net-work in loosely-coupled systems.

Question:

Consider a sphere of radius a which has a charge q evenly distributed on its surface. What is the electric field outside the sphere?

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/Users/wenhuchen/Documents/Crawler/Physics/D17-0568.htm

Solution:

Due to the fact that the charge is uniformly distributed over the surface of the sphere (and therefore symmetric), we realize that the electric field must have the same strength at any point a distance r away from the center of the charged sphere. Furthermore, we expect the lines of E^\ding{217} to begin on the positive surface charges and emanate radially from the surface of the sphere. Consider a spherical surface of radius R and with the same center as the charged sphere. If the field at this surface is E, then the total electric flux through the surface is \varphi_E = \intE^\ding{217} \textbullet dS^\ding{217} = \int E dS because E^\ding{217} and dS^\ding{217} are parallel. Hence, \varphi_E = 4\piR^2E where 4\piR^2 is the surface area of the spherical surface. Since the charge enclosed by the surface is q, then by Gauss's Law, \varphi_E = q/\epsilon_0 . Therefore, 4\piR^2E = q/\epsilon_0 from which E = [1/(4\pi\epsilon_0)] (q/R^2) = K_-E (q/R^2) Thus the field outside the sphere is just as if all the charge q were located at the center of the sphere.

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Question:

A sports car weighing 1200 lb and traveling at 60 mph fails to stop at an intersection and crashes into a 4000-lb de-livery truck traveling at 45 mph in a direction at right angles to it. The wreckage becomes locked and travels 54.7 ft before coming to rest. Find the magnitude and direction of the constant force that has produced this deceleration.

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/Users/wenhuchen/Documents/Crawler/Physics/D06-0317.htm

Solution:

Let the sports car be traveling in the positive x-direction and the truck in the positive y-direction. After the collision at the origin, the combined mass travels in a direction inclined at \texttheta\textdegree to the positive x- axis with a velocityV. Momentum must be conserved in both the x- and y- directions. Therefore, referring to the diagram, m_1v_1 = (m_1 + m_2)V sin \texttheta(1) andm_2v_2 = (m_1 + m_2)V cos \texttheta(2) Dividing equation (1) by (2), [(m_1 + m_2) V sin \texttheta] / [(m1+ m_2) V cos \texttheta] = (m_1v_1)/(m_2v_2) tan \texttheta = (m_1v_1)/(m_2v_2) = (m_1gv_1)/(m_2gv_2) = (4000 Ib × 45 mph)/(1200 Ib × 60mph) = 2.5 \therefore \texttheta = 68.2\textdegree Squaring equations (1) and (2) and then adding them, we get (m_1 + m_2)^2 v^2 sin^2 \texttheta + (m_1 + m_2)^2 V^2 cos2\texttheta = m^2 _1v^2 _1 + m^2 _2v^2 _2 V2( sin^2 \texttheta + cos^2 \texttheta) = V^2 = [(m^2 _1v^2 _1^ + m^2 _2v^2 _2)] / [(m_1 + m_2)^2] = (m^2 _1g^2v^2 _1 + m^2 _2g^2v^2 _2) / (m_1g + m_2g)^2 = [(4000 Ib)^2 × (45 mph)^2 + (1200 Ib)^2 × (60mph)^2] / [(4000 Ib + 1200 Ib)^2] = 1389.9 (mph)^2. \therefore V = 37.3 mph = 54.7 ft/s, which is the velocity of the combined mass immediately after impact. The wreckage comes to rest in 54.7 ft. Apply the equation for constant acceleration, v^2 = v^2 _0 + 2as. Here V_0 is the initial velocity of the locked mass, s is the distance it travels, and a is the applied acceleration. Hence, when v = 0, s = 54.7 ft and 0 = (54.7 ft/s)^2 + 2a × 54.7 ft. \thereforea = - 27.35 ft/s^2. The deceleration due to friction is thus 27.35 ft/s^2, and since the mass affected is (1200 + 4000)/32 slugs, the magnitude of the frictional force is F = [(5200)/(32)] slugs × 27.35 ft/s^2 F = [(5200)/(32)] slugs × 27.35 ft/s^2 = 4443 l = 4443 l b. This decelerating force must act in a direction opposite to that in which the wreckage is traveling in order to bring it to rest. Thus it is a force of 4443 lb acting at an angle of 68.2\textdegree to the negative x-axis. Note that momentum is conserved only during the collision, for, at that time, the collision forces are much greater than the external forces acting (friction), and the latter may be neglected.

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Question:

Most aquatic animals have a special sensory system used fordetecting water vibrations. Describe the anatomy of this system.

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/Users/wenhuchen/Documents/Crawler/Biology/F13-0331.htm

Solution:

This system, peculiar to most aquatic vertebrates, consists of sensorycells arranged in a linear fashion. In cyclostomes, amphibians, andsome other forms, the lines of sensory cells are exposed on the body surface. Usually they are embedded in canals, which may remain open as agroove (e.g., in primitive sharks), or more often are closed over, with poresat intervals. In most bony fish, the canals run through the scales. These canals are the canals of the lateral-line system and consist of the mainlateral-line canal and four main branches on the head. The head canalsare supplied by branches accompanying the facial nerve. The lateral-linesystem is found in cyclostomes, aquatic larvae of amphibians, andin most other fishes. The sense organs of the lateral line system consist ofneuromasts, whichare clusters of sensory and supporting cells. The sensory cells have aterminal hair projecting into the lumen of the canal and are innervated by nervesof the system. The sensory cells functionally resemble those of the internalear in terrestrial vertebrates.

Question:

Calculate the frequency of oscillation of the configura-tion shown in the figure. All surfaces are frictionless.

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/Users/wenhuchen/Documents/Crawler/Physics/D09-0365.htm

Solution:

When the mass m moves, one spring is always stretched, and the other is always compressed by the same length. Thus: ∆x_1 = -∆x_2 where ∆x_1 is the distance that the spring with force constant k_1 stretches, and ∆x_2 that of the other spring. Note that a negative stretching distance represents a distance compressed. We denote the distance of the mass to the right of the origin 0 by ∆x. Thus: ∆x_1 = ∆x ∆x_2 = -∆x taking positive displacement as pointing to the right. The force on the mass at any time is therefore: F= -k_1 ∆x_1 - (-k_2 ∆x_2) = -k_1 ∆x + k_2 (-∆x) = -(k_1 + k_2) ∆x = -k' ∆x where we let k' = k_1 + k_2 Since the frequency of an oscillator having force constant k ׳ is: \nu = 1/(2\pi)\surd(k'/m) then:\nu = 1/(2\pi)\surd{(k_1 + k_2)/m}

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Question:

store the product in Y.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G20-0491.htm

Solution:

First remember that the sum of two vectors is obtained by A + B = (A_1 + B_1 , A_2 + B_2,....,A_N + B_N), where N is the total number of vector elements. The scalar (or dot) product of vector A by scalar D is D\bulletA = (DA_1, DA_2, DA_3,...,DA_N). Finally, the length of a scalar is written as \vertA \vert =\surd(A^2_1 + A^2_2 + A^2_3 + .....+A^2_N). All three calculations can be made in one program segment: CFIND VECTOR SUM, SCALAR PRODUCT, SCALAR NORM ALENG =0.0 DO 20 I = 1,N X(I) = A(I) + B(I) Y(I) = D\textasteriskcenteredA(I) ALENG = ALENG + A(I)\textasteriskcentered\textasteriskcentered2 20CONTINUE ALENG = SQRT (ALENG) WRITE (5,100) X(I),Y(I),ALENG 100FORMAT (1X,'VECTOR SUM =', F10.4, 'SCALAR PRODUCT =', Note that N should be defined in the program before the above segment occurs.

Question:

Two balloons at the same temperature of equal volume and porosity are each filled to a pressure of 4 atmos-pheres, one with 16 kg of oxygen, the other with 1 kg of hydrogen. The oxygen balloon leaks to a pressure of 1/2 atmosphere (atm) in 1 hour. How long will it take for the hydrogen balloon to reach a pressure of 1/2atm?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E03-0099.htm

Solution:

This problem deals with effusion, the escape of molecules through a hole, and is therefore an application of Graham's Law: Rate of effusion of a gas is inversely proportional to the square root of its density rate_1 / rate_2 = \surdd_2 / \surdd_1 From this we know that large molecules effuse less rapidly than small molecules, since they move more slowly. This allows us to predict that hydrogen effuses more rapidly than oxygen. By applying Graham's Law, we can determine how much faster. rate_hydrogen/rate_oxygen = \surd16 kg per unit volume / 1 kg per unit volume = 4. There, hydrogen effuses 4 times more quickly, which is in agreement with our prediction. Since it takes 1 hour for oxygen to leak to a pressure of 1/2atm, it requires 1/4 hr. = 15 min for hydrogen to decrease to the same pressure inside the balloon.

Question:

What controls the process of molting incrustacea?

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/Users/wenhuchen/Documents/Crawler/Biology/F12-0315.htm

Solution:

Most crustaceans have successive molts. With each molt, they go through a gradual series of immature, larval stages, until they finally reach the body form characteristic of the adult. The lobster, for example, molts several times early in life, and at each molt it gets larger and resembles the adult more. After it reaches the adult stage, additional molts with only changes in size allow growth of the animal to continue. Molting is under the control of a hormone released from nerve cells. Secretion by nerve cells rather than glands, is known as neurosecretion . An "antimolting" hormone is produced by a cluster of nerve cell bodies located near the eyestalk, called the X-organ. The axons of these cells are expanded at their tips, and it is in these expanded tips that the hormone is stored. The bundles of axons together constitute the sinus gland (note that the sinus gland is actually misnamed for it is not a glandular tissue) . Proof that the sinus gland secretes a hormone inhibiting molting comes from surgical removal of this gland, which resulted in repeated molting by the crustacean. Theantimoltinghormone prevents the secretion of & hormone by the Y-organ, a gland composed ofectodermalcells located at the base of themandibularmuscles. When the sinus gland hormone falls below a certain level in the blood, the Y-organ is stimulated to release its hormone. This hormone initiates the event of molting. Just before molting, epidermal cells, stimulated by the Y-organ hormone, secrete enzymes which digest the chitin and proteins of the inner layers of the old cuticle. A soft, flexible, new cuticle is then deposited under the old one, with folds to allow for growth. At this point, the crustacean seeks a protected retreat or remains in its burrow. In order to rid itself of its old shell, a crustacean will take in enough water to swell its body up to three to four times its normal size. The resulting pressure which builds up on the old, rigid shell causes it to burst. The underlying new shell, how-ever, is pliable enough to accommodate the larger size. Eventually, theintaken water is replaced by the growth of new tissue. Water and air are taken in and the old skeleton swells up and bursts, as he new skeleton expands. The animal then extricates itself from the old skeleton. The epidermis secrete enzymes to harden the new cuticle by oxidizing some of the compounds and by adding calcium carbonate to the chitin. The new exoskeleton is com-pleted after subsequent secretion of additionalcuticulelayers.

Question:

The female menstrual cycle last roughly 28 days. Trace the The female menstrual cycle last roughly 28 days. Trace the ovarian and hormonal changes which occur during a normal menstrual cycle.

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/Users/wenhuchen/Documents/Crawler/Biology/F21-0556.htm

Solution:

Under the influence of primarily FSH from the pituitary, a single ovarian follicle containing an ovum reaches maturity in about two weeks. During the second week the follicle cells and other ovarian cell types are stimulated by FSH to increase their secretion of estrogen. Near the middle of the cycle, or about the fourteenth day, the heightened level of estrogen in the blood reaches a critical value which stimulates the hypothalamus to signal the pituitary to release a surge of LH. This surge of LH induces rupture of the ovarian follicle. This releases the mature ovum into the fallopian tube of the uterus, a process called ovulation. Estrogen secretion decreases for several days following ovulation, perhaps because of the negative feedback effect of excess estrogen on FSH, which stimulates its production. The ruptured follicle is rapidly transformed into the corpus luteum, which secretes progesterone and, in lesser amounts, estrogen. The estrogen secreted by the corpus luteum raises its level in the blood, which inhibits further FSH secretion. Progesterone suppresses additional LH production, and thus prevents additional ovulation from occurring. The combined influence of progesterone and estrogen thickens and prepares the uterus for implantation of the embryo (in the event of fertilization). Failure of oocyte fer-tilization leads to the degeneration of the corpus luteum during the last few days of the cycle. The disintegrating corpus luteum is unable to maintain its secretion of estrogen and progesterone, and their blood concentrations drop rapidly. The marked decrease of estrogen and proges-terone lead to degeneration of the uterine wall followed by sloughing off of the uterine surface tissue. It also results in the removal of inhibition of FSH secretion.The blood concentration of FSH begins to rise, follicle and ovum development are stimulated, and the cycle begins anew. The cycle is usually broken up into three general phases. The first phase, encompassing follicle development up to ovulation, is called the follicular phase and lasts about 9 days FSH and estrogen dominate during the phase. The second phase, lasting from ovulation until the beginning of the disintegration of the uterine lining, is called the luteal phase and lasts approximately 14 days. Progesterone dominates during this phase. Following this phase is the flow phase which lasts 5 days, and is characterized by bleeding due to the sloughing off of the uterine lining.

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Question:

(a) \rule{1em}{1pt}3^\rule{1em}{1pt}2(b) (\rule{1em}{1pt}3)^\rule{1em}{1pt}2(c) (\rule{1em}{1pt}3)/(4^\rule{1em}{1pt}1)

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/Users/wenhuchen/Documents/Crawler/Chemistry/E33-0948.htm

Solution:

(a) Here the exponent applies only to 3. Since x^\rule{1em}{1pt}y = (1/x^y), \rule{1em}{1pt}3^\rule{1em}{1pt}2 = \rule{1em}{1pt} (3^\rule{1em}{1pt}2) = \rule{1em}{1pt} (1/3^2) = \rule{1em}{1pt} (1/9). (b) In this case the exponent applies to the negative base. Thus, (\rule{1em}{1pt}3)^\rule{1em}{1pt}2 = [1/(\rule{1em}{1pt}3)^2] = [1/{(\rule{1em}{1pt}3) (\rule{1em}{1pt}3)}] = 1/9. (c) (\rule{1em}{1pt}3) / (4^-1) = [(\rule{1em}{1pt}3) / {(1/4)^1}] = [(\rule{1em}{1pt}3) / (1^1/4^1)] = [(\rule{1em}{1pt}3) / (1/4)] Division by a fraction is equivalent to multiplication by that fraction's reciprocal, thus [(\rule{1em}{1pt}3) / (1/4)] = \rule{1em}{1pt} 3 \bullet (4/1) = \rule{1em}{1pt} 12, and[(\rule{1em}{1pt}3) / (4)^-1]= \rule{1em}{1pt} 12.

Question:

Write a FORTRAN program to compute the standard deviation of N real numbers. Again, you may assume that these numbers have already been read into an array X having members X(1), X(2),...,X(N).

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G19-0481.htm

Solution:

We apply the formula \varphi= \surd[^N\sum_N=1 (X_i -X)^2 / (N - 1)] \varphi \varphi = \surd[{(X_1 -X)^2 + (X_2 -X)^2 +...+ (X_N -X)^2} / (N - 1)]. To further simplify the calculation we will assume thatX, the mean, has already been calculated as the variable AVG from the preceding problem (#12). Let SIGMA, stand for\varphi. The program would then be: \varphi DIMENSION X(N), Y(N) READ, N SUMSQ = \O.\O DO 5\O I = 1,N Y(I) = X(I) - AVG SUMSQ = SUMSQ + Y(I) \textasteriskcentered\textasteriskcentered 2 5\OCONTINUE A = FLOAT (N) SIGMA = SQRT (SUMSQ / (A-1.0)) STOP END

Question:

Indicate the equilibrium equation and constant expression for theautoprotolysisof liquid ammonia. If KNH_3 = 10^-22 , how many molecules of ammonia are ionized in 1 mole of ammonia? Assume a density of 0.771g/ml for ammonia.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E10-0337.htm

Solution:

Autoprotolysisis that phenomenon whereby an ammonia molecule can donate a proton to another NH_3 molecule to form posi-tive and negative charged species. The equation of theautoprotolysiscan be written NH_3 + NH_3 \rightleftarrows NH+_4 + NH^-_2 or (i) 2NH_3 \rightleftarrows NH+_4 + NH^-_2 . To find the constant expression, consider the equilibrium con-stant expression for the reaction: K = [{(NH+_4) (NH-_2)}/{(NH_3)^2}] . Note, though, that the concentration of NH_3 in pureamnoniais always constant. By analogy with theautoprotolysisof water (where the K_W expression is written [OH^-] [H_3O^+] - without [H_2O]^2 in the denominator), the constant expression for theautoprotolysisof NH_3 is (ii) K_NH3 = [NH+_4] [NH-_2] . To find the number of molecules of ammonia ionized in 1 mole of ammonia, use the equation K = [{(NH+_4) (NH-_2)}/{(NH_3)^2}] . Let x be the number of moles of ammonia ionized. Then the NH_3 remainingnonionizedis 1-x moles. Since each 2 ammonia molecules must ionize to produce one NH+_4 and one NH-_2 , the number of NH+_4 = number of NH^-_2 = (x/2). Let V be the volume of one mole of ammonia. The concentration of NH^+_4 and NH-_2 can be rewritten as [(x/2)/V], and the concentration ofnonionizedNH_3 is [(1 - x)/V] . The equation for K can be rewritten as (iii)K = [({(x/2)/V} {(x/2)/V})/({(1 - x)/V}^2)] = [(x^2/4)/{(1 - x)^2}] \textbullet [(1/V^2)/ (1/V^2)] = [(x^2)/{4(1 - x^2)}] . x, the number of moles of ionized ammonia, can be calculated if K is known. To solve for K , consider a more general case of the equation K = [{[NH+_4] [NH-_2]}/{[NH_3]^2}] . The numerator [NH^+_4] [NH-_2] is the constant expression for the auto-protolysisof NH_3 and must always equal K_NH3 \textbullet K_NH3 is given as 10^-22. To find [NH_3]^2 , use the fact that the density of ammonia is 0.771g/ml. The mole weight of ammonia is 17.03g. Thus, 0.771g is [(0.771g)/(17.03g/mole)] = 0.0453 moles of ammonia; and the density of ammonia is 0.0453 moles/ml. = 45.3 moles/liter. Thus, [NH_3] = 45.3M. Substitute these results in K = [{[NH^+_4] [NH-_2]}/{(NH_3)^2}] (iv)K = (10^-22/45.3^2) = 4.9 × 10^-26 . Substitute this value of K into (iii) (v)4.9 × 10^-26 = [(x^2)/{4(1 - x^2)}] . To simplify the problem, note that the dissociation of ammonia is very small and thus x << 1. Thus, approximate 1-x^2 as 1. Then, (v) becomes (vi)(x^2/4) = 4.9 × 10^-26 . Solve to obtain x = 4.42 × 10^-13 . This is the number of moles of NH_3 that ionized. To find the number of molecules, remember that 1 mole = 6.02 × 10^-23 molecules. Thus, No. of molecules ionized = (4.42 × 10^-13moles ) (6.02 × 10^23 molecules/mole) = 2.66 × 10^11 molecules. 2.66 × 10^11 molecules of ammonia are ionized.

Question:

A plate of glass 1.00 cm thick is placed over a dot on a sheet of paper. The dot appears 0.640 cm below the upper surface of the glass when viewed from above through a microscope. What is the index of refraction of the glass plate?

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Solution:

The eye records the direction from which a light ray enters it as though it has traveled in a straight line from the source. It cannot compensate for refraction of the light. Therefore, the observer sees the dot 0.640 cm below the upper surface of the glass when in actuality it is 1.00 cm below the sur-face. Assuming the index of refraction of air to be one (n_air = 1.00029), we have from Snell's law, n_glass sin \texttheta_glass = n_air sin \texttheta_air(1) n_g = (n_g/n_a) = (sin \texttheta_a/sin \texttheta_g)since n_a \approx 1(2) The sine function can be expressed as sin \texttheta = tan \texttheta cos \texttheta(3) We know from that figure that tan \texttheta_a = (x / 0.64 cm) tan \texttheta_g = (x / 1.00 cm) From the trigonometric relation cos \texttheta = \surd(1 - sin^2 \texttheta) we have cos \texttheta_a = \surd(1 - sin^2 \texttheta_a) cos \texttheta_g = \surd(1 - sin^2 \texttheta_g)(4) From equation (1), Sin \texttheta_g = (n_a/n_g) sin \texttheta_a(5) Substituting equation (5) into equation (4), Cos \texttheta_g = \surd[1 - (n_a/n_g)^2 sin^2 \texttheta_a Using the values found for the cosines and tangents of the angles and equation (3), substitute into equation (2) n_g = (sin \texttheta_a/sin \texttheta_g) = (tan \texttheta_a cos \texttheta_a/tan \texttheta_g cos \texttheta_g) = {(x / 0.64 cm) \surd(1 - sin^2 \texttheta_a)} / {(x / 1.00 cm) \surd[1 - (n_a/n_g)^2 sin^2 \texttheta_a]}(6) The observer looks at the dot through a microscope. This means that he looks at the dot from directly above and \texttheta_a must be very small. We can then make the approximation sin \texttheta_a \approx sin 0\textdegree = 0 Substituting this value in equation (6), we find n_g = {(x/0.64 cm)\surd(1 - 0)} / {(x/1.00 cm)\surd[1 - (n_a/n_g)^2 (0)]} = (1.00/0.64) = 1.57

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Question:

If 40 liters of N_2 gas are collected at 22\textdegreeC over water at a pressure of .957 atm, what is the volume of dry nitrogen at STP (Standard temperature and pressure)?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E03-0089.htm

Solution:

The key to this problem is the recognition that PV/T = k, where P is the pressure, V the volume, T the absolute temperature and k a constant. If you let the STP conditions (1-atm. and 0\textdegreeC), be represented by [P_1V_1]/T_1 and the conditions of the gas collected over the water by [P_2V_2] / T_2,then [P_1V_1] / T_1 = [P_2V_2] / T_2 since both equal the same constant, k. Therefore, to answer the question, you have to substitute the known values into the equation and solve for the unknown, V_1. However, there is one precaution. Over water, you have vapor pressure. At 22\textdegreeC, water has a vapor pressure of .026 atm. As such, this makes the initial pressure of N_2 .957 - .026, or .931 atm. Add 273\textdegree to the Celsius temperature to obtain the temperature in Kelvin. Therefore, T_2 = 22\textdegreeC + 273\textdegree= 295\textdegreeK and T_1 = 0\textdegreec + 273 = 273\textdegreeK. Therefore, by substitution [(1)(V_1)] / 273 = [(.931)(40)] / 295 . Now, solving for V_1, the volume at STP, you obtain 34.4 liters.

Question:

A boy bailing a boat lifts 2.0 kg of water to a height of 1.2 m above the bilge as he pours the water over the side, (a) How much work does he perform? (b) How much work is done by the force of gravity?

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Solution:

(a) The boy does work against gravity. Therefore, the force he must exert on the water is just equal to its weight mg. Work equals the product of force and the distance the force acts over. W = Fs = 2.0 kg [(9.8nt)/(kg)] × 1.2 m = 23.5 joules The boy's work is converted to potential energy, which is then converted to the kinetic energy of the falling water. (b) If the direction of the upward displacement is called positive, then the gravitational force is in the negative direction, and W = Fs = (-19.6nt)(1.2 m) = - 23.5 joules The negative sign means that work was done against gravity.

Question:

A) ProgramB) Heuristic programmingC) Time sharing D) Computer languageE) Data processing

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Solution:

A)Program: A set of instructions telling the computer what to do. When the program is written in an HLL (High Level Language) such as Pascal or COBOL, the instructions are referred to as statements. B) Heuristic programming: The term "heuristic" refers to short-cut approximate solutions to problems that are otherwise very difficult to solve (e.g. intractable) . Heuristic program-ming may be used in Artificial Intelligence software in order to guide the computer toward a practical (common sense) solution to a given problem orsubproblem. C) Time sharing: The best way to understand the concept of time sharing is to contrast it with multiprogramming. Multipro-gramming allows multiple programs that are ready for execution to reside in primary memory. While one program is doing some-thing unrelated to CPU, such as input-output, another one can run on CPU doing computations. This way, I/O activity involving slower devices is overlapped by computations involving fast CPU in order to increase the utilization of resources. Typically, a program (defined as a process by the operating system) is sus-pended whenever it needs to do I/O, and the CPU is assigned to another program. After finishing I/O, the suspended program is again readied for CPU assignment. Timesharing requires multiprogramming and not vice versa. In time sharing, in addition to multiprogramming where a process is suspended upon, say I/O request; each process (program) is regularly suspended after its time quantum (in milliseconds) expires. The purpose is to let multiple users and their programs to share the CPU by taking turns at quick time intervals (usu-ally in round robin fashion). Although the CPU can do only one thing at a time (e.g. can run one program at a time), the users feel as if they are simultaneously using the CPU. This idea is sometimes referred to as "virtual machines," where if there are ten users on a fast CPU, each feels as if he/she is using a machine ten times as slow (on the average, of course!). D) Computer language : A language that computer understands as opposed to natural language that homo sapiens can understand. These languages come in families belonging to various hierarchi-cal levels. From bottom to top, the hierarchy looks like: Ma-chine languages, Assembly languages, High-level languages, and Very high-level languages. E) Data processing : Data processing involves manipulation of data that represents facts, events, actions, or concepts in order to achieve a desired and useful result in terms of new data. Although not essential, data processing makes a heavy use of computer systems.

Question:

A horizontal force of 5 N is required to maintain a velocity of 2 m/sec for a box of mass 10 kg sliding over a certain rough surface. How much work is done by the force in 1 min?

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Solution:

First, we must calculate the distance traveled: s = vt = (2 m/sec) × (60 sec) = 120 m. Then, W = Fs cos \texttheta, where \texttheta is the angle between the force and the distance. In this case \texttheta = 0\textdegree so we can write, W = Fs = (5 N) × (120 m) = 600 N - m = 600 J

Question:

Write a program which generates the first 12 rows of Pascal's Triangle.

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Solution:

Pascal's triangle yields the coefficients ofx^kandy^kin the expansion of (x+y)^n . The first few rows are: (x+y)^0 : 1 11 121 1331 14641 15101051 (x+y)^1 : (x+y)^2 : (x+y)^3 : (x+y)^4 : (x+y)^5 : In order to find a formula for the coefficients of the n^th degree expan-sion, consider Newton's binomial theorem: (x+y)^n = [x^n/0!] + [(nx^n-1) / 1!]y + {[{n(n-1)} / 2!] x^n-2y^2} + {[n(n-1)(n-2) x^n-3] / 3!}y^3 + {[n(n-1) (n-2) (n-3)] / 4!}x^n-4 y^4 +... + {[n(n-1) ... (n-r+1)] / r!}x^n^-r y^r +...+ nxy^n-1 +y^n. The coefficient of the (n-r)-thpower of x is {[n(n-1) ... (n-r+1)] / r!} . But [n(n-1) ... (n-r+1)] = [n! / (n-r)!] =^np_r. Hence ther^thcoefficient where 1 \leqq r \leqq N is given by [n! / {r! (n-r)!}]. But this is the formula for the number of combinations of n things taken r at a time (a combina-tion is an arrangement where order is immaterial, e.g., ABC = BAC). Thus the binomial coefficients are given by [n! / {r! (n-r)!}] =^nC_r. 1\O FOR N = \O TO 11 2\O FOR R = \O TO N 3\O LET C = 1 4\O FOR X = N TO N - R + 1 STEP -1 5\O LET C = C \textasteriskcentered X/(N-X+1) 6\O NEXT X 7\O PRINT C; 8\O NEXT R 9\O PRINT 1\O\O NEXT N 11\O END

Question:

The most common constituent of gasoline is iso-octane. It is a hydrocarbon, composed by weight of 84.12 % carbon, and 15.88 % hydrogen. Given that it contains 5.27 × 10^21 molecules per gram, what is its molecular formula?

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Solution:

The molecular formula for iso-octane is C_nH_m ,where n is the number of moles of C and m is the number of moles of H present in 1 mole of iso-octane. One can find n and m from the molecular weight of the compound. It is given that there are 5.27 × 10^21 molecules/g present in one gram of the compound. One knows that there are 6.02 × 10^23 molecules per mole. Thus, MW = (6.02 × 10^23 molecules/mole) / (5.27 × 10^21 molecules/g) = 114.23 g/mole The C in this compound weighs 84.12 % of 114.23 g and the H weighs 15.88 % of 114.23 g. weight of C = .8412 × 114.23 g/mole = 96.09 g/mole weight of H = .1588 × 114.23 g/mole = 18.14 g/mole. The number of moles of carbon present, n, is equal to 96.09 g divided by 12.01 g/mole, its atomic weight. n = (96.09 g) / (12.01 g/mole) = 8.00 moles The number of moles of H, m, can be calculated in a similar manner. m = (18.14 g) / (1 g/mole) 18.14 mole , where 1 g/mole is the atomic weight of hydrogen. The formula is C_8H_18, 18.14 is rounded off to the nearest whole number.

Question:

(1) Two lead balls whose masses are 5.20 kg and .250 kg are placed with their centers 50.0 cm apart. With what force do they attract each other? (2) At the surface of the earth g = 9.806 m/s^2. Assum-ing the earth to be a sphere of radius 6.371 × 10^6 m, compute the mass of the earth.

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Solution:

(1). The force of gravitational attraction between two bodies with masses m_1 and m_2 separated by a distance s is F = Gm_1m_2/s_2 = [6.67 × 10^-11 {(nt- m^2)/kg^2}][(5.20 kg × .250 kg)/(.500m)^2] = 3.46 × 10^-10nt (2). The only force acting on a body of mass m near the surface of the earth is the gravitational force. Hence, using Newton's Second Law F = ma =Gmm_e/r^2 where r is the distance of m from the earth's center. At the surface of the earth, a = g and r = R_e mg =Gmm_e/R2_e whencem_e = gR2_e/G = [{(9.806 m/s^2)(6.371× 10^6 m)^2}/(6.670 × 10^-11 nt\bulletm^2/kg^2)] = 5.967 × 10^24 kg

Question:

A single body of mass M in free space is acted on by a constant force F^\ding{217} in the same direction in which it is moving. Show that the work done by the force is equal to the increase in kinetic energy of the body.

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Solution:

This is a case of motion in a straight line with constant acceleration. In the figure suppose that at the zero of time the body is at the origin and is mov-ing with a velocity v_0^\ding{217}. Suppose that at time t it has moved through a distance x^\ding{217} and its velocity has changed to v^\ding{217}. If a^\ding{217} is the constant acceleration, then we can derive an equation relating v to x. From the equations v = v_0 + atandx = v_0t + (1/2)at^2 we must eliminate t. Since t = (v - v_0)/a ,then x = v_0 [(v - v_0)/a] + (1/2) a [(v - v_0)/a]^2 = (v^2 - v^2_0)/2a The work done by the force F^\ding{217} in moving the body from the origin through a distance x^\ding{217} is Work Done = Fx = [F(v^2 - v^2_0)]/(2a) But from Newton's second law F = Ma. So work done = [Ma(v^2 - v^2_0)] / [2a] = (1/2)M (v^2 - v^2_0) = (1/2) Mv^2 - (1/2)MV^2_0 Work done = Final kinetic energy - initial kinetic energy.

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Question:

The formula Z = [(e^ax- e ^-ax)/ 2] sin (x + b) + a log [(b + x) / 2] is to be evaluated for all combinations of x: 1.0(0,1) 2.0 a: 0.10(0.05)0.80 b: 1.0(1.0) 10.0 where x: 1.0(0.1) 2.0 means x = 1.0, 1.1, 1.2,...,2.0 and so on. (There are 11 × 15 × 10 = 1650 combinations.) For each combination a line giving x, a, b, and Z values is to be written. Write a program containing three DO loops to carry out this computation.

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Solution:

We need to obtain integral subscripts for our DO loops. This is no problem for b. For x, though, note that 10{_\ast}X takes on values 10., 11.,...,20. Thus if I = 10, 11,...,20, X= FLOAT(I)/10. Similarly, 100{_\ast}A takes on values 10.0, 15.0, 20.0,...,80.0 in steps of 5. Let J = 10, 15, 20, ,..,80. So A = FLOAT (J)/100. [Remember that the library function FLOAT changes an integer number into a real number.] Hence we obtain the program below. DO 100 I = 10,20 X = FLOAT (I)/10.0 DO 100 J = 10,80,5 A = ELOAT(J)/100.0 DO 100 K = 1,10 B = FLOAT(K) Z = 0.5{_\ast}(EXP(A{_\ast}X) - EXP(-A{_\ast}X)){_\ast}SIN(X+B) 1+ A{_\ast}AL0G ((B+X)/2.) WRITE (6,10) A,B,X,Z 10FORMAT (4X,3(2X,F6.3), E13.7) 100CONTINUE STOP END

Question:

The diagram below is an example of a phase diagram for a pure substance. To what phases do the regions A, B, and C correspond?

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Solution:

Following the 1 atm constant pressure line from left to right, we are proceeding from low values of the temperature to high values. Therefore, we will intersect the three regions in the order solid-liquid-vapor. The regions A, B, C hence correspond to the solid, liquid, and vapor phases, respectively. Point a denotes the normal freezing (melting) point of the substance and the point b denotes the normal boiling point.

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Question:

Find the length of a meter stick (L_0 = 1 m) that is moving lengthwise at a speed of 2.7 × 108m/s.

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Solution:

The length of the stick seems to be smaller when it is moving with respect to the observer. The relativistic formula giving the contracted length L, is L = [ L_0 \surd{1- (v2/ c^2 )}] where v is the speed of the stick with respect to the observer . Hence , L = (1m) × \surd[1 - {(2.7 × 10^8 m/s) / (3 × 108m/s)}^2 ] L = (1m) × 0.44 = 0.44m .

Question:

Let the label for an assembly language statementbegin in column 1 and consist of at most eight characters, the first of which is a letter followed by letters or digits. The label is followed by at least one blank. A statement without a label begins with a blank. Write a program that reads in assembly language statements and prints those statements with invalid labels.

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Solution:

LETTER = 'ABCDEFGHIJKLMNOPQRSTUVWXYZ' DIGIT ='0123456789' LABEL = ''\vert ANY(LETTER) (SPAN(LETTER DIGIT)@N \textasteriskcenteredLE(N,8) +\vertNULL) '' \vert &NCHOR = 1 RCARD = INPUT:F(END) CARDLABEL:S(R) OUTPUT = CARD:(R) END The first two statements are regular assignment statements, assigning the string of letters to LETTER and a string of digits to DIGIT. The third assignment statement contains alternation - the expression where different patterns are separated by symbol \vert (with at least one blank on each side). The value of the expression is a pattern structure that matches any string specified by any of the alternative patterns. The first alternative pattern in the third statement is ''- a regular blank, while the second pattern includes a few new features. The first of them is a primitive function ANY. ANY and the opposite function NOTANY are primitive functions whose values are pattern structures that match single characters. ANY matches any character appearing in its argument, following the function in parentheses. NOTANY matches any character not appearing in its argument. Thus the pattern structure ANY (LETTER) matches any letter of the alphabet (the structure ANY('ABCD...XYZ') would have the same meaning). How ANY LETTER then concatenates with another alternative structure needs a detailed explanation. The SPAN function returns a matched run of characters, included in concatenated strings LETTER and DIGIT (any other character met with indicates the end of the run). Matching a pattern structure against a subject string, in SNOBOL IV, is done by a procedure called the SCANNER. The scanner has a pointer (control) called the CURSOR which is positioned to the left of the character that the scanner must match. As the scanner matches one by one the characters of the string, the cursor correspondingly moves to the right. These concepts are important in understanding the function of the next variable of the second alternative structure in line 3 - @N. The unary operator @ is called the cursor position operator. At any moment of the pattern matching process, it assigns an integer, representing the cursor position at that moment, to a variable following it (in our case, N). For example, in statement 'EXAMPLE' @JUNK 'L' The cursor, at first, is located to the left of E. No match occurs; the value 1 is assigned to the variable JUNK, and the cursor moves to X. The process goes on until the cursor hits L. By that time JUNK has the value of 5. Letter L matches, execution goes to the next statement, and variable JUNK is stored with value 5. Thus, in our problem, N indicates the position of the cursor after the scanner matched all characters of the first string on the card up to the first blank. In other words, N counts the number of characters in the label. Next comes the comparison of N. The unary operator \textasteriskcentered (asterisk) postpones the evaluation of its operand. Thus, LE(N,8) is an unevaluated expression until it appears as part of a pattern LABEL in statement 6. This way the expression is evaluated only when needed. An alternative expression for concatenation in the second part of statement 3 is a string NULL which, in SNOBOL IV indicates a string of length zero. It is different from the strings 'O' and '', each of which has a length one. The NULL string consists of no characters, no blanks, no length. A blank finishes the pattern. It is now possible to follow through the function of the pattern LABEL in the sixth statement. If the statement on the input card starts with a blank, the pattern matches at once, and the execution returns to the previous statement CARD = INPUT :F(END), reading in the next card. If, however, the statement on the card starts with a letter, the characters following that letter are examined. All the characters (including the initial letter) are counted up to the first blank (specified in pattern LABEL by the blank at the end). Also all those characters have to belong either to LETTER or to DIGIT. If, in fact, the cursor hits a symbol which belongs to neither of those two strings, the matching fails, and the card is printed out as having an invalid label. Meanwhile, the integer produced by the counting procedure is assigned to N. If all the characters are legitimate, N is compared with 8. If N appears to be less than or equal to 8, the match succeeds, and the next card is considered. In case N is greater than 8, the match fails, and the card is printed out. If the first letter is followed by no other characters (NULL string), just a blank, the match succeeds, and the next card is considered. The program terminates after consideration of the last card.

Question:

What is the basic mechanism of digestion and what digestive processes take place in the mouth and stomach?

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Solution:

The process of digestion is the breakdown of large, ingested molecules into smaller, simple ones that can be absorbed and used by the body. The breakdown of these large molecules is called degradation. During deg-radation, some of the chemical bonds that hold the large molecules together are split. The digestive enzymes cleave molecular bonds by a process called hydrolysis. In hydrolysis a water molecule is added across the bond to cleave it. Within living systems, chemical reactions require specific enzymes to act as catalysts. Enzymes are very specific, acting only on certain substrates. In addition, different enzymes work best under unlike conditions. Digestive enzymes work best outside of the cell, for their optimum pHs lie either on the acidic (e.g., gastric enzymes) or basic side (e.g., intestinal and pancreatic enzymes). The cell interior, however, requires an almost neutral (about 7.4) pH constantly. Digestive enzymes are secreted into the digestive tract by the cells that line or serve it. Digestion begins in the mouth. Most foods contain polysaccharides, such as starch, which are long chains of glucose molecules. Saliva (and the intestinal secretions) contain enzymes that degrade such molecules. Salivary amylase, an enzyme that is also called ptyalin, hydrolyzes starch into maltose. (Compounds whose names end with -"ase" are enzymes, and those with the suffix "-ose" are sugars.) Glucose is eventually absorbed by the epithelial cells lining the small intestine. The saliva has a pH of 6.5 - 6.8. This is the optimal range for salivary enzyme activity. Food spends a relatively short amount of time in the mouth, and eventually enters the stomach. The stomach is very acidic, with a pH of 1.5-2.5. The acid is secreted by special cells in the lining of the stomach called parietal cells. The low pH is required for the activity of the stomach enzyme pepsin. Rennin coagulates milk proteins in the infant's stomach, making them more susceptible to enzyme attack. Pepsin is a proteolytic enzyme (protease): it degrades proteins. Pepsin starts the protein digestion in the stomach by splitting the long proteins into shorter fragments, or peptides, that are further digested in the intestine. Pepsin will split any peptide bond involving the amino acids tyrosine or phenylalanine. There are 20 different kinds of amino acids that can make up a protein and some proteins are thousands of amino acids long. The body needs the amino acids it obtains from digestion to synthesize its own proteins.

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Question:

How does mating behavior serve reproduction? Give examples. Define and give examples of courtship behavior.

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Solution:

Mating behavior is part of the synchronization of sexual activity. For most animals there exists a particular period or season of the year when it is most advantageous to mate. Therefore the production of sperm and release of eggs must be synchronized. Often environ-mental cues will initiate mating behavior. Changes in length of daylight, temperature, rainfall, tidal and lunar cycles are some stimuli that can elicit mating behavior. Examples of mating behavior are the migration of the gray whale to Baja California, and the migration of salmon upstream to breed. Courtship behavior is a specific form of reproductive synchronization. It has two main purposes: to decrease aggression between two potential mates and to aid them in identifying one another. The song patterns of birds and the enlargement of colorful pouches in some fowls are two examples of courtship behavior. Courtship brings two potential mating partners together so that copulation, insemina-tion and fertilization can follow for the production of offspring.

Question:

What characteristics of arthropods have been of primary importance in their evolutionary success?

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Solution:

Arthropodais an extraordinarily large and diverse phylum. There are about a million different arthropod species. Approximately 80 percent of all animal species belong to this phylum. The ClassInsecta, contain-ing more than 750,000 species, is the largest group of animals; in fact, it is larger than all other animal groups combined. The development of achitinousexoskeleton has been an important factor in the evolutionary success of the arthropods. Anexoskeletalplate covers each body segment, and movement is possible because the plates of each segment are joined by a flexible, thinarticularmembrane. In most arthropods, there has been a reduction in the total number of body segments, fusion of many segments, and se-condary subdivision of certain segments. This has re-sulted in body parts specialized for different functions, with movable joints between the parts.Infoldingof the embryonic ectoderm has also resulted in an internal skeleton, providing sites for muscle insertions. Arthropod musculature is complex, and quite unlike that of most other invertebrates. The exoskeleton also lines both the anterior and posterior portions of the digestive tract. The exoskeleton is termed a cuticle and is composed of protein and chitin. The cuticle consists of anepicuticle, anexocuticle, and an endocuticle . The outermostepicuticleoften contains wax, as it does in insects, and serves to prevent water loss. Theexocuticleis tougher than the innermostendocuticle. Theepi-cuticleandexocuticleare absent at joints to provide for flexibility. The cuticle of the crustaceans is further strengthened by deposition of calcium salts. Where theepicuticleis absent, the cuticle is relatively perme-able to gas and water. Pine pore canals in the cuticle permit elimination of secretions from ducts. The support and locomotion provided by the elaborate exoskeleton are major advantages that arthropods have over all other in-vertebrates, and the ability for modification has enabled the cuticle to serve many functions. Another evolutionary advantage of the arthropods is the adaptation of well-developed organ systems. The digestive, respiratory, and blood vascular systems are all complex, and often show improvements over other in-vertebrates' systems. The reproductive structures are well developed in many arthropods. Internal fertilization occurs in all terrestrial forms, and in many arthropods, the eggs and the young are brooded and "nursed" by adults. The excretory system is well-developed in terrestrial arthropods. The complex organization of the nervous tissues in the arthropods is yet another key factor in the evolution-ary success. The brain in most arthropods is large, and many complex sense organs are present. Motor innervation to the muscles is precise and allows for many different movements, speeds, and strengths. The compound eyes of insects and crustaceans result in a wide visual field, and a great ability to detect movement. Other sensory receptors are sensitive to touch, chemicals, sound, position and movement. Arthropods often have complex behavioral patterns, many of which are under hormonal control.

Question:

The molecular diameter of an N_2 molecule, as deduced from the VanderWaals b parameter, is 3.15 × 10^-8 cm. The density of liquid nitrogen is 0.8081 g/cm^3. On a hard - sphere model, what fraction of the liquid volume appears to be empty space?

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Solution:

To find the fraction of the liquid volume that appears to be empty space, one must subtract the volume of spheres from the total volume of the liquid. One can find the volume of the spheres that constitute one mole of N_2 liquid by multiplying the volume of one sphere by 6.02 × 10^23, the number of spheres in one mole. The volume of a sphere is equal to 4/3\pir^3 where r is the radius of the sphere. The diameter of the N_2 sphere is given as (3.15 × 10^-8) cm, thus the radius is half of length or (1.58 × 10^-8 cm). Solving for the volume of 1 sphere : volume of 1 sphere= (4 / 3 \pi) (1.58 × 10^-8 cm)^3 = 1.652 × 10^-23 (cm^3 / sphere) Volume of 1 mole of spheres = [(6.02 × 10^23 )(spheres / mole)] × [(1.652 × 10^-23 (cm^3 / sphere)] = 9.945 (cm^3 / mole) One can find the total volume of 1 mole of liquid N_2 by dividing the molecular weight by the density. (MW = 28) total volume = [28 (g / mole)] / [0.8081 (g / cm^3)] = (34.65 cm^3 / mole) The volume of the empty space is equal to the volume of the spheres subtracted from the total volume. volume of empty space per mole= (34.65 cm^3) - (9.945 cm^3) = 24.70 cm^3 The percent of the volume taken up by empty space is equal to the volume of the space divided by the total volume and multiplied by 100. % of empty space = [(24.70 cm^3) / (34.65 cm^3)] × 100 = 71.30 %.

Question:

The average bromine content of sea water is 0.0064% . (a)How much sea water, in cubic feet, would be required to How much sea water, in cubic feet, would be required to obtain one pound of bromine? (b) What volume of chlorine gas, measured at STP, would be required to liberate the bromine from one ton of sea water? One cubic foot of sea water weighs about 63 pounds.

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Solution:

(a) The weight of bromine in one cubic foot of sea water is found by multiplying 63 pounds (the weight of 1 cu. ft. of sea water) by 0.000064. This is because bromine compose 0.0064 %of the weight of sea water. weights of Br in 1 cubic ft = .000064 × 63 lb = 4.032 x 10^-3 lb One can find the number of cubic feet of sea water necessary to extract 1 lb of Br by dividing 1 lb by the number of pounds of Br in one cubic foot. no. of cubic ft =1 lb / (4.032 × 10^-3 lb/cubic ft) = 248.02 cubic feet. (b) Since chlorine is more active than bromine, the latter may be liberated from its salt by treatment with chlorine. 2Br ^- + Cl_2 \ding{217} 2Cl^- + Br_2 One mole of CI_2 will liberate 2 moles of Br^-. Therefore, to find the volume of Cl_2 gas necessary to liberate the bromine in one ton of sea water, one must first calculate the amount of Br^- present in 1 ton of sea water. One is given that 1 cubic foot of sea water weighs 63 pounds. Thus the number of cubic feet of sea water in 1 ton can be found. (1 ton = 2000 lbs). no. of cubic feet = 2000 lbs / (63 lbs / cubic feet) = 31.75 cu. ft. In the previous section one found that each cubic foot contains 4.032 × 10^-3 lb of Br^-, thus one finds the amount of Br^- in 31.75 cu. ft. by multiplying the number of cubic feet by the weight of one cubic foot. weight of Br^- in 1 ton sea water = = 31.75 cu. ft. × 4.032 × 10^-3 lb/cu. ft. = .128 lb. One finds the number of moles present by dividing .128 lb by the molecular weight in pounds. (MW of Br^- = 80gr/mole.) There are 454 g in 1 lb, therefore grams are con-verted to pounds by multiplying the number of grams by 1 lb / 454 \alpha. MW of Br^- in lb = 80 g/mole × (1 lb / 454g) = .1762 lb/mole One can now find the number of moles of Br^- present in one ton. no. of moles = (weight of Br^-) / (MW in lbs) no. of moles = (.128 lbs) / (.1762 lbs /mole) = .73 moles . From the equation one knows that 1/2 as many moles of Cl_2 are needed as Br^-. Therefore, the amount of Cl_2 used is equal to 1/2 of .73 moles or .365 moles. The volume of one mole of gas at STP (Standard Temperature and Pressure, 0\textdegreeC and 1atm) is defined to be 22.4 liters. Therefore, one can find the volume of Cl_2 gas required for the reaction by multiplying the number of moles of gas present by 22.4 liters. volume of Cl_2 = .365 × 22.4 liters = 8.18 liters.

Question:

For the complex ion equilibrium for the dissociation of Cu(NH_3)^2_4^+ , calculate (1) the ratio of Cu^2+ ion to Cu(NH_3)^2_4^+ in a solution 1 M in NH_3; (2) the equilibrium concentration of NH_3 needed to attain 50 % conversion of Cu^2+ to Cu(NH_3)^2_4^+. K_a for Cu(NH_3)^2_4^+ = 2 × 10^-13.

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Solution:

(1) The equilibrium constant for the reaction A \leftrightarrowsnB+mC is defined as K = {[B]^n[C]^m} / [A] where K is the equilibrium constant, [B] is the concentration of B, n is the number of moles of B formed, [C] is the con-centration of C, m is the number of moles of C produced and [A] is the concentration of A. The reaction in this problem is Cu(NH_3)^2_4^+ \leftrightarrows Cu^2+ + 4NH_3 This indicates that the equilibrium constant can be stated K_d= {[Cu^2+][NH_3]^4} / [Cu(NH_3)^2_4^+] = 2 × 10^-13 Here, one is trying to find the ratio of Cu^2+ ion to Cu(NH_3)^2_4^+ in the solution or stated in another way [Cu^2+] / [Cu(NH_3)^2_4^+] . To obtain this ratio, one can use the equation forK_d. In the problem, one is told that the concentration of NH_3 is 1 M. Therefore, [Cu^2+] / [Cu(NH_3)^2_4^+] can be found. K_d= {[Cu^2+][NH_3]^4} / [Cu(NH_3)^2_4^+] = 2 × 10^-13 {[Cu^2+][1.0]^4} / [Cu(NH_3)^2_4^+] = 2 × 10^-13 [Cu^2+] / [Cu(NH_3)^2_4^+] = [(2 × 10^-13)/{(1.0)^4}] = 2 × 10^-13 (b) To find the equilibrium concentration of NH_3 needed to convert 50 % of the Cu^2+ ion present to Cu(NH_3)^2_4^+ ion, the equation for the equilibrium constant will again be used. K_d= {[Cu^2+][NH_3]^4} / [Cu(NH_3)^2_4^+] = 2 × 10^-13 When 50 % of Cu^2+ is converted to Cu(NH_3)^2_4^+, the concentration of Cu^2+ will equal that of Cu(NH_3)^2_4^+. This is true because there is one Cu^2+ ion used in the formation of each Cu(NH_3)^2_4^+. ion. If there were 50 Cu^2+ ions in the solution and 25 of them were converted to Cu(NH_3)^2_4^+, there would be 25 Cu^2+ ions left and 25 Cu(NH_3)^2_4^+ ions formed. This means that for a 50% conversion [Cu^2+] / [Cu(NH_3)^2_4^+] = 1 One can now solve for the [NH_3] by using the equation forK_d. K_d= {[Cu^2+][NH_3]^4} / [Cu(NH_3)^2_4^+] = 2 × 10^-13 (1)[NH_3]^4 = 2 × 10^-13 [NH_3] = ^4\surd(2 × 10^-13) [NH_3] = 6.7 × 10^-3 M A concentration of 6.7 × 10^-3 M of NH_3 will cause a 50 % conversion of Cu^2+ to Cu(NH_3)^2_4^+.

Question:

It has been suggested that a 10\textdegree rise in temperature results in a twofold response in the rate of a chemical reaction. What is the implied activation energy at 27\textdegreeC? ; at 1000\textdegreeC?

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Solution:

The activation energy, E_A, is the difference between the heat concentration of the active molecules and the inert molecules. The activation energy can be related to the rate constants of the reaction at two different temperatures by one form of the Arrhenius equation: (i) log (k_2 / k_1) = -(E_a / 2.303R)[(1/T_2) - (1/T_1)] (where R = universal gas constant; k_1 is the rate constant at temperature T_1 ;k_2 is the rate constant at T_2). This equation can be rewritten as (ii) E_a = [2.303R / {(1/T_1) - (1/T_2)}] log (k_2 / k_1) Thus to solve for E_a , you find T_1, T_2, and k_2/k_1 andsubstitute. Let T_1 = initial temperature. If there is a 10\textdegree rise in temperature, then T_2 = T_1 + 10\textdegreeK. You are told that when the temperature is increased, the original rate is doubled. Thus, if k_1 = original rate, k_2 = 2k_1. Thus, log k_2/k_1 = log 2 = .301. For the case, T_1 = 27\textdegreeC = 300\textdegreeK, T_2 = 310\textdegreeK you have E_a = [(-2.303R) / {(1/T_2)( 1/T_1)}] log (k_2 / k_1) = [{(-2.303)(1.987)(1 k cal / 1000cal)} / {(1/310) -(1/300)}](.301) = 12.8 k cal / mole For the case with T_1 = 1000\textdegreeC = 1273\textdegreeK, T_2 - 1283\textdegreeK, You have, E_a = [{(-2.303)(1.987)(1 k cal / 1000cal)} / {(1/1283) -(1/1273)}] (.301) = 226 k cal/mole (In both cases, 1 k cal/1000 cal is a conversion factor.)

Question:

Evaluate \surd400.

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Solution:

400= 4 × 100 Thus,\surd400= \surd(4 × 100) Since\surdab= \surda \surdb, \surd400= \surd4 \surd100 = 2 \bullet 10 = 20 Check: If \surd400 is 20, then 20^2 must equal 400, which is true. Hence, 20 is the solution.

Question:

The rate law for the reaction H_2(g) + Br_2(g) \rightarrow 2HBr(g) is {d[HBr]} / dt = {k[H_2 ] [Br_2 ]^1/2} / {1 +k' [HBr]/[H_2]}\textbulletDerive this law from the reaction mechanism, which consists of chain-starting, chain-propagating, and chain-terminating steps. Use steady- state approximation.

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Solution:

A chain-reaction comes about when an inter-mediatespecies that is consumed in one step is regenerated in a later step. As a consequence of this, a sequence of steps is set up such that they endlessly repeat themselves until the chain is terminated or the starting material is used up. By various experiments and observations, the mechanism has been found to be the following: Chain - Starting step: (1) Br2k_1\rightarrow 2Br \textbullet Chain - propagating step (2,3,4): (2) Br_2 \textbullet + H2k_2\rightarrow HBr + H \textbullet (3) H \textbullet + Br2k_3\rightarrow HBr + Br \textbullet (4) H \textbullet + HBrk_4\rightarrowH_2 + Br \textbullet Chain - propagating step: (5) 2Br \textbulletk_5\rightarrow Br_2 To derive the rate law for this reaction, the formation of HBr, use these individual reactions: Notice, first, that only those reactions that dealt with HBr in some way were included. In other words,(1) + (5) did not have this molecule, so that they cannot be included. The whole idea of writing the rate in this fashion comes from the fact that when, say A + B \rightarrow products, rate = {-d[A]} / dt = {-d[B]} / dt =k [A] [B]. This is called the expression for the rate law of a second order reaction. By order, you mean the number of atoms or molecules whose con-centrations directly determine the reaction rate, in other words, for (2), d[HBr]/dt = k_2[Br][H_2]. You can write similar expressions for (3) and (4) . The overall rate of HBr formation is the sum of the rates of formation of the HBr in each of these reactions. In (2) and (3), you are making HBr, so that their rate law expressions are added. In (4), however, HBr is being eliminated, so that you subtract. To repeat, you now have {d[HBr]} / dt = k_2[Br] [H_2] + k_3[H] [Br_2] - k_4 [H] [HBR] To obtain the desired formula, you need to find [Br] and [H]. You can do this by writing overall rate expressions for the formation of [H] and [Br]. To do this, use the same reasoning employed for [HBr]. You find that {d[H]} / dtHHh= k_2[Br] [H_2] - k_3[H] [Br_2] - k_4[H] [HBr] HHh Note: Only (2), (3), and (4) dealt with H, which explains why only they are used. You subtract both (3) and (4) from (2), since they consume H instead ofproducing it as (2) does. For Br, you find {d[Br]} / dt= 2k_1 [Br_2] - k_2[Br] [H_2] + k_3[H] [Br_2] + k_4[H] [HBr] - 2k_5[Br]^2 . Again, use the same type of reasoning and technique to obtain this ex-pression. You are told steady-state conditions exist. This means that the concentrations of the reactive intermediates that are regenerated do not change with time. This means, therefore that d[Br]/dt = 0 and d[H]/dt = 0, since each is being regenerated. Thus, you can equate the above ex-- pressions with zero. You obtain {d[Br]} / dt = 0 = 2k_1[Br_2] - k_2[Br] [H_2] + k_3[H] [Br] + k_4[H] [HBr] - 2k_5[Br]^2 . {d[H]} / dt = 0 = k_2[Br] [H_2] - k_3[H] [Br_2] - k_4[H] [HBr]. These two equations can be solved simultaneously to find [H] and [Br]. If you perform this operation, you find [Br] = [k_1/k_5 [Br]]^1/2and [H] = k_2 [{(k_1/k_5)^1/2 [H_2] [Br_2]^1/2} / {k_3 [Br_2] + k_4[HBR]}] . Recalling, now, that {d[HBr]} / dt = k_2[Br] [H_2] + k_3[H] [Br_2] - k_4[H] [HBr],you can substitute these values for [H] and [Br] to obtain {d[HBr]} / dt = {2 k_3 k_2 k_4^-1 k_1^-1/2 k_5-1/2[H_2][Br]^1/2} / {k_3 k_4^-1 + [HBr] [Br_2]^-1}. Now , if you letk = 2k_2 k_1^1/2 k_5^-1/2 and k' = 1 / k3k_4^-1 , you obtain {d[HBr]} / dt = {k[H_2][Br_2]^1/2} / {1 + k' [HBr] / [H_2]} \textbullet

Question:

A compound of vanadium and oxygen is analyzed and found to contain 56.0 % vanadium. What is the equivalent weight of vanadium in this compound?

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Solution:

To find the equivalent weight of vanadium one can use the Law of Definite Proportions. This law states that when elements combine to form a given compound, they do so in a fixed and invariable ratio by weight. This means that the ratio of the weight of vanadium to the weight of the oxygen that reacts is equal to the ratio of the equivalent weight of the vanadium to the equivalent weight of the oxygen. (weight of V) / (weight of O) = (equivalent weight of V) / (equivalent weight of O) Solving for the equivalent weight of vanadium in this compound, one assumes he has 100 g of the compound for calculations. Because 56 % of this compound is vanadium, it means that 56 g of it is vanadium. This indicates that (100 % - 56 %) = 44 % of the compound is oxygen and that in 100 g of the compound there are 44 g of oxygen. Here one will assume that the equivalent weight of oxygen is its atomic number, 8. Solving for the equivalent weight of vanadium: (56 g) / (44g) = (equivalent weight of V) / (8) equivalent weight of V = (8 × 56g) / (44 g) = 10.2 The equivalent weight of vanadium in this compound is 10.2 .

Question:

If 25 ml of .11 M sucrose is added to 10 ml of .52 M sucrose, what would be the resultingmolarity?

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Solution:

In this problem one mixes two solutions of different concentrations and wishes to find the resultingmolarity. The answer is obtained by first calculating the total number of moles present and then dividing this by the total volume. Molarityis defined as moles \div volume. Therefore, moles =Molarity× Volume For the .11 M solution, no. of moles= .11 moles/liter × .025 liter = .00275 moles For the .52 M solution, no. of moles= .52 moles/liter × .010 liter = .00520 moles total no. of moles = .00275 + .00520 = .00795 total no. of liters = .025 + .010 = .035 FinalMolarity= (.00795/.035) = .23 M.

Question:

A steam engine is operated reversibly between two heat reservoirs, a primary reservoir and a secondary reservoir at 35\textdegreeC. Determine the efficiency of the engine.

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Solution:

By definition, the primary reservoir is the reservoir from which the engine absorbs heat and the secondary reservoir is the reservoir to which the engine releases heat. Since the engine is steam operated, the primary reservoir must be a source of steam (such as a boiler) operating at a temperatureT_p= 100\textdegreeC (T_p= 100 + 273 = 373\textdegreeK). The secondary reservoir operates at a temperature T_s = 35 + 273 = 308\textdegreeK. The efficiency of such an engine is defined as efficiency = (T_P - T_s)/T_p where T_P and T_s are in degrees Kelvin. Thus, the efficiency of the engine is efficiency = (T_P - T_s)/T_p= (373\textdegreeK - 308\textdegreeK)/( 373\textdegreeK) = 0.174 or 17.4%.

Question:

Obtain expressions for the exclusive-OR and the equivalence functions of two variables A and B.

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Solution:

The Exclusive-OR function is defined as a function which gives an output 1 whenever either A or B is 1, but gives an output 0 whenever A and B are both 1 or both 0. The equivalence function of two variables A and B is defined as a function which gives an output 1 whenever both A and B are equal (both 1 or both 0), but gives an output 0 whenever A and B are unequal. These two functions can be represented by a Truth Table as follows: If A and B are binary variables, they can be represented in 4 different ways, giving us the 4 rows in the Truth Table. The Truth Table is shown in figure 1. Col. 1 Col. 2 Col. 3 VARIABLES AB EXCLUSIVE - OR A×B EQUIVALENCE A \equiv B 00 01 10 11 0 1 1 0 1 0 0 1 We note from the Truth Table of fig. 1 that the exclusive-OR: and the equivalence functions are complements of each other. Also, from the Truth Table, we can write the equations: a)Expressions for exclusive-OR A × B = A' \bullet B + A \bullet B' . (This is a sum of products form of the expression obtained by considering entries of column 2 which are equal to 1). Also, A × B = (A+B) \textbullet (A '+B') . (This is a product of sums form, obtained by considering entries of column two which are equal to 0.) b)Expressions for Equivalence: A \equiv B = A'\bulletB' + A\bulletB (Sum of products form) A \equiv B = (A+B') \textbullet (A'+B). (Product of sums form) We shall now realize the exclusive-OR and the equivalence functions, using a) NAND gates only, and, b) NOR gates only

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Question:

What mass of calcium (Ca, atomic mass = 40.08 g/mole) mustbe combined with 1.00 g of phosphorus (P, atomic mass= 30.97 g/mole) to form the compound Ca_3p_2?

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Solution:

From the formula of the compound we are trying to form, Ca_3p_2, we seethat the ratio of moles of calcium to moles of phosphorus must be 3/2. By calculating the number of moles of P in 1.00 g, we can determine the re-quirednumber of moles of Ca and then convert this to a mass. The number of moles of P in 1.00 g is molesP = (mass P) / (atomic mass P) = 1.00g / 30.97 g/mole = 0.0322 mole. Then, using the ratio of (moles Ca / moles p) = 3/2 or moles Ca = 3/2 molesP, the number of moles of Ca required to combine with 0.0322 mole P in a 3/2 ratio is molesCa =3/2 molesP = 3/2 × 0.0332 mole = 0.0483 mole. To convert this to a mass we multiply by the atomic mass, obtaining massCa = moles Ca × atomic mass Ca = 0.0483 mole × 40.08 g/mole = 1.94 g Ca .

Question:

In aquatic animals, external respiration is carried out by specialized structures called gills. How do these gills operate?

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Solution:

Fish, molluscs and many arthropods (except insects) possess gills for respiratory purposes (see figure). Organisms such as these require a mechanism to keep a fresh supply of water flowing about them. This is necessary because gas exchange occurs between the blood vessels in the gills and the water which contains dissolved oxygen. Continuous bathing of the gills ensures that there will always be enough oxygen available for the organism. It also ensures that the carbon dioxide diffusing from the gills into the water is removed from the organism. This is important in that the CO_2 concentration gradient is maintained, so diffusion of CO_2 will continue to occur from the gills to the water. A fish opens its mouth and receives a quantity of water. It then closes its mouth and forces the water out past its gills by contracting its oral cavity. Gills have thin walls, are moist, and are well supplied with blood capillaries. Oxygen, which is dissolved in the water, diffuses though the gill epithelium and into the blood capillaries. Simultaneously, carbon dioxide diffuses in the opposite direction in accordance with the CO2 concentration gradient. The amount of oxygen dissolved in sea water is relatively constant, but the amount in freshwater ponds may fluctuate greatly. A fish would suffocate in water that is lacking sufficient amounts of dissolved oxygen. It is interesting to note that the direction of blood flow through a gill is opposite to the direction of water flow over the gill. This counter-current system maximizes the amount of gas exchange that can take place. If both the blood flow and water current were in the same direction, the CO_2 and O_2 concentration gradients across the gills would decrease. This would result in a slower rate of diffusion, and the amount of gas exchange would be reduced.

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Question:

What arePackedarrays? How do they differfrom conventional arrays?

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Solution:

The reserved word Packed instructs the compiler to store the values of the array in memory in the most com-pact way. Usually, values of types char andbooleando not occupy the entire memory location provided for them. Since each element of an array of such types occupies a separate memory location, this results in a considerable waste of memory space. When PACKED ARRAY representation is used, the values are packed together as closely as possible, ignoring the borders between memory location. Therefore, a packed array takes up much less memory space than one that is not packed. On the other hand, the time required to access each individual element of an array is greatly increased , since the computer not only has to locate the memory location storing the desired value, it must also separate it from other values sharing the same memory location. Thus, what is gained in termsof memory space is lost in terms of in-creased processing time. This trade-off varies considerably from one machine to another, and even from one Pascal imple-mentation to another. So the decision whether or not to use packed arrays should take that into account, as well as the requirements of the specific problem under consideration.

Question:

If the ∆H' of water is 40,600 (J / mole at 100\textdegreeC, what is the boiling point of water at a pressure of .750 atm.

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Solution:

The boiling point is defined as the temperature at which the vapor pressure of a liquid equals that of the surrounding atmosphere. The normal orstandardboiling point is that temperature where the vapor pressure of a gas equals one atmosphere. For water, this is 100\textdegreeC. In this problem, therefore, you have two pressures for water, one temperature, and ∆H', which equals the heat required to transform one mole of liquid to the ideal-gas state. These parameters are used to solve for the non-standard boiling point in the Clausius-Clapeyron equation, which states log(P_1 / P_2) = (∆H')(19.15) [(1 / T_2) - (1 / T_1)], where p_1 and p_2 are vapor pressures and T_1 and T_2 are their respective temperatures in degrees Kelvin. If you let T_2 be the boiling point of water at .750 atm, sub-stituting into the equation: log(1.00 / .750) = (40,600 / 19 .15) [(1 / T_2) - (1 / 373)] Solving for T_2, you obtain 365\textdegreeK or 92\textdegreeC, which is the boiling point of water at .750 atm.

Question:

What is the meaning and significance of the Pauling electronegativity scale?

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Solution:

In any chemical bond, electrons are shared between the bonding atoms. In covalent bonds, the valence electrons are shared almost equally - the shared electrons spend about the same amount of time with each atom. In ionic bonds, the valence electrons are monopolized by one atom. The degree to which a bond will be ionic or covalent is dependent upon the relative electron-attracting ability of the bonding atoms. The Paulingelectronegativityscale provides a measure for the relative electron-attracting abilities orelectronegativityof each element. The most electronegative element, fluorine, is assigned the highest number, 4.0. The least electron attracting elements (and consequently, those most willing to lose electrons), cesium and francium, are assigned the lowest number, 0.7. Numbers are assigned to the remaining elements so that no element has a higher number than an element more electronegative than it. Furthermore, the numbers are assigned so that the differ-ences between theelectronegativitiesof two elements is indicative of the ionic quality of the bond that forms between them. In particular, if the difference is 1.7 or greater, then the shared electrons are monopolized by the more electronegative atom to such a great extent that the bond is said to be ionic. If the difference is less than 1.7, the bond is covalent. Thus, when given a compound, the Paulingelectro-negativityscale can be used to determine the species that acts as the electron donor, the species that acts as the electron acceptor, and the degree of electron polarization of the bond.

Question:

A small amount of arsenic is present in a germanium crystal. If the energy E of the fifth electron may be represented approximatelyby the relationship E = [(-1 × 10^-2eV)/n^2] howmuch energy is required to free the electron? Does thermalvibra-tion of the ions provide sufficient energy to free thearsenic's fifth electron? At room temperature the thermal energyis 0.04eV.

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Solution:

Notice that the relationship E = [(-1 × 10^-2eV)/n^2] hasthe same form as the hydrogen-energy equation giving the energy of thenth orbit: E_n = [(-13.6eV)/n^2] According to this equation, one has to supply 13.6eVto free the electron fromthe ground level (n = 1), and less energy to free it from the other levels. Similarly, the energy needed to raise the arsenic electron to the conductionband where the electron is practi-cally free, is E = 1 × 10^-2eV. The average thermal energy of the silicon ions vibrating in the crystal at roomtemperature is 0.04eV, so sufficient energy could be trans-ferred duringa collision to free the fifth electron of an arsenic atom and allow it to migrate through the crystal.

Question:

Describe the pathway of electron transport in the respiratory chain.

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Solution:

A chain of electron carriers is responsible for transferring electrons through a sequence of steps from molecules such as NADH (reduced nicotinamide adenine dinucleotide) to molecular oxygen. NADH collects electrons from many different substrates through the action of NAD+ - linked dehydrogenases. For example, NADH is produced from the oxidation of glyceraldehyde-3-phosphate in glycolysis, and by several dehydrogenations in the citric acid cycle. In the electron transport system, NADH is oxidized to NAD+ with the corresponding reduction of flavin mono-nucleotide (FMN) to FMNH_2. (see Figure 1). Some of the energy released is used for the production of one ATP molecule. Then FMNH_2 is oxidized back to FMN, and the electrons of FMNH_2 are transferred to ubiquinone or coenzyme Q. It is at this site where the electrons of FADH_2 are funneled into the electron transport chain. As ubiquinone is oxidized, the first of a series of cytochromes, cytochrome b, is reduced. The cytochromes are large heme proteins (see Fig. 2). A heme group consists an iron atom surrounded by a flat organic molecule called a porphyrin ring. It is the iron atom of the cytochrome which is oxidized (Fe^3+) or reduced (Fe^2+) . Cytochrome b reduces cytochrome c_1; the energy released is used for the formation of a sec-ond ATP molecule. Cytochrome c_1 reduced to cytochrome c (not shown in figure) which in turn reduces cytochrome a (a complex of cytochromes a and a_3, plus two copper atoms), cytochrome a is oxidized in the last step, with molecular oxygen being the final electron acceptor. As oxygen is reduced to H_2O, a third and last ATP molecule is synthesized. Thus for each NADH molecule entering the respiratory chain, 3 ATP molecules are produced. Therefore, eight NADH molecules give rise to 24 ATP per glucose molecule in the citric acid cycle. Only 4 ATP are produced by the 2 FADH_2 molecules, since they enter the electron transport system at ubiquinone, thus bypassing the first site of ATP synthesis. The two cytoplasmic NADH from glycolysis produce only four ATP instead of six because one ATP is expended per cytoplasmic NADH in order to actively transport NADH across the mitochondrial membrane. By means of the respiratory chain, 32 ATP are produced per glucose mole- cule. The other 4 ATP molecules are produced during substrate-level phosphorylation in glycolysis and the TCA cycle. Thus there is a net yield of 36 ATP per glucose molecule oxidized. (Recall that "substrate-level" indicates reactions not involving the electron transport system.)

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Question:

What is the output of the following program? What is the significance of the numbers produced? What algo-rithm is being used to produce them? PROGRAM MYSTERY (output); VAR A : ARRAY[1..30]of integer; J, K, FLAG: Integer; BEGIN FOR J: = 1 to 30 DO A[J]: = 1; FLAG: = 1; J: = 1; WHILE FLAG = 1 DO BEGIN J: = J + 1; while A[j] = 0 DO J: = J + 1; IF J = 30 then FLAG = 0 ELSE BEGIN write1n (J); K:= J + J; while K<30 DO BEGIN A[K]:= 0; K:= K + J END END END.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G16-0404.htm

Solution:

In order to determine the output of a program, it must be carefully simulated. The first executable statement of the above program, sets all elements of the array A to 1. Then, entering the WHILE Loop, J becomes 2. A[J] is not 0, so we go on to the next statement. J is not 30, and the ELSE part is executed. The value of J, 2, is printed out, K be-comes 4. Entering the WHILE K<30 DO...Loop, A [4] becomes 0, K becomes 6, which is less than 30, so again A[6]=0, K=8, A[8]=0, etc. In other words, elements in all even positions starting with 4 are set to 0. When K becomes 30, the loop will terminate, returning control back to first loop. Flag is 1, so it enters the loop again, J becomes 3, A[3] is not 0, J is not 30, so 3 is printed out, K becomes 6, and similarly to the previous iteration, all muliples of 3 become 0. When K reaches 30, the control is transferred back to Loop 1. Flag is still 1, J becomes 4. A[4] is 0, J is incremented to 5, A[5] is not 0, so the second loop terminates, and the ELSE part of the next statement is executed. 5 is printed out, K becomes 10, and now all multiples of 5 are set to 0. By now, it should be clear that the program is printing out prime numbers, using the following algorithm: Set all ele-ments of the Array to 1. Starting with J=2 if A[J] is 0, then J .is a prime. Print it out and set all multiples of it to 0. Increment J by 1 and repeat the same procedure.

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Question:

Using electron-dot notation, show for each of the following the outer shell electrons for the uncombined atoms and for the molecules or ions that result: (a) H+H\ding{217} Hydrogen molecule (b) Br+Br\ding{217} bromine molecule (c) Br+Cl\ding{217} bromine chloride (d) Si+F\ding{217} silicon fluoride (e) Se+H\ding{217} hydrogen selenide (f)Ca+O\ding{217} calcium oxide

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Solution:

When electrons are transferred from one atom to another, ions are formed, which gives rise to ionic bonding. Two atoms, both of which tend to gain electrons, may combine with each other by sharing one or more pairs of electrons. These two atoms form a covalent bond. To solve this problem, one must know the number of valence electrons, in each of the atoms in the equations. The valence number reflects the combining capacity of an atom. Next, one must know which atoms combine to form ionic bonds and which form covalent bonds. The only ionic bond formed in these equations is for Ca + O; the other bonds are covalent, and electrons are shared to form an isoelectronic electron cloud such as a noble gas. Thus,

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Question:

What are the defects that produce myopia, hypermetropia and astigmatism? What corrective measure can be taken for each defect?

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Solution:

The most common defects of the human eye are nearsightedness (myopia) , which is the inability to see distant objects clearly; farsightedness (hypermetropia), which is the inability to see nearby objects distinctly; and astigmatism, a defect owing to an improperly shaped eyeball or irregularities in the cornea. In the normal eye (Figure A) the shape of the eyeball is such that the retina is located at a proper distance behind the lens for light rays to converge in the fovea. In a nearsighted eye (Figure B) , the eyeball is too long, and the retina is too far from the lens. The light rays converge at a point in front of the retina, and are again diverging when they reach it, resulting in a blurred image. This defect can be cor-rected by placing a concave lens in front of the eye, which diverges the light rays before they reach the lens, making it possible for the eye to focus these rays properly on the retina. In a farsighted eye (Figure C), the eyeball is too short and the retina is thus too close to the lens. Light rays strike the retina before they have converged, again resulting in a blurred image. Convex lenses, when placed in front of the defective eyes, correct for the farsighted condition by causing the light rays to converge farther forward, so that they can come to a focal point on the retina. In contrast to length of the eyeball as the cause for these visual disorders, a myopic eye may have a lens which is too strong - it bends light rays too much. A hypermet-ropic eye has a lens which is too weak. Whether the cause be from eyeball length or lens strength, a myopic eye will fo-cus the image in front of the retina, while a hypermetropic eye will focus the image behind the retina. In astigmatism (Figure D), the cornea is curved unequally in different planes, so that light rays in one plane are focused at a different point from those in another plane. To correct for astigmatism, a cylindrical lens is used which bends light rays going through certain irregular parts of the cornea.

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Question:

The distance between the sun and earth is 1.5 × 10^11 m and the earth's orbital speed is 3 × 10^4 m/s. Use this information to calculate the mass of the sun.

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Solution:

Since the earth's orbit around the sun is very nearly circular, it is assumed that it is exactly circular. The radius of the circle is equal to the distance d between the earth and sun. The centripetal acceleration of the earth is then a = v^2/d Newton's second law, F = ma, may then be written as F = (m_1v^2)/(d) where m_1 is the mass of the earth. F is the force acting on the earth and is responsible for its centripetal acceleration. This force is also the gravitational force of attraction between the two objects. It is described by the law of universal gravitation, F = (Gm_1m_2)/(d^2) where m_2 is the mass of the sun. Equating these two expressions (Gm_1m_2)/(d^2) = (m_1v^2)/(d)Upon rearranging and substituting the known quantities, the mass of the sun is m^2 = (v^2d)/G = [(3 × 10^4 m/s)^2 (1.5 ×10^11 m)]/(6.67 × 10^-11N m^2/kg^2) = 2.02 × 10^30 kg The mass of the sun is more than 300,000 times as large as the mass of the earth.

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Question:

(a) Explain what is meant by even parity and odd parity. (b) Decode the following under the odd parity 8421 redundant (8421 R) code and determine if an error has occurred. (i) 00011 (ii) 01011 (iii) 01001 . (c) Decode the following four-digit number under the horizontal and vertical odd parity 8421 R coding scheme and determine if it had been correctly transmitted: 01101 00101 01011 00100 1101

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Solution:

(a) The even parity system adds a 1-bit to a data word if the word has an odd number of 1-bits or adds a 0-bit if the word has an even number of 1-bits. Thus, in an even parity system every legal code word has an even number of 1-bits. The odd parity system adds a 1-bit to a data word if the word has an even number of 1-bits, or adds a 0-bit if the word has an odd number of 1-bits. Thus, in an odd parity system every legal code word has an odd number of 1-bits. In an odd parity system it is also true that every legal code word has at least one 1-bit and at least one 0-bit, thus yielding additional information on whether or not the transmission line is operative. (b) When a parity bit is added to an 8421 code, it is sometimes known as the 8421 redundant (8421 R) code. The parity bit here-is called a redundancy bit. In an odd parity system, this parity bit would be 1 if the 8421 code word had an even number of 1's and 0 if the 8421 code word had an odd number of 1's. Thus, with the addition of the parity bit, every number in the odd parity 8421 R code has either one 1-bit or three 1-bits. If a received word has two or no 1-bits, an error has occurred. In decoding, the rightmost bit is not used, and the other four bits are decoded as regular 8421 code representation. (i) Since 00011 has two 1-bits, an error has occurred. (ii) Since 01011 has three 1-bits the word has been correctly trans-mitted. 01011 in 8421 R = 0101 in 8421 = decimal 5 . (iii) Since 01001 has two 1-bits, an error has occurred. (c) With odd parity, the presence of an even number of 1-bits indicates an error, but the precise faulty bit is not known. This can be over-come by checking for parity in columns as well as rows. This pro-cedure is called horizontal and vertical odd parity check. Thus, if the received message is: 01101 \leftarrow first row 00101\leftarrowsecond row 01011 \leftarrow third row 00100 \leftarrow fourth row first column\rightarrow 1101 \leftarrow fourth column the rightmost bits of the first four rows are row parity bits, and the last row consists of column parity bits. The first, third and fourth rows contain an odd number of 1-bits, so no error has occurred in those rows. But the second row has two 1-bits, indicating an error in that row. To determine which bit in that row is faulty, check the column parity bits. The first, second and third columns contain an odd number of 1-bits, so no error has occurred in those columns. But the fourth column has two 1-bits, indicating an error in that column. Therefore, the bit in the fourth column and the second row is in error. Note that this system not only has error- detecting capabilities, but error correcting capabilities as well, since, once the faulty bit is located, it can be corrected by changing it to its complement. (Recall that a bit in error is the complement of the input value and hence, complementing it again would yield the original input value, by the involution law.) The amount of redund-ancy in an error-correcting code must be much higher than in a mere error-detecting code. Also, note that although the regular parity (8421 R) system cannot pick up two errors occurring in the same data word (since the first error would result in an even number of 1-bits and the second error would result in an odd number of 1-bits, thus allowing the received word to pass the parity check), the horizontal and vertical parity system will indicate an error if a double error has occurred. The ability to detect the faulty bit is lost, however. Again, this encoding, decoding and error-correcting scheme is auto-matically implemented by the machine hardware.

Question:

For the system described by the following diagram, is the forward reaction exothermic or endothermic?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E14-0478.htm

Solution:

By observing the diagram, we see that the products are at a higher potential energy than the re-actants, thus energy must have been absorbed during the course of the reaction. The only mechanism by which this can occur is absorption of heat and subsequent conversion of heat into potential energy. Hence, the reaction absorbs heat and is therefore endothermic.

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Question:

Two rough planes A and B, inclined, respectively, at 30\textdegree and 60\textdegree to the horizontal and of the same vertical height, are placed back to back. A smooth pulley is fixed to the top of the planes and a string passed over it connecting two masses, the first of 0.2 slug resting on plane A and the other of mass 0.6 slug resting on plane B. The coefficient of kinetic friction on both planes is 1/\surd3. Find the acceleration of the system.

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/Users/wenhuchen/Documents/Crawler/Physics/D04-0123.htm

Solution:

There are four forces acting on each of the two masses: the weight acting downward, the normal force exerted by the plane at right angles to the plane, and the two forces acting along the plane, the tension in the string, and the retarding force due to kinetic friction. In each case, resolve the weight into components along the plane and at right angles to it, as shown in the figure. Since there is no tendency for either mass to rise from the plane, the normal force and the component of the weight at right angles to the plane must be equal and opposite. Further, if \mu is the coefficient of kinetic friction between mass and plane, the frictional force in each case is \mu times the normal force (\muN). The larger mass on the steeper plane will descend. The frictional force is opposite to the motion, i.e., up the plane, and therefore, by Newton's second law (F_net = m'a) m'g sin 60o - T - F' = m'g sin 60o - T - \mu m'g cos 60o = m'a. where a is the acceleration of m', and N = m'g cos 60\textdegree. Since the pulley is smooth, the tension is the same at all points in the string. For the other mass, motion is up the plane and thus the frictional force acts down the plane. Thus T - mg sin 30o - F = T - mg sin 30o -\mu mg cos 30o Adding the two equations obtained, one has m'g sin 60\textdegree - mg sin 30\textdegree - \mu(m'g cos 60\textdegree + mg cos 30\textdegree) = (m + m')a. Therefore one can write, for the acceleration of the system, a = [{m'g sin 60o - mg sin 30o -\mu (m'g cos 60o + mg cos 30o)} / {m + m'}] sin 60o = \surd3/2sin 30o = 1/2 cos 60o = 1/2cos 30o = \surd3/2 Then a = [({0.6 slug × (\surd3/2) - 0.2 slug × 1/2} - (1/\surd3){0.6 slug × (1/2) + 0.2 slug × (\surd3/2)}) / (0.8 slug)] g = [{(2 \surd3/10) - 0.2}/0.8] × 32 ft/sec^2 = 5.9 ft/sec^2. Had we guessed that mass m would descend, the above analysis would have lead to a negative accelera-tion indicating that m ascends.

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Question:

Design a synchronous sequential circuit with one input and one output that recognizes the input sequence 01. For ex-ample, if the input sequence is x = 0101000111101 then the output sequence will be Z = 0101000100001

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G04-0090.htm

Solution:

The first step in the design procedure is the construction of a state diagram which represents the input - output behavior described. The diagram is constructed as follows: First it is assumed that the machine is in some starting state A and the first input is a 1. Since a 1 is not the first element in the input string to be considered, the machine remains in state A and yields an output z = 0. This is shown in fig. 1. If the machine is in the initial state A and the in-put is 0, then because this input is the first input symbol in the string to be recognized, the machine moves to a new state B and yields an output z = 0, shown in figure 2. Now the machine is in state B and the next input sym-bol is a 0. Because this is not the second symbol in the sequence 01 the machine remains in state B and gives an output z = 0, as shown in figure 3. Finally, if the machine is in state B and the next input symbol is a 1, then the machine moves to state A and produces an output z = 1, shown in figure 4. The relationship between the number of states (N_s) and the number of flip-flops (N_FF) is given by the ex-pression 2^(N)FF(-1)_ < N_s\leq 2^(N)FF State Assignment is arbitrarily chosen to be A = 0 B = 1. The state table, figure 5, now can be drawn as the transi-tion table, Figure 6. Transition tables: y^k, denotes the present state of the circuit which is the current output of the flip-flop. y^k+1 denotes the next state of the machine, which is the output of the flip-flop after a transition has occurred. Using the clocked SR flip-flop input tables, figure 7, the excitation tables for the problem are derived, figures 8 and 9. Input Present State Next State SR y^K yk+1 00 0 0 00 1 1 01 0 0 01 1 0 10 0 1 10 1 1 11 0 Not used 11 1 Not used As an example of derivation, consider the upper left hand corner of the transition table given in Fig. 6. Here y^k = 0, x = 0 and y^k+1 = 1. To effect a state change from y^k = 0 to y^k+1 = 1, the signals which must appear on the Set and Reset lines, found from figure 7, are S = 1 and R = 0. Hence, these signals appear in the corresponding positions in the excitation maps of Figs. 8 and 9. Next consider the state transition in the upper right hand corner of the transition table where y^k = 0, x = 1, and y^k+1 = 0. Since no change in state must occur, the signal, found from figure 7, on the set line must be 0, while the reset line doesn't matter. From the state diagram of figure 4 the Karnaugh map of the output z is constructed, as shown in fig. 10. The combinational circuit of figure 11 is created from the simplified equations s =x, R = x, z = xy^k.

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Question:

Using the pedigree below, determine the method of inheritance of the trait and, as far as possible, fill in the genotypes of each individual. Assume that the trait is recessive.

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/Users/wenhuchen/Documents/Crawler/Biology/F26-0687.htm

Solution:

Let B represent the normal allele and b rep-resent the allele for Mongolian spot. Using the system established in the previous question, we can say that P_1, F_1,2, F1,3,F_1,4- ,F_1,5 ,F_2,2 and F_1,4 (see pedigree) are genotypically bb, because they all express the recessive Mongolian spot trait. Since all of the progeny of P_1 and P_2 carry the trait, the genetic make-up of P_2 who does not express the trait but must carry the recessive gene, is Bb. In the cross between F_1,1 and F_1,2, we see that one of their children (F_2,3) shows the trait. Since F_1,2 must have contributed one recessive gene, the other re-cessive gene must have come from F_1,1. But F_1,-1 is normal, so her genotype must be Bb. F_2,1 is also Bb. To determine the genotypes of F_3 and P_4, we have to examine the cross between F_1,5 and F_1,6. One of the offspring in this cross (F_2,4) shows Mongolian spots. This implies that both parents must have a copy of b. We already know that F_1,5 is bb. Therefore F_1,6 must be Bb, since he is normal F_2,3 is also Bb because he can receive only the recessive allele from F_1,5. Since F_1,6 is he-terozygous, one of his parents must have allele b. We cannot assume which parent carries the b allele, because the information is not available for us to be certain. Conceivably, both parents can be Bb, the number of their progeny being too small to necessarily include a homozygous recessive. Or one parent may be BB, in which case all their offspring would show the normal phenotype. The genotype of F_1,7 can thus be either BB or Bb. We can summarize the genotypes of the individual in the following diagram:

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Question:

A small body of mass 1 slug is rotated in a vertical circle at the end of a string 2 ft. long. If the tension in the string just vanishes at the top of the circle, what is the velocity of the body and the tension in the string (a) when the string is horizontal, and (b) when the body is at its lowest point?

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/Users/wenhuchen/Documents/Crawler/Physics/D05-0244.htm

Solution:

At each of the positions of the body in its rotation, only two forces act on it, the weight mg^\ding{217} acting downward and the tension of the string , T^\ding{217}, acting toward the center of the circle. At the top of the swing, the two forces act in the same direction and together provide the centripetal force necessary to keep the body in its circular path. Thus (see figure) by Newton's Second Law, mg + T = (mv^2)/r. When T is just zero at the top, v^2 = rg = 2 ft × 32 ft/s^2 = 64 ft^2/s^2. \thereforev = 8 ft \textbullet s^-1. (a) When the string is horizontal, the body has lost potential energy and gained a corresponding quantity of kinetic energy. If we refer to the diagram we have from the principle of conservation of energy. (1/2) mv^2_1 = (1/2)mv^2 + mg × r or v^2_1 = v^2 + 2gr = 3 gr. \thereforev_1 = 8 \surd3ft/s = 13.86 ft/s. Here, the potential energy is taken to be zero at the height of the center of the circle. Further, T_1^\ding{217} is the only force acting radically. Hence T1= (mv^2_1)/r = 3 mg = 3 × 1 slug × 32 ft \textbullet s^-2 = 96 lb. (b) When the body is at its lowest point, similar arguments about gain of kinetic energy and loss of potential energy apply. Thus (1/2) mv^2_2 - mgr = (1/2) mv^2 + mgr \therefore v^2_2 = v^2 + 4gr = 5gr.\thereforev_2 = 8\surd5 ft/s = 17.9 ft/s Although T_2^\ding{217} still acts radically, it is now opposed by the gravitational force which equals the weight of the body. The resultant force provides the necessary centripetal force, and, by Newton's Second Law T_2 - mg = (mv^2_2)/r Then T_2 = (mv^2_2)/r + mg = 5mg + mg = 6mg = 6 × 1 slug × 32 ft/s^2 =192 lb.

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Question:

What is the difference between a hypha and a mycelium and between an ascus and a basidium?

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/Users/wenhuchen/Documents/Crawler/Biology/F06-0167.htm

Solution:

A few fungi are unicellular, but most have multicellular bodies made up of tubular branching filaments called hyphae. A mass of hyphae is called a mycelium. The hyphae of the algai fungi, or Phycomycetes (which con-tain the Oomycetes and the Zygomycetes), are not divided by cross walls between adjacent nuclei - they are coenocytic hyphae. These fungi are thus multinucleate. The Ascomycetes and Basidiomycetes have hypae divided by cross walls - they are septate; these fungi are multicellular. The mycelium may appear as a cobweb-like mass of fibers, as in bread mold, or may be fleshy and compact, as in truffles. A basidium and an ascus are both reproductive structures. Ascus is Latin for sac, and fungi whose life cycle includes an ascus are termed sac fungi or ascomycetes. The ascus is formed from a single parent cell. The zygote is an elongated cell, and its nucleus divides meiotically to produce four haploid spores. These usual-ly divide meiotically to produce eight small spore cells. Spores are released when the ascusruptures. ruptures. The basidiomycetes, or club fungi, have reproductive structures called basidia. Mushrooms are members of this group. The basidium differs from the ascus in that spores are formed on the outside of the parent cell rather than within it. The zygote nucleus within the elongated, club- shaped basidium divides meiotically to produce four haploid nuclei. These nuclei migrate to protuberances which develop at the tip of the basidium. Each protuberance then buds off to form a spore. The spores may fall from the basidium or they may be ejected. Both the ascus and the basidium develop from a single zygote.

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Question:

Describe the body plan of hydra.

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/Users/wenhuchen/Documents/Crawler/Biology/F11-0276.htm

Solution:

Hydra belong to the animal phylum called the Coelenterata (or Cnidaria). This phylum includes also such animals as the jelly fish, sea anemones and corals. Most are marine except for Hydra which is freshwater. These animals display radial symmetry, a digestive cavity, which opens into a mouth, and a circle of tentacles sur-rounding the mouth. The body wall consists of three basic layers: an outer epidermis; an inner layer of cells lining the digestive cavity or gastrovascular cavity called the gastrodermis; and between these two, a layer called the mesoglea. This last layer is usually devoid of cells, containing rather a gelatinous matrix. Though all coelenterates are basically tentaculate and radially symmetrical, two different structural types are encountered within the phylum. One type which is sessile, is known as the polyp. The other form is free-swimming and is called the medusa. The hydra exists only in the polypoid stage. The body of Hydra is divided into several specialized regions (see Figure). The first region is called the hypostome, or manubrium. This is a conical elevation located at the oral end of the body, the apex of which is the mouth. Surrounding the hypostome is a second region composed of tentacles. In Hydra, there are usually 5 or 6 hollow and highly extensile tentacles. Along the arms of the tentacles are located batteries of stinging cells called cnidocytes. These specialized cells, which are unique to and characteristic of all coelenterates, contain stinging structures called nematocysts. Nematocysts need not be restricted to the tentacles but may be found else-where in the hydra. A combination of physical and chemical stimuli causes the release of the nematocyst to take place. Undifferentiated cells in the epidermal layer, called interstitial cells, become modified to form the cnidocyte cells. These interstitial cells may also form generative cells, resulting in germ cell-producing bodies as well as buds. The body of the hydra is in the shape of a cylindrical tube. The tube is more or less divided into a stomach or gastric region and a stalk region that terminates in the region of the pedal disk. The gastrodermis in the gastric region contains enzyme-secreting gland cells in addition to flagellated, amoeba-like cells. The cells in the stalk region do not produce enzymes, but are highly vacuolated, indicating intracellular absorption. The pedal disk marks the attaching end of the body. An adhesive substance for attaching the body to some object is secreted by the epidermal cells. These cells can also produce a gas forming a bubble inside the adhesive secretion, which allows the organism to float to the surface of the water.

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Question:

What are the roles of ATP in the cell? How is\simP produced, \sim stored, and utilized?

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Solution:

One of the roles of ATP in the cell is to drive all of the energy- requiring reactions of cellular metabolism. Indeed, ATP is often referred to as the "energy currency" of the cell. The hydrolysis of one mole of ATP yields 7 kcal of energy: Because of their larger free energies of hydrolysis, the first and second bonds broken in ATP are called high energy phosphate bonds and can be written: adenosine - P\simP\simP . \sim \sim A second role of ATP is to activate a compound prior to its entry into a particular reaction. For example, the biosynthesis of sucrose has the following equation: Glucose + fructose\rightleftharpoonssucrose + H_2O The forward reaction is very unfavorable in a plant cell because of a preponderance of H_2O compared to glucose and fructose (recall Le Chatelier's principle). A great deal of energy would be needed to help the forward reaction occur. The cell alleviates this problem by first acti-vating glucose with ATP: glucose + ATP \textemdash\textemdash> glucose -1- phosphate + ADP The phosphate bond attached to glucose is a high energy bond; the cleavage of this bond yields enough energy to enable the following reaction to proceed with the formation of sucrose: glucose -1-phosphate + fructose \textemdash\textemdash>sucrose + phosphate This series of reactions shows how activation by ATP permits a thermodynamically unfavored anabolic process to occur. To produce energy-rich phosphate groups (\simP) , energy from the \sim complete oxidation of glucose in the process of respiration is utilized. The energy is used to add inorganic phosphate (P_i) to ADP to form ATP. The energy-rich phosphate groups are stored in the form of ATP. They are utilized during the hydrolysis of ATP to ADP and P_i. This energy-yielding reaction is coupled to energy-requiring reactions, allowing the latter to occur. The amount of ATP present in a cell is usually small, so muscle cells, which require much energy during a brief period of time (during contraction), may run short of it. Therefore, an additional substance, creatine phosphate, serves as a reservoir of\simP. The terminal phosphate \sim of ATP is transferred by an enzyme to creatine to yield creatine phosphate and ADP. The phosphate bond is also high energy but the\simP must be \sim transferred back to ADP to form ATP in order for it to be used in an energy-requiring reaction. That is, creatine phospate replenishes the cell's ATP supply by donating its\simP to ADP. \sim

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Question:

Which of the following compounds have optical isomers?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E20-0749.htm

Solution:

A carbon atom to which are attached four different atoms or groups is called an asymmetric carbon atom. A molecule containing an asymmetric carbon cannot be superimposed on its mirror image molecule so that every point will coincide. Therefore, the two forms are not identical, since they are not superimposable. Isomers whose molecules differ only in the positioning of groups around asymmetric carbon atoms are called optical isomers. To solve this problem, one must find which carbon is asymmetric, if any. If there is a carbon that has four different groups attached to it, then it must be asymmetric and there will be a possible geometric isomer. In (a), 2-butanol, there exists an optical isomer, since there are four different groups attached to a central carbon atom. In (b), 2-chloropropane, the compound has no optical isomer since two of the four groups are the same. In (c), \alpha-hydroxybutanal, the compound has an optical isomer, since it has four different groups attached to a central carbon atom. In (d), 2-hydroxymethylcyclohexanol, the compound has no optical isomer, since two of the four groups attached to the central carbon atom are the same.

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Question:

A 65-lb horizontal force is sufficient to draw a 1200-lb sled on level, well-packed snow at uniform speed. What is the value of the coefficient of friction?

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Solution:

If the sled moves at constant velocity, it experiences no net force. Therefore, the applied force must be equal to the frictional force. F_applied = 65 lb = Ffriction Since the frictional force is proportional to the normal force F_friction = \mu_k N Applying Newton's Second Law, F = ma, to the vertical forces acting on the block, we find N - mg = ma_y where a_y is the vertical acceleration. In this problem, a_y = 0 because the sled doesn't rise off the surface upon which it slides. Hence N = mg andF_friction = \mu_k N = \mu_k mg ButF_friction = 65 lb Therefore(65 lb) = \mu_k(mg) \mu_k = 65 lb/1200 lb = .054

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Question:

(1) A reaction proceeds five times as fast at 60\textdegreeC as it does at 30\textdegreeC. Estimate its energy of activation. (2) For a gas phase reaction with E_A = 40,000 cal/mole, estimate the change in rate constant due to a temperature change from 1000\textdegreeC to 2000\textdegreeC.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E13-0470.htm

Solution:

The actuation energy E_A can be related to the rate constants k_1 (at temperature T_1) and k_2 (at temperature T_2) by the Arrhenius equation: log (k_2 / k_1) ={-(E_a / 2.303R) } {(1/T_2) - (1/T_1)}, where R = universal gas constant. (1) You are told a reaction proceeds five times as fast at 60\textdegreeC as it does at 30\textdegreeC. Therefore, if k_1= rate constant at 30\textdegreeC = 303\textdegreeK with T_1 = 303\textdegreeK, then k_2 = 5k_1 at 60\textdegreeC = 333\textdegreeK with T_2 = 333\textdegreeK. You are given R. Sub-stitute these values into the Arrhenius equation, andsolve for E_A . Rewriting and substituting, E_a = [(-2.303R) / {(1/T_2) - (1/T_1)}]log (k_2/k_1) = [{(- 2.303) (1.987)(1 k cal / 1000cal)} / {(1/333) - (1/303)}] log 5 =(15.4 kcal / mole)(.699) = 10.8 k cal / mole. [Note: 1 k cal/1000 cal is a conversion factor to obtain the correct units.] To answer (2) find k_2/k_1 from the Arrhenius equation. Rewriting and substituting, k_2 / k_1= antilog [(E_a / 2.303R) {(1/T_2) - (1/T_1)}] = antilog [{(-4000) / (2.303)(1.987)} {(1/2,273) - (1/1,273)}] =antilog 3.02 = 1.05 × 10^3 . That is, the rate should be about 1050 times as great at 2000\textdegreeC as at 1000\textdegreeC.

Question:

Write a program in the PDP-8 computer Assembler Language to find the factorial of numbers. The program should use Sub-routine jumps The numbers are stored in an array starting at location 300_8. The results, i.e., the values of the factorials are to be stored in another array starting at location 200_8. The end of the numbers array is indicated by the occurrence of a negative number. The program is to start at location 400_8.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G09-0217.htm

Solution:

The mathematical background for this problem is as follows: Factorial 0 = 0! = 1 Factorial 1 = 1! =1 Factorial 5 = 5! = 5 × 4 × 3 × 2 × 1 (= 120). Factorial n = n! = n × (n -1) × (n - 2)×......× 3 × 2 × 1. The strategy to be used for solving this problem is as follows: A number will be obtained first from the array. It will be tested to see if it is negative. If it is indeed negative, then it will indicate an end of the array, and the program will stop. Next, the number will be tested to see if it is equal to zero. If it is indeed zero, then the value of its fac-torial will immediately be entered in the results-array as equal to one. After this, the number will be tested to see if it is equal to 1. If it is indeed equal to 1 then the value of its factorial will be written down as 1 also . If the number was neither negative, nor zero nor equal to one, then it means that it is a positive number n (\geq2). In this case we jump to a subroutine which does some calcu-lations and returns a value of n! Of course, the way the program is written for this ex-ample, it will be seen that the factorial subroutine itself calls a multiplication, subroutine to do the various multi-plications required to be done in obtaining the value of n ! The program is written as follows. It should be noted that the main program starts at location 400_8, a factorial subroutine starts at location 440_8 and a multiplication subroutine starts at location 450_8. Each instruction is to be punched on one card, starting at column 2. A comment is indicated by a slash (/). Everything on the card follow-ing a / is regarded as a comment. \textasteriskcentered400 /INDICATES TO THE COMPUTER THE STARTING LOCATION OF THE PROGRAM LOOP1,ISZ ARYADR /LOOP1 IS A LABEL TO THIS STATEMENT. ARYADR IS THE NAME OF A LOCATION WHICH STORES A VALUE (277_8), i.e., THE STARTING ADDRESS (LESS 1) OF THE NUMBER-ARRAY. ISZ INCREMENTS THE CONTENT OF 277_8 BY 1 TO 300_8 AT THE START. ISZ RESADR /RESADR IS THE NAME OF A LOCATION WHERE AN ADDRESS IS STORED. THIS ADDRESS IN TURN IS THE LOCATION WHERE THE RESULT OF A FACTORIAL COMPUTATION IS TO BE STORED CLA CLL /INITIALLY CLEAR THE AC AND THE LINK L DCA I RESADR /CLEAR THE CONTENT OF RESADR. NOTE THAT INDIRECT ADDRESSING IS TO BE USED BECAUSE THE LOCATION OF THIS STATEMENT IS 403_8, i.e., ON PAGE 2. BUT THE LOCATION OF RESADR IS ON A DIFFERENT PAGE. RESADR IS IN LOCATION 200_8 WHICH IS ON PAGE 1. TAD I ARYADR /LOAD THE NUMBER INTO THE AC SPA /CHECK IF THE NUMBER IS \geq 0. SKIP IF IT IS. HLT /HALT IF THE NUMBER IS NOT \geq 0 SNA /SKIP IF THE AC CONTENT IS NOT 0. JMP LOOP2 /THE PROGRAM COMES TO THIS STEP IF THE AC IS 0. THEN IT JUMPS TO LOOP 2. DCA A /THE NUMBER IS STORED IN LOCATION A. AS A IS ON THE SAME PAGE DIRECT ADDRESSING CAN BE USED RATHER THAN INDIRECT ADDRESSING REQUIRED FOR ARYADR. DIRECT ADDRESSING SAVES TIME. TAD A /THE NUMBER IS AGAIN LOADED INTO THE AC. CMA IAC /THE TWO'S COMPLEMENT OF THE NUMBER IS FORMED. IAC /THE TWO'S COMPLEMENT IS INCREMENTED BY 1. SNA /CHECK IF THE INCREMENTATION RESULTS IN A ZERO. THIS WOULD HAPPEN IF THE NUMBER WAS ORIGINALLY1. JMP LOOP2 /JUMP TO LOOP 2 IF THE VALUE IS ZERO. DCA B /IF THE NUMBER ISN'T NEGATIVE, OR 0 OR 1 THEN STORE THE NEGATIVE OF THE NUMBER INTO B. JMS FCTRL /GO TO THE SUBROUTINE TO CALCULATE FACTORIALS. TAD A /LOAD THE AC WITH THE FACTORIAL OF THE NUMBER FROM A. DCA I RESADR /STORE THE VALUE OF THE FACTORIAL INTO RESADR. JMP LOOP 1 /GO BACK TO LOOP1 TO FETCH A NEW NUMBER. LOOP2, IAC /COMES TO THIS STEP IF THE NUMBER IS 0 OR 1. THE AC IS 0. THEREFORE IAC MAKES AC = 1. DCA I RESADR /THE VALUE 1 GETS STORED IN RESADR IN THIS CASE. JMP LOOP 1 /GO BACK TO LOOP 1 TO FETCH A NEW NUMBER. \textasteriskcentered430 /INDICATES START OF A NEW MEMORY BLOCK TO BE USED BY THE PROGRAM. ARYADR, 277 /SHOWS THAT ARYADR LOCATED IN 430_8 CONTAINS A VALUE 277_8. RESADR, 177 /RESADR, LOCATED IN 431_8 CONTAINS 177_8. A, 0 /A, IN LOCATION 432_8 CONTAINS 0 B, 0 /B HAS 0. C, 0 /C HAS 0 \textasteriskcentered440 /START OF A NEW BLOCK FCTRL, 0 /FIRST LOCATION OF THE SUBROUTINE FCTRL. THIS LOCATION HAS A VALUE 0. WHEN THE SUBROUTINE WILL BE CALLED, THE RETURN ADDRESS WILL BE STORED IN THIS LOCATION. LOOP3, TAD B /THE NEGATIVE OF THE NUMBER IS LOADED INTO THE AC FROM B. DCA /THE SAME VALUE IS STORED IN C. JMS MLTPLY /GO TO A SUBROUTINE TO MULTIPLY 2 NUMBERS. ISZ B /INCREMENT B AND CHECK IF RESULT IS ZERO. JMP LOOP 3 /REPEAT LOOP 3 IF B IS NOT ZERO. JMP I FCTRL /IF B IS ZERO, PROGRAM COMES TO THIS STEP, WHICH INDICATES THAT THE VALUE OF THE FACTORIAL IS READY. THEREFORE A JUMP IS PERFORMED TO THE CALLING PROGRAM VIA THE ADDRESS STORED IN LO-CATION CALLED FCTRL. \textasteriskcentered450 /START OF A NEW MEMORY BLOCK. MLTPLY, 0 /FIRST LOCATION IN THE SUBROUTINE MLTPLY. LOOP 4, TAD A /ADD THE NUMBER FROM A INTO AC. ISZ C /INCREMENT C AND CHECK IF IT IS ZERO. JMP LOOP 4 /IF C IS NOT ZERO THEN GO BACK TO LOOP 4 TO ADD A ONCE AGAIN TO AC. DCA A /WHEN C IS ZERO, IT SIGNIFIES THAT THE MULTI-PLICATION IS OVER. AND THEREFORE STORE THE CONTENTS OF AC INTO A. JMP I MLTPLY /RETURN TO THE CALLING FCTRL ROUTINE, VIA THE ADDRESS STORED IN MLTPLY. \textasteriskcentered300 /START OF A NEW BLOCK. +6 /A NUMBER +6, STORED IN LOCATION 300_8. THIS IS THE FIRST NUMBER IN THE NUMBERS-ARRAY. 0 /ANOTHER NUMBER IN THE NUMBERS-ARRAY. 1 /ANOTHER NUMBER IN THE NUMBERS-ARRAY. -2 /A NEGATIVE NUMBER IN THE NUMBERS-ARRAY TO INDICATE THE END OF THE NUMBERS-ARRAY. $ /INDICATES END OF PROGRAM.

Question:

An object is 4 inches from a concave mirror whose focal length is 12 inches. Where will the image be formed?

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/Users/wenhuchen/Documents/Crawler/Physics/D27-0847.htm

Solution:

First make a ray-diagram. Solving mathematically, D_0 is 4 inches and f is 12 inches. Substitution in (1/D_0) + (1/D_I) = (1/f) (1/4 in.) + (1/D_I) = (1/12 in.) whence D_I = - 6 in. which means that since D_I is negative the image is 6 inches from the mirror on the same side on the mirror as the object and is virtual.

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Question:

What structures are present in both the root and the stem of a higher vascular plant? Describe the function of each structure.

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/Users/wenhuchen/Documents/Crawler/Biology/F10-0247.htm

Solution:

Both the stem and the root of a plant are covered on the outside by a layer of rectangular cells, known as the epidermis (see figures 1 and 2). The cells of the epidermis of the root are thin-walled and non-cutinized, in contrast to those epidermal cells of the stem which are heavily cutinized. If a waxy cuticle were present in the root epidermis, it would undoubtedly interfere with the absorption of water. The epidermis of the root gives rise to root hairs which increase the surface area for absorption. Just inside the epidermis is the cortex. The cortex of the root is a wide area composed of many layers of large, thin-walled, nearly spherical parenchymal cells. The cortex of the stem is narrower than that of the root. The cortical cells differ from those of the root in that the former is photosynthetic while the latter is not. In addition, there is an outer layer of thick-walled collenchymal cells in the stem, which serve as supportive tissue. Collenchymal cells are not present in the root. Inside the ring of cortex is a one-called thick layer called the endodermis. The endodermis is interrupted at some points by passage cells through which water and minerals can pass. Immediately adjacent to the endodermis is another layer knows as the pericycle. The pericycle of the root consists of thin-walled parenchymal cells which can be transformed into meristematic cells that later give rise to lateral roots. The pericycle of the stem, however, has thick-walled cells for the purpose of support. On the inner border of the pericycle is the vascular system, composed of xylem and phloem tissues. In the root, the vascular tissues occupy the central longi-tudinal portion. The cells of the root xylem, tracheids and vessels, are usually arranged in the shape of a star, or like the spokes of a wheel. They are thick-walled and elongated tubular cells which conduct water and dissolved mineral salts. Between the arms of the xylem star are bundles of phloem cells, which are smaller and thinner- walled than the xylem. Phloem cells function in the conduction of organic substances. In the stem, the xylem and phloem tissues are arranged in scattered bundles in the monocots (see figure 3) or arranged in a ring in the dicots. The stem contains a central axial core of tissue, called the pith, which is composed of colorless parenchy-mal cells that serve in the function of nutrient storage.

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Question:

Describe the characteristic differences between the monotreme, marsupial and placental mammals.

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/Users/wenhuchen/Documents/Crawler/Biology/F13-0343.htm

Solution:

The class Mammalia is divided into three sub-classes: Prototheria (Monotremes), Metatheria (Marsupials), and Eutheria (placental mammals). Of these three, the mono-tremes can be considered the most primitive. They are an early offshoot of the main mammalian groups. The Marsupials and Placentials appeared at about the same time. Major characteristics of mammals include the possession of mammary glands and hair. Most mammals have sweat glands. All mammals are homeotherms. Monotremes are peculiar in that they possess mammalian as well as reptilian traits. They are classified as mammals, however, because the number of mammalian traits far exceed the number of reptilian traits. Some of the more outstanding mammalian traits include: a layer of hair covering the body and the secretion of milk by mammary glands. There are only two surviving species of monotremes: the duck-billed platypus and the spiny anteater, both of which are found in Australia. The trait that most clearly distinguishes monotremes from other groups of mammals is their ability to lay eggs. In marsupials and placentals, embryonic development occurs in the uterus of the female. The characteristic difference between the marsupials and the placentals is the time of embryonic development within the uterus. Marsupial embryos undergo a short developmental period before they leave the uterus. Embryonic development is completed in an abdominal pouch of the mother where the embryo is attached to a nipple. In contrast to this is the placentals. In this group, the embryo develops completely in the uterus of the mother. In both marsupial and placental mammals, the young are born alive.

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Question:

You have just received the output from one of your Basic Assembly Language (BAL) programs, but to your dismay, theoutput is not what you had hoped for. To debug this program, your first reaction is to check the contents of your storagelocations. What procedures would you follow inthis processof debugging?

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G09-0211.htm

Solution:

It is the default of the computer's operating system to provide a Cross Reference Table and a P/P Storage Dump with each program listing. Among other things, the Cross Reference Table yields the names ofall variables and labels used in the program and their corresponding locationsin memory. The variables and labels of this table are gen-erally displayedin alphabetical order. Thus, finding any member of this table is comparableto obtaining a word from the page of a dictionary. When this variableis found the table provides us with the amount of space it occupiesin memory (LEN), where it is defined in the listing (DEFN), where itis referenced in the listing, and most important, the location of the variablein memory (value). The order of the information in the table will notnecessarily be as stated above. For our present purpose all we need to concern our-selves with is thevalue (address) that the table indicates for the symbol. Going to the P/P Storage Dump, we match our symbol value with the address we are searchingfor and also the contents of this address. By the use of this procedurewe are able to check the contents of any symbol in our program. There are two points to be noted: 1. 1. Depending on the computer, the Cross Reference Table will probably producemore information than presented here. 2. 2. Some computers will refer to the P/P Store Dump by a slightly different name, for example, Post Execution Storage Dump.

Question:

(a) What is the magnitude of the electric field at a distance of 1\AA (= 10^\rule{1em}{1pt}8 cm) from a proton? (b) What is the potential at this point? (c) What is the potential difference, in volts, between positions 1 and 0.2 \AA from a proton?

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/Users/wenhuchen/Documents/Crawler/Physics/D18-0582.htm

Solution:

(a) From Coulomb's law E = e / r^2 \approx [(5 × 10^\rule{1em}{1pt}10statcoulomb) / (1 × 10^\rule{1em}{1pt}8 cm)^2 ] \approx 5 × 10^6statvolts/cm \approx (300)( 5 ×10^6) V/cm \approx 1.5 × 10^9V/cm Here, e is the unit of electronic charge, and r is the distance between the proton and the point at which we calculate the field. We have also used the fact that 1statvolt/ cm = 300 (V / cm) The field is directedradiallyoutward from the proton. (b) The electrostatic potential at a distance r from the proton is \Phi(r) = e / r \approx (5 × 10^\rule{1em}{1pt}10statcoulomb) / (1 × 10^\rule{1em}{1pt}8 cm) \approx 5 × 10^\rule{1em}{1pt}2statvolts \approx 15V from the conversion factor given above. (c) The potential at 1 × 10^\rule{1em}{1pt}8 cm is 15 V; at 0.2 × 10^\rule{1em}{1pt}8 cm it is 75 V. The difference 75 \rule{1em}{1pt} 15 = 60 V.

Question:

Assume that the Gross National Product (GNP) is now 1 ×1012dollars, and is increasing by 4% each year. Also assumethat the gross output of the computing industry (GOC) is now 6 × 10^9 dollars, and is increas-ing by 17% each year. If these growth rates are continuous, starting from this year, in how many years will GOC exceed GNP?

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G24-0570.htm

Solution:

The answer to this problem is a fictional one, since GOC will never exceedGNP in reality. Projections such as these are interesting brain-teasers. The logic is straightforward, and the program is presented below : 10 REMGNP VS.GOC 20 READ GNP, GOC 25 YEAR = 0 30 NUP = 0.0 40 NUC = 0.0 50 NUP = GNP\textasteriskcentered .04 60 GNP = GNP + NUP 70 NUC = GOC\textasteriskcentered .17 80 GOC = GOC + NUC 90 YEAR = YEAR + 1 100 IF GOC \leq CNP THEN 5\varnothing 110 PRINT YEAR 120 DATA 1000, 6 130 END

Question:

A damper (or dashpot) is connected to the mass M of the previous problem. This could represent air resistance. The entire system could be a simple model of an automobile wheel suspension system (assuming the automobile body immobile in a vertical direction). Then the damper acts as a shock absorber. As before, the system is displaced and released and x(t_0) = x_0 and v(t_0) = v_0 . It can be shown that the motion of the system Is described by the following dif-ferential equation: Mẋ + Dẋ + Kx(t) = 0 where D is the damping factor of the dashpot and ẋ = v(t) \equiv velocity at time t . Model and simulate the motion of the system from time t = t_0 to t = t_f , using a digital computer program, FIG. 1

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G12-0304.htm

Solution:

Repetition of the procedure used in the previous problem leads to step 6, where the Euler predictor-corrector method will be used again. Transforming the differential equation to a simultaneous system of first-order equations: ẋ = v(t)(1b) ṿ = -[D{_\ast}v(t) + K{_\ast}x(t)]/M(2b) The relation can be diagrammed as shown in FIG.2 The only change in the program needed is substitution of the new expression for acceleration. Since this expression appears in two lines and is fairly long, a statement function is used. This has the additional effect of making the program more general. It can be applied to any second-order differential equation. Only the statement function definition and the I/O statements vary. The program: REAL K,M,N F(Y,Z) = -(D{_\ast}Y + K{_\ast}Z)/M READ K,M,N,T,TFIN,XCOR,VCOR,ACCUR,D DT = (TFIN - T)/N 50X = XCOR V = VCOR A = F(V,X) PRINT T,X T = T + DT IF(T.GT.TFIN) STOP XPRED = X + V {_\ast} DT VPRED = V + A{_\ast} DT 100XDOT = VPRED VDOT = F(V,XPRED) XCOR = X + 0.5 {_\ast}(V + XDOT) {_\ast} DT VCOR = V + 0.5 {_\ast}(A + VDOT) {_\ast} DT XDIF = ABS (XCOR - XPRED) VDIF = ABS(VCOR - VPRED) IF (XDIF .LE .ACCUR. AND. VDIF{_\ast} LE. ACCUR) GO TO 50 XPRED = XCOR VPRED = VCOR GO TO 100 END

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Question:

In the popular nursery rhyme "Mary Had a Little Lamb", her lambwould follow Mary wherever she went. Discuss a possi-blebiological explanation.

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/Users/wenhuchen/Documents/Crawler/Biology/F31-0813.htm

Solution:

The bond established between Mary and her lamb might have resultedfrom a form of learning called imprint-ing. The concept of imprintingwas first formulated byKonradLorenz, an Austrianethologist (ethologyis the study of behavior). Lorenz discovered that new-born geesewill follow the first moving object they see, especially if it makes noise. They form a strong and lasting bond to it, adopting the object as theirparent. Usually, the first moving object the bird sees is its mother. Thus imprinting serves to establish a bond between the young and their motherin nature. This type of response has important survival values, suchas keeping the young near their parent and establishing species recognitionand inter-action. Imprinting is characterized by having a short timeperiod during which it can occur. This critical period lasts only about 32 hours in ducklings. Lorenz demonstrated imprinting by allowing goslings (young geese) tosee him as the first moving creature. They then followed Lorenz as if he weretheir parent. Young birds of most species become independent when they attain sexualmaturity. If birds are reared by a mother of a different species, they attemptto mate with birds of the foster mother's species when sexually mature. Although imprinting was first demonstrated in birds, it is known to occurin sheep, goats, deer and other mammals. Orphan lambs reared by humansfollow them and show little interest in other lambs. That is why "every-wherethat Mary went, her lamb was sure to go."

Question:

Describe the interrelations between the Basic Assembler Language (BAL) Cross Reference Table and its Post Execution Storage Dump.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G09-0209.htm

Solution:

Along with other types of information, the Cross Reference Table provides the location of all the symbols used within the program. By finding this address in the Post Execution Storage Dump, the contents of a symbol can be easily found. In the Cross Reference Table, the address of the symbols are coded in hexadecimal (HEX). Not only are these addresses coded in hexadecimal in the Post Executional Storage Dump, but so is all other data found in this Dump. The leftmost column of the Dump contains addresses that consecutively increase by 32 bytes; here it is impor-tant to remember the dump is coded in HEX. To the right of the addresses are rows of 32 bytes. These bytes represent the contents of 32 consecutive bytes in core. The first byte of each row represents the content of the location listed in the address column to the left of it. Let's represent a location in the address column by the letters 'DCA'. By hexadecimally adding 1 to 'DCA' we can determine the content of the second byte in this specific row. For example so DCACONTAINS90 DCA + 1CONTAINSEC DCA + 2CONTAINSDO DCA + 30CONTAINS5A and DCA + 31CONTAINSCO Thus, by adding the required amount, the contents of any byte in the row can be determined. It should be noted again that the addition is a hexadecimal one. In summation, we can determine the contents of a symbol by obtaining its address and length from the Cross Reference Table and then locating the content of this symbol in the Post Execution Storage Dump. When the address is found in the Dump, the number of bytes to be observed is dictated by the length of the symbol. It should be noted that in the Dump the addresses are generally denoted by six hexadecimal digits, but on most occasions we need only concern ourselves with the last three. Also, to the right of the dump we usually find a stan-dard (English) translation of each hexadecimal row of the Dump. This translation confirms the contents of the Dump and makes it more comprehensible.

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Question:

A parachutist, after bailing out, falls 50 meters without friction. When the parachute opens, he decelerates downward 2.0 meters/sec^2. He reaches the ground with a speed of 3.0 meters/sec. (a) How long is the parachutist in the air? (b) At what height did he bail out?

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/Users/wenhuchen/Documents/Crawler/Physics/D04-0142.htm

Solution:

(a) The parachutist, starting at rest, falls 50 m at an acceleration equal to g, the accelera-tion of gravity. Since this is constant s = v_0t + (1/2)at^2, where v_0 is the initial velocity. Here v_0 = 0 and 50 = (1/2)gt^2, t = \surd(100/g) = 10/g^1/2 = 10(3.13) = 3.2 sec He then decelerates at 2.0 m/sec2 until he reaches a final velocity, v_f of 3 m/sec^2. When he begins his deceleration he has reached a speed: v_f= v_0 + at,v_0 = 0 =gt= 9.8(3.2) = 31.3 m/sec Thus:v_f' = v_0' +a't' 3 = 31.3 - 2t';2t' = 31.3 - 3;t' = 14.2 total time of flight = t + t' = 3.2 sec + 14.2 sec = 17.4 sec (b) We know that the parachutist falls 50 meters before the parachute opens. Thus, the problem reduces to one in which we must find the distance s the para-chutist travels until he is decelerated to a speed of 3.0 m/sec, having started at velocity v = 31.3 m/sec. We know that: S =vt' + (1/2) at'^2 S = (31.3 m/sec)(14.2 sec) + 1/2(-2.0 m/sec^2)(14.2 sec)^2 = 444.5 - 201.6 = 242.9 meters. Hence the parachutist bailed out from a height of 50 + 242.9 = 292.9 meters.

Question:

What is the difference between the number of carbon atoms in 1.00 g of C-12 isotope (atomic mass = 12.000 g/mole) and 1.00 g of C-13 isotope (atomic mass = 13.003 g/mole

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/Users/wenhuchen/Documents/Crawler/Chemistry/E04-0111.htm

Solution:

The difference in the number of carbon atoms in each sample is equal to the difference in the number of moles times Avogadro's number 6.02 × 10^23 . Hence , wemust begin by calculating the number of moles of C-12 and of C-13 in 1.00 g samples of each. The number of moles is equal to the mass divided by the atomic weight. Therefore moles C-12 = (mass C-12) / (atomic mass C-12) = 1.00g /{12.000(g/mole)} = 0.083 mole and moles C-13 = (mass C-13) / (atomic mass C-13) = 1.00g /{13.003(g/mole)} = 0.077 mole. The difference in the number of moles between the two samples is (moles C-12) - (moles C-13) = 0.083 - 0.077 = 0.006 mole. Multiplying by Avogadro's number gives the difference in the number of carbon atoms in the two samples: number of carbon atoms = 0.006 mole × 6.02 × 10^23(molecules / mole) = 3.61 × 10^21 molecules.

Question:

Trace the path of blood through the human heart.

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/Users/wenhuchen/Documents/Crawler/Biology/F15-0374.htm

Solution:

The heart is the muscular organ that causes the blood to circulate in the body. The heart of birds and mammals is a pulsatile four-chambered pump composed of an upper left and right atrium (pi., atria) and a lower left and right ventricle. The atria function mainly as entryways to the ventricles, whereas the ventricles supply the main force that propels blood to the lungs and throughout the body. Depending on where the blood is flowing from, it would enter the heart via one of the two veins: the superior vena cava carries blood from the head, neck and arms; the inferior vena cava carries blood from the rest of the body. The blood from these two veins enters the right atrium. When this chamber is filled with blood, the chamber contracts and forces the blood through a valve called the tricuspid valve and into the right ventricle. Since this blood has returned from its cir-culation in the body's tissues, it is deoxygenated and contains much carbon dioxide. It therefore must be transported to the lungs where gas exchange can take place. The right ventricle contracts, forces the blood through the pulmonary semilunar valve into the pulmonary artery. This artery is unlike most arteries in that it carries de-oxygenated blood. The artery splits into two, with one branch lead-ing to each lung. The pulmonary arteries further divide into many arterioles, which divide even further and connect with dense capil-lary networks surrounding the alveoli in the lungs. The alveoli are small sac-like cavities where gas exchange occurs. Carbon dioxide diffuses into the alveoli, where it is expelled, while oxygen is picked up by the hemoglobin of the erythrocytes. The capillaries join to form small venules which further combine to form the four pulmonary veins leading back to the heart. The pulmonary veins are unlike most veins in that they carry oxygenated blood. These veins empty into the left atrium, which contracts to force the blood through the bicuspid (or mitral) valve into the left ventricle. When the left ventricle, filled with blood, contracts, the blood is forced through the aortic semilunar valve into the aorta, the larg-est artery in the body (about 25 millimeters in diameter). The aorta forms an arch and runs posteriorly and inferiorly along the body. Before it completes the arch, the aorta branches into the coronary artery, which carries blood to the muscular walls of the heart itself, the carotid arteries, which carry blood to the head and brain, and the subclavian arteries, which carries blood to the arms. As the aorta runs posteriorly, it branches into arteries which lead various organs such as the liver, kidney, intestines and spleen and also the legs. The arteries divide into arterioles which further divide and become capillaries. It is here that the oxygen and nutrients dif-fuse into the tissues and carbon dioxide and nitrogenous wastes are picked up. The capillaries fuse to form venules which further fuse to become either the superior or inferior vena cava. The entire cycle starts once again. The part of the circulatory system in which deoxygenated blood is pumped to the lungs and oxygenated blood returned to the heart is called the pulmonary circulation. The part in which oxygenated blood is pumped to all parts of the body by the arteries and deoxygenated blood is returned to the heart by the veins is called the systemic circulation.

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Question:

To obtain 500 kg of glass composed of equimolar proportions of Na_2SiO_3 and CaSiO_3, what weights of Na_2CO_3, CaCO_3, and SiO_2 should be used?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E27-0898.htm

Solution:

When a mixture of limestone, sodium carbonate, and sand is melted together, a clear, homogeneous mixture of sodium and calcium silicates is produced. When the fused liquid is cooled, it becomes more and more viscous and finally hardens to a transparent rigid mass called glass. From the stoichiometry of this equation, an equal number of moles of CaCO_3 and Na_2CO_3 plus twice as many moles of SiO_2 are consumed to produce an equal number of moles of CaSiO_3 and Na_2SiO_3. To solve this problem, determine the number of moles of glass substituents produced; this number will be equal to the number of moles of CaCO_3 and Na_2CO3 and half the number of moles of SiO_2. From the number of moles and each constituent's molecular weight, the actual weights can be determined. The molecular weights are MW of Na_2SiO_3 = 122 g / mole;MW of CaSiO_3 = 116 g / mole; MW of CaCO_3 = 100 g / mole;MW of Na_2CO_3 = 106 g / mole; MW of SiO_2 =60 g / mole. Since the number of moles of each product is unknown (but equal), the number of moles is set equal to x and (x) (122 g / mole) + (x) (116 g / mole) = 500,000 g of glass 238 x = 500,000 g the number of moles = 2101 moles. Thus, the number of moles of the reactants needed are CaCO_3 = 2101 moles; Na_2CO_3 = 2101 moles; SiO_2 = 4202 moles; and the weights of each are CaCO_3 : (2101 moles) (100 g / mole)= 210,100 g = 210.1 kg Na_2CO_3 + (2101 moles) (106 g / mole)= 222,706 g = 222.7 kg SiO_2 : (4202 moles) (60 g / mole)= 252,120 g = 252.1 kg.

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Question:

Write a procedure which reads a matrix of non-negative integers and stores it in an array of type MATRICES = ARRAY [1. .XSIZE,1 .. YSIZE] of NONNEGINTEGER where XSIZE and YSIZE are appropriately defined con-stants and NONNEGINTEGER is an appropriately defined type.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G16-0415.htm

Solution:

To guard against the possibility of the input data being incomplete (orthere being too much input data), the matrix elements are assumed to be followed by a negative end- marker. The procedure checks every datum and, if it detects the end-marker prematurely, it writes a warning message and stops immediately. PROCEDURE READMATRIX (VAR MATRIX: MATRICES; VAR VALID: BOOLEAN); LABEL 99; VAR DATUM: INTEGER; X:1.. XSIZE; Y:1.. YSIZE; BEGIN VALID: = FALSE; (\textasteriskcenteredbeing pessimistic!\textasteriskcentered) FOR X: = 1ToXSIZE DO FOR Y: = 1ToYSIZE DO BEGIN READ (DATUM); IF DATUM> = o THEN MATRIX [X, Y] := DATUM ELSE(\textasteriskcenteredtheendmarkerhas been read \textasteriskcentered) BEGIN WRITELN ('DATA INCOMPLETE'); GOTO 99; (\textasteriskcentered leave the procedure immediately \textasteriskcentered) END END; READ (DATUM);(\textasteriskcentered..... this should be theendmarker\textasteriskcentered) IF DATUM>=o THEN WRITELN ('TOO MUCH DATA') ELSE VALID: =TRUE; 99: END. (\textasteriskcentered READMATRIX \textasteriskcentered)

Question:

(a) isobutane and isobutylene (b) sec-butyl alcohol and n-heptane.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E21-0768.htm

Solution:

To distinguish between two compounds, try to select for a characterization test. Begin by writing the structures of the compounds involved. You see that isobutylene is an alkene, and that isobutane is an alkane. Thus, you should be able to distinguish isobutylene from isobutane by an addition reaction. Addition reactions are characteristic of alkenes and other unsaturated compounds. When Br_2/CCI_4 , a reddish brown solution, is in contact with an alkene, the following reaction occurs: When the alkene consumes the Br_2, the red color disappears. Thus, alkenes can be detected by their ability to decolorize the red solution of Br_2/CCI_4 . You can also use a cold, dilute, neutral permanganate solution, the Baeyer test, instead of Br_2/CCI_4 to detect the presence of an alkene. Permanganate solution is purple colored, upon reaction with an alkene, the purple disappears and is replaced by a brown manganese dioxide precipitate. The reaction can be written Alkenes do give this reaction. Thus, you now have two simple tests to distinguish isobutane from isobutylene. These two can be distinguished by using a chromic anhydride reagent. Secondary alcohols, such as sec-butyl alcohol, are oxidized by chromic anhydride, CrO_3 , in aqueous sulfuric acid. The clear orange solution turns opaque, blue-green when the alcohol is added. In other words, N-heptane is an alkane, and, as such, does not turn the solution blue green. Thus, you can distinguish sec- butyl alcohol from n-heptane by adding chromic anhydride.

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Question:

Describe the changes accompanying the ripening of a fruit. Why do fruits kept in plastic bags or in the refrigerator ripen more slowly?

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/Users/wenhuchen/Documents/Crawler/Biology/F09-0232.htm

Solution:

Ripening in fruit involves a number of changes that convert the fruit from a seed-manufacturing to a seed-dispersing organ. One transformation that frequently occurs is a color change from green to yellow, orange, red, or blue; this increases the fruit's chance of being spotted by potential dispersal agents such as animals and insects. Chlorophylls are broken down and other pigments such as the red and blueanthocyaninsare synthesized. Another change is the conversion of starch, organic acids, and oils to sugar so that the fruit becomes sweet. Simultaneously, the fleshy part of the fruit softens as a result of enzymatic digestion of pectin, the principle component of the middle lamella of the cell wall, which gives the unripe fruit its toughness. When the middle lamella is weakened, cells are able to slip past one another. Volatile flavor components are synthesized, and, together with the conspicuous coloring of the fruit, they serve to attract fruit eaters. During the ripening of a fruit, there is a large increase in cellular respiration, evidenced by a rapid uptake of oxygen. Cellular respiration can be suppressed or retarded by a decrease in available oxygen. Keeping fruits in plastic bags lowers the amount of oxygen available to the fruits and thus delays ripening. Cold also suppresses cellular respiration; this is why we put fruits in plastic bags and/or refrigerate them in order to slow down their rate of ripening.

Question:

B^10 is bombarded with neutrons, and \alpha particles are observed to be emitted. What is the residual nucleus?

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/Users/wenhuchen/Documents/Crawler/Physics/D34-1026.htm

Solution:

Only \alpha particles (helium nuclei) are observed to be emit-ted in the reaction. The reaction can be described as follows: ^10_5B + 1 0 n \rightarrow ^A_Z(X) + ^4_2He where the superscript gives the mass number A. It is the total number of protons and neutrons in that nucleus. In a nuclear reaction the total nucleon number and the total charge is conserved, therefore the mass number A of the unknown nucleus must be such that 10 + 1 = A + 4. Hence A = 7. The subscripts refer to the atomic numbers, total number of protons in each nucleus. Since the above reaction involves protons and neutrons only, the protons carry the total charge. The conservation of total electric charge in that case reduces to the conservation of the total number of protons. Therefore, 5 + 0 = Z + 2 orZ = 3. The nucleus with A = 7, Z = 3 is^7_3Li.

Question:

State the basic laws of Boolean algebra.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G03-0039.htm

Solution:

If A, B and C are Boolean variables, then the basic laws of Boolean algebras are (where + and \textbullet are binary operators representing OR (inclusive) and AND respectively): 1A:A + A = A [Idempotent law for + ] 1B:A \bullet A = A [Idempotent law for \textbullet ] 2A:A + B = B + A [Commutative law for + ] 2B:A \bullet B = B \bullet A[Commutative law for \textbullet ] 3A:A + (B+C) = (A+B) + C [Associative law for + ] 3B:A \bullet (B\bulletC) = (A\bulletB) \textbullet C [Associative law for \textbullet ] 4A:A \bullet (B+C) = (A\bulletB) + (A\bulletC) [Distributive law for \textbullet over + ] 4B:A + (B\bulletC) = (A+B) \bullet (A+C) [Distributive law for + over \bullet ] 5A:A + 1 = 1 (Law of Union) 5B:A \bullet 0 = 0 (Law of Intersection) 6A:A \bullet 1 = A [1 is the identity element for \textbullet ] 6B:A + 0 = A [0 is the identity element for + ] The law of \backsim : 7: \textasciitilde(\textasciitilde A) = A [Double Negative Law or Involution Law] 8A: A + \backsimA = 1 │Law of complement 8B: A \bullet \backsimA= 0 │ 9A: \backsim(A+B) = \backsimA \bullet \backsim B [deMorgan's law] 9B: \textasciitilde(A \textbullet B) = \textasciitilde A + \textasciitilde B [deMorgan's law] 10: 1^\backsim = 0 and 0\backsim= 1 Any system obeying these laws is known as Boolean algebra. A set with its subsets and the subset operations union and intersect (+ and \bullet , respectively) is a Boolean algebra. The one-to-one correspondence between Boolean expressions and switching circuits suggests that a switch-ing algebra is a Boolean algebra. In fact, 1A - 3B obviously holds true for switching circuits. The switching analog of 4A claims that Fig. 1, and Fig. 2 are equivalent circuits. Since both circuits conduct unless A is open or both B and C are open, they are, in fact, equivalent. Therefore, 4A holds true for switching circuits. And similarly, 4B - 10 can be shown to hold for switching circuits by exhibiting the switching circuits representing the left and right sides of the equation and show-ing that the two are equivalent. Once it is shown that all 18 laws hold for switching circuits, they can be used for simplifying circuits. For example, 4A and 4B reduce four-switch circuits to three-switch cir-cuits. Actually all 18 laws are not a minimal set of Boolean algebra laws, since laws 1A, 1B, 3A, 3B, 5A, 5B, 7, 9B and 10 can be derived from the other laws. For example, once this fact is established, only 8 properties need be satisfied to determine a Boolear algebra, since the other ten follow immediately. Since the gate elements OR, AND and NOT are representations of the operations +, \textbullet and \textasciitilde respec-tively, it follows that a gating algebra is a Boolean algebra. This fact can be used for simplifying gating circuits. For instance 4A and 4B reduce 3-gate circuits to 2-gate circuits.

Question:

(a) CH_3CHOHCH_2CH_2CH_2CH_3

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/Users/wenhuchen/Documents/Crawler/Chemistry/E20-0738.htm

Solution:

All of these compounds are alcohols; they fit into the general formula R-OH, where R is any alkyl group. In the systematic naming of any alcohol, the following rules should be followed: (1) The longest chain that contains the hydroxyl group (OH) is considered the parent compound. (2) The ^-e ending of the name of this carbon chain is replaced by - ol. (3) The locations of the hydroxyl and any other groups are indicated by the smallest possible numbers. Thus, compound (a) has a 6-carbon chain and an hydroxyl group on carbon number 2. The name of this compound is 2-hexanol. Compound (b) has a 5-carbon chain, because this is the longest chain that contains the hydroxyl group. There is a methyl group on carbon number 3 , and an hydroxyl group on carbon number 1. The name of this compound is, therefore, 3-methyl-1- pentanol. Compound (c), has a 6-carbon chain that contains the OH group, two methyl groups on carbons 2 and 3, and an hydroxyl group on carbon number 3. Thus, the name is 2,3-dimethyl-3- hexanol. Compound (d), has a 5-carbon chain, two methyl groups on carbons 2 and 4 , and an hydroxyl group on carbon 1. Thus, the name is 2, 4-dimethyl-1-pentanol.

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Question:

Consider a pendulum which is oscillating with an amplitude so large that we may not neglect the \texttheta^3 term in the expansion of sin \texttheta, as we do for a harmonic oscillator. What is the effect on the motion of the pendulum of the term in \texttheta^3? This is an elementary example of ananharmonicoscillator. Anharmonic or nonlinear problems are usually diffi-cult to solve exactly (except by computers), but approximate solutions are often adequate to give us a good idea of what is happening.

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/Users/wenhuchen/Documents/Crawler/Physics/D09-0390.htm

Solution:

The expansion of sin \texttheta (for small \texttheta) to terms of order \texttheta^3, commonly expressed as "expansion to O (\texttheta^3)," is sin \texttheta = \texttheta - (1/6) \texttheta^3 + ... , so that the equation of motion of a simple oscillator becomes, to this order, (d^2\texttheta/dt^2) + \omega0^2\texttheta - (\omega0^2/6) \texttheta^3 = 0,(1) where \omega_0 ^2 denotes the quantity g/L. This is the equation of motion of an anharmonic oscillator. We see if we can find an approximate solution to (1) of the form \texttheta = \texttheta_0 sin\omegat+ \epsilon\texttheta_0 sin 3\omegat,(2) It is now evident that (2) can only be an approximate solution at best. It remains for us to determine \epsilon, and also \omega; while \omega must reduce to \omega_0 at small amplitudes, it may differ at large amplitudes. For simplicity we suppose that \texttheta = 0 at t = 0. An approximate solution of this type to a differential equation is called a perturbation solution, because one term in the differential equation perturbs the motion which would occur without that term. As you have seen, we arrived at the form of (2) by guided guesswork. It is easy enough to try several guesses and to reject the ones which do not work. We have from (2) \texttheta= - \omega^2\texttheta_0 sin\omegat- 9\omega2\epsilon\texttheta_0 sin 3\omegat; \texttheta^3= - \texttheta0^3 (sin^3\omegat+ 3\epsilon sin^2\omegatsin 3\omegat + ...), where we have discarded the terms of order \epsilon^2 and \epsilon^3 because of our assumption that we can find a solution with \epsilon << 1. Then the terms of (1) become \omega2_0\texttheta = (\omega0^2/6)\texttheta^3 - \textthetä \omega02\texttheta_-0 sin \omega_0t + \epsilon\omega02\texttheta_0 sin 3\omegat \approx (\omega0^2/6)\texttheta_03× (sin^3\omegat+ 3\epsilon sin^2\omegatsin 3\omegat) +(\omega^2\texttheta_0sin\omegat+ 9 \omega2\epsilon \texttheta_0 sin 3\omegat). The coefficient of the term sin^2\omegatsin 3\omegat in the previous equation is small by 0(\epsilon) or by 0(\texttheta0^2), compared with the other terms in the equation. Since \epsilon and \texttheta_0 are small quantities, we neglect this term. Using the trigonometric identity sin^3\omegat = [(3 sin\omegat- sin 3\omegat)/4]we get \omega_02\texttheta_-0 sin \omega_0t + \epsilon\omega_02\texttheta_0 sin 3\omegat \approx{(1/8) \omega02\texttheta_-0 + \omega2\texttheta_-0} × sin\omegat+ {9 \omega^2\epsilon\texttheta_0 - (1/24) \omega0^2\texttheta0^3} sin 3\omegat(3) In order for this equation to hold, the corres-ponding coefficients of sin\omegatand sin 3\omegat on each side of the equation must be equal: \omega0^2 = (\omega_0 ^2\texttheta0^2/8) + \omega^2(4) \epsilon\omega0^2= - (1/24)\omega0^2\texttheta_0 ^2 + 9\epsilon\omega^2(5) From (4),\omega^2\approx \omega_02{1 - (1/8) \texttheta0^2} or\omega\approx \omega0{1 - (1/8) \texttheta0^2}^1/2 \approx \omega_0 {1 - (1/16) \texttheta0^2} using the binomial equation for the square root. We see that as \texttheta_0 \ding{217} 0, \omega \ding{217} \omega_0. The frequency shift ∆\omega is ∆\omega = \omega - \omega_0 = (\omega_0/16) \texttheta0^2 In (5), we make the substitution to \omega^2 \approx \omega0^2 and obtain an expression for \epsilon. \epsilon \omega0^2\approx -(1/24) \omega0^2\texttheta_0 ^2 + 9\epsilon\omega0^2 \epsilon\approx \texttheta0^2/(8 × 24) = (1/16) \texttheta0^2 We think of \epsilon as giving the fractional admixture of the sin 3\omegat term in a solution for \texttheta dominated by the sin\omegatterm. Why did we not include in (2) a term in sin 2\omegat? Try for yourself a solution of the form \texttheta = \texttheta_0 sin\omegat+ n\texttheta_0 sin 2\omegat, and see what happens. You will find n = 0. The pendulum generates chiefly third harmonics, i.e., terms in sin 3\omegat, and not second harmonics. The situation would be different for a device for which the equation of motion included a term in \texttheta^2. What is the frequency of the pendulum at large amplitudes? There is no single frequency in the motion. We have seen that the most important term (the largest component) is in sin\omegat, and we say that \omega is the fundamental frequency of the pendulum. To our approximation \omega is given by (4). The term in sin 3\omegat is called the third harmonic of the fundamental frequency. Our argument following (2) suggests that an infinite number of harmonics are present in the exact motion, but that most of these are very small. The amplitude in (2) of the fundamental component of the motion is \texttheta_0; the amplitude of the third harmonic component is \epsilon\texttheta_0.

Question:

What are nested records? How is the WITH statement used with them? Illustrate by designing the data structure for the following problem: A school wishes to send to each of its 1,000 students the end-of-semester report, showing the courses taken and the grades received. Assume each student takes 4 courses, the name of the course is at most 80characterlong, and the grades are integers between 0 and 100. Each re-port must also include student's full name, address and the date of first attendance. Show how the information can be accessed in the program.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G16-0410.htm

Solution:

Nested Records are Records, whose fields are other records. Such records are needed to design complex data structures, such as for this problem . We know that the record must contain information onstudent's name , address, starting date, and grades. These are obvious choices for the fields of the record. Each of these fields must also contain sub-fields. Thus, under name, we must specify first, middle, and last name. The ADDRESS field must contain num-ber and street, city, state and zip code. Starting date must include the date, which is an integer in the range of 1 to 31, the month and the year. The GRADEREPORT FIELD must con-tain the description of 4 courses and the grade associated with each. So we would need an array of 4 records - one for each course. Then the record might look like this: TYPE STUDENTREC = RECORD NAME = RECORD FIRST, Middle, Last: packed array [1. .15] of char; END; ADDRESS = RECORD Number: integer; Street, City, State: packedarray[1..20] of char; Zip: integer END; STARTDATE = RECORD DATE: 1..31; month : (jan,feb, mar,apr, may,jun,jul,aug, sep ,oct,nov,dec); year : integer END; GRADEREPORT:ARRAY[1..4] of RECORD COURSENAME = packed array [1..80] of char; GRADE: 1..100 END END; Since there are 1,000 such records, declare VAR ALLRECORDS : ARRAY[1..1000] of STUDENTREC. Here's how to initialize the complete record of, say, 66th student: With ALLRECORDS [66] DO BEGIN NAME.FIRST:='JOHN'; NAME.LAST: ='DOE'; NAME.MIDDLE: ='PETER'; With ADDRESS DO BEGIN NUMBER: = 10; STREET: = 1PUDDLE LANE'; CITY: = 'New York'; STATE: = 'New York'; ZIP: = 10010; END Start date. DATE: = 1; Start date.month:=Jan; Start date.year:=1981; GRADEREPORT[1]. COURSENAME: = 'MATH 101'; GRADEREPORT[ 1].GRADE: = 85; \bullet \bullet etc . END; The with statement is used instead of writing the field designators all the time . Thus, without the first WITH statement, we would have to write ALLRECORDS [66]. foreach item.

Question:

If the vapor pressure of methyl alcohol, CH_3OH, is 0.0526 atm at 5.0\textdegreeC and 0.132 atm at 21.2\textdegreeC, what do you predict the normal boiling point will be?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E07-0262.htm

Solution:

The normal boiling point of a liquid is the temperature at which the vapor pressure of the liquid is equal to the prevailing atmospheric pressure. Atmospheric pressure is defined as 1 atm. The pressure is related to the temperature by the equation 1n p = [(-∆H) / (RT)] + C where p is the vapor pressure, ∆H is the heat required to transform one mole of liquid to the ideal-gas state; R is the gas constant, (8.314 J / mole \textdegreeK); T is the absolute tem-perature; and C is a constant dependent on the liquid. One can solve for ∆H by using the values obtained for p at 5\textdegreeC and 21.2\textdegreeC. For p = 0.0526 atm T = 5.0\textdegreeC + 273 = 278\textdegreeK R = 8.314 (J / mole \textdegreeK) In (0.0526) = [-∆H / (8.314 J / mole \textdegreeK)(278\textdegreeK)] + C a)-2.945 = [(-∆H) / (2.311 × 10^3 J / mole)] + C Forp = 0.132 atm T = 21.2\textdegreeC + 273 = 294.2\textdegreeK R = 8.314 (J / mole \textdegreeK) In(0.132) = (-∆H) / [(8.314 J / mole \textdegreeK)(294.2\textdegreeK)] + C b)-2.025 = [-∆H / 2.446 × 10^3 (J / mole)] + C One can solve for ∆H by subtracting equation a from equation b. -2.025 = [-∆H / (2.446 × 10^3 J / mole)] + C -{-2.945 = [-∆H / (2.311 × 10^3 J / mole)] + C} .920 = -∆H / 2.446 ×10^3 J / mole + ∆H / 2.311 × 10^3 (J / mole) (2.446 × 10^3 J / mole)(2.311 × 10^3 J / mole) (.920) = (-∆H / 2.446 × 10^3 J / mole) + (∆H / 2.311×10^3 J / mole)(2.446 × 10^3 J / mole) × (2.311 × 10^3 J / mole) 5.200 × 10^6J^2 / mole^2 = -(2.134 × 10^3 J / mole) ∆H + (2.446 × 10^3 J / mole) ∆H 5.200 × 10^6J^2 / mole^2 = (.135 × 10^3 J / mole) ∆H (5.200 × 10^6J^2 / mole2) / (.135 × 10^3 J / mole) = ∆H 3.85 × 10^4 J / mole = ∆H Using a similar method the normal boiling point can be found now that one knows ∆H. Forp = 0.0526 atm T = 5.0\textdegreeC + 273 = 278\textdegreeK R = 8.314 (J / mole \textdegreeK) ∆H = 3.85 × 10^4 (J / mole) In 0.0526 = [{(-3.85 × 10^4 J/mole)}/{(8.314 J/mole \textdegreeK) (278\textdegreeK)}] + C c)-2.945 = -16.657 + C Forp = 1.0 atm T = ? R = 8.314 (J /mole \textdegreeK) ∆H = (3.85 × 10^4 J / mole) In 1.0 = [{(-3.85 × 10^4 J /mole)} / {(8.314 J / Mole \textdegreeK)T}] + C, 0 = [(4.63 × 10^3 0K) / T ] + C d)Subtracting equation d from equation c one can obtain T. -2.945 = -16.657 + C -[0 = -(4.63 × 10^3\textdegreeK) / T ] + c -2.945 = -16.657 + [(4.63 × 10^3\textdegreeK) / T] 13.712 = [(4.63 × 10^3\textdegreeK)/ T] T = [(4.63 × 10^3\textdegreeK) / (13.712)] = 337.7\textdegreeK Boiling point in \textdegreeC = 337.7\textdegreeK -273 = 64.7\textdegreeC.

Question:

A biology student observes filamentous appendages on a bacterium and concludes they are flagella. However, he is told that the bacterium is non-motile. He therefore concludes that the flagella are non-operational. Why are both ofhis conclusions incorrect?

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/Users/wenhuchen/Documents/Crawler/Biology/F05-0120.htm

Solution:

It is rare for an organism to have a structure which serves no function . The filamentous appendages observed by the student are not flagella , butpili(orfimbriae).Piliarehairlikestructures found on some Gram-negative bacteria. They are smaller, shorter, and more numerous than flagella. They are found on non-motile as well as motile bacteria, but do not function in motility. There are several different kinds ofpili. The sexpilus(singular) is involved in bacterial conjugation (the transfer of part of a chromosome from one bacterial cell to another). Mostpiliaid the bacterium in adhering to the surfaces of animal and plant cells, as well as to inert surfaces such as glass. Thepilienable the bacteria to fix themselves to tissues from which they can derive nutrients.

Question:

A steel producer uses three different processes to produce different qualities of steel. It takes three tons of raw material (iron, coal, oil and minor trace elements) to produce a ton of steel in any process. The percentages of iron, coal and oil used in each process vary and are given below: Process A Process B Process C Iron Coal Oil 40 50 10 45 25 30 25 60 15 Write a program in Basic to print the number of tons of raw material needed to make 100, 150, 200, 250, 300 tons of steel if the ratio of process usages A to B to C is as 1:3:6.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G18-0451.htm

Solution:

We can treat the composition of each process as a column in an input matrix. The input matrix is there-fore │404525│ │502560│ │103015│ The ratio of processes A, B, C is always 1 to 3 to 6. This means that for 100 tons of steel, Method A produces10 tons, Method B 30 tons, and Method C 60 tons. In fact, since the ratios total 10, A will always be 10%, B 30%, and C 60% of total output. To find the number of tons of iron, coal and oil used in the three processes we post multiply (1) by the column vector │A│ │B│ │C│where A, B, C, Will vary according to total tonnage of output (100, 150, 200, 250). A coarse flow-chart for the program is as shown in Fig. 1. 10PRINT "STEEL PROCESSING\textquotedblright 20DIM A (2, 2) B (2, 0) C (2, 0) 30PRINT: PRINT "IRON," "COAL," "OIL," "TOTAL TONS" 40MAT READ A 50FOR J = 100 TO 250 STEP 50 60X = J/10 70LET B (0, 0) = X 80LET B (1, 0) = 3\textasteriskcenteredX 90LET B (2, 0) = 6\textasteriskcenteredX 100MAT C = A\textasteriskcenteredB 110MAT C = (3/100)\textasteriskcenteredC 120LET S1 = C (0, 0) + C (1, 0) + C (2, 0) 130PRINT C (0, 0), C (1, 0) , C (2, 0), S1 140NEXT J 150STOP 160DATA 40, 45, 25, 50, 25, 60, 10, 30, 15 170END

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Question:

From which group of organisms did the amphibians develop? What two characteristics of the amphibian's ancestors were preadaptations for life on land?

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/Users/wenhuchen/Documents/Crawler/Biology/F13-0334.htm

Solution:

It is believed that the lobe-finned fish were the ancestors of the land vertebrates. This group was thought to be extinct for about 75 million years. In 1939, however, a living lobe-finned fish of the superorder Crossopterygii, the coelacanth, was caught by a commercial fisherman from waters off the coast of South Africa. Since that time, many more of these fish have been caught. The coelacanths are not the particular lobe- finned fish thought to be the ancestors of land vertebrates, but they resemble those ancestral forms in many ways. In addition to the presence of lungs used for breathing, these fish had another important preadaptation for life on land - the large fleshy bases of their paired pectoral and pelvic fins. These fleshy bases had a bony support through them which allowed for greater support. It is these supportive bones which later form the leg bones of the amphibians. At times, especially during droughts, lobe-finned fish probably used these leg-like fins to pull themselves onto sand bars and mudflats. They may even have managed to crawl to a new pond or stream when the one they previously occupied had dried up. During ancestral times, before the vertebrates came onto land, the land was already filled with an abundance of plant life. Now, should any animal have the ability to survive on land, that animal would have a whole new range of habitats open to it without competition. Any lobe-finned fish that had appendages slightly better suited for land locomotion than those of its companions would have been able to exploit these habitats more fully. Through selection pressure exerted over millions of years, the fins of these first vertebrates to walk, or crawl, on land would slowly have evolved into legs. Thus, over a period of time, with a host of other adaptations for life on land evolving at the same time (e.g., atmospheric changes), one group of ancient lobe-finned fishes must have given rise to the first amphibians.

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Question:

When a group of chickens from different flocks are placed together there is much more fighting than is observed in flocks which have been together for some time. Explain.

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/Users/wenhuchen/Documents/Crawler/Biology/F31-0818.htm

Solution:

The group of chickens fight when first placed together in order to establish dominance-subordination relationships within the new flock. The relationships are formed through a series of aggressive encounters between each pair of chickens. Chicken A may be larger, stronger, or just more temperamental than all the chickens it encounters and thus may win all its fights. It becomes the dominant bird, while all the others become subordinate to it. Chicken B may be dominant over all the birds except A. For example, A may peck at B without B pecking back. Yet B may peck at C, D, E, etc. This social hierarchy is sometimes called a pecking order. Pecking orders can get complex. For example, chicken Z may be particularly strong in its encounter with X, yet it may have lost its fight with Y, who is subordinate to X. This pecking order can be diagrammed as follows: Among the various factors that determine an animal's position in a hierarchy are: age, sex, size, strength, temperament, physiological condition (hormones), and seniority in the group. Once established, a "peck-right system," common in chickens, becomes rather fixed and rigid, with very few fighting encounters. Thus, an old, weak chicken may still be dominant because the sub-ordinate chickens do not retaliate when pecked. However, in a "peck-dominance system," common in pigeons, sub-ordinate birds often retaliate and thus are more apt to discover any weakness of the dominant pigeons. This system thus promotes more social mobility than the "peck-right system." The adaptive value of the establishment of social hierarchies is the promotion of order and stability to the relationships in the group. The less fighting that occurs, the more time can be allocated to finding food, avoiding predators, and mating.

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Question:

A cylinder contains 40 g He, 56 g N_2, and 40 g Ar. (a) What is the mole fraction of each gas in the mixture? (b) If the total pressure of the mixture is 10atm, what is the partial pressure of He?

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Solution:

(a) The mole fraction of a component in a system is defined as the number of moles of that com-ponent divided by the sum of all the moles present in the system. Here, one must first calculate the number of moles present of each component, then the mole fractions can be found. The number of moles of each gas can be found by dividing the number of grams present of the gas by the molecular weight. (MW He = 4, MW N_2 = 28, MWAr= 40.) no. of moles= no. of grams / MW no. of moles of He = 40 g / (4 g/mole) = 10 moles no. of moles of N_2 = 56 g / (28 g/mole) = 2 moles no. of moles ofAr= 40 g / (40 g/mole) = 1 mole. The total number of moles of gas in the system is the sum of the number of moles of the three gases, Thus, there are 13 moles of gas in the system. The mole fraction can now be found for each gas by dividing the number of moles of each gas by 13, the total number of moles mole fraction = no. of moles / total no. of moles in system mole fraction of He = 10 moles / 13 moles = .77 mole fraction of N_2 = 2 moles / 13 moles= .15 mole fraction ofAr= 1 mole / 13 moles= .08 (b) In a system where various gases are present, the partial pressure of each gas is proportional to the mole fraction of the gas. The relationship between the partial pressure of a particular gas and the total pressure is partial pressure = total pressure × mole fraction In this problem, one is given the total pressure of the system and one has found the mole fraction of He. One can now find the partial pressure of He partial pressure of He = 10atm× .77 = 7.7 atm.

Question:

Design a synchronous sequential circuit with one input line and one output line that recognizes the input string x = 1111. The circuit is also required to recognize overlapping sequences such that if the input string is x = 1101111111010 then the output string should be z = 0000001111000

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Solution:

The first step in the design procedure is to draw a state diagram which represents the behavior of the circuit, which in this case happens to be a sequence 1111 detector. First you assume that the machine is in some starting state A and the first input is a 0. Since 0 is not part of our sequence, the machine remains in state A with an output of 0, 'waiting' for a 1. As soon as a 1 occurs at the input, the machine jumps into state B. The output is still 0, because our sequence has not yet been completed. See fig. 1. If an input of 0 occurs while the machine is in state the 1111 sequence is interrupted and the machine must go back to state A to start all over again. See fig. 1. However, if an input of 1 occurs, the machine jumps from state B to state C, meaning that the second '1' of the sequence was detected. See fig. 2. If the next input is a 0, the 1111 sequence is again interrupted, and the machine must go from C back to A to start all over again; also in fig. 2; but if the input is a 1, while in state C, the machine jumps into state D, meaning that it has just detected the third '1' of the 1111 sequence; see fig. 3. If the next input is a 0, the sequence is again interrupted, and the machine must go back to state A; how-ever, if the input is a 1, the machine has just detected the last '1' of the 1111 sequence, and the output finally becomes 1. See fig. 3 again. Note in fig. 3 that when the machine is in state D, the '1/1' loop will continue giving outputs of 1 as long as the inputs remain 1; thus ensuring the detection of overlapping 1111 sequences. Fig. 3 is the complete state diagram. The next step is to set up a state table from the state diagram. See fig. 4. Note, for example, that the top, right-hand box of the state table means that when the present state is A and in-put becomes a 1, the next state becomes B and the output is 0. Now we choose arbitrary state assignments as follows: A = 00 A = 00 B = 01 C = 10 D = 11 This results in a transition table, and an output table (which is simply extracted from the state table). See fig. 5. Note that states D and C have been interchanged to facilitate the use of K-maps. The next step is to pick a memory element. We shall pick the JK flip-flop. Note that we will need two memory elements, since Log_2 (no. of states)\leq No. of memory elements. Y^K Y^K+1 J K NOTE: d MEANS "DON'T CARE" FIG. 6 0 0 0 d 1 1 d 0 0 1 1 d 1 0 d 1 To work with flip-flop logic, an application table is ex-tremely helpful. This table is easily derived from the truth table of the particular flip-flop. See fig. 6. Next we form JK excitation tables using the JK application table, and the table, and the transition table. See fig. 7a, b, c, d. Note that J_1K_1 corresponds to the y_1^K+1 column on the transition table. Now we solve for the logic equations using the excitation tables as K-maps: J_1 = Y_2XK_1 =X J_2 = XK_2 =X+Y_1 Z = XY_1Y_2 From the logic equations we can now build the actual cir-cuit which will perform the desired function; in this case a 1111 sequence detector. See Fig. 8.

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Question:

A house near a lake is kept warm by a heat engine. In winter, water from beneath the ice covering the lake is pumped through the heat engine. Heat is extracted until the water is on the point of freezing when it is ejected. The outside air is used as a sink. Assume that the air temperature is -15 \textdegreeC and the temperature of the water from the lake is 2\textdegreeC. Calculate the rate at which water must be pumped to the engine. The efficiency of the engine is one-fifth that of a Carnot engine and the house requires 10 kW.

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Solution:

The thermal efficiency of a heat engine is defined as the ratio of the heat converted to mechanical work by the engine to the heat Q2 supplied to it. If Q_1 is the heat rejected to the reservoir, efficiency is, n = (Q_1 - Q_2)/Q2 The efficiency of a Carnot engine operating between two reservoirs at temperatures T_2 > T_1,given by the ratio, n_c= (T_2 - T_1)/T_2 Hence, for the practical heat engine of the problem we have, n = (Q_1 - Q_2)/Q_2= (1/5)n_c= (T_2 - T_1)/T_2 Heat is taken from the lake water as it cools from 2\textdegreeC to 0\textdegreeC before ejection. The mean temperature of the hot-temperature source is thus 274 \textdegreeK. If m is the mass of water flowing through in time t, the heat taken in at the hot reservoir in unit time is Q_2/t = (m/t)c × 2\textdegreeC, where c is the specific heat capacity of water. Heat is rejected to the air as sink at a temperature of -15\textdegreeC = 258\textdegreeK, the amount of air available being assumed infinite so that the temperature remains constant. Further the work done (Q_2 - Q_1) is given as 10 kW = 10^4 J \textbullet s^-1. Thus, from the first equation, we have, [10^4 J/s] / [(m/t) × 4.18 J/g \textdegreeC × 2C deg] = (1/5) [{(274 - 258)\textdegreeK}/{274\textdegreeK}] \therefore m/t = [(5 × 274 × 10^4)/(2 × 4.18 ×16)]g/s = 102.4 × 10^3 g/s. The rate of water flow necessary is thus 102.4 liters/s.

Question:

Why is it that in most ecological successions the last stageis longlasting?

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Solution:

If no disruptive factors interfere, most successions eventually reach astage that is much more stable than those that preceded it. The communityof this stage is called the climax community. It has much less tendencythan earliersuccessionalcommunities to alter its environment in amanner injurious to itself. In fact,itsmore complex organization, larger organicstructure, and more balanced metabolism enable it to control its ownphysical environment to such an extent that it can be self- perpetuating. Consequently, it may persist for centuries without being replacedby another stage so long as climate,physiographyand other majorenvironmental factors remain essentially the same. However, a climaxcommunity is not static. It does slowly change and will change rapidlyif there are major shifts in the environment. For example, fifty years ago, chestnut trees were among the dominant plants in the climax forests ofmuch of eastern North America, but they have been almost completely eliminatedby a fungus. The present-day climax forests of this region are domi-natedby other species.

Question:

Explain, by means of examples, the following functions used in the handling of matrices: 1) Reversal function. 2) Rotation function.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G14-0366.htm

Solution:

APL is ideally suited for matrix manipulation. The above functions help in the storage and processing of data which is in matrix form. Suppose we wish to represent the first 12 whole numbers as a 3 × 4 matrix. Key in M \leftarrow 3 4 \rho ʅ 12(1) A matrix is entered by using \rho(rho), the shape operator. The iota function ( ʅ ) tells the computer to list the first 12 positive integers. The general form of (1) is: M \leftarrow m n \rho x_1 x_2 .... x_mn(2) Comparing (1) and (2) we see that m, the number of rows, is 3, and n, the number of columns, is 4. The matrix of (1) is: \mid1234\mid M =\mid5678\mid(3) \mid9101112\mid a) Reversal function: The general form of the revers-al function is \textphi [K]M. Here K = 1 or K = 2. When K = 1, (3) becomes \mid9101112 \mid M' =\mid5678\mid(4) \mid1234\mid Thus, \textphi [1] M reverses the order of the rows. When K = 2, the order of the columns is reversed. Applying \textphi [2] M to (3): \mid 4321\mid M" =\mid 8765\mid(5) \mid1211109\mid If we apply \textphi [2] M to (4): \mid1211109\mid \mid 8765\mid(6) \mid 4321\mid While \textphi [1] M on (5) yields \mid1211109\mid \mid 8765\mid.(7) \mid 4321\mid Matrices (6) and (7) are equal. b) Rotation function: This is a dyadic function of the form S \textphi [K] M. Here S is a vector with as many compon-ents as there are elements in the dimension being rotated. Suppose we want to shift i) all first elements in each row of (1) one ele-ment. from top to bottom. ii) all second elements by two elements from top to bottom and iii) all other elements remain the same. Then, applying the rotation function to (1) 1 2 0 0 \textphi [1] M \mid91034\mid \mid5278\mid \mid161112\mid To illustrate rotation of columns apply -2 1 0 \textphi [2]M. The minus indicates a rotation from right to left. The effect of this function is as follows: i) all the first elements in each column to be cyclically shifted two elements from left to right. ii) all second elements in every column to be cyc-lically shifted one element from right to left. iii) a zero shift in the third elements. The upshot of all this on M is \mid3412\mid \mid6785\mid \mid9101112\mid Thus, APL is useful for rotating matrices.

Question:

During the courtship of the common tern, the male presents afish to another tern. Account for this behavior and discuss whycourtship in general may benecesaryfor reproduction.

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Solution:

Courtship is an important means of communication between some animals. Courtship involves a multi-tude ofprecopulatorybehavior patterns whichserve a specific purpose for the animal that performs them. The courtshipof the common tern, a type of gull, exemplifies the purposes of courtshipbehavior. During the mating season of the common tern, the males will catch afish and present it to another tern. If the recipient tern is a sexually receptivefemale, she will probably accept it. The male tern then bows to herand prepares to make a nest in the sand where they will mate. If the maleoffers the fish to another male, the recipient will reject it with agressivebehavior. Why would the male offer the fish to another male? In thecommon tern, both sexes look so similar that this courtship behavior is theonly way of differentiating males from females. Thus, one function of courtshipbehavior is sexual identification. Courtship is necessary for reproduction in some cockroaches and otherselected insects. It serves to decrease aggressive tendencies long enoughfor the two animals to mate. In these insects, the male secretes a substanceon its back which the female feeds on during copulation. This divertsher attention so mating may occur. Courtship is also necessary for some animals to produce eggs. A femalelyrebird will only lay eggs if she can see a male displaying his feathersand singing. However, eggs can also be laid if the neck of the femaleis tickled in a way to simulate the tickling done by the male during courtship, even if no male is around. Finally, courtship is necessary to establish species identification. Many birds have species-specific songs and courtship displays which only havesignificance for others of the same species. Such behavior also servesto ward away other males of the same species from the displaying male'sterritory. If courtship behavior did not exist, much energy would be wasted byanimals trying to mate with others of the same sex or of different species.

Question:

Calculate the average over time of the kinetic and potential energy of a harmonic oscillator.

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/Users/wenhuchen/Documents/Crawler/Physics/D09-0388.htm

Solution:

Assume that the harmonic oscillator we are dealing with is a mass on a spring. The position of the mass is then given by x = A sin (\omega_0t + \varphi) where A is the amplitude of the motion of the mass, and \varphi is a phase angle defining the initial position of the mass (att = 0). The kinetic energy is then k = (1/2) Mẋ^2 = (1/2)M[\omega_-0Acos(\omega_0t + \varphi)]^2, The time average of the kinetic energy over one period T of the motion is defined as< k > = [{^T\int_0 (kdt)}/T ] But \omega_0 is the angular frequency and \omega_0 = 2\pif where f is the frequency of motion of the mass. Since f = 1/T \omega_0 = 2\pi/T and T = 2\pi/\omega_0. Then = (1/2)M\omega^2 _0A^2 [{ ^2\pi/(\omega)0\int0cos^2(\omega_0t +\varphi)dt}/{2\pi/\omega_0}] \varphi)dt}/{2\pi/\omega_0}] Lety = \omega_0t + \varphi(1) anddy= \omega_0dt since \varphi is a constant. We then have = (1/2)M\omega^2 _0A^2 [{\varphi+2\pi\int\varphicos^2 y(dy/\omega_0)}/{2\pi/\omega_0}](2) \varphi+2\pi \varphi {2\pi/\omega_0}](2) where the limits are obtained by noting that, from (1), y = \varphi at t = 0, and at t = 0, and y = 2\pi+ \varphi at t =2\pi/\omega_0.Rewriting(2) + \varphi at t = Rewriting = (1/2) [(M\omega^2 _0A^2)/(2\pi)] (\varphi+2\pi\int\varphicos^2 ydy) \varphi+2\pi \varphi Butcos^2 y = (1/2) (1 +cos2y) and = [(M\omega^2 _0A^2)/(4\pi)] [\varphi+2\pi\int\varphi(1/2) (1 +cos2y)dy] \varphi+2\pi \varphi (1/2) (1 + = [(M\omega^2 _0A^2)/(8\pi)] [\varphi+2\pi\int\varphidy+^\varphi+2\pi\int_ \varphicos2ydy] \varphi+2\pi \varphi dy+^\varphi+2\pi\int_ \varphi = [(M\omega^2 _0A^2)/(8\pi)] [\varphi+2\pi\int\varphidy+ (1/2)^\varphi+2\pi\int_ \varphicos2y(2dy)] \varphi+2\pi \varphi dy+ (1/2)^\varphi+2\pi\int_ \varphi = [(M\omega^2 _0A^2)/(8\pi)] [2\pi + (1/2) {sin(2\varphi + 4\pi) - sin (2\varphi)}] \varphi + 4\pi) - sin (2\varphi)}] Since sin(2\varphi + 4\pi) = sin 2\varphi = [(M\omega2_0A^2)/(4)] The potential energy is, with x = A sin (\omega_0t + \varphi) U = (1/2) Cx^2 = (1/2)CA^2 sin^2 (\omega_0t + \varphi) where C is the spring constant. Now, = [( ^T\int_0 Udt)/(T)] = [ ^2\pi/\omega(0)\int_0 (1/2)CA^2 sin^2 (\omega_0t + \varphi) at]/[2\pi/\omega_0] Using(1) = (1/2)CA^2 [{ ^\varphi+2\pi\int_\varphi sin^2 y (dy/\omega_0)}/{2\pi/\omega_0}] = [(CA^2)/(4\pi)]^\varphi+2\pi\int_\varphi sin^2 ydy Because sin^2 y = (1/2) (1 -cos2y) = [(CA^2)/(4\pi)] [^\varphi+2\pi\int_ \varphi (1/2) (1 -cos2y)dy] = [(CA^2)/(8\pi)] [^\varphi+2\pi\int_ \varphidy-^\varphi+2\pi\int\varphicos2ydy] = [(CA^2)/(8\pi)] [^\varphi+2\pi\int_ \varphidy- (1/2)^\varphi+2\pi\int\varphicos2y (2dy)] = [(CA^2)/(8\pi)] [2\pi - (1/2) {sin(2\varphi + 4\pi) - sin (2\varphi)}] Since sin (2\varphi + 4\pi) = sin(2\varphi) = CA^2/4 But, \omega^2 _0 = C/M, by definition of the frequency of the simple harmonic oscillator. Therefore = (M\omega2_0A^2)/4 Thus = and the total energy of the harmonic oscillator is E = + = (1/2) m\omega2_0A^2. Note that E = because the total energy is a constant of the motion. The equality of the average kinetic energy and potential energies is a special property of the harmonic oscillator. This property does not hold in general for enharmonic oscillators. It does hold for weakly damped oscillators.

Question:

lf the atomic weight of carbon 12 is exactly 12amu, find the mass of a single carbon-12 atom.

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Solution:

To solve this problem, one must first define the mole concept. A mole is defined as the weight in grams divided by the atomic weight (or molecular weight) of the atom or compound. mole = weight in grams / atomic or molecular weight If one has a mole of carbon 12, then there are 12 grams of it present. One mole of any substance contains 6.022 × 10^23 particles. The mass of a single carbon atom is found by dividing 12 g / mole by 6.022 × 10^23 atoms / mole. mass of 1 C atom= (12 g / mole) / (6.022 × 10^23 atoms / mole) = 2.0 × 10^-23 g / atom.

Question:

Sometimes there is more than one way to write a program. Maybe one method is more efficient than another. Demonstrate this by rewriting the program which uses a characterstring of up to 200 parentheses as input. The programis to deter-mine whether the input string is 'well formed', has too many right parentheses, or, too many left parentheses.

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Solution:

The program can be written as follows: STRING:PROCOPTIONS(MAIN); DCL (R,L,T,M) FIXED (3) ,INIT(0), PAR CHAR (200) VAR; /\textasteriskcentered R IS THE COUNT OF THE NUMBER OF RIGHT PARENTHESES, L IS THE COUNT OF THE NUMBER OF LEFT PARENTHESES, T REPRESENTS THE LENGTH OF THE CHARACTER STRING PAR, M IS A VARIABLE TO BE USED IN THE PROGRAM \textasteriskcentered/ PUTLIST('STRING','MESSAGE'); PUT LIST (' ----' ,'-------') SKIP (0) ; PUTSKIP(2); DOWHILE(1 = 1); GETLIST(PAR); PUTLIST(PAR); DO WHILE(R\lnot>L&T\lnot = 1) ; T =LENGTH(PAR) ; IFINDEX(PAR,')1) = 1 THEN R = R + 1; ELSE L = L + 1; IF T\lnot = 1 THEN PAR=SUBSTR(PAR,2) ; END; M = L-R IF M<0 THEN PUT LIST ('UNMATCHED RIGHT PARENTHESES'); ELSE IF M>0 THEN PUT LI ST (M, ' TOO MANY LEFT PARENTHESES'); ELSE PUTLIST('WELL FORMED'); END; PUTSKIP(2); END STRING;

Question:

What is the change in entropy of a gas if the temperature increases from 100\textdegreeK to 101\textdegreeK when heat is added and the volume is kept constant?

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Solution:

Consider a system containing a large number of particles. When heat is added to this system, the average kinetic energy of the particles will increase. This is reflected as an increase in the temp-erature of the system. The system will have a higher internal disorder as a result of increased thermal motion of its constituents. The entropy of a system is a measure of the tendency of a system to increase its internal disorder. Therefore, as heat is added, entropy increases. In our problem, let the increase in entropy be ∆s when the system reaches a new equilibrium after its temperature increases by ∆T = 1\textdegreeK Since ∆T << T = 100\textdegreeK, the amount of the heat added must be very small, and the entropy change is ∆s = Q/T where Q is the quantity of heat added. The heat added to a gas is equal to the gas' increase in internal energy plus the work done on the gas while expanding. The volume is kept constant, therefore the mechanical work done is zero . Using N for Avogadro's number and k for Boltzmann's constant, we have (for an ideal monatomic gas) q = ∆E = (3/2)Nk∆T where ∆E is the increase in the internal energy of the gas. Hence, ∆s = (3/2)Nk(∆T/T) = (3/2)[{(6.02 × 10^23 mole^-1) × (1.38 × 10^-23J/\textdegreek)1\textdegreek}/{100\textdegreeK}] = 0.125 Joule/mole\textdegreeK

Question:

A transmission line delivers power at a potential of 240,000V to a transformer designed to step the potential down to 2,400 V. If the primary coil has 10,000 turns, how many turns should the secondary coil contain? Suppose that the transformer delivers 500 A of current at the secondary. What would be the current in the primary coil?

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/Users/wenhuchen/Documents/Crawler/Physics/D23-0773.htm

Solution:

The voltage output of a transformer depends on the input voltage and on the turns ratio according to the relationship (See figure), (V_1/V_2) = (N_1/N_2) Substituting, we find the turns N_2 in the secondary coil to be, N_2 = [(N_1 V_2)/V_1] = {[(1 × 10^4 turns) (2.4 × 10^3 V)]/(2.4 × 10^5 V)} = 1 × 10^2 turns If we assume that power is transferred from primary to the secondary of this transformer at 100 per cent efficiency, the power in equals power out because of the principle of conservation of energy. Therefore, V_1 I_1 = V_2 I_2 or I_1 = [(V_2 I_2)/V_1] {[(2.4 × 10^3 V) (5 × 10^2 A)]/(2.4 × 10^5 V)} = 5 A

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Question:

The specific cell processes accelerated or decelerated by hormones are numerous and varied, but most of them fit into one of two general categories. One is the alteration of the rate of membrane transport of substances, and the other is the alteration of the rate of enzyme activity. Explain.

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Solution:

The effect of many hormones is to facilitate or inhibit the transport of substances across the membrane into the cell. Insulin, for example, somehow affects cell membranes so as to increase the rate of glucose transport into the cells. Other hormones, such as glucagon (an antagonist of insulin), inhibit glucose transport. A similar pattern involving hormone-mediated inhibition and stimulation also operates for the membrane transport of amino acids and other sugars. The transport of ions and water into and out of kidney cells is also under the influence of hormones. The hormonal regulation of enzymatic activity usually involves catalysis for which there are different enzymes for the forward and reverse reactions. An example of this is the relationship that exists between glucose and glycogen in the liver: In this particular example, the activity of enzyme 2 is increased by the hormone insulin, and that of enzyme 1 is increased by the hormones glucagon (a pancreatic hormone) and epinephrine (an adrenal medullary hormone). The rate of either the forward or reverse reaction can be increased by production of more of the appropriate enzyme. But there is another way which does not change the total number of enzyme molecules? it does not have any effects on enzyme synthesis. Most enzymes exist in both an active and inactive state. The active state is the functional form of the enzyme. It is believed that the amount of active enzyme can be increased by converting inactive enzyme molecules into active ones. Thus, a hormone may function by converting a certain specific inactive enzyme into its active form, thereby augmenting the enzyme activity and thus the reaction rate. The advantage of this mechanism over synthesis of new enzyme molecules is that protein synthesis requires some time, whereas the activation of enzymes within the cell requires only a few minutes. A hormone may also function by inactivating certain enzymes to bring about a specific physiological change. Note again that in this functional mechanism no new enzymes are being formed, although there is a change in enzyme activity and hence protein production.

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Question:

Write a FORTRAN program that outputs the customer's name and account number, the number of message units used, the arrears, and the total bill. The rate schedule should be 10 cents per message unit up to 75, and 8 cents per message unit beyond 75.

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Solution:

This is a telephone billing scheme which excludeslong-distance calls. We read in the cards, each of which contains the following information: Variable description Number of spaces per card NAME NUMCUS MSGS ARREAR customer name customer number message units arrears 12 7 5 6 Program control is managed by a loop through which each card is read. If a blank space appears in the first space of NAME, control passes out of the loop, and the program terminates. The rest of the logic is a straightforward computation of the billing rates described above. Appropriate messages and headings will be contained in the output. Integer arithmetic is used for simplicity's sake. INTEGER NAME (12), NUMCUS, MSGS, BLANK, 1TOTBIL BLANK = '\square' 5READ (2,101)NAME, NUMCUS, MSGS, ARREAR 101PORMAT(12A1,I7,2X,I5,7X,I6) CIF FIRST COLUMN IS BLANK, ENDPROGRAM IF (NAME (1).EQ.BLANK)GO TO 25 C10 CENTS PER MESSAGE UNIT PLUS ARREARS TOTBIL = 10{_\ast}MSGS + ARREAR CDISCOUNT FOR MSGS OVER 75 IF (MSGS.GT.75) TOTBIL = TOTBIL - 8{_\ast}(MSGS - 75) WRITE (6,102) NAME, NUMCUS, MSGS, ARREAR, TOTBIL 102FORMAT (1X, 12A1,2X, 17/17,'MSG UNITS',3X, I6, 1'CENTS ARREARS'/'AMOUNT DUE', 17,'CENTS'/) GO TO 5 25CONTINUE STOP END

Question:

Suppose you are given 3 sides of a triangle. Write a program inPL/I to compute the perimeter, the largest side, the samllestside, the area, the inscribed radius, the circum- scribedradius, and the three angles.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G13-0330.htm

Solution:

We need some facts from plane geometry. Suppose the 3 sides of thetriangle are denoted by S_1 S_2, and S_3 . Then the perimeter is given by P = S_1 + S_2 + S_3. Let h = P/2. Then the area is given by Area =\surd[h(h - S_1) (h - S_2) (h - S_3)]. The radius of the inscribed circle is given by \rho = Area/h. The radius of the circumscribed circle is given by R = S_1S_2S_3/4a. The three angles are determined from the trigonometricrela-tionships : tan\alpha_1 = \rho/(h - S_1) tan\alpha_2 = \rho/(h - S_2) tan\alpha_3 = \rho/(h - S_3) In the computer program we denote the sides of the tri-angle by S1, S2, and S3; the perimeter by P; the auxiliary variable by H; the area by A; theradius of the inscribed circle by RI; the radius of the circumscribed circleby RC; and the three angles by A_1, A_2, A_3. The program uses the built-in functions MAX and MIN to get the longestand shortest sides of the triangle. The above formulas are used to computeinscribed and circumscribed radii, and the built-in function ATAND computes the angle from the tangent in degrees. Formatting of the output is handled through the PUT EDIT statement. TRIANGLE: PROC OPTIONS (MAIN) RPT:GETLIST(S1,S2,S3); P = S1 + S2 + S3; H = P/2; B = MAX (S1,S2,S3); S = MIN (S1,S2,S3); A = SQRT [H\textasteriskcentered(H - S1)\textasteriskcentered(H - S2)\textasteriskcentered(H - S3)]; RI = A/H; RC = S1\textasteriskcenteredS2\textasteriskcenteredS3/(4\textasteriskcenteredA); A1 = 2\textasteriskcenteredATAND[RI/(H - S1)]; A2 = 2\textasteriskcenteredATAND[RI/(H - S2)]; A3 = 2\textasteriskcenteredATAND[RI/(H - S3)]; PUT EDIT('SIDES OF TRIANGLE ARE: ',S1,S2,S3, '3 ANGLES ARE:'A1,A2,A3, 'THE PERIMETER IS:',P, 'THE AREA IS:',A, 'INSCRIBED RADIUS:',RI, 'CIRCUMSCRIBED RADIUS:',RC, 'THE BIGGEST SIDE IS:',B, 'THE SMALLEST SIDEIS:',S) (2(A(30),3E(15,5),X(25),6(A(30)),E(15,5),SKIP)); PUT SKIP (2); GO TO RPT; END TRIANGLE;

Question:

In figure (a), block A of weight w_1 rests on a frictionless inclined plane of slope angle \texttheta. The center of gravity of the block is at its center. A flexible cord is attached to the center of the right face of the block, passes over a frictionless pulley, and is attach-ed to a second block B of weight w_2. The weight of the cord and friction in the pulley are negligible. If w_1 and \texttheta are given, find the weight w_2 for which the system is in equilibrium, that is, for which it remains at rest or moves in either direction at constant speed.

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Solution:

The free-body diagrams for the two blocks are shown in figure (b) and to the right of fig. (a). The forces on block B are its weight w_2 and the force T\ding{217} exerted on it by the cord. Since it is in equi-librium, the block has no acceleration and T - w_2 = 0 by Newton's Second Law. Block A is acted on by its weight w_1\ding{217} the force T\ding{217} exerted on it by the cord and the force N\ding{217} exerted on it by the plane. We can use the same symbol (T\ding{217}) for the force exerted on each block by the cord, because these forces are equivalent to an action-reaction pair and have the same magnitude. The force N\ding{217}, if there is no friction, is perpendicular or normal to the surface of the plane. Since the lines of action of w_1 and T\ding{217} inter-sect at the center of gravity of the block, the line of action of N\ding{217} passes through this point also. It is simplest to choose x- and y-axes parallel and perpend-icular to the surface of the plane, because then only the weight w_1\ding{217} needs to be resolved into components. The conditions of equilibrium for block A give, since it isn't accelerated, \sumF_x = T - w_1 sin \texttheta = 0 \sumF_y = N - w_1 cos \texttheta = 0. Thus, if w_1 = 100 lb and \texttheta = 30\textdegree,we have from the first equation above w_2 = T = w_1 sin \texttheta = 100 lb × 0.500 = 50 lb, and from the second equation above N= w_1 cos \texttheta = 100 lb × 0.866 = 86.6 lb. Note carefully that in the absence of friction the same weight w2of 50 lb is required whether the system remains at rest or moves with constant speed in either direction. This is not the case when friction is present.

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Question:

The following reaction using hydrogen and oxygen is carried out in a bomb calorimeter: 2H_2 (g) + O_2(g) \rightarrow 2H_2O(l). The following data are recorded: Weight of water in calorimeter = 2.650 kg., Initial temperature of water = 24.442\textdegreeC., Final temperature of water after reaction = 25.635\textdegreeC., Specific heat of reaction vessel is 0.200 Kcal/\textdegree\rule{1em}{1pt}kg, the weight of the calorimeter is 1.060 kg, and the specific heat of water is 1.00 Kcal/\textdegree\rule{1em}{1pt}kg, calculate the heat of reaction. Assuming that 0.050 mole of water was formed in this experiment, calculate the heat of reaction per mole of liquid water formed. Neglect the specific heat of the thermometer and stirrer.

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Solution:

The heat emitted during the reaction must be equal to the heat absorbed by the system, including the water and the reaction vessel. Therefore, the heat of re-action can be found by calculating the heat absorbed by the water and the heat absorbed by the reaction vessel. 1) To find the heat absorbed by the water one must be concerned with the specific heat of water, the amount of water present and the number of degrees the temperature is raised. The specific heat is defined as the number of kilocalories absorbed when a kilogram of mass is raised one degree. The specific heat of water is 1 Kcal/kg \rule{1em}{1pt}\textdegree. To find the total heat absorbed by the water, the weight of the water and the number of degrees that the temperature is raised must be multiplied by the specific heat. Number of Kcal absorbed by water = 2.650 kg × 1 kg/\textdegree\rule{1em}{1pt} kg × (25.635\textdegree \rule{1em}{1pt} 24.442\textdegree) = 3.161 Kcal. 2) To determine the number of Kcal absorbed by the reaction vessel a similar procedure is used. The specific heat of the vessel is multiplied by the weight of the vessel and the number of degrees the temperature was raised. Number of Kcal absorbed by the reaction vessel = = 1.060 kg × 0.200 Kcal/kg\rule{1em}{1pt}\textdegree× (25.635\textdegree \rule{1em}{1pt} 24.442\textdegree) = 0.253 Kcal. 3) The heat of reaction is the sum of the heat absorbed by the water and the reaction vessel. heat of reaction = (3.161 + 0.253)Kcal = 3.414 Kcal. A ratio can be set up to determine the heat of reaction per mole of H_20(l) if 3.414 Kcal is the heat of reaction of 0.05 mole of water. The heat of reaction of 1 mole of water will be related to one as 3.415 Kcal is related to 0.05 moles. Let x = the heat of reaction of 1 mole of water. (3.415 Kcal) / (0.05 mole) = x/1 mole x = 68.3 Kcal. Therefore the heat formed when one mole of water is formed is 68.3 Kcal.

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Question:

It is known that energy from proteins is derived by breaking off the amino group as NH_3. How would you compare the energy derived from oxidation of a protein with energy obtained from oxidation of carbohydrates?

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Solution:

The amount of energy derived from oxidation of an organic compound depends to a great extent on the number of bonds broken. You can compare the energy re-leased upon oxidation of a carbohydrate and protein by noting how many bonds are broken in the process. You are told that the amino group comes off as NH_3. The general formula for amino acids, of which proteins are composed is When the amino group is broken off, only the C\rule{1em}{1pt}N bond is broken. The bonds between the nitrogen and hydrogen are not broken when the proteins are oxidized. In carbo-hydrates, which are simple sugars or materials that can be hydrolyzed to simple sugars, all bonds are broken. Thus, because only one bond is broken in the protein's component, the energy released from a protein molecule is not as great.

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Question:

A silver dollar weighs about 0.943 ounces. Express this weightin grams, in kilograms, and in milligrams.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E01-0012.htm

Solution:

One ounce is equal to 28,35g; thus, to convert from ounces to grams, one multiplies the number of ounces by the conversion factor, 28.35 g/1 ounce. no. of grams = 0.943 ounces × 28.35 g/1 ounce = 26.73 g. There are 1,000 g in 1 kg; therefore, to convert from grams to kilograms, onemultiplies the number of grams by 1 kg/1,000 g. no. of kg = 26.73g × 1 kg/1000 g = .02673 kg. There are 1,000 mg in one gram; thus, to convert from grams to milligrams, multiply the number of grams by the conversion factor, 1000 mg/1g. no. of mg = 26.73 g × 1000 mg/1 g = 26,730 mg.

Question:

Describe the entire digestive process in the crayfish.

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Solution:

The crayfish is a member of the class Crustacea order Decapoda. Most crustaceans are marine animals. The crayfish are the most successful freshwater decapods. Most are about 10 cm long, but some are as large as lobsters, another decapod. Crayfish are mostly bottom- dwellers, while some may be semi-terrestrial. Crayfish are omnivorous. The feeding appendages of the decapods consist of a pair of mandibles, two pairs of maxillae, and three pairs of maxillipeds. Each pair of feeding appendages is a different modification of the leg-like processes of the primitive body segments. The mandibles are short and heavy with opposing grinding and biting surfaces. The accessory appendages serve to hold the food, tear it, or push it to the mandibles The first pair of crayfish legs is larger and heavier than the other pairs, and function as claws or pincers. Food is caught or picked up by these pincers, called chelipeds, and then pushed to the third pair of maxillipeds which pushes the food between the other mouth parts. A piece is bitten off by the mandibles, while the remainder is pulled from the mouth by the maxillae and maxillipeds. The bitten piece is pushed into the pharynx and another bite is taken. When the crayfish is not eating, the third pair of maxillipeds covers the other mouth parts. In the digestive tract, a short esophagus leads to a large cardiac stomach. The cardiac stomach is separated by a constriction from a smaller pyloric stomach. A long in-testine extends from the pyloric stomach through the abdomen to the anus. The anus is located on the ventral surface of the last abdominal segment, the telson. Opening into the midgut (the posterior part of the pyloric stomach in crustaceans)' is a single blind sac which functions as a digestive gland or hepatopancreas. It also functions in food absorption and storage. The hepatopancreas is composed of ducts and secretory tubules. Absorption of food occurs only in the tubules of the digestive gland and in the walls of the midgut. Food is consumed in large pieces by crayfish and other decapods. The cardiac stomach serves to mix and break up these pieces. The walls of the cardiac stomach have a number of chitinous ridges and calcareous ossicles, that function as teeth to grind and tear the food. Digestive enzymes, secreted by the hepatopancreas, are passed forward to the cardiac stomach. The cardiac stomach has a complex musculature which controls movement of the stomach walls. These factors enable the cardiac stomach to reduce food to a very fine consistency. The pyloric stomach has setae, chitinous bristles, on complex folded walls. The setae act to filter the food and direct it to the hepatopancreas. Only fluid and the smallest particles are able to pass through the duct of the digestive gland. The musculature of the pyloric stomach serves to force food through the filters, and also directs it to the intestine. Food that is not absorbed is eliminated via the anus.

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Question:

A simple pendulum has a period of 2.40 sec at a place where g = 9.810 m/sec^2. What is the value of g at another place on the earth's surface where this pendulum has a period of 2.41 sec? (See figure).

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Solution:

The period of oscillation is given by T = 2\pi\surd(l/g). Therefore, two different periods, T1and T_2 will corresponds to two different gravitational constants, g_1 and g_2, as follows. T_1/T_2 = [{2\pi\surd(l/g_1)}/{2\pi\surd(l/g_2)}] = \surd(g_2/g_1) or(2.40)/(2.41) = \surd[(g_2)/(9.810m/sec^2)] g_2 = 9.729 m/sec^2

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Question:

Cytochrome C is a molecule with an iron-porphyrin head connected to a protein tail. Analysis shows the molecule is 0.45 percent iron by weight. Calculate the molecular weight of cytochrome C.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E22-0803.htm

Solution:

From the accompanying figure, one can see that there is one iron atom in each cytochrome molecule. Thus, there is 1 mole of Fe atoms in 1 mole of cytochrome C. If each molecule is 0.45 percent iron by weight, then in 1 mole of cytochrome C, 0.45 percent of the total mo-lecular weight of cytochrome C is equal to the weight of one mole of Fe. MW of Fe = (4.5 × 10^-3) (MW of cytochrome C) = 55.8 g/mole MW of cytochrome C = [55.8 (g / mole)] / [4.5 × 10^-3] = 12,400 g/mole.

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Question:

A common method for commercially peeling potatoes is to soakthem for 1 - 5 minutes in a 10 - 20% solution ofNaOH (molecularweight = 40.0 g/mole) at 60 - 88\textdegreeC, and to spray offthe peel once the potatoes are removed from solution. As aneconomy measure, a manufacturer titrates theNaOH solutionwith standardized H_2SO_4 (molecular weight = 98.0 g/mole) at the end of each day to determine whether the solutionis still capable of peeling potatoes. If, at the end of oneday, he finds that it takes 64.0 ml of a 0.200 M solution ofH_2SO_4 to titrate a 10.0 ml sample ofNaOHsolution to neutrality, what concentration ofNaOHdid he find?

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Solution:

At the neutralization point, the number of equivalents ofNaOHis equalto the number of equivalents of H_2SO_4. But since the number of equivalentsis equal to the product of the normality, N, and the volume, V, thiscondition may be stated as N_NaOHV_NaOH= N_H2SO4V_H2SO4 . To solve this problem, we must solve forN_NaOH, N_NaOH= [(N_H2SO4 V_H2SO4)/(V_NaOH)] Since H_2SO_4 has twoionizableprotons, its normality is equal to twiceitsmolarity, or N_H2SO4 = 2 × 0.200 M = 0.400 N.NaOHhas only one hydroxylgroup, so that its normality is equal to itsmolarity. Then N_NaOH= [(N_H2SO4 V_H2SO4)/(V_NaOH)] = [(0.400 N × 64.0 ml)/(10.0 ml)] = 2.56 N. The concentration ofNaOHsolution is therefore 2.56 N, or 2.56 M.

Question:

The reaction A + B \rightarrow C was studied kinetically and the following data was obtained. Experiment A B Rate(mole/litter - min) 1 1.0M 1.0M 0.15 2 2.0M 1.0M 0.30 3 1.0M 2.0M 0.15 Determine the rate expression.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E13-0442.htm

Solution:

The rate of reaction is equal to some rate constant, k, multiplied by the concentrations of A and B raised to the appropriate powers. That is, rate = k [A]^m [B]^n where the exponents m and n are to be determined. Comparing experiments 2 and 1, we see that holding [B] constant while doubling [A] doubles the rate of re-action (from 0.15 to 0.30). Hence, the rate is directly proportional to [A] and m = 1. Thus, if we hold [B] constant and triple [A], the rate triples. Comparing experiments 3 and 1, we see that holding [A] constant and changing [B] (from 1.0M to 2.0M) has no effect on the rate. Hence, the rate is independent of [B] and n = 0 (so that [B]\textdegree = 1 and [B] does not appear in the rate expression). Substituting m = 1 and n = 0 into the rate expression gives rate = k [A]^1 [B]\textdegree, or, rate = k [A].

Question:

A wine has an acetic acid (CH_3 COOH, 60 g/formula weight) content of 0.66% by weight. If the density of the wine is 1.11 g/ml, what is the formality of the acid?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E08-0275.htm

Solution:

This problem involves the correct inter-pretation of percent by weight. 0.66% by weight means 0.66 × 10^-2 g acid per 100 g wine or 6.6 × 10^-2 g acid per 1000 g wine. To convert 6.6 × 10^-2 g acid to formula weight we divide by 60 g/formula weight and obtain 6.6 × 10^-2 g/60 g/formula weight = 1.1 × 10^-3 formula weight. To convert 1000 g wine to volume, we divide by the density of the wine and obtain 1000 g/1.11 g/ml = 900 ml = 0.90 liter. The formality is then formality = [(formula weights)/{volume (liters)}] = [(1.10 × 10^-3 formula weights)/(0.90 liter)] = 1.2 × 10^-3 formal = 1.2 × 10^-3 F.

Question:

Determine the equilibrium constant for the following re-action at 25\textdegreeC: Mg (s) + Sn^2+ \rightarrow Mg^2+ +Sn(s)

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Solution:

The equilibrium constant, k, is related to the cell potential, ∆E\textdegree, by means of the following two formulas: ∆G\textdegree = - n F ∆E\textdegree,and∆G\textdegree = - 2.303 RT log k, where n is the number of electrons transferred, F the Faraday constant, R the gas constant, ∆G\textdegree the free energy, and T the absolute temperature. ∆E\textdegree will be determined from the half-cell potentials, E\textdegree, First, the cell potential is determined from thehalf-cellreactions: Mg (s) \rightarrow Mg^2+ + 2e-E\textdegree = + 2.37 v Sn^2+ + 2e- \rightarrowSn(s)E\textdegree = -0.14 v. Adding the two half-cell reactions and their respective values for E\textdegree, Mg (s) + Sn^2+ + 2e- \rightarrow Mg^2+ 2e- +Sn(s) ∆E\textdegree = (2.37 v) + (- 0.14 v) or,Mg (s) + Sn^2+ \rightarrow Mg^2+ +Sn(s)∆E\textdegree = 2.23 v. The cell potential for the reaction is therefore 2.23 v. Equating the two expressions ∆G\textdegree = - n F ∆E\textdegree and∆G\textdegree = - 2.303 RT log k gives- 2.303 RT log k = - n F ∆E\textdegree, or,log k = [(n F ∆E\textdegree)/(2.303 RT)]. Since two electrons are transferred from Mg(s) to Sn^2+, n = 2. The absolute temperature is T = 25\textdegreeC + 273 = 298\textdegreeK. Hence log k = [(n F ∆E\textdegree)/(2.303 RT)] = [{(2) (2.30609 × 10^4 cal/mole-v) (2.23 v)} /{(2.303) (1.987 cal/mole-deg) (298\textdegreeK)}] = 75.6, or,k = 107 5 \bullet 6= 4 × 10^75. The equilibrium constant for this reaction is therefore 4 × 107 5.

Question:

(a) Suppose that the statement IF(I - J) 5,6,6 is replaced by IF(I - J) 6,5,5 in the program of the previous problem. Explain the effect of this change. (b) Explain what happens if the same statement is replaced byIF(J - I) 6,5,5 (c) Explain what happens if the statement GO TO 7 is omitted in the program of the previous problem.

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Solution:

a)If we use the statement IF(I - J) 6,5,5, then whenever the value of I - J is negative the computer will go to statement number 6, which causes the output of the value of I, followed by the end of the program (i.e., STOP statement). On the other hand, if the value I - J is zero or positive, statement number 5 will be executed. Therefore, the program is written to find the smallest of two numbers. b)If the statement is replaced by IF(J - I) 6,5,5 execution will happen exactly as it did in a), except that the values of I and J will be interchanged. At the end of the output will be the largest of two numbers. c)If the GO TO 7 statement is omitted, there are two possible cases: I.If I - J < 0 (i.e., I < J), the computer will go to statement 5 and print the value of J. II.If I - J \geq 0 (i.e., I \geq J), the computer will go to state-ment 6 and first print the value of I, and then the value of J.

Question:

Ammonia is synthesized commercially according to the equation (1/2) N_2 (g) + 1(1/2) H_2 (g) \rightarrow NH_3 (g), under a constant total pressure of 50 atm. At 350\textdegreeC the equilibrium constant has been determined as 0.0278. The heat of reaction for this process is ∆H = -11.04 Kcal/mole. Determine the value of equilibrium constant at 450\textdegreeC.

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Solution:

For a reaction with equilibrium constant k_1 at absolute temperature T_1 and equilibrium constant k_2 at temperature T_2, log (k_2 /k_1) =[(- ∆H\textdegree)/(2.303 R)] [(1/T_2 ) - (1/T_1 )] where R is the gas constant. Let T_1 = 350\textdegreeC = 623\textdegreeK, k_1 = 0.0278, and T_2 = 450\textdegreeC = 723\textdegreeK. ∆H\textdegree = - 11.04 Kcal/mole = - 11, 040 cal/mole. Then log (k_2 /k_1) =[(- ∆H\textdegree)/(2.303 R)] [(1/T_2 ) - (1/T_1 )] log [(k_2 )/(0.0278)] = [(- 11, 040 cal/mole)/(2.303× 1.987 cal/mole - deg)] [{(1)/(723\textdegreeK )} - {(1)/(623\textdegreeK)}] = - 0.535. Solving for k_2, log [(k_2 )/(0.0278)] = - 0.535 log [(k_2 )/(0.0278)] = - 0.535 [(k_2 )/(0.0278)] = 10-0 .5 3 5= 0.292 k_2 = 0.0278 × 0.292 = 0.00812. Thus, the value of the equilibrium constant at 450\textdegreeC. is 0.00812.

Question:

To produce a minimum reflection of wavelengths near the middle of the visible spectrum (550 m\mu), how thick a coating of MgF_2 (n = 1.38) should be vacuum-coated on a glass surface?

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Solution:

Consider light to be incident at near-normal incidence. We wish to cause destructive interference be-tween rays r_1 and r_2 that maximum energy passes into the glass. A phase change of 1/2\lambda occurs in each ray, for at both the upper and lower surfaces of the MgF_2 film, the light is reflected by a medium at greater index of refrac-tion. When striking a medium of lower index refraction, the light is reflected with no phase change. Since in this problem both ray 1 and 2 experience the same phase shift, no net change of phase is produced by these two reflections. Hence the only way a phase change can occur is if the 2 rays travel through different optical path lengths. (the optical path length is defined as the product of the geo metric distance a ray travels through and the refractive index of the medium in which the ray is travelling). For destructive interference, the 2 rays must be out of phase by an odd number of half wavelengths. Therefore, the op-tical path difference needed for destructive interference is, 2nd = (2N + 1) \lambda/2N = 0, 1, 2,... for minima Note that 2nd is the total optical path length that the rays traverse. WhenN = 0 , d = (1/2\lambda)/(2n) = \lambda/4n = [(550 m\mu) / (4 × 1.38)] = 100 m\mu = 1.0 × 10^-5 cm

Question:

Write a FORTRAN program containing four subroutines to initialize a list, insert a name in alphabetical order, delete a name, and output the list in order. You will need to use pointers to create this data structure, known as a chained or linked list.

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Solution:

For this project, you should construct the main program so that it has a case structure, enabling the user to choose whether he would like to insert or delete a name. The main program should consist only of type declarations, a DO-WHILE loop to choose the case the user wants, and statements to call the subroutines. Since this part is perhaps the easiest, we will do it first and then explain the meaning of the variables. Maximum name length is five letters. INTEGER CHOICE, PTR, DATA, AVAIL, N, NAME DIMENSION DATA (10), PTR (10) N = 10 CALL INIT (PTR, N, AVAIL) CDO WHILE CHOICE NOT EQUAL TO 'STOP' 10IF (CHOICE.EQ.'STOP') GOTO 99 WRITE (5, 100) 100FORMAT (5X,'TYPE INSRT, DELET, OR STOP') READ (5,101) CHOICE 101FORMAT (A5) CIF CHOICE IS INSRT, THEN IF (CHOICE.NE.'INSRT') GO TO 20 WRITE (5, 102) 102FORMAT (5X, 'TYPE THE NAME, PLEASE') READ (5, 103) NAME 103FORMAT (A5) CALL INSERT (DATA, PTR, N, AVAIL, NAME) CALL PRINT (DATA, PTR, N) GO TO 10 CIF CHOICE IS DELET, THEN 20IF (CHOICE.NE.'DELET') GO TO 30 WRITE (5, 102) READ (5, 103) NAME CALL DELETE (DATA, PTR, N, AVAIL, NAME) CALL PRINT (DATA, PTR, N) 30GO TO 10 CEND DO-WHILE 99STOP END DATA (N) is the array which contains the names to be put in order. PTR (N) is the array containing pointers which can be manipulated to accommodate insertions and deletions. The mechanics of pointers will be explained as the problem de-velops, but for now remember that pointers permit the access-ing of data in memory without performing an actual search- and-sort routine. Continuing with the main program, the variable CHOICE can take on three possibilities: INSRT if you want to in-sert a name, DELET if you want to delete a name, STOP to end the session. The program will keep looping back to the beginning, allowing you to adjust the list until you type in STOP. Notice also that after each insertion or deletion, the subroutine PRINT allows you to see the elements of the list. Subroutine INIT sets up the linked list. N is the to-tal number of names (in this case, 10). AVAIL is the ad-dress of the first available location in the storage area DATA. PTR (1), known as the list head, is set to zero. This means that DATA (l) is not used, a necessary precaution that avoids complications in the special case of an empty list. PTR (2) through PTR (9) are initialized such that PTR (2) = 3, PTR (3) = 4, and so on until PTR (9) =10. Finally, PTR (10) is also set to zero. Let us now begin to fill DATA with names to illustrate the activity of the pointers. (See Figure 1.) Subroutine INSERT will go down the list, search for the first available storage location in DATA, insert the name, and most important, rearrange the pointers so that they will cause the list to be printed in alphabetical order. The comparison for alphabetizing is done at statement 23, but the ordering of the pointers is accomplished by the state-ments following statement 33. Let us "walk through" these statements to discover what is happening at each step: J becomes 3 AVAIL becomes PTR (3), which is 4 PTR (3) becomes PTR (1), which is 1 DATA (3) becomes 'CANDY' PTR (1) becomes 3 Now, let us turn to the case where we want to delete a name, say BILLY. The list starts off as it is in Figure 1, after the addition of 'CANDY'. (See Figure 2.) The storage location at DATA(2) is saved in the variable AVAIL; only the pointers are rearranged. In subroutine DE-LETE, statement 14 does the table lookup of the name to be deleted. If it is present, control is passed to statement 34, where the pointers are rearranged to eliminate the de-sired name. A "walk through" yields the following: JD becomes 2 PTR (1) becomes PTR (2), which is 3 PTR (2) becomes 4 AVAIL becomes 2 SUBROUTINE INIT (PTR, N, AVAIL) INTEGER I, N, PTR (N), AVAIL CINITIALIZE LIST HEAD AND START OF FREE LIST AVAIL = 2 PTR (1) = 0 CSET POINTERS FOR FREE LIST DO 12 I = 2, N - 1 PTR (I) = 1+1 12CONTINUE CSET NULL POINTER AT END OF FREE LIST PTR (N) = 0 RETURN END SUBROUTINE INSERT (DATA, PTR, N, AVAIL, NAME) INTEGER II, J, N, PTR (N), DATA (N), NAME IF (AVAIL.NE.0) GO TO 13 WRITE (5, 104) 104FORMAT (5X, 'NO STORAGE LEFT') GO TO 53 13II = 1 CSEARCH LIST FOR INSERTION POINT CDO-WHILE PTR (II) NOT EQUAL TO ZERO CAND DATA (PTR (II)) LESS THAN NAME 23IF (PTR(II).EQ.O.OR.DATA(PTR(II)).GE.NAME) 1GO TO 33 II = PTR (II) GO TO 23 33CONTINUE CEND DO-WHILE CALLOCATE SPACE FROM FREE LIST FOR CNEW ENTRY AND INSERT IT FOLLOWING CENTRY II BY SETTING POINTERS J = AVAIL AVAIL = PTR (J) PTR (J) = PTR (II) DATA (J) = NAME PTR (II) = J 53CONTINUE RETURN END SUBROUTINE DELETE (DATA, PTR, N, AVAIL, NAME) INTEGER ID, JD, N, PTR (N), AVAIL, DATA(N) ID = 1 CFIND NAME TO BE DELETED CDO WHILE PTR (ID) NOT EQUAL TO ZERO AND CDATA (PTR (ID)) NOT EQUAL TO NAME 14IF (PTR (ID).EQ.O.OR. DATA (PTR (ID)).EQ. NAME) 1GO TO 24 ID=PTR (ID) GO TO 14 24CONTINUE CEND DO-WHILE IF (PTR (ID).NE.O) GO TO 34 WRITE (5, 105) NAME 105FORMAT (5X, A5, 'NOT PRESENT IN LIST') GO TO 54 CENTRY FOLLOWING ID NOW CONTAINS CNAME. DELETE IT FROM DATA LIST AND CPUT IT ON FREE LIST. 34JD = PTR (ID) PTR (ID) = PTR (JD) PTR (JD) = AVAIL AVAIL = JD 54CONTINUE RETURN END SUBROUTINE PRINT (DATA, PTR, N) INTEGER IP, N, PTR (N), DATA (N) IP = PTR (1) CDO WHILE IP NOT EQUAL TO ZERO 15IF (IP.EQ.O) GO TO 25 WRITE (5, 107) DATA (IP) 107FORMAT (1X, A5) GO TO 15 25CONTINUE CEND DO-WHILE RETURN END On some systems, the DO-WHILE constructs in INSERT and DELETE may cause problems because of a reference to DATA (0). To amend that difficulty, you may substitute the following: (for INSERT) 23IF (PTR (II).EQ.O) GO TO 33 IF (DATA (PTR (II)).GE.NAME) GO TO 53 (for delete) 14IF (PTR (ID).EQ.O) GO TO 24 IF (DATA (PTR (ID)) .EQ.NAME) GO TO 54

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Question:

The force required to start a mass of 50 kilograms moving over a rough surface is 343 Nt. What is the coefficient of starting friction?

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Solution:

The coefficient of starting friction is given by the relation F =\mu_stN where F is the force of friction,\mu_stis the coefficient of starting friction, and N is the force normal to the direction of travel. Since we assume the object is travelling on a horizontal plane, the normal force is simply the force of gravity, by Newton's Second Law. This force is N = mg N = 50 kg (9.80 m/s^2) = 490 Newton Therefore 343nt=\mu_st× 490nt \mu_st= 0.70.

Question:

Calculate the osmotic work done by the kidneys in secreting 0.158 moles of Cl^- in a liter of urine water at 37\textdegreeC when the concentration of Cl^- in plasma is 0.104 M, and in urine 0.158 M.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E26-0881.htm

Solution:

Work done on a system is measured by the change in free energy of the system. To calculate the osmotic work done by living systems in transferring solutes from regions of lower concentration to those of higher, the following equation is used: ∆G = G_2 - G_1 = n RT 1n C_2/C_1 Where R = gas constant, T = temperature, n = number of moles, G = free energy, and C = concentration. This equation gives the free energy change (∆G) corresponding to a change in the concentration of one component (solute). For n moles, if C_2 is the final con-centration and C_1 the initial concentration, then, from the equation, an increase in concentration will cause the solution to gain free energy with respect to that particular component (∆G will be positive), and that dilution with respect to the same component will cause the solution to lose free energy (∆G will be negative). A loss in free energy represents a loss in the capacity to do work. For this problem, R = 1.987 cal/mole \textdegreeK, T = 273\textdegree + 37\textdegreeC = 310\textdegreeK n = 0.158 moles Cl^-, C_2 = 0.158 M, C_1 = 0.104 M. Thus, osmotic work = ∆G = (0.158) (1.987) (310) In [(0.158 M) / (0.104 M)] = (2.303) (0.158) (1.987) (310) log 1.52 = 40.7 cal.

Question:

Write a BASIC program to find and print the difference, product, and quotient for pairs of complex numbers which are inputted from a ter-minal. Again, treat a single complex number as an ordered pair of real numbers.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G10-0224.htm

Solution:

We use the same notation developed in the previous problem for adding complex numbers. If Z_1 = (a, b) and Z_2 = (c, d), then the difference, product, and quotient of Z_1 and Z_2 are respectively Z_1 - Z_2 = (a-c, b-d) Z_1 \textasteriskcentered Z_2 = (ac -bd, ad + be) Z_1 / Z_2 = [(ac +bd) / (c^2 + d^2), (bc- ad) / (c^2 + d^2)] 1\O PRINT \textquotedblleftTHIS PROGRAM FINDS THE DIFFERENCE,"; 15 PRINT "PRODUCT, AND QUOTIENT OF TWO"; 17 PRINT "COMPLEX NUMBERS IN A,B FORM"; 2\O PRINT 3\O PRINT "TYPE IN THE FIRST NUMBER" 4\O INPUT A,B 5\O IF A = 999 THEN 999 6\O PRINT "TYPE IN THE SECOND NUMBER" 7\O INPUT C,D 8\O PRINT "THE DIFFERENCE IS ("; A-C;",";B-D;")" 9\O PRINT "THE PRODUCT IS ("; 95 PRINT A\textasteriskcenteredC - B\textasteriskcenteredD;",";A\textasteriskcenteredD+B\textasteriskcenteredC;")" 96 PRINT "THE QUOTIENT IS ("; 98 PRINT (A\textasteriskcenteredC+B\textasteriskcenteredD)/(C_\uparrow2+D_\uparrow2); ","; (B\textasteriskcenteredC-A\textasteriskcenteredD)/(C_\uparrow2+D_\uparrow2); ")" 99 GO TO 20 999 END

Question:

The disease diabetes mellitus is caused by an insulin deficiency. Can this disease be inherited? What happens if a person with diabetes goes untreated? What is the treatment given for diabetes mellitus?

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/Users/wenhuchen/Documents/Crawler/Biology/F21-0538.htm

Solution:

It is believed that diabetes mellitus probably results from an insufficiency in insulin produc-tion. This may happen either as a genetic abnormality in which insulin synthesis is impaired, or as an acquired disease in which insulin-secreting cells of the pancreas are hypofunctional or destroyed. In fact, the latter may also have a genetic origin. Thus in many cases of diabetes mellitus, the disease can be inherited. The children of the person with diabetes may tend to exhibit the symptoms of diabetes. The disease, however, may develop slowly or to a varying degree, for diabetes is not an "all-or-none" disease. Moreover the overt symptoms of diabetes mellitus may all but disappear .with appropriate measures such as a change in diet. Thus it would be more valid to say that a tendency toward diabetes mellitus can be inherited. The untreated person with diabetes, suffering from a deficiency of insulin, will have an excess of glucose in the blood. A large amount of glucose will be excreted and lost from the body in the urine. The person with diabetes will therefore fail to acquire enough energy by glucose metabo-lism alone. He will have to obtain energy by way of triglyc-eride catabolism. The breakdown of triglyceride results in an elevation of fatty acid and glycerol levels in the blood. If further energy is needed, the diabetic body will convert his protein reserve into glucose, which would be accompa-nied by an accumulation of ketones. The products of these two metabolic pathways (i.e. fatty acids and ketones) will upset the delicately balanced pH system of the body, and thus may be considered toxic. Because insulin also has the effect on hindering glycogen formation from glucose, the diabetic individual has very little carbohydratere-serves. Most of the glucose consumed in food, unavailable to the cells for use, simply passes out of the body. In short, the untreated person with diabetes will face the paradox of cell starvation in the presence of an elevated plasma glu-cose concentration. Only the brain cells are spared the glu-cose deprivation since their uptake of glucose is not hormonally regulated. The consequence of these catabolic processes is progressive loss of weight despite any increase in food intake, and a lowered pH of the body fluids. Another- well known consequence of untreated diabetes is dehydration of the body tissues. The large amount of glucose excreted into the kidney tubules induces water to passively follow suit (osmosis), and prevents it from being reabsorbed into the body cells. For some complicated reason, sodium reabsorption is also reduced. The net effect is the marked excretion of water and sodium, resulting in a decreased plasma volume and hence blood pressure. The lowered blood pressure may reduce blood flow to the brain, leading to brain damage, coma and possibly death. Associated with diabetes are arteriosclerosis, small- vessel and nerve disease, susceptibility to infection, and a variety of other complications. At one time it was thought that these problems resulted only from insulin deficiency. However, this view is now being questioned in light of the evidence which has shown that these problems can be delayed but not abolished by administration of insulin. Many researchers now believe that excessive hyperglycemia or possibly even hypoglycemia due to inappro-priate insulin administration contribute to the occurrence of complications. In treating persons with diabetes, the aim is to main-tain the plasma glucose at a near normal value. The admi-nistration of insulin is the major therapy, and must be given through injection. This is because insulin, being a protein hormone, would be broken down by gastrointestinal enzymes if given orally. The dosage must be carefully determined since an overdose would drastically lower the plasma glucose concentration, thereby depriving brain cells of needed glucose and ultimately causing death. Non-insulin drugs which stimulate the islets of Langerhans to produce more insulin have been used in some cases, but are not effective therapeutic measures. There are actually two types of diabetes. Juvenile dia-betes (Type I) is the insulin-deficient diabestsjust de-scribed. These patients require insulin and show some of the more severe effects of the disease. Adult-onset diabetes (Type II) is much more common and less threatening. This type of diabetes is seen more frequently in older often obese patients. Their bodies produce and release insulin yet the receptors for insulin are less sensitive. The receptors are "down-regulated." Type II patients are not insulin-de-pendent. Improvement may occur with proper diet and exer-cise.

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Question:

What is instinct? What is instinct?

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/Users/wenhuchen/Documents/Crawler/Biology/F31-0790.htm

Solution:

The term "instinct\textquotedblright is difficult to define since its meaning has been so broadly applied. Instinct has been used to designate behavior that is stereotyped; that is, behavior which is rigidly executed and therefore highly predictable when the appropriate stimuli are presented. For example, when the eggs of herring gulls and other ground-nesting birds accidentally fall out of the nest area, the parent gulls use their bills to roll them back into the nest. This is a stereotyped response and can be repeated by purposely displacing the eggs. The eggs serve as releasing stimuli to elicit the egg-rolling behavior. When investigators placed abnormally large wooden models of eggs adjacent to the gulls' own dis-placed eggs, the gulls preferred to roll the larger models back to the nest. Any such stimuli having a quality that is preferred over that of the natural stimulus encoun-tered in the life of an animal - in this case abnormally large size - is referred to as a supernormal stimulus. Instinct is often believed to consist solely of innate responses whichare determined by set neural path-ways of the brain and not altered by experience. How-ever, recent study has shown that learning may play a role in stereotyped instinctive behavior. For example, ducks and other ground-nesting birds either cry out alarm calls or crouch in fear when a bird of prey flies overhead (see Figure). It was discovered that the basic shape of the flying bird is the releasing stimulus for fear behavior; particular anatomical features are not crucial. Thus, when an investigator passed different shapes of cardboard silhouettes over the heads of ducklings, only the bird shape having a head shorter than its tail evoked the fear response (See Figure). When this same silhouette was passed over the ducklings in the reverse direction, so that the figure appeared to have a long head and short tail, no fear response was evoked. In the first instance, the bird's silhouette represented a hawk, while the second silhouette represented another duck or goose. This fear response was initially thought to be a strictly inherited response to a sign stimulus. Further study, however, has indicated elements of learning in this behavior. Ducklings are initially afraid of all overhead flying objects. Eventually they stop reacting to ones they see repeatedly, such as other, ducks and falling leaves. Hawks, however, are generally scarce, and the ducklings never become accustomed to their peculiar shape. Learning therefore plays a role in what seems to be a totally innate stereotyped response. Hence, it is difficult to make a distinction between instinct and other kinds of behavior, since some instincts can be shown to contain learned elements.

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Question:

At 1000\textdegreeK, K = 2.37 × 10-^14 for the reaction, N_2 (g) + 3H_2 (g) \rightleftarrows 2NH_3 (g) . If you inject one mole of each N_2 and H_2 in a one-liter box at 1000\textdegreeK, what per cent of the H_2 will be converted to NH_3 at equilibrium?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E09-0320.htm

Solution:

To answer this question, determine the concentration of H_2 at equilibrium. Once this is known, subtract it from the initial con-centration. The difference yields the amount that reacted to produce NH_3 . By multiplying the quotient of the difference divided by the original amount of H_2 by 100, you obtain the per cent. To find [H_2] at equilibrium, employ the equilibrium constant expression. It states that K, the equilibrium constant, is equal to the concentration ratio of the products to reactants, each raised to the power of its coefficient in the chemical reaction. For this re-action, then, K = 2.37 × 10-^3 = [NH_3]^2 / {[H_2 ]^3 [N_2] } . To find [H_2 ], proceed as follows. Let x = amount of N_2 that dis-sociates. The initial concentrations of all species are 1M, sincemolarity= moles/liter, and 1 mole of each is placed in a one-liter box. If the N_2 's initial concentration is one and x moles/liter dis-sociates, then, at equilibrium, there is (1\Elzbar x)moles/liter left of N_2 . In other words, at equilibrium, [N_2 ] = 1\Elzbar x. Now, from the chemical equation, it is seen that for every mole of N_2 that reacts, 3molesof H_2 are necessary. Thus, when x moles/liter of N_2 dissociate, 3x moles/Iiterof H_2 are required. The initial concentration is 1, so that, at equilibrium [H_2 ] = (1\Elzbar 3x) moles/Iiter. Notice, also, that for every mole/liter of nitrogen that dissociates, 2(mole/liter) of ammonia is obtained. Substituting these values into the equilibrium constant expression, one obtains 2.37 × 10-^3 = [NH_3 ]^2 / { [N_2 ] [H_2 ]^3 } = (2x)^2 / { (1\Elzbarx) (1\Elzbar3x)^3 } . Solving for x, one obtains x = .0217 moles/liter of N_2 that dis-sociate. Thus, [H_2 ] = 1\Elzbar 3x = .935 moles/liter at equilibrium. The initial concentration was 1 mole/liter. The difference is the amount that dissociated, i.e., the H_2 . The difference = 1\Elzbar .935 = .065, Thus, the percent that dissociated equals [(.065M) / (1M)] × 100 = 6.5% .

Question:

An efficient method of organizing an unordered set of re-cords is by means of a HASH (or SCATTER) FILE. A HASH FILE works as follows: A specified set of arithmetic operations are performed on a data element (also called a key) of a record. The result of this operation is then used as the HASH address for that key. Using this HASH address we can store and locate the record. It should be noted that dif-ferent records may have the same HASH address. Our program will analyze how well a HASH function (the arithmetic oper-ations on a key) performs for a sample input file. We will use as our key element the ID number of the record. The arithmetic operation will be to divide the key by 50, and use the remainder as the HASH address. Thus, our addresses will range from 0 to 49 (possible remainders). Finally, we will set up and print the following tables:

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Solution:

1 HASHFILE START 0 2 SAVE (14,12) 3 BALR 3,0 4 ST 13, SAVE+4 5 LA 13, SAVE \textasteriskcentered THE LOADING OF TABLE 2 \textasteriskcentered BY INCREMENTING REGISTER 5 BY 1 AND STORING \textasteriskcentered THE RESULT IN CONTIGUOUS FULLWORDS. 9 LA 6, ASHTABLE 10 L 5, = F '0' FIRST HASH-ADDRESS OF TABLE 11 L 10, = F '50' LOOP CONTROL FOR REMAINDERS 12 LOAD CVD 5, CONVERT 13 PACK 0 (2, 6), 0 (2, 6) 14 ZAP 0 (2, 6), CONVERT + 6 (2) STORING POSSIBLEREMAINDERS 15 PACK 2 (2, 6), 2 (2, 6) 17 LA 6, 4 (6) GOTO NEXT WORD OF ASHTABLE 18 A 5, = F '1' NEXT REMAINDER FOR ASHTABLE 19 CR 5, 10 DO WE HAVE REMAINDERS 0 - 49? 20 BL LOAD \textasteriskcentered PROCESSING THE KEY AFTER READING IT IN 22 GETKEY READ CARD CARDINPA, CRDEOF 23 CLC CRDIDNO, = C '9999' END OF FILE CARD 24 BE PRTHEAD 2 25 PACK WRKKEY, CARDIDNO 26 ZAP CONVERT, WRKKEY 27 CVB 9, CONVERT REGISTER 9 HOLDS ID# PRIOR TO DIVISION 28 SR 8, 8 CLEAR REGISTER 8. 29 D 8, FINDHASH REMAINDER DETERMINES HASH ADDRESS 30 CVD 8, CONVERT CONVERT REMAINDER IN REGISTER 8 TO DECIMAL \textasteriskcentered PRINTING THE ID NUMBER AND ITS HASH - ADDRESS 32 MVC PRTKEY, CRDIDNO 33 ZAP REM, CONVERT STORE REMAINDER 34 MVC PRTHASH1, PATT1 35 ED PRTHASHI, REM 36 PRINTLIN PRTDETL, 133 \textasteriskcenteredINCREMENTING THE COUNT FOR AN ADDRESS 38 LA 6, ASHTABLE 39 LR 9, 8 40 SR 8, 8 41 M 8, POSITION 42 AR 6, 9 GOTO REMAINDER IN ASHTABLE 43 ZAP CONVERT, 2 (2, 6) STORE COUNT IN 'CONVERT' 44 CVB 7, CONVERT PLACE COUNT IN REGISTER 7 45 A 7, = F '1' INCREMENT COUNT 46 CVD 7, CONVERT 47 ZAP 2 (2, 6) , CONVERT + 6 (2) 48 B GETKEY 49 PRTHEAD2 PRINTLIN HEADING2, 133 \textasteriskcenteredPRINTING HAS\rule{1em}{1pt}ADDRESS AND THEIR COUNTS 51 LA 6, ASHTABLE 52 LA 5, 50 53 PRINTHASH MVC PRTHASH2,PATT1 54 ZAP TEMPHOLD, 0 (2, 6) 55 ED PRTHASH2, TEMPHOLD 56 MVC PRTCOUNT,PATT1 57 ZAP TEMPHOLD,2 (2, 6) 58 ED PRTCOUNT, TEMPHOLD 59 PRINTLIN PRTTABLE. 133 60 LA 6, 4 (6) 61 BCT 5 , PRINTASH 62 CRDEOF L 13, SAVE + 4 63 RETURN (14,12) \textasteriskcenteredDATA DEFINITION FOR CARD INPUT AREA 64 CARDINPA DS \varphi CL80 65 CRDIDNO DS CL5 66 DS CL75 \textasteriskcenteredDATA FOR KEY AND ITS ADDRESS 67 HEADING1 DS \varphi CL133 68 PRTCC1 DS C '1' CARRIAGE CONTROL-RESULTS IN PRINTING ON NEW PAGE 69 DC 20C'' 70 DC C 'KEY HASH-ADDRESS' 71 DC 92C' ' DATA DEFINITION FOR PRINT AREA 72 PRTDETL1 DS \varphi CL133 73 DC C '0' CARRIAGE CONTROL 74 DC 19C ' ' 75 PRTKEY DS CL5 76 DC 8C' ' 77 PRTHASH1 DS CL4 78 DC 96C ' ' \textasteriskcenteredDATA DEFINITION FOR HEADING OF HASH-ADDRESS AND COUNT 79 HEADING2 DS \varphi CL133 80 PRTCC2 DC C '1' 81 DC 20 'C' 82 DC C 'HASH-ADDRESS COUNT' 83 DC 92 'C' \textasteriskcenteredDATA DEFINITION FOR ASHTABLE 84 PRTTABLE DS \varphi CL133 85 DC C '0' 86 DC 25C ' ' 87 PRTHASH2 DS CL4 88 DC 10C ' ' 89 PRTCOUNT DS CL4 90 DC 89C ' ' \textasteriskcenteredDATA DEFINITION FOR WORK AREAS 91 SAVE DS 18F 92 ASHTABLE DC 50F '0' 93 WRKKEY DS PL4 94 POSITION DS F '4' 95 CONVERT DS D 96 FINDHASH DC F '50' THE DIVISOR 97 REM DS PL2 98 TEMPHOLD DS PL2 99 PATT1 DC X' 4020212020' 100 END HASHFILE In general, the format of our program is as follows: The table to contain the hash-addresses, and the amount of records hashed to each address (count), is defined as having a length of 50 fullwords. By use of Register 5, the set of all possible remainders (0-49) are placed in the first two bytes of the words, each of the words containing one possible remainder, increasing from one to 50. The latter two bytes of each word will then store the 'count' for each corresponding hash-address. As each address is found, it is printed along with its key. This constitutes the first table of the program. Af-ter printing the key and its address, we then increment the count of this address in the 'address- count' table (ASH-TABLE) . When no more key elements are found (keys 99999), the contents of the second table are printed. It is to be observed that, in the case of 'ASHTABLE', we manipulate 2 bytes of each word at a time and not the complete word. We first load all the remainders in the first two bytes of each word in the 50-word area. This is accomplished by means of: 13.PACK0 (2, 6), 0 (2, 6) and14.ZAP0 (2, 6), CONVERT + 6 (2) The pack instruction of Line 13 simply converts the contents of the location addressed by 0 (2, 6) to packed decimal. The location addressed by 0 (2, 6) is a part of ASHTABLE, and since ASHTABLE is defined as having zeroes, this location also contains zeroes. The contents of this location are then replaced by a possible remainder. This is done by the ZAP instruction of Line 14. Using the LA instruction of Line 17 we can address the next word of ASHTABLE. LA 6, 4 (6): Coded in such a manner, this instruction adds a displacement of 4 bytes to the present address in Register 6. Hence, if Register 6 contains 1080, the new address after LA 6, 4(6) would be 1084. This technique en-ables us to use just the two instructions of Lines 13 & 14 to load the table. Incidentally, Operand-2 of Line 14 is an example of the combination of implicit and explicit ad-dressing. Since the ZAP instruction proceeds from right to left, this coding was not really necessary, but it - indicates how the two types of addresses can be combined. ZAP 0(2,6),CONVERT+6(2) tells the computer to place the last two bytes of the doubleword 'CONVERT' into Operand- 1. The following figure indicates 'CONVERT+6(2)'. As is always the case, the location of the first byte of a field represents the address of the field in main mem-ory. By 'ADDING' integers to the field (symbol), we can reference a particular byte in the field. Furthermore, if we also specify a length factor, we are able to reference a particular set of bytes in the field. So in the combination of both types of addresses, we have SymbolAdditionLengthFactor \downarrow\downarrow\downarrow CONVERT+6(2). CONVERT+6(2). Note, you should be careful not to make references out-side the Note, you should be careful not to make references out-side the field, since this will result in error messages from the computer or field, since this will result in error messages from the computer or unintended executions. unintended executions. Interestingly enough, we can also reference previous bytes in Interestingly enough, we can also reference previous bytes in memory by "subtracting" integers from our symbol. For example, ZAP memory by "subtracting" integers from our symbol. For example, ZAP ANY, SYM-6. If 'SYM' were at location 1848, then the contents of location ANY, SYM-6. If 'SYM' were at location 1848, then the contents of location (1848-6) 1842 would be placed in 'ANY'. Once again, you should check to (1848-6) 1842 would be placed in 'ANY'. Once again, you should check to be certain that your intentions are carried out. be certain that your intentions are carried out. Let us now backtrack our steps a little. The PACK in-struction of Let us now backtrack our steps a little. The PACK in-struction of Line 15 prepares the latter two bytes of each of the 50 words for the count. Line 15 prepares the latter two bytes of each of the 50 words for the count. In Lines 38-47, each count is incremented when it is found. The technique In Lines 38-47, each count is incremented when it is found. The technique involved in the incrementing of the table is one that should be given some involved in the incrementing of the table is one that should be given some thought. If our HASH-address (remainder) is 37 after dividing, we cannot thought. If our HASH-address (remainder) is 37 after dividing, we cannot simply increment the count at the 37th position in the table. The count to simply increment the count at the 37th position in the table. The count to be incremented will not be at this position, but at the location of the 37th be incremented will not be at this position, but at the location of the 37th fullword of the table. So in order to get to the 37th count, we must multiply fullword of the table. So in order to get to the 37th count, we must multiply 37 (HASH-address) by 4, the length of a fullword. By adding this result to 37 (HASH-address) by 4, the length of a fullword. By adding this result to the address of ASHTABLE, we obtain the desired count. This task is the address of ASHTABLE, we obtain the desired count. This task is per-formed by the following instructions: per-formed by the following instructions: 41.M8, POSITION 41.M8, POSITION 42.AR6, 9 42.AR6, 9 Remember that two registers are needed for a multiplication in fixed point Remember that two registers are needed for a multiplication in fixed point binary arithmetic and that the result is us-ually small enough to be stored in binary arithmetic and that the result is us-ually small enough to be stored in one (the odd) register. one (the odd) register. We will now turn our attention to the LA instruction. The general We will now turn our attention to the LA instruction. The general format of this instruction is: format of this instruction is: (a)LAR, SYMBOL (a)LAR, SYMBOL or(b)LAR, D (L,B) or(b)LAR, D (L,B) Let us concentrate on Case (b). Operand-1 is a register, and Operand-2 Let us concentrate on Case (b). Operand-1 is a register, and Operand-2 specifies a displacement D, a length L, and base register B. Coded like specifies a displacement D, a length L, and base register B. Coded like this, the instruction loads the address specified by Operand-2 into the this, the instruction loads the address specified by Operand-2 into the register. Now, if we were to code the instruction with the length (L) and register. Now, if we were to code the instruction with the length (L) and base register (B) absent (Line 52), it takes on a somewhat different base register (B) absent (Line 52), it takes on a somewhat different meaning. Instead of loading an actual address, the instruction loads the meaning. Instead of loading an actual address, the instruction loads the number specified as a displace-ment into the register. For example, Line number specified as a displace-ment into the register. For example, Line 52, 52, LA5, 50 LA5, 50 The number 50 is loaded into Register 5. Such a method provides The number 50 is loaded into Register 5. Such a method provides us with a simple way of loading numbers in-to registers. Many us with a simple way of loading numbers in-to registers. Many programmers prefer this technique since it is easier to code than that of programmers prefer this technique since it is easier to code than that of the L (load) instruction. the L (load) instruction. Our program coded here is a good example of calcula-tions which Our program coded here is a good example of calcula-tions which involve the combination of packed decimal and fixed point binary involve the combination of packed decimal and fixed point binary arithmetic. Incidentally, if you are not clear on the matter, fixed point binary arithmetic. Incidentally, if you are not clear on the matter, fixed point binary arithmetic is the name given to calculations in which any of the sixteen arithmetic is the name given to calculations in which any of the sixteen general purpose registers are used. Many times, the oper-ands involved in general purpose registers are used. Many times, the oper-ands involved in these calculations must be registers or fullwords. As our program mostly these calculations must be registers or fullwords. As our program mostly worked on two bytes at a time, we found it rather difficult to directly worked on two bytes at a time, we found it rather difficult to directly increment the elements of ASHTABLE using fixed point binary arithme-tic. increment the elements of ASHTABLE using fixed point binary arithme-tic. Thus, like most computer features, binary arithmetic can be restrictive at Thus, like most computer features, binary arithmetic can be restrictive at times. times.

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Question:

Given that the H-to-H distance in NH_3 is 0.1624 nm and N-H distance is 0.101 nm, calculate the bond angle H-N-H.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E17-0637.htm

Solution:

Because the NH distances are equal, an isosceles triangle can be drawn as shown. The altitude NX is drawn to form two right triangles. The magnitude of the angle HNX can be found trigonometrically from its sine. The sine of an angle is equal to the length of the opposite side divided by the hypotenuse. Sin HNX = (0.0812 nm / 0.101 nm) = 0.804. Using a table of trigonometric functions, one can find the angle whose sin is 0.804. sine^-1 0.804 = 53.50\textdegree. The angle H-N-H is twice HNX. As such, the measure of angle H-N-H is 2 × 53.50\textdegree or 107\textdegree. It is known that H-N-H is twice HNX because the altitude of an isosceles triangle bisects the vertex angle.

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Question:

Explain how an unconditional branch instruction is specified in an APL program.

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Solution:

An unconditional branch instruction is specified in APL by means of a right pointing arrow followed by a number or a label. The number represents the program seg-ment to which a jump is desired. If a label is used in-stead of a number, the label specifies the statement in the program which is thuslabelled, and to which the transfer should be made. For example , in [4] \rightarrow 2, the statement num-ber 4 of a certain program instructs the computer to branch unconditionally to the statement number 2 of the same pro-gram.

Question:

What is an essential element in plants? Name all the macronutrients and micronutrients of plants. What are their respective functions in plant growth?

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Solution:

An element is considered essential (1) if the plant fails to grow normallyand complete its life cycle when deprived of the element and no otherelement can replace the missing one or (2) if the element can be shownto be part of a molecule clearly essential in the plant's structure or function. Essential elements are usually separated into two categories - themacronut-rients, each of which comprises at least .1% of the dry weight ofthe plant, and micronutrients, which are present in as little as a few partsper million of the dry weight of the plant. The macronutrients include predominantly nitrogen and other mineral elements such as potassium, calcium, phosphorous, magnesium and sulfur. Nitrogen is an essential element in the synthesis of amino acids and thus proteins; nucleotides such as ATP, ADP, NAD, and NADP; chlorophyll (and other similar organic molecules with complex ring structures ); the nucleic acids, DNA and RNA; and many vitamins, such as the vitamin B group. Potassium regulates the conformation of some proteins , and affects some enzymatic reactions in the synthesis of biomolecules . It also affects water and ion balance by its' osmotic effects. Calcium is a structural component of the cell wall - it combines withpectic acid in the middle lamella of the plant cell wall. It also plays a part in cell growth and division, and is a cofactor for some enzymes. Phosphorous occurs in the sugar phosphate backbone of DNA and RNA, in nucleotides such as ADP and ATP, and in the phos-pholipids of the cell membrane. Magnesium is a structural part of the chlorophyll molecule and an activator of many enzymes. It functions in formation of amino acids, vitamins, fats and sugars.Sulferoccurs in coenzyme A and in some amino acids and proteins . The micronutrients include iron, chlorines, copper, manganese, zinc , molybdenum, and boron. Iron is a structural component of the electron carriers, thecytoch-romesandferredoxin, and is thus extremely important in the processes of photosynthesis and cellular respiration. Chlorine regulates osmosis and ionic balance in the cell. It may play a role in root and shoot growth. Copper, manganese and zinc are currently found to be activators of some enzymes. Copper is used in carbohydrate and protein metabolism, and with manganese, it functions in chlorophyll synthesis . Zinc is used in the formation ofauxin, chloroplasts and starch. Molybdenum is believed to be involved in nitrogen metabolism. Boron influences calcium ion uptake and utilization by the cell, and is suggested to be involved in carbohydrate transport as well. It affects cell division, flowering, pollen germination and nitrogen metabolism. As can be readily seen, minerals fill a wide range of basic cell needs and are involved in a variety of fundamental biological and biochemical processes. The effects of mineral deficiencies are therefore widespread , affecting a number of structures and functions in the plant body .

Question:

Write a FORTRAN program to multiply two matrices.

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Solution:

It Is desired to compute the matrix product AB = C, where A is dimensioned L1 × L2, B is dimensioned L2 × M2, and C is dimensioned L1 × M2. Note that the column dimension of A must equal the row dimension of B. This program illustrates the concept of nested DO loops. In terms of subscripts, C_ij= ^L2\sum_K=1A_ik×B_kj Thus, subscript k forms the innermost loop, whileiand j are the subscripts of the outer loops. DIMENSION A (L1, L2), B (L2, M2), C (L1, M2) DO 100 I = 1, L1 DO 100 J = 1, M2 C (I, J) = 0.0 DO 100 K = 1, L2 C (I, J) = A (I, K)\textasteriskcenteredB (K, J) + C (I, J) 100CONTINUE STOP END

Question:

A block of mass m, initially at rest, is dropped from a height h onto a spring whose force constant is. K. Find the maximum distance y that the spring will be compressed. See figure.

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Solution:

The general procedure used in solving any problem in mechanics is to calculate all the forces acting on the system and then derive the equation of motion of the system. An easier way to do mechanics problems involves the use of conservation principles. These laws are not applicable to all problems, but when they are, they simplify the calculation of the solution tremendously. In this problem, we may use the principle of conservation of energy. We relate the energy of the block before it was released to the block's energy at the point of maximum compression (see figure). At the moment of release; the kinetic energy is zero. At the moment when maximum compression occurs, there is also no kinetic energy. As shown in the figure, the reference level for gravitational potential energy is the surface S. The initial gravitational potential energy of m is mgy_1. At the point of maximum compression, the potential energy of m is mgy^2. However, at this point, the spring is compressed a distance y and also has elastic potential energy 1/2 ky^2. Hence, equating the energy at the point of release to the energy at the point of maximum compression, mgy_1 = mgy_2 + (1/2) ky^2 mg(y_1 - y_2) = (1/2) ky^2 But y_1 - y_2 = h + y and mg(h + y) = (1/2) ky^2 mg(h + y) = (1/2) ky^2 y^2 = [(2 mg)/(k)] (h + y) y^2 - [(2 mg)/(k)] y - [(2 mgh)/(k)]= 0 Therefore, using the quadratic formula to solve for y, y = (1/2) [(2 mg)/(k)\pm \surd{(2 mg/k)^2 + (8 mg/k)} ].

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Question:

a) 3652_8 × 24_8 b b ) 11.01_2 × 1.01_2 ) 11.01_2 × 1.01_2 c) 5AB_16 × F5A_16

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G02-0033.htm

Solution:

a) When doing multiplication in a base other than base 10, there aretwo common approaches. The first is to multiply the numbers together withbase 10 multiplication, get an answer in base 10, and then encode theanswer into the desired base. The other method is a direct one; you needto know the multiplication tables for the particular base. In this example, we have each line as a partial product. After the products are obtained, we add them together as base 8 numbers. Thus, 3652 ×24_8 10 24 30 14 124 14 ___6____ 114510_8 b) Binary numbers can also be multiplied via partial products. 11. 01_2 ×1. 01_2 1101 0000 1101 100.0001_2 As in base 10, the total number of digits in the multiplier and multi-plicand tothe right of the binary point must equal to the number of digits to the rightof the binary point in the product. c) Hexadecimal multiplication is often confusing because letters are used tosignify numbers. Although it is possible to use the partial product method, it is far simpler to convert the hexadecimal numbers to decimal numbersand then perform the multiplication. 5AB_16 = (5 × 16^2) + (A × 16^1) + (B × 16^0) Remembering that A_16 = 10_10 and B_16 = 11_10, you can write = (5 × 16^2) + (10 × 16^1) + (11 × 16^0) = 1280 + 160 + 11 5AB_16 = 1451_10 5AB_16 = 1451_10 F5A_16 = (F × 16^2) + (5 × 16^1) + (A x 16^0) Also recall thatF_16 = 15_10, so that F5A_16 = (15 × 16^2) + (5 × 16^1) + (10 × 16^0) = 3840 + 80 + 10 F5A_16 = 3930_10 Now the multiplication is simple: 1451_10 × 3930_10 43530 13059 4353 5702430_10 To convert back to hexadecimal, you can subtract powers of 16. The value of16^5 is 1,048,576. Multiply this by5,and you will get 5,242,880, which is lessthan 5,702,430. Subtract and look at the answer. Check the value of 16^4 and do the same as above. Continue the process until all digits have beenaccounted for. The example will show the correct procedure. 16^516^416^316^216^116^0 1048576655364096256161 5702430Digits of Hex - 5242880(= 5 × 16^5)Number 4595505 - 458752(= 7 × 16^4)7 798 -000(16^3 is too big)0 798 -768(= 3 × 16^2)3 30 -16(= 1 × 16^1)1 14 -14(= 14 × 16^0)D The answer is 57031D_16.

Question:

How is an array declared inpascal? Example using examples of one-dimensional and multi-dimensional arrays.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G16-0397.htm

Solution:

An array is defined as follows: TYPE A = ARRAY [indextype]ofbasetype; whereindextypeis the type of the array index, andbasetypeis the type of the array elements. Indextype can be any scalar type excepttypesin-teger and real. Onlysubrangesof integer are allowed.Basetypecan be any type whatsoever . example :TYPE A =ARRAY[1. .100]ofreal; This declaration only defines type A as an array of 100 real numbers . Therefore, any variable of type A will be interpreted as an array of 100 real numbers . Alternatively, any variable could be declared as an array by: VAR table : ARRAY[1. .100]ofreal; Multidimensional arrays are essentially arrays of arrays. Thus, having declared TYPE A =ARRAY[1. .100] of real; we can declare a matrix of size 150 × 100 by the following: VAR matrix :ARRAY[1. .150]ofA; This, however, is an unconventional way to declare multi-dimensional arrays . Usually, the same variable is declared by: VAR matrix : ARRAY[1. .150,1. .100]ofREAL; The first index parameter always refers to the number of rows, and second refers to the number of columns. The index parameters do not have to be of the same data type.Higher-dimensionalarrays can be declared similarly .

Question:

What is the role of transpiration in plants? Describe the mechanism of transpiration.

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Solution:

The leaves of a plant exposed to the air will lose moisture by evaporation unless the air is saturated with water vapor. This loss of water by evaporation, mainly from the leaves (but also from the stems) is called transpiration. Transpiration plays a very important role in water transport in plants. It directs the upward movement of water and minerals from the soil to the leaves, keeps the air spaces of leaves moist for carbon dioxide to dissolve in, concentrates initially dilute leaf cell solutions of minerals that have been absorbed by the roots, and contributes to the cooling of the plant body by removing heat during the vaporization of water (540 calories of heat are needed to convert each gram of water to water vapor). Transpiration is responsible for the great amount of water that passes through a plant body per day. About 98 per cent of the water absorbed by the roots is lost as vapor. As water is lost by evaporation from the surface of amesophyllcell, the concentration of solutes in the cell sap increases, causing water to pass into it from neighboring cells that contain more water, that is, lower solute concentration. These neighboring cells in turn receive water from thetracheidsand vessels of the leaf veins, which ultimately obtain water from the soil via root hair cells. Thus, during transpiration, water passes from the soil via the xylem system of the roots, stem, and leaf veins, and through the intervening cells, to themesophylland finally into air spaces in the leaves, where most of it vaporizes and escapes.

Question:

Assuming ideal gas behavior, what is the molecular weight of a gas with a density of 2.50 g/liter at 98\textdegreeC and .974 atm?

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Solution:

You must employ the ideal gas law to answer this question. This law, also called the equation of state for an ideal gas, relates pressure, volume, temperature, and moles to each other quantitatively, it states PV = nRT, where R is called the universal gas constant, P = pressure, V = volume, T = temperature in Kelvin, and n = number of moles. You can also write this equation as n/V = P/RT to determine moles per liter. By substituting, [(.974 atm)/{(.0821 liter-atm/mole-\textdegreeK) (3716K)}] = .0320 moles/liter. Therefore, according to the above calculation, you know there are .0320 moles in one liter. However, the density is 2.50 g/liter. Hence, you can compare the two, and obtain the fact that .0320 moles weighs 2.50 g. From this, the molecule weight determination follows via a proportion. If 0.032 moles weighs 2.50 g, then 1 mole weighs x grams, which equals the molecular weight. In other words, [(.032)/(2.50)] = (1/x). Solving for x, we obtain x = 78.1 g.

Question:

Compare the reactions involved in the oxidation of lactic and succinic acids.

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Solution:

Lactic acid is first dehydrogenated enzymatically by lactate dehydrogenase to form pyruvic acid. [Refer to Fig. 1.] The two hydrogen atoms and two electrons are transferred to a pyridine nucleotide, nicotinamide adenine dinucleotide (NAD+). NAD+ serves as an electron acceptor in dehydrogenation reactions involving substances (in this case, lactate) with an arrangement as follows: (Secondary alcohols, such as in lactic acid, get oxidized to ketones, such as in pyruvic acid). Following this dehydrogenation reaction is the decarboxy-lation (loss of CO_2) of pyruvic acid to form acetaldehyde. Addition of coenzyme A to acetaldehyde forms a compound which undergoes another dehydrogenation to form acetyl coenzyme A. The oxidation of succinic acid is part of the TCA cycle. [Refer to Fig. 2.] Succinic acid is first dehydrogenated enzymatically by succinic dehydrogenase to form fumaric acid. The two hydrogen atoms and their electrons are not transferred to NAD+, as in lactic acid oxidation, but are transferred to flavin adenine dinuc-leotide (FAD). FAD is an electron acceptor in dehydroge-nation reactions involving substances (in this case, acid. In this case, NAD+ accepts the electrons (with a H) to form NADH. Oxaloacetic acid will either combine with acetyl coenzyme A to form citric acid in the TCA cycle, or be decarboxylated to form pyruvic acid. In most cases though, the oxoloacetata will combine with acetyl coen-zyme A and continue the Krebs cycle. The reader should take note that in both the oxidation of lactate and suc-cinate, dehydrogenation reactions occur, but different electron acceptors are used in each case: NAD+ for lac-tate and FAD for the direct oxidation of succinate. The reducing equivalent NADH and FADH_2 are formed.

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Question:

Categorize the various procedural verbs used in COBOL.

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Solution:

Procedural Verbs: Verbs specify action to be performed. In COBOL, each verb built into the system causes a specific series of events to occur when the program is running. The various verbs in COBOL are categorized in the following manner. Input / OutputOPEN CLOSE READ WRITE ACCEPT DISPLAY ArithmeticADD SUBTRACT MULTIPLY DIVIDE CAMPUTE Data movement and ManipulatedMOVE EXAMINE Sequence ControlGO TO ALTER PERFORM STOP CompilerDirectingENTER NOTE EXIT

Question:

A chemist has a piece of foil that is approximately 5 × 10^4 atoms thick. If an alpha particle must come within 10^-12 cm of a nucleus for deflection to occur, what is the probability that an alpha particle will be deflected, assuming the nuclei are not directly behind one another? Assume that the area of one atom is 4 × 10^-16 cm^2.

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Solution:

To calculate the chance of deflection occurring in any one layer of the foil, assume that each layer is one atom thick. The foil is, therefore, 5 × 10^4 layers thick. If the number of deflections in any one layer is known, it can be multiplied by 5 × 10^4 to find the total number of possible deflections as an alpha particle moves through the foil. To find the chance of a deflection occurring in one layer, assume that the atoms are spherical. At 10^-12 cm from the center of the circle, a deflection can occur. The radius of the circle for which a deflection will occur is 10^-12 cm. The area of a circle is equal to \pir^2 where \pi = 3.14 and r = radius. The area of one atom in which deflection can occur is equal to 3.14 × (10^-12)^2 or 3.14 x 10^-24 cm^2. The area of each atom is 4 × 10^-16 cm^2. The thickness of each layer is one atom. Therefore, the chance of getting a deflection in any given layer of atoms is [(3.14 × 10^-24 cm^2)/(4 × 10^-16 cm^2)] = 7.85 × 10^-9 There are 5 × 10^4 layers. Thus, a chance of deflection in one of the 5 × 10^4 layers is 5 x 10^4 × 7.85 × 10^-9 = 4 × 10^-9 or 1 in 2500.

Question:

Phosphorus (atomic weight = 30.97) combines with another elements such that 1g of phosphorus requires 0.7764g of the other element. If the atomic ratio of phosphorus to the other element is 4:3 , what is the atomic weight of the unknown element?

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Solution:

First determine how many moles of phosphorus reacted . From this , one can calculate the number of moles of the unknown element that reacted using the atomic ratio. This ratio represents the relative mole amounts. If the number of moles of the unknown element is determined, the atomic weight can be calculated because the number of moles is equal to number of grams/atomic weight. (At. wt. of P = 30.97). One can now determine the atomic weight of the unknown element. The number of moles of phosphorus that reacted = (1.0g) / (30.97 g/mole) = 0.03229 moles. Let x = the number of moles of the unknown element. The atomic ratio of phosphorus to the unknown element is 4 : 3. Thus, 4atoms / 3atoms = (.03229 moles / x) . Solving, x = 0.02422 moles unknown element. Therefore, the atomic weight of the unknown element is equal to (0.7764) / (0.0242moles) =32.06 g/mole .

Question:

In the reaction _5B^11 + _2He^4 \ding{217} _7N^14 + _0n^1 the masses of the Boron and nitrogen atoms and the \alpha - particle are 11.01280 amu, 14.00752 amu, and 4.00387 amu, respectively. If the incident \alpha - particle had 5.250 MeV of kinetic energy and the resultant neutron and nitrogen atom had energies of 3.260 HeVand 2.139MeV, respectively, what is the mass of the neutron?

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Solution:

We may apply the principle of conservation of total energy to this reaction. The total energy of the reactants is the sum of their rest mass energies (given by Einstein's mass-energy relation, E_rest = mc^2) and their kinetic energies. Hence, E_react= m_Bc^2 + m_\alphac^2+T_\alpha(1) where B and \alpha stand for the boron and \alpha particle respectively . m stands for rest mass, and T for kinetic energy. (Note that the boron is initially at rest). Similarly, the total energy of the product is E_prod = m_Nc^2 + m_nc^2+ T_N + T_n(2) where N and n stand for the nitrogen and neutron, respectively. Then, equating (1) and (2) by the principle of conservation of total energy m_Bc^2 + m_\alphac^2+T_\alpha= m_Nc^2 + m_nc^2+ T_N + T_n Solving for mn {(m_B+m_\alpha-m_N)c^2 + (T_\alpha- T_N -T_n)} / c^2 = mn m_n = (m_B+m_\alpha-m_N) + {(T_\alpha_ - T_N - T_n ) / c2} Substituting the given data m_n= (11.01280 + 4.00387 - 14.00752)amu + (5.250 - 2.139 - 3.260)(MeV / c^2)(3) Now, 1 MeV = 10^6eV = (10^6) (1.602 × 10^-19 J/eV) 1 MeV = 1.602 × 10^-13 J And 1MeV / c^2 =(1.602 × 10^-13 J) / (9 × 1016m^2s^2) = 1.78 × 10^-30kg Since 1 kg = 6.024 × 10^26 amu 1MeV / c2= (1.78 × 10^-30 kg) {6.024 × 10^26 (amu/kg)} 1MeV / c2= 1.072272 × 10^-3 amu Then, (3) becomes m_n = 1.00915 amu - (.149) ( 1. 072272 × 10^-3amu) m_n = 1.00899 amu

Question:

A group of mountain climbers set up camp at a 3 km altitude and experience a barometric pressure of 0.69 atm. They discover that pure water boils at 90\textdegreeC and that it takes 300 minutes of cooking to make a "three-minute" egg. What is the ratio of the rate constantk_100\textdegreec and k90\textdegreec?

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Solution:

Since we do not know the rate expression for cooking an egg, we will assume one of the form rate = k [A]^m [B]^n ... where k is the rate constant, A, B, ... are the reactants, and the overall order of the reaction is m + n + ..., We will write the rate equations at the normal boiling point of water (100\textdegreeC) and at 90\textdegreeC as rate_100\textdegreeC= k_100\textdegreeC [A]^m[B]^n ... and rate_90\textdegreeC= k_90\textdegreeC [A]^m[B]^n ... Dividing the first of these by the second gives (rate_100\textdegreeC) / (rate_90\textdegreeC) = (k_100\textdegreeC [A]^m[B]^n ...) / (k_90\textdegreeC [A]^m[B]^n ...) = (k_100\textdegreeC) / (k_90\textdegreeC) Since the egg cooks 100 times faster at 100\textdegreeC than at 90\textdegreeC (300 min/3 min = 100), rate_100\textdegreeC /rate k_90\textdegreeC = 100. Hence , (rate_100\textdegreeC) / (rate_90\textdegreeC) = 100 = [(k_100\textdegreeC) / (k_90\textdegreeC)] , or [(k_100\textdegreeC) / (k_90\textdegreeC)] = 100.

Question:

A potato peeling solution was found to be 2.56 M inNaOH (formula weight = 40.0 g/mole) at the end of the day. To operate, the solution must be at least 10 %NaOHby weight (100 gNaOHper 1000 g of solution) . What weight percent corresponds to 2.56 M? The density of a 2.56 M solution of NaOHis about 1.10 g/ml.

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Solution:

To solve this problem, the concentration 2.56 M must be converted to a weight-weight basis. 2.56 M = 2.56 molesNaOH/1 liter solution = 2.56 molesNaOH/1000 ml solution. To obtain the mass corresponding to 2.56 moles ofNaOH, we multiply by the formula weight ofNaOH, or, 2.56 moles × 40.0 g/moles = 102.4 gNaOH. To obtain the mass ofNaOHcontained in 1000 ml of solution, we multiply by the density of the solution, 1000 ml × 1.10 g/ml = 1100 g. Hence, 2.56 M = [(2.56 molesNaOH)/(1000 ml solution)] = [(102.4 gNaOH)/(1100 g solution)] ; [(102.4 gNaOH)/(1100 g solution)] × 100 % = 9.3 % by weight. Since this is less than 10 % by weight, the solution is no longer capable of peeling potatoes.

Question:

A 1.00-liter reaction vessel containing the equilibrium mixture CO + Cl_2 \rightleftarrows COCL_2 was found to contain 0.400 mole of COC1_2 , 0.100 mole of CO, and 0.500 mole of Cl_2 . If 0.300 mole of CO is added at constant temperature, what will be the new concentrations of each component at equilibrium?

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Solution:

If a stress is placed on a system in equilibrium, whereby the equilibrium is altered, that change will take place which tends to re-lieve or neutralize the effect of the added stress. Thus, in this re-action when more CO is added after equilibrium has been established more C0Cl_2 will be formed to re-establish the equilibrium. One is given the concentrations of the components at equilibrium, thus the equilibrium constant (Keq) can be calculated. The equilibrium con-stant for this reaction can be stated: Keq= [COCl_2] / {[CO] [Cl_2]} , the ratio of products to reactants, where [ ] indicate concentration. One can solve for theKeqby using the concentrations given for the first equilibrium , Keq= [COCl_2] / {[CO] [Cl_2]}[COCl_2] = 0.400 mole/liter [CO] = 0.100 mole/liter [Cl_2] = 0.500 mole/liter Keq= (0.400) / { (0.100) (0.500)} = (0.400) / (0.05) = 8.0 . One knows, from the definition ofKeq, that when more CO is added to the mixture that the concentrations of the components rearrange so that the equation forKeqis equal to 8. Using LeChatelier'sprinciple, stated in the first sentence of this explanation, one knows that when more CO is added to this mixture more COCl_2 is formed. Let new concentrations of COCl_2 = [0.400 + x] . From the equation, one knows that for each mole of C0Cl_2 formed one mole of Cl_2 is used, thus the new concentration of Cl_2 = [0.500 - x]. One mole of CO is also used for each mole of COCl_2 formed, thus the new concentration of CO will be equal to x subtracted from the sum of the number of moles originally present and the number added. The new concentration of CO = [0.100 + 0.300\Elzbarx]. Because one knows that Keq=8, one can now solve for x. Keq= [COCl_2 ] / { [CO ] [Cl_2 ] } =8where [COCl_2 ] = (0.400 + x) moles [CO] = (0.100 + 0.300\Elzbar x) moles [Cl_2] = (0.500\Elzbar x) moles Solving, 8 = {0.400 + x} / {(0.100 + 0.300\Elzbar x) (0.500\Elzbar x)} (0.400 + x) / (0.2\Elzbar 0.9x + x^2 ) = 8 (0.400 + x) = 8(0.2-\Elzbar 0.9x + x^2) (0.400 + x) = (1.6 - \Elzbar 7.2x + 8x^2) 0 = 1.2\Elzbar 8.2x + 8x^2 . One uses the quadratic equation to solve for x: x = {\Elzbarb \pm \surd(b^2\Elzbar 4ac) } / {2a} ,where 0 = c +bx+ ax^2 x = [8.2 \pm \surd{(\Elzbar8.2)^2\Elzbar 4(8 × 1.2)}] / [2 × 8] x = (8.2 \pm 5.37) / (16) x = (8.2 + 5.37) / 16 = 0.85 or x = (8.2\Elzbar5.37) / 16 = .177 . One cannot use x = .85 in this problem because when it is used, the concentrations of CO and Cl_2 will be negative values. Thus x = .177. The new concentrations can now be found. [COCl_2 ] = (0.400 + x) = .577 moles . [CO] = (0.400\Elzbar x) = .223 moles [Cl_2] = (0.500\Elzbar x) = .323 moles

Question:

Ethanol, C_2H_5OH, reacts withHBr, hydrogen bromide, to giveethyl bromide, C_2H_5Br, and water. Ethanol, however, doesnot react withNaBr. Explain. Suggest a mechanism for thereaction withHBr.

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Solution:

This problem entails an understanding ofnucleophilic displacement. In such reactions, a strongerBrӧnstedbase, i.e. one that is ӧ aproton acceptor, dis-places the weakerBrӧnstedbase. This is why the leavinggroups are usually extremely weak bases, such as I-, Br-, andCl-. With this in mind, notice thatNaBris a salt. This means that in solution,NaBrhas dissociated into Na+ and Br- ions. WhenNaBris present, then, you are asked why Br- cannot replace OH- to formethyl bromide. Recall, displacement only occurs if the base which youdisplace is a weaker base. Br- is a weak base, as mentioned. It is a weakerbase than OH-. Thus, Br- cannot replace OH-. You can suggest a mechanismthat occurs for the reaction withHBrby realizing thatHBris an acid. Thus, in solution, you have H+ and Br- ions. The H+protonatesthe OH group. Br - then proceeds to displace the water. Why? While Br- is a weaker base than OH-, it is a stronger base than H_2O (water), a very weak base. In other words, anucleophilicdisplacement occurs. Overall, it can be written as C_2H_5OH +HBr\rightarrowC_2H_5OH_2+ + Br- \rightarrowC_2H_5Br +H_2O .

Question:

A sample of a radioactive material contains one million radioactive nuclei. The half-life is 20 sec. How many nuclei remain after 10 sec?

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/Users/wenhuchen/Documents/Crawler/Physics/D34-1013.htm

Solution:

We will employ 2 methods of solution. One will be very physical, and the other will be more mathematical. Suppose that after every 10 sec period, a fraction f of the number of nuclei present at the beginning of the period remains at the end of the period. Then, at time t = 0 sec there are 10^6nuclei. At time t = 10 sec there are 10^6f nuclei. At the end of the next 10 sec a fraction f of these remain. Therefore, at timet = 20 sec there are 10^6f^2nuclei. But 20 sec is the half- life, during which half the nuclei disintegrate. Therefore f^2 = 1/2 f = 0.707. The number of nuclei remaining after 10 sec is 10^6 × 0.707 or 707,000. Now, we set up a differential equation describing the process of radioactive decay. It is observed that the rate of decay of nuclei is proportional to the number of nuclei present, or dN/dt=\Elzbar\lambdaN(1) where the minus sign indicates that the number of nuclei is decreasing with time. We may write (1) as dN /N =\Elzbar\lambdadt. Integrating both sides of this equation, lnN =\Elzbar\lambdat+ c. Taking the exponential of both sides: N = e\Elzbar\lambdat+ c= e\Elzbar\lambdate^c. Becausee^cis only a constant, we replace it by A N(t) =Ae\Elzbar\lambdat(2) Assume that at t = 0, we have N_0 nuclei. Then N(t = 0) =Ae\textdegree = A = N_0 N_0 = A. Therefore, from (2), N(t) = N_0 e\Elzbar\lambdat.(3) We also know that after 20 sec, we have 1/2 the original number of nuclei, N_0 /2. Hence N(20 sec) = N_0 e\Elzbar(20sec)\lambda= N_0 / 2 ore =\Elzbar(20sec)\lambda= 1/2 2 = e^(20sec)\lambda . Taking the logarithm of both sides, ln2 = (20sec)\lambda. Therefore \lambda = [1/20sec]ln2. Substituting this in (3), we obtain N(t) = N_0 e\Elzbar[(1/20sec)ln2] t Hence, after 10 sec, assuming N_0 = 10^6nuclei, N(10sec) = (10^6 nuclei)e\Elzbar[(1/20sec)ln2] 10sec N(10sec) = (10^6 nuclei)e\Elzbar(1/2)ln2 N(10sec) = (10^6 nuclei)e(ln2)\Elzbar(1/2) Bute^ln^ x = x. Then N(10sec) = (10^6 nuclei) [(2)\Elzbar(1/2)] N(10sec) = (10^6 nuclei) / \surd2 = 0.707 × 10^6 nuclei.

Question:

Find the equation for theautoprotolysisof water. Indicate which species is the acid, the base, the conjugate acid, and the conjugate base.

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Solution:

One can begin by definingautoprotolysis. It may be defined as the donation of a proton from a molecule of one specie to another molecule of the same specie to produce positive and negative ions. Thus, for water, the equation is H_2O + H_2O \rightarrow H_3O^+ + OH^-. An acid is defined as a specie that donates protons. A base is a substance that accepts protons. From the equation, one sees that either water (H_2O) molecule can be the base or acid. A conjugate base is a specie obtained by abstracting a proton (H^+). If one abstracts a proton from water, one obtains OH^-. Thus, OH^- is the conjugate base. The conjugate acid is defined as the base plus a proton. It was stated that either H_2O molecule could be the base. If one adds a proton to one of them, one obtains H_3O^+. Thus, H_3O^+ is the conjugate acid.

Question:

A cubical tank of water of volume 1 m^3 is kept at a steady temperature of 65\textdegreeC by a 1-kW heater. The heater is switched off. How long does the tank take to cool down to 50\textdegree C if the room temperature is 15\textdegreeC?

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Solution:

While the heater is operating, the heat supplied by it, 1 kW = 240 cal\bullets^-1, is just sufficient to make up for the heat loss that would take place according to Newton's law of cooling. dt dt /d\cyrchar\cyrt= -k(t -t_s)(1) /d\cyrchar\cyrt= -k(t -t_s)(1) where t is the temperature of the cooling body at \cyrchar\cyrt = 0,t_sis the ambient temperature, and k is a constant. Furthermore,dt/d\cyrchar\cyrtis the rate of change of temperature with time. Before using this law, we must evaluate k for the case at hand. In order to do this, note that the definition of the specific heat of a substance is c =dQ/mdt where m is the mass of the substance,dQis an increment of heat energy, anddtis an increment of temperature. Then dt =dQ/mc(2) anddt/d\cyrchar\cyrt=mcd\cyrchar\cyrt(3) Butdt/d\cyrchar\cyrt< 0 in (1) if (t -t_s) > 0, anddt/d\cyrchar\cyrt> 0 in (3) ifdQ> 0. But, if dQ > 0, the temperature changedt> 0, as in (2). Hence, there is an in-consistency in the sign ofdt/d\cyrchar\cyrtbetween (1) and (3). For consistency, replacedt/d\cyrchar\cyrtin (2) by -dt/d\cyrchar\cyrt. Nowdt/d\cyrchar\cyrthas the same meaning in (1) and (2). Taking account of this, and inserting (2) in (1) - (dQ/mcd\cyrchar\cyrt) = - k(t -t_s) ormck= [1/(t -t_s)] (dQ/d\cyrchar\cyrt) ButdQ/d\cyrchar\cyrtis the heat power supplied to the tank. Thus mck = [(240 cal\bullets^-1)] / [(65 - 15)\textdegreeC] = 4.8 cal\bullets^-1\bullet C deg^-1. But the mass of 1 m^3 of water is 10^6 g and the specific heat capacity of water is 1 cal \bullet g-1\bullet C deg^-1. Hence, k = 4.8 × 10^-6 s^-1. When the heater is switched off, the tank cools according to the equationdt/d\cyrchar\cyrt= -k(t -t_s). \therefore^50\textdegreeC\int_ 65\textdegreeC [dt/(t -t_s)] = - k \int^\cyrchar\cyrt_0d\cyrchar\cyrt. Where t = 65\textdegreeC at \cyrchar\cyrt = 0 and t = 50\textdegree C at \cyrchar\cyrt = \cyrchar\cyrt. Then ^50\textdegreeC\int65\textdegreeCdt/(t -t_s) = -k\cyrchar\cyrt ln\mid t -t_s^50\textdegreeC\mid_65\textdegreeC = -k\cyrchar\cyrt In \vert50\textdegreeC - 15\textdegreeC\vert - 1n \vert65\textdegreeC - 15\textdegreeC\vert = -k\cyrchar\cyrt In \vert65\textdegreeC - 15\textdegreeC\vert - 1n \vert50\textdegreeC - 15\textdegreeC\vert =k\cyrchar\cyrt ln [\mid50\textdegreeC\mid / \mid35\textdegreeC\mid] = k \cyrchar\cyrt \cyrchar\cyrt = (1/k) 1n \mid10/7 \mid \therefore \cyrchar\cyrt = (10^6/4.8) 1n (10/7)s = [(10^6 × 0.3567)/4.8] s = 74343.5 s = 20.64 hr.

Question:

Describe the primary functions of the nervous system. What other systems serve similar functions?

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Solution:

The human nervous system, composed of the brain, spinal cord, and peripheral nerves, connects the eyes, ears, skin, and other sense organs (the receptors) with the muscles, organs, and glands (the effectors). The nervous system functions in such a way that when a given receptor is stimulated, the proper effector responds appropriately. The chief functions of the nervous system are the conduction of impulses and the integration of the activities of various parts of the body. Integration means a putting together of generally dissimilar things to achieve unity. Other systems involved in similar functions are the endocrine system and the regulatory controls intrinsic in the enzyme systems within each cell. Examples of the latter are inhibition and stimulation of enzymatic activities. The endocrine system utilizes substances, knows as hormones, to regulate metabolic activities within the body.

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Question:

Describe the steps by which simple inorganic substances mayhave undergone chemical evolution to yield the complexsystem of organic chemicals we recognize as a livingthing. Which of these steps have beenduplicated experi-mentally?

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Solution:

Life did not appear on earth until about three billion years ago. This wassome two billion years after the formation of the earth, either from a portionbroken off from the sun or by the gradual condensation of interstellardust. The primitive atmosphere before the appearance of any formof life is believed to have contained essentially no free oxygen; all the oxygenatoms present were combined as water or as oxides. Deprived of freeoxygen, it was thus a strongly reducing environment composed of methane, ammonia, and water which originated from the earth's interior. At that time there were obviously no organic compounds on earth. Reactions by which organic substances can be synthe-sized from inorganicones are now well known. Originally, the carbon atoms were presentmainly as metallic carbides. These could have reacted with water toform acetylene, which could subsequently have polymerized to form largerorganic compounds. That such reactions occurred was suggested byMelvin Calvin's experiment in which solutions of carbon dioxide and waterwere energetically irradiated and formic, oxalic, andsuccinicacids wereproduced. These organic acids are important because they are inter- mediatesin certain metabolic pathways of living organisms. After the appearance of organic compounds, it is believed, simple aminoacids evolved. How this came about was demonstrated by Urey andMiller, who in 1953 exposed a mixture of water vapor, methane, ammoniaand hydrogen gases to electric charges for a week. Amino acids suchasglycineandalanineresulted. The earth's crust inprebiotictimes probablycontained carbides, water vapor, methane, ammonia and hydrogengases. Ultraviolet radiation or lightning discharges could have providedenergies analogous to the Urey-Miller apparatus, and in this manner, simple organic compounds could have been produced. Most, if not all, of the reactions by which the more complex organic substanceswere formed probably occurred in the sea, in which the inorganicprecursors and organic products of the reaction were dissolved andmixed. These molecules collided, reacted, and aggregated in the sea toform new molecules of increasing size and complexity. Intermolecular attractionprovided the means by which large, complex, specific molecules couldhave formed spontaneously. Once protein molecules had been formed, they acted as enzymes to catalyze other organic reactions, speedingup the rate of formation of additional molecules. As evolution progressed, proteins catalyzed the polymerization of nucleic acids, giving riseto complex DNA molecules, the hereditary materials and regulators of importantfunctions in living organisms. Enzymes also probably catalyzed thestructural combination of proteins and lipids to form membranes, permittingthe accumulation of some molecules and the exclusion of others. With DNA and a membrane structure, the stage was set for life to beginsome three billion years ago.

Question:

Write a program that

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Solution:

Before analyzing this problem the concept of pointers and address of variables must be made clear. Pointers : The declaration format for a pointer is type \textasteriskcenteredvariable _ name; The \textasteriskcentered informs the compiler that the variable is actually a pointer,eg. int \textasteriskcenteredptr; Addresses : The ampersand (&) is used to specify an address (actual memory location) of a variable. In pointers, we have created the variable ptr . Now we can assign to it the address of the variable integer _ variable. i.e. ptr = &integer _ variable; In the above assignment,ptrdoes not contain the value that is contained in the variable, integer _ variable, but it contains the location of integer _ variable in memory i.e. ptr = address of integer _ variable in memory To reference the value stored in this address throughptrwe can use it as \textasteriskcenteredptr. The \textasteriskcentered (asterisk) informs the computer to get the value contained at the location pointed by the pointer i.e.,ptr. Now if we make the following comparison that if (\textasteriskcenteredptr== integer _ value) the result of the above expression is true because both variables reference the same value main ( ) {intx = 4; /\textasteriskcenteredinitial value of x \textasteriskcentered/ inty = 9; /\textasteriskcenteredinitial value of y \textasteriskcentered/ /\textasteriskcenteredpass addresses of x and y to change _ values\textasteriskcentered/ change _ values (&x, &y) /\textasteriskcenteredpass addresses of x and y to square _ of _ number\textasteriskcentered/ square _ of _ number (&x, &y) /\textasteriskcenteredprint the squares of x and y \textasteriskcentered/ printf("Square of x = %d and y = %d\textbackslashn", x, y); } } change _ values (xptr,yptr) int\textasteriskcenteredxptr, +yptr; { /\textasteriskcenteredmodify the values of variables x and y\textasteriskcentered/ \textasteriskcenteredxptr= \textasteriskcenteredxptr+ 1; \textasteriskcenteredyptr= \textasteriskcenteredyptr+ 1; } square _ of _ number (ptrx,ptry) int\textasteriskcenteredptrx, \textasteriskcenteredptry; { /\textasteriskcenteredcalculate the squares \textasteriskcentered/ \textasteriskcenteredptrx= \textasteriskcenteredptrx\textasteriskcentered \textasteriskcenteredptrx; /\textasteriskcenteredsquare of variable x \textasteriskcentered/ \textasteriskcenteredptry= \textasteriskcenteredptry\textasteriskcentered \textasteriskcenteredptry /\textasteriskcenteredsquare of variable y \textasteriskcentered/ }

Question:

The equivalence (also called biconditional) function of x_1, x_2 written x_1 \equiv x_2 (or x_1 \Leftrightarrow x_2), is defined as: x_1 \equiv x_2 = (\textasciitildex_1 + x_2) \textbullet (x_1 + \textasciitilde x_2)(1) (a) The + operator denotes the inclusive OR function here; \textbullet represents AND and \textasciitilde representes the negation operator. Set up the truth table for the equivalence function. (b) Use AND, OR and NOT gates to evaluate the equivalence function.

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Solution:

(a) The truth table is set up by first specifying the pos-sible combinations of values of x_1 and x_2 . There are 2^2 = 4 of them and they represent the numbers 0 to 3 in the binary system. The next step is to evaluate the terms of statement (1): x_1 x_2 \backsimx_1 \backsimx_2 \backsimx_1 + x_2 x_1 + \backsimx_2 0 0 1 1 0 1 0 1 1 1 0 0 1 0 1 0 1 1 0 1 1 0 1 1 Now transfer columns 1, 2, 5, and 6 to a truth table evaluating statement (1): x_1 x_2 \backsimx_1 + x_2 x_1 + \backsimx_2 x1\equiv x2 = (\backsimx_1 + x_2) \bullet ( x_1 + \backsimx_2) 0 0 1 1 0 1 0 1 1 1 0 1 1 0 1 1 1 0 0 1 Note that x_1 \equiv x2= 1 if and only if x1= x_2, which is why it is called the equivalence function . (b) Recall that variables which are operands of the + operation are inputs to an OR gate and a variable that is the operand of the \textasciitilde operation is the input to a NOT gate. Thus, the gate representations of the terns in parentheses are as shown in Fig. 1. Since each term is an operand of the \bullet operation, each of the above gate circuits is an input to an AND gate. Thus, the gating circuit of statement (1) is as shown in Fig. 2. Note that the output of this network is 1 if and only if = x1= x_2. Thus, the network acts as a comparing device. Specifically, it is the circuit for an equality comparator. An equality comparator compares one bit, x_1, with another bit, x_2. If two strings of bits are to be compared, they could be compared serially. That is, compare a bit in the first position of the first string with a bit in the first position of the second string, then compare a bit in the next position of the first string with a bit in the next position if the second string and continue until either a zero is output, indicating the strings are not equal, or all positions have been compared, indicating the strings are equal. Or they could be compared in parallel. That is, a parallel comparing device that inputs two strings of length n would have n equality comparators. An output of a string of ones indicates that the two inputs are equal. The advantage of parallel comparing is its higher speed compared with series comparison. It will always require k time units to compare strings where k is the operation time of the equality comparator. The series comparator requires n\bulletk time units at worst and k time units at best. The advantage of the series comparator is its economy. But since gates can be manufactured inexpensively, no com-mercial computer would use a series comparator, since every instruction must be identified by using the comparator and hence, there would be a serious loss of speed.

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Question:

A trait has the inheritance pattern shown below for four different families, a) Determine whether the trait is dominant or recessive b) Based on your answer to a), determine the possible genotypes for each individual in the four families. (Solid blocks or circles indicate individuals who express the trait).

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Solution:

a) Let us look at what the pattern within each individual family tells us about the trait. This does not really tell us anything about the trait, for each a pattern could result from the inheritance of either a dominant or a recessive trait. offspring. This is definite evidence that the trait must be recessive. We know that since the trait is present in the offspring, then one or both of the parents must carry the gene for that trait. Yet neither parent expresses the trait. Therefore, the gene must be present, but masked, and so it must be recessive. We need go no further since we have estab-lished that the trait is recessive. However, let us see what we might have been able to learn about the trait from the remaining pedigrees. 3. In this family, one parent expresses the trait, yet none of the offspring express it. This is inconclusive, because such a pattern could result from either dominant or recessive inheritance. If the trait were dominant, the parent expressing it could be heterozygous, and so carrying some recessive allele. Although it is not highly probable, it is possible that only the recessive allele was given to the offspring, and none received the domi-nant allele. If the trait were recessive, the parent not expressing the trait would carry at least one dominant gene which could have been inherited by all the offspring. This gene would mask any recessive alleles the offspring may have. 4. Again this pattern is compatible with both dominant and recessive inheritance, assuming that one of the parents is homozygous and the other heterozygous. b) Since we know from family two's pedigree chart that the inherited trait is recessive, we can determine the possible genotypes of all the individuals in all the families. Let A represent the dominant trait (whose inheritance is not shown: ,0) and a the recessive trait (whose inheritance 0 is shown: , ) the genotypes are: 1P^1 ( ) = aa2P^1 ( ) = Aa P^2 ( ) = aaP^3 ( ) = Aa F_1 ( ) = aaF^31( , ) = AA or Aa F^1,2_1() = aa 3P^1 ( ) = aa4P^1 ( ) = aa P^2 ( ) = AA or AaP^2 ( ) = aa F1 - 5( , ) = AaF1 ,2(, ) = aa F3 ,4(0,0) = Aa

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Question:

Develop a sorting subroutine that compares elements that are separated by a distance K. In your algorithm, try to make exchanges between elements such that each exchange moves an element close to its final position.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G07-0158.htm

Solution:

This method is known as a Shell sort. Compari-sons are made between an element X_i. and element xi + k. For the bubble and insertion sorts, K is always 1. With the shell sort, you set K at the largest power of two that is less than N, N being the total number of elements. There-fore 2^k < N \leq 2k+1 Once K is chosen, a pass is made through the array. Each successive pass results in the division of K by 2. Finally, for the last pass, K = 1 and the procedure becomes an ordinary bubble sort. This variety of exchange sort functions best when the size of the array is relatively small, viz., less than 25 elements. However, if you are nottoconcerned about speed and efficiency, this sort is conceptually simple and short. SUBROUTINE SHELL (X, N) DIMENSION X (N) MIDPT = N 1\OMIDPT = MDPT / 2 IF (MIDPT.LE.O) RETURN LIMIT = N - MIDPT DO 20 J = l, LIMIT IJ = J 3\OIM = IJ+MIDPT IF (X (IM).GE.X (IJ)) GO TO 20 TEMP = X (IJ) X (IJ) = X (IM) X (IM) = TEMP IJ = IJ - MIDPT IF (IJ.GE.1) GO TO 30 20CONTINUE GO TO 1\O END

Question:

How many grains of sulfuric acid are contained in 3.00 liters of 0.500 N solution? (MW of H_2SO_4 = 98.1.)

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Solution:

The number of grams of sulfuric acid contained in this solution can be determined by using the definition of normality. Normality = [(grams of solute)/(equivalent weight × liters of solution)] Here, one is given the normality (N) and the number of liters of solution. The equivalent weight for an acid is the molecu-lar weight divided by the number of replaceablehydrogens, equivalent weight = [(MW)/(no. of replaceable H)] The number of replaceablehydrogensis determined by the number ofhydrogensthat will ionize when the acid is placed in solution. This number is 2 for H_2SO_4 , as shown in the following equation: H_2SO_4 \leftrightarrows 2H^+ + SO-_4 Once the number of replaceable hydrogen is known, the equivalent weight can be found. equivalent weight = [(MW)/(no. of replaceable H)] equivalent weight of H_2SO_4 = [(98.1 g)/(2 equiv)] = 49 g/equiv. At this point the number of grams of solute can be determined. grams of solute = Normality × equiv weight × liters of solution grams of H_2SO_4 = 0.500 equiv/l × 49 g/equiv × 3.0 liter = 73.5 g.

Question:

For a single year, the motor vehicles in a large city produced a total of 9.1 × 10^6 kg of the poisonous gas carbon monoxide (CO) .How many moles of CO does this correspond to?

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Solution:

The number of moles of a substance is equal to the quotient of the mass (in grams) of that substance and its molecular weight (in g/mole), or moles = [{mass (g)}/{molecular weight (g/mole)}] The mass of CO is 9.1 × 10^6 kg = 9.1 × 10^6 kg × 1000 g/kg = 9.1 × 10^9 g. The molecular weight of CO is the sum of the atomic weight of C and the atomic weight of O, or molecular weight of (CO) = atomic weight (C) + atomic weight (0) = 12 g/mole + 16 g/mole = 28 g/mole. Hence, moles = [{mass (g)}/{molecular weight (g/mole)}] = [(9.1 × 10^9 g)/(28 g/mole)] = 3.3 × 10^8 moles.

Question:

Draw a circuit realization of a clocked JK flip-flop. Ex-plain the operation of the device, give timing diagram, and truth tables for clocked, and static operations. What is the logic symbol for a clocked JK flip-flop?

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Solution:

The clocked JK flip-flop, for which the logic symbol is given in Fig. 1, has five input signals J, K, S, R, and C. J, C, and K are used for synchronous, clocked operation of the device. In this mode the set (S), and reset (R) inputs are held at zero. To initiate a state change in the device a 0 \rightarrow 1 \rightarrow 0 clock pulse must occur, because internally this flip-flop consists of two flip-flops in a master-slave configuration. This property is shown in the circuit realization of the JK flip-flop in Fig. 2. The first clock transition, 0 \rightarrow 1, causes the master flip-flop to operate, while the second, 1 \rightarrow 0, transfers data from the master to the slave. The outputs Q andQcome from the slave flip-flop. The truth table for clocked operation (when S = R = 0) is shown in figure 3. Inputs clock Paresent State J(t) K(t) C(t) Q(t) Q(t + E) 0 0 0\rightarrow1\rightarrow0 0 0 No change 0 0 0\rightarrow1\rightarrow0 1 1 0 1 0\rightarrow1\rightarrow0 0 0 Reset 0 1 0\rightarrow1\rightarrow0 1 0 1 0 0\rightarrow1\rightarrow0 0 1 Set 1 0 0\rightarrow1\rightarrow0 1 1 1 1 0\rightarrow1\rightarrow0 0 1 Complement 1 1 0\rightarrow1\rightarrow0 1 0 Fig. 3 Truth table - synchronous operation (S = R = 0) Note that when either S or R are NOT EQUAL to zero there will be a STATIC OPERATION. The truth table for static operation is shown in figure 4. S(t) R(t) Q(t) Q(t + E) 0 0 0 ?Clocked 0 0 1 ? Operation 0 1 0 0Reset 0 1 1 0 1 0 0 1Set 1 0 1 1 1 1 0 d Not allowed 1 1 1 d Fig. 4 Truth Table - static operation Note: J and K inputs don't matter in static operation. The clocked JK flip-flop timing diagram is shown in Figure 5.

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Question:

A thin square metal sheet of side 6 cm is suspended vertically from a balance so that the lower edge of the sheet dips into water in such a way that it is parallel to the surface. If the sheet is clean, the angle of con-tact between water and metal is 0\textdegree, and the sheet appears to weigh 4700 dynes. If the sheet is greasy, the contact angle is 180\textdegree and the weight appears to be 3000 dynes. What is the surface tension of water?

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Solution:

The contact angle \texttheta is a measure of the curvature of the liquid surface adjacent to the metal sheet (see figure). In either case there are three forces acting on the sheet: the tension in the suspension which gives the apparent weight, the actual weight of the sheet, and the total surface- tension force. In the first case the angle of contact is zero and the surface tension force acts downward since it tries to restore the liquid to its original level. Thus since the sheet is in equilibrium F_1 = W + 2T, the factor 2 being necessary since thereare two sides to the sheet. In the second case the angle of contact is 180\textdegree and thus the surface-tension force is acting upward. Hence F_2 = W - 2T. The surface tension \Upsilon is defined as the ratio of the surface force T, to the length, l, along which the force acts. Each force T acts along one side of the sheet, the thickness of the sheet assumed negligible with respect to the length of the sheet. This length is perpendicular to T and is given to be 6 cm. We have T = \Upsilonl Subtracting F_2 from F_1, we get F_1 - F_2 = (W + 2T) - (W - 2T) = 4T F_1 - F_2 = 4\Upsilonl Therefore \Upsilon = (F_1 - F_2)/(4l) = [(4700 - 3000)dynes]/[4 × 6 cm] = 70.8 dynes \bulletcm^-1.

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Question:

In what respects are viruses living things? How are they unlike living things?

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Solution:

All living things possess organization. The fundamental unit of organization is the cell. Living cells are variably bounded by a membrane, which regulates the movement of substances into and out of the cell. Viruses do not have any membranes because they have no need to take in or expel material. Viruses lack all metabolic machinery while cells possess this machinery in order to extract energy from theenviromentto synthesize their components. All cells produce ATP but viruses do not. Cells utilize ATP to build complex materials and to sustain active in- teractions with their environment. Viruses do not perform energy-requiring processes . Cells are capable of growth in size but viruses do not have this capacity . Most living things respond in complex ways to physical or chemical changes in their environment. Viruses do not. Most importantly, livingthinqspossess the cellular machinery necessary for reproduction. They have a complete system for transcribing and translating the messages coded in their DNA. Viruses do not have this system . Although they do possess either RNA or DNA (cells possess both ) they cannot independently reproduce, but must rely on host cells for reproductive machinery and components. Viruses use their host's ribosomes , enzymes, nucleotides, and amino acids to produce the nucleic acids and proteins needed to make new viral particles. They cannot reproduce on their own. However, viruses are unlike non-living things in that they possess the potentiality for reproduction. They need special conditions , such as the presence of a host, but they are able to duplicate themselves . Since non-living things do not contain any nucleic acids, they cannot duplicate themselves. This is the critical argument of those who propose viruses to be living things. The difficulty in deciding whether viruses should be considered living or non-living reflects the basic difficulty in defining life itself.

Question:

What are the expected types of offspring produced by a cross between a heterozygous black, short-haired guinea pig and a homozygous white, long-haired guinea pig? Assume black color and short hair are dominant charac-teristics.

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Solution:

This is an example of a dihybrid cross. A dihybrid cross is one involving parents who differ with respect to two different traits. The principles of the cross are the same as those of monohybrid crosses. Let B be the gene for black color, b the gene for white color, S the gene for short hair and s the gene for long hair. The parental genotypes are Bb Ss (hete-rozygous black, short-haired) and bbss (homozygous white, long-haired). The gametes produced are obtained, as they are in monohybrid crosses, by Mendel's Second Law, such that the genes segregate independently. Note that each gamete formed can contain only one of the allele from each allelic pair. Thus there are only four possible gametes from the heterozygous parent, as shown below: B bS s BSBsbSbs The homozygous parent can produce only one gamete type, bbss. Doing the cross: Thus the offspring are: 1/4 black, short-haired (BbSs), 1/4 black, long-haired(Bbss), 1/4 white, short-haired (bbSs),and 1/4 white, long-haired(bbss).

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Question:

"Hard" water contains small amounts of the salts calcium bicarbonate (Ca(HCO_3)_2) and calcium sulfate (CaSO_4, molecular weight = 136 g/mole). These react with soap before it has a chance to lather, which is responsible for its cleansing ability. Ca(HCO_3)_2 is removed by boiling to form insoluble CaCO_3. CaSO_4, is removed by reaction with washing soda (Na_2 CO_3, molecular weight = 106 g/mole) according to the following equation: CaSO_4 + Na_2 CO_3 \rightarrow CaCO_3 + Na_2 SO_4. If the rivers surrounding New York City have a CaSO_4 concentration of 1.8 × 10^-3 g/liter, how much Na_2 CO_3 is required to "soften" (remove CaSO_4) the water consumed by the city in one day (about 6.8 × 10^9 liters)?

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Solution:

We must determine the amount of CaSO_4, present in 6.8 × 10^9 liters and, from this, the amount of Na_2 CO_3 required to remove it. The number of moles per liter, ormolarity, of CaSO_4 corresponding to 1.8 × 10^-3 g/liter is obtained by dividing this concentration by the molecular weight of CaSO_4. Multiplying by 6.8 × 10^9 liters gives the number of moles of CaSO^4 that must be removed. Hence, moles CaSO_4= [(concentration (g/liter)/(molecular weight of CaSO_4)] × 6.8 × 10^9 liters = [(1.8 × 10^-3 g/liter) / (136 g/mole)] × 6.8 × 10^9 liters = 9.0 × 10^4 moles. From the equation for the reaction between CaSO_4, and Na_2 CO_3, we see that one mole of CaSO_4 reacts with one mole of Na_2 CO_3. Hence, 9.0 × 10^4 moles of Na_2 CO_3 are required to remove all the CaSO_4. To convert this to mass, we multiply by the molecular weight of Na_2 CO_3 and obtain mass Na_2 CO_3= moles Na_2 CO_3 × molecular weight Na_2 CO_3 = 9.0 × 10^4 moles × 106 g/mole = 9.5 × 10^6 g = 9.5 × 10^6 g × 1 kg/1000 g = 9.5 × 10^3 kg, which is about 10 tons.

Question:

Vinegar is classified as an acid. What does this tell you about the concentration of the hydrogen ions and the hydroxide ions?

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Solution:

Compounds can be classified as acidic or basic depending upon the relative concentrations of hydrogen and hydroxide ions formed in aqueous solution. In an aqueous solution, acidic compounds have an excess of hydrogen ions, and basic compounds have an excess of hyroxide ions. Since vinegar is considered to be an acid, it will have an excess of hydrogen ions with respect to hydroxide ions when in solution. Thus, it can be deduced that the pH of an aqueous solution of vinegar will be within the range of 0 to 7. Acetic acid is the constituent that gives vinegar its acidic properties. Acetic acid has a pH of 2.37 at a one molar concentration.

Question:

When J.F. Piccard made a stratosphere flight in a balloon, the balloon seemed to be only half filled as it left the ground near Detroit. The gas temperature was about 27\textdegreeC, the pressure 700 mm, and the volume of gas in the balloon 80000 cubic feet. What was the gas volume at high alti-tude where the temperature was - 3\textdegreeC, and the pressure 400 mm?

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Solution:

The solution to this problem necessitates the use of the combined gas law. It is stated as follows: For a given mass of gas, the volume is inversely pro-portional to the pressure and directly proportional to the absolute temperature. The general formula for the combined gas laws may be written (P_1 V_1 )/T_1= (P_2 V_2 )/T_2orV_2 = V_1 × (P_1 /P_2 ) × (T_2 /T_1 ) in which P_1 and T_1 are pressure and absolute temperature of the gas at volume V_1; and P_2 and T_2 are pressure and absolute temperature at volume V_2. Thus, if P_1 = 700 mmandP_2 = 400 mm V_1 = 80,000cu.ft.T_2 = 273\textdegree - 3\textdegree = 270\textdegreeK T_1 = 27\textdegree + 273\textdegree = 300\textdegreeK then V_2 can be determined by substitution in the general formula above. V_2 = (80,000) [(700 mm)/(400 mm)] [270\textdegreeK)/(300\textdegreeK)] = 125,000 cu. ft.

Question:

The sulfate ion consists of a central sulfur atom with four equivalent oxygen atoms in a tetrahedral arrangement. Keeping in mind the octet rule, draw the electronic struc-ture for the ion. What should the internal O-S-O bond angle be?

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Solution:

The formula for the sulfate ion is SO_4^=. To write the electronic structure of the ion in this problem, one must consider the definition of valence and the octet rule. Valence electrons refer to those outer electrons that participate in chemical bonding. The octet rule states that for stability, there can be no more than eight electrons in the outer orbit of an atom, either as a result of transfer or sharing. Thus, after determining the number of valence electrons present, arrange them so that they obey the octet rule. The electronic configuration of sulfur is 1s^22s^22p^63s^23p^4. It has 6 valence electrons; they are found in the 3s and 3p orbitals. The electronic con-figuration of oxygen is 1s^22s^22p^4, which means it also has six valence electrons; the first orbital contains only 2 electrons. The electronic structure of each of these atoms can be represented as shown in figure a and figure b, where x's and dots indicate valence electrons. The electronic structure of the ion can be pictured as figure c. Here, one can see that the sulfur and two of the oxygen atoms are surrounded by eight electrons. Because two of the oxygen atoms are only surrounded by seven electrons, each of these atoms possesses a negative charge. This structure may also be drawn as shown in figure d. For greatest stability the oxygen atoms must take positions as far apart as possible. It has been shown that when four atoms surround a fifth the most stable arrangement is tetrahedral. In this arrangement, the O-S-O bonds form a 109.5\textdegree angle.

Question:

Lunar soil contains,ilmenite, a mineral with the composition FeO\bulletTiO_2. It has been suggested that moon explorers might obtain water and oxygen from lunar soil by reducing it with hydrogen (brought from earth) according to the following reactions. Beginning with soil heated to 1300\textdegreeC in a solar furnace: FeTiO_3 + H_2-----------------\rightarrow H_2O + Fe + TiO_2 Electrolysis 2H_2O-----------------\rightarrow 2H_2O + O_2 How much water and oxygen could be obtained from 50 kg of soil if the soil is taken to be 5 per centilmenite?

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Solution:

From thestoichiometryof these equations, 1 mole of FeO\bulletTiO_2 is consumed to form 1 mole of H_2O and 2 moles of H_2O is consumed to form 1 mole of O_2. If 5 % (0.05) of the soil is FeO\bulletTiO_2 then only (50,000 g) (0.05) = 2,500 grams areilmenite. The molecular weight ofilmeniteis 152 g/mole. Thus, number of moles ofilmeniteis [(2,500 gilmenite) / (152 g / moleilmenite)] = 16.4 moles. Thus, there must be 16.4 moles H_2O and the weight of H_2O is (16.4moles) (18.02 g/mole) = 296 g H_2O. If there are 16.4 moles of H_2O, then there are 8.2 (or half 16.4 moles) moles of O_2. Thus, (8.2moles) (32 g / mole) = 262 g O_2.

Question:

The protein human plasma, albumin, has a molecular weight of 69,000. Calculate the osmotic pressure of a solution of this protein containing 2 g per 100 cm^3 at 25\textdegreeC in (a) Torr and (b) millimeters of water. The ex-periment is carried out using a salt solution for solvent and a membrane permeable to salt.

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Solution:

When a semipermeable membrane separates a solution of a protein in water from pure water, osmosis will occur. The concentration - or the thermodynamic activity - of water molecules in the protein solution is less than in pure water, and the system will compensate for this difference by net movement of the water from the pure water compartment into that containing the protein solution, until the concentration of the water is the same on both sides of the membrane. Osmotic pressure is the force that must be applied to just prevent such osmotic flow. Osmotic pressure can be determined by using the following equation. \pi = (C R T) / M, where \pi is the osmotic pressure, c is the concentration in (g/liter), R is the gas constant (0.82 liter atm/mole \textdegreeK), T is the absolute temperature and M is the molecular weight. Solving for \pi. \pi =[(20 g/liter) (.082 liter atm/mole \textdegreeK)(298\textdegreeK)] / [69,000 g/mole] = 7.08 × 10^-3 atm. Note: 2(g/100 cm^3) corresponds to 20(g/liter). (a) There are 760 Torr in 1 atm. \pi in Torr = 7.08 × 10^-3 atm × (760Torr/atm ) = 5.38 Torr (b) 5.38 Torr is equivalent to 5.38 mmHg. This can be converted to mm H_2O by multiplying 5.38 mmHg by the density of Hg divided by the density of H_2O, i.e., 5.38 mmHg × (13.534 g/ml) / (0.997 g/ml) = 73.0 mm H_2O.

Question:

A certain farsighted person has a minimum distance of distinct vision of 150 cm. He wishes to read type at a distance of 25 cm. What focal-length glasses should he use?

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Solution:

The principal axis is the line passing through the centers of curvature of the faces of the lens. The op-tical center is the point in the lens through which light can pass without being bent. All rays of light parallel to the principal axis pass through F, the point of principal focus. The distance f between F and the optical center is called the focal length of the lens. It is positive for converging, convex lenses, negative for concave, diverging lenses. Since the person cannot see clearly objects closer than 150 cm, the lens must form a virtual image at that dis-tance. Since the image is formed on the same side of the lens as the object, its distance q from the lens is negative. p = 25 cm q = -150 cm. Substituting in the general equation for lenses, 1/p + 1/q = 1/f (1/25cm) + (1/-150cm) = 1/f f = 30 cm. Since f is positive, this lens is converging and is convex.

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Question:

Compare the electrostatic and gravitational forces that exist between an electron and a proton.

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Solution:

The electrostatic force law and the gravitational force law both depend on 1/r^2 F_G = G(m_1m_2 / r^2 ) where G = 6.67 × 10^\rule{1em}{1pt}8 (dyne\rule{1em}{1pt}cm^2 / g^2 ), and r is the distance between masses m_1 and m_2 . Furthermore, F_E = q_1q_2 / r^2 , in thec.g.s. system, where r is the distance between q1and q_2 . Therefore, the ratio F_E/F_G is independent of the distance of separation: F_E / F_G = (q_1q_2) / (Gm_1m_2) For the case of an electron and a proton this becomes F_E / FG= [(e)(e)] / (Gm_em_p) = (e)^2 / (Gm_em_p) Substituting the values of the quantities, we find F_E / F_G = [{(4.80 × 10^\rule{1em}{1pt}10statC)^2} / {(6.67 × 10^\rule{1em}{1pt}8dyne\rule{1em}{1pt}cm2/ g^2) × (9.11 × 10^\rule{1em}{1pt}28g) × (1.67 × 10^\rule{1em}{1pt}24 g)}] =2.3 × 10^39 Thus, the electrostatic force between elementary particles is enormously greater than the gravitational force. Therefore, only the electro-static force is of importance in atomic systems. In nuclei, the strong nuclear force overpowers even the electrostatic force but not to the extent that electrostatic forces are completely negligible. Many im-portant nuclear effects are the result of electrostatic forces.

Question:

We consider the vector A\ding{217} = 3x̂ + \^{y} + 2ẑ (a) Find the length of A\ding{217}. (b) What is the length of the projection of A\ding{217} on the xy plane? (c) Construct a vector in the xy plane and perpendicular to A\ding{217}. (d) Construct the unit vector B˄. (e) Find the scalar product with A\ding{217} of the vector C\ding{217} = 2x̂. (F) Find the form of A\ding{217} and C\ding{217} in a reference frame obtained from the old reference frame by a rotation of \pi/2 clockwise looking along the positive z axis. (g) Find the scalar product A\ding{217} \textbullet C\ding{217} in the primed coordinate system. (h) Find the vector product A\ding{217} × C\ding{217}. (i) Form the vector A\ding{217} - C\ding{217}. The primed reference frame x', y', z', is generated from the unprimed system x, y, z, by a rotation of +\pi/2 about the z axis.

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Solution:

(a) When a vector is given in the form A_xx̂ + A_y\^{y} + A_zẑ, its length is given by \surd(A_x^2 + A_y^2 + A_z^2). This can be seen from diagram 1. Vector A\ding{217} has components in the x, y and z directions. The x and y components form the legs of a right triangle. By the Pythagorean theorem the length of the hypotenuse of this triangle is \surd(A_x^2 + A_y^2). But this line segment whose length is \surd(A_x^2 + A_y^2) is one leg in a right triangle whose other leg is A_zz and whose hypotenuse is vector A\ding{217}. Applying the Pythagorean theorem again, we find that the length of A\ding{217} is \surd(A_x^2 + A_y^2 + A_z^2). Substituting our values we have \surd(3^2 + 1^2 + 2^2) = \surd14. (b) We refer again to diagram 1. The projection of A\ding{217} on the xy plane is simply the dotted line which is the vector A_xx̂ + A_y\^{y}. Its length is \surd(A_x^2 + A_y^2) by the Pythagorean theorem. In our problem, the length is \surd(3^2 + 1^2) = \surd10. (c) Construct a vector in the xy plane and perpendicular to A. We want a vector of the form B = B_xx̂ + B_y\^{y} with the property A \textbullet B = 0 (since A\ding{217} \textbullet B\ding{217} = \mid A\ding{217} \mid \mid B\ding{217} \mid cos \textphi where \textphi is the angle between A\ding{217} and B\ding{217}). Hence (3x̂ + \^{y} + 2ẑ) \textbullet (B_xx̂ + B_y\^{y} ) = 0. On taking the scalar product we find 3B_x + B_y = 0 or(B_y/B_x) = - 3. The length of the vector B is not determined by the specification of the problem. We have therefore deter-mined just the slope of vector B, not its magnitude. See diagram 2. (d) The unit vector B is the vector in the B direction but with the magnitude 1. It lies in the xy plane, and its slope (B_y/B_x ) is equal to - 3. Therefore, B˄ must satisfy the following two equations: B˄_x^2 + B˄_y^2 = 1 (B˄_y/B˄_x) = - 3 Solving simultaneously we have: B˄_x^2 + (- 3B˄_x)^2 = 1 or B˄_x = 1/\surd10 and B˄_y = - 3/\surd10. The vector B is then: B˄ = (1/\surd10)x̂ - (3/\surd10)\^{y} (e) Converting the vectors into coordinate form and computing the dot product (scalar product): (3x̂ + \^{y} + 2ẑ) \textbullet (2x̂ + 0\^{y} + 0ẑ) = 6 + 0 + 0 = 6 (f) Find the form of A\ding{217} and C\ding{217} in a reference frame obtained from the old reference frame by a rotation of \pi/2 clockwise looking along the positive z axis. The new unit vectors x˄ , y˄ , z˄ are related to the old x˄, y˄, z˄ by (see fig. 3) x˄' = y˄; y˄' = - x˄; z˄' = z˄ . Where x˄ appeared we now have - y˄' ; where y˄ appeared, we now have x˄' so that A = x̂ - 3\^{y}' + 2ẑ' ;C = - 2\^{y}' . (g) Using the results of part (f), we convert the vectors A\ding{217} and C\ding{217} into coordinate form in the primed co-ordinate system, giving us the following dot product: A\ding{217} \textbullet C\ding{217} = (x̂' - 3\^{y}' + 2ẑ' ) \textbullet (0x̂' - 2\^{y}' + 0ẑ') = 0 + 6 + 0 = 6 This is exactly the result obtained in the unprimed system. (h) Find the vector product A\ding{217} × C\ding{217} In the unprimed system A\ding{217} × C\ding{217} is defined as x̂\^{y}ẑ 312 200 = 4\^{y} - 2ẑ. (i) Form the vector A\ding{217} - C\ding{217}. We have A\ding{217} - C\ding{217} = (3 - 2)x̂ + \^{y} + 2ẑ = x̂ + \^{y} + 2ẑ.

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Question:

A car of mass 1 × 10^3 kg is traveling at a speed of 10 m/s. At acertain instant of time we measure its position with a precisionof 1 × 10^-6 m. What is the theoretical limitin precisionfor measuring the speed of the car?

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Solution:

The uncertainty principle will be used to find the limit of precision of onevariable when the precision of a complementary variable is given. If \Deltaxis the uncer-tainty (i.e., the limit of precision) in the position of a particle, andif\Deltap=m\Deltavis the uncertainty of the momen-tum of the particle, then, \Deltax\Deltap=\Deltaxm\Deltav\cong (h/4\pi) whereh is Planck's constant. Substitution of the given values for m, and \Deltaxin the equation gives \Deltav= [h/{(\Deltax) 4\pim}] = [(6.6 × 10^-34 J s)/[(1 × 10^-6 m) (4\pi) (1 × 10^3 kg)}] = 5.3 × 10^-32 m/s This is a very small uncertainty in the speed of the car and could never be detectedby available measurement tech-niques.

Question:

From the following thermochemical data, determine which is more stable and why: cis-2-buteneH_2\ding{217}n-butane ∆H\textdegree = 28.6 Kcal / mole trans-2-buteneH_2\ding{217}n-butane ∆H\textdegree = -27.6 Kcal / mole

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Solution:

One method of finding the relative stabilities between two organic compounds is by a comparison of their heats of hydrogenation. This is exactly what the problem presents to you. Both cis and trans-2-butene, upon hydro-genation, yield the same product. Yet, the trans-2-butene evolves 1 Kcal/mole less heat and therefore the trans-form must be stabilized to that extent relative to the cis -isomer. To find out why trans isomer is 1 Kcal/mole more stable than the cis -isomer, you must write their structures. Notice, in the cis configuration, the methyl groups (CH_3) are located on the same side of the double bond, while in trans, they are found on opposite sides. The cis -isomer will exhibit steric interference, which means that because of the size of the methyl groups, they will tend to "crowd" each other. To avoid this "crowding", there will be repulsion between the methyl groups which, of course, decreases the stability of the molecule. The trans -isomer has no crowding problem as the methyls are located on opposite sides. As such, its stability will not be reduced by steric effects, which explains why it is more stable than the cis -isomer.

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Question:

Describe in general the use of the Data Control Block (DCB) in basic assembler language of IBM/360-370 computers.

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Solution:

In writing a basic assembler language program, files have to be used. Each file used in the program must be defined in a Data Control Block (DCB). These files can be output files, input files, and so forth. In coding the Data Control Block, the mnemonic DCB is used to indicate the operation code. Like any other operation code this is written in columns 10-14. The operands following it are the file specifications. These operands are keywords which designate the specific characteristics of the file. Many keywords may be required, which means continuation cards may be used, if the program is punched on cards. The gen-eral format of a DCB statement (file specification) is: Col 1-910-1216+ FILENAME DCBKEYWORD = SPECIFICATION, KEYWORD = ... The keywords used are determined by the kind of input/output device to be used (card reader, printer, magnetic tape, disc, optical character reader, etc.) and by the organization of the records in this file (sequential, random, or otherwise). To illustrate a DCB we will code one for a program punched on cards. The card and printer files are sequential files, which means processing of these files is done in the order in which they are written. Example \textasteriskcenteredTHE CARD FILE DEFINITION CARDIN DCB DSORG = PS, RECFM = F, X MACRF = GM, BLKSIZE = 80, X LRECL = 80, DDNAME = CRDIN, X EODAD = CRDEOF In this example we have used 4 cards to code the DCB. The X, which is in column 72, is a continutation character. The label CARDIN is the filename, it is generally up to 8 alphanumeric characters in length. DCB is the instruction code. DSORG = PS indicates that the data set is sequential (actually physical sequential). This entry determines what kind of access method will be used for the file and what input/output modules ('SUBPROGRAMS') will be required. RECFM = F: RECFM, which stands for record format, indicates that the records (cards) are of fixed length. (A card's length is always fixed). MACRF (macro form) = GM indicates what form of input or output instruction will be coded for this file. To read cards, the GET instruction (GM) is used. BLKSIZE and LRECL indicate block and record size. Each 80 byte (one byte for each of the 80 columns) card forms a LRECL, therefore, both are 80. DDNAME gives the name of the file, which was previously written as the label of the DCB instruction. Finally, we have EODAD which means end of data address. EODAD = CRDEOF: CRDEOF is the label in the pro-gram which the computer must branch to when no more data is found. The above is a general description of the format of a DCB. The keywords of the instruction will vary with the input/output devices and the files being used. But as you can observe, the DCB precisely defines a file to be used by a program.

Question:

A chemist has a mixture of KCIO_3, KHCO_3, K_2 CO_3, and KCl . She heats 1,000 g of this mixture and notices that the following gases evolve: 18 g of water (H_2 O), 132 g of CO_2, and 40 g of O_2 according to the following reactions: 2KC1O_3\rightarrow 2KC1 + 3O_2 2KHCO_3\rightarrow K_2 O + H_2 O + 2CO_2 K_2 CO_3\rightarrow K_2 O + CO_2 TheKClis inert under these conditions. Assuming complete decomposition, determine the composition of the original mixture.

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Solution:

The solution of this problem involves the use of the mole concept and the ability to employ it using chemical (balanced) equations. You need to determine the number of moles of the gases generated from the masses given and their molecular weights. A mole = mass in grams /molecular weight. Once this is known, you can calcu-late the number of moles of substances in the mixture that had to exist to produce the given quantities. You proceed as follows: The only source of the 18 g of H_2 O is from 2KHCO_3 \rightarrow K_2 O + H_2 O + 2CO_2. The molecular weight of H_2 O = 18. Thus, you have 18 g/18 g/mole = 1 mole of H_2 O. But the equation states that for every mole of H_2 O, you originally had 2 moles of KHCO_3. Therefore, the mixture must have had 2 moles of KHCO_3. The molecular weight of KHCO_3 is 100.1 g. There-fore, the weight in grams of it was 2(100.1) = 200.21 g. The O_2 is generated from only the reaction 2KClO_3 \rightarrow 2KCl + 3O_2. You have 40 g of O_2 evolved. Molecular weight of O_2 = 32. Thus you have 40/32 = 1.25 moles of O_2. According to the equation, for every 3molesof O_2 produced, there existed 2 moles of KClO_3. Thus, you have 2/3 (1.25) or .833molesof KClO_3. Molecular weight of KClO_3 = 122.6. Therefore, the number of grams = (122.6) (.833) = 102.1 g The CO_2 gas has two sources : 2KHCO_3 \rightarrow K_2 O + H_2 O + 2CO_2andK_2 CO_3 \rightarrow K_2 O + CO_2. From the water evolved, you already know that you have 200.21 g of KHCO_3. Its molecular weight is 100.1 g. Thus, the number of moles of it is, 200.2/100.1 = 2. From the equation, however, you see that for every 2 moles of KHCO_3, you obtain 2 moles of CO_2. Maintaining this one to one ratio, 2molesof CO_2 must be generated. The molecular weight of CO_2 is 44. Thus, from KHCO_3, 2(44) = 88 g of CO_2 produced. The total number of grams CO_2 evolved was given as 132 g. This means, therefore, that the other source of CO_2, K_2 CO_3, must give off 132 - 88 = 44 grams of CO_2. You have K_2 CO_3 \rightarrow K_2 O + CO_2 This equation shows a 1 : 1 mole ratio between K_2 CO_3 and CO_2. You have 44 grams of CO_2 released or 1 mole, since the molecular weight of CO_2 is 44. Therefore, you must have 1 mole of K_2 CO_3 in the mixture. The molecular weight of K_2 CO_3 is 138.21.Thus,thenumber of grams is (1) (138.21) for K_2 CO_3 = 138.21 g. In summary, you have 200.14 g of KHCO_3, 102.12 g of KCIO_3 and 138.21 g of K_2 CO_3. The total mass of these substances = 440.47 g. The original mixture was 1000 g. Thus, 1000 - 440.47 = 559.53 g is the mass of the inertKCl.

Question:

A paramecium is introduced to a medium rich in yeast. How do the yeast cells get into the food vacuole of the paramecium? What is the fate of the yeast in the food vacuole?

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Solution:

Paramecia are ciliates (organisms covered with cilia over their entire surface) . On one side of the paramecium there is a slight depression, the oral groove, which serves as a mouth. A paramecium feeds by creating a current (by the beating of the cilia around the oral groove) which sweeps the food particles into the oral groove. In this case, the food particles are the yeast cells. Under pressure from the current, the yeasts and fluids are pushed into the cytoplasm at the base of the groove, called the cytopharynx, and a food vacuole is formed. The vacuole breaks away from the groove and circulates around the cell, diminishing in size as it moves. As the vacuole circulates enzymes made in the cytoplasm are secreted into it. These enzymes break down the yeasts into simple biomolecules (sugars, amino acids, etc.) which diffuse out of the vacuole into the cytoplasm to be used later for the organism's metabolic processes. When the vacuole reaches a tiny specialized region of the cell surface called the anal pore, it becomes attached to the pore and then ruptures, expelling to the outside any remaining bits of indigestible material. The sequence of events is reviewed in the figure below. Since the enzymatic hydrolysis of foodstuff occurs inside the cell of the paramecium, digestion in this organism is of the intracellular type, known as intracellular diges-tion.

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Question:

I1I2I3\cdot\cdot\cdot \cdot\cdot\cdot J1J2J3\cdot\cdot\cdot \cdot\cdot\cdot \cdot\cdot\cdot K1K2K3\cdot\cdot\cdot \cdot\cdot\cdot \cdot\cdot\cdot ::: : : :: : : where I_s, J_s, K_s ... are integers, and all rows, columns, and diagonals have the equal integer sums. Write a program to create three 3 × 3 magic squares using a random number generator. Integers may range from 0 to 100. Use BASIC.

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Solution:

The statement INT(100\textasteriskcenteredRND(X)) gives the programmer random integer numbers between 0 and 100. This statement is repeated 3 times to generate the three random integers L, K, J. The magic square is built then on the basis of those numbers by arranging them in the following manner: J + LL - (J+K)K + L K+L - JLJ+L - K L - KJ+K + LL - J Now, if you sum the numbers in rows, columns, and diagonals it will be seen that all the sums are the same and equal to 3L! For example, row 1: (J+L) + L-(J+K) + (K+L) = J + L + L - J - K + K + L = L + L + L = 3L The program looks as follows: 10 LET N = 1 15 LET J = INT(100\textasteriskcenteredRND(X)) 20 LET K = INT(100\textasteriskcenteredRND(X)) 25 LET L = INT(100\textasteriskcenteredRND(X)) 30 LET P1 = J + L 35 LET Q1 = L - (J+K) 40 LET R1 = K +L 45 LET P2 = K + L -J 50 LET Q2 = L 55 LET R2 = J + L - K 60 LET P3 = L - K 65 LET Q3 = J + K + L 70 LET R3 = L - J 75 PRINT P1, Q1, R1 80 PRINT P2, Q2, R2 85 PRINT P3, Q3, R3 90 LET N = N + 1 95 IF N < = 3 THEN 15 100 STOP 105 END

Question:

Write a program to sort an integer array into ascending order using a selection sort.

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Solution:

SORT: PROC OPTION (MAINS); DCL (ARRAY (N), SMALL, N) FIXED (3.0); / \textasteriskcentered IT IS ASSUMED THE CONTENTS OF ARRAY AND ITS DIMENSION ARE GIVEN \textasteriskcentered / K = 1; DO WHILE K

Question:

You are given the following absolute entropies, S\textdegree, for the reactionCuO(s) + H_2 (g) \rightarrow Cu (s) + H_2 O (g) at 25\textdegreeC; CuO(s)10.4 cal/mole H_2 (g)31.2 cal/mole Cu (s)8.0 cal/g - atom H_2 O (g)45.1 cal/mole Is the ∆S favorable for this reaction, assuming standard conditions?

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Solution:

To find whether the reaction will proceed spontaneously, calculate the change in entropy, ∆S\textdegree, for the reaction. When ∆S\textdegree is positive the reaction will proceed spontaneously. Remember, entropy is a measure of randomness and the natural order of the universe is for things to become more random, which necessitates a positive S\textdegree. ∆S\textdegree = S\textdegree _products - S\textdegree _reactants . Thus for this reaction. ∆S\textdegree =S\textdegree_Cu(s)+ S\textdegreeH2 O (g)-S\textdegree_CuO(s)- S\textdegree_H2 (g) = 8 + 45.1 - 10.4 - 31.2 = + 11.5 cal/deg - mole. Since ∆S\textdegree is positive, the reaction will proceed spontaneously.

Question:

Before the advent of pH meters,aurologist collected 1.3 litersof urine from a hospitalized patient over the course of a day. In order to calculate what the pH was a laboratory techniciandetermined the number of equi-valents of acid presentin the sample. A given sample of urine contains 1.3 × 10^-6 equivalents of dissociated acid.What is the pHof thissample?

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Solution:

To solve this problem, we have to note that 1 equivalent of H^+ is the sameas 1 mole of H^+. We then determine [H^+] and from this thepH . 1.3 × 10^-6 equivalent of H^+ is the same as 1.3 × 10^-6 mole of H^+. The concentration of H^+ in the sampleis .then [H^+] = moles of H^+/volume = 1.3 × 10^-6 mole/1.3 liter = 10^-6 M. The pH is defined as pH = - log [H^+], hence pH= - log [H^+] = - log (10^-6) = - (- 6) = 6, which is the pH of normal urine.

Question:

If the hydrolysis constant of Al^3+ is 1.4 × 10^-5, what is the concentration of H_3O^+ in 0.1 M AlCl_3?

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Solution:

Hydrolysis refers to the action of the salts of weak acids and bases with water to form acidic or basic solutions. Consequently, to answer this question, write out the reaction, which illustrates this hydrolysis, and write out an equilibrium constant expression. From this, the concentration of H_3O^+ can be defined. The net hydrolysis reaction is AlCl_3 \rightarrow Al^3+ + 3Cl^- Al^3+ + 2H_2O \rightleftarrows AlOH^2+ + H_3O^+ K_hyd= 1.4 × 10^-5 = {[H_3O^+][ AlOH^3+]}/[ Al^2+] . Water is excluded in this expression since it is considered as a constant. Let x = the moles/liter of [H_3O^+]. Since H_3O^+ and AlOH^2+ are formed in equal mole amounts, the concentration of [AlOH^2+] can also be represented by x. If one starts with 0.1 M of Al^3+, and x moles/liter of it forms H_3O^+ (and AlOH^2+), one is left with 0.1 - x at equilibrium. Substituting these repre-sentations into theK_hydexpression, [(x \textbullet x)/(0.1 - x)] = 1.4 × 10^-5. If one solves for x, the answer is x = 1.2 × 10^-3 M, which equals [H_3O^+].

Question:

An animal has adipoidnumber of 8. During meiosis, how many chromatids are present a)inthe tetrad stage? b) In latetelophaseof the first meiotic division? c) In metaphase of the second meiotic division?

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Solution:

In doing this problem, one must remember that meiosis involves both the duplication of chromosomes and the separation of homologous pairs . a) In the tetrad stage of meiosis, homologous chromosomes synapse , or pair. But prior to this, every chromosome had been duplicated . Synapsis therefore results in a tetrad, a bundle of 4 chromatids (2 copies of each one of the homologous chromosomes.) The number of tetrads equals the number of haploid chromosomes. Therefore, there are (1/2) × 8 or 4 tetrads. Since each tetrad has 4 chromatids, there are a total of 4 × 4 or 16 chromatids in the tetrad stage. b) In latetelophaseof the first meiotic division, the homologous chromosomes of each pair have separated. But each chromosome is still double and composed of two daughter chromatids. So there are the haploid number (4) of doubled chromosomes, or a total of 8 chromatids. c) In metaphase of the second meiotic division, the doubled chromosomes have lined up along the equator of the cell, but daughter chromatids have not yet separated. So the number of chromatids is still 4 × 2 or 8.

Question:

Using the information supplied in the accompanying table, determine the comparative heats of combustion of water gas (CO + H_2) and of methane (CH_4) per unit weight and per unit volume. The products are CO_2 (g) and H_2 O(g) in both reactions. Enthalpies of formation in kilojoules per mole from the elements for various compounds at 298\textdegreeK and 1 atm pressure. Compound ∆H CO (g) - 110 H_2O (g) - 242 CO_2 (g) - 394 CH_4 (g) - 75

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Solution:

Write out the chemical reactions and calculate the enthalpy change. Enthalpy change is a measure of the amount of heat absorbed or released in the reaction. For the water gas, the reaction is CO + H_2 + O_2 \rightarrow CO_2 (g) + H_2O(g). The overall enthalpy is the sum of the en-thalpies of formation of the products minus the sum of the enthalpies of formation of the reactants. Thus, ∆H\textdegree = - 394 KJ + (- 242 KJ) - (- 110 KJ) = - 526 KJ. Notice that the ∆H's of H_2 and O_2 are not included. Elements are considered to have ∆H's of zero. The question asks for ∆H by unit weight and volume. Therefore, ∆H by unit weight = (∆H\textdegree) / (mol.wt.CO + mol.wt.H_2) = (-526 KJ) / (30 g) = - 17.5 KJ/g. The sum of the molecular weights of CO + H_2 is 30. To find ∆H by molar volume, note that two molar volumes are involved, one from H_2 and one from CO, which means ∆H by volume= (∆H) / (no. of molar volumes) = (-526 KJ) / (2 molar volumes) = - 263 KJ/molar volume. An analogous method is used for methane. The reaction is CH_4 + O_2 \rightarrow CO_2(g) + 2H_2 O(g). ∆H\textdegree = - 394 + 2(- 242)(- 75) = - 803 KJ The molecular weight of CH_4 = 16. Therefore, ∆H per unit weight = (-803 KJ) / (16 g) = - 50.2 KJ/g. The only molar volume is that of methane. Then (∆H) / (molar volume) = (- 803 KJ) / (1) The comparative heats can thus be related as (263) / (803) = (.33) / (1) by volume and (17.5) / (50) = (.35) / (1) by weight.

Question:

A 50 ml solution of sulfuric acid was found to contain 0.490 g of H_2SO_4. This solution was titrated against a sodium hydroxide solution of unknown concentration. 12.5 ml of the acid solution was required to neutralize 20.0 ml of the base. What is the concentration of the sodium hydroxide solution.

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Solution:

At the neutralization point, the number of equivalents of acid in the 12.5 ml volume is equal to the number of equivalents of base in the 20.0 ml volume. Since the normality is defined as the number of equivalents per liter of solution, the number of equivalents is equal to the normality times the volume, at the neutralization point we have N_aV_a=N_bV_b where N_a = normality of acid,V_a= volume of acid,N_b= normality of base, V_b= volume of base. The normality of the 50.0 ml (0.050 l) sulfuric acid solution is N_a = [(number of equivalents in 0.050 l)/(0.05 l)] = [(mass of acid/gram equivalent weight)/(0.05 l)] The gram equivalent weight of sulfuric acid is 49.0 g/equivalent, because there are 2 equiv per molecule. The MW of H_2SO_4 is 98 g/mole. N_a = [(mass of acid/gram equivalent weight)/(0.05 l)] = [(0.490 g/49.0 g/equivalent)/(0.05 l)] = 0.200 equivalent/l = 0.200 N The normality of the base is then found as follows: N_aV_a=N_bV_b N_b= [(N_aV_a)/V_a] = [(0.200 N × 12.5 ml)/(20.0 ml)] = 0.125 N. Therefore, the sodium hydroxide solution is 0.125 N, which because there is 1ionizableOH^- inNaOH, is equal to 0.125 M.

Question:

Skeletal muscle is the largest tissue in the body, accounting for 40 to 45 per cent of the total body weight. What are the structural characteristics of skeletal muscle?

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/Users/wenhuchen/Documents/Crawler/Biology/F19-0462.htm

Solution:

The term muscle, as it is commonly used, refers to a number of muscle fibers bound together by connective tissue. Skeletal muscle fibers are multinuc-leated cylindrical cells, 10 to 100\mum in diameter, and may be up to 1 ft. long. Generally each end of an entire muscle is attached to a bone by bundles of collagen fibers known as tendons. Some tendons are very long, and the site of attachment of the tendon to the bone is far removed from the muscle. For example, some of the muscles which move the fingers are found in the lower portion of the arm, between the elbow and the wrist. The most striking feature of muscle fibers is the series of transverse light and dark bands forming a regular pattern along the fiber. Both skeletal and cardiac muscle exhibit such banding patterns; smooth muscle does not. Though the pattern appears continuous across a fiber, the fiber is actually composed of a number of independent cylindrical elements in the cytoplasm of the fiber known as myofibrils. Bundles of these fibrils are enclosed by the muscle cell membrane or sarcolemma. Each myofibril is about a micron in diameter. Between the myofibrils are large numbers of mitochondria, which are to be expected in cells that have such a high energy requirement. The myofibrils show the same pattern of cross striations as the fibers of which they are a part. When viewed with the electron microscope, the structures responsible for the banding patterns become evident. The myofibrils consist of smaller myofilaments which form a regular repeating pattern along the length of the fibril. One unit of this repeating pattern is known as a sarcomere, and is the functional unit of the contractile system of the muscle. Each sarcomere (Figure 1) contains two types of myofilaments: thick and thin. In the central region of the sarcomere are the thick myofilaments, and they appear as a dark band. This band is termed the A band. These thick filaments contain the protein myosin. The thin myofilaments contain the protein actin and are attached at either end of the sarcomere at a structure known as the Z line. The limits of a sarcomere are defined by two successive Z lines. The thin filaments extend from the Z lines toward the center of the sarcomere where they overlap with the thick filaments. Two more bands are distinguishable in the muscle. The I band represents the region between the ends of the A bands of two adjoining sarcomeres. In this region only thin filaments are present; there is no overlap of the thick and thin filaments. Because it contains only thin filaments, the I band usually appears light. The H Zone corresponds to the space between the ends of the thin filaments; only thick filaments are present. The thick and thin filaments are arranged hexagonally with respect to one another. Each thick filament is surrounded by six thin filaments, and each thin filament is surrounded by three thick ones. (See Figure 2) An average muscle fiber contains about 15 billion thick and 65 billion thin filaments. Under extremely high magni-fication, the gap between thick and thin filaments in the region of the A band appears to be bridged by projections at intervals along the filaments. It appears that these projections, or cross bridges, are arranged in a spiral around the thick filament. These projections are thought to play an active role in the contraction of the muscle. (See following question.) An important muscle cell organelle is the sarcoplasmic reticulum, a continuous system of tubules extending throughout the cytoplasm, forming a closely meshed canal network around each myofibril. This organelle corresponds to the endoplasmic reticulum of other cell types, but in muscle it is largely devoid of ribosomes and exhibits a highly specialized repeating pattern. The tubules of the reticulum overlying the A bands have a prevailing longitudinal orientation but anastomose freely in the region of the H band. At regular intervals along the length of the myofibrils the longitudinal tubules of the sarco-plasmic reticulum, called sarcotubules, come together with transversely oriented channels of larger caliber called terminal cisternae. (See Figure 3) Pairs of parallel terminal cisternae run transversely across the myofibrils in close apposition to a slender intermediate element, the transverse tubule, commonly called the T tubule. These three transverse structures, namely the two parallel terminal cisternae and one T tubule, constitute the triads of skeletal muscle. In man, there are two triads to each sarcomere, situated at the junctions of each A band with the adjacent I bands. The cavity of the T tubule does not open into the adjacent cisternae and, strictly speaking, is not part of the sarcoplasmic reticulum. It is merely a slender invagination of the sarcolemma. The lumen of the T tubule is thus continuous with the extracellular medium surroun-ding the muscle fiber. The T tubules are involved in transmitting electrical signals from neurons on the cell surface to deep within the muscle fiber.

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Question:

How many alpha particles per second would be emitted from 4 × 10-^12 g of ^210Po (t_(1/2) = 138 days)?

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Solution:

When given the mass and the half-life of a sub-stance, one can find the number of disintegrations per second (dN/dt) by using the following equation: dN/dt= - [(ln2/t_(1/2)) × N], where t_(1/2) is the half-life in second, N is -the original number of nuclei. (1) Solving for the half-life in seconds: t_(1/2) = 138 days × 24 hr/day × 60 min/hr × 60 sec/min = 1.192 × 10^7 sec (2) Determining the number of nuclei present. One is given that there was originally 4 × 10^-12 g of ^210Po present. The number of nuclei present is found by multiply-ing the number of moles byAvogrado's number (6.02 × 10^23), the number of nuclei in one mole. The number of moles is found by dividing 4 × 10^-12 g by the molecular weight of ^210Po. (MW = 210). no. of moles= [(4 × 10^-12 g) / (210 g/mole)] = 1.905 × 10^-14 moles no. of nuclei= 1.905 × 10^-14 moles × 6.022 × 10^23 nuclei/mole = 1.147 × 10^10 nuclei (3) Solving fordN/dt. [dN/dt] = - [(ln^2/t_(1/2)) × N]N= 1.147 ×10^10 nuclei t_(1/2) = 1.192 × 10^7 sec ln2 = 0.693 (dN/dt)= -{.693 / (1.192 × 10^7 sec)} × 1.147 × 10^10 nuclei = - 666.94 nuclei / sec Because (dN/dt) is negative, it means that the change in the number of nuclei per unit time is caused by disinte-grations of the nuclei. This is the disintegration rate at t = 0.

Question:

In a photoelectric effect experiment it is found that for a certain metal surface the kinetic energy of an electron ejected by blue light of wavelength 4.1 × 10^-7 m is 3.2×10^-19 J. (a) What is the work function of the electrons in the metal? (b) What is the longest wavelength of light that will eject electrons from this surface?

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Solution:

(a) Light energy is composed of quanta or photons of energy hf. When photons strike the surface of a metal, they transmit this energy completely to the elec-trons in the metal. The work functionE_wis the minimum energy an electron must acquire if it is to leave the metal's surface. The maximum energy of the electrons leaving the surface is carried by the surface electrons and, by conservation of energy ishf-E_w. Non-surface electrons have less than this amount because of energy losses as they cross the surface. The energy the elec-trons have after they leave the metal is in the form of kinetic energy. Therefore, K =hf-E_w To find the frequency of the photon, we use, f = (c/\lambda) = [(3 × 10^8 m/s)/(4.1 × 10^-7 m)] = 7.3 × 10^14 Hz Substituting into the first equation, K =hf-E_w 3.2 × 10^-19J = (6.6 × 10^-34 J s) (7.3 × 10^14Hz) -E_w E_w= 4.8 × 10^-19 J - 3.2 × 10^-19 J = 1.6 × 10^-19 J (b) The limiting case occurs when the electron kinetic energy is zero. Therefore we set K = 0 to obtain the limiting frequency. hf-E_w= 0 f = (E_w/h) = [(1.6 × 10^-19J)/(6.6 × 10^-34Js)] = 2.4 × 10^14 Hz This corresponds to a wavelength \lambda given by \lambda = (c/f) = [(3 × 10^8 m/s)/(2.4 × 10^14 Hz)] = 1.25 × 10^-6 m

Question:

A sound with a frequency of 1,000 cycles per second is produced when the air temperature is 15\textdegree centigrade . What is the wavelength of this sound?

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Solution:

Velocity (v), frequency (n) and wavelength (\lambda) are related by the relation v =n\lambda. The speed of sound at sea level at 0\textdegreeC is 1090 ft/sec. This speed increases by 2 ft/sec for every degree centigrade above 0\textdegreeC. Hence, Velocity in ft/sec = 1090 + (2 × 15) = 1090 + 30 = 1120 Since v =n\lambda, \lambda = v/n = (1120 ft/sec) / (1000 cycles/sec) = 1.12 ft/cycle.

Question:

The ionization constant for NH_4OH is 1.8 × 10^-5. (a) Calculate the concentration of OH^- ions in a 1.0 molar solution of NH_4OH.

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Solution:

The ionization constant (K_b) is defined as the concentration of OH^- ions times the concentration of the conjugate acid ions of a given base divided by the concentration of unionized base. For a base, BA, K_b = {[B^-][A^+]}/[BA], where K_b is the ionization constant, [B^-] is the concentration of ionized base ions, [A^+] is the concentration of the conjugate acid, and [BA] is the concentration of unionized base. The K_b for NH_4OH is stated as K_b = {[NH_4^+][OH^-]} /[NH_4OH] = 1.8 × 10^-5 When NH_4OH is ionized, one NH_4^+ ion is formed and one OH^- ion is formed, NH_4OH \rightleftarrows NH_4^+ + OH^- Thus, the concentrations of each ion are equal. [NH_4^+] = [OH^-] The concentration of unionized base is decreased when ionization occurs. The new concentration is equal to the concentration of OH^- subtracted from the concentration of NH_4OH. [NH_4OH] = 1.0 - [OH^-] Since [OH^-] is small relative to 1.0, one may assume that 1.0 - [OH^-] is approximately equal to 1.0 [NH_4OH] = 1.0 - [OH^-] \cong 1.0 Using this assumption, and the fact that [OH^-] = [NH_4^+], K_b can be rewritten as K_b = {[OH^-][OH^-]}/[1.0] = 1.8 × 10^-5 Solving for [OH^-] : {[OH^-][OH^-]}/[1.0] = 1.8 × 10^-5 [OH^-]^2 = 1.8 × 10^-5 [OH^-] = \surd(1.8 × 10^-5) = 4.2 × 10^-3

Question:

A 600\rule{1em}{1pt}ohm resistor is in series with a 0.5 henry inductor and a 0.2 \muf capacitor. Compute the impedance of the circuit and draw the vector impedance diagram (a) at a frequency of 400 cycles/sec, (b) at 600 cycles/sec.

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Solution:

The formula for the total impedance is Z = \surd{R^2 + (X_L - X_C)^2} (a) At 400 cycles/sec, X_L = 2\pi × 400 × 0.5 = 1256 ohms, X_C = {1/(2\pi × 400 × 0.2 × 10^\rule{1em}{1pt}6)} = 1990 ohms, X = X_L - X_C = 1256 - 1990 = \rule{1em}{1pt} 734 ohms, Z = \surd{(600)^2 + (\rule{1em}{1pt} 734)^2} = 949 ohms. {see figure, part (a).} (b) At 600 cycles/sec, X_L = 2\pi × 600 × 0.5 = 1885 ohms, X_C = {1/(2\pi × 600 × 0.2 × 10^\rule{1em}{1pt}6)} = 1328 ohms, X = X_L - X_C = 1885 - 1328 = 557 ohms, Z = \surd{(600)^2 + (557)^2 = 818} ohms. {see figure, part (b).}

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Question:

29.7 % copper, 15% sulfur, 2.8 % hydrogen, and 52.5 % oxygen. Determine the empirical formula of this hydrate from these percentages.

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Solution:

A hydrate is the chemical combination of water with another compound. For example, copper sulfate combines with water to form the hydrate of the composition CuSO----_4 . 5H_2O CuSO-_4 + 5H_20 \ding{217} CuS0_4 \textbullet 5H_2O One can find the empirical formula for a compound, when given the weight percents of the various elements making up the compound. This is done by finding the number of moles of each element in 100 g of the compound. The weight of each element is equal to the percent weight. The number of moles is equal to the weight divided by the molecular weight of the element. In 100 g of this compound, there are 29.7 g Cu, 15.0 g S, 2.8 g H, and 52.5 g 0.One can now determine the number of moles of each element present. number of moles = (number of grams) / MW MW of Cu = 63.5, MW of S = 32, MW of H = 1, MW of O = 16. number of moles of Cu =(29.7 g) / (63.5 g/mole) = 0.47moles number of moles of S = (15.0 g) / (32.0 g/mole) = 0.47 mole number of moles of H = (2.8 g) / (1 g /mole) = 2.8 moles number of moles of O = (52.5 g) / (16 g/mole) = 3.28 moles. To determine the empirical formula for this hydrate one must look at the ratio of Cu : S : H : O. The ratio of the number of moles of these elements is Cu : S : H : O .47 : .47 : 2.8 : 3.28 To find the empirical formula, these numbers should be made into integers. This is done by making the lowest number equal to 1 and solving for the other three. .47 / 1 = .47 / xx = .47 / .47 = 1 .47 / 1 = 2.8 / xx = 2.8 / .47 = 6 .47 / 1 = 3.28 / xx = 3.28 / .47 = 7 The ratio now becomes Cu : S : H : O 1 : 1: 6 : 7 The empirical formula for the compound is CuSH_6-O_7 or CuSO_4 \textbullet 3H_2O .

Question:

Deuterons with a mass of 3.3 × 10-27kg may have a velocity of 5 × 107(m/s) and an orbit radius of 0.8 m in cyclotron . (a) Find the frequency at which the accelerating fieldmust change. (b) What is the energy of the deuterons in MeV .

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Solution:

As shown in the figure , the particles move in the circular orbit in the dees under the influence ofthe magneticfield and are accelerated across the gap by the electric field . Therefore , the polarity of the electric field varies with time in such a way that each time the particle enters the gap , it is accelerated rather than decelerated . Deuteronstravel during each cycle a distance D = 2 \pi r = 2 × \pi × ( 8 × 10^-1 m ) = 5.0 m in the cyclotron . The frequency of the electric field should be equal to the frequency of the deuteron's revolutions of the dee . f = (1/period) = [(particle speed)/(distance travelled during one revolution)] = [{5 × 10^7 (m/s)}/5m] = 1 × 107 Hz = 10 MHz. b)Since the deuteron speed is much less than the speed of light we can use the non-relativistic expression for kinetic energy : K = (1/2)mv^2 = (1/2) (3.3 × 10^-27 kg) (5 × 10^7 m/s)2 = 4.13 × 10^-12 J Since 1 ev equals 1.6 × 10^-19 J, the kinetic energy may be written K = [(4.13 × 10^-12J)/{1.6 × 10^-19 (J/eV)}] = 2.58 × 10^7 eV = 25.8 MeV

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Question:

Find the distance between a plane passing through the centers of one close-packed layer of spheres and an-other plane passing through the centers of an adjacent close- packed layer. You may assume that all spheres are identical with a radius of .200 nm.

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Solution:

The small fraction of a space lattice, which is the pattern of points that describes the arrangement of atoms or molecules in a crystal, is a unit cell. The unit cells contain only points that locate atomic or molecular centers. since atoms are space-filling entities, their structures can be described as resulting from the packing together of representative spheres. The most efficient packing together of equal spheres is called closest packing. In this problem, you are asked for a distance between two planes, which is just the spacing between the layers of spheres. A face- centered cube, a type of unit cell, is a close packed structure. From the accompanying figure, you can see that the body diagonal of a face- centered cube is perpendicular to the close-packed stacking layers. Thus, if you stand it on a corner, the close packed layers will be parallel to the floor. To see this more clearly, you can use models. From this, you can relate the body diagonal to the spacing between layers. Namely, body diagonal = 3 times the spacing. Thus, you need to find the length of the body diagonal. Body diagonal = \surd3 (edge length). Edge length= 1/\surd2 (face diagonal) . The face diagonal = 4 times the radii of the spheres, as seen from the space lattice. You are told the radius of a sphere is .200 nm. Thus, face diagonal = 4(.200 nm) = .800 nm. Thus, edge length = (1/\surd2)(.800) = .5657 nm and body diagonal = (\surd3)(.5657) = .9798 nm. Recall, body diagonal is 3 times the spacing, which means the distance between the planes is .9798 nm / 3 = .327 nm.

Question:

How many mercury atoms would there be in a 100 g piece of swordfishsaid to contain 0.1ppm(part per million by weight) ofmercury?

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Solution:

One can find the number of atoms present by multiplying the numberof moles of mercury by the number of atoms in 1 mole . The numberof atoms in one mole is equal to Avogadro'sNumber or 6.02 × 10^23 atoms /mole .To solve for the number of mercury atoms present find: (1). the amount of mercury present (2). the number of moles of mercury (3). the number of atoms of mercury Solving : (1). Because there is 0.1ppmof mercury in theswordfish ,one multiplies0.1 × 10^-6 by the amount of fish present. amountof mercury = 0.1 × 10-6× 100g = 1.0 × 10^-5 g. (2).Thenumber of moles is found by dividing the weight of the mercurypresent by the weight of one mole , the molecular weight (MW =200.6 ) no. of moles = (1.0 × 10^-5 g) / (200.6 g / mole) = 4.99 × 10^-8 moles. (3).Thenumber of atoms is found by multiplying the number of molesby the number of atoms in one mole, 6.02 × 10^23 atoms. no. of atoms = 4.99 × 10^-8 moles × 6.02 × 10^23atoms/mole = 3.00 × 10^16 atoms.

Question:

The equation for the burning of naphthalene is C_10H_8 (s) + 12O_2(g) \rightarrow1OCO_2 + 4H_2 O (l). For every mole of C_10H_8 burned, - 1226.7 Kcal is evolved at 25\textdegree in a fixed- volume combustion chamber. ∆H\textdegree for H_2O(l) = - 64.4 Kcal/mole and H_2O (g) = - 57.8 Kcal/mole. Calculate (a) the heat of reaction at constant temperature and (b) ∆H for the case where all of the H_2O is gaseous.

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Solution:

This problem deals with the heat evolved when a compound is heated with oxygen to form carbon dioxide and water (the heat of combustion) . The heat of reaction, or combustion in this case, ∆H, is given by the formula ∆H = ∆E + ∆nRT, where ∆E = amount of heat released per mole, ∆n = change in moles, R = universal gas constant and T = temperature in Kelvin (Celsius plus 273\textdegree). Thus, to answer (a) substitute these values and solve for ∆E. ∆n is moles of gas produced - moles of gas reacted, which is 10 moles - 12 moles= -2 moles based on the coefficients in the equation. T = 25\textdegree + 273 = 298\textdegreeK. Use R in terms of kilocalories. As such ∆H = - 1226.7 kcal + (- 2 moles)(1.987 cal/mole \textdegreeK) (298\textdegreeK) (1 Kcal/1000 cal) = - 1226.7 - 1.2 = - 1227.9 Kcal/mole. To find (b), note that the reaction is the. same, except that H_2O is gaseous not liquid. You have, there-fore, 4H_2O(liq) \rightarrow 4H_2O(g). The ∆H, change in enthalpy, for this conversion is 4∆H\textdegree(liq) - 4H\textdegree∆(g) of H_2O = 4(- 57.8) - 4(- 68.4) = 42.4 Kcal/mole. It follows, then, that the resulting ∆H = - 1227.9 + 42.4 = - 1185.5 Kcal/mole.

Question:

Why is the phenomenon of diffusion important tomovement ofmaterials in living cells?

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/Users/wenhuchen/Documents/Crawler/Biology/F02-0059.htm

Solution:

In a living cell, chemical reactions are con-stantly taking place to producethe energy or organic com-pounds needed to maintain life. The reactingmaterials of chemical reactions must be supplied continuously to theactively metabolizing cell, and the products distributed to other parts of thecell where they are needed or lower in concentration. This is extremely importantbecause if the reactants are not supplied, the reaction ceases, andif the products are not distributed but instead accumulate near the site ofreaction,LeChatelier'sPrinciple of chemical reactions operates to drive thereversible reaction backward, diminishing the concentration of the products. Thus, in order to maintain a chemical reaction, the re-actants mustbe continuously supplied and the products must move through the cellmedium to other sites. Diffu-sion is how these processes occur. When a certain chemical reaction is operating in the cell, some reactingsubstance will be used up. The concen-tration of this substance is necessarilylower in regions closer to the site of reaction than regions fartheraway from it. Under this condition, a concentration gradient is established. The concentration gradient causes the movement of moleculesof this substance from a region of higher concentration to a regionof lower concentration, or the reaction site. This movement is called diffusion. Thus, by diffusion, molecules tend to move to regions in the cell wherethey are being consumed. The products of the reaction travel away fromthe reaction site also by this process of diffusion. At the reaction site, thecon-centration of the products is highest, hence the products tend to moveaway from this region to ones where they are lower in concentration. The removal of products signals the reaction to keep on going. When the productconcen-tration gets too high, the reaction is inhibited by a built-in feedbackmechanism. Thus diffusion explains how movement of chemical sub-stances occursinto or out of the cell and within the cell. For example, oxygen moleculesare directed by a concentra-tion gradient to enter the cell and movetoward the mitochondria. This is because oxygen concentration is nec-essarilythe lowest in the mitochondria where oxidation reactions continuallyconsume oxygen. Carbon dioxide is produced when an acetyl unitis completely oxidized in the citric acid cycle. The CO_2 will then travel awayfrom the mitochondria, where it is produced, to other parts of the cell, or out of the cell into the bloodstream where it is lower in concentration.

Question:

Compute the force of gravitational attraction between the large and small spheres of a Cavendish balance, if m = 1 gm, m' = 500 gm, r = 5 cm

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/Users/wenhuchen/Documents/Crawler/Physics/D04-0138.htm

Solution:

Two uniform spheres attract each other as if the mass of each were concentrated at its center. By Newton's Law of Universal Gravi-tation, the force of attraction between 2 masses m and m' separated by a distance r is F =Gmm'/r^2 = {(6.67 × 10^-8dyne\bulletcm^2/gm^2) × (1 gm) × (500 gm)}/(5 cm^2) = 1.33 × 10^-6dyne, or about one-millionth of a dyne.

Question:

PROGRAM EXAMPLE (output); VAR A, B, C, D: integer; PROCEDURE PI (a, b, c,d :integer); BEGIN A: = A + B + C; B: = C + A END; PROCEDURE P2 (a, b: integer; VAR C:integer); BEGIN A: = A + B \textasteriskcentered C; B: = B - A; C: = A; END; PROCEDURE P3 (VAR a, b: integer); VARC :integer; BEGIN C: = 10; A: = A + C; B: = B + D; C: = A - B; END; BEGIN{main part of the program} A: = 1; B: = 2; C: = 3; D: = 4; P1 (a, b, c, d); write1n (a ,b , c ,d); P2 (a,b ,c); write1n(a , b , c ,d); P3 (a, b);write1n(a , b ,c ,d) END.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G16-0401.htm

Solution:

The key to understanding what this program doesis understanding the difference between calls by reference and calls by value. When procedure P1 is invoked, param-eters A,B,C and D are passed to it. Since these param-eters are not preceded byVARin the procedure declaration, these are not variable parameters. This is known as call by value - i.e. only values of these parameters are passed, to the procedure, and if their values are changed during the execution of that procedure, these changes do not become permanent. Therefore, executing procedure P1 doesn't really change the values of A, B, C, and D, and the first write1n statement prints 1 2 3 4. In the second procedure P2, C is variable parameter- meaning that its memory location is passed to the procedure, and consequently, any changes in its value become permanent. This is an example of a call by reference . During the execution of the procedure, A becomes equal to 7, B to -5, and C to 7. Since C is variable parameter, it passes this value back to the variable C in the main program. The output after second WRITELN statement is then 1274. In the third procedure P3, C is declared as a local variable, and in all subsequent references to it within that procedure it is treated as such, having nothing to do with the C in the main program. Executing the statements of the procedure, we get: C:=10; A:= 11, B:= 6; C:= +5. A and B are variable parameters, their values are passed back to the calling variables. C is a local variable, which is "lost" after the procedure is executed , the value of C in the main program remains 7, and D retains its original value of 4. So the output of the final write1n statement is 11 6 7 4.

Question:

Mrs. Doe and Mrs. Roe had babies at the same hospital at the same time. Mrs. Roe brought home a baby girl and named her Nancy. Mrs. Doe received a baby boy and named him Richard. However, she was sure she had had a girl and brought suit against the hospital. Blood tests showed that Mr. Doe was type 0, Mrs. Doe was type AB, and Mr. and Mrs. Roe were both type B. Nancy was type A and Richard type 0. Had an exchange occurred?

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/Users/wenhuchen/Documents/Crawler/Biology/F25-0664.htm

Solution:

Inheritance of blood groups is an example of a trait controlled by multiple alleles. The term multiple alleles is applied to three or more genes that can control a single trait; different combinations of any two genes may be present in a gene pair determining that trait in an individual. Any individual in the population may have any two of the possible alleles, but never more than two, because only two genes for a particular trait can be carried by an individual. Any gamete may have only one of the possible alleles. However, in the population as a whole, three or more different alleles will occur. The blood types of man, 0, A, B and AB, are coded for by multiple alleles. Gene I^A provides the code for the synthesis of a specific protein, agglutinogen A, in the red cells. Gene I^B leads to the production of a different protein, agglutinogen B. Gene i produces no agglutinogen. Gene i is recessive to the other two genes, but neither gene I^A nor I^B is dominant to the other. The symbols I^A, I^B and i are used to emphasize that all three alleles are alleles at the same locus. Individuals with genotypes I^A I^A and I^A i make up blood group A, and produce agglutinogen A. Those with genotypes I^B I^B and I^B i compose blood group B, and produce agglutinogen B. Blood group 0 individuals have genotype ii, and produce no agglutinoqens. When an individual has the genetic make-up of I^A I^B, he has both agglutinogens A and B and he belongs to blood group AB. These blood types are genetically determined and do not change during an individ-ual's lifetime. In reference to the problem given, it is possible to determine if an exchange occurred by comparing the blood-types of the babies with the possible bloodtypes that could be found in any offspring of each set of parents. Inother words, we can cross the genotypes of each set of other words, we can cross the genotypes of each set of parents and determine what genotypes are possible for their offspring. Then we can see if the bloodtypes of the babies that each family brought home is compatable with the possibilities, and if the parents could then have indeed produced a child with that given bloodtype. We are told that Mrs. Doe is type AB. Therefore, her genotype must be I^A I^B. Mr. Doe is type 0, so his genotype is ii. The possible bloodtypes of any Doe family offspring are obtained as follows: Thus, Mr. and Mrs. Doe can only produce offspring having bloodtypes A or B. In the Roe family, we are told that Mr. and Mrs. Roe are both type Roe are both type B. Therefore, each parent has one of two possible genotypes,. I^B I^B or I^B i. Taking each possiblility: Thus, the Roes can produce only offspring having bloodtype B or 0. Now let us look at the bloodtypes of the babies that each family brought home. The Does brought home Richard, who is type 0. But, as we have seen, the Does can produce only offspring of type A or B. The Roes brought home Nancy, who is type A. But it would be impossible for the Roes to have a child with bloodtype A, for the only possible bloodtypes for their offspring are B and 0. Therefore Richard, who is type 0, must be their child, and Nancy who is type A, must be the Doe's daughter. We see that an exchange did indeed take place.

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Question:

A curling stone of mass 1 slug is sent off along the ice and comes to rest after 100 ft. If the force of kinetic friction between stone and ice is 1/2 lb, with what velocity did the stone start, and how long did it take to come to rest?

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/Users/wenhuchen/Documents/Crawler/Physics/D04-0128.htm

Solution:

The force decelerating the stone is 1/2 lb and the stone has a mass of 1 slug. Using the equation F = ma, the deceleration is a = -(1/2) ft/s^2. Apply the equation of uniform motion, v2_0 + 2a (x - x_0) = v, where v_0 is the stone's initial velocity and x - x_o is the distance it travels. We obtain v2_0 = 2 × (1/2) ft/s^2 × 100 ft \therefore v_0 = 10 ft/s, which is the initial velocity of the stone. If we now apply the further equation v =v_o+ at, the time of motion is t = -(v_0/a) = [10 ft/s] / [1/2 ft/s^2] = 20 s.

Question:

What is the half-life of an unstable substance if 75% of any given amount of the substance decomposes in one hour?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E13-0460.htm

Solution:

The half-life is defined as the time it takes for one half of the amount of a substance to decompose. When given the time elapsed and the percent of decompo-sition, one can find the half-life. One knows that 1/2 of the substance decomposes in the time equal to the half- life. This leaves 1/2 of the substance, 1/2 of this decom-poses in the next span of timeelapsed equal to the half- time. This leaves 1/4 of the substance. 75% of ithas de-composed after two half-life have elapsed. Thus 2 half- lifesequalone hour or the half-life of the substance is 1/2hour.

Question:

Give the set of arithmetic operators used in PL/I. Explain whatis meant by the priority rule. Explain how the expres-sionA + B - C \textasteriskcentered D/F \textasteriskcentered\textasteriskcentered G is evaluated. Also, find the valueof the above expression if A = 7, B = 2, C = 3, D = 8, F = 2, and G = 3.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G13-0314.htm

Solution:

Every language has some fixed set of arithmetic operators. The normalalgebraic operations are all available in PL/I. Arithmetic operators aresymbols used to indicate these operations. The following operators areavailable: a)Infix Operators: +addition- subtraction \textasteriskcenteredmultiplication/ division \textasteriskcentered\textasteriskcenteredexponentiation b)Prefix Operators +plus -negationor minus The operators are shown in two groups above. Operators in group a)arecalled infix operators because they operate on two operands. For example , A+B, 2\textasteriskcentered3, X\textasteriskcentered\textasteriskcenteredY, 1 - 1, u/v, etc. The operators in group b) are called prefix operators because they operateon single operands. For example, + 10, - 52, - (A + B) Notice that in the - (A + B), the + sign is for an infix opera-tor, between two operandsA and B. But, the - sign is a pre-fix operator. The complete parentheses, (A + B), acts as a single operand for the prefix - operator. Now consider the given expression: A + B - C\textasteriskcenteredD/F\textasteriskcentered\textasteriskcenteredG. Notice that if parentheses are omitted as in the above, the expression will evaluateto different answers, depending on the order in which the differentoperations are carried out. To avoid any ambiguity on the evaluation process, PL/I has a set of Precedence Rules.The Precedence Rule says that an operator at a higherlevel of priority is executed first, and then an operator at a lower levelof priority is execut-ed. The various operators are set at different prioritylevels. An operator having a higher priority level is exe-cuted first. The priority set for the operators are asfol-lows : First level \textasteriskcentered\textasteriskcentered, Negation(highest priority) Second level\textasteriskcentered ,/(second highest priority) Third level +, -(lowest priority). If two operators at the same level of priority are encounter-ed, then PL/I appliesa left to rightassociativity, e.g. A-B-C is considered as (A - B) - C. Arithmetic expressions are analyzed as follows: 1. The expression is examined from right to left in search of the two firstlevel operators. The operations specified by them are performed as theoperators are encoun-tered. 2. All subsequent scans take place from left to right. During the first ofthese left to right scans, \textasteriskcentered and / op-erations are performed as they occur. 3. The final left to right scan carries out calcula-tions involving infix + and- operations. Using the above rules of Precedence andAssociativity, ambiguity is eliminated. Hence, consider the given expression, viz., A + B - C\textasteriskcenteredD/F\textasteriskcentered\textasteriskcenteredG. In the above, it can be seen that there is an exponentiation operator \textasteriskcentered\textasteriskcentered, whichhas the highest priority from amongst all the operators in the given expression. Hence, F\textasteriskcentered\textasteriskcenteredG is eval-uated first. The expression therefore reduces as shown below: A + B - C\textasteriskcenteredD/(F\textasteriskcentered\textasteriskcenteredG) . The highest priority for the remaining operators belongs to \textasteriskcentered and /. Now, both\textasteriskcentered and / are on the same level of prior-ity. Hence, the left to right AssociativityRule is applied. Thus, C\textasteriskcenteredD is evaluated first. The expression thusreduces as shown below: A + B - (C\textasteriskcenteredD)/(F\textasteriskcentered\textasteriskcenteredG). Now, the / operation is performed. The expression then re-duces as shown below: A + B - (C\textasteriskcenteredD/F\textasteriskcentered\textasteriskcenteredG). Now remains the + and - operators. Both are again at the same level of priority. Hence, the left to rightAssociativ-ityrule is again applied, and A+B iscarried out first. This reduces the expression to that shown below: (A + B) - (C\textasteriskcenteredD/F\textasteriskcentered\textasteriskcenteredG) . Now, the subtraction is finally carried out, completing the evaluation of the expression. Using the numeric values of the various variables, the expression becomesas follows: 7 + 2 - 3\textasteriskcentered8/2\textasteriskcentered\textasteriskcentered3. = 7 + 2 - 3\textasteriskcentered8/(2\textasteriskcentered\textasteriskcentered3) = 7 + 2 - 3\textasteriskcentered8/(8) = 7 + 2 - (3\textasteriskcentered8)/(8) = 7 + 2 - (24)/(8) = 7 + 2 - (24/8) = 7 + 2 - (3) = 7 + 2 - 3 = (7 + 2) - 3 = (9) - 3 = 9 - 3 = 6

Question:

Distinguish betweenCebidae(New World monkeys) and Cercopithecidae (Old World monkeys).

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Solution:

The first members of the suborderAnthropoi-deadiverged from the prosimians during the Oligocene era. Theprosimiansare the earliest known primates in evolutionary history. Two lines of anthropoid monkeys (CatarrhinesandPlatyrrhines) , probably arose at about the same time from closely relatedprosimianancestors. One of the lines (Platyrrhines) gave rise to the super familyCeboideaand the other line (Catarrhine) gave rise to the twosuperfamiliesCercopithecoideaandHominoidea. TheCebidaeis a family grouping in the super familyCeboidea, and is therefore a New World monkey. In addition to theCebidae, there is another family in thesuperfamilyCeboidea, theCallithricidaeor marmosets. The familyCercopithecidaeis in thesuperfamily Cercopithecoidea and are thus Old World monkeys. The family from which man descended is theHominidaewhich is in thesuperfamilyHominoidea, the second lineage ofcatarrhinemonkeys. The members of the familyCebidaeare New World monkeys, with the adjective "New World" referring to their presence in South America and Central America. Theceboids, isolated in South America, underwent evolution independent of the members of theCercopithecoideaand Hominoidea . Among the livingceboidsare the howler monkey, the marmoset (squirrel monkey), the capuchin monkey (organ grinder's monkey) and the spider monkey. All these monkeys are found in tropical forest habitats and show a wide variety of adaptations to arboreal life, some of which parallel those of the Old World monkeys. Thecercopithe- coids are Old World Monkeys, "Old World" referring to their widespread distribution in Africa and Southeast Asia. The large group of Old World monkeys includes the macaque, mandrill,mangabey, baboon,langur, proboscis monkeys and others. They live in a variety of environments, both arboreal and terrestrial, from grassland savannas to jungle forests. The New World and Old World monkeys differ in many ways and most differences are readily observable. Most New World monkeys, have a prehensile tail that they use almost like another hand for grasping objects and hanging from trees. The prehensile tail has a tactile pad at the tip with ridge patterns similar to fingerprints. The tail of the Old World monkeys is not prehensile, but is used instead for balance. The nostrils of theCeboidea(New World monkeys) are separated by a wide partition, and are flat and facing in a lateral (outward) position. This condition is known asplatyrrhini, hence the nameplatyrrhinimonkey. Theceboidsare allplatyrrhinimonkeys. The nostrils of thecercopithecoidsand hominoids are closer together and are directed downward and forward. This condition is known ascatarrhini, and thus primates with nostrils of this type are calledcatarrhinemonkeys. Thecercopithecoidsand hominoids are all classified ascatarrhinemonkeys. New World monkeys have no adaptations for terrestrial life styles. Old World monkeys tend to sit upright, and have buttocks with a hardened, insensitive sitting pad called the ischial callosities. Theischialcallosities is usually very brightly colored, being either red or blue. Old World monkeys possess distinctive molars (cheek teeth), which can be used to distinguish them from the New World monkeys. These are known as thebilophodentmolars. Furthermore Old World monkeys have a bony auditory tube which New World monkeys do not have. This may aid New World monkeys in distinguishing the direction of sounds.

Question:

A physicist is trying to locate a sub-atomic particle. Using diffraction theory he obtains information as to where his plotted coordinates are in relation to the particle. For example, suppose the particle is at (11, 9, 32) and the physicist bombards (2, 16, 21). His experimental data then tells him that he is to the North-west and too low. The physicist (player) specifies the dimension of the search area and the number of trials allowed before the particle disintegrates without being observed.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G23-0564.htm

Solution:

In the program, the dimension of the search area is left variable, as are the number of trials. The direc-tions are given according to the orientation. (0,0,0) is the origin of the grid and all coordinates are positive. 10PRINT "QUIRKY QUARKS" 20INPUT "DIMENSION OF SEARCH AREA"; S: PRINT 30N = INT (LOG(S)/LOG(2)) + 1 40PRINT "YOU ARE AN EXPERIMENTAL PHYSICIST,"; 50PRINT "TRYING TO LOCATE A SUB-ATOMIC PARTICLE." 60PRINT "YOU HAVE";N;"TRIES. SPECIFY GUESS OF 70PRINT "LOCATION WITH A TRIO OF INTEGERS." 80A = INT (S\textasteriskcenteredRND (1) ) : B = INT (S\textasteriskcenteredRND(1)): C = INT (S\textasteriskcenteredRND (1)) 90FOR D = 1 TO N: PRINT: PRINT "TRIAL #";D;: INPUT X1,X2, X3 100IF ABS (X1 - A) + ABS (X2 - B) + ABS (X3 - C) = 0 THEN 150 110GOSUB 190: PRINT: NEXT D 120PRINT: PRINT "PARTICLE HAS DISINTEGRATED.DON RADIATION"; 130PRINT "SUIT" 140PRINT "THE PARTICLE WAS AT" A;",\textquotedblrightB;",";C 150PRINT: PRINT "YOU FOUND THE PARTICLE IN" ;D; "TRIES" 160PRINT: PRINT: INPUT "ANOTHER GAME(Y OR N)" ;A$ 170IF A$ = "Y" THEN 80 180PRINT "OK.SEE YOU IN OSLO": GO TO 290 190PRINT "CALCULATIONS SHOW, THAT YOUR TRY IS"; 200IF X2 > B THEN PRINT "NORTH"; 210IF X2 < B THEN PRINT "SOUTH"; 220IF X1 > A THEN PRINT "EAST"; 230IF X1 < A THEN PRINT "WEST"; 240IF X2 <> B OR X1 <> A THEN PRINT "AND"; 250IF X3 > C THEN PRINT "TOO LOW." 260IF X3 < C THEN PRINT "TOO HIGH." 270IF X3 = C THEN PRINT "HEIGHT OK" 280RETURN 290END

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Question:

How many gallons of a liquid that is 74 percent alcohol must be combined with 5 gallons of one that is 90 percent alcohol in order to obtain a mixture that is 84 percent alcohol?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E33-0943.htm

Solution:

If we let x represent the number of gallons needed of the first liquid and remember that 74 percent of x is 0.74x, then the table (see table) shows all the data given in this problem. Number of gallons Percentage of alcohol Number of gallons of alcohol First liquid X 74 0.74x Second liquid 5 90 0.90 (5) = 4.5 Mixture x + 5 84 0.84 (x + 5) We are told that we are combining the number of gal-lons of alcohol in the 74 percent alcohol (0.74x) with the number of gallons of alcohol in the 90 percent alcohol (4.5) to obtain the number of gallons of alcohol in the 84 percent alcohol [0.84 (x + 5)]. Thus .74x + 4.5 = .84(x + 5) Multiplying both sides by 100, 74x + 450 = 84 (x + 5) 74x + 450 = 84x + 420 30 = 10x x = 3 Therefore, 3 gallons of liquid that is 74 percent alcohol must be combined with 5 gallons of one that is 90 percent alcohol to obtain a mixture of 84 percent alcohol.

Question:

A solution of hydrogen peroxide is 30% by weight H_2O_2. Assuming a density of 1.11 g/cm^3 and a dissociation constant of 1.0 × 10^-12 for H_2O_2, what is the pH of the solution?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E12-0432.htm

Solution:

To obtain the pH of any solution, calculate [H^+], since pH = - log [H^+]. To find [H^+], write a balanced chemical equation that expresses the reaction, and from this, set up a dissociation constant expression. Begin by writing the reaction. H_2O_2 \rightleftarrows H^+ + HO_2^-. The dissociation constant for this reaction, K, equals 1.0 × 10^-12. K also = {[H^+] [HO_2^-]} / [H_2O_2]. Equating, one obtains {[H^+] [HO_2^-]} / [H_2O_2] = 1.0 × 10^-12. Solving this expression for [H^+], one obtains thepH.To do that it is necessary to know [H_2O_2]. Given the density, one can assume a volume of 1 liter solution. This means that the mass of the solution is 1.11 g/cm^3 × 1000 ml (or 1 liter) = 1110 g/l. However, the percent by weight of H_2O_2 is 30%. Therefore, there is only 0.30 1110 g/l of H_2O_2 or 333 g/l. The molecular weight of H_2O_2 is 34 g/mole. Themolarityof the H_2O_2 solution is, then, (333 g/liter) / (34 g/mole) = 9.79 M H_2O_2, sincemolarityequals moles per liter. If one lets x = [H^+], then [H_2O_2] = 9.79 - x, since, at equilibrium, the original amount of H_2O_2 must be decreased by the amount of H^+ (or O_2H^-) formed. The equation, H_2O_2 \rightleftarrows H^+ + HO_2^- indicates that H^+ will be formed in equal mole amounts with HO_2^-. Thus, [HO_2^-] can equal x. Substituting these values into the expression: {[H^+] [HO_2^-]} / [H_2O_2] = 1.0 × 10^-12, (x \bullet x) / (9.79 - x) = 1.0 × 10^-12. Solving for x, x = 3.13 × 10^-6 = [H^+]. Since pH = - log [H^+], pH = 5.5.

Question:

What are the chief differences between plant andanimal cells?

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/Users/wenhuchen/Documents/Crawler/Biology/F02-0033.htm

Solution:

A study of both plant and animal cells reveals the fact that in their mostbasic features, they are alike. However, they differ in several importantways. First of all, plant cells, but not animal cells, are surroundedby a rigid cellulose wall. The cell wall is actually a secretion fromthe plant cell. It sur-rounds the plasma membrane, and is responsible forthe maintenance of cell shape. Animal cells, without a cell wall, cannot maintaina rigid shape. Most mature plant cells possess a single large central fluid sap, the vacuole. Vacuoles in animal cells are small and frequently numerous. Another distinction between plant and animal cells is that many of thecells of green plants contain chloro-plasts, which are not found in animalcells. Thepresence of chloroplasts in plant cells enablegreen plantsto beautotrophs, organisms which synthesize their own food. As is generallyknown, plants are able to use sunlight, carbon dioxide and water togenerate organic substances. Animal cells, devoid of chloroplasts, cannotproduce their own food. Animals, therefore, areheterotrophs, organismsthat depend on other living things for nutrients. Some final differences between plant and animal cells are in the processof cell division. In animal cells, undergoing division, the cell surfacebegins to constrict, as if a belt were being tightened around it, pinchingthe old cell into two new ones. In plant cells, where a stiff cell wall interfereswith this sort of pinching, new cell membranes form between the twodaughter cells. Then a new cell wall is deposited between the two new cellmem-branes. During cell division, as the mitotic spindle apparatus forms, animal cells have two pairs ofcentriolesattached to the spindles at oppositepoles of the cell. Even though plant cells form a spindle apparatus, most higher plants do not containcentrioles.

Question:

Write a program in APL to compute the gross pay of any num-ber of employees.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G14-0379.htm

Solution:

Although APL is slanted towards mathematical problems, it may also be used in clerical functions that involve repetition. The program below lists the name of each employee and his/her gross pay. Hours worked and rate of pay are input but do not form part of output . \nablaPAYDAY [1]NAME\leftarrow'' [2]GRS \leftarrow '' [3]'ENTER THE NUMBER OF EMPLOYEES' [4]N \leftarrow [] [5]I \leftarrow 0 [6]REPEAT: 'ENTER NAME' [7]NAME \leftarrow NAME, 20\uparrow[] [8]'ENTER HOURS AND RATE' [9]GRS \leftarrow GRS, ×/[] [10]I \leftarrow I + 1 [11]\rightarrow REPEAT ×ʅ(I < N) [12]NAME \leftarrow (N,20) \rho NAME [13]GRS \leftarrow (N,1) \rho GRS [14]'' [15]'NAME -------------------- GROSS PAY' [16]NAME, 7 2 DEF GRS [17]\nabla Note the following points about the program: Statements [6] and [7] ensure that all names are entered. Statement [11] corresponds to the CONTINUE statement in FORTRAN. Names will continue to be entered and gross pay computed while (I < N) . When I = N,ʅ(I < N) is 0. But since there is no statement numbered 0, the program comes to a halt. Statement [14] produces a line of blank spaces between the input and output phases of the program.

Question:

The current from a dc supply is carried to an instrument by two long parallel wires, 10 cm apart. What is the magnetic flux density midway between the wires when the current carried is 100 A?

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/Users/wenhuchen/Documents/Crawler/Physics/D21-0703.htm

Solution:

The magnetic field due to each wire in the diagram at the point midway between them will be into the paper. This may be seen by use of the right hand rule. If the thumb of the right hand points in the direction of current through the wire, then the fingers will curl in the direction of the magnetic field (or magnetic flux density) created by the current. Applica-tion of this rule to both current carrying wires in-dicates that the field of each is into the page (see figures A and B). The effects due to the wires are therefore additive at that point and the total effect is twice the effect of either alone. Hence, midway between the wires the magnetic field due to one wire is B = (\mu_0 /2\pi) (I/r) where the permeability \mu_0 = 4\pi × 10^-7 N - A^-2 , I is the current through the wire, and r is the distance from the point being considered to the wire . Thus B = 2 × 10^-7 N - A^-2 × [(100 A)/(0.05 m)] = 4 × 10^-4 × - m^-2 The magnetic field due to both wires is then B_T = 2B = 8 × 10^-4 Wb - m^-2 .

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Question:

Of the several ways in which the supersonic transport (SST) acts as a pollutant, one is by the elimination of carbon dioxide (CO_2). During normal flight, CO_2 is released at a temperature of about 627\textdegreeC and at a press-ure of 0.41 atm at an estimated rate of 6.6 × 10^4 kg per hour. To what volume does this amount of gas correspond?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E02-0069.htm

Solution:

This problem is an application of the ideal gas law, PV = nRT, where p = pressure, V = volume, n = number of moles, R = gas constant, and T = absolute tem-perature. Solving for V, V = (nRT/P) We know that R = 0.082 (liter - atm/mole -\textdegreeK), T = 627\textdegreeC = 900\textdegreeK, and P = 0.41 atm. n is obtained by dividing the mass of CO_2 (6.6 × 10^4 kg = 6.6 x 10_4 kg × 10^3 g/kg = 6.6 × 10^7 g) by the molecular weight of CO_2. The molecular weight of CO_2 is equal to atomic mass_C + 2 atomic mass_O = 12 g/mole + 2 (16 g/mole) or 44 g/mole.Hence n = [(mass CO_2 )/(molecular weight CO_2 )] = [(6.6 × 10^7 g)/(44 g/mole) = 1.510^6moles Substituting the values of n, R, T, and P into the equation for V Gives V = (nRT/P) = [(1.5 × 10^6 moles × 0.082 liter-atm/mole-\textdegreeK× 900\textdegreeK) /(0.41 atm)] = 2.7 × 10^8 liters.

Question:

In the early 1900's, many fruit growers made a practice of ripening fruits by keeping them in a room with a kerosene stove. They believed it was the heat that ripened the fruits. Was this belief based on correct facts? Explain your answer.

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/Users/wenhuchen/Documents/Crawler/Biology/F09-0233.htm

Solution:

Fruit growers long believed that the heat generated from a kerosene stove had the effect of ripening fruit. However, as experiments showed, it was actually the incomplete combustion products of the kerosene that were responsible for the ripening effect. The most active component of the incomplete combustion products was iden-tified as ethylene. As little as 1 part per million of ethylene in the air will speed the onset of fruit-ripening. The onset of fruit-ripening occurs when there is a large increase in cellular respiration, as evidenced by a rapid uptake of oxygen. This increase in cellular res-piration is triggered by the sudden production of ethylene, now thought to be a natural growth regulator in plants. The sudden production of ethylene, in turn, is induced byauxin. It is believed that some of the effects on fruits once attributed toauxinare related actually toauxin'seffects on ethylene production. When a fruit ripens rapidly, its production of ethylene may increase many times over the rate initially necessary to reach the critical triggering level. Thus, "one bad apple in a barrel" producing ethylene can produce enough ethylene to trigger the ripening of the rest. Also, the ethylene produced by the mold on an orange may be enough to set off undesirable changes in the rest of the oranges. Unfortunately, the exact mechanism of ethylene in fruit-ripening is not known, but is a subject currently under much investigation.

Question:

Some solidCaOin a test tube picks up water vapor from the surroundings to change completely to Ca(OH)_2 (s). An ob-served total initial weight (CaO+ test tube) of 10.860 g goes eventually to 11.149 g. What is the weight of the test tube?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E05-0181.htm

Solution:

The equation for the reaction is CaO+ H_2 O \rightarrow Ca (OH)_2 This means that one mole of H_2 O reacts with one mole ofCaO. The difference in the weights of the test tubes is the weight of the H_2 O that the CaO absorbed. weight of H_2 O = 11.149 - 10.860 = .289 g. One should now solve for the number of moles of H_2 O because the number of moles of water equals the number of moles ofCaOpresent. From this one can find the weight of theCaOand the test tube. The number of moles equals the number of grams divided by the molecular weight (MW of H_2 O = 18.0). number of moles of H_2 O = [(.289 g) / (18.0 g/mole) = .0161 moles Therefore, .0161molesofCaOwere originally present in the test tube. One finds the number of grams by multiplying by the molecular weight of CaO (MW ofCaO= 56.08). number of grains ofCaO= .0161 moles × 56.08 g/mole = .900 g. The weight of the test tube is equal to .900 g sub-tracted from the original weight of the test tube and material. weight of test tube = 10.860 - .900 = 9.960 g.

Question:

What exactly is a seed? What tissues are present and what are their respective functions?

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/Users/wenhuchen/Documents/Crawler/Biology/F08-0198.htm

Solution:

A seed is actually a matured ovule, and consistsof a seed coat surrounding a core of nutritive tissue in which the embryo is embedded. The seed is an interesting structure in that it is composed of tissues from three generations. The embryo consisting of 2n cells derived from the fusion of egg and sperm, is the new sporophyte generation and functions in the continuation of the species by developing into a new reproducing plant. The nutritive tissue, or endosperm, is derived from the female gametophyte. In gymnosperms it is haploid, but in angiosperms it is triploid, resulting from the fusion of both polar bodies with the sperm. Its high starch content provides a source of nourishment for the growing embryo. The seed coat differentiates from the outer layer of the ovule, known as the integument, and as such is 2n and belongs to the old sporophyte generation. The seed coat encloses the endosperm and embryo, and owing to its tough, resistant properties, protects the seed from heat, cold, desiccation, and parasites.

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Question:

Write possible sets of quantum numbers for electrons in the second main energy level.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E18-0649.htm

Solution:

In wave mechanical theory, four quantum numbers are needed to describe the electrons of an atom. The first or principal quantum number, n, designates the main energy level of the electron and has integral values of 1, 2, 3, ... . The second quantum number, l, designates the energy sublevel within the main energy level. The values of l depend upon the value of n and range from zero to n - 1. The third quantum number,m_l, designates the particular orbital within the energy sublevel. The number of orbitals of a given kind per energy sublevel is equal to the number ofm_l values (2l + 1) . The quantum numberm_lcan have any integral value from + l to - l including zero. The fourth quantum number, s, describes the two ways in which an electron may be aligned with a magnetic field (+ 1/2 or - 1/2). The states of the electrons within atoms are described by four quantum numbers, n, l,m_l, s. Another important factor is the Pauli exclusion principle which states that no two electrons within the same atom may have the same four quantum numbers. To solve this problem one must use the principles of assigning electrons to theirorbitals. If n = 2; l can then have the values 0, 1;m_lcan have the values of + 1, 0 or - 1, and s is always + 1/2 or - 1/2. Thus, the answer is: n= 2 l= 0, 1 m_l = + 1, 0, - 1 s= + 1/2, - 1/2 n l m_l m_s 2s 2 2 0 0 0 0 +1/2 -1/2 2p 2 2 1 1 +1 +1 +1/2 -1/2 2 2 1 1 0 0 +1/2 -1/2 2 2 1 1 -1 -1 +1/2 -1/2

Question:

There are twelve pairs of cranial nerves in man. Give the type(s) of fibers found in each and briefly discuss the function(s) of each of them.

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/Users/wenhuchen/Documents/Crawler/Biology/F20-0511.htm

Solution:

Cranial nerves leave the brain at different regions and serve different functions. They connect the brain to variouseffectororgans, primarily the sense organs, muscles and glands of the head. Some cranial nerves contain only motor fibers, some contain only sensory fibers, and some contain both types of fibers. In addition, some are composed of parasym-pathetic fibers (which are exclusively motor), the action of which is involuntary. There are twelve pairs of cranial nerves in man. The olfactory nerve (cranial nerve I) is composed en-tirely of afferent fibers carrying impulses for the sense of smell from the olfactory epithelium to the base of the brain. The optic nerve (II) is also entirely sensory and contains afferent fibers running from the retina to the visual center of the brain. The occulomotor (III) andtrochlear(IV) nerves both have afferent and efferent branches which connect the midbrain to the muscles of the eye. Together they are responsible forproprioceptionof the eye muscles, movement of the eyeball,accomodationof the eye, and constriction of the pupil. The trigeminal nerve (V) contains both afferent and efferent fibers running between theponsand the face and jaws. It functions mainly in stimulating movement of the muscles of the jaws involved in chewing. The fibers of theabducensnerve (VI) run between theponsand muscles of the eye. This nerve conveys the sense of position of the eyeball to the brain, and aids the III and IV nerves in effecting mo-vement of the eyeball. The facial nerve (VII) has both afferent and efferent fibers that innervate muscles of the face, mouth, forehead and scalp. It also functions in the transmission of impulses for the sense of taste from the anterior part of the tongue to the brain. The fibers of the auditory nerve orvestibulocochlear(VIII) exclusively afferent, run from the inner ear to the junction of theponsand medulla. These fibers convey the senses of hearing and equilibrium (movement, balance and rotation) to the appropriate centers in the brain. Connecting the medulla to the epithelium and muscles of the pharynx, and to the salivary gland and tongue are the fibers of theglossophar-yngeal nerve (IX) This nerve is responsible for the sense of taste from the posterior part of the tongue and from the lining of the larynx. It is also responsible for the reflex-ive act of swallowing. The tenth cranial nerve, the vagus , has both sensory and motor branches, but the motor fibers are parasympathetic autonomic fibers and thus their action is involuntary. Its sensory fibers originate in many of the internal organs - lungs, stomach, aorta, larynx, to name a few - and its motor (parasympathetic) fibers run to the heart, stomach, small intestine, larynx and esophagus. Be-sides the vagus , the III, VII, and IX nerves also contain parasympathetic fibers, but in smaller amounts. The efferent and afferent branches of the accessory nerve (XI) connect the medulla to the pharynx and larynx and muscles of the shoulder, which they innervate. Afferent fibers of the last cranial nerve, the hypoglossal, convey the sense ofproprio-ceptionfrom the tongue to the medulla, and the efferent fibers stimulate movement of the tongue. The cranial nerves are numberedanteriorlytoposteri-orly. I and II originate in the olfactory epithelium and retina respectively. Ill and IV originate in the midbrain; V through VIII (vestibular branch) originate in the pons ; and VIII (cochlear branch) through XII originate in the medulla.

Question:

Calculate the enthalpy change, ∆H\textdegree, for the reaction N_2 (g) + O_2 (g) = 2NO (g), given the equilibrium constants 4.08 × 10^-4 for a temperature of 2000\textdegreeK and 3.60 × 10^-3 for a temperature of 2500\textdegreeK.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E15-0543.htm

Solution:

The effect of temperature on chemical equilibrium is determined by ∆H\textdegree(enthalpy or heat content); over moderate ranges in temperature, ∆H\textdegree is relatively independent of temperature. If, as in the case of this problem, ∆H\textdegree is in dependent of temperature then, (1)∆G\textdegree = - RT In k,where ∆G\textdegree = standard free energy change, R = gas constant, T = absolute temperature and k the equilibrium constant. (2)∆G\textdegree = ∆H\textdegree - T∆S\textdegree (3)∆H\textdegree - T∆S\textdegree = - RT In Kor - [(∆H\textdegree - T∆S\textdegree)/(RT)] = In k, ∆S\textdegree = the change in entropy or randomness of system. (4)[(- ∆H\textdegree)/(RT)] + [(∆S\textdegree)/(R)] = In k For two different temperatures, T_1 and T_2, equation (4) becomes In k_2 - In k_1 = [{(- ∆H\textdegree)/(RT_2)} + {(∆S\textdegree)/(R)}] - [{(- ∆H\textdegree)/(RT_1)}] + {(∆S\textdegree)/(R)}] or In (k_2 /k_1) = [(- ∆H\textdegree)/(RT_2)] + [(∆H\textdegree)/(RT_1)] In (k_2 /k_1) = [{∆H\textdegree (T_2 - T_1)}/(RT_2 T_1)]or, finally, log (k_2 /k_1) = [{∆H\textdegree (T_2 - T_1)}/{(2.303) R T_2 T_1}] Thus,log [(k_2500\textdegreeK)/(k_2000\textdegreeK)] = log [(3.60 × 10^-3 )/(4.08 × 10^-4)] = {∆H\textdegree (2500 - 2000)\textdegreeK} /{(2.303) (1.987 cal\textdegreeK^-1 mole ) (2500)\textdegreeK (2000)\textdegreeK} log 8.82 = .945 log 8.82 = .945 = [{∆H\textdegree (500\textdegreeK) = [{∆H\textdegree (500\textdegreeK) }/{(2.303) (1.987) (5 × 10^6) cal\textdegreeK^-1 mole^-1 \textdegreeK^2)}] [{(0.945) (2.303) (1.987) (5 × 10^6)}/{(500) mole cal [{(0.945) (2.303) (1.987) (5 × 10^6)}/{(500) mole cal -1 }] = ∆H\textdegree ∆H\textdegree = 43, 240 cal/mole.

Question:

Give some examples of valid and invalid variable names in BASIC. Ex-plain briefly the differences.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G10-0221.htm

Solution:

Variable names in BASIC can be represented by a single letter of the alphabet,or a single letter followed by a single digit, For example: VALID INVALID C 5N (starts with a digit) F BG4 (too many letters) K1 M28 (too many digits) L9

Question:

The binomial distribution is given by b(x) = (^n_x)p^xq^n^-x where (^n_x) = n!/{x! (n-x)!} for integers n and x, and q = 1-p. Write a function that computes b for given n, x, and p. Feel free to use the factorial function developed in chapter 19.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G19-0475.htm

Solution:

A straightforward answer is based on the notion of a function. FUNCTION B(P,N,X) INTEGER X Z = P \textasteriskcentered\textasteriskcentered X \textasteriskcentered ((1.0-P) \textasteriskcentered\textasteriskcentered (N-X)) B = Z \textasteriskcentered FACT(N) / (FACT(X) \textasteriskcentered FACT(N-X)) RETURN END

Question:

An electron is released from rest in a uniform electric field E = 1 N/coul. What velocity will it acquire in traveling 1 cm? What will then be its kinetic energy? How long a time is required? (Neglect the gravitational force.)

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/Users/wenhuchen/Documents/Crawler/Physics/D19-0622.htm

Solution:

The force on the electron is constant, there-fore the electron moves with a constant acceleration (see the figure). By Newton's Second Law a = F/m = eE/m =1.60 x 10^\rule{1em}{1pt}15 n/9.1 × 10^\rule{1em}{1pt}31 kg = 1.8 × 10^15(m/sec^2). Its velocity after traveling 1 cm, or 10^-2 m, is v^2 = v0^2 + 2a(x - x_0) where v_0 and x_0 are the initial velocity and position of the electron. Both of these are zero in this problem. Hence, v = \surd2ax = \surd[(2)(1.8 x 10^15 m/s^2)(10^\rule{1em}{1pt}2 m)] = 6.0 x 106m/sec. Its kinetic energy is 1/2 mv^2 = (1/2) (9.11 x 10^\rule{1em}{1pt}31 kg) (36 x 10^12 m^2/ s^2) = 1.6 x 10^\rule{1em}{1pt}17 joules. and it represents the work done by the electric force F^\ding{217} on the electron. The time is found from v = v_0 + at . But v_0 = 0 hence t = v/a = 3.3 × 10^\rule{1em}{1pt}9 sec.

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Question:

Describe the processes by which a root absorbs water and salts from the surrounding soil.

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/Users/wenhuchen/Documents/Crawler/Biology/F10-0243.htm

Solution:

The movement of water from the soil into the root can be explained by purely physical principles. The water available to plants is present as a thin film loosely held to the soil particles and is called capillary water. The capillary water usually contains some dissolved inorganic salts and perhaps some organic compounds, but the concentration of these solutes in capillary water is lower than that inside the cells of the root. The cell sap in the root hair of the epidermal cells has a fairly high concentration of glucose and other organic compounds. Since the plasma membrane of this cell is semi-permeable (it is permeable to water but not to glucose and other organic molecules), water tends to diffuse through the membrane from a region of higher concentration (the capillary water of the soil) to a region of lower concentration (the cell sap of the root hair). This movement of water is controlled by a process called osmosis. As an epidermal (root hair) cell takes in water, its cell sap now has a lower solute concentration than that of the adjacent cortical cell. By the same process of osmosis, water passes from the root hair to the cortical cell. Because of the osmotic gradient, water will continue to diffuse inward toward the center of the root. In this way, water finally reaches the xylem and from there, water is transported upwards to the stem and leaves by a combination of root pressure and transpiration pull.

Question:

Classify the six kinds of APL function.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G14-0373.htm

Solution:

In APL, function are classified on the basis of two main criteria: i) The number of arguments contained in the header (the header is the name of the program; it follows the \nablaop-erator) . ii) Whether the function is limited or unlimited. A func-tion is limited if it cannot be used as in a compound expression. Per contra, an unlimited function can be used within another function without causing an error condition. i) APL functions can contain zero, one or two arguments are called niladic , monadic and dyadic functions respectively. Thus we see that there are six kinds of APL functions: three kinds, depending on the number of arguments and the choice of "limited" or "unlimited" for each kind. Examples of each type follows: \nablaINFLATION [1]'ENTER BASE-YEAR INDEX' [2]CHEAP \leftarrow [] [3]'ENTER PRESENT INDEX' [4]TOOMUCH \leftarrow [] [5]PERCENT \leftarrow 100 × TOOMUCH \div CHEAP [6]"DISTRESS:" PERCENT [7]\nabla The function INFLATION isniladicbecause it has no arguments. It is limited because it is not used in the main body of the program. b)Niladicand unlimited: \nablaH \leftarrow HEIGHT [1]H \leftarrow H \div 2.0 [2]\nabla HEIGHT is a program concerning the effects of wearing padded shoes. The function is unlimited because its results will be stored in H which is contained in both the header and main program. c) Monadic and limited: \nablaSQUARE N [1]N \textasteriskcentered 2 [2]\nabla Here the function SQUARE which computes the square of a number is a function of one variable. Thus it is monadic. It is limited since it cannot be used in the program. d) Monadic and unlimited: \nablaC \leftarrow CUBE N [1]C \leftarrow N \textasteriskcentered 3 [2]\nabla The function CUBE which computes the cube of a number is a function of one variable and therefore monadic. Now CUBE can be used as an argument in other APL functions. Hence it is unlimited. e) Dyadic and unlimited: \nablaR \leftarrow N ROUNDOFF D [1]M \leftarrow N + 5 × 10\textasteriskcentered - (D + 1) [2]M \leftarrow LM × 10\textasteriskcenteredD [3]R \leftarrow M × 10\textasteriskcentered - D [4]\nabla Theroundofffunction rounds off a given integer to a required number of decimal places D. It is a function of both N and D and is therefore dyadic, f) Dyadic and limited: \nablaR CYVOL H [1]'THE VOLUME OF THE CYLINDER IS: ' [2]V \leftarrow 3.14159 × H × R\textasteriskcentered2 [3]V [4]\nabla Here CYVOL is a function of H and R but it does not oc-cur in the body of the program.

Question:

Find log_10(10^2 \bullet10^\rule{1em}{1pt}3 \textbullet 10^5).

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/Users/wenhuchen/Documents/Crawler/Chemistry/E33-0958.htm

Solution:

Recall log_x(a \bullet b \bullet c) = log_xa + log_xb + log_xc. Thus log_10(10^2 \textbullet 10^\rule{1em}{1pt}3 \textbullet 10^5) = log_1010^2 + log_1010^\rule{1em}{1pt}3 + log_1010^5Recall log_bb^x = x, since b^x = b^x; therefore, log_1010^2 + log_1010^\rule{1em}{1pt}3 + log_1010^5 = 2 +(\rule{1em}{1pt}3) + 5 = 4. Thus log_10(10^2 \textbullet 10^\rule{1em}{1pt}3 \bullet 10^5) = 4. Another method of finding log_10(10^2 \textbullet 10^\rule{1em}{1pt}3 \textbullet 10^5) is to note 10^2 \textbullet 10^\rule{1em}{1pt}3 \textbullet 10^5 = 10^2+(\rule{1em}{1pt}3)+5 = 10^4 (because a^x \textbullet a^y \textbullet a^z = a^x+y+z). Thus log_10(10^2 \textbullet 10^\rule{1em}{1pt}3 \textbullet 10^5) = log_10 10^4 = 4.

Question:

One of the functions of blood is to carry oxygen to the cells of the body. Exactly how are the oxygen-molecules carried, and why aren't the carriers found free in the plasma?

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/Users/wenhuchen/Documents/Crawler/Biology/F14-0350.htm

Solution:

The respiratory and circulatory systems have evolved to provide the cells with a continuous and adequate supply of oxygen. There are limitations however: Because oxygen is highly insoluble in water, it has a very low solubility in plasma, since plasma is 90% water. A major adaptation is the evolution of oxygen-carrying molecules. The oxygen carrying molecule found in the blood is hemoglobin. Hemoglobin is contained in the erythrocytes of vertebrates. Human erythrocytes, or red blood cells, are small, biconcave, disc-shaped cells (See Figure 1). During the maturation of red blood cells in mammals, they lose their nuclei, mitochondria, Golgi apparatus and other organelles. It is believed that the evolutionary loss of the nuclei in mammals has the adaptive advantage of leaving more space for hemoglobin. Similarly, since the red blood cell functions to transport oxygen, it would be counterproductive to have mitochondria which utilize oxygen. The mature ery-throcyte does not need these organelles since its main function is to carry hemoglobin, which in turn transports oxygen from the lungs to the tissues. Each human erythrocyte contains about 280 million hemoglobin molecules. A hemoglobin molecule consists of four polypeptide chains, two \alpha chains and two \beta chains. The capacity of hemoglobin to bind oxygen depends on the presence of a non--protein unit called a heme group. The heme group consists of an organic portion, called protoporphyrin, and an inor-ganic iron atom in the center (See Figure 2). The iron atom is capable of binding reversibly to a molecule of oxygen. Since each polypeptide chain contains a heme group, each hemoglobin molecule can combine with a maximum of four oxygen molecules. When all four binding sites contain oxygen molecules (Hb + 4O_2), the resulting compound is called oxyhemoglobin, as opposed to deoxyhemoglobin. Oxy and deoxyhemoglobin differ in quaternary structure- that is, oxygenation changes the three-dimensional configuration of the four polypeptide chains, it is this change in struc-ture which causes oxyhemoglobin to appear a brighter red than deoxyhemoglobin. This, in turn, accounts for the slight change in color between oxygen-rich blood and oxygen-poor blood. While some invertebrates also utilize hemoglobin in oxygen transport, others have different oxygen-carrying molecules. Hemocyanin is a protein in which a molecule of oxygen is carried between two copper atoms. When this mo-lecule is oxygenated, it is a pale blue color (deoxyhemocyanin is colorless). Hemocyanin is the oxygen carrying pig-ment in molluscs, crustaceae and arthropods. It was mentioned earlier that most of the hemoglobin molecules are contained within the red blood cells and not freely flowing in the plasma. By concentrating the hemoglobin molecules, the affinity for oxygen is increased, and the entry of oxygen from the outside to the lungs and from the lung tissues to the circulatory system is made faster and more efficient. Also, if the hemoglobin mo-lecules were found in the plasma, the increased concentration of protein would thicken the blood. This increased vis-cosity would require that the heart perform more work in pumping the blood to peripheral organs and particularly to the brain. Erythrocytes thus act as hemoglobin containers for more efficient physiological functioning.

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Question:

Differentiate between homologous and analogous structures. Give examples of each.

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/Users/wenhuchen/Documents/Crawler/Biology/F28-0729.htm

Solution:

Homologous structures are structures derived from a similar evolutionaryorigin. They may have diverged in their functions and phenotypicappearance but their relationships to adjacent structures and embry-onicdevelopment are basically the same. Homologous structures, suchas a seals front flipper, a bat's wing, a cat's paw, a horse's front leg, andthe human hand and arm, all have a single evolutionary origin, but havediverged in order to adapt to the different methods of locomotion requiredby different lifestyles. Analogous structures are similar in function and often in superficial appearancebut, in direct contrast to homologous structures, they are of differentevolutionary origins. The wings of robins and the wings of butterfliesare examples of analogous structures. Although both are for the samepurpose, namely flying, they are not inherited from a common ancestor. Instead, they evolved independently from different ancestral structures.

Question:

It is known that AgCI has the same structure as NaCl. X-ray measurements of AgCl show a unit-cell edge length of .55491 nm. The density of the crystal is found to be 5.561 g/cm^3. Assuming 10^-7cm/nm, find the percentage of the 7 sites that would appear empty. Assume that lattice vacancies are the only defects.

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Solution:

The key to solving this problem is to compute the theoretical density of the unit cell from its X-ray determined cubic edge length. Then, compare this value with the actual density to find the percent of the sites that would appear to be empty. Theoretical density = mass of 1 unit cell / volume of 1 unit cell Each unit cell possesses 4 ions of each Ag^+ and Cl^-. This is determined from the fact that each corner ion contributes 1/8 of its volume to each of 8 adjacent unit cells, each face ion contributes 1/2 of its volume to the two adjacent cells, each edge ion contributes 1/4 of its volume to each of 4 adjacent cells and each enclosed ion contributes its entire volume. As seen in the accompany-ing figure, there are 8 corner ions, 12 edge ions, 6 face ions and 1 totally enclosed ion. The molecular weight of AgCl is 143.32. Thus, the mass of 1 mole of unit cells is 4 times 143.32 g/mole of unit cells or 573.28 g/mole of unit cells. Since the unit cell is a cube, volume = (cubic-edge length)^3 . Cubic edge length = .55491 nm or .55491 × 10^-7 cm. Thus, volume becomes equal to (.55491 × 10^-7)^3 = 1.708 x 10^-22 cm^3 . As such, theoretical density = {(573.28 g/mole of unit cells) / (6.022 × 10^23 unit cells/mole)} / (1.708 × 10^-22 cm^3) / (1.708 × 10^-22 cm^3) = 5.571 g/cm = 5.571 g/cm 3 The given density is 5.561 g/cm^3. Thus, the actual figure is [(5.571 - 5.561) / 5.571] × 100% = .2%too small. This is the percent of lattice that is vacant.

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Question:

Write a PL/I program to simulate the computer dating service. The following factors will be used in the selection process: A)The sex of a person looking for a date. B)Interest of the person in the following fields: 1)Theatre2)Sports3)Politics MoviesDancingSocial events Opera The minimum and maximum acceptable ages of the person(s) selected.

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Solution:

This data must be available for each person on record and for the person looking for a date. The following rules are used for selecting a possible date. 1)The person selected must be of an opposite sex from the person looking for a data. 2)They must have at least one common interest in the first set of activities (Theatre, Movies, Opera). 3)They must have at least one common interest in either the second or the third set of activities. 4)The age of the partner selected must be greater than or equal to a minimum age, and less then or equal to a maximum age, specified by the original client looking for a date. The program procedure looks as follows: DATING : PROCEDURE OPTIONS (MAIN) ; /\textasteriskcentered READ THE PERSONAL DATA OF THE PERSON LOOKING FOR DATE \textasteriskcentered/ GET LIST (/\textasteriskcenteredGENERAL INFO \textasteriskcentered/ IDENTIFICATION, INF_SEX, /\textasteriskcenteredSPECIAL INTERESTS \textasteriskcentered/THEATRE, MOVIE, OPERA, SPORTS, DANCING POLITICS, SOC-EVENTS, /\textasteriskcentered THE AGE RANGE \textasteriskcentered/ MDLAGE, MAX_AGE ); /\textasteriskcentered THE HEADINGS FOR THE OUTPUT ARE CONSTRUCTED BY THE NEXT THREE STATEMENTS \textasteriskcentered/ PUT LIST('THESE ARE SELECTED PARTNERS FOR', IDENTIFICATION) PAGE LINE (5); PUT LIST ('IDENTIFICATION', 'SEX OF PARTNER1, 'AGE OF PARTNER') SKIP (2); T = 0; /\textasteriskcentered T IS VARIABLE USED TO TEST FOR THE END OF INPUT STREAM\textasteriskcentered/ NEXT: IF T = 1 THEN GO TO TERMINATE; /\textasteriskcentered THE VARIABLE T WILL ONLY BE CHANGED BY THE VERY LAST PART OF THE INPUT STREAM. IN OTHER CASES, TWO COMMAS (,,) IN THE INPUT STREAM ENSURES THAT THE VALUE OF T IS NOT ALTERED \textasteriskcentered/ /\textasteriskcentered ACQUIRE THE DATA OF THE INDIVIDUAL KEPT ON FILE \textasteriskcentered/ GET LIST (IDENTIFICATIONS, INF_SEX_P, THEATREJP, MOVIE_P,OPEFA_P, SP0RTS_P, DANCING_P, POLITICS_P, SOC_EVENTS_P, INDIVIDUAL.AGE, T); IF INF_SEX = INF_SEX_P THEN GO TO NEXT; IF THEATRE \lnot = THEATRE_P THEN IF MOVIE \lnot = MOVIE_P THEN IF OPERA \lnot = OPERA_P THEN GO TO NEXT; /\textasteriskcentered AT LEAST ONE OF THE NEXT FOUR INTERESTS MUST BE SHARED \textasteriskcentered/ IF SPORTS \lnot = SPORTS_P THEN IF DANCING \lnot = DANCING_P THEN IF POLITICS \lnot = POLITICS_P THEN IF SOC EVENTS \lnot = SOC_EVENTS_P THEN GO TO NEXT; /\textasteriskcentered NOW TEST WHETHER THE PERSON IN OUR FILE FITS INTO THE PROPER AGE GROUP. A SELECTED CANDIDATE IS PRINTED IN THE OUTPUT \textasteriskcentered/ IF INDIVIDUAL_AGE> = MIN_AGE THEN IF INDIVIDUAL_AGE < = MAX_AGE THEN PUT LIST (IDENTIFICATION_P, INF_SEX_P, INDIVIDUAL_AGE) SKIP; GO TO NEXT; TERMINATE : END EATING; For identification of a person, the integer numbers that may represent driver's license numbers or social security numbers are used. The input data for persons on file consists of the following information: where nnnn indicates identification number, S shows the sex of the person (the digit 1 indicates female and 0 - male), specification of personal interest in activities follows, ZZ is the age of potential partner, and m is used to indicate the end of file. The person looking for a date is described by the following input format: Last two items specify the desired minimum age (Z1) and the maximum age (Z2) of the future date.

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Question:

A bricklayer is supplied with bricks by his mate who is 10 ft below him, the mate tossing the bricks vertically upward. If the bricks have a speed of 6 ft/s when they reach the bricklayer, what percentage of the energy used up by the mate serves no useful purpose?

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Solution:

Once the bricks leave the mate's hands, the only force which acts on them is the gravitational force. Since this produces a constant acceleration (a = - g =- 32 ft/s^2), the kinematics equation v^2 = v^2_0 - 2a (x - x_0) can be used to describe its motion. The initial velocity v_0 of the bricks is found by substituting known values in the above equation, (x - x_0 is the distance travelled by the bricks) v^2_0 = v^2 + 2g (x - x_0) = 36ft^2/s^2 + 2 × 32 ft/s^2 × 10ft =676 ft^2/s^2 .\therefore v_0 = 26 ft/s. The kinetic energy given each brick, and supplied by the bricklayer's mate, is E_1= (1/2)mv^2_0 = m × 338 ft^2/s^2 If the bricklayer's mate supplied only just enough energy to the bricks for them to reach the required level and no more, the initial velocity being u, they would have zero velocity at the level of the bricklayer. Hence u^2 = 0 + 2g(x - x_0) = 2 × 32 ft/s^2 × 10 ft. \thereforeu= 8 \surd(10) ft/s. The kinetic energy supplied by the mate in this case is E_2 = (1/2)mu^2 = m × 320 ft^2/s^2. The mate supplies an energy equal to E_1 when he only needs to expend energy equal to E_2. Therefore he wastes an amount of energy E_1 - E_2. The percentage of energy wasted is [(E_1 - E_2)/E_1] × 100 = [(338 - 320)/338] × 100 = (18/338) × 100 = 5.3%.

Question:

Ecologists refer to the ocean as the marine ecosystem. How are the major marine communities classified and where are the regions of their habitation?

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Solution:

The ocean is a very complex ecosystem with many different kinds of communities. Temperature, salinity, availability of light, and depth all affect the nature of these communities and present major barriers to the free movement of organisms that have specific adaptability to certain environmental conditions. The major currents, which keep the waters of the ocean in continous circulation, also affect the nature of marine communities, as do waves and tides. Another important phenomenon, called upwelling, occurs where the winds move surface waters away from the steep coastal slopes, bringing cold water to the surface from the depths which is rich in such nutrients as phosphates and nitrates; the regions of such upswellings are often among the most productive regions of the ocean. The major marine habitats are shown in the accom-panying diagram. Marine organisms in general are divided into two groups, depending upon whether they live in the open water or upon or beneath the bottom. Those organisms that live suspended or swimming in the open waters are said to be pelagic. Bottom dwellers, called benthic organisms, may live on the surface or beneath the surface of the ocean floor. The benthic organisms include both attached organisms and those organisms that move in or on the substratum. Bottom dwellers are generally distinct for each of the neritic, or coastal, regions and the type found in each region is largely dependent on the type of ocean floor, whether it is sand, rock, or mud. Many animals are adapted for living in the sand and this group includes representatives of virtually every phylum of animals. Pelagic organisms are of two types. The microscopic plants and animals that are unable to move against the current are collectively called the plankton. The larger, free swimming animals that are able to move at will are called nekton. The planktonic plants constitute the phytoplankton. The phytoplanktonic algae are the most important producers throughout the ocean. The phytoplank-ton are eaten by a great array of invertebrate animals, ranging in size from protozoans to large medusae. These latter are the primary consumers. These, in turn, are consumed by other, larger invertebrates and vertebrates. As would be expected, plankton attains its greatest density in the upper lighted zone where there is sufficient light for photosynthesis to occur. In productive waters, planktonic organisms may occur in such enormous numbers that the water appears turbid. The animals that are permanent inhabitants of the lower dark zones are carnivorous, suspension, or detritus (dead organic matter) feeders and depend ultimately on the photosynthetic activity of the microscopic algae in the upper, lighted regions. Therefore, vertical distribution of marine organisms is largely controlled by the depths of light penetration. Light sufficient for photosynthesis to exceed respiration penetrates only a short distance below the surface, with this distance being dependent upon the turbidity of the water. Below this is a region where some photosynthesis can occur, but here the production of organic matter is less than the loss through respiration.

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Question:

Explain the need for STOP JOB CARDS, and list the differentmethods used in computers to bring about a proper endof the program.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G13-0331.htm

Solution:

The computer is just a machine without any intel-ligence of its own. It only does exactly what is has been programmed to do, i.e., told, to do. When an instruction is given to the computer to get some kind of data, say, from a punched card, the computer keeps on getting the data till itis explicitly told again to stop getting any more data. When a computer is taking in data as part of a program, there must besome means of telling the computer how many data items to expect, especiallywhen the appropriation of data is part of an iterative loop. If the computertries tocontinouslyread in data it will eventually not find any moredata. The computer now assumes there is an error, and either terminatesthe program, or gives out an error message. To eliminate this problemwe must have some ways of telling the computer just how much datato expect or what to do (other than bombing the program!) when the dataruns out. There are normally four ways to do this: 1) Header Card method,2) Trailer Card method, 3) End-File condition method,4) Count and compare method

Question:

Describe sexual reproduction in mosses.

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Solution:

The moss plant has a life cycle characterized by a marked alternation between the sexual and asexual generations. The sexual gametophyte generation is the familiar, small, green, leafy plant with an erect stem held to the ground by numerous rhizoids. When the gametophyte has attained full growth, sex organs develop at the tip of the stem, in the middle of a circle of leaves and sterile hairs called paraphyses. The male organs are sausage-shaped structures, called the antheridia. Each antheridium produces a large number of slender, spirally-coiled swimming sperm, each equipped with two flagellae. After a rain or in heavy dew, the sperm are released and swim through a film of moisture to a neighboring female organ, either on the same plant or on a different one. The female organ, the archegonium, is shaped like a flask and has one large egg at its broad base, the ventor. The archegonium releases a chemical substance that attracts the sperm and guides them in swimming down the archegonium to the ventor. Here one sperm fertilizes the egg. The resulting zygote is the beginning of the diploid, asexual sporophyte generation. The mature sporophyte is composed of a foot embedded in the archegonium and a leafless, spindle-like stalk or seta which rises above the gametophyte. The sporophyte is nutritionally dependent on the gametophyte, absorbing water and nutrients from the archegonium via the tissues of the foot. A sporangium, called the capsule, forms at the upper end of the stalk. Within the capsule each diploid spore mother cell undergoes meiotic division to form four haploid spores. These spores are the beginning of the next gametophyte generation. When the capsule matures, it opens and releases spores under favorable conditions in a mechanism specific to its kind. In the Sphagnum, for example, the mature capsule shrinks, bursts, pushes away the lid, and exposes the spores to the wind. If a spore drops in a suitable place, it germinates and develops into a protonema, a green, creeping, filamentous structure. The protonema buds and produces several leafy gametophytes, thereby completing the moss life cycle.

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Question:

What is the cost of operating for 24 hr a lamp requiring1.0 amp on a 100-volt line if the cost of electric energy is $0.50/kw-hr?

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Solution:

The power P developed by the lamp resistance is P = (energy dissipated/time) = VT The energy dissipated by the lamp in 24 hours is then E =Vlt= (100 volts)(1.0 amp)(24 hr) = 2400 watt-hr = 2.4kw-hr But each kilowatt-hour of electrical energy costs $0,050. Therefore the cost of 2.4kw-hr is Cost=2.4 (kw-hr) × ($0.050/kw-hr) = $0.12. = $0.12.

Question:

A force of 100ntis required to stretch a steel wire 2.0mm^2 in cross-sectional area and 2.0 m long a distance of 0.50 mm. How much work is done?

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Solution:

In this problem we will make use of Hooke's Law which states that the force needed to stretch a material a distance y is proportional to this distance; i.e., F =ky, where k is called the spring constant. Therefore, to find k, we divide the force by the distance y, k = (100nt)/(5.0 × 10^-4m) = 20 × 10^4nt/m The work done is given by, W = \int F^\ding{217} \bulletdy^\ding{217}(1) Since the force and the displacement are in the same direction, (1) may be simplified to W = \intFdy(2) but F =kyso (2) becomes W = \intkydy= (1/2)ky^2 = (1/2)(20 × 10^4nt/m)(5.0 × 10^-4 m)^2 = 0.025nt- m.

Question:

Sodium concentration in the plasma is largely determined by sodium excretion by the kidney. Describe how sodium excretion is regulated.

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Solution:

The control of sodium excretion depends upon two variables of renal function: theglomerularfiltration rate and sodium reabsorptionby the kidney tubules. The plasma sodium level is intimately related to the plasma volume because of the principle of osmosis. If the sodium concentration in the blood increases, water enters the plasma from the surrounding tissue cells or inter-stitial fluid, raising the plasma volume. Conversely, if the plasma sodium concentration decreases, theosmolarityof the blood decreases. Water then leaves the plasma by osmosis, reducing the blood volume. Thus the sodium level in the body is able to be regulated in the same way that blood volume is regulated via theglomerularfiltration mechanism. For instance, if the sodium level falls low in the blood, the blood volume drops due to osmosis of water out of the circulatory system. Decreased blood volume causes a fall in blood pressure, and consequently a drop inglomerularcapillary pressure. The amount glomerularfiltrate declines, and more sodium is conserved in the body. Decreased blood volume also reduces filtration via an autonomic feedback mechanism involvingbaroreceptorsin arterial walls and the central nervous system. The net effect is that water and sodium are retained in body to maintain a normal physiological state. The second renal mechanism regulating sodium ex-cretion occurs in the distal convoluted tubule. It involvesaldosterone, a hormone produced by the adrenal cortex, and a group of factors leading to the synthesis and release ofaldosterone. (See previous question) .Aldosteroneacts on the epithelium of the distal convoluted tubule, increasing its permeability to sodium, and thus enhancing sodiumreabsorptionby the renal tubular cells. It is important to note that that as sodium is reabsorbed by the distal convoluted tubules under the influence ofaldosterone, water is also reabsorbed. This occurs because as sodium is removed from the tubular space, an osmotic gradient is established. This gradient causes water to leave the tubular space and reenter the blood. Therefore, an increased sodiumreabsorptionresults also in an in-creased waterreabsorption. This is whyaldosteroneis stimulated when blood volume falls; thealdosterone released stimulates sodium and therefore waterreabsorption. The water enters the bloodstream and helps restore the blood volume to normal.

Question:

Two blocks, A and B, are attached by a cord wrapped around a frictionless, massless pulley, C. The coefficient of friction between each block and the horizontal platform is 0.3. Block A weighs 10 lb, block B weighs 20 lb. If the platform rotates about the vertical axis shown, determine the angular speed at which the blocks start to slide radi-ally.

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Solution:

The two blocks will start sliding radially when the centrifugal force on B is sufficient to overcome (a) the force of friction between block B and the table and (b) the tension in the cord which is equal to the centrifugal force on block A plus the force of friction between block A and the table. Using the formula for centrifugal force: F_C = m v^2 / r = W v^2 / gr = W\omega^2 r^2 / gr = W\omega^2 r / g and for friction:F_f = \mu W, where W is the weight of a block, \omega is the angular velocity, r the radius, g = 32 ft / sec ^2, the acceleration due to gravity, and \mu the coefficient of static friction, we can write the desired equation as: W_B\omega^2 r_B / g = \muWB+ \muW_A + W_A\omega^2 r_A / g where the subscripts A and B refer to the two blocks. (W_B\omega^2 r_B / g) \rule{1em}{1pt} (W_A\omega^2 r_A / g) = \mu(W_A + W_B) [\omega^2(W_B r_B \rule{1em}{1pt} W_A r_A)] / g = \mu(W_A +W_B) \omega^2 = [\mug(W_A +W_B)] / (W_B r_B \rule{1em}{1pt} WAr_A) \omega^2 = \surd[{(\mug)(W_A + W_B)} / (W_B r_B \rule{1em}{1pt} WAr_A)] . Substituting the numerical values, yields \omega = \surd[{.3 × 32 (10 + 20)} / {(20 × 3) \rule{1em}{1pt} (10 × 2)} ] = \surd(7.2) =2.68 radians/sec.

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Question:

A newborn infant is examined by a doctor, who determines that the child has syphilis. Explain how the child con-tracted syphilis .

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Solution:

Syphilis is a venereal disease ofman, thatis included in the more general category of contact dis-eases. Contact diseases of man usually result from the entry of an infectious agent into the individual through the skin or mucous membranes. Contact may be direct (through wounds and abrasions ) or indirect (through vectors mediating transmitters such as insects ). In the case ofveneraldiseases, transmittance is usually through direct genital contact during sexual activity. The organism in-volved is a spirillum calledTreponemapallidum. However, direct sexual contact is not the only way in which syphilis is transmitted. An infected mother can transmit the organism by placental transfer to the fetus during the first four months of pregnancy. Contraction of the disease is also possible as the fetus passes through the infected vagina during birth. The disease usually requires an incubation period of 3 to 6 weeks after infection. The disease produces lesions called chancres which resemble ulcerous sores. In the late stages of the disease, the cardiovascular and central ner-vous systems may be affected, with possible paralysis. The disease, if promptly detected, can be treated with penicillin . There is no known means for immunization againstT. pallidum infection . Persons who have recovered from a syphilitic infection are just as likely to contract it upon subsequent exposure to the organism. Preventive measures include avoiding carriers and using local prophylactic measures such as condoms.

Question:

Given, for acetic acid that ∆H_fus= 2592 cal/mole at its melting point, 16.6\textdegreeC and ∆H_VAP = 5808 cal/mole at its boiling point, 118.3\textdegreeC, calculate the change in entropy that takes place when 1 mole of the vapor is condensed at its boiling point and changed to a solid at its melting point, all under constant pressure, taken as 1 atm. Assume that the molar heat capacity of acetic acid is 27.6 cal/deg - mole.

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Solution:

Entropy measures the randomness of a system. In this problem, there are two phase changes. As such, there exist three changes in entropy to calculate. The entropy change from vapor to liquid at the boiling point, the cooling entropy change from the liquid at the boiling point to liquid at the freezing point, and the entropy change from the liquid to solid. Measure these by employ-ing the following equations: ∆S = (q_rev/T), which states that for a reversible reaction, the change in entropy (∆S) equals the heat absorbed or released by a system (q_rev) divided by the temperature in Kelvin (Celsius plus 273\textdegree) . This equation can be used for the condensation and freezing. To measure cooling, use the fact that ∆S = 2.30 3 C_P log (T_2 /T_1 ). Proceed as follows: ∆S _condensation = - ∆S_VAP = -∆H_VAP /T. Condensation is the opposite of vaporization. Thus, condensation = - vaporization. Since ∆H is the only heat in this reaction, its value becomesq_rev. Sub-stitute in values to obtain ∆S _condensation = -[(5808 cal/mole)/(391.5\textdegreeK)] = - 14.84 cal/deg - mole. For ∆S _cooling ∆S = 2.303 C_P log (T_2 /T_1 ) = (2.303) (27.6 cal/deg - mole) log [(289.8\textdegree)/(391.5\textdegree)] = - 8.30 cal/deg - mole. [Notes:T_1 = b. p. in Kelvin, T_2 = m. p. in Kelvin.] For freezing, ∆S _freezing = -∆S_fus_- = - ∆H_fus/T. Fusion is the opposite of freezing. Thus, freezing = - fusion. ∆H is the only heat involved, so its value becomesq_rev. Therefore, ∆S = - (2592 cal/mole) /298.8\textdegreeK = - 8.67 cal/deg - mole. The total entropy change is ∆S _condensation + ∆S _cooling + ∆S _freezing = - 14.84 - 8.30 - 8.67 = - 31.81 cal/deg - mole.

Question:

Calculate the hydrolysis constants of the ammonium and cyanide ions, assuming K_W = 1 × 10^-14 and K_a = 4.93 × 10^-10 for HCN and K_b = 1.77 × 10^-5 for NH_3. For each, determine the percent hydrolysis in a .1M solution.

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Solution:

To find the hydrolysis constant, you must know what it defines. Hydrolysis is the process whereby an acid or base is regenerated from its salt by the action of water. The hydrolysis constant measures the extent of this process. Quantitatively, it is defined as being equal to [K_W/(K_a or K_b)] , where K_W = the equilibrium constant for theautodissociationof water, K_a = dissociation of acid, and K_b = dissociation of base. You are givenK_w, K_a and K_b. Thus, the hydrolysis constants can be easily found by substitution. LetK_h= hydrolysis constant. For cyanide ion:K_h= (K_W/K_a) = [(1 × 10^-14)/(4.93 × 10^-10)] = 2.02 × 10^-5 For ammonium ion:K_h= (K_W/K_b) = [(1 × 10^-14)/(1.77 × 10^-5)] = 5.64 × 10^-10 To find the percent hydrolysis in a .1M solution, write the hydrolysis reaction and express the hydrolysis constant just calculated in those terms. After this, represent the concentrations of the hydrolysis products in terms of variables and solve. For cyanide ion: The hydrolysis reaction is CN^- + H_2O \rightleftarrows HCN + OH^-. Therefore, K_h= {[HCN][OH^-]}/[CN^-]. But you calculated thatK_h= 2.02 × 10^-5 . Equating, 2.02 × 10^-5 = {[HCN][OH^-]}/[CN^-]. You start with a .1M solution of CN^-. Let x = [HCN] formed. Thus, x = [OH^- ] also, since they are formed inequimolaramounts. If x moles/liter of substance are formed from CN^-, then, at equilibrium you have .1 - x moles/liter left. Substituting these values, 2.02 × 10^-5 = [(x \textbullet x)/(1 - x)] Solving, x = 1.4 × 10^-3M. The percent is just 100 times (x/.10M) since the initial concentration is .10M, so that you have 1.4% hydrolysis in a .1M solution. For ammonium ion: The hydrolysis reaction is NH_4^+ + H_2O \rightleftarrows NH_3 + H_3O^+. K_hfor this reaction = {[NH_3][H_3O^+]}/[NH_4^+]. The calculatedK_h= 5.6 × 10^-10. Equating, 5.6 × 10^-10 = {[NH_3][H_3O^+]}/[NH_4^+]. From this point, you follow the same reasoning as was used with the cyanide. Solving: Let x = [NH_3] 5.6 × 10^-10 = [{(x) (x)}/(.1 - x)] x^2 = (5.6 × 10^-11) - (5.6 × 10^-10)x x^2 + (5.6 × 10^-10)x - 5.6 × 10^-11 = 0 Using the quadratic formula one can solve for x, where ax^2 +bx+ c = 0 x = [{- b\pm \surd(b^2 - 4 ac)}/2a] x = ([-5.6 × 10^-10 \pm \surd{(5.6 × 10^-10)^2 - 4(1)(-5.6 × 10^-11)}] / {2 (1)}) x = [{-5.6 × 10^-10 \pm \surd(2.24 × 10^-10)} / (2)] x = [(-5.6 ×10^-10 \pm 1.50 × 10^-5) / (2)] x = [(1.50 × 10^-5)/2] or x = [(1.50 × 10^-5)/2] x cannot be negative, because concentration cannot be negative. Thus, x = 7.5 × 10^-6 Solving for the percent: [(7.5 × 10^-6)/(.1)] × 100% = 7.5% Thus, you find that the percent hydrolysis is 7.5 × 10-3%.

Question:

Simulate a game of American roulette with a basic program.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G23-0553.htm

Solution:

American roulette is played on a wheel with 38 compartments numbered 1 to 36, 0 and 00. Odd and even num-bers alternate around the wheel as do red and black. In the simulation bets can range from $5 to $500 and you may bet on red or black, odd or even, first or second 18 numbers, a column or a single number. Place bets according to the following payoffs: 1) The numbers 1 to 36 signify a straight bet on that num-ber and the payoff is 35:1 (i.e. 35 dollars return for a $1 bet). 2) If you bet that the ball will land in the column or on groups of numbers 1 - 12 or 13 - 24 or 25 - 36, the bets are 2:1. In the program, the bet numbers are 1 - 36 for single numbers and 37 for 1 - 12, 38 for 13 - 24, 39 for 25 - 36. 3) If you bet that the ball will land on i) a number from groups 1 - 18 or 19 - 36 ii) even or odd iii) red or black the payoff is 1:1 ($1 additional for every $1 bet). The even money bet numbers are 43 for 1 - 18, 44 for 19 - 36, 45 for even, 46 for odd, 47 for red and 48 for black. 4) 0 and 00 pay off 35:1 and count only when they are expli-citly betted upon: The bet number for 0 is 49 and 50 for 00. 10PRINT "ROULETTE" 20PRINT "MINIMUM BET IS $5, MAXIMUM IS 500" 30PRINT "WHEN I ASK FOR EACH BET, TYPE"; 40PRINT "THE NUMBER AND THE AMOUNT, SEPARATED BY A"; 50PRINT "COMMA. FOR EXAMPLE: TO BET $500 ON BLACK,"; 60PRINT "TYPE 48,500" 70DIM B(100), C(100), T(100), X(38) 80DIM A(50) 90FOR I = 1 TO 38: X(I) = 0 : NEXT I 100REM MAT X = ZER 110P = 1000 120D = 100000 130PRINT "HOW MANY BETS"; 140INPUT Y 150IF Y < 1 OR Y <> INT(Y) THEN 130 160FOR I = 1 TO 50: A(I) = 0 : NEXT I 170FOR C' = 1 TO Y 180PRINT "NUMBER"; C; 190INPUT X,Z 200B(C) = Z: T(C) = X 220IF X < 1 OR X > 50 OR X <> INT(X) THEN 180 230IF Z < 1 OR Z <> INT(Z) THEN 180 240IF Z < 5 OR Z > 500 THEN 180 250IF A(X) = 0 THEN 280 260PRINT "YOU ALREADY MADE THAT BET" 270GOTO 180 280A(X) = 1: NEXT C 300PRINT "SPINNING" 310PRINT: PRINT 330S = INT(100\textasteriskcenteredRND(1)) 340IF S = 0 OR S > 38 THEN 330 350X(S) = X(S) + 1 360IF S < 37 THEN 420 370IF S = 37 THEN 400 380PRINT "00" 390GOTO 520 400PRINT "0" 410GOTO 520 420RESTORE 430FOR I = 1 TO 18 : READ R: IF R = S THEN 500: NEXT I 470A$ = "BLACK" 480PRINT S; A$ 490GOTO 520 500A$ = "RED" 510GOTO 480 520PRINT 530FOR C = 1 TO Y: IF T(C) < 37 THEN 1210 540ON T(C) - 36 GOTO 590, 690, 720, 750, 800, 850, 900, 970, 1000 560ON T(C) - 45 GOTO 1030, 1060, 1130 570GOTO 1210 580STOP 590REM 1 - 12(37) 2 : 1 600IF S < = 12 THEN 650 610PRINT "YOU LOSE"; B(C); "DOLLARS ON BET"; C 620D = D + B(C) 630P = P - B(C) 640GOTO 680 650PRINT "YOU WIN"; B(C)\textasteriskcentered2; "DOLLARS ON BET"; C 660D = D - B(C)\textasteriskcentered2: P = P + B(C)\textasteriskcentered2 680GOTO 1310 690REM 13-24(38) 2 : 1 700IF S > 12 AND S < 25 THEN 650 710GOTO 610 720REM 25 - 36(39) 2 : 1 730IF S > 24 AND S < 37 THEN 650 740GOTO 610 750REM FIRST COLUMN (40) 2 : 1 760FOR I = 1 TO 34 STEP 3 : IF S = I THEN 650 : NEXT I 790GOTO 610 800REM SECOND COLUMN (41) 2 : 1 810FOR I = 2 TO 35 STEP 3: IF S = I THEN 650 : NEXT I 840GOTO 610 850REM THIRD COLUMN (42) 2 : 1 860FOR I = 3 TO 36 STEP 3 : IF S = I THEN 650: NEXT I 890GOTO 610 900REM 1 - 18 (43) 1 : 1 910IF S < 19 THEN 930 920GOTO 610 930PRINT "YOU WIN"; B(C); "DOLLARS ON BET"; C 940D = D - B(C) 950P = P + B(C) 960GOTO 1310 970REM 19 - 36(44) 1:1 980IF S < 37 AND S > 18 THEN 930 990GOTO 610 1000REM EVEN (45) 1 : 1 1010IF S/2 = INT(S/2) AND S < 37 THEN 930 1020GOTO 610 1030REM ODD (46) 1 : 1 1040IF S/2 <> INT(S/2) AND S < 37 THEN 930 1050GOTO 610 1060REM RED (47) 1:1 1070RESTORE 1080FOR I = 1 TO 18 1090READ R: IF S = R THEN 930: NEXT I 1120GOTO 610 1130REM BLACK (48) 1 : 1 1140RESTORE 1150FOR I = 1 TO 18: READ R: IF S = R THEN 610: NEXT I 1190IF S > 36 THEN 610 1200GOTO 930 1210REM 1 TO 36, 0,00 (1 - 36,49,50) 35 : 1 1220IF T(C) < 49 THEN 1260 1230IF T(C) = 49 AND S = 37 THEN 1280 1240IF T(C) = 50 AND S = 38 THEN 1280 1250GOTO 610 1260IF T(C) = S THEN 1280 1270GOTO 610 1280PRINT "YOU WIN"; B(C)\textasteriskcentered35; "DOLLARS ON BET"; C 1290D = D - B(C)\textasteriskcentered35: P = P + B(C)\textasteriskcentered35 1310NEXT C 1330PRINT "TOTALS:", "ME", "YOU" 1340PRINT " ", D, P 1350IF P > 0 THEN 1380 1360PRINT "YOU JUST SPENT YOUR LAST DOLLAR" 1370GOTO 1470 1380IF D > 0 THEN 1420 1390PRINT "YOU BROKE THE HOUSES" 1410GOTO 1480 1420PRINT "AGAIN"; 1430INPUT Y$ 1440IF LEFT $ (Y$,1) = "Y" THEN 130 1450DATA 1,3,5,7,9,12,14,16,18,19,21,23,25,27,30,32,34,36 1470PRINT "THANKS FOR YOUR MONEY" 1480END.

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Question:

Discuss and illustrate the meaning of absolute and relative addressing.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G09-0190.htm

Solution:

One does not need an in depth knowledge of com-puter science to realize that any program loaded into the main memory of a computer is stored at a particular address. In basic assembler language, by means of a BALR instruction, the address of the first instruction of the program is al-ways placed in a base register. We can safely say that this address never has location "zero." This is so because the first section of the computer's memory is reserved for the operating system. If we call the address at which the pro-gram is loaded its ENTRY address, then, in general, to cal-culate the address of any of the program's symbols, we must add the distance (number of bytes) that is between the sym-bol and the ENTRY address to the ENTRY address. For example if the ENTRY address' location is 925 and the symbol is 48 bytes away, then the address of the symbol is 925 + 48 = 973; the symbol is at location 973. The ENTRY address and the process of adding the displacement dictate what is known as the absolute address of the program and its symbols. In fact, the absolute address of a symbol is its actual loca-tion in main memory. In relative addressing however, we are only interested in the positions which statements occupy relative to the beginning of the program. As a result of this we do not concern ourselves with the contents of the base register (since it is constant for all statements in the program) in calculating this address. The relative address is merely the distance of a particular statement from the beginning of the program. Both relative and absolute addressing are illustrated by the example in Fig. 1.

Question:

A soap bubble consists of two spherical surface films very close together, with liquid between. A soap bubble formed from 5 mg of soap solution will just float in air of density 1.290 g \textbullet liter when filled with hydrogen of density 0.090 g \textbullet liter. The surface tension of soap solution is 25 dynes \textbullet cm^-1. What is the excess pressure in the bubble?

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/Users/wenhuchen/Documents/Crawler/Physics/D10-0425.htm

Solution:

When the bubble is floating in air, the weight of soap solution plus the weight of the hydrogen must just be balanced by the upthrust due to the displaced air. The buoyant force of the air is equal to the volume of air displaced by the bubble multiplied by the weight density of air. Hence, if the bubble has a volume of y, g(5 × 10^-3 grams) + (0.09 × 10^-3 grams \textbullet cm^-3) gy = (1.29 × 10^-3 grams \bullet cm^-3) gy. \thereforey = (5 cm^3)/(1.2) = 25/6 cm^3. But the bubble is spherical and of radius r. Thus y = (4/3) \pir^3 = 25/6 cm^3. \thereforer^3 = (25 × 3 cm^3)/(24\pi)or r = 1 cm. Consider half of the bubble, as shown in the figure. The other half exerts a force to the left equaling twice the surface tension, \Upsilon, multiplied by the perimeter or F_left = (2\Upsilon)(2\pir) We use twice the surface tension since the soap bubble has both an inner and an outer surface producing tension. The thickness of the bubble is assumed small in comparison with its radius letting us use the average value of r for both inner and outer surfaces. The force on the bubble to the right equals the pressure difference, P, between the outer and inner surfaces of the bubble times the area of the bubble in the direction being considered. This area is obtained by projecting the half-bubble on a plane perpendicular to this direction, as shown in the figure. The projected area of a sphere on a plane is a circular area and is equal to \pir^2. Then, F_right = (P) (\pir^2) Since the half-bubble is in equilibrium, we have from the first condition of equilibrium that F_left = F_right and4\piry = P\pir^2 yielding P = 4\Upsilon/r for a soap bubble. The excess pressure in the bubble is then p = 4\Upsilon/r = (4 × 25 dynes \bullet cm^-1)/(1 cm) = 100 dynes \bullet cm^-2.

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Question:

Show the mechanism for the transesterification of methyl benzoate with ethanol in an acidic solution.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E21-0782.htm

Solution:

Begin this problem by defining transesterification. An ester is an organic compound of general formula , where R and R' are alkyl groups that can be the same or different. In transesteri-fication, the alcoholic portion of the ester is exchanged for another alcoholic group. You are asked to give the mechanism for the reaction shown in figure A. You will note that you exchange the methyl alcohol portion with ethyl alcohol. From radioactive isotopes, the following mechanism shown in figure B has been established. Due to the electronegativity of oxygen, the oxygen of the carbonyl group, , has a slightly negative charge and the carbon has a slightly positive charge. The proton from the acid is attracted to the oxygen as in figure C. These two are in equilibrium with each other. The carbon atom is positively charged and seeks electrons. They can be obtained from the unpaired electrons on the oxygen atom in the ethanol molecule. Thus, the oxygen bonds to that carbon; you still have the positive charge, though. A proton comes off of the oxygen from ethyl and is transferred to the oxygen of methyl. See figure D. Methanol (CH_3OH) leaves, allowing the compound shown in figure E to expel a proton, which leads to the formation of ethylbenzoate. The acid expelled acts as a catalyst for the transesterification of another compound in the solution.

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Question:

Draw the open-chain structure of glucose (C_6H_12O_6) and number the carbon atoms. Is this the correct structure of glucose in actuality?

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/Users/wenhuchen/Documents/Crawler/Biology/F01-0025.htm

Solution:

The open-chain aldehyde form of a glucose molecule has the following structure and numbering system: Numbering starts with the carbon of the carbonyl functionality. Until the early 1890's this open-chain form had been widely accepted as the only structure of glucose. This structure became questionable as it could not account for numerous experimental observations, among which was the fact that glucose fails to undergo the reactions typical of an aldehyde. In 1895 a cycle structure for glucose was proposed: This cyclic form can be looked upon as the result of an intramolecular reaction involving the aldehyde group at C_1 and the hydroxyl group at C_5 of the open-chain structure. Actually, in an equilibrium mixture, there exist two forms of glucose which differ with respect to the configura-tion at C_1. These two forms, called anomers, are designated \alpha (alpha) and \beta (beta): These cyclic anomers of glucose exist in equilibrium with the open- chain form, but the latter occurs to a very minor extent. The following expression depicts their relationship: \alpha-glucose \leftrightharpoons open-chain aldehyde form \leftrightharpoons\beta-glucose \beta-glucose Experimentally it has been found that an equilibrium solution of glucose contains about 36% of the \alpha form and 64% of the \beta form, with the open- chain form occurring less than 0.5% of the time.

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Question:

If a tunnel were drilled through the earth along one of its diameters and if a stone were dropped into it from one end, how long would it be before the stone returned? Compare the answer with the period of an earth satellite in an orbit of minimum radius and comment on the two values. Assume the earth to be of uniform density, and make use of the information that a body inside the earth at a distance r from the center has a gravitation-al force acting on it due only to the portion of the earth of radius r.

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/Users/wenhuchen/Documents/Crawler/Physics/D09-0389.htm

Solution:

Consider the regions I and II of the earth constructed out of the conical sections with pivot m (see the figure). Region I contains a smaller mass than region II. It is, however, closer to m than region II. Since the gravitational force due to a mass is directly proportional to the mass, and inversely proportional to the square of distance, then it appears that regions I and II provide equal and oppos-ite forces on m, due to the tradeoff between mass and distance. A detailed mathematical analysis verifies that regions I and II do neutralize each other and have no resultant effect on m. This is true for any such regions constructed by the above procedure, and the net effect of the portion of the earth outside the sphere of radius r is zero. At any distance r from the center the stone is acted on by a force of magnitude F = - [(GM'm)/(r^2)] = - [(GM)/(r^2)]×(4/3)\pi\rhor^3 = -(4/3)\pi\rho Gmr, where m is the mass of the stone, \rho the density of the earth, and M' the mass of that portion of the earth of radius r. We also used the fact that M' = \rhoV = \rho(4/3 \pir^3). The negative sign indicates that the force acts in a direction that opposes an increase in separa-tion of the two bodies (it's attractive). Therefore, by Newton's second law, F = ma = - [(4/3) \pi\rho Gm] r orF = - kr where k = 4/3 \pi\rho Gm. The form of this equation indicates that the mass is undergoing simple harmonic motion about the center of the earth. The period of any simple harmonic oscillator of mass m is T = 2\pi\surd(m/k) where k is the effective spring constant. Therefore, in this case, T = 2\pi/\surd{(4/3) \pi\rhoG} But at the surface of the earth the gravitational force of attraction on the mass m (i.e. its weight) is W = mg = GMm/R^2 = (4/3) \pi\rho GmR, by Newton's Law of Universal Gravitation. R is the radius of the earth and M is now the total mass of the earth. Thus g/R = (4/3)\pi\rhoG HenceT = 2\pi\surd(R/g) = 2\pi\surd[(6.370 ×103m)/(9.8 m\bullets^-2)] = 5061.6 s = 84.8 min. The time period of an earth satellite in a circular orbit is T' = 2\pid^3/2 / R\surdg, where d is the distance of the satellite from the center of the earth. If the height of the satellite above the earth's surface is negligible in comparison with the radius of the earth, then to a first approximation d = R, and T' = 2\pi\surd(R/g), the same period as that of the stone under-going simple harmonic motion along the tunnel. This is not surprising, since any simple harmonic motion may be considered the projection on a diameter of the motion of a point in a circle with constant speed. Thus we should expect the period of an earth satellite near the earth's surface to be the same as the period of a particle undergoing simple harmonic motion along a diameter of the circle.

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Question:

(a) a 10,000 kg truck moving at 27.8 m/sec, (b) a 50 mg flea flying at 1 m/sec,and(c) a water molecule moving at 500 m/sec. The student proceeded to calculate the wave-length in centimeters of each object. What were these wavelength? h = 6.626 × 10^-34 J sec.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E18-0676.htm

Solution:

The answer to this question requires the use of de Broglie's equation to solve for the wavelength of a moving particle. It states that \lambda = h/mv, where \lambda = wave-length, m = mass, v = velocity, and h = Planck's constant. One is given h, m, and v. Substitute these values into the equation and solve for \lambda. 1 cm = 10^-2 m. (a)\lambda = (h/mv) = [(6.626 × 10^-39 J sec)/(10,000 kg \textbullet 27.8 m/sec)] = 2.4 × 10^-34 meters = 2.4 × 10^-37 cm (b)\lambda = (h/mv) = [(6.626 × 10^-34 J sec)/(5 × 10^-5 kg \textbullet 1 m/sec)] = 1.3 × 10^-29 meters = 1.3 × 10^-27 cm (c)A mole of water has a mass of 18 g. Thus, \lambda = (h/mv) = [(6.626 × 10^-34 J sec) / {(.018 kg/mole) /(6.02 × 10^23 molecules/mole)] 500 m/sec} = 4.43 × 10^-11 m = 4.43 × 10^-9 cm

Question:

Write a program which converts a number in base ten [with value\leq 2^20 ] into the corresponding representation in base two.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G02-0035.htm

Solution:

The method upon which our solution is based is to subtract off the highestpower of two contained in the given number. The remainder then becomesthe new number and we continue subtracting off powers of two untilthe exponent of the power of two subtracted off is zero. Thus, if N, the originalnumber, is given by N = I \textasteriskcentered (2\textasteriskcentered\textasteriskcenteredE) + R (I = integer quotient, R = remainder), thenwe can calculate I from I =INT(N/(2\textasteriskcentered\textasteriskcenteredE)), sothat R = N - I\textasteriskcentered(2\textasteriskcentered\textasteriskcenteredE). RbecomesN in the next iteration. 1\OREAD N 2\OPRINT N; \textquotedblleftBASE TEN= \textquotedblright 3\OFOR E = 20TO0STEP - 1 4\OLET I=INT(N/2\uparrowE) 5\OPRINT I; 6\OLET R = N - I\textasteriskcentered2\uparrowE 7\OLET N = R 8\ONEXT E 85PRINT \textquotedblleftBASE TWO\textquotedblright 86PRINT 9\OGO TO 10 1\O\ODATA999999.,16 11\OEND

Question:

What is thegyrofrequencyof an electron in a magnetic field of 10 kilogauss, or 1 × 10^4 gauss? (A field of 10 to 15 kilogauss is typical of ordinary laboratory iron core electromagnets.)

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/Users/wenhuchen/Documents/Crawler/Physics/D21-0718.htm

Solution:

The centripetal force keeping the electron in its orbit is provided by the magnetic field. This is given by F^\ding{217} = (q/c) v^\ding{217} × B× Since the velocity is perpendicular to the magnetic field, this reduces to F = (q/c)vB In circular motion, v =R\omega. Therefore F = (q/c)R\omegaB Newton's third law tells us that F = ma, and since in circular motion a = \omega^2 R, we have F = m\omega^2 R. Equating this with the above expression for force we have: q/cR\omegaB= m\omega^2 R \omega = \omega = (qB/mc) Substituting for our values: \omega = [{4.8 × 10^-10esuf(1 × 10^4 gauss)} / {.911 × 10^-27 gm (3 × 10^10 cm/sec)}] = {4.8 × 10^-6esu[(gm - cm)/(sec^2esu)} / {2.7 × 10^-17 gm (cm/sec)} = 1.8 × 10^11 sec^-1 Dividing the angular velocity by 2\pi we arrive at the frequency v v = [(1.8 × 10^11 sec^-1)/(2\pi)] = 2.8 × 10^10 cps Electromagnetic wavelength is given by \lambda = (c/v) . Therefore, this gyrofrequency gyrofrequency is equivalent to an electro-magnetic is equivalent to an electro-magnetic wavelength of wavelength of \lambda = [{3 × 10^10 (cm/s)}/(2.8 × 10^10 s^-1 )] \cong 1 cm .

Question:

Describe experimental evidence which indicates that it is not the kind ofribosomesbut the kind of mRNA which determines the type of proteins which will be produced.

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/Users/wenhuchen/Documents/Crawler/Biology/F24-0627.htm

Solution:

Most DNA is found in the nucleus. Transcription of the DNA gives rise to mRNA. So when the nucleus is removed from a cell, the source of mRNA is removed. Such an enucleated cell can live for a time, but growth and enzyme production will soon stop. For example, an enuclea-ted ameoba(anameobawith its nucleus removed)- will crawl around for a time, it may engulf food but it will eventually be unable to digest it, growth stops and in time, the organism dies. The above experimental observation can be explained in the following manner. The removal of the nucleus of the organism prevents further transcription of mRNA. The supply of digestive enzymes normally found in the organism becomes depleted because thesoruceof mRNA is cut off and the translation of mRNA into enzyme cannot occur. Therefore, the food engulfed cannot be digested. This explanation shows that mRNA manufactured in the cell nucleus is essential for protein production of the cell. In order to test what effect the type of mRNA has on the type of proteins produced, one can perform experiments where the nucleus of one species, producing certain types of mRNA, is introduced into an enucleated cell of a different species. For example, the nucleus of a mammalianreticulocyte, a cell that synthesizes hemoglobin, can be extracted and then introduced into an enucleated cell, such as an amphibianoocyte. This egg cell with a transplanted nucleus survives and produces mammalian hemoglobin instead of amphibian hemo-globin. This is good evidence showing that mRNA controls the type of proteins produced because the transplanted nucleus is the nucleus of a mammalian cell involved in the production of hemoglobin, the mRNA transcribed in the nucleus, when translated on theribosomes, produces mammalian hemoglobin. This experiment also implies that it is not the kind ofribosomesthat determines the type of proteins produced . If this were true, then the hemoglobin produced would be amphibian hemoglobin, but this does not happen.

Question:

Write an APL program to solve the following system of equations in three unknowns using Cramer's Rule. x + y + z = 6 x + y -2z = -3 2x + y - z = 1 Also, show an easier method of solving the above system using the symbol [+] from APL.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G20-0505.htm

Solution:

The determinant A of the form \mida11a_12a_13 \mid A=\mida_21a_22a_23 \mid \mida_31a_32a_33 \mid can be evaluated as follows: \vertA\vert = a11a22a_33 + a12a_23 a_31 + a13a_21 a_32 -a11a23a_32 -a12a_21 a_33 -a13a_22 a_31 According to Cramer's Rule, the system a_1x + b_1y + c_1z = d_1 a_2x + b_2y + c_2z = d2 a_3x + b_3y + c_3z = d_3 has the following solutions in term of determinants: d_1b_1c1a_1d_1c_1a_1b1d_1 d_2b_2c2a_2d_2c_2a_2b2d_2 x =d_3b_3c_3,y =a_3d_3c_3,z =a_3b3d_3 a_1b1c_1a_1b1c_1a_1b1c_1 a_2b2c_2a_2b2c_2a_2b2c2 a_3b3c_3a_3b3c_3a_3b3c3 Thus, a defined function for the evaluation of a determinant must be developed first: \nabla Z\leftarrow DeterminantX [1]A \leftarrow 33\rhoC [2]Z \leftarrow ((A [1; 1] × A [2; 2] × A [3; 3]) + (A [1; 2] × A [2; 3] × A [3; 1]) + (A [1; 3] × A [2; 1] × A [3; 2]) [3]Z \leftarrow Z-(A[1; 1] × A [2; 3] × A [3; 2]) + (A_\nabla [1; 2] × A [2; 1] × A [3; 3]) + (A [1; 3] × A [2; 2] × A [3; 1]) Note that this defined function comes directly from the method of evaluation of the determinant explained previously. The following function computes the values of x, y, z. \nabla Solution [1]'Enter the coefficients and constant' [2]So \leftarrow 34\rho [] [3]D \leftarrow E \leftarrow F 33\uparrow So [4]R \leftarrow DeterminantD [5]\rightarrow (R = 0) / END [6]X \leftarrow [Determinant D [; 1] \leftarrow So [; 4]] \div R [7]Y \leftarrow [Determinant E [; 2] \leftarrow So [; 4] \div R [8]Z \leftarrow [Determinant F [; 3] \leftarrow So [; 4] \div R [9]'The Solution is: ' ; x, y, z [10]\rightarrow 0 [11]END: "No unique solution " [12]\nabla Note that the function solution uses the function Determinant in steps [4], [6], [7] and [8]. Applied to the given system of equations x + y + z = 6 x + y - 2z = -3 2x + y - z = 1 The function solution works as follows:The coefficient and values of system's equations 111611-2-321-11. are entered first; Step [2] reshapes the vector: \mid 1116 \mid So =\mid 11-2-3 \mid \mid 21-11 \mid Step [3] takes the 3 × 3 matrix from So : \mid 111 \mid D =\mid 11-2 \mid \mid 21-1 \mid Step [4] computes the determinant of D. Steps [6], [7], and [8] are all very similar. The fourth column of So is replaced with the appropriate column of D. [For x, it is the first column, for y, it is the second column, for z, it is the third column.] Dividing the determinants of D by R will yield the appropriate solutions. The symbols A [\div] B are used in APL for solving systems of equations. A is a vector that contains the constant term (value) from each equation in the system. B is a matrix that contains all the coefficients of the system's variables. ([\div] is formed by typing in [], backspacing, and typing in \div ). To solve the given system, (1), one just types the following statements: A \leftarrow 6-31 B \leftarrow 33\rho 11111-2 2 1 -1 A[\div] B.

Question:

With what force will the feet of a passenger press down-ward on the elevator floor when the elevator has an acceleration of 4 ft/sec^2 upward if the passenger weighs 160 lb?

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/Users/wenhuchen/Documents/Crawler/Physics/D04-0127.htm

Solution:

This example illustrates a problem that is frequently encountered, in which it is necessary to find a desired force by first computing the force that is the reaction to the one desired, and then using Newton's third law. That is, we first calculate the force with which the elevator floor pushes upward on the passenger P; the force desired is the reaction to this. The figure shows the forces acting on the passenger. We may use Newton's Second Law, F = ma, to relate the net force on the man to his acceleration. The resultant force is P - W. The mass m of the passenger is his weight, mg, divided by 32 f/s^2, or 5 slugs, and his acceleration is the same as that of the elevator, \sumF = ma P - 160 lb = 5 slugs × 4 (ft/sec^2) = 20 lb, .P = 180 lb. The passenger exerts an equal and opposite force downward on the elevator floor.

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Question:

When mixed at 700\textdegreeK, H_2(g) and I_2 (g) react to produce HI(g). The reaction is first order in H_2 and first order in I_2. Suppose at time t = 0, one mole of H_2 and one mole of I_2 are simultaneously injected into a 1-liter box. One second later, before the reaction is complete, the contents of the box are examined for the number of moles of HI. What would be the probable effect on this number if each of the following changes were made in the initial conditions? (a) Use two moles of H_2 instead of one. (b) Use two moles of I_2 instead of one. (c) Use a 2-liter box. (d) Raise the temperature to 750\textdegreeK. (e) Add a platinum catalyst, (f) Add enough neon gas to double the initial pressure.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E13-0458.htm

Solution:

When a reaction is said to be first order in a particular reactant, it means that the rate of the re-action is proportional to the concentration of that re-actant. Thus, for this reaction the rate law can be written Rate = k [H_2][I_2], where k is the rate constant and [ ] indicates concentration. When 1 mole of each H_2 and I_2 are used, one can solve for the rate by substituting these values into the rate law. Rate = k (1)(1) = 1k Solving for the rate when [H_2] is 2 moles Rate = k[2][1] = 2k Thus, when the concentration of H_2 is doubled the rate is doubled. (b)One can solve for the rate when [I_2] = 2 moles by substituting into the rate law. Rate = k(1) (2) = 2 k Thus when [I_2] doubled the rate is also doubled. (c)Boyle's law states that the pressure is inversely proportional to the volume. Thus, when this reaction which originally takes place in a 1-liter box is moved to a 2- liter box, the pressure of the reactants is halved. Solving for the rate: Rate = k (1/2) (1/2) = 1/4 kor.25 k. Thus, for this reaction, when the pressure of the reactants is halved, the rate is 1/4 of the original rate. (d)Product is formed when an atom of H collides with an atom of I, thus the rate of the reaction will increase with a rise in temperature because the atoms will move more quickly creating a greater chance for collisions. (e)By definition, a catalyst will increase the rate of the reaction by lowering the amount of energy needed for the reaction to proceed. (f)When neon gas is added to increase the pressure in the 1-liter box, the rate of the reaction does not change because the concentrations of H2and I_2 remain unchanged. When the pressure in the system is doubled by decreasing the volume by 1/2 the rate is increased because the con-centrations of H_2 and I_2 are increased. If one starts out with one mole per liter of H_2 at 1 atm and then increases the pressure to 2 atm by adding neon, the con-centration of H_2 is still 1 mole per liter. When the pressure in a container containing 1 mole per liter of H_2 is increased from 1 to 2 atm by halving the volume of the container, the concentration of H_2 is now 1 mole per 1/2 liter or 2 moles per liter. Thus when neon is added to the system in this problem, the rate is unchanged.

Question:

Some scientists object to the Darwin-Wallace theory of naturalselection on the basis that it cannot explain the presenceof many apparently useless structures in some organisms. These vestigial structures seem to defy the processesof natural selection which tend to provide successivegenerations with better adaptations to their environment. How can you defend the theory ofnatural selection?

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Solution:

While certain structures may be useless to the organism in which theyoccur, they may be present because they are controlled by genes thatare closely linked to other genes controlling characteristics important forsurvival. Such closely linked genes will be inherited together in most cases, and so the vestigial characteristic will tend to be inherited along withthe more important trait. Other vestigial structures may have physiologicaleffects of survival value which are not visible to the observer. They may represent intermediate stages in the evolution of some importantadaptive structure, or be an incidental by-product of an importantbut invisible physiological trait. They may also be present becausethey have become fixed in a population by chance or genetic drift. Genetic drift is important in the evolution of small, isolated populations. Within a small breeding population, heterozygous gene pairs tendto become homozygous for one allele or the other by chance rather thanselection. This may lead to the expression and accumulation of certainphenotypes of no particular adaptive value. If the phenotypes are notlethal, they will remain in the popu-lation. Because natural selection canact only when there is a certain level of variation among the individuals, it loses its significance in a small breeding population. Under thiscondition, chance becomes the important factor in evolution. Thus, the presence of apparently useless traits does not necessarilyinvalidate the theory of natural selection. It should be noted thatnatural selection and the presence of vestigial structures are mutually exclusivephenomena, and therefore, the inheritance of vestigial traits shouldnot be governed by selection processes. Natural selection is definedas the process by which a beneficial trait is selected for, such that moreorganisms possessing the gene for this trait will be present in the population. These are the organisms producing the most offspring, and thuscontributing the most genes to the population's gene pool. Vestigial structuresare defined as those structures which no longer have any biologicalsignificance, neither advantageous, nor deleterious to the organismpossessing them. Since natural selection operates to select for advantageoustraits by eliminating deleterious traits, and vestigial structuresare neither beneficial nor harm-ful, these structures will neither beselected for nor eliminated. However, it should be noted that many vestigialtraits are remnants of what were once important structures. While thegenes for the vestigial traits were not directly selected for, their precursortraits were, and so more individuals than not will tend to possessthe vestigial trait. Its inheritance in future generations, however, willnot be governed by natural selection.

Question:

What would happen if, following ovulation, no corpus luteum was formed?

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Solution:

After rupture of the follicle and discharge of the ovum during ovulation, a transformation occurs within the remaining follicle in the ovary, giving rise to a yellowish glandlike structure called the corpus luteum. If pregnancy does not occur, the corpus luteum degenerates in 10 days. If pregnancy does occur, the corpus luteum grows and persists until near the end of pregnancy. The corpus luteum, which is composed of follicular cells, continues to secrete estrogen as the follicular cells did prior to ovulation. In addition, the corpus luteum cells also secrete progesterone. Progesterone functions in preparing the uterus to receive the embryo. This hormone acts on the uterine lining, causing maturation of the complex system of glands in the lining. Estrogen also acts on the uterus to cause a thickening of the lining. This, however, occurs primarily during the stage of follicle formation prior to ovulation, called the follicular phase. The luteal phase follows ovulation, and during this phase progesterone has the primary influence on the development of the uterine lining. This hormone also stimulates breast growth, particularly of glandular tissue. If a female failed to develop a corpus luteum following ovulation, a hormonal imbalance would result. There would be a lower than normal concentration of estrogen and progesterone in the circulation. As the levels of these two hormones become low, the feedback inhibition of FSH and LH production is removed. The FSH concentration begins to rise in the plasma, and when it reaches a sufficiently high level, follicle and oocyte development are stimulated. Thus a new menstrual cycle would begin shortly after ovulation if the corpus luteum did not develop. Even worse would be the consequence if fertilization occurred and the corpus luteum did not form. The uterus would then fail to mature and implantation of the fertilized egg would be prevented. Even if implantation did occur, subsequent development would be abnormal because of incomplete placental development.

Question:

Two very long straight parallel wires carry currents i_1 and i_2 and are a distance r apart. Calculate the force on a length I_2 of the wire with current i_2 Verify that the force is attractive when the currents are in the same direction, but repulsive when they are in opposite directions .

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Solution:

Figure a is the case of the two currents flowing in the same direction. The magnetic lines of forces due to i_1 are circles that go around clockwise for an observer looking in the direction of i_1 . The magnetic field produced by i_1 at a point on the other wire is (in the CGS system) B = (2i_1 )/(cr)(1) This field is perpendicular to i_1 and to the radius r drawn perpendicular to i_1 and to the radius drawn perpendicularly to both wires. Since B also perpendicular to i_2 , \O = 90 o , sin \O = 1. The force exerted by the field on lenth I_2 of the wire with current i_2 is given by F_m = (1/c) i_2 l_2^\ding{217} × B^\ding{217} . I^\ding{217} and B^\ding{217} are perpendicular to each other (see the diagram). Therefore, F_m = [(i_2 I_2 B)/(c)](2) Combining equations (1) and (2), the required force is F_m = [(2i_1 i_2 I_2 )/c^2 r)] In figure a the magnetic field is into the plane of the paper. Applying the right hand rule the thumb would point to the left, toward i_1 . Figure b is the case of currents flowing in opposite directions, i_1 is still upward and B is still into the plane of the paper. Using the right hand rule. The thumb would point to the right, so the force is repulsive.

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Question:

What is the electric field intensity at a point 30 centi-meters from a charge of 0.10 coulombs?

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Solution:

The electrostatic force on the test charge q, due to the charge Q is, by Coulomb's law, (see figure) F = (kQq)/(r^2) The electric field intensity at point B is defined as: \epsilon = (F/q)[ = {(kQq)/(r^2)/q}] = (kQ)/(r^2) In the problem we are presented with, we have Q = 0.10 coulomb, r = 30 cm = 3.0 × 10^-1 m Then \epsilon = 9 × 10^9 (nt - m^2)/(coulombs^2) × (0.10 coulombs)/(3.0 × 10^-1)^2 m^2 = 10^10 nt/coulombs

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Question:

A non-uniform bar of weight W\ding{217} is suspended at rest in a horizontal position by two light ropes as shown in the figure. One rope makes an angle of 30\textdegree with the vertical and the other an angle of 60\textdegree. If the length l of the bar is 10 m compute: (a) the tension in each rope and (b) the distance X from the left-hand end of the bar to the center of gravity.

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Solution:

For the bar to be in equilibrium the sum of the forces that act on it and the sum of the torques must equal zero. (a) We will treat the forces in terms of their horizontal and vertical components. The sum of the components in these directions must equal zero. The sum of the components in the horizontal direction is: T_2 sin 60\textdegree - T_1 sin 30\textdegree = 0sin 60\textdegree = \surd3/2 and sin 30\textdegree = 1/2. Then (\surd3/2) T_2 = (1/2)T_1 T_1 = \surd3 T_2 where T_1 is the tension of the wire making an angle of 30\textdegree with the vertical and T_2 is the tension of the other wire. Forces pointing to the right are taken as positive and those pointing to the left as negative. The sum of the components in the vertical direction is: T_2 cos 60\textdegree + T_1 cos 30\textdegree - W = 0 W = \surd3/2 T_1 + (1/2) T_2 = \surd3/2 (\surd3 T_2) + (1/2) T_2 = (3/2)T_2 + (1/2) T_2 = 2 T_2 T_2 = (1/2)w T_1 = \surd3 (1/2 W) = (\surd3/2) W where forces pointing upward are taken as positive and those pointing downward as negative. (b) To calculate X, we set the sum of the torques equal to zero. Torque is defined as: \cyrchar\cyrt\ding{217} = r\ding{217} × F\ding{217} \cyrchar\cyrt\ding{217} = r F sin \texttheta where r is the distance of the point of action of the force F from an arbitrary reference point. The sum of the magnitudes of the torques about the point of appli-cation of the force W is: T_2 (l - x) sin 150\textdegree - T_1 X sin 120\textdegree = 0 sin 50\textdegree = 1/2 , sin 120\textdegree = 0.866 = \surd3/2. Then (1/2)T_2(l - x) = (\surd3/2)T_1X (1/2){(1/2)W} (10 m) - (1/2){(1/2)W}X = \surd3/2 [(\surd3/2)W]X (4/4) WX = (5/2)Wm, x = (5/2)m.

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Question:

The density of ice at 0\textdegreeC is .917 g/cm^3 and has an entropy of 37.95 J mol^-1 deg^-1. The density of liquid water at this temperature is .9998 g/cm^3 and has an entropy of 59.94 J mol^-1 deg^-1. Given these data, calculate the change of entropy ∆S, change of enthalpy ∆H, and the change of energy ∆E for the conversion of one mole of ice to liquid water at the normal melting point.

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Solution:

The solution of this problem involves the ability to relate energy, free energy, entropy, and enthalpy. One must first calculate the change in entropy. You are told the entropies of both ice and water. The change in entropy is their difference or ∆S =S_liq-S_solid= 59.94 - 37.95 = 21.99 J mol^-1 deg^-1 . To calculate the change in enthalpy at the melting point use the fact that ∆G = 0 at the melting point of a substance. ∆G = ∆H -T_Mp∆S, where ∆H = change in enthalpy,T_Mp= temperature of melting point in Kelvin and ∆S = change in entropy. Since ∆G = 0 at the melting point, ∆H =T_Mp∆S. From the previous calculation, you know ∆S.T_Mpis given. Thus, ∆H = T∆S ∆H = (273.15K) (21.99 J mol^-1 deg^-1) = 6010 J/mole. To calculate the ∆E for the conversion of one mole of ice to liquid water at the normal melting point, employ the equation ∆E = ∆H - P∆V, where P = pressure, V = volume. ∆H = 6010 J/mole. To determine P∆V, note that ∆V isV_liq-V_solid. You are given the densities of ice and water. Recalling that density = mass/volume, and that the molecular weight (mass) of ice or water is 18.015 grams/mole, you have forV_liq: 18.015/.9998. ForV_solid, You have 18.015/.917.Therefore ∆V =V_liq-V_solid= [(18.015 g/mole)/(.9998 g/cm^3)] - [(18.015 g/mole)/(.917 g/cm^3)] = - 1.63 cm^3 /mole = - .00163 liter/mole. Since one is asked to make these calculations for 1 mole of H_2 O, the change in volume, ∆V, equals - .00163 liter. Because you are asked to calculate at the normal melting point, the pressure must be 1atm(by definition, the "normal" is considered to be under a pressure of one atm.) Therefore, P∆V = (1.00atm) (-0.00163 liter) = -0.00163 liter - atm. Since 1 liter -atmequals 101.3 J, P∆V is equal to (-0.00163 liter -atm) (101.3 J liter^-1 atm^-1) = - 0.1651 J. Therefore, ∆E = ∆H - P∆V = 6010 + .1651 = 6010.1651 J.

Question:

What is meant by a gene "repressor"? A "corepressor"? An "inducer"? How are they used to regulate protein synthesis?

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Solution:

It is an established fact that all the cells of a given organism contain the same kind and number of chromosomes. However, different cells perform different functions. For example, muscle cells are involved in lo-comotion, whereas nerve cells form the basis of perception and response. The variation in function of different cell types implies that each is using a different portion of its total store of genetic information. In addition, in both simple and complex organisms, different cell stages, environ-mental conditions, and available nutrients require the functioning of different enzyme systems at different times. It would be wasteful indeed for a cell to be manufacturing certain proteins at a time when they are not needed. Thus, both cell differentiation and function require the presence of a control system to regulate the expression of the genes in the cells. The agents for these control systems are repressors, corepressors, and inducers. A repressor is a regulatory protein molecule that "turns off" or represses the transcription of a gene and hence inhibits the synthesis of the protein (enzyme) which that gene specifies. Remember that enzymes play a direct role in regulating reactions at the metabolic level. However, the production of that enzyme depends on the trans-cription of the corresponding gene. In the absence of transcription of that gene, the cell cannot perform the certain reaction which requires that corresponding enzyme, hence the activity of the cell is regulated. Repressors prevent transcription by binding to the operator of an operon, thereby preventing the transcription of its genes by RNA polymerase. There are two distinct systems by which repressors operate to regulate transcription. In an inducible system, the repressor protein is produced in an active form and as such is able to bind to the operator. However, in the presence of an inducer molecule, the rep-ressor is no longer able to bind to the operator, and the gene is "turned on". The inducer molecule may be a protein, a given substrate or a metabolite whose presence indicates a need for the enzymes that the inducible genes code for. Consider an inducible system such as the lac operon. When E. coli is grown on a glucose medium, there is no need to produce the enzymes which metabolize lactose and the corresponding genes are not transcribed. However, if E.coli is grown on a lactose medium, the lac operon must be "turned on" in order for the lactose to be metabolized. Here, the lactose molecule functions as an inducer. The lac repressor protein has two active sites: one that interacts with the operator, and one that can bind lactose. The repressor can interact with lactose while it is bound to the operator. However, the interaction with lactose produces a confor-mational change in the repressor protein (the repressor is an allosteric enzyme) and this change prevents the binding of the repressor with the operator. Transcription by RNA polymerase can now occur. The second mode of regulation is found in a repressible system. An example of a repressible system is the histidine biosynthetic pathway. Here, the repressor protein is produced in an inactive form which is unable to bind to the operator. This allows production of the enzymes which synthesize histidine. The repressor protein has a binding site for histidine, the corepressor molecule. As the level of histidine in the cell increases, histidine binds to the repressor protein, forming an active repressor - corepressor complex which can bind to the operator and prevent transcription. Histidine is called the corepressor, since in its absence there is no repression. Repressible systems are widely employed in biosynthetic pathways with the end products acting as corepressors to repress their own synthesis when they reach sufficient levels in the cell. As seen in E. coli, the synthesis of a repressor protein may be constitutive. In some systems, however, the synthesis of the repressor is itself under control. In these systems, the repressor protein, in addition to binding to the operator of a given operon, can bind to the operator of its own gene or genes. It thus, acts as a repressor of its own synthesis when its level in the cell reaches a given quantity. Such a system is depicted in the accompanying figure. It is important to differentiate between repression and feedback (end product) inhibition. Repression operates at the level of transcription to prevent enzyme production. Feedback inhibition acts at the enzyme level to prevent enzyme activity. In feedback inhibition, an end product actsto inactivate the enzymes responsible for the early steps of its synthesis. Regulation of cellular functions is much quicker by feedback inhibition than by repression, since repression merely prevents further enzyme production, leaving the enzymes already present in a cell completely active. It is through feedback inhibition that a cell can "turn off" the enzymes already present. One can see that by using a combination of systems, a cell can execute rather fine control over its metabolic activities.

Question:

All organic matter is made up entirely or mostly of the basic elements carbon and hydrogen. In view of this, why is there such a diversity of organic compounds present?

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Solution:

The diversity of organic compounds is so vast that these organic compounds have been divided into families, such as alkanes, alkynes and aromatic compounds, which have no counterparts among inorganic compounds. The tremendous variety of these compounds is made possible by the unique properties of carbon. To be able to understand how carbon can form such a huge number of compounds with a great variety of properties, the way in which the atoms in these molecules are bonded together must be looked at. Carbon has a valence number (the number of bonds an atom of an element can form) of four. This means that each carbon atom always has four bonds which can be either bonded to four other atoms, as in methane; to three other atoms, as in formaldehyde; or to 2 other atoms as in hydrogen cyanide. In other words carbon is capable of forming single, double and triple bonds: But the diversity this element possesses does not stop here. It is capable of bonding to other carbon atoms in an almost unique variety of chain and ring structures. This property is called catenation and accounts for the tremendous variety seen in the following compounds: These examples show the use of a minimum number of carbon atoms. The diversity possible is apparent when one considers the fact that these different types of bond can be joined together in infinite numbers. To exemplify this, look at the compound: As it can be seen, a multitude of different variations could be made with the only constraining factor being that carbon must have four bonds. The freedom allowed carbon is apparent when contrasted with the halogens (iodine, chlorine, bromine, and flourine) which are monovalent and thus can only form one bond, for example: H-ClBr-Br hydrochloric acidbromine Another feature of the compounds of carbon that contributes to their variety is the existence of isomers (compounds composed of the same number and kind of atoms, but with the atoms arranged differently in space). Look, for example, at the compound C_4H_10. This can exist as: Although these compounds have the same chemical formulas they have different chemical, physical and biological properties due to their arrangement in space.

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Question:

The "pill" is a birth control device available to women, and is one of the most effective and widely used contraceptives in today's society. How does it function? Are there any side effects when taking the pill?

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Solution:

The pill contains a small dose of synthetically produced female sex hormones, primarily estrogen and progesterone. These two substances act together to suppress ovulation, thus disrupting the normal reproductive process. The progesterone-estrogen pill acts on the cervix of the uterus to increase mucus secretion which impairs the motility of sperms. It acts upon the uterine lining, preventing implantation because the fertilization of the egg is not in synchrony with the development of the uterus. But most importantly, the pill prevents FSH and LH secretions by direct effect on the pituitary, causing the uterus to be sustained in its post-ovulatory state. It achieves this by acting as a feedback inhibitor on the anterior pituitary, so that FSH and LH secretions are suppressed. With mi-nimal amounts of FSH and LH, follicular maturation and ovulation cannot occur. The pill must be taken daily for 20 or 21 days of the 28-day menstrual cycle, starting with the fifth day after the beginning of menstruation. When taken each day without fail, it is virtually 100% effective. There are, however, some side effects when using the pill. Up to 25% of the women taking the pill develop some of the outward symptoms of pregnancy: swelling and tenderness of the breasts, nausea, irritability, etc. Some users of the pill also develop increased susceptibility to inflam-mation of the veins combined with problems ,of blood clotting. The principle concern over the use of the pill is that the levels of the synthetic estrogens in certain pills are dangerous, since these synthetic estrogens had been shown to be carcinogenic in experimental animals. There are two major points that should be considered in the use of the pill. In the first place, do oral contra-ceptives significantly threaten the users with blood clot problems? Secondly, is prolonged use of the pill by humans going to result in a marked rise in breast cancer? It is the view of many endocrinologists that the evidence in experimental animals suggests that such a threat may exist among women presently planning to use oral contra-ceptives .

Question:

What are the types of simple conditional expressions and compound conditional expressions? Give their formats and any conditions if necessary.

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Solution:

The Simple Conditional Expression: A simple condition reducing to the value true or false can be ex-pressed in the following types of simple conditional ex-pression: i) a relation ii) a condition - name iii) a sign condition iv) a class condition v) a switch status condition. Any of the above are used in a decision-making opera-tion to select different paths of control in a program. 1) The Relational Condition: This condition causes a comparison of magnitude between two quantities. The general form is: Note: means choose one. The first quantity is called the subject and the second is referred to as the object of the condition. A relational operator must be preceded and followed by a space. The list of relational operators are GREATER, LESS, EQUAL. The word, NOT, before an operator specifies the opposite of what the operator normally specifies. e.g.,AGE IS GREATER THAN 21 SUM IS NOT LESS THAN 100 A+B IS EQUAL TO A+M The words IS and THAN are optional and do not alter the meaning of the expression. e.g.,AGE GREATER 21 is equivalent to AGE IS GREATER THAN 21. 2) The Condition-Name condition: A condition name is a name, assigned in the Data Division, to one of the values a conditional variable assumes. The variable is tested to de-termine whether or not its value is equal to a value asso-ciated with a particular condition name. Using an IF statement, the test to determine marital status may appear as follows: IF MARRIED THEN......... which is equivalent to IF MARITAL-STATUS IS EQUAL TO 2 THEN...... NOT is used to specify the opposite of what the condition- name test would normally specify. e.g.,IF NOT SINGLE ................ is equivalent to IF MARITAL-STATUS IS NOT EQUAL TO 1............. (3) Sign Condition: This condition determines whether a numeric quantity is less than, equal to, or greater than zero. The general format is: Note: [ ] means optional e.g., A + B {_\ast} CIS POSITIVE The identifier in a numeric status test must always represent a numeric value. 4) Class Condition: The class condition test deter-mines whether a quantity is purely numeric or alphabetic. The general format of this conditional expression is as follows: The test must be consistent with data description of the item being tested. That is, only the NUMERIC test may be used for data which has been described as numeric and only the ALPHABETIC test for alphabetic data either may be used for alpha numeric data. The usage of the identifier must be defined, either explicitly or implicitly, as DISPLAY. 5) The Switch Status Condition: This condition is used to determine whether a particular hardware switch is off or on. The Compound Conditional Expression A conditional expression having a single condition is referred to as a simple conditional expression. One con-taining more than one condition is referred to as a compound conditional expression. The various conditions are connected by the logical operators AND and OR and the format is: For example: A IS EQUAL TO B OR C IS LESS THAN D AND E IS GREATER THAN F Implied Subject: In a compound conditional expression it is possible for several consecutive conditions to have the same subject. e.g., SUM IS GREATER THAN 100 OR SUM IS LESS THAN 1000 ie equivalentto SUM IS GREATER THAN 100 OR LESS THAN 1000. Implied Operator:Operators, as well as subjects may be im-plied in a compound condition consisting of a consecutive series of relational expressions. e.g., SALES IS EQUAL TO 25 OR EQUAL TO 66 OR EQUAL TO 85 is equivalent to SALES IS EQUAL TO 25 OR 66 OR 85 Implied operators must only be used in expressions where the subjects are also implied.

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Question:

Discuss why one would place Cro-Magnon man higher on the evolution ladder than Neanderthal man.

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Solution:

During the period ranging from 40,000 to 10,000 year ago, members of our species - Homo sapiens - continued to live as small-band hunters and gatherers. They spread to various parts of the world where they encountered and adapted to a variety of environmental conditions. Due to their burial of the dead, several nearly complete Cro-Magnon skeletons have been preserved. The bones of Cro-Magnon remains indicate a robust build, but not as robust as the Neanderthal's. Restorations reveal that Cro-Magnon man had smaller eye ridges, a less receding forehead, smaller and shorter face, narrower and higher-bridged nose, and a more prominent chin than Neanderthal man. His brain size resembled that of modern man more than that of the Neanderthals. Generally, therefore, Cro-Magnon man had a more modern appearance. Aside from having more modern physical features, Cro-Magnon man had a culture that was more refined and di-verse than the one of the Neanderthals. Due to the great variety of environments entered, there was a great variety of habitations. Besides rock shelters and caves which were widely inhabited, Cro-Magnon populations also lived in oval huts constructed out of large bones, tree branches and animal skins. TheCro- Magnons inherited and continued many of the Neanderthal tool technology. They also manufactured finer, more complex stone tools and other delicate implements. For example, knifelike blades of high quality were produced, and fishing hooks and needles with an eye were invented. The Cro-Magnons were painters as well as skilled craftsmen. They also domesticated the dog and developed fur clothing. The most important achievement of the Cro-Magnon culture is probably the development of a fully articulate language with speech. A refined communication system can lead to a refined cultural assemblage, which in turn leads to greater adaptability and powers of exploiting the environment. Another great evolutionary advantage of Cro-Magnon man over Neanderthal man is the ability to make notational counts enabling him to record events for the future. Hence, the culture of the Cro-Magnon man is seen to involve a greater degree of higher level functioning, which may reflect an increase in intelligence in this hominid group.

Question:

A pendulum which has a period of 1 s in London, where g = 32.200 ft\bullets^-2, is taken to Paris, where it is found to lose 20 s per day. What is the value of g in Paris?

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Solution:

Since the period of the pendulum is 1 s in London, the number of oscillations it performs per day is [(60 sec)/(1 min)] × [(60 min)/(1 hr) × [(24 hr)/(1 day)] = 86,400 oscillations/day In Paris itlooses20 s per day, i.e., it makes only 86,380 complete oscillations per day. In all pendulum formulas the period is T = k/\surdg, where k is a constant depending on the shape and poss-ibly the mass of the pendulum. Thus, in London, T = k/\surdg, and in Paris, T' = k/g'. Since T = 1/f, where f is the number of oscillations in a given time interval, T/T' = [(1/f)/(1/f ')] = f'/f = (86,380 oscillations/day)/(86,400oscillations/day) = \surd(g'/g) org'= [(8638)/(8640)]^2 g = [(8640 - 2)/(8640)]^2 g = [1 - (2/8640)]^2 g \cong [1 - (4/8640)] g since (2/8640) << 1. Hence,g'\approx [1 - (4/8640)] × 32.200 ft\bullets^-2 = 32.185ft\bullets^-2. Neglecting rotational effects, Paris is slightly farther away from the center of the earth than London.

Question:

A 10^6 liter tank of seawater contains 16,600 kg of chlorine (Cl^-), 9200 kg of sodium (Na^+) and 1180 kg of magnesium (Mg^++). Calculate themolarityof each. Is all the charge accounted for?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E08-0283.htm

Solution:

You are given the volume and weight of the materials in seawater and asked to calculate themolarity. Sincemolarity= no. of moles/liters, you need to calculate the number of moles ofCl^-, Mg^++ and Na^+ present to solve this problem. A mole = (no. of grams/atomic weight) . ForCl^- , then, the no. of moles = [(1.66 × 10^7 g)/(35.45 g/moles)] = 4.68 × 10^5 moles Therefore, itsmolarityis [(4.68 × 10^5)/10^6] = .468 MCl^- For Mg^++ , the no. of moles = [(1.18 × 10^6 g)/(24.3 m/g)] Itsmolarityis, thus, [{(1.18 × 10^6 g)/(24.3 m/g)} / (10^6 liters)] = .048 M For Na^+ , you have [(9.200 × 10^6 g)/(23 m/g)] = no. of moles of Na^+. Itsmolarity= [{(9.200 × 10^6 g)/(23 m/g)} / (10^6)] = .400 M. The total positive charge is .400 + 2(.048) = .496 M. The concentration of Mg^++ must be doubled (in total positive charge) since each ion has 2 charges instead of one, as in Na^+. The total negative charge stems only fromCl^-, which must equal .468 M. Thus, 0.496 moles/liter - .468 moles/liter = .028 moles/liter negative charge is unaccounted for, since all opposite charges must cancel each other out.

Question:

HO - OH(g) \rightarrow 2OH (g) ∆H\textdegree_diss= 51 Kcal/mole From this value and the following data calculate: (a) ∆H\textdegree_fof OH(g). (b) C-O bond energy; ∆H\textdegree_dissin CH_3OH(g). ∆H\textdegree_fof H_2O_2(g) = \rule{1em}{1pt} 32.58 Kcal/mole. ∆H\textdegree_fof CH_3(g) = 34.0 Kcal/mole. ∆H\textdegree_fof CH_3OH(g) = \rule{1em}{1pt} 47.96 Kcal/mole.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E14-0480.htm

Solution:

(a) ∆H\textdegree_dissfor this reaction is equal to the ∆H\textdegree_fof the reactants subtracted from the ∆H\textdegree_fof the products, where ∆H\textdegree equals enthalpy or heat content. ∆H\textdegree_diss= 2 × ∆H\textdegree_fOH \rule{1em}{1pt} ∆H\textdegree_fHO-OH = 51 Kcal/mole 2 × ∆H\textdegree_fOH = 51 Kcal/mole + ∆H\textdegree_fHO-HO ∆H\textdegree_fOH = (51 Kcal/mole - 32.58 Kcal/mole)/ 2 ∆H\textdegree_fOH = (18.42 kcal/mole)/2 = 9.21 Kcal/mole. (b) CH_3OH dissociates by breaking the C-O bond: CH_3 OH(g) \rightarrow CH^+_3 (g) + OH^\rule{1em}{1pt}(g) Therefore, the bond energy of the C-O bond equals ∆H\textdegree_diss\bullet∆H\textdegree_dissequals the sum of the ∆H's of the products minus the ∆H's of the reactants. Thus, bond energy of C\rule{1em}{1pt}O = ∆H\textdegree_fof CH^+_3 (g) + ∆H\textdegree_fof OH^\rule{1em}{1pt} (g) \rule{1em}{1pt} ∆H\textdegree_fof CH_3OH (g). bond energy of C\rule{1em}{1pt}O = 34.0 Kcal/mole + 9.21 Kcal/mole + 47.96 Kcal/mole = 91.17 Kcal/mole.

Question:

By direct calculation, determine the value of the electric intensity at any distance from an infinite plane sheet of uniformly distributed charge. Show that the result follows at once from an application of Gauss's law.

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/Users/wenhuchen/Documents/Crawler/Physics/D17-0571.htm

Solution:

Consider any point X at a perpendicular distance h from the plane sheet of charge density \rho. See figure A. Drop the perpendicular from the point to the sheet, cutting the latter at 0, and draw two circles, centered at 0, at radii of y and y + dy. Take a small portion of the annulus so formed, of length dl, and consider the electric intensity dE' at the point X due to the small (almost point-like) element of charge. Then the electric field due to a point charge is dE' = dq/4\pi\epsilon_0r^2 Since the charge density \rho = dq/dA = dq/dldy then, dE' = (\rho dl dy)/(4\pi\epsilon_0 r^2). The direction of dE' is the same as that of r, and dE' may therefore be resolved into components along OX and at right angles to it. The component along OX has the same value; no matter what position on the annulus dl occupies. But the element of the annulus diametrical-ly opposite dl produces a component perpendicular to OX equal but opposite to that produced by dl. (See figure B). These two components thus cancel out, as do all components from diametrically opposite elements. The electric intensity from the whole annulus is thus perpendicular to the sheet and has magnitude dE = \oint dE' sin \texttheta = [(\rho dy)/(4\pi\epsilon_0r^2)] sin \texttheta \oint dl = (\rho dy)/(4\pi\epsilon_0r^2) \bullet (h/r)2\piy = [h\rhoy dy] / [2\epsilon_0(h^2 + y^2)^3/2 ]. We used the fact that sin \texttheta = h/r (geometric considerations in figure A) and that the sum of all the infinitesimal elements of length dl about the whole ring is equal to the circumference of the ring. Also used was the fact that r = (h^2 + y^2)^1/2. For the whole sheet of charge the electric intensity is the sum of contributions due to all the annulii of radius y = 0 to y = \infty (for the infinite plane sheet). Or E = \int dE = ^\infty\int_0 [h\rhoy dy] / [2\epsilon_0(h^2 + y^2)^3/2 ] = (h\rho/2\epsilon0) ^\infty\int_0[y dy] / [(h^2 + y^2)^3/2] = \rule{1em}{1pt} [h\rho/2\epsilon_0] [(h^2 + y^2)-1/2]^\infty_0 = \rule{1em}{1pt} [h\rho/2\epsilon_0] [0 - (1/h)] = \rho/2\epsilon_0. To apply Gauss's law to the same problem, construct a cylinder of small and uniform cross-sectional area dA at right angles to the sheet and bisected by the sheet (see figure C). Since the sheet is infinite and the charge uniformly distributed, the electric intensity must be the same at all points equidistant from the sheet, and thus by symmetry must be everywhere per-pendicular to the sheet. Hence E is everywhere parallel to the sides of the cylinder and thus the flux of E from the cylinder through its sides is zero. The magni-tude of E at each end of the cylinder will be the same if the cylinder is bisected by the sheet, and E will be perpendicular to each end. Hence, applying Gauss's law, we obtain \int E^\ding{217} \bullet dA^\ding{217}= \int[(E^\ding{217} \bullet dA^\ding{217})_top + (E^\ding{217} \bullet dA^\ding{217})_side + (E^\ding{217} \bullet dA^\ding{217})_bottom] Since dA^\ding{217} is small, \intE^\ding{217} \bullet dA^\ding{217}\approx E^\ding{217} \bullet dA^\ding{217} and \intE^\ding{217} \bullet dA^\ding{217}= [EdA + 0 + EdA = 2EdA]= dq/\epsilon_0 ; but \rho = dq/dA and 2EdA = (\rhodA/\epsilon_0). Therefore E = \rho/2\epsilon_0. Thus E is everywhere perpendic-ular to the sheet and has the same value \rho/2\epsilon_0 at all points.

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Question:

The following BASIC program is supposed to calculate the sum of the squares of the first 10 digits. Find the syntactical and logical errors. 10PROGRAM TO CALCULATE SUM OF SQUARES OF 20FIRST TEN DIGITS. 30LET SUM = 0 40LET J = 1 50LET S = J 60IF J = 10 THEN GO TO 90 70LET J = J + 1 80GO TO 50 90PRINT S 100END.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G25-0596.htm

Solution:

Syntax errors are errors in the grammar of the programming language being used. They are detected by the compiler and hence the program is returned even before it is executed, i.e., no output is printed. Statements 10 and 20 are comment lines. But they are not prefaced by REM (the obligatory method of informing the BASIC compiler that a remark is forthcoming) . Hence the compiler treats 10 and 20 as instructions, which cannot be done. Compilation stops and the program is ejected at line 10. Statement 30 initializes the variable SUM. Unfortunately BASIC allows only two character variable names (of the form letter digit). Hence SUM is an invalid variable name and compilation ceases. Statement 60 is: 60IF J = 10 THEN GO TO 90(1) Now when the compiler begins translation of this statement into machine language it classifies 60 as of the form 60IF X = 10 THEN 90(2) Comparing (2) and (1), the place where 90 should be is taken by GO. Thus the compiler reads 60IF X = 10 THEN GO Since GO is not a statement number the instruction is nonsense. The program remains uncompiled. Finally in 100, END is succeeded by a period. The compiler is aware that END signifies that compilation is over and execution can begin but END. is not recognized. (Note that this last is an error only if, when the compiler has started on a line, it works through to the last column. If the compiler stops translating a line after receiving a sound construc-tion, then the period will make no difference.) After all the above errors have been corrected the program is ready for execution. It looks like this: 10REM PROGRAM TO CALCULATE SUM OF SQUARES OF 20REM FIRST TEN DIGITS 30LET S = 0 40LET J = 1 50LET S = J 60IF J = 10 THEN 90 70LET J = J + 1 80GO TO 50 90PRINT S 100END The program is executed, but the output is 10. The program gives a wrong result. Logical errors are errors in the meaning of the program. The program is grammatically sound (otherwise the compiler would have rejected it). But its meaning is not what the programmer intended. As an analogy consider the English language sentence The selfish house slept furiously. The statement has nouns and verbs in the right order and the descriptive words are acceptable too. The punctuation is accurate. But the total sentence is almost meaningless. ('Almost' since it could appear in a poem or modern novel). Examining the program for logical flaws, observe that squares have not been calculated. Thus the program must be corrected if the desired result is to be obtained: Statement 50 should be as follows: 50LET S = S + J\uparrow2 or 50LET S = S + J\textasteriskcenteredJ .

Question:

Suppose you discovered a new species of bioluminescent worm . How could you prove that it was the worm itself and not some contaminating bacteria that was producing the light ?

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/Users/wenhuchen/Documents/Crawler/Biology/F05-0145.htm

Solution:

There are many bioluminescent organisms but sometimes it is difficult to distinguish whether it is the actual organism that is emitting light or whether the light is due to luminous bacteria living in the organism. Some species of fish have light organs under their eyes where light- emitting bacteria live. One possible way of deter-mining if a bacterium is the source of the light is to take some of the light-producing substance and to place it in complete growth media (containing all possible nutrients). If the luminescent material proliferates, then a bacterium is most likely the causal factor. Another way would be to physically examine the light- emitting sub-stance under a microscope. This could be done by scraping out some light-emitting substance from a bioluminescent organism and transferring it to a glass slide. If we see individual bacterium emitting light then we can conclude that the bioluminescence is caused by bacteria.

Question:

Consider the tire on a car wheel, outer radius R = 0.36 m, as the car accelerates uniformly from rest to a maximum speed of 27 m/sec in a time of 30 sec. Calculate the acceleration of the car as well as the angular velocity and angular acceleration of the tire.

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/Users/wenhuchen/Documents/Crawler/Physics/D04-0155.htm

Solution:

Since the acceleration is constant, a_c= (v_Final - v_Initial)/∆t = [(27){m/(sec-0)}]/(30 sec) = (0.9)(m/sec^2 ) The angular velocity vector \omega, which is constant for uniform motion, is The angular velocity vector \omega, which is constant for uniform motion, is perpendicular to the plane of the circle. This will be the linear acceleration of the axle on which the wheel is supported. Imagine that you are sitting on the axle. You will observe the road moving by with velocity V and the tire spinning with a vel-ocity such that, at the maximum speed, the angular velocity \omega= V/R = [27(m/sec)]/(0.36 m) = 75 sec^-1 when the acceleration is constant, the angular acceleration is given by \alpha= (\omega_Final - \omega_initial)/(∆t) = (75 sec^-1 - 0)/(30 sec) = 2.5 sec^-2 and we can see that a_c = \alphaR . A point on the circumference of the wheel experiences an acceleration, directed toward the axle, of magnitude a= V^2/R =\omega^2R = (75 sec^-1)^2(0.36 m) = 2.0 × 10^3 (m/sec^2) since this point is travelling in a circular path.

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Question:

Compare the open and the closed type of circulatory systems. Describemolluscancirculation in a bivalve, such as a clam. What is ahemocoel?

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/Users/wenhuchen/Documents/Crawler/Biology/F12-0294.htm

Solution:

A closed type of circulatory system is one in which the blood is always contained in well-defined vessels. In an open circulatory system, there are some sections where vessels are absent and the blood flows through large open spaces known as sinuses. All vertebrates have closed circulatory systems, as do the annelids (earth-worms and their relatives). All arthropods (insects, spiders, crabs, crayfish, and others) as well as mostmolluscshave open circulatory systems. Movement of blood through an open system is not as fast, orderly or efficient as through a closed system. Aquaticmolluscsand most aquatic arthropods respire by means of gills. In bivalves, blood in large open sinuses bathes the tissues directly. The blood drains from the sinuses into vessels that go to the gills, where the blood is oxygenated. The blood then goes to the heart, which pumps it into vessels leading to the sinuses. A typical circuit is heart \rightarrow sinuses \rightarrow gills \rightarrow heart. In arthropods the blood sinuses are termedhemocoels. Thehemocoelis not derived from thecoelom, but from the embryonicblastocoel.

Question:

If the electrode spacing is 1 x 10^\rule{1em}{1pt}3 m and the voltage is 1 × 10^4 V, how large is the Coulomb force that is respons-ible for the separation of dust particles from the air? Assume that the particle's charge is + 200 quanta.

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/Users/wenhuchen/Documents/Crawler/Physics/D18-0598.htm

Solution:

The electric field \epsilon existing between two plates is \epsilon = V/d = (1 × 10^4V) / (1 × 10\rule{1em}{1pt}3m) = 1 × 10^7 V / m Quanta represent the number of unit charges present on the dust particle. A unit charge is the charge present on a proton or 1.6 x 10^\rule{1em}{1pt}19 coul. Therefore, the electro-static force experienced by the particle is F = q\epsilon = (2 × 10^2)(1.6 × 10^\rule{1em}{1pt}19 C)(1 × 10^7 V/m) =3.2 × 10^\rule{1em}{1pt}10 N This is approximately 10 to 100 times larger than the gravitational force on the particle.

Question:

A barometer tube extends 89.4 cm above a free mercury sur-face and has air in the region above the mercury column. The height of the column is 74.5 cm at 25\textdegreeC when the reading on a true barometer is 76 cm. On a day when the temperature is 11 \textdegreeC it reads 75.2 cm. What is the true atmospheric pressure?

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/Users/wenhuchen/Documents/Crawler/Physics/D14-0519.htm

Solution:

Since there is a fixed mass of air at all times in the top of the barometer tube, the gas law (pV/T) = const may be applied to it directly. In the first case when the temperature is 25\textdegreeC = 25 + 273 = 298\textdegreeK, the barometric height is 76 cm of mercury. This is the pressure exerted at the free surface of mercury and thus, by the laws of hydrostatic pressure, at the same horizontal level inside the barometer tube. At points in the tube higher than this, the pressure drops off with height. If there were true vacuum instead of air above the mercury column in the tube, the mercury would rise to a height of 76 cm, which is the atmospheric pressure. How-ever, the rise of the mercury is resisted by the pressure of the air trapped in the tube (see the figure) , therefore the pressure of the trapped air must be equal to that ex-erted by (76 - 74.5) = 1.5 cm of mercury. The volume of the trapped air is (89.4 - 74.5)A cm^3 = 14.9A cm^3, where A cm^2 is the cross-sectional area of the tube. In the second case, when the temperature is 11\textdegreeC = 284\textdegreeK, the pressure of the trapped air is (p_0 - 75.2 cm of mercury), where p_0 is the atmospheric pressure on that day. The volume of trapped air is (89.4 - 75.2)A cm^3 = 14.2A cm^3 . Hence (1.5 cm × 14.9A cm^3)/(298\textdegreeC) = [(p^0 - 75.2cm)14.2A cm^3] / [284\textdegreeC] \therefore p_0 - 75.2 cm = 1.5 cm or p_0 = 76.7 cm of mercury.

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Question:

Glucose-1-phosphate, essential to the metabolism of carbohydrates in humans, has a molecular weight of 260 g/mole and a density of about 1.5 g/cm^3. What is the volume occupied by one molecule of glucose-1-phosphate?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E05-0201.htm

Solution:

In general, volume = mass/density. Hence, in order to determine the volume, we must determine the mass of one molecule of glucose-1 -phosphate. One mole of glucose-1-phosphate weighs 260 g (this is the meaning of a molecular weight of 260 g/mole), Since there is an Avogadro's number of molecules in a mole (6.02 × 102 3molecules/mole), the mass of one molecule of glucose-1-phosphate is given by mass = [(molecular weight) / (Avogadro's number)] = [(260 g/mole) / (6 × 102 3molecules/mole)] = 4.3 × 10^22 g/molecule. Hence, one molecule of glucose-1-phosphate weighs 4.3 × 10^-22 g and has a volume of volume = [(mass) / (density)] = [(4.3 × 10^-22 g) / (1.5 g/cm^3) \cong 2.9 × 10^-22 cm^3 density.

Question:

A compound subjected to analysis was found to have the following composition by weight: 69.96 % carbon (atomic weight = 12.0 g/mole), 7.83 % hydrogen (atomic weight =1.01 %g/mole), and 22.21 % oxygen (atomic weight = 16.0 g/mole). If the molecular weight of this compound is 360 g/mole, what is its molecular formula?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E04-0143.htm

Solution:

Molecular formula may be defined as the formula stating the actual number of each type atom in a particular compound. Problems of this sort are solved by assuming a sample of some convenient mass and then calculating the number of moles of each atom in this sample. Once the number of moles has been calculated for each component, these numbers are divided by the greatest common factor in order to obtain the proportions in which the various components appear in the molecule (empirical formula). From this, we check to see if the elements, in those proportions, give the actual molecular weight. If they do, we have the actual molecular formula. If they do not, we multiply by a factor until their weights do give the molecular weight. There is no error in assuming a sample of definite mass, since the size of a sampleof molecules does not affect the composition of the molecules. Assume a sample weighing 100 g. Then the masses of C, H, and O in the sample are: mass C = 69.96 % × 100 g = 69.96 g mass H = 7.83 %× 100 g =7.83 g mass O = 22.21 % × 100 g = 22.21 g We convert the mass of each element to the correspond-ing number of moles by dividing by the atomic weight of that element. Thus, we obtain the following number of moles of each element: moles C = (69.96 g) / (12.0 g/mole) = 5.83 moles moles H = (7.83 g)/ (1.01 g/mole)= 7.75 moles moles O = (22.21 g)/ (16.0 g/mole) = 1.39 moles The greatest common factor is 1.39 moles. Dividing the number of moles of each element by 1.39 moles, we obtain (moles C) / (1.39moles) = (5.83moles) / (1.39moles) \allequal 4.2. (moles H) / (1.39moles) = (7.75moles) / (1.39moles) \allequal 5.6. (moles O) / (1.39moles) = (1.39moles) / (1.39moles) \allequal 1. Multiplying these numbers by 5 in order to get whole numbers we obtain: Proportion of C atoms = 4.2 × 5 = 21 Proportion of H atoms = 5.6 × 5 = 28 Proportion of O atoms =1 × 5 = 5. Thus, the molecular formula of this compound is C_21H_28O_5. As a check, we determine the molecular weight of this compound to be molecular weight = (21 × atomic weight of c) + (28 × atomic weight of H) + (5 × atomic weight of O) = (21 × 12.0 g/mole) + (28 × 1.01 g/mole)+ (5 × 16.0 g/mole) \allequal 360 g/mole. which is the experimentally determined molecular weight.

Question:

A neutron decays into an electron, a proton and an antineutrino. Calculate the kinetic energy shared by the electron and the antineutrino.

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Solution:

The neutrino and antineutrino are particles which have a negligibly small and possibly zero mass. They can carry energy and momentum, and presumably travel with the speed of light. The neutrino is different from the photon in that the photon has spinhand the neutrino spinh/2. We will make the calculation in atomic mass units. 1 kg = 6.024 × 1026amu m_P= 1.67 ×10\Elzbar27kg = (1.67 ×10\Elzbar27kg ) [ 6.024 × 10^26 (amu/ kg)] = 1.007593amu Similarly, m_n= 1.008982amu m_e = 0.000549amu where m_p ,m_n, m_e are the masses of the proton, neutron and electron, respectively. The reaction occurring is n^0 \rightarrow e\Elzbar1+ p^+1 +\Upsilon^0 where n^0 , e\Elzbar1, p^+1and\Upsilon0are the symbols for a neutron, electron, proton and antineutrino, respectively. Since we want to calculate the kinetic energy shared by the electron and antineutrino, we apply the principle of con-servation of relativistic energy and obtain m_nc^2 = m_e c^2 + m_p c^2 + m( \Upsilon )0c^2 + T_e +T_p+ T(\Upsilon )0 where T_e ,T_pand T(\Upsilon )0are the kinetic energies of the electron, proton and antineutrino respectively, andm_nc^2 , m_e c^2 , m_p c^2 , m( \Upsilon )0c^2are the rest energies of the neutron, electron, proton and neutrino. But m(\Upsilon )0= 0. Also, the proton remains at rest after the decay, and, therefore,T_p= 0. Hence, (m_n\Elzbar m_e\Elzbar m_p )c^2 = T_e + T(\Upsilon )0= T Using the previous mass calculations T = (.000840amu) (9 × 10^16 m^2/s^2) = (.000840amu) (1.66 × 10\Elzbar17kg/amu) (9 × 10^16 m^2/s^2) = 1.26 × 10\Elzbar13Joules Since 1 Joule = 6.24 × 10^18eV T = 7.86 × 10^5eV and this is shared by the electron and antineutrino.

Question:

A coil of 60 ohms inductive reactance in series with a resistance of 25 ohms is connected to an a-c line of 130 volts. What will be the current through the circuit?

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/Users/wenhuchen/Documents/Crawler/Physics/D22-0747.htm

Solution:

X_L = 60 ohms,R = 25 ohms, and E = 130 volts The impedance of an a-c circuit is given by Z = \surd[R^2 + (X_L - X_C )^2], Where the minus sign results because X_C and X_L are 180 o out of phase . Then Z = \surd[(25 ohms)^2 + (60 ohms)^2 ] = 65 ohms From ohm's law for an a-c circuit I = E/Z, we have I = 130 volts / 65 ohms = 2.0 amp.

Question:

Develop a FORTRAN program which solves a system of linear simultaneous equations according to the GAUSS SEIDEL method.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G20-0508.htm

Solution:

To apply the Gauss-Seidel method, the system of linear equations AX = C is rewritten in the form: x_1 = 1/ a_11 (c_1 \rule{1em}{1pt} a_12x_2 \rule{1em}{1pt} a_13x_3 \rule{1em}{1pt} ... \rule{1em}{1pt} a_1nx_n)(1) x_2 = 1/ a_21 (c_2 \rule{1em}{1pt} a_22x_1 \rule{1em}{1pt} a_23x_3 \rule{1em}{1pt} ... \rule{1em}{1pt} a_2nx_n)(2) .. .. .. x_n= 1 /a_nn(c_n\rule{1em}{1pt} a_n1x_1 \rule{1em}{1pt} a_n2x_2 \rule{1em}{1pt} ... \rule{1em}{1pt} an, n\rule{1em}{1pt}1x_n\rule{1em}{1pt}1)(n) Next, a set of starting values x^0_1, x^0_2,...,x^0_n is chosen. While the program developed will take 0 as starting values, x^0_i- =c_i/a_iiare also frequently used as starting values. Substituting x_2 = x_3 = ...=x_n= 0 into (1) yields an approximation for x_1. This approximation is used in (2) where x_3 = x_4 ... =x_n= 0. Now we have approximations for x_1 and x_2 . Continuing, observe thatx_nis found by substituting the approximations for x_1,x_2,...,x_n\rule{1em}{1pt}1, into (n). This completes the first iteration. For the k-thiteration: x^k+1_1 = 1/ a_11 (c1\rule{1em}{1pt} a_12x^k_2 \rule{1em}{1pt} a_13x^k_3 \rule{1em}{1pt} ... \rule{1em}{1pt} a_1nx^k_n) x^k+1_2 = 1/ a_22 (c1\rule{1em}{1pt} a_21x^k+1_1 \rule{1em}{1pt} a_13x^k_3 \rule{1em}{1pt} ... \rule{1em}{1pt} a_2nx^k_n) .. .. .. x^k+1_n = 1/a_nn(c_n_ \rule{1em}{1pt} a_n1x^k+1_1 \rule{1em}{1pt} a_n2x^k+1_2 \rule{1em}{1pt} ... \rule{1em}{1pt} an, n\rule{1em}{1pt}1x^k+1_n\rule{1em}{1pt}1). A sufficient condition for convergence (and the one which will be assumed to hold) is that \sum_j\not =1 \verta_ij\vert < \verta_ii\vert fori= 1, 2, ...,n As an illustrative example, let x_1 = (1/10) (9 \rule{1em}{1pt} 2x_2 \rule{1em}{1pt} x_3) x_2 = (1/20) (\rule{1em}{1pt}44 \rule{1em}{1pt} 2x_1 \rule{1em}{1pt} 2x_3) x_3 = (1/10) (22 \rule{1em}{1pt} 2x_1 \rule{1em}{1pt} 3x_2). These equation correspondto the determinant system \mid1021 \mid\mid x_1 \mid\mid 9\mid \mid 220\rule{1em}{1pt}2 \mid\mid x_2 \mid=\mid\rule{1em}{1pt}44\mid \mid\rule{1em}{1pt}2310 \mid\mid x_3 \mid\mid 22 \mid For x^(0)_i= 0 we have x^(1)_1 = 9 / 10 x^(1)_2 = (1/20) (\rule{1em}{1pt}44 \rule{1em}{1pt} 2x^(1)_1 + 3x^(0)_3) = + (1/2) (\rule{1em}{1pt}44 \rule{1em}{1pt} (18 /10)) = \rule{1em}{1pt} 2.29 x^(1)_3 = (1/10) (22 + 2x^(1)_1 \rule{1em}{1pt} 3x^(1)_2) = + (1/10) (22 + 18 /10 \rule{1em}{1pt} 3 (\rule{1em}{1pt}2.29)) = 3.067 etc. The program itself is given below. CGAUSS-SEIDEL ITERATION OF SIMULTANEOUS EQUATIONS DIMENSION A(30,30), X(30), Y(30) 1READ 999, N, ITLAST, ((A(I, J), J = 1,N), Y(I),I = 1,N) PUNCH 996, N,((A(I,J),J = 1,N), I = 1,N) PUNCH 995, (Y(I),I = 1,N) DO 10 I = 1,N 10X(I) = 0. IT = 1. 20PUNCH 994, IT DO 60 I = 1,N P = Y(I) DO 50 J = 1,N IF (I - J) 40,50,40 40P = P - A(I,J)\textasteriskcenteredX(J) 50CONTINUE X(I) = P/A(I, I) 60PUNCH 998, I, X(I) IT = IT + 1 IF (IT - ITLAST) 20,2D ,1 994FORMAT(/24X, 9HITERATI0N 12) 995FORMAT(/22X,15HC0NSTANT VECTOR / (3E18.7)//22X, 115HSOLUTI0N VECTOR) 996FORMAT(//21X, 15HMATRIX OF ORDER 12 //(3E18.7)) 998FORMAT(20X, I2, E16.7) 999FORMAT(2I5/(8F10.0)) END

Question:

Write a FORTRAN program to load and add two given m × n matrices.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G20-0498.htm

Solution:

An m × n matrix is an ordered array of real numbers containing m rows and n columns. Let A = [a_i, _j ], B = [b_i, _j ] (i= 1,...,m; j = 1,...,n) be the two matrices to be added. Their sum C = [c_i, _j] = A + B = [a_i, _j + b_i, _j ]. To load the matrices, a dimension statement reserving two sets ofmn places each is required. Also, one more set ofmnplaces must be dimensioned for the finally obtained array. Then, two nested DO-loops must be constructed to read in the values ofa_i, _j and b_i, _j. After that, the two arrays are added using two nested DO-loops again. Finally, the resulting array is printed out by the similar method. Note, that the format statements 20 and 30 are left out because they depend on the actual data used. DIMENSION A (10,15), B (10,15), C (10,15) CTHE ARRAY C (10,15) WILL CONTAIN THE SUM COF ARRAYS A (10,15) AND B(10,15). CLOAD THE ARRAYS DO 50 I = 1,10 DO 50 J = 1,15 READ (5,20) A (I,J), B (I,J) 50CONTINUE CADD THE ARRAYS DO 60 I = 1,10 DO 60 J = 1,15 C (I,J) = A (I, J) + B (I, J) 60CONTINUE COUTPUT THE RESULT DO 70 I = 1,10 DO 70 J = 1,15 WRITE (6,30) C (I,J) 70CONTINUE STOP END

Question:

Write a program which uses as input character strings of up to200 characters, consisting of left parentheses and right parenthesesonly (assume there are no blanks separating them), and determine whether the strings are \textquotedblleftwell formed formed\textquotedblright . A well-formed set of parentheses is one where 1)thereare equal numbers of left and right parentheses. 2) For every rightparentheses, there must be a left parentheseswhich appears earlier in the string. Print out the following message as applicable. (a) WELL FORMED (b) UNMATCHED RIGHT PARENTHESES (c) TOO MANY LEFT PARENTHESES.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G13-0349.htm

Solution:

This problem deals with the manipulation of char-acter strings. SUBSTR, INDEX and LENGTH are built-in func-tions used for solving problemson character strings. The program for the above problem has logical steps as 1) Read in parentheses 2) Find the Index of right parenthesis, i.e. position of first right parenthesisin string of parentheses. 3) If index (say) N = 1 then increment the count of right parenthesis; elseincrement count of left parenthesis. 4) Compare: IfR_count>L_countthen there is an un-matched right parenthesis. 5) Otherwise find the length L, of string parentheses. 6) If L = 1 compareR_count&L_count. If they are equal, then the stringis well-formed. IfL_count>R_count, then the string has too many leftparentheses. 7) If L \not = 1 remove the first left or right parenthesis from the characterstring of parentheses and call this new reduced character string asparentheses itself using PAR = SUBSTR[PAR,2]. Loop back to (2). The program is as follows. STRING:PROCEDUREOPTIONS(MAIN); ONENDFILE(SYSIN) STOP; /\textasteriskcentered R_COUNT COUNTS RIGHT PAR, L_COUNT COUNTS LEFT PAR, L_LENGTH, N_INDEX NUMBER \textasteriskcentered/ DCL(KEY,R_COUNT,L_COUNT,L,M,N) FIXED DEC(3),PAR CHAR(200) VAR; PUTLIST('STRING' ,'MESSAGE'); PUTLIST(' ----' ,'------') SKIP (0) ; PUTSKIP(2); LOOP_1:DOWHILE(1 = 1); GETLIST(PAR); PUTLIST(PAR); R_COUNT,L_COUNT,KEY = 0; LOOP_2:DOWHILE(KEY = 0); N =INDEX(PAR,')'); /\textasteriskcentered FINDS POSITION OF FIRST RIGHT PARENTHESIS \textasteriskcentered/ IF N = 1 THEN R_C0UNT = R_C0UNT + 1; ELSE L_C0UNT = L_C0UNT + 1; /\textasteriskcentered INCREMENTING R_COUNT OR L_COUNT DEPENDING ON VALUE OF N \textasteriskcentered/ IF R_COUNT>L_COUNT THEN DO; PUTLIST('UNMATCHED RIGHT PARENTHESES ') ; KEY = 1; END; L =LENGTH(PAR); IF L = 1 THEN K = 1; LOOP_3:DO WHILE (L\lnot = 1) ; PAR =SUBSTR(PAR, 2) ; L = 1; END LOOP_ 3; END LOOP_2; M = IL-IR; IF IR = IL THEN PUT LIST ('WELL FORMED'); IF IR

Question:

What mass of calcium bromide CaBr_2, is needed to prepare 150 ml of a 3.5 M solution? (M.W. of CaBr_2 = 200.618 g/mole)

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/Users/wenhuchen/Documents/Crawler/Chemistry/E08-0284.htm

Solution:

M is themolarityof the solution and is defined as the number of moles of solute per liter. The solute is the substance being added to solution. A I M solution con-tains 1 mole of solute per liter (1000 ml) of solution. We are asked to calculate the mass, however, no term for mass appears in themolarityequation. Therefore, a connection must be found between mass and another variable. The connection is the mole equation (moles = mass/M.W.) . After substitution, themolarityequation reads M = [{grams (mass) / M.W.} / (liters)] grams = liters (M.W.)(M) = . 15l (200.618 g/mole) (3.5 moles/l) = 105.32 g of CaBr_2.

Question:

In some underdeveloped nations dung is used as fuel for cooking. This is wasteful because it deprives the soil of valuable nutrients. It should be possible to use solar cookers in some of these countries. Generally, the "stove" consists of a curved aluminum mirror which focuses the heat energy on a collector plate (see figure). Cal-culate how long it takes to raise the temperature of 1 liter of water from 293\textdegreeK to the boiling point, 373\textdegreeK. Assume that the diameter of the mirror is 1 m and that 70 percent of the incident solar energy is actually avail-able for heating the water. To raise the temperature of1 liter of water 1\textdegreeK, 4.186 × 10^3 J of thermal energy is required. The power radiated by the sun at the surface of the earth is 5.5 × 10^2 W/m^2.

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/Users/wenhuchen/Documents/Crawler/Physics/D16-0548.htm

Solution:

The collection area A of the mirror is, A = \pir^2 = \pi(5 × 10^-1 m)^2 = 7.9 × 10^-1 m^2 The total power P incident on the reflector is, P = (5.5 × 10^2W/m^2)(7.9 × 10^-1m^2) = 4.345 × 10^2 W Since the conversion efficiency is only 70 percent, the power converted to thermal power H is, H = P(7.0 × 10^-1) = (4.34 × 10^2W) (7 × 10^-1) = 3.04 × 10^2 J/s The temperature must be increased by 80\textdegree K (from 293 to 373\textdegree K). The total thermal energy Q required is then, Q = (4.19 × 10^3 J/\textdegreeK)(8 × 10^1 \textdegreeK) = 3.35 × 10^5 The time would be the amount of thermal energy needed divided by the rate at which thermal energy is produced. Hence, t = Q/H = (3.35 × 10^5J)/(3.04 × 10^2J/s) = 11.0 × 10^2 s = (11.0 × 10^2s)/(60s/min) = 1.84 × 10^1 min.

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Question:

What are the functions of the nucleus? What is the evidence that indicates the role of the nucleus in cell metabolism and heredity?

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/Users/wenhuchen/Documents/Crawler/Biology/F02-0040.htm

Solution:

If the cell is thought of as a miniature chemical plant designed to carry out all the processes of life, then the nucleus can be compared to a central computer that controls a network of sophisticated and highly complicated biochemical machinery. This is because the nucleus contains the chromosomes, which bear the genes, the ultimate regulator of life. The genes comprise a library of programs stored in the nucleus: programs that specify the precise nature of each protein synthesized by the cell. The nucleus monitors changing conditions both inside the cell and in the ex-ternal environment, and responds to input of information by either activating or inhibiting the appropriate genetic programs. Control of protein synthesis is the key to control-ling the activities and responses of the cell, since a tremendeous array of important biological and biochemical processes are regulated by enzymatic proteins. By switch-ing particular genes on and off, the cell controls not only the kinds of enzymes that it produces, but also the amounts. Both qualitative and quantitative control of enzyme synthesis are crucial to the proper functioning of the cell and the whole organism. Experiments with the large single-celled alga Acetabularia. Cells without nuclei have a very limited range of function. The mammalian red blood cell does not contain a nucleus and is unable to reproduce; it is functional only for a relatively short period. Egg cells, from which nuclei have been experimentally removed, may divide for a while, but the products of division never differentiate into specialized cell types, and eventually die. Fragments without a nucleus, cut from such large unicellular organisms as the amoeba or the unicellular algae Acetabularia, survive temporarily; but ultimately they die, unless nuclei from other cells are transplanted into them. These experiments demonstrate that the nucleus is essential to long term continuation of life processes, and to structural and functional differentiation of cells. The nucleus is of central importance also in the transmission of hereditary information. The nucleus carries the information for all the characteristics of the cell. This information is found in the chromosomes, which consist of protein and deoxyribonucleic acid (DNA). DNA is the hereditary material which contains all the information for the growth and reproduction of the cell. When the cell divides, the nuclear information is trans-mitted in an orderly fashion to the daughter cells by replication of the chromosomes and division of thenucleus. Thus, in this type of division, the daughter cells each possess a single complement of genetic information iden-tical to that of the mother cell. The hereditary importance of the nucleus can be demonstrated experimentally (see Figure). If a fragment containing the nucleus is cut from an Acetabularium of one species, characterized by a given appearance, the fragment will regenerate a whole cell of that species. This regenerative ability permits experiments of the type shown in the accompanying figure, in which nuclei of one species are grafted onto the cytoplasm of a different species. The observation from these experiments shows that the appearance of the regenerated cell eventually resembles that of the organism from which the nucleus is taken. Hence, we conclude that the nucleus, as a con-trolling center, produces information that controls the cytoplasm and participates in the regulation of cell growth and structure. Since the crucial information comes from the nucleus, the regenerated cell will resemble the organism from which the nucleus is taken.

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Question:

With what force does the Earth attract the moon?

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/Users/wenhuchen/Documents/Crawler/Physics/D04-0136.htm

Solution:

By the Universal Law of Gravitation, we have F_G = G(m_mm_E/r2_m) wherer_mis the distance between the earth and moon, and m_m and m_e are the masses of the moon and the earth respectively. G = 6.67 × 10^-8 (dyne - cm^2)/g^2 r_m= 3.84 × 10^10 cm m_m = 7.35 × 10^25 g m_e = 5.98 × 10^27 g F_G = (6.67 × 10-8dyne-cm^2/g^2) × [{(7.35 × 10^25 g) ×(5.98×1027g)} / (3.84 × 10^10 cm)^2}] = 2.0 × 10^25 dynes.

Question:

What are the major forces of evolution?

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/Users/wenhuchen/Documents/Crawler/Biology/F27-0720.htm

Solution:

Evolution is the result of the interaction of four major forces. These are1) mutation, 2) genetic drift, 3) migration, and 4) natural selection. These four forces have one thing in common - each can bring about evolutionby changing the allele frequencies in the gene pool of a populationover time. Mutations are random events that occur at a very low rate of approximatelyone out of every 10^6 genes. If evolution depended on the occurrenceof mutations alone, life would not exist as we know it today. However, mutation is important because it introduces variety into a population. Natural selection can act upon these variations to increase the frequenciesof the few adaptive mutations and decrease the frequencies ofthe many more maladaptive ones over time. Genetic drift acts in the evolutionary process by causing chance fluctuationsin gene pool frequencies. Although this has essentially no effectin large popu-lations, it is probably the major cause of evolution in smallpopulations. In these populations, chance - including accidents, matingpreferences and other unpredictable events - can cause an allele's frequencyto fluctuate widely from one generation to the next. This can resultin either the fixation of that allele or its elimination from the gene pool. Genetic drift can also interact with the processes by which small populationsbecome established to produce major evolutionary changes. Two processes of population establishment are the 1) Founder Effect and 2)isolation. In the Founder Effect, a small group of individuals may leave alarge population and re-establish in another location. The main populationhas certain gene frequencies - for example, the frequency of geneA may be .6, and that of gene a may be .4. It is very possible that the smallgroup will, entirely by chance, have allele frequencies different than thoseof the main population - perhaps A = .2 and a = .8. Thus, the new populationbeing founded will begin with a different assortment and proportionof alleles than the parent population. If the new population survivesgenetic drift over time may result in its becoming drasti-cally differentfrom the original population. This has been shown to be especiallytrue on small isolated islands that have been settled by a few animals. These same principles may be applied to small, naturally occurringpopulations that have been isolated by events in nature. For example, after a severe dry season, marine organisms formerly inhabiting onelarge lake may become isolated from one another as the lake dries up intomany small pools. This could result in a variety of small populations havingvarying allele frequencies and gene pool composition. Migration occurs when individuals from one breeding population leaveto join another. Migration can be either inward (immigration) or outward(emigration). The effect of migration on allele frequency in a populationis a function of the size of that population in relation to the size ofthe migration. In a small population, the frequency of many genes can besignificantly altered by a small migration, while large populations can usuallyabsorb large amounts of migration before a noticeable change in frequenciesoccurs. Migration may lead to either more variation in a population, due to the introduction of new genes through immigration, or lessvariation due to the loss of genes through emigration. Natural selection ensures that the changes in allele frequency causedby mutation, migration, and genetic drift are adaptive. As such, evolutionwill more likely give rise to a species better adapted to live in its environment. Although the previously discussed forces are non-directional , natural selection is a directional process through which populationsadapt to their parti-cular environments by changing their allele frequenciesin response to environmental or selection pressure. More specifically, natural selection operates through dif-ferential reproduction, whichis said to have occurred when certain individuals are able, by survivingand/or reproducing at a higher rate, to preferentially propagate andtransmit their respective genes over those of other individuals. Natural selectionis a very gradual process, with the changes accumulating and alteringthe physical appearance of a population over very long periods of time.

Question:

Using COBOL language write a DATA DIVISION. There are four independent data items in the program, namely FIRST- NAME, eight characters; LAST-NAME, ten characters? TEL- NO, twelve characters; and STATE, twenty characters.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G11-0254.htm

Solution:

DATA-DIVISION is the third division of the COBOL program. It describes the data to be used in the procedure division. Input data, output data, and working data must all be described. DATA-DIVISION has two sections 1) File Section,2) Working- Storage Section. When files are not used one can omit the File-Section. In the Working-Storage section independent and subdi-vided items can be described. An item which is not subdi-vided is called an independent item. Social security number is a good example of an independent item. It can not be subdivided. On the other hand ADDRESS is a subdivided, group item. An address consists of a combination of elements such as street, house number, etc. Independent data items which are not subdivided are al-ways given level number 77, which must be entered in the A margin. Any other level number can be used to describe subdivided items. In COBOL, describing data means specifying a level num-ber, giving a data-name, and giving a PICTURE (PIC) descrip-tion. The PICTURE clause tells the computer how many charac-ters are in the data item, and what kind of characters they are. The most widely used picture character is X; this char-acter is used to tell the computer that the corresponding data item might be any character in the standard character set. A Standard character could be a number, letter, $, or any of the special characters. When PICTURE XX., or PICTURE X(2). is written, its meaning to the computer is that the data item is two charac-ters long and both of them are standard characters. Given in Pig. 1 is the DATA DIVISION for the program described.

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Question:

A stone of mass 5 kg drops through a distance of 15 m under the influence of gravity. Draw graphs of the work done by the stone and the power of the stone as a function of time.

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/Users/wenhuchen/Documents/Crawler/Physics/D05-0273.htm

Solution:

This is a case of constant acceleration with a = g = 9.8 m/sec^2. The equation d = v_0t + (1/2)at^2 can be used to find the total time the stone is in motion. The initial velocity is zero, thus d = (1/2)gt^2 where d = 15 n. The time required for the stone to fall through 15 m is t^2 = (2 × 15)/(9.8) = 3.06 t = 1.75 sec. The force acting on the stone is constant at F = mg = 5 kg × 9.8 m/sec^2 = 49 newtons. The work done by the stone is equal to W = Fd. To find the power of the stone, note that P = Fv. The velocity can be found from v = v_0 + at = gt. Below are the calculated values of d, v, W, and P for intervals of 0.25 seconds and the graphs requested. t v = gt d = (1/2)gt^2 W = Fd P = Fv 0 0m/sec 0m 0J 0W 0.25 2.45 0.31 15.2 120 0.5 4.9 1.23 60 240 0.75 7.35 2.76 135 360 1.0 9.8 4.90 240 480 1.25 12.3 7.66 376 600 1.50 14.7 11.0 540 720 1.75 17.2 15.0 735 840

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Question:

What is your understanding of an addition reaction mechanism? Illustrate your answer with a specific example.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E21-0756.htm

Solution:

An important difference between saturated and unsaturated hydrocarbons is the general type of reaction that each will undergo. Alkanes react by substitution, while alkenes react by addition reactions (as well as substitution). In an addition reaction, the pi bond of the double bond is destroyed and two sigma bonds are formed, one to each of two carbon atoms. A mixture consisting of only hydrogen and an alkene undergoes no detectable addition reaction even at high temperature and pressure. However, if a small amount of finely divided metal, such as platinum, palladium, or nickel is added as a catalyst, addition of hydrogen to the alkene occurs quite readily even at room temperature. (The same is true for alkynes, i.e. unsaturated compounds containing a triple bond.) According to the commonly accepted view, alkenes and alkynes, with relatively loosely held electrons in their pi bonds, act as electron-pair donors, that is, Lewis bases, and protons from addition reagents, HX (where X is H, Cl, Br, ...), act as Lewis acids, that is, e- lectron-pair acceptors. The first step, then, in the addition of HX to an alkene is the addition of a proton in a Lewis acid-base reaction. The process involves the formation of alkyl cations and chlorine anions as re-action intermediates. Such alkyl cations are called carbonium ions; they are groups of atoms that contain a carbon atom having only six electrons. An example of an addition reaction would be

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Question:

A radio receiving set is tuned to a certain station by the use of a .25 millihenry (2.5 × 10^\rule{1em}{1pt}4 henry) inductance, and a 32.2 picofarad (32.2 × 10^\rule{1em}{1pt}12 farad) condenser. What is the frequency of the station? What is its wave length?

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/Users/wenhuchen/Documents/Crawler/Physics/D23-0778.htm

Solution:

The heart of the receiving set of the radio station is the tank or tuning circuit as shown in figure (A). The circuit is more 'responsive to certain frequencies than to others. At a frequency \omega0 , denoted the resonance frequency, the current through the circuit is much greater than the current response of the circuit at any other frequency. This resonance peak is shown in figure (B). This resonant frequency is determined by the condition that the impedance of the capacitance equal the impedance of the inductance. Or (1/\omega0 C) = \omega0 L \omega0 = 2\pif = \surd(1/LC) f = (1/2\pi) \surd(1/LC) f = (1/2\pi) \surd{1/(2.5 × 10^\rule{1em}{1pt}4 h × 32.2 × 10^\rule{1em}{1pt}12 f)} = (1/2\pi) \surd(10^16/81) f = (10^8/2\pi9) = (10^8/56.5) = 1.77 × 10^6 The velocity of radio waves is the same as that of the speed of any electromagnetic wave. It is therefore, 3 × 10^10 cm/sec, andv = f\lambda for all waves \therefore \lambda = (v/f) = (3 × 10^10)/(1.7710^6) = 1.7 × 10^4 = 17,000 cm = 170 m

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Question:

What is meant by the term basal metabolic rate? What are some conditions that cause the rate to change?

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/Users/wenhuchen/Documents/Crawler/Biology/F17-0414.htm

Solution:

The "basal metabolic rate" refers to the amount of energy expended byonesbody under resting conditions. For example, sitting at rest in a room at a comfortable temperature requires 100 kcal. each hour for an "average" 70 kg. man. If another man who weighs 70 kg. requires only 95 kcal. per resting hour, his metabolic rate is 5% below the norm. The basal metabolic rate is the measure of the amount of energy utilized just to keep alive, when no food is being digested and no muscular work is being done. The average minimum energy requirement for young adult men is 1600 kcal. each day, and is about 5% less for women. To measure the basal rate, the subject is rested for 12 hours after the last meal, and reclined for 30 minutes at room temperature (18\textdegree- 26\textdegreeC). Pure oxygen is inhaled for an hour. It is known that the body absorbs only one fifth of the oxygen inhaled and that the body produces 4.8 kcal of energy for each liter of oxygen absorbed. Therefore, the basal metabolic rate can be calculated by measuring a person's oxygen consumption over a given time span. The amount of heat produced can be determined from the amount of oxygen consumed. By directly measuring the heat given off by a person in an insulated chamber, one can also measure the basal metabolic rate directly, but this is a more awkward process. The energy, requirement per square meter of body surface can also be calculated, and is found to be directly proportional to one's basal metabolic rate (BMR). The average energy expenditure is 40 kcal. per square meter. The BMR is a useful indicator of the state of one's metabolism. This is important for the diagnosis of certain hormonal disorders such as diabetes or goiter. Several factors influence one's metabolic rate. Children have high metabolic rates and aging adults have low rates. The female metabolic rate is below that of a male of similar size but it increases during pregnancy and lactation. Digestion requires about one tenth of the metabolic energy. Any infection or disease raises the metabolic rate and causes fever. Fever, in turn, raises the rate further and results in a loss of weight. Hormones and environmental conditions also play a major role in determining one's metabolic rate. Mental tension increases adrenalin production, raising the metabolic rate, blood pressure and heart rate. Adrenalin production can only continue for a few hours. The body will adjust to the presence of the stimulant and the sudden termination in its production drastically lowers the metabolic rate far below normal. There is a slowing of the heart rate, a drop in blood pressure, a shifting of blood to the intestines, and a decrease in blood sugar level. The largest influence on metabolism is muscular activity. During hard work, one's metabolic rate will double and may even reach fifteen times the BMR in brief bursts of activity.

Question:

An observer is at a distance r from a point light source whose power output is P_0. Calculate the magnitudes of the electric and the magnetic fields. Assume that the source is monochromatic, that it radiates uniformly in all directions, and that at distant points it behaves like the traveling plane wave of Figure 1.

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/Users/wenhuchen/Documents/Crawler/Physics/D37-1081.htm

Solution:

The power that passes through a sphere of radius r is (S)(4\pir^2), whereSis the average value of the Poynting vector at the surface of the sphere. This power must equal P_0, or P_0 =S4\pir^2. From the definition of S S=[(1/\mu_0)EB]. Using the relation E = cB to eliminate B leads to S= [(1 / (\mu_0c)]E^2. The average value of E^2 over one cycle is (1/2)E2_m , since E varies sinusoidally. This leads to p_0 = [(E2_m) / (2\mu_0c)] (4\pir^2), or E_m = (1 / r)\surd[(P_0\mu_0c) / 2\pi ] . For P_0 = 10^3W and r = 1.0m this yields E_m = [1/(1.0m)] \surd[{(10^3W) (4\pi × 10^-7 Wb / A\bulletm) (3 × 10^8m / s)} / (2\pi)] = 240 V/m. The relationship E_m = cB_m leads to B_m = (E_m / c) = [(240 V / m) / (3 × 10^8 m / s)] = 8 × 10^-7T. Note that Emis appreciable as judged by ordinary laboratory standards but that B_m (= 0.008 gauss) is quite small.

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Question:

A rocket, when unloaded, has a mass of 2000 kg, carries a fuel load of 12,000 kg, and has a constant exhaust velocity of 5000 km \textbullet hr^-1, what are the maximum rate of fuel consumption, the shortest time taken to reach the final velocity, and the value of the final velocity? The great-est permissible acceleration is 7g. The rocket starts from rest at the earth's surface, and air resistance and variations in g are to be neglected.

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Solution:

The problem can be approached using Newton's second law for a system of variable mass. m (dv^\ding{217}/dt) = F^\ding{217}_ext + v^\ding{217}_rel dm/dt(1) F\ding{217}_ext is the external force acting on the rocket of mass M. In this case, it is the attractive gravitational force due to the earth (i.e. the weight of the rocket Mg). v\ding{217} is the velocity of the rocket relative to a stationary observer O and v\ding{217}_rel is the relative velocity of the ejected mass with respect to the rocket. The last term is the rate at which momentum is being transferred out of the system by the mass that the system has ejected. F\ding{217}_ext = -mg, the negative sign indicating that the force acts downward. Also v_rel = -v_r where v_r is the velo-city of the rocket relative to that of the material ejected i.e., the reverse of the exhaust velocity. Regarding dm as a very small bit of material ejected in a time dt, and the resulting small increase in velocity of the rocket by dv, then equation (1) becomes m(dv/dt) = -v_r (dm/dt) - mg Upon division of both sides of this equation by m, and multiplication of both sides by dt, dv = -v_r(dm/m) - gdt(2) Therefore the acceleration at any time is a = dv/dt = -(v_r/m)(dm/dt) - g. The velocity v_r is constant and dm/dt must be constant also (for any change in the rate at which mass is ejected would necessarily lead to a change in the velocity of the mass ejected), so that a varies only with m. The smallest value of m gives the greatest value of a; but a cannot exceed 7g. Therefore 7g = -{(5 × 10^6m - hr^-1)/60min - hr^-1 × 60s - min^-1)} × [1/(2 × 10^3kg)] (dm/dt) - g. \therefore (dm/dt) = -{(8g × 72 × 10^5kg)/(5 × 10^6m\bullet s^-1)} = - {(8 × 9.8m \bullet s^-2 × 72 × 10^5 kg)/(5 × 10^6m \bullet s^-1)} = -112.9 kg \bullet s^-1, where -dm/dt is the maximum rate of fuel consumption. This rate of consumption then equals the total fuel load divided by the time taken to reach the final velocity (T). Thus T = {(12 × 10^3kg)/(112.9kg \bullet s^-1)} = 106.3 s = 1.77 min. Integrating equation (2) on has v = -v_r 1n m - gt + c. But if m_0 is the total load at time t = 0, when v = 0, then O = -v_r 1n m_O + C. C = v_r1n m_O and v = -v_r 1n m - gt + v_r 1n m_O \therefore v = v_r 1n(m_O/m) - gt. for 1n(m_O/m) = 1n m_O - 1n m The final velocity is thus v = {(5 × 10^6m \bullet hr^-1) / (60 × 60 s \bullet hr^-1)} 1n(14000 kg/2000 kg) - 9.8 m \bullet s^-2 × 106.3 s = (2703 - 1042)m \bullet s^-1 = 1661 m \bullet s^-1 = 5980 km \bullet hr^-1.

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Question:

There are 2N electrons in a one dimensional infinite square potential well of size L as shown in the figure. What is the energy of the last filled state (the Fermi energy) at T = 0\textdegreeK. The exclusion principle forbids the occupation of the same energy level by more than two electrons.

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Solution:

We can find the momenta of the electrons in the well from the Schroedinger equation. If \psi (x) is the wave function of the electron in the well, then it must satisfy the equation [ - (h^2/2 m) (d^2/dx^2) + V(x) ] \psi (x) = E \psi (x)(1) where E is the energy of the electron. \psi(x) is non-zero only between x = 0 and x = L, since the electron cannot escape from an infinitely large potential well. The solution of (1) can be obtained if we write it as (d^2\psi/dx^2) + (2 m/h^2)E \psi = 0 for 0 < x < L.(2) the solution of (2) is \psi (x) = A cos kx + B sin kx where k = \surd(2mE/h^2) . Now, we utilize the condition that \psi (x) vanishes at the boundries of the potential. \psi (x = 0) = \psi (x = L) = 0. This gives \psi (0) = A = 0 and \psi (L) = A cos kL + B sin kL = B sin kL = 0 The last equation shows that sin kL = sin n\pi = 0, n = 1, 2, ... . Therefore the allowed values of k are such that kL = n\pi k_n = (n\pi/L) The corresponding energy values are E^2_n = (h^2k^2)/(2m) = [h^2/{(4\pi^2)2m}] [(n^2\pi^2)/L^2] = (h^2n^2/8mL^2) (Here we made use of the definitionh= h/2\pi). The energy levels are shown in the figure. We can put only 2 electrons in each level, therefore at T = 0\textdegreeK, 2N electrons fill the lowest N levels, starting from n = 1. Notice that the level n = 0, which corresponds to zero energy is not allowed on the basis of the uncertainty relation. Since \Deltax \Deltap\geqh, \geq for \Deltax = L, the momentum of the electron cannot be strictly zero, but has a spread of \Deltap\geq(h/L) \geq This corresponds to a non-zero energy even for the lowest level. The energy of the highest level is E_F = (N^2h^2/8mL^2) = [(h^2/8m)(N/L)^2] . We see that the Fermi energy is a function of the number of electrons per unit length, N/L.

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Question:

How much work is done in joules when a mass of 5 kilograms is raised a height of 2 meters?

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Solution:

Mechanical work is given by the product of the force applied to a body, and the distance for which it is applied (W = Fs when force is constant and force and line of travel are in the same direction). The force of gravity on the 5 kilogram weight is equal to the force exerted against gravity (by Newton's Third Law) and is given by: F = mg F = 5 kg (9.80 m/sec^2) = 49.0 [(k-m)/(sec^2)](newtons) and the work is given by W = 49.0newtons(2m) = 98 joules.

Question:

Based on your knowledge of plant nutrition, explain how a photosynthetic plant growing in iron - and magnesium - deficient soil would look.

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Solution:

Magnesium is an essential component in the synthesis of chlorophyll in green plants (see structures in Problem 9-3). A deficiency of magnesium results in an inability of the plant to manufacture enough chlorophyll, and the plant, deficient in green pigment, will look pale. Since chlorophyll is of vital importance in the light re-actions ofphotsynthesis, very little if any radiant ener-gy can be captured and hence little food can be manufac-tured. One obvious result is that the plant will cease to grow. It may survive on the limited food reserves for some time, but it will eventually die. Iron is of particular importance in the synthesis of electron carriers ofphotophosphorylationand the electron transport chain. The cytochromesandferredoxin, two important electron carriers, are both iron - containing proteins. An iron - deficient plant will suffer from a defective electron transport system. Thus, in the photosynthetic process, the stream of electrons activated by absorbed light energy will pass down an inefficient electron acceptor chain. Most of the energy of these electrons will be lost rather than stored in ATP and NADPH as would occur in a healthy plant. Similarly, because the iron-containingcytochromesare also involved in the process of oxidativephosphorylation, cellular respiration in an iron-deficient plant yields little usable energy stored in ATP and much energy would be lost as heat. The iron - deficient plant, therefore, not only fails to make enough ATP and NADPH in the light reactions, but also fails to generate ATP efficiently in the oxidation of food reserves. In summary, a plant growing in soil deficient in both magnesium and iron contents will be pale, small, short, and weak, if itsurviesat all.

Question:

"A virulent organism is as good as dead if it is not communicable ." Explain.

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Solution:

A virulent pathogen will eventually die if it is restricted to its original host . The host produces substances which either prevent the growth and spread of the pathogen or actually destroy it. Even if the host fails to do this , a virulent pathogen will bring about its own destruction by killing the host that sustains it. For the pathogen to survive and cause disease in a number of organisms, it must find new hosts to infect. Communicable pathogens are transferred from one host to another by a number of means. Pathogens of the respira-tory tract, such as Diplococcus pneumoniae, can leave the body in discharges from the mouth , nose and throat. Sneezing and coughing expedite the spread of these or-ganisms. Enteric (intestinal) pathogens, such as those which cause typhoid fever, leave the host in fecal excre-tions and sometimes in the urine. The bacteria in these wastes contaminate the food and water which a subsequent host may ingest. Other pathogens cannot live long outside a host and are transmitted by direct contact, most commonly through breaks in the skin or contact of mucous mem-branes. The final common means of transmittance is by certain insect vectors, although these insects may be carried further by other organisms such as rats and man . For example, ticks transmitRickettsiaericketsiito man, causing Rocky Mountain spotted fever. Some pathogens are transmitted by specific means, and enter the new host by a specific portal of entry. For ex-ample, enteric bacteria have a special affinity for the alimentary tract and are able to survive both the enzymatic activity of the digestive juices and the acidity of the stomach. We thus see that the success of a pathogen depends on its successful transmittance to a new host via air, food, water, insects, or by contact.

Question:

What is themolalityof a solution in which 49 g of H_2SO_4, (MW 98) is dissolved in 250 grams of water?

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Solution:

Molalityis defined as the number of moles of solute per kilogram of solvent. Molality= (moles of solute/no. of kg. of solvent) Here, the solute is H_2SO_4 and the solvent is water. In this problem, one is given the number of grams of solute and the number of grams of solvent. One must calculate the number of moles of solute and the number of kilograms of solvent. The number of moles of solute (H_2SO_4) is found by dividing the number of grams available by the molecular weight. no. of moles = (no. of grams/MW) no. of moles of H_2SO_4 = [49 g/(98 g/mole)] = 0.5 moles Grams can be converted to kilograms by multiplying the number of grams by the conversion factor 1 kg/1000 gm. For the water, 250 × 1 kg/1000 gm = .250 kg. Themolalitycan now be found. Molality= [(no. of moles of H_2SO_4)/(no. of kg. of H_2O)] Molality= (0.5 moles/.250 kg) = 2.0 moles/kg.

Question:

Assuming all other orbitals to have zero net spin, what would be the net spin and the multiplicity of an atom having each of the following outer shell configurations; 2p^1, 2p^3, 2p^5, 3d^1, 3d^3, 3d^5; Rank these net spins in order of increasing paramagnetism.

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Solution:

We proceed by first defining ''total spin" and "multiplicity". Let the number of electrons having "spin up" be denoted by n_\uparrow and the total number of electrons having "spin down" be denoted by n_\downarrow. Assigning a value of (+ 1/2) to spin up and (- 1/2) to spin down, the total spin s is given by the absolute value of n_\uparrow (+ 1/2) + n_\downarrow (- 1/2) = n_\uparrow/2 + n_\downarrow/2. The multiplicity is then defined as m = 2s + 1. The configuration with the highest spin (highest multi-plicity) will have the greatest paramagnetism. (.Substances that are weakly attracted to magnets are paramagnetic.) The configuration with the lowest spin (lowest multiplicity), will be the least paramagnetic. The p orbital can contain six electrons and the d orbital can contain 10 electrons. When these orbitals are filled, electrons are sequentially placed in each sub-orbital (e.g. the p_x or the d_xy) until there is one electron in each suborbital, all electrons having the same spin. Only when there is one electron in each suborbital does another electron enter a suborbital already containing an electron, this time with a spin opposite to that of the electron already in the suborbital (the spins are then "paired") in accordance with the Pauli Exclusion Principle. In the accompanying table we show the orbital filling corresponding to each configuration. Counting n_\uparrow and n_\downarrow, we then calculate s and m. The configurations p^1, p^5, and d^1, have the same electron paramagnetism [all have spin (1/2)]. The next most paramagnetic configurations are p^3 and d^3 (spin 3/2) and the most paramagnetic configuration is d^5 (spin 5/2).

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Question:

An axon can conduct impulses in both directions simultaneously. However, an impulse normally propagates down an axon in one direction only, either axon to dendrite or dendrite to axon. Explain how this unidirectional con-duction of impulses along the neuron is possible.

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Solution:

Normally, an impulse travels from the dendrites to the cell body, then along the axon to the synaptic junction. However, impulses can be made to travel along an axon to the dendrites. But these impulses are not relayed to the next neuron since dendrites do not con-tain the chemical substance (neurotransmitter) required in the transmission of an impulse across a synapse. Although two directions of transmission along an axon are potentially possible, once an impulse is started at either end of a neuron, it will pass on through the neuron toward the opposite end. This phenomenon is due to the existence of a recovery period, known as the refractory period in the neuronal membrane. After a region of the axon is stimulated, a refractory period must elapse before it can be excited again. To understand the relation between the refractory period and the one-way conduction of impulses, one must recall that a region undergoing an action potential is positive with respect to other points inside the axon. Since that region is of higher potential than the surrounding region, a flow of Na^+ ions spreads from that local region to the adjacent regions on both sides of it. This is confirmed by observations which indicate that within a neuron, an impulse can travel in either direction from its point of origin. However, because the region that has just been stimulated cannot be made to depolarize until it has recovered, the impulse moves in one direction only (see Figure). By the time that region has regained the ability to be depolarized, the impulse has traveled too far away from it, and hence cannot depolarize it. It should be remembered that the unidirectional propagation of impulses is determined by the stimulus location (i.e., at which end the stimulus is given) rather than an intrinsic inability of the axon to conduct in the opposite direction. If action potentials are initiated near the middle of a neuron, impulses propagate from this region toward the two ends. In most nerves, however, action potentials are initiated at one end of the cell and impulses therefore propagate in only one direction toward the other end of the cell.

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Question:

The van der Waal equation is a modification of the ideal gas equation. It reads [P + (an^2 / V^2)] (V - nb) = nRT where P = pressure, V = volume, n = number of moles, R = gas constant, T = absolute temperature, and (a) and (b) are constants for a particular gas. The term an^2 /V^2 corrects the pressure for intermolecular attraction and the term - nb corrects the volume for molecular volume. Using this equation, determine whether a gas becomes more or less ideal when: (a.) the gas is compressed at constant temperature; (b.) more gas is added at constant volume and temperature; and (c.) The temperature of the gas is raised at constant volume.

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Solution:

As the behavior of a gas approaches ideality, PV/nRT approaches 1. Hence, we must manipulate the van der Waal equation into a form in which it can be de-termined whether PV/nRT approaches 1 as the variables are changed. Thus, [P + (an^2 /V^2 )] (V - nb) = nRT PV - Pnb + (an^2 /V^2 ) V - (an^2 /V^2 ) nb = nRT PV - Pnb + (an^2 /V) - (an^3 b/V^2 ) = nRT PV = nRT + Pnb - (an^2 /V) + (an^3 b/V^2)]. Dividing by nRT we obtain (PV/nRT) = 1 + (Pb/RT) - (an/RTV) + (an^2 b/RTV^2 ) (a.)When the gas is compressed at constant tem-perature, volume decreases and pressure increases. Hence, the last three terms on the right side increase, and PV/nRT deviates from 1. The gas thus becomes less ideal. (b.)If more gas is added at constant volume and temperature, both n and P increase, hence the last three terms on the right side increase and the gas becomes less ideal. (c.)As the temperature of the gas is increased at constant volume, the pressure increases. The last two terms on the right side become smaller while the second term (Pb/RT) increases slightly. Hence, the gas comes closer to being ideal.

Question:

Describe the flow of data from the Program Counter to the Accumulator in a PDP-8 computer.

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Solution:

The Program Counter (PC) contains the address of the next instruction. So after executing the previous instruction the PC starts as follows: The contents of the PC are transferred to the memory ad-dress register (MAR).The PC is then incremented in anticipation of the next instruction tobe executed. Now, the MAR has the address where an instruction is stored. Therefore, theinstruction corresponding to this address is read from the memory into thememory buffer register (MBR). Next, the contents of the MBR are treated as follows: The first three bits (bits 0 to 2) are sent to the IR. These 3 bits represent theOP code. Now, bit No. 4 of the MBR is examined. If this bit is 0, then the MAR is clearedto 0. (All bits of the MAR become 0.) But, if this bit is 1, then the MAR is not cleared. After this, bits Nos. 5 to 11 of the MBR are transferred to bits Nos. 5 to 11 ofthe MAR. As a result, at this stage the MAR contains information which iseither one of the following: Bits 0 to 4:all 0's;Bits 5 to 11:same as Bits 5 to 11 of MBR or,Bits 0 to 4:same as originally in the MAR;Bits 5 to 11: same as Bits 5 to 11 of MBR. Next, Bit No. 3 of the MBR is examined. This bit sig-nifies whether the instructionuses direct or indirect addressing. 1)If Bit No. 3 is 0, the instruction utilizes direct addressing, and the contentsof the memory location cor-responding to the new value in the MARareread into the MBR. The MBR now contains an operand. Hence, thecon-tents of the MBR can be transferred to the AC to carry out the operationcorresponding to the op-code stored in the IR. 2)If Bit No. 3 is 1, then the contents of the memory location correspondingto the new value in the MAR are read into the MBR. Now, thecontents of the MBR are transferred again to the MAR, and the contentsof the memory location, corresponding to the latest MAR contents, are read into the MBR. (This mode of addressing is called Indirect Addressing.) The contents of the MBR are now finally transferred to the AC to be processedaccording to the op-code stored in the IR.

Question:

A diver descends to a depth of 15.0 m in pure water (density 1.00 g/cm^3). The barometric pressure is 1.02 standard atmospheres. What is the total pressure on the diver, expressed in atmospheres? (b) If, at the same barometric pressure, the water were the Dead Sea (1.20 g/cm^3), what would the total pressure be?

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Solution:

Pressure is defined as force per unit area. Atmospheric is measured by using a barometer (usually a mercury barometer; see figure). It is constructed by inverting a tube longer than 76 cm filled with mercury into a dish of mercury. The atmosphere will support only that height of mercury which exerts an equivalent pressure; any excess mercury will fall into the reservoir and leave a space with zero air pressure above it. The pressure exerted on the diver from above is equal to the sum of the pressure exerted by the sea water and the pressure exerted by the atmosphere. One standard atmosphere equals the pressure exerted by exactly 76 cm (= exactly 760 mm) of mercury at 0\textdegreeC (density Hg = 13.5951 g/cm^3) and at standard gravity, 980.665 cm/s^2. Thus, 1 standard atm equals 13.5951 g/cm^3 × 76 cm (exactly) × 980.665 cm/s^2 = 1.01325 × 10^6 dynes/cm^2. The pressure exerted by the water is found similarly: pressure of water = density × height × standard gravity = 1.00 g/cm^3 × 1.5 × 10^3 cm × 980.665 cm/s^2 = 1.47 × 10^6 dynes/cm^2. However, the problem states that atmospheric press-ure is 1.02 standard atmospheres and, therefore, equals (1.02) (1.01325 x 10^6 dynes/cm^2) = 1.03 × 10^6 dynes/cm^2. To this atmospheric pressure, the pressure of the water is added to yield a total pressure of (1.03 × 10^6 dynes/cm^2) + (1.47 × 10^6 dynes/cm^2) = 2.50 × 10^6 dynes/cm^2. This answer expressed in atmospheres gives [(2.50 × 10^6 dynes/cm^2 ) / (1.01 × 10^6 dynes/cm^2 /atm)] = 2.48 atm. (b)This part is very similar to part (a). The total pressure exerted is the pressure of the water plus the pressure of the atmosphere. The pressure of the atmosphere from part (a) is 1.03 × 10^6 dynes/cm^2. The pressure of the water must be calculated: pressure of water = density × height × standard gravity = 1.20 g/cm^3 × 1.5 × 10^3 cm × 980.665 cm/s^2 = 1.77 × 10^6 dynes/cm^2. The total pressure is 1.77 × 10^6 dynes/cm^2 + 1.03 × 10^6 dynes/cm^2 equals 2.80 × 10^6 dynes/cm^2 . This answer expressed in atmospheres is [(2.80 × 10^6 dynes/cm^2 )/{1.01 × 10^6 (dynes/cm^2 )/atm}] = 2.77 atm.

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Question:

Consider a system of N particles each of which can exist in only two energy levels + \epsilon and - \epsilon. If the probability of occupying an energy level at temperature T is given by P_E = e- E/KT where E is the energy of the level, and K is Boltzmann's constant, calculate the internal energy of the system.

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Solution:

The ratio of the number of particles in levels E_1 = \epsilon and E_2 = - \epsilon is given by (N_1/N_2) = (P_1/P_2) = [(e- E1/KT)/(e- E2/KT)] = [(e- \epsilon/KT)/ [(e ^\epsilon/KT)] = e- 2\epsilon/KT(1) The total number of particles, N, in the system is constant though each particle can make transitions from one level to the other. Substituting the result N = N_1 + N_2 in (1), we get [(N - N_2)/N_2] = e- 2\epsilon/KT (N/N_2) = 1 + e- 2\epsilon/KT Giving (N/N_2) = [1/(1 + e- 2\epsilon/KT)] = [(e ^2\epsilon/KT)/(1 + e ^2\epsilon/KT)] . where we multiplied both the numerator and the de-nominator by e^2\epsilon/KT for the last step. Similarly, for N_1/N, we have (N_1/N) = (N_1/N_2) (N_2/N) = e- 2\epsilon/KT[(e ^2\epsilon/KT)/(1 + ^e2\epsilon/KT)] = [1/(1 + e ^2\epsilon/KT)] The internal energy of the system is given by the sum of the energies of the particles in each level, U = N_1E_1 + N_2E_2 = N_1 \epsilon + N_2 (- \epsilon) = \epsilon (N_1 - N_2) =\epsilonN(N_1/N - N_2/N) =\epsilonN[(1 - e ^2\epsilon/KT)/( 1 + e ^2\epsilon/KT)] =\epsilonN[(e ^- \epsilon/KT - e ^\epsilon/KT)/(e ^\epsilon/KT + e ^- \epsilon/KT)] =\epsilonNtanh(- \epsilon/KT) = -\epsilonNtanh(\epsilon/KT) As T \rightarrow 0,tanh(\epsilon/KT) \rightarrow 1, and we have U (T \rightarrow 0) = -\epsilonN This result shows that all the particles occupy the lower level at T = 0\textdegreeK. As T \rightarrow \infty,tanh(\epsilon/KT) \rightarrow 0 and U(T \rightarrow \infty) = 0 which shows that both levels are occupied with equal probability at T \rightarrow \infty\textdegreeK, i.e. N_1 = N_2.

Question:

Describe the mechanism involved in the maintenance of an erect posture.

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Solution:

As long as an individual is conscious, all his muscles are contracted slightly; this phenomenon is known as tonus. It is by the partial contraction of the muscles of the back and neck and of the flexors and extensors of the legs that posture is maintained. When a person stands, both the flexors and extensors of the thigh must contract simultaneously so that the body sways neither forward nor backward. The simultaneous contractions of the flexors and extensors of the shank lock the knee in place and hold the leg rigid to support the body. When movement is added to posture, as in walking, a complex coordination of the contraction and relaxation of the leg muscles is required. Receptors in the joints and associated ligaments and tendons, along with their pathways through the nervous system, play an important role in the unconscious control of posture and movement and also give rise to the conscious awareness of the position and movement of joints. Joint receptors are accurate indicators of movement and position. Skin receptors sensing contact of the body with other surfaces also play a role in the regulation of body posture.

Question:

A particular type of bacteriophage attaches to and pen- etrates a number of bacterial cells. After 20 minutes, some bacterial cells lyse and release many new viruse. Other bacteria remain intact and reproduce normally. After exposure to ultraviolet light, the remaining bac-teria lyse within an hour. Explain.

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Solution:

Viruses such as the T_2 bacteriophage always multiply when they enter a host bacterial cell. Lysis eventually occurs and the newly synthesized progeny are released from the cell. Some viruses such as the \lambda (lambda) phage do not always multiply and lyse their host. They are referred to as the temperate phages, in contrast to the virulent phages which always kill their host. The temperate virus must be present in some in-active state within the bacterial cell. In this special relationship between the virus and the bacterium, called lysogeny, the viral DNA is incorporated into a specific section of the host's chromosome. The viral chromosome becomes an integral part of the host chromosome, and is duplicated along with it during each cell division. The phage DNA can be transferred from one cell to another during bacterial conjugation. When the viral chromosomeis integrated within the host chromosome it is called the prophage. Bacteria containing prophages are called lysogenic bacteria and viruses whose chromosomes can be-come prophages are called lysogenic viruses (in contrast to lytic viruses). The integration of the viral chromosome occurs by a crossing over between the host chromosome and a circular form of the viral chromosome. Both chromosomes break and rejoin with the viral chromosome attaching to the broken ends of the bacterial chromosome. The point of attachment and crossing over occurs at a specific region of the chromosome. The insertion can be shown as follows: The prophage can remain integrated within the bacterial chromosome for many generations. Under certain natural conditions, the viral DNA becomes active and is removed from the host's chromosome. Once independent of the host chromo-some, the lysogenic virus replicates and eventually lyses the cell. This entrance into the lytic cycle can be arti- ficially induced by ultraviolet light or certain chemicals. It is thought that while the viral DNA is inserted as a prophage, the transcription (the process where DNA is used to code for a complementary sequence of bases in an RNA chain, i.e. production of mRNA) of almost all phage genes is blocked. The only protein made by the \lambda phage DNA is the \lambda repressor, which causes the blockage of viral- specific mRNA transcription. Since no viral components can be made, the viral DNA remains inactive within the host chromosome. However, when the \lambda repressor is inactivated by an agent such as UV light, mRNA transcription occurs and the virus begins replication. Lysis eventually results with the re-lease of new \lambda phages. When a lysogenic virus is in the prophage state, its genes can produce phenotypic effects on the host bacterium. It can modify the cell wall, affect the production of en-zymes and antigens, and possibly confer toxigenicity upon the lysogenic bacteria. Certain lysogenic phages enable the bacteriumCorynebacteriumdiphtheriaeto produce the toxin that causes the disease diphtheria. The lytic and lysogenic cycles can be summarized in the following diagram:

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Question:

A lab technician prepared a calcium hydroxide solution by dissolving 1.48 g of Ca(OH)_2 in water. How many milliliters of 0.125 N HCI solution would be required to neutralize this calcium hydroxide solution?

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Solution:

For a neutralization reaction to occur, the number of equivalents of acid must equal the number of equivalents of base. Therefore, one can solve this problem by computing how many equivalents are present in 1.48 g of Ca(OH)_2; the equivalents of acid (HCI) must equal this number. An equivalent is defined as the weight of a substance in grams that releases one mole of protons (H^+.) or hydroxyl ions. When one mole of Ca(OH)_2 dissolves in water, it pro-duces 2 moles of hydroxyl ions (OH^-) according to the equation Ca(OH)_2 \rightarrow Ca^+2 + 2OH^-. The weight of 1 mole of Ca(OH)_2 is equal to its molecular weight. Therefore, Ca(OH)_2 has a weight of 74 grams. Thus, 74 grams of Ca(OH)_2 produces 2 moles of OH^- ions. Recalling the definition of equivalency, 74/2 = 37 grams is the equivalent weight of Ca(OH)_2. The number of equivalents of Ca(OH)_2 is : Ca(OH)_2 = [(weight in grams)/(equivalent weight)] = (1.48 g/37 g) = 0.04 equiv. The number of equivalents of HCI must also be .04 for neutralization to occur. The normality of HCI is given as 0.125. Normality is defined as the number of equivalents divided by liters of solution, i.e., N = equivalents/volume. This means that 0.125 N = 0.04 equivalents/volume for HCI. The volume in liters required for neutralization can be obtained by solving this equation. liters = (0.04/0.125) = 0.32 l To convert to milliliters multiply by the conversion factor 1000 ml/l. One has 0.32 l × 1000 ml/l = 320 ml.

Question:

In order for a photon of light incident upon metallic potassium to eliminate an electron from it, the photon must have a minimum energy of 4.33eV(photoelectric work function for potassium). What is the wavelength of a photon of this energy? Assume h = 6.626 × 10^-27 erg-sec and c = 2.998 × 10^10 cm/sec.

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Solution:

This problem illustrates two fundamental relationships. The first is that the product of wavelength \lambda and frequency ѵ is the speed of light, c =\lambdaѵ(or ѵ = c/\lambda). The second is the relationship between frequency and energy E, E =hѵ, where h is Planck's constant. Combining these equations we obtain E =hѵ= h (c/\lambda) =hc/\lambda,or\lambda =hc/E. Before applying this equation, we must convert the energy E from electron volts (eV) to the more convenient unit of energy, the erg. Since 1eV= 1.602 × 10^-12 erg, E = 4.33eV= 4.33eV× 1.602 × 10^-12 erg/eV = 6.94 × 10^-12 erg. The wavelength of light corresponding to 4.33eVis then \lambda = (hc/E) = [(6.626 × 10^-27erg-sec × 2.998 × 10^10 cm/sec) / (6.94 × 10^-12 erg)] = 2.864 × 10^-5 cm = 2.864 × 10^-5 cm × 10^8 \AA/cm = 2864 \AA.

Question:

Sulfurylchloride decomposes according to the equation, SO_2CI_2(g) \rightleftarrows SO_2(g) + CI_2(g) . Given that at 300\textdegree K and 1atmpressure, the degree of dissociation is .121, find the equilibrium constant,K_p. What will be the degree of dissociation when pressure = 10atm?

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Solution:

To find the equilibrium constant,K_p, one uses the expression for K_p. This is the product of the pressures of the products divided by pressure of the reactant. These pressures are brought to the powers of their coefficients. From this, employ Dalton's law concerning partial pressures to help solve forK_p. At p = 10atm, assumeK_pis pressure-independent because of ideal gas behavior. This, then, allows the determination of the degree of dissociation from theK_pexpression. Thus, one commences by the definition ofK_p. K_p= [(P_SO2 P_CI2)/(P_SO2 CI2)] To findK_p, evaluate these partial pressures. To do this, useRaoult'slaw, which states partial pressure of gas = mole fraction of gas in mixture times total pressure, i.e., p =Xp_T. Thep_Tis given as 1 atm. Thus, the key is to determine or represent the mole fraction X. Let "a" = the degree of dissociation of SO_2Cl_2 . From the chemical equation, it is seen that SO_2 and Cl_2 are formed inequimolaramounts. Thus, at equilibrium there are "a" moles of both SO_2 and Cl_2 . Starting with 1 mole of SO_2Cl_2 and "a" moles dissociated, then, at equilibrium, there remain 1 - a moles. In the mixture, therefore, total number of moles = (1 - a) + a + a = (1 + a). Thus, P_SO2CI2 = X_SO2CI2 P_T = [(1 - a)/(1 + a)] P_T , P_SO2 = P_CI2 = [a/(1 + a)]P_T . Therefore, K_p= [(P_SO2 P_CI2)/(P_SO2 CI2)] = ([{a/(1 + a)} P_T {a/(1 + a)}P_T] / [{(1 - a)/(1 + a)} P_T]) This can be simplified to K_p= [a^2/(1 - a^2)]P_T . Given the degree of dissociation, a, is equal to .121 and P_T = 1atm;K_p therefore, equal [{(.121)^2}/{1 - (.121)^2}] (1) = 1.49 × 10^-2 . At 10atm, then, K_p= [a^2/(1 - a^2)] (10) = 1.49 × 10^-2 . Solving for a, a = .0386.

Question:

A yule log is being dragged along an icy horizontal path by two horses. The owner keeps the log on the path by using a guide rope attached to the log at the same point as the traces from the horses. Someone in the adjacent woods fires a shotgun, which causes the horses to bolt to opposite sides of the path. One horse now exerts a pull at an angle of 45\textdegree, and the other an equal pull at an angle of 30\textdegree, relative to the original direction. What is the minimum force the man has to exert on the rope in order to keep the log moving along the path?

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Solution:

The figure shows the forces exerted on the log by the horses at the moment they bolt. These forces can be resolved into components along the path and at right angles to the path. Thus the total forces in the x- and y-directions are \sumF_x\ding{217} = F cos 45\textdegree \^{\i} + F cos 30\textdegree \^{\i} and\sumF_y = F sin 45\textdegree \^{J} - F sin 30\textdegree \^{\i} where F is the magnitude of the force that each horse exerts. To keep the log on the path the man must counteract the unbalanced force in the y-direction, \sumF_y\ding{217}, by an equal and opposite force -\sumF_y\ding{217}. We can see that any force he may have exerted to keep the log moving along the path, exerted in other than the y-direction of the figure, would not have been the minimum force possible. Any otherwise directed force would have an x-component as well as -\sumF_y\ding{217}. But the latter alone could keep the log moving along the path. The magnitude of the result-ant force would then have a greater magnitude than -\sumF_y\ding{217}. Hence the minimum force he must exert has magnitude P_y = F(sin 45\textdegree - sin 30\textdegree) = F [(1/\surd2) - (1/2)] = 0.207 F, and must be directed in the negative y-direction, i.e., at right angles to the path. In the above analysis, frictional forces have been ignored. The frictional force acting along the path does not affect the solution. The frictional force trying to prevent motion at right angles to the line of the path reduces the magnitude of the force the man need apply. It is, however, assumed that on an icy path this frictional force is small in comparison with F, and its effect is therefore ignored.

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Question:

Calculate the standard enthalpy change, ∆H\textdegree, for the combustion of ammonia, NH_3(g), to give nitric oxide, NO(g), and water H_2O(l). The enthalpies of formation, ∆H\textdegree_f, are - 68.32 Kcal/mole for H_2O(l) , - 11.02 Kcal/mole for NH_3 (g) , and 21.57 Kcal/mole for NO(g).

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Solution:

The enthalpy change, ∆H, refers to the change in heat content between the products and the reactants of a chemical reaction. The symbol ∆H\textdegree is the enthalpy change for a reaction in which each reactant and product is in its standard state at a specified reference temperature (the common reference temperature is 25\textdegreeC) . The standard enthalpy of formation, ∆H\textdegree_f, of a sub-stance is defined as the change in enthalpy for the re-action in which one mole of the compound is formed from its elements at standard conditions. For any chemical reaction, the change in enthalpy, ∆H, may be expressed as ∆H ∆H reaction = \sumH_products - \sumH_reactants (where the symbol \sumH means the summation of the enthalpies of each substance). To solve this problem, one must first write and balance the equation for the combustion of NH_3(g): 4 NH_3 (g) + 5O_2 (g) \rightarrow 4NO(g) + 6 H_2O(l) From the balanced equation, one knows that 4 moles of NH_3(g) reacts with 5 moles of O_2 (g) to form 4 moles of NO(g) and 6 moles of H_2O(l). For the combustion of 4 moles of NH_3(g): 4 \bullet ∆H\textdegree_f, _NH3(g)= 4 [- 11.02 (Kcal/mole)]= - 44.08 Kcal 5 \bullet ∆H\textdegree_f, _O2(g)= 5 (0)= 0 4 \bullet ∆H\textdegree_f, _NO(g)= 4 [21.57(Kcal/mole)]= 86.28 Kcal 6 \bullet ∆H\textdegree_f, _H2O(l)= 6 [- 68.32 (Kcal/mole)]= - 409.92 Kcal Note: The ∆H\textdegree_f of all elements is zero. As such ∆H\textdegree_f,O2 (g)= 0. Thus , ∆H\textdegree = [4\bullet∆H\textdegree_f, _NO(g) + 6\bullet∆H\textdegree_f, _H2O(l)] - [4\bullet∆H\textdegree_f,NH3 (g)+ 5\bullet∆H\textdegree_f,O2 (g)] = [(86.28 Kcal) + (-409.92 Kcal)] - [(-44.08 Kcal) + 0] = - 279.56 Kcal. However, since this result is the standard enthalpy change for the combustion of 4 moles of NH_3 (g) , one must divide - 279.56 Kcal/mole by 4 to give the combustion of 1 mole of NH_3, ∆H\textdegree = (- 279.56 Kcal) / (4 moles) = - 69.89 Kcal/mole. = - 69.89 Kcal/mole.

Question:

(a)Water flows into a tank at a constant rate Q. Q is determined by the position of a valve connected to the inflow pipe. Write a FORTRAN program to simulate this system, using Euler'smethod, from time t = 0 to t = t_f if V(0) = V_0 and the capacity of the tank is C units of volume. Compare the Eulerapproximations with the exact values. (b)A hole is made in the bottom of the tank allowing water to flow out at a constant rate, Q_1. Modify the program of part (a) to simulate this system.

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Solution:

(a)Changes in this system's state can be viewedthrough changes In the volume of water in the tank, V. Note that once the valve position is set, Q is determined and remains unchanged for the length of the simulation. Therefore the valve is external to the system and does not effect the system during the time we are observing it in any given simulation run. The behavior of the system can be diagrammed as shown in Fig. 3 meaning V is the accumulation of the rate of change, Q(when Q has units of volume/time). If V = dV/dt = constant, Q, then the complete mathematical model looks as shown in Fig. 4, meaning V̇ = Q or dV/dt = Q(1) Using Euler's method to simulate this system means observing changes in V for small, discrete changes in t: V_new = V_old + V_added = V_old + Q \textasteriskcentered(t_new - t_old)(2) The exact solution of equation (1) is : V(t) = Q\textasteriskcenteredt + V_0(3) Note that substituting zero for t_old and V(0) = V_0 for V_old inequation (2) yields equation (3). The following program uses TFIN for t_F and DT for \Deltat: CPROGRAM WATERFLOW DATA T/0.0/ READ,N, TFIN,Q,VO,C REALN = N DT = TFIN/REALN VOL = V0 VEXACT = V0 DO 10 I = 1,N IF(VEXACT.NE.O.) ERROR = ABS((VEXACT-VOL)/ VEXACT)\textasteriskcentered 100. PRINT, T,VOL,VEXACT,ERROR T = T + DT VOL = VOL + Q\textasteriskcenteredDT IF(VGL.GT.C) GO TO 999 VEXACT = VO + Q\textasteriskcenteredT 10CONTINUE ERROR - ABS( (VEXACT-VOL)/VEXACT)\textasteriskcentered 100. PRINT,T, VOL, VEXACT,ERROR GO TO 99 999PRINT,'OVERFLOW AT TIME', T 99STOP END (b)Since Q1is the rate of flow out of the tank, it tends todecrease V, decrease V, and so it has a negative sign. V is the accumulation of the net flow, Q + (-Q_1) (see Fig. 5) From this, the following equation can be written: v̇ = Q - Q_1(1) Since both Q and are constant, Q - Q_1 is constant and equa-tion (1) can be written: V̇ = Q_2 where Q_2 = Q - Q1 Hence, the programmer needs only to substitute the value Q_2 of for Q in hisinput data for the program of part (a).

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Question:

Classify the following graphs according to the following criteria: a) directed or undirected b) connected or unconnected c) planar or nonplanar

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Solution:

Figure 1 is an example of an undirected, con-nected, planar graph. Undirected graphs contain no arrows: there is no particular direction associated with any of the branches. This graph is connected, since each node has at least one branch leading to it. Also, the graph is planar, because we can draw it on a plane so that none of the branches overlaps. Figure 2 is an undirected, unconnected, nonplanar graph. It is unconnected because node F is not attached to any branch. In addition, it is a nonplanar graph because of the overlapping of the various branches. There is no way to redraw this graph on a plane without running into over-laps. Finally, Figure 3 is a connected, directed, and planar graph. Notice the arrow along the branches, giving direc-tionality to it. We say it is a digraph, an abbreviated form of directed graph.

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Question:

Besides geographic isolation, what mechanismsmaintain Besides geographic isolation, what mechanismsmaintain isolationand differentiation?

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Solution:

Geographic isolation is an extrinsic mechanism of speciation. There are, in addition, a variety of intrinsic isolating mechanisms which establish andmaintain the differentiation of species. Inecogeographicisolation, two populationsinitially separated by some extrinsic barrier may in time becomeso specialized to their res-pective environmental conditions that evenif the original extrinsic barrier was removed, they would not be able tointerbreed with each other. In effect, they have evolved genetic differencesgreat enough to maintain them as two distinct species. In habitatisolation, two populations occupying different habitats which are withinreach of each other find their individuals mating only with members oftheir own population simply due to a greater frequency of encounter. In timethey may evolve into two different species. In behavioral isolation, twopopulations diverge genetically as a result of some behavioral differencessuch as courtship patterns and mate preference, that prevent themfrom interbreeding. In mechanical isolation, structural differences betweentwo closely related groups make it physically impossible for matingsbetween males of one and females of the other to occur. In this waythe two populations are prevented from interbreeding. In seasonal isolation, two groups cannot interbreed even when brought together becausethey breed during different seasons of the year. Ingametic isolation, even if indi-viduals of two groups succeed in mating with each other, actual fertilization fails to take place. In developmental isolation, evenwhen fertilization occurs, the development of the embryo is abnormal andmay cease before birth. In hybridinviability, hybrids are often weak andmalformed, and frequently die before they reach their reproductive age. In hybrid sterility, the hybrids are normal struc-turally but are sterile. In selectivehybrid elimination, the hybrids are normal and fertile, but their offspringare often less well adapted, and may soon be eliminated. Finally, inpolyploidy speciation (which occurs in plants,)polyploidscarrying an even-numberedmultiple of chromo-somal complements (4N, 6N, and so on) are unable to cross with the normal diploid individuals, but can breed amongthemselves. Thepolyploidsare genetically distinctive and reproductively isolated, they are usually more vigorous than the parental diploid form and areable to maintain themselves as a separate species.

Question:

A person exposed to gamma ray radiation from a nuclear bomb blast is likely to sustain complete destruction of his bone marrow. Why would this person become anemic?

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Solution:

The first step in the solution is to define anemia. Anemia is a deficiency of red blood cells, caused either by rapid loss or slow production of these cells. This results in decreased oxygen transport and thus im-pairment of the metabolism of all tissues. Gamma-ray radiation causes bone marrowaplasia, or a lack of functioning bone marrow. This disease is lethal if all the bone marrow is destroyed, because the bone marrow is the site of red blood cell production. During the middle part of development, the fetus produces erythrocytes in the liver, spleen and lymph nodes. But later in gestation and after birth, red blood cells are produced almost ex-clusively by the bone marrow. Until adolescence, erythro-cytes are produced in almost all bone marrow. In the adult, however, only the marrow in the ends of the long bones and the shafts of the flat bones, like those of the skull and ribs, form erythrocytes. Red blood cells are derived from cells,hemocytoblasts, which arise from unspecialized primordial stem cells in the bone marrow. The hemocytoblasts have a nucleus but no hemoglobin. They divide to form precursor cells which gradually transform into mature erythrocytes, having no nucleus and containing hemoglobin. Although mature erythrocytes have no nucleus, they already contain all the enzymes and machinery necessary for metabolism and energy production. When red blood cells are released into the circulatory system from the bone marrow, they normally circulate about 120 days. As the cells become older, they become more fragile and the membrane tends to rupture, killing the cell. In the walls of blood vessels in the spleen and liver arephagocyticcells which engulf and destroy old red blood cells. The hemoglobin released from these cells is degraded: the iron atoms are recovered and returned to the bone marrow to be used in erythrocyte synthesis; thehemeportion is degraded and excreted by the liver as bilirubin , a bile pigment. Under normal circumstances, the rate of formation of new erythrocytes in the bone marrow equals the rate of destruction of old erythrocytes in the liver and spleen. About 2.5 million erythrocytes are made and destroyed each second. Thus, the total number of circulating erythrocytes remains constant.

Question:

Calculate the pH of a 0.2 M NH_3 solution for which K_b = 1.8 × 10^-5 at 25\textdegreeC. The equation for the reaction is NH_3 + H_2O \rightleftarrows NH_4^+ + OH^-.

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Solution:

pH is defined in terms of [H^+], (pH = - log [H^+]), but this reaction shows the production of OH^-. However, [H^+] and [OH^-] are related by the definition, [H^+] = 10-14/[OH^-]. Therefore, one can find [H^+] and pH after solving for [OH^-]. One solves for [OH^-] by using the fact that K_b is equal to {[NH_4^+] [OH^-]} / [NH_3]. [H_2O] is excluded from this expression because it is assumed to be constant. In solving for [OH^-], one can assume that 1 l of the 0.2 M NH_3 solution is present. This means that there were originally 0.2 moles of NH_3 present. For each NH_3 that dissociates, 1 mole of OH^- and 1 mole of NH_4^+ is formed. Let x = [OH^-] = [NH_4^+] and 0.2 - x = [NH_3] K_b = {[NH_4^+] [OH^-]} / [NH_3] = 1.8 × 10^-5 = [(x) (x)] / [0.2 - x] One can assume that x is negligible compared to 0.2. Therefore, 0.2 - x = 0.2. Using this assumption, one can now solve for x. 1.8 × 10^-5 = (x^2 / 0.2) x^2 = 3.6 × 10^-6 ;x = 1.9 × 10^-3. [NH_4^+] = 1.9 × 10^-3 M, [OH^-] = 1.9 × 10^-3 M and [NH_3] = 0.2 - 1.9 × 10^-3 \approx 0.2. One can now solve for [H^+] using [OH^-]. [H^+] = 10^-14 / [OH^-] = (10^-14) / (1.9 × 10^-3) = 5.26 × 10^-12 M Using this value one can solve for pH. pH = - log [H^+] = - log (5.25 × 10^-12) = - (.72 - 12) = 11.28.

Question:

A room is heated by a heating system controlled by an on-off thermostat, which will turn the heat on when the room temperature drops h degrees below its set point, T_s, and cuts heat off at a temperature h degrees higher than T_s . The system heats at a constant rate Q (thermal units per time unit), and the volume of the room is V, so, if the room were perfectly insulated, the heating system would raise the temperature of the room at the rate KVQ (degrees per time unit), where K is a con-stant of proportionality giving the change in temperature per thermal unit per unit volume. However, there is a heat loss to the outside that is proportional to the difference in temperature between the room and the outside. Using the modified Euler method, write a FORTRAN program to simulate this system, if the room temperature at time is T(t), T(0) = T_0, and the outside temperature, T_out' is given by: Case (1) : Toutis constant< T_s Case (2) : T_out =▕A sin(\omegat + \varphi) ▏, reflecting daily cycles. A is \omega temperature range, 2\pi/\omegais the period of a cycle, and \varphi is \omega the time lag (or lead).

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Solution:

First, note that a typical thermostat consists of a bimetallic strip, whose elastic reaction to transient heating or cooling is used to control a triggering mechanism controlling the heating system. Thus, the internal components of the thermostat make it a dynamic element, and it can be considered a subsystem. The internal workings of the thermostat could be included in the simulation. Here we will ignore the internal operations that determine the action of the thermostat and take a "black box approach". That is, we know that information fed into the thermostat will determine whether or not it will switch, but we do not carehowthat mechanism operates: Note that the thermostat's behavior will be nonlinear. An initial picture of the system is shown in figure #1. Since the rate of heat loss, L, acts as a valve, and the thermostat acts as a switch, we can draw them as shown in figure #2. Now L = C(T-T_out) where C is a constant of proportionality which out represents change in heat per degree per time unit. Note that Tout> T \Rightarrow L < 0 and T_out < T \Rightarrow L > 0 . Let U(T) be the heating function: U(T) =\midQ , if switch is on \midO , if switch is off Then the net rate of heat flow is H-L, and the net rate of room temperature change is given by: T = KV(U-L) = KV(U-C(T-T_out)) Arbitrarily, the switch is considered off initially, U changes when T(t) deviates from T_s by h degrees. Otherwise, U remains the same. Also, the programs use TEMP for T(t), TMPOUT for T0ut, TMPSET for T_s and TEMPO for T_0 : Case (1) : REAL T/0.0/,K,V,H,C,TEMP,TMPOUT,TEMPO, ACCUR,TFIN,DT,TMPSET,Q,U/0.0/ INTEGER I,N COMMON U, TMPOUT,K,V,C READ,N,TFIN,ACCUR,TEMPO,K,V,H,C,TMPOUT, TMPSET,Q PRINT T, TEMPO REALN = N DT = TFIN/REALN TEMP = TEMPO DO 10 I = 1,N T = T + DT IF (TEMP.LE.(TMPSET-H))U = Q IF (TEMP.GE.(TMPSET+H))U = 0.0 CALL MEULER (T, TEMP, ACC UR, DT) PRINT T, TEMP 10CONTINUE STOP END FUNCTION G(W) COMMON U, TMPOUT,K,V,C REAL K,V,C, U,TMPOUT, W,G G = K\textasteriskcenteredV\textasteriskcentered(U-C\textasteriskcentered(W-TMPOUT)) RETURN END Case (2) :Using OMEGA for\omega, and PHI for \varphi : \omega REAL T/0.0/,K,V,H,C,TEMP,TMPOUT,TEMPO,ACCUR, TFIN,DT,TMPSET,Q,A,OMEGA,PHI, U/0,0/ INTEGER I,N COMMON U,TMP0UT,K,V,C READ,N,TFIN,ACCUR,TEMPO,K,V,H,C,TMPOUT,TMPSET, Q,A, OMEGA,PHI PRINT,T,TEMPO REALN = N DT = TFIN/REALN TEMP = TEMPO DO 10 I = 1,N T = T + DT IF (TEMP.LE. (TMPSET-H))U = Q IF (TEMP.GE.(TMPSET+H))U = 0.0 TMPOUT = A\textasteriskcenteredS IN ( OMEGA. \textasteriskcenteredT+PHI) CALL MEULER (T,TEMP,ACCUR,DT) PRINT T,TEMP 10CONTINUE STOP END FUNCTION G(W) COMMON U, TMPOUT,K,V,C REAL K,V,C,U,TMPOUT,W,G G = K\textasteriskcenteredV\textasteriskcentered(U-C\textasteriskcentered(W-TMPOUT)) RETURN END

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Question:

An economist suspects that a leading indicator in the business cycle has the form xcosx. He wishes to know when the indicator will peak, i.e. achieve a maximum, in different time intervals. Write a FORTRAN program to find the maximum of xcosx in interval [a,b].

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Solution:

There are many methods of computing the maximum of a function over its domain or a subset of its domain (local maximum). A method particularly suited for computer applications is the elimination scheme. To illustrate the method, let a = o, b = \pi and suppose we wish to Maxxcosxx\epsilon [o,\pi].(1) 1) Place two search points close together at the centre of the interval. Let the distance between the points be \epsilon > o. 2) Evaluate f(x) at X_L and X_R and call the results f(X_L), f(X_R). If f(X_L) \geq f(X_R) Max f(X) lies between o and X_R and the segment [X_R \pi] can be discarded. 3) Place two search points close together at the center of the remaining interval and repeat step (2). 4) Suppose f(X_(R)1 ) \geq f (X_(L)1). Then Max f(X) lies in the interval [X_(L)1 ,X_R] and the segment [0, X_(L)1 ] can be discarded. 5) The process continues until an interval less than 2\epsilon is obtained. Since the search cannot continue, the maximum is assumed to occur at the center of this interval. In the program, a statement function is used to define xcosx so that it can be referred to at any further point in the program. Also, if the search points pass a toler-ance limit, the program halts. CDEFINE THE FUNCTION Y(X) Y(X) = X\textasteriskcenteredCOX(X) READ XL, XR, EPSI I = 1 CCALCULATE INTERIOR POINTS 10XL1 = XL + . 5\textasteriskcentered (XR - XL - EPSI) XR1 = XL1 + EPSI YL1 = Y(XL1) YR1 = Y(XR1) PRINT YL1, YR1, XL, XL1, XR1, XR IF (YL1 - YR1) 20, 50, 30 20XL = XL1 GO TO 40 30XR = XR1 CTEST FOR END OF SEARCH 40IF (I.GE.100) GO TO 60 I = I + 1 IF (XR - XL.GT.3\textasteriskcenteredEPSI) GO TO 10 50XMAX = .5\textasteriskcentered(XL1 + XR1) YMAX = . 5\textasteriskcentered (YL1 + YR1) PRINT YMAX, XMAX GO TO 70 CPRINT OUTPUT TERMINATED BECAUSE OF MAX- CIMUM ITERATION COUNT 60WRITE (6, 300) 300FORMAT (54HO THE SOLUTION HAS NOT CONVERGED AFTER 100 ITERATIONS. TERMINATE PROGRAM 70STOP END

Question:

Describe some of the method which insure cross-pollination.

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/Users/wenhuchen/Documents/Crawler/Biology/F08-0205.htm

Solution:

Cross-pollination involves pollen transfer between plants having different genetic constitutions. Since land plants are immobile, they must evolve specia-lized features to allow them to mate over distances. The seed plants have evolved structures which promote cross-pollination by wind, insects, bats, birds, and other animals. Wind-pollination is common in plants having inconspicu-ous flowers, such as grasses and sedges. Flowers of grasses are dull-looking, and most of them lack both odor and nectar, making them unattractive to insects. More-over, their pollen is light, dry and easily carried by the wind. Their stigmas are feathery and expose a large surface area to catch pollen. Their stamens are well exposed to the action of the wind. Wind-pollinated flowers, such as those of the grasses, probably represent a special adaptation to colder climates, where insects are less prevalent. Insect-pollination is the method of pollen dispersal in plants with colorful, showy flowers. These flowers also produce a sweet nectar and volatile compounds having unique odors which attract insects. As an insect approaches a flower, floral architecture ensures transfer of pollen onto the insect's body. In this way, pollen of one flower is carried to the stigma of another as the insect goes from flower to flower. Flower-visiting bats are also an agent of cross-pollination. Bats are attracted to the flowers largely through their sense of smell. Bat-pollinated flowers characteristically have very strongformetitingor fruit-like odors. Bats fly from plant to plant, eat floral parts, and carry pollen on their fur. Birds may also serve as pollinators. Birds have keen vision, and most bird-pollinated flowers are colorful. Some birds regularly visit these flowers to feed on nectar, floral parts, and flower-inhabiting insects. Pollen transfer is also aided by other animals with fur as they pass from plant to plant. Certain species of plants suremonoecious, meaning they have Certain species of plants suremonoecious, meaning they have both sexes contained within the same plant. In order to effect In order to effect cross-pollination and avoid inbreeding within the same plant, specific tactics have been devised. In some species, such as the goldenrod In some species, such as the goldenrod (Solidago), the male and female organs of the same plant mature at slightly different times, making self-pollination unlikely. In some other species, the pollen is unable to germinate on the same plant. In plants such as these, bisexuality has an advantage in that a pollinator can both pick up and deliver pollen at the same time.

Question:

Give the name of the compound formed by the addition of HCI to (a) isobutene; (b) 3-methyl-2-pentene.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E21-0758.htm

Solution:

When a halogen acid, such as HCI, is added to alkenes, an addition reaction occurs. Reactions with the addition of an acid to unsymmetrical molecules (where the two doubly bonded carbon atoms have different numbers of hydrogen atoms attached to them) usually follow a predictable course: the hydrogen atom of the acid adds to that carbon atom which has attached to it the greater number of hydrogen atoms. The acid anion then adds to that carbon atom which has the lesser number of hydrogen atoms. This is known as Markovnikov's rule. To solve this problem, write the structural formulas, set up the reaction equation, and then follow the rules just laid out to find the product. Once the product is known in terms of its structure, the next step is to name the compound using the IUPAC system. Thus,

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Question:

A man stands at rest on frictionless roller skates on a level sur-face, facing a brick wall. He sets himself in motion (backward) by pushing against the wall. Discuss the problem from the work-energy standpoint. Fig. A) Man pushes against wall with force F^\ding{217}; Wall exerts equal and opposite force R^\ding{217}. (B) Free-bogy diagram of forces acting on the man, including his weight W^\ding{217} and the normal force N^\ding{217} of the ground on the man.

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Solution:

The external forces on the man are his weight, the upward force exerted by the surface, and the horizontal force exerted by the wall. (The latter is the reaction to the force with which the man pushes against the wall.) The definition of the work done by a force F^\ding{217} is w = \intF^\ding{217} \bullet ds^\ding{217} where ds^\ding{217} is an element of the path traversed by the object acted on by F^\ding{217}. No work is done by the first two forces because they are at right angles to the motion. No work is done by the third force because there is no motion of its point of application. The external work is therefore zero and the internal work (of the man's muscular forces) equals the change in his kinetic energy.

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Question:

An aircraft is climbing with a steady speed of 200 m/sec at an angle of 20\textdegree to the horizontal (see figure). What are the horizontal and vertical components of its velocity?

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/Users/wenhuchen/Documents/Crawler/Physics/D01-0013.htm

Solution:

Using trigonometric relations for right triangles, the velocity can be broken down into two components perpendicular to each other. Horizontal component = 200 cos 20\textdegree Vertical component = 200 sin 20\textdegree. Trigonometric tables tell us that cos 20\textdegree = 0.9397 and sin 20\textdegree = 0.3420 Therefore, horizontal component= 200 × 0.9397 = 187.94 m/sec Vertical component= 200 × 0.3420 = 68.40 m/sec. Notice that the sum of 187.94 and 68.40 is not 200, but you can check that (187.94)^2 + (68.40)^2 = (200)^2. This occurs because the horizontal and vertical components, V\ding{217}_x and V\ding{217}_y, of the velocity are vectors and must be added accordingly. Since they are pernependicular to each other, forming a right triangle with V\ding{217} as the hypotenuse, V^2_x + V^2_y = V^2

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Question:

The frames in a home movie must be magnified 143 times before the picture formed on a screen 12 ft from the projection lens is large enough to please the family watching. What distance must the film be from the lens and what is the focal length of the lens?

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/Users/wenhuchen/Documents/Crawler/Physics/D28-0883.htm

Solution:

The magnification produced by the lens is given by m = - (s'/s) = -(12ft/8) = - 143. where s' and s are the image and object distance, re-spectively. (The image must be inverted that is, the magnification is negative, since s must be positive [see figure]). Thus the film-to-lens distance is s = (12/143) ft = (144/143) in. = 1.007 in. Applying the lens formula, we can obtain the focal length, since 1/f = 1/s + 1/s' = (143/12ft) + (1/12ft) = 144/12ft f = (12/144)ft = 1in.

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Question:

How much pressure is needed to raise water to the top of the Empire State Building, which is 1250 feet high?

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/Users/wenhuchen/Documents/Crawler/Physics/D10-0411.htm

Solution:

Pressure is given by height time density

Question:

How much energy is required to break up a C^12 nucleus into three \alpha particles?

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/Users/wenhuchen/Documents/Crawler/Physics/D34-1016.htm

Solution:

This reaction is C^12 \rightarrow He^4 + Be^8 Be^8 \rightarrow 2He^4 Energy is required since the mass of the three \alpha particles is greater than the mass of the C^12 nucleus. This is so because the C^12 nucleus can be considered to be a system of bound \alpha particles, and work must be done (energy must be added) to break up a bound system into its constituent parts. The additional mass of the products comes from energy-mass conversion. To find the additional energy required, calculate the change in mass. By definition, the atomic mass of C^12 is 12 AMU (exactly). And, 3 × m(He^4 ) = 3 × (4.002603 AMU) = 12.007809 AMU Therefore, the energy required is \epsilon = [3 × m(He^4 )\Elzbar m(C^12)]× C^2 = (12.007 809 AMU\Elzbar 12 AMU) × C^2 = (0.007 809 AMU) × (931.481MeV/AMU) = 7.274MeV.

Question:

In one type of mass spectrometer the charged particles pass through a velocity selector before entering the magnetic field. In another the particles pass through a strong electric field before entering the magnetic field. Compare the ratio of the radii of singly charged lithium ions of masses 6amuand 7amuin the two cases.

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/Users/wenhuchen/Documents/Crawler/Physics/D21-0714.htm

Solution:

In the magnetic field , an ion moves in a circle , the centripetal force necessary being provided by the magnetic force on it . If the velocity v \ding{217} of the ion is perpendicular to the field of magnetic induction B \ding{217} , then the magnetic force on the ion of charge q , is F = q v b(1) Since this is the requiredcentrepetalforce , F = (mv^2 /R)(2) Where m is the ion's mass , and R is the radius of the circle traversed by the ion . Equating (1) and (2) , q v B = (mv^2 /R) orv = (q B R)/(m)(3) When the ions have passed through a velocity selector , both lithium ions have the same velocity in the field. Further , they have the same charge and are in the same magnetic flux density. Thus, using (3), (R_6 / m_6 ) = (R_7^ / m_7) (R_6 /R_7 ) = (m_6 /m_7 )= (6/7) = 01857 . If the ions have passed through a strong electric field , they have both acquired the same kinetic energy. But, from equation (3) we have (1/2) mv^2 = [(q^2 B^2 R^2 )/(2m)] Therefore , since q and B are the same for both isotopes, (R^2_6 /m_6 ) = (r^2_7 /m_7 )or(r_6 /r_7 ) = [\surd(m_6 /m_7 )] = 0 . 926.

Question:

Certain connoisseurs can recognize hundreds of varieties of wine by tasting small samples. How is this possible when there are only four types of taste receptors?

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/Users/wenhuchen/Documents/Crawler/Biology/F20-0499.htm

Solution:

Taste buds on the tongue and the soft palate are the organs of taste in human beings. Each taste bud contains supportive cells as well as epithelial cells which function as receptors. These epithelial cells have numerous microvilli that are exposed on the tongue surface. Each receptor is innervated by one or more neurons, and when a receptor is excited, it genera-tes impulses in the neurons. There are four basic taste senses: sweet, sour, bitter, and salty. The receptors for each of these four basic tastes are con-centrated in different regions of the tongue-sweet and saltly on the front, bitter on the back, and sour on the sides (see Figure). The sensitivity of these four regions on the tongue to the four different tastes can be demonstrated by placing solutions with various tastes on each region. A dry tongue is insensitive to taste. Few substances stimulate only one of the four kinds of receptors, but most stimulate two or more types in varying degrees. The common taste sensations we ex-perience daily are created by combinations of the four basic tastes in different relative intensities. Moreover, taste does not depend on the perception of the receptors in the taste buds alone. Olfaction plays an important role in the sense of taste. Together they help us distinguish an enormous number of different tastes. We can now understand how a connoisseur, using a combination of his taste buds and his sense of smell, can recognize hundreds of varieties of wine.

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Question:

The sensitivity of the silver halides to light is the basis of photography. The change from white silver chloride to a gray purple solid is the result of photochemical decomposition, The ∆H of this reaction is 30.362 Kcal/mole. From this, determine the frequency of light needed to decompose 1 molecule of AgCl.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E30-0924.htm

Solution:

To solve this problem (1) determine the energy consumed per molecule AgCl by dividing the ∆H given by Avogadro's number. (2) Convert the energy per molecule of AgCl from units of Kcal/molecule to units of J/molecule using the factors 1 Kcal = 1000 cal and 1 cal = 4.18 J. (3) Use the expression E = hѵ where E is the energy in J/molecule, h is Planck's constant, and ѵ is the frequency of light. Thus, (1)E = [{30.362 (Kcal/mole)} / {6.02 × 10^23 (molecules / mole)}] = 5.04 × 10-^23 (Kcal/molecule AgCl) (2)E = [5.04 × 10-^23 (Kcal/molecule )] [1000(cal/Kcal)] [4.18 (J/cal)] = 2.11 × 10-^19 J/molecule AgCl (3)E = hvorѵ = E/h ѵ = (2.11 × 10-^19 J/molecule) / (6.63 × 10-^34 J-s) = 3.18 × 10^14 cycles/sec per molecule AgCl.

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Question:

The photoelectric work functions for several metals are listed below. Calculate the threshold wavelengths for each metal. Which metals will not emit photoelectrons when illuminated with visible light? W- (Joules) Cs 3.2 × 10^-19 Cu 6.4 ×10^-19 K 3.6 ×10^-19 Zn 5.8 ×10^-19

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Solution:

When light with a frequency above some definite level (called the threshold frequency) illuminates certain metals, it is observed that electrons are emitted. The energy of an incident photon is entirely given up to one electron of the metal. If this is an interior electron, it will travel towards the surface of the metal and, in the process, loose energy due to collisions with atoms of the metal. In addition, the electrons will loose energy because they must overcome the attractive force of the atoms of the metal in order to escape from the surface. (The energy needed to overcome this attractive force is called the metal's work function, W). Electrons near the surface of the metal, however, don't experience the collisions de-scribed above, and they can't lose energy due to this cause. As a result, these surface electrons will be emitted with a higher kinetic energy than the interior electrons. Hence, using the principle of conservation of energy, we may write, for surface electrons, (1/2) mv^2_max + W =hu, where (1/2)mv^2_max is the kinetic energy of these electrons. At the threshold frequency, \Elzpscrv_0, the emitted electrons will have no kinetic energy. Therefore, h \Elzpscrv_0 = 0 + W or\Elzpscrv_0 = (W/h) The threshold wavelength \lambda_0 is \lambda_0\Elzpscrv_0 = c, thus \lambda_0 = (hc/w) hc= (6.63 × 10^-34 J\textbullets) (3 × 10^8 m/s) = 1.99 × 10^-25 J-m \lambda_0 (meters) \lambda_0 (\AA) Cs 6.2 × 10^-7 6200 Cu 3.1 ×10^-7 3100 K 5.5 ×10^-7 5500 Zn 3.4 ×10^-7 3400 The calculated values of \lambda_0 are shown in the table. To convert to Angstrom units (\AA), we note that 1\AA = 10^-10 m. Hence, dividing each value of \lambda_0 (in units of meters) by 10^-10 m will yield \lambda_0 in units of \AA. Since the range of the visible spectrum is 4,000 \AA - 7,600 \AA, we note that Cu and Zn will not emit photoelectrons if illuminated by visible light.

Question:

Graham's law states that the rate at which gas molecules escape through a small orifice (rate of effusion) is inversely proportional to the square root of the density of the gas. Derive Graham's law from the following assumptions: (a) temperature is directly proportional to the average kinetic energy of the molecules; (b) the rate of effusion is directly proportional to the root mean square speed of the molecules;(c) the density of a gas at constant temperature and pressure is directly proportional to the molecular mass.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E03-0107.htm

Solution:

Consider the assumptions individually. The first states that temperature, T, is proportional to the average kinetic energy 1/2 mu^2, where m = molecular mass, u = speed, and the bar over u^2 indicates the average. Hence T = k_1 × 1/2 mu^2 where k_1 is a constant. The second assumption states that the rate of effusion is directly proportional to the root mean square speed of the molecules, \surdu^2 ,or rate = k_2\surdu^2- where k_2 is a constant. The third assumption states that the density, p, is directly proportional to the molecular mass . or p = k_3 m where k_3 is a constant. Thus, T = k_1 × (1/2)mu^2 = k_1 ×(1/2) × (P / k_3) ×u^2. Taking the square root of both sides we obtain T^1/2 = [k_1/2k_3]^1/2 P^1/2\surdu^2 . Multiplying by [2k_3 / k_1]^1/2 this becomes [2k_3 / k_1]^1/2 T^1/2 = P^1/2\surdu^2. At constant temperature, the expression [2k_3 / k_1]^1/2 T^1/2 is a constant, say k_4. Then, k_4 = P^1/2\surdu^2 = P^1/2 × rate/k_2. Multiplying by k_2 / P^1/2 and letting k_2 × k4= k_5, a constant, we obtain: rate = (k_2 × k_4)/ P^1/2 = k_5 / P^1/2 which states that the rate of effusion is inversely proportional to the square root of the density, which is Graham's law.

Question:

Find log 0.0364.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E33-0964.htm

Solution:

0.0364 = 3.64 × 10^\rule{1em}{1pt}2. Therefore, the characteristic, the power of 10, is \rule{1em}{1pt}2. From a table of logarithms, the mantissa for 3.64 is 0.5611. Therefore, log 0.0364 = \rule{1em}{1pt}2 + 0.5611 = \rule{1em}{1pt}1.4389.

Question:

What are lichens? Describe the relationship that exist in a lichen. How do they reproduce?

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/Users/wenhuchen/Documents/Crawler/Biology/F06-0170.htm

Solution:

Lichens are composite organisms consisting of algae and fungi. They grow on tree bark, rocks and other substrates not suitable for the growth of plants. Lichens may be found in low-temperature environments characteristic of polar regions and very high altitudes. Structurally, a lichen can be likened to a fungal "sandwich" whose hyphae entwine a layer of algal cells (See figure). Structures known as rhizoids, which are short twisted strands of fungal hyphae, serve to attach the bottom layer to the substrate. Not all species of algae or all species of fungi can enter into a lichenlike relationship. Most of the fungi found in lichens are Ascomycetes, although there are a few of the Basidiomycetes. Algae of the lichens are from the Cyanophyta (blue-green algae) or the Chlorophyta (green algae) phylum. Each lichen thallus consists of a single species of fungus associated with a single species of alga. There are two types of lichens: fruticose and crustaceous. The fruticose have an erect, shrublike morphology. The crustaceous are very closely attached to the substrate or may even grow within its surface. Lichens are the product of a relationship called symbiosis, in which each partner of the association derives something useful from the other for its survival. The algae provides the fungus with food, especially carbohydrates produced by photosynthesis, and possibly vitamins as well. The fungus absorbs, stores and supplies water and minerals required by the alga, as well as providing protection and a supporting framework for the alga. Lichens appear to reproduce in a variety of ways. Fragmentation may occur, in which bits of the thallus are broken off from the parent plant and produce new lichens when they fall on a suitable substrate. Lichens may produce "reproductive bodies" called soredia which are knots of hyphae containing a few algal cells. In addition, the algal and fungal components of a lichen may reproduce independently of each other. The fungal com-ponent produces ascospores and the algal component re produces by cell division or infrequently via sporulation. Some species of lichens in the Arctic have been alive for 4,500 years which suggests a very well balanced association between the symbionts. Lichen grow very slowly because of their low metabolic rate. They are also very resistant to heat and desiccation. Lichens produce many interesting organic products. Unusual fats and phenolic compounds make up from two to twenty percent of the dry weight of the lichen body. Litmus, the pigment indicator, and essential oils used in perfumes are obtained from lichens.

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Question:

The ballistic pendulum (see figure) is a device for mea-suring the velocity of a bullet. The bullet is allowed to make a completely inelastic collision with a body of much greater mass. Find the bullet's velocity before the collision.

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/Users/wenhuchen/Documents/Crawler/Physics/D06-0326.htm

Solution:

The momentum of the bullet-block system immediately after the collision equals the original momentum of the bullet since the block is initially at rest. Although the ballistic pendulum has now been super-seded by other devices, it is still an important labora-tory experiment for illustrating the concepts of momentum and energy. In the figure, the pendulum, consisting perhaps of a large wooden block of mass m', hangs vertically by two cords. A bullet of mass m traveling with a velocity v, strikes the pendulum and remains embedded in it. If the collision time is very small compared with the time of swing of the pendulum, the supporting cords remain practically vertical during this time. Hence no external horizontal forces act on the system during the collision, and the horizontal momentum is conserved. Then, if V represents the velocity of bullet and block immediately after the collision, mv = (m + m')V, by the principle of conservation of momentum. Hence V = mv/(m + m')(1) The kinetic energy of the system, immediately after the collision, is E_k = (1/2) (m + n')V^2. The pendulum now swings to the right and upward until all of its kinetic energy is converted to gravitational potential energy. (Small frictional effects can be neglected.) Hence (1/2) (m +m') V^2 = (m + m')gy' V^2 = 2gy By (1) (m^2v^2)/(m+m')^2 = 2gy v^2 = [2gy(m + m')^2] / [m^2] v = [(m + m')/m] \surd(2gy). By measuring m, m', and y, the original velocity v of the bullet can be computed. It is important to remember that kinetic energy is not conserved in the collision. The ratio of the kinetic energy of bullet and pendulum, after the collision, to the original kinetic energy of the bullet, is [(1/2) (m + m')V^2] / [(1/2) mv^2] = m/(m + m'). Thus if m' = 1000 gm and m = 1 gm, only about one- tenth of one percent of the original energy remains as kinetic energy; 99.9% is converted to heat.

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Question:

Define buffer and interrupt. What is the distinctionbetween thetwo?

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G05-0106.htm

Solution:

Inboth inputand output of data or physical records, the memory hasreserved buffers. These are areas of core memory outside the main processor'smemory used to compensate for the discrepancy between the rela-tivelyquick speed of the processor and the slower electromechanical I/O devices such as printers or card readers. An economical use of buffers is through logical switching. Say we have twofixed-length buffers, one each for input and output. We want to output somerecords that are presently sitting in an input buffer by using an outputbuffer. If the status of the output buffer is idle, and if it has adequate length, a simple change of machine addresses, giving the output buffer theinput buffer's address, will send the input into the output stream. There is no physical movement of data, across memory space, but only an addressswitch, hence the name logical switch-ing. This concept is used in multiprogrammingsystems. An interrupt is an internal control signal that diverts the machine's attention awayfrom one task and causes it to perform another. These interrupts are associatedwith various priorities within the CPU. If the interrupt facility is disabled, the interrupt request will be ig-nored. If the output of an important programis expect-ed, the other programs in the system will be temporarily halteduntil the important one is completed. It is through the interrupt that thistask of halting is per-formed . Buffers and interrupts are both features of time-sharing and multiprogrammingsystems. The reason for their implementation is so that nopart of the CPU or I/O accessories will be left idle.

Question:

A chemist discovered a new compound and wished to determine its molecular weight. He placed 2.00 grams of the unknown compound in 10 grams of water and found that the normal freezing point of the water was depressed to-1.22\textdegreeC. What is the molecular weight of the new compound?

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Solution:

To answer this question, you must know the quantitative relationship for the depression of the freezing point, ∆T_f. This relationship can be expressed as ∆T_f=k_f× m, where ∆T_fis the actual depression, k_f=molalfreezing point constant (-1.86 deg mole^-1 for water) , m =molalityof the solution.Molalityis defined as the number of moles of solute per 1 kilogram of solvent. In this problem, water is the solvent and the unknown compound is the solute. You can express the freezing point relationship as ∆T_f=k_f× [(g.solute/mol.wt.) / (g. solvent)] × 1000, where the numerator is the number of moles of solute. You have all of the unknowns, except the molecular weight. Solving for the molecular weight of the unknown: -1.22\textdegree = -1.86 × [(2 / mol. wt.) / (10)] × 1000 mol. wt. = (305 g /mole).

Question:

It is desired to move a small, 50 kg, space vehicle by a lamp which emits 100 watts of blue light (\lambda = 4700 A) . If the vehicle is in free space what will be its acceleration?

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Solution:

According to quantum theory, light may be viewed as having wave-like characteristics as well as particle characteristics. In this problem, it is the particle nature of light which determines the be-havior of the space vehicle. Since the emitted light carries away momentum and because the latter is conserved, the rocket will be propelled to the left in the figure. We will suppose that N photons/sec are emitted, with each photon energy hf. Then the power of the lamp is p = [{(energy of one photon) (# of photons)}/time] p = Nhf By definition, f = (c/\lambda), Hence p = (Nhc/\lambda) Thus N = (p\lambda/hc) = [(100 watt × 4.70 × 10^-7 m)/(6.63 × 10^-34 j-sec × 3 × 10^8m/sec)] = 2.4 × 10^20 Each photon will have momentum, by the de Broglie relations p = (h/\lambda) = [(6.63 × 10^-34 j.s) / (4700 \AA)] = [(6.63 × 10^-34 j.s) / (4.7 × 10^-7 m)] = 1.4 × 10^-27 N-sec The total force on the vehicle will be, by Newton's second law, F = [d(np)/dt] Where n is the number of photons emitted and p is the momentum of each photon. Since p is constant, we then have, photon. Since p is constant, we then have, F = p(dn/dt) = pN for N is just the number of photons emitted per unit time . F = 1.4 × 10^-27 N-sec 2.4 × 10^20 sec^-1 = 3.4 × 10^-7N Thus the acceleration has the very small value a = (F/m) = [(3.4 × 10^-7 N)/50 Kg] = 6.8 × 10^-9 m/sec^2

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Question:

Calculate the mole fractions of ethyl alcohol, C_2H_5OH, and water in a solution made by dissolving 9.2 g of alcohol in 18 g of H_2O. M.W. of H_20 = 18, M.W. of C_2H_5OH = 46.

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Solution:

Mole fraction problems are similar to % composition problems. A mole fraction of a compound tells us what fraction of 1 mole of solution is due to that particular compound. Hence, mole fraction of solute = moles of solute / (moles of solute + moles of solvent) The solute is the substance being dissolved into or added to the solution. The solvent is the solution to which the solute is added. The equation for finding mole fractions is: molesA/ (molesA+ moles B) = mole fraction A Moles are defined as grams/molecular weight (MW) . Therefore, first find the number of moles of each compound present and then use the above equation. moles of C_2H_5OH= 9.2 g / (46.0 g/mole)= .2 mole moles of H_2O= 18 g / (18 g/mole)= 1 mole mole fraction of C_2H_5OH = .2 / (1 + .2)= .167 mole fraction of H_2O = 1 / (1 + .2) = .833. Note, that the sum of the mole fractions is equal to 1.

Question:

What would be the final partial pressure of oxygen in the following experiment? A collapsed polyethylene bag of 30 liters capacity is partially blown up by the addition of 10 liters of nitrogen gas measured at 0.965atmand 298\textdegreeK. Subsequently, enough oxygen is pumped into the bag so that, at 298\textdegreeK and external pressure of 0.990atm, the bag contains a full 30 liters, (assume ideal behavior.)

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/Users/wenhuchen/Documents/Crawler/Chemistry/E03-0086.htm

Solution:

One should first find the number of moles of gas that fill a volume of 30 liters at .990 atm. One then solves for the number of moles of N_2 that was pumped in and the number of moles of O_2 that are needed to fill the bag. The partial pressures of these gases are equal to the mole fraction of the gas times the total pressure. To solve for the partial pressure of O_2 one should find: 1) the total number of moles of gas that will fill 30 liters at .990 atm. 2) the number of moles of N_2 present 3) the partial pressure of N_2 4) the partial pressure of O_2. Solving: 1) One uses the Ideal Gas Law to solve for the number of moles of gas needed to fill 30 liters at 298\textdegreeK and .990 atm. The Ideal Gas Law can be stated as PV =nRTorn = PV / RT where n is the number of moles, P is the pressure, V is the volume, R is the gas constant (.082 liter-atm/mole-\textdegreeK) and T is the absolute temperature. Solving for n in this system P = .990atm V = 30 liters R = .082 liter-atm/mole-\textdegreeK T = 298\textdegreeK n = (.990atm× 30 liters) / (.082liier-atm/mole- 0K × 2980K) = 1.22 moles 2) One also uses the Ideal Gas Law to find the number of moles of N_2 already present. P = .965atm V = 10 liters. R = .082 liter-atm/mole-\textdegreeK T = 298\textdegreeK n = (.965atm× 10 liters) / (.082 liter-atm/mole-0K × 2980K) n = 0.395 moles. 3) The partial pressure is related to the total pressure by the following P_X = P_OX where P_X is the partial pressure, P_O is the total original pressure and X is the mole fraction of the gas. The mole fraction is defined as the number of moles of the gas present divided by the total number of moles in the system. Solving for the partial pressure of N_2, P_N2 , P_N + P_O XP_(N)2 = ? P_O= .990atm X = .395 / 1.22 P_(N)2 = .990atm× .395 / 1.22 P_(N)2 = .321 atm. 4)The total pressure of the system is equal to the sum of the partial pressures. Thus, .990atmis the sum of P_(N)2and P_(O)2. One has already found P_(N)2, therefore, one can now solve for P_(O)2 P_(O)2=P_total-P_(N)2P_total= .990atm P_(N)2= .321atm P_(O)2= .990atm- .321atm P_(O)2= .669 atm.

Question:

Where would an object have to be located in front of a concave mirror in order to have a virtual image formed? Precisely where would it be located if the radius of the mirror were 20 cm and the image were 20 cm behind (i.e., to the right of) the mirror?

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Solution:

Formula Method A virtual image is one which is uninverted and cannot be focused on a screen, and in the case of a concave mirror, occurs to the right of the mirror. Referring to the figure, we reason back to the inter-section of two light rays "emitted" from the tip of the image. The ray reflected back through the center of curvature must have originated there, since the angle of incidence is equal to the angle of reflection, and the angle of incidence is zero. The ray reflected through the focal point must have originated as a ray parallel to the axis. The intersection of these two rays is between the mirror and the focal point (see figure). To find the exact position from the values specified in the problem, we use the mathematical relationship: (1/0) + (1/i) = (2/R) Substituting, we have (1/0) - (1/20 cm) = (2/20 cm) (1/0) = (3/20 cm) 0 = 6(2/3) cm Note that i is negative because the image is located behind the mirror. Ray-Diagram Method To form a virtual image with a concave mirror, the object must lie between the focal point and the mirror. For an image to be 20 cm to the right of the mirror, the ray approaching the head end of the image from the focal point must have been reflected through the focal point, whereupon it must have been parallel to the axis before reflection. The ray approaching the head end of the image from the center of curvature must have experienced no change in direction. Consequently, by tracing these two rays back to their origin, the head end of the object is located as in the diagram, at approximately 6 2/3 cm in front of the concave mirror. The diagram checks the calculations.

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Question:

If a ring is cut through the bark all the way around a tree, down to the wood, the tree will live for a while, then die. Explain why.

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Solution:

When a ring is cut through the bark down to the wood of a tree, the vascular system is being separated into two halves. Water moves continuously upward in the xylem in response to osmotic pressure build-up in the roots, transpiration and capillary action. Severing the xylem will prevent the upper region of the tree from obtaining water from the roots by the processes mentioned. The upper half of the tree, deprived of further supply of water could however continue to carry on photosynthesis using as much of the remaining water and minerals as it has stored in the upper half of the bisected xylem. It will inevitably suffer rapid water loss as a result of evaporation at the surface of the leaves. As a consequence of the water loss at the leaf surface and the failure to receive water from the roots, the cells in the upper half of the tree gradually lose theirturgor. This reduced turgidity causes the guard cells to collapse, closing the stomata in an attempt to prevent further water loss through the stomata. What this indicates is that even though photosynthesis could theoretically be maintained in the upper region temporarily on stored nutrients and water, this usually does not occur. Photosynthesis is suspended as the stomata close to avoid desiccation. Since the phloem is also discontinued at the cut, organic substances traveling down the phloem ooze out of the cut, and cannot be delivered to the lower portion of the tree. The lower half of the stem and the root system can nevertheless sustain a low level of respiration for a short while, using the food reserve, such as starch, stored in the cells. They will reduce their energy expenditure by reducing cell division and enlarge-ment, which are energy requiring processes. Thus, after the cut, the tree will strive to survive with minimal growth for a while. But once the water supply is exhausted in the upper half of the tree, and the food reserve is used up in the lower half, both halves cannot generate any more ATP to carry out the vital biological activities, and the tree will eventually die.

Question:

The tentacles of hydra are armed with stinging cells. What is the structure of these cells, how are the stingers fired and of what value are they to the animal?

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Solution:

The stinging cells of hydra are called cnidocytes and are located throughout the epidermis. These cells contain stinging structures called nematocysts (see Figure). A cnidocyte is a rounded or ovoid cell with a short, stiff, bristle-like process, called a cnidocil, at one end. The cnidocil is exposed to the surface. The interior of the cell is filled by a capsule containing the nematocyst, which is a coiled tube, and the end of the capsule that is directed toward the outside is covered by a cap or lid. Supporting rods run the length of the cnidocyte. The nematocysts are discharged from the cnidocyte and are used for anchorage, for defense, and for capture of prey. The discharge mechanism apparently involves a change in the permeability of the capsule wall. Under the combined influence of mechanical and chemical stimuli, which are initially received and conducted by the cnidocil, the lid of the nematocyst opens. Water pressure within the capsule everts the tube, and the entire nematocyst explodes to the outside. A discharged nematocyst consists of a bulb representing the old capsule, and a thread- like tube of varying length, which may be spiked. From a functional standpoint, nematocysts can be divided into three major types. The first, called a volvent, is used to entangle prey. When discharged, the volvents wrap around the prey animal. The second type is called a penetrant. The tube of a penetrant is open at the tip and frequently armed with barbs and spines. At discharge, the tube penetrates into the tissues of the prey and injects a protein toxin that has a paralyzing action. The nematocysts of hydra do not have this effect on man; but the larger marine Cnidaria can produce a very severe burning sensation and irritation. The third type of nematocyst is a glutinant, in which the tube is open and sticky and is used in anchoring the animal under certain conditions. The cnidocyte degenerates following discharge of its nematocyst, and new cnidocytes are produced from inter-stitial cells.

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Question:

Assuming the ionization potential of sodium is 5.1eVand the electron affinity of chlorine is 3.6eV, calculate the amount of energy required to transfer one electron from an isolated sodium (Na) atom to an isolated chlorine (Cl) atom.

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Solution:

Ionization potential is the amount of energy required to pull an electron off an isolated atom. Electron affinity is the amount of energy released when an electron is added to an isolated neutral atom. In this problem , one must add energy to remove an electron from Na and energy will be released upon the addition of an electron to Cl. 5.1eVare needed to expel an electron from Na and 3.5eVare released whenClaccepts an electron . Thus, the amount of energy required for the overall process to occur is the difference between the ionization potential of Na and the electron affinity of Cl. The energy necessary to be added to the system for this reaction to occur is 5.1eV-s 3.6eVor 1.5eV.

Question:

In Drosophilia, black body and vestigial wings are both recessive traits, while gray body and normal wings are dominant. How would you determine if the genes for black body and vestigial wings are on the same chromosome? Show the expected results if they are on the same chromosome, and the results if they are on different chromosomes.

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Solution:

To determine whether or not the genes for body color and wing type are on the same chromosome, we can do a testcross between an individual heterozygous for both traits and a double recessive individual. A testcross with a double recessive allows the genotype of the other parent to be expressed in the offspring in the same proportions as the gametes produced by that parent. This is because the double recessive parent has no genes that will obscure the genotype of the other parent. We use a heterozygote to determine if linkage is present because such a parent carries all the genes involved. If the offspring of the testcross give a 1:1:1:1 phenotypic ratio, then the genes have assorted independently and are not linked. If they give some other ratio, then linkage is indicated, and they are probably located on the same chromosome. This can best be illustrated by doing the crosses. Let G be the gene for gray body, g be the gene for black body, N be the gene for normal wings, and n be the gene for vestigial wings. We can obtain the heterozygote by crossing a known homozygous dominant with a homozygous recessive. Then we do the test cross. If the genes are not linked we have: F_2 GN Gn gN gn gn GgNn Ggnn ggNn ggnn Phenotypically: GgNnisgray-normal, Ggnnisgray-normal, ggNnisgray-normal, ggnnisgray-normal, According to the Law of Independent Segregation, the gametes formed from the GgNn parent (GN, Gn, gN, gn) are in equal numbers. Thus the proportion of different F_2 offspring are equal; that is, 1 : 1 : 1 : 1. If the genes are located on the same chromosome, the heterozygous parent will have the genotype. (Note: the genes are written on a barGNto show linkage) Since the genes are linked, there is a chance that crossover will occur between them during meiosis. The heterozygous parent can thus form four kind of gametes in the following manner: Note that since their presence in the gametic population depends on the occurence of a crossover event, the frequency of the recombinants (those arising from cross-over) will be lower than that of the parental types (those having the same gene combination as the parent). In the test cross we obtain: F_2 GN Gn gN gn gn GN gn Gn gn gN gn gn gn Phenotypically: Although the same phenotypes are produced in this cross as in the cross between genes on separate chromo-somes, the ratios will not be the same. Because their frequency as gametes is much higher than the recombinant-types, the parental types (gray-normal and black-vestigial) will occur with much higher frequency in the offspring. The frequency of the recombinant types will depend on the distance between the two genes. If the genes are far apart, there is more chance of a crossover event between them, and the occurence of recombinants will be higher than if the genes were very close together on the chromo-some. Also, since the formation of one recombinant type necessitates that the other type also be formed (refer to the diagram of the crossover), their frequencies will be about equal, as will the frequencies of the two parental types. Thus whenever the observed ratio in a test cross varies from the predicted Mendelian ratio, and resembles the ratios outlined above, linkage is usually indicated.

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Question:

Write a FORTRAN program to compute the arithmetic mean of N real numbers. You may assume that the numbers have been read into an array X having members X(1), X(2),..., X(N).

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Solution:

We apply the formula X= (^N\sum_i_=1 X(i)) / N = (X(1) + X(2)+...+X(N)) / N . The variable SUM is used to accumulate the value of the sum of the X(i)'s. The variable AVG (the mean value) is set equal to SUM/N. DIMENSION X(N) SUM = 0.0 READ (5,20) N 2\OFORMAT (2X,I5) DO 50 I = 1,N 5\OSUM = SUM + X(I) AVG = SUM/FLOAT (N) STOP END

Question:

(a) A moving particle makes a perfectly elastic collision with a second particle, initially at rest, along their line of centers. Find the ratio of the masses which makes the kinetic energy transferred to the second particle a maximum. (b) If the ratio of the masses is not that calculated above, show that the amount of energy transferred can be increased by inserting a third particle between the first two. For optimal transfer, the mass of the third particle is the geometric mean of the other two.

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Solution:

(a) Let the energy of the incoming particle be E, and refer to figure (1) for the system of notation. Since the collision is perfectly elastic, both energy and momentum are conserved. Therefore E = EA+ EBandm_Au = m_AvA+ m_Bv_B. ButE = (1/2) m_Au^2 = (m ^2Au^2) / (2m_A)orm_Au = \surd(2m_AE), and similarly for the other kinetic energies. The second equation is therefore \surd(2m_AE) = \surd(2m_AE_A) + \surd(2m_BE_B)or \surdE = \surd(E_A) + \surd(xE_B), Wherex = m_B/m_A. Then E = [\surd(E_A )+ \surd(xE_B)]^2 = E_A + xE_B + 2\surd(xE_AE_B) \thereforeEA+ E_B = E = E_B + xE_B + 2\surd(xE_AE_B) Transposing, E_A + EB- E_A - xEB= 2\surd(xE_AE_B) (1 - x) EB= 2\surd(xE_AE_B) Squaring both sides and using E_A = E - E_B, (1 - x)^2E_B = 4xE_A = 4x(E - E_B). \thereforeE_B/(E - E_B) = 4x /[(1 - x)^2] Inverting, (E - E_B)/E_B = (E/E_B)- 1 = [(1 - x)^2]/4x E/EB= [{(1 - x)^2}/ 4x] + 1 = [{(1 - x)^2}/ 4x] + (4x/4x) = [{(1 - x)^2 + 4x}/ 4x] Inverting once more so as to get the needed ratio of E_B to E, E_B/ E = 4x/[(1 - x)^2 + 4x] = 4x/[1 + x^2 - 2x + 4x] = 4x/[1 + 2x +x^2] = 4x/(1 + x)^2 For maximum energy transfer, E_B should be as large a proportion of E as For maximum energy transfer, E_B should be as large a proportion of E as possible and (d/dx) (E_B/E) should be zero. Thus, for maximum energy transfer, (d/dx)(E_B/ E) = [4/(1 + x)^2] - [8x/(1 + x)^3 = 0 or4(1 + x) = 8x\therefore x = 1,[i.e. (m_B/m_A) = 1] Thus, the two masses should be equal when all the energy is transferred to the second particle. (b) If x has a fixed value not equal to 1, insert a further mass m_c between m_A and m_B. Then in the first collision, we have from part (a), E_C/E = 4y/(1 + y)2, where y = m_C /m_A [see Fig. (2)]. Now m_c collides with m_B and E_B/E_C = 4z/(1 + z)2, Where, z = m_B/m_C. but yz = x, and therefore E_B/E_C = [4(x/y)]/[1 + (x/y)]^2. E_B/E = (E_B/E_C) \bullet (E_C/E) = 16x/[(1 + y)^2{1 + (x/y)}^2]. (d/dy)(E_B/E) = - [32x/{(1 + y)^3[1 + (x/y)]^2}] + [(32x^2/y^2)/{(1 + y)^2[1 + (x/y)]}] For maximum energy transfer to the final mass, this quantity must be zero, and so 32x[1 + (x/y)] = 32x^2/ y2(1 + y) Multiplying by y_2 and cancelling the term 32x, y^2 + xy = x +xyory^2 = x. \thereforem ^2C/ m ^2_A= m_B/mAorm_C = \surd(m_Am_B). For maximum energy transfer the intermediate particle must have a mass which is the geometrical mean of the other two. But the first term in the final inequality is the maximum transfer of energy when three particles are involved, using the necessary relation y^2 = x. The second term is the energy transfer when only two particles are involved. Therefore, not only is maximum energy transferred in the three- particle case when m_C = \surd(mAm_B), but the energy acquired by the particle of mass m_B is greater than it is when only two particles are involved.

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Question:

You have decided to build up your body, so you have a dinner that is heavy in protein content, followed by a quart of milk. Trace the movement of the food from the mouth to the duodenum. What happens to the food in the stomach and small intestine?

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Solution:

When food is introduced into the mouth, it is mixed with saliva and masticated by the teeth into small pieces to facilitate swallowing and digestion (by increasing the surface area exposed to digestive enzymes). With the help of the tongue, the chewed food is rolled into a ball, or bolus, which is then thrust into the back of the mouth for swallowing. As soon as the food enters the esophagus, it is moved down the tract by rhythmic muscular contractions known as peristaltic waves. Peristalsis is the alternate contraction and relaxation of the smooth muscles lining the digestive tract, and helps to move the bolus along. Movement of the bolus is also assisted in this region by salivarymucin. As the food nears the stomach, the sphincter muscle controlling the opening of the stomach relaxes and allows the food to enter the stomach. In the stomach, it is acted upon by the gastric juice, which works optimally in an acidic medium. Gastric juice contains: (i) hydrochloric acid, which acidifies the medium, in order for the proteolyticenzymes to work optimally; (ii) pepsin, aproteolyticenzyme that cleaves long protein molecules into shorter fragments called peptides; and (iii) rennin, an enzyme that solidifies casein, a milk protein, so that it can be retained in the stomach long enough for pepsin to act upon it. After about four hours in the stomach, the semi--digested food, or chyme, is released into the duodenum through the pyloric sphincter. The duodenum receives pancreatic juice from the pancreas and bile from the gallbladder. The proteasestrypsinandchymotrypsin, found in the pancreatic juice, further degrade peptides into smaller fragments. The digestion of protein is then completed in the ileum, the last part of the small intestine. Here, the remaining peptide fragments are further digested to free amino acids bycarboxypeptidasesandarainopeptidases, which split off amino acids from the carboxyl and amino ends of the peptide chains, respectively. The free amino acids are then actively transported across the intestinal wall into the bloodstream.

Question:

In attempting to map out the climatic conditions of the biosphere, would a climatologist find it helpful to correspond withan ecologist who has mapped the vegetation formations orbiomes of the biosphere?

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Solution:

If you traveled on the earth several hundreds of miles at a time, you wouldfrequently encounter vegetation types that were very different from oneanother. You would also get the impression of a world set out into differentblocks of vegetation. These blocks are termed formations. The successionof formations between the equator and the arctic is roughly mirroredin both the Northern and Southern Hemispheres. But why should the vegetations of different parts of the world look sodifferent? Why should enormous stretches of land have vegetation of a specialform? Also, why should the plants of different formations respect each other'sterritory? Some aspects of the climate seem to be the controllingfactors, with the most important two being temperature and the availabilityof water. Mapping a thing such as climate is a very hard task, but mapping vegetationtypes is easy by comparison. So climatologists could use vegetationmaps as a base for climatic maps. The earth is parceled out intoblocks of different types of plants in a corresponding manner to the waydifferent climatic conditions are distributed. The distinctive appearanceof each formation results from the adaptations of local plants tocommon climatic and other environmental conditions. Thus, it is correct tosay that vegetation-formation mapping does provide a basis for climatic mapping. If formations merge, climatic conditions should converge. Similarly, if formations diverge, climates should differ as well. The distribution of animals is closely linked with the distribution of thevegetation in which they live and it is best to accept the formation boundariesand to include the animals in the formation descriptions. Biologists find it convenient to recognize a limited number of major formationscalled biomes. Thus the grassland of the western United States isone biome, and the nearby desert is a second biome. A grassland and a desertalso exist in South America, but the species that comprise these biomesare almost entirely different from their counterparts in the United States.We speak of these biomes that resemble one another in physical appearancebut differ in species composition as comprising a worldwide biome-type.

Question:

The dissociation sequence of thepolyproticacid H_3PO_4 showsthreeBronsted-Lowry acids. Rank them in order of decreasingstrengths.

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Solution:

Polyproticacids are ones which furnish more than one proton per molecule. From its molecular formula, H_3PO_4, it is observed that there are 3 available hydrogen atoms available for release.In general, the equation fora dissociation reaction is HA + H_2O \rightarrow H_3O^+ + A^-, whereHA is the acid and water is acting as a weak base. With this in mind, one can write the 3 dissociation reactions as (1)H_3PO_4 + H_2O\rightarrowH_3O^+ + H_2PO_4^- (2)H_2PO_4^- + H_2O\rightarrowH_3O^+ + HPO_4^-2 (3)HPO_4^-2 + H_2O\rightarrowH_3O^+ + PO_4^-3 From this, one can see that the three acids areH_3PO_4 ,H_2PO_4^- , andHPO_4^-2 . The acids decrease in strength in the order ofH_3PO_4 , H_2PO_4^- ,HPO_4^-2 . Ranking can be explained by noting that equivalent H-O bonds are beingbroken to give off H^+. The second and third protons that dissociate leavea progressively more negative ion. This means it is more difficult for theion to dissociate in order to produce additional H^+ ions. This stems fromthe fact that the increased negativity results in increased attraction forthe proton (H^+). In summary, as the negative charge of the acid increases, the weaker the acid becomes.

Question:

What is meant by arteriosclerosis and what is the cause of thiscondition?

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Solution:

Arteriosclerosis, commonly known as \textquotedbllefthardening of the arteries,\textquotedblright is adisease characterized by a thickening of the arterial wall with connective tissueand deposits of cholesterol. Though it is not really clear how this thickeningoccurs, it is known that smoking, obesity, high-fat diet, and nervoustension predispose one to this disease. The suspected relationshipbetween arteriosclerosis and blood concentrations of cholesteroland saturated fatty acids has received widespread attention. Many studies are presently trying to evaluate the hypothesis that high bloodconcentration of these lipids increase the rate and the severity of the arterioscleroticprocess. Physiologically, cholesterol is an important substancebecause it is the precursor of certain hormones and the bile acids. The major dietary source of cholesterol is animal fats. However, the liveris capable of producing large amounts of cholesterol, particularly from saturatedfatty acids; even if the intake of cholesterol is markedly reduced, theblood cholesterol level will not be lowered considerably since the liver respondsby producing more. It may be the high content of saturated fatty acids, rather than cholesterol, which causes the ingestion of animal fat to makeone more susceptible to arteriosclerosis. Arteriosclerosis in the coronary artery increases the risk of heart attackand death. Coronary artery arteriosclerosis is estimated to cause 500,000 deaths per year.

Question:

Explain the physical changes which take place during inspiration.

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Solution:

Just prior to inspiration, at the conclusion of the previous expiration, the respiratory muscles are relaxed and no air is flowing into or out of the lungs. Inspiration is initiated by the contraction of the dome-shaped diaphragm and the intercostal muscles. When the diaphragm contracts, it moves downward into the abdomen. Simultaneously, the intercostals muscles which insert on the ribs contract, leading to an upward and outward move-ment of the ribs. As a result of these two physical changes, the volume of the chest cavity increases and hence the pressure within the chest decreases. Then, the atmospheric pressure, which is now greater than the intrathoracic pressure, forces air to enter the lungs, and causes them to inflate or expand. During exhalation, the intercostals muscles relax and the ribs move downward and inward. At the same time, the diaphragm relaxes and resumes its original dome shape. Consequently, the thoracic volume returns to its pre-inhalation state, and the pressure within the chest increases. This increase in pressure, together with the elastic recoil of the lungs, forces air out of the lungs causing them to deflate. The role of the diaphragm in breathing can be demonstrated by the figure below.

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Question:

In the 1930s F.T. Bacon designed hydrogen fuel cells with extremely good performance. Using his concepts Pratt and Whitney has constructed hydrogen fuel cells that produce a potential of 0.9 V and generate a current of about 2.3 × 10^3 A for each square meter of anode area. If a Bacon battery is to deliver 100 W of power, how large must the anode be?

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Solution:

Recall that the power dissipated (or pro-duced) by a device is, P = IV Where I is the current going through a device, and V is the voltage across its terminals. In this example P = 1 × 10^2 W and V = 0.9V. Therefore, the current, required to deliver 100W of energy is, I = (P/V) = {(1 × 10^2 W)/(9 × 10^\rule{1em}{1pt}1 V) = 1.1 × 10^2 A The surface area A required is, A = {(1.1 × 10^2A)/(2.3 × 10^3A/m^2)} = 4.8 × 10^\rule{1em}{1pt}2 m^2 If the anode is designed to have a square shape, then the area A is equal to d^2 , where d is the length of one edge. We find that, d^2 = 4.8 × 10^\rule{1em}{1pt}2 m^2 d = 2.2 × 10^\rule{1em}{1pt}1 m A very compact energy supply may be constructed using a hydrogen fuel cell.

Question:

Differentiate between adaptive radiation and convergent evolution . Diagram illustrating the difference between divergent and convergent evolution. In adaptive radiation, a single stock may branch to give many diverging stocks (in the given diagram,1 stockdiverges \rightarrow 2 stocks). In convergent evolution, many stocks that are originally quite different can come to resemble each other more and more; as time passes, they converge. In the given diagram, 2 stocks are shown converging.

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Solution:

Because of the constant competition for food and living space, a group of organisms will tend to spread out and occupy as many different habitats as pos-sible. This process, by which a single ancestral species evolves to a variety of forms that occupy somewhat different habitats, is termed adaptive radiation. Adaptive radiation is clearly an advantageous process in evolution in that it enables the organisms to tap new sources of food or to escape from predators. A classical illustration of adaptive radiation is the great variety of finches found on the Galapagos Islands west of Ecuador. These finches, derived from a single common ancestor, exhibit diversity in beak size and structures, as well as in feeding habits, all of which are mutually related. Some of these birds feed on seeds, others feed mainly on cacti, and still others live in trees and eat insects. The diversity of food sources on the island has allowed each of the many forms of finches to survive in its particular habitat, and thus, prevent intra specific competition for food and space. Such adaptive radiation is often called divergent evolution, since its result is a diversity of adaptive forms evolving from a common ancestor. An opposite process is convergent evolution, which also occurs quite frequently. Convergent evolution is the process by which two or more unrelated groups become adapted to a similar environment, and in doing so, develop characteristics that are more or less similar. For example, wings have evolved hot only in birds but also in mammals (bats), reptiles (pterosaurs), and insects (grasshoppers). Likewise, the shark (a cartilaginous fish), the dolphin, and the porpoise (both mammals) have developed marked superficial similarity because of their adaptation to a similar environment.

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Question:

It is observed that the chemical transmission occurring at a synapse is unidirectional that is, the neurotransmitter always travels from thepresynapticaxon to the post-synaptic dendrite . Explain why.

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Solution:

Because there are physical separations (synaptic clefts) between neurons , an impulse traveling down the axon of one neuron must first be converted from an electrical form into a chemical form at the synapse, which can diffuse across the physical separation. This chemical form of transmission utilizes a chemical substance secreted by thepresynaptic axon terminal, known as a neurotransmitter. This neurotransmitter has the ability of changing the permeability properties of the post-synaptic cell membrane , causing it to depolarize and an action potential to be generated . The action potential then propagates down the axon of the postsynaptic neuron, forming a wave of depolarization. This new impulse, when it reaches the axon terminal, again can induce a neuro-transmitter to be released at the new synapse, and sub-sequently another neuron can be excited . Hence, by this mechanism, there can be continuous conveyance of neural passages by neurons, despite the presence of physical disconnections between them. The chemical transmission that occurs at a synapse is unidirectional since the neurotransmitter is present in the axonal end of a neuron only. This chemical is contained within small, sac-like structures called synaptic vesicles. The synaptic vesicles fuse with the membrane of the axon, induced by an electric impulse that reaches the axon terminal. They then break open, discharging the contents into the synaptic cleft. Because of the fact that no synaptic vesicle is present inside the dendrites of a neuron, chemical transmission is always from the axonal end to the dendritic terminal, and not the reverse.

Question:

Knowing that the K_sp for AgCl is 1.8 × 10-^10 , calculate E, the electrode potential, for a silver-silver chloride electrode im-mersed in 1M KCI. The standard oxidation potential for the (Ag, Ag^+) half reaction is -0.799 volts.

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Solution:

The silver-silver chloride electrode process is a special case of the (Ag, Ag^+) couple, except that silver ions collect as solid AgCI on the electrode itself. However, solid AgCl has some Ag^+ in equilibrium with it in solution. This [Ag^+] can be calculated from the K_sp equation; [Ag^+ ] [Cl^- ] = K_spor[Ag+ ] = [K_sp ] / [Cl^- ] [Ag_+] = (1.8 × 10^-10) / (1) = (1.8 × 10^-10) . This value for [Ag^+ ] can be inserted into the Nernst equation for the (Ag, Ag^+) half-reaction , Ag \rightarrow Ag^+ + e-E\textdegree = -0.799 The Nernst equation is E = E\textdegree - [(0.05916 log Q) / n] , where E\textdegree = standard electrode potential (25\textdegreeC and 1M), n = number of electrons transferred, and Q = the mass-action expression or equilibrium constant. For this reaction, one sees that one electron is transferred. Hence, n = 1. Also, Q =[Ag] / [Ag^+], but, since Ag is solid, its concentration is assumed to be constant; it can be removed from the expression, so that Q = [Ag+] . Thus, E = E\textdegree - {0.05916 log [Ag^+] / (1)} = -0.799 - 0.05916 log(1.8 × 10^-10) = -0.799 + 0.576 = 0.223 volts .

Question:

By how much will 50 grams of water have its freezing point depressed if you add 30 grams (molecular weight 80) of glucose to it?

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Solution:

The addition of any substance to water will alter its boiling or freezing point. To determine the amount of change, you must know the concentration of solute (the substance dissolved) in the solvent (water). This information is required because the freezing point depression, ∆T_f, equals themolalfreezing point constant, k_f, times themolality(m). ∆T_f=k_f(m). The concept ofmolalityrefers to the number of moles of solute per 1 kilogram of solvent. The solute, glucose, weighs 30 grams. Therefore, the number of moles of glucose is (30 g / 180) (g / mole) = (1 / 6moles). You have 50 grams of water. However, molality refers to moles per thousand grams. As such, a conversion is required; namely (50 / 1000). Hence, themolalityis [(1 / 6)(moles)] / [(50) / (1000)g] = 3.33m Recall that the amount of depression of the freezing point is defined as ∆T_f= (k_fm).k_fis given for water as - 1.86 deg mole^-1. You calculated m. The temperature de-pression is thus ∆T_f=k_f× m = - 1.86\textdegree × 3.33 m = - 6.2\textdegree.

Question:

What is the inductive reactance of a coil of 20millihenries inductance when it is connected to a 60 cycle line?

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/Users/wenhuchen/Documents/Crawler/Physics/D22-0737.htm

Solution:

Here inductance L = 20millihenries= 0.020henryand f = 60 cps. Then the inductive reactance of the coil = X_L = 2\pifL = 2\pi × 60 cps × 0.020 h = 7.5 ohms

Question:

A chemistry student wrote on an exam that "Complete oxidation of a mole of glucose releases more energy than oxidation of a mole of lactose." Why did this student not receive full credit for his answer?

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Solution:

Oxidation is the addition of oxygen to organic compounds to produce water and carbon dioxide plus heat. The more carbon present in the molecule, the more energy released. Also, more energy will be released if the percentage of oxygen in the compound is low. Consider the structures of glucose and lactose, which are carbohydrates or simple sugars. Hydrolysis is the reaction of a compound with dilute acid or base and water to break down into its simple components. Notice that if you hydrolyze lactose by adding dilute acid, you can cleave the bond to form a glucose molecule and a galactose molecule. Now, if the lactose molecule already contains a glucose molecule plus galactose, it would yield more energy upon oxidation.Thus, this student's statement is false.

Question:

Realize a circuit which acts as a full adder and a full sub tractor.

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Solution:

The truth table of a full adder is shown in Figure 1. A and B are the bits to be added, C_i-1 is the carry bit from a previous addition. A B C_i-1 SUM CARRY 0 0 0 0 0 0 0 1 1 0 0 1 0 1 0 0 1 1 0 1 1 0 0 1 0 1 0 1 0 1 1 1 0 0 1 1 1 1 1 1 Fig. 1 The truth table shown in Figure 2 is that of a full sub-tractor. B is the bit subtracted from A, C is the borrow bit from a previous subtraction. A B C Diff Borrow 0 0 0 0 0 0 0 1 1 1 0 1 0 1 1 0 1 1 0 1 1 0 0 1 0 1 0 1 0 0 1 1 0 0 0 1 1 1 1 1 Fig. 2 The Karnaugh Maps for sum, carry, borrow and difference are shown in Figure 3. The Sum and Carry functions of a Full Adder are given by: SUM = AB 'C' + A' BC' + ABC + A'B'C = (AB' +A'B) C' + (AB+A'B') C SUM = (A+B) C' + (A + B) 'C = A + B + C CARRY = A \bullet B + A \bullet C + B \bullet C = A \bullet[B+C] + B \bullet C The Difference and Borrow functions of a full subtractor are given by: Difference = A + B + C (same as sum) Borrow = A' \bullet B + A' \bullet C + B \bullet C = A' \bullet [B+C] + BC It can be easily seen that Sum and Difference functions are the same and that the Carry and Borrow functions re-quire little manipulation. The combinational circuit is shown in Figure 4. An alternate manipulation of Borrow and Carry gives further reduction in the above circuitry. First let us redraw the Karnaugh Maps of the Borrow and Carry functions, as shown in Figure 5. Carry = BC + AB'C + ABC'Borrow = BC + A'B'C + A'BC' = BC + A [B+C]= BC + A' [B+C] The combinational circuit implementing the simplified expressions is shown in Figure 6.

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Question:

A particle is projected horizontally with a velocity of 10^4 m \bullet s^-1 in such a direction that it moves at right angles to a horizontal magnetic field of induction, of magnitude 4.9 × 10^-5Wb\textbullet m^-2 . The particle, which carries a single electronic charge, stays in the same horizontal plane. What is its mass?

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Solution:

The upward acting magnetic force is F^\ding{217} = q(v^\ding{217} × B^\ding{217}), where q is the charge of the particle, v^\ding{217} its velocity, and B^\ding{217} the magnetic induction. Since the motion is at right angles to the direction of the magnetic in-duction, it follows that the magnitude of F^\ding{217}, \vertF^\ding{217}\vert = q v B sin 90\textdegree = q v B. Since the particle stays in the same horizontal plane during the motion, the magnetic force on it must, by Newton's Second Law, just balance its weight (= mg). Thus, mg =qvBor m =qvB/g . \thereforem = [(1.6 × 10^-19 C × 10^4 m\bullets^-1 × 4.9 × 10^-5 wb\bulletm^-2 )/(9.8 m \bullet s^-2 )] = 8.0 × 10^-21 kg.

Question:

A 2.0-ton elevator is supported by a cable that can safely support 6400 lb. What is the shortest distance in which the elevator can be brought to a stop when it is descending with a speed of 4.0 ft/sec?

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Solution:

The maximum force that can be used to stop the elevator without breaking the cable is 6400 lb - 4000 lb = 2400 lb upward (4000 lb. is the weight of the elevator). This maximum force gives the shortest distance in which the elevator can be stopped since it provides a maximum deceleration a_max = F_max/m. m = W/g = 4000 lb/32 ft/sec^2 = 125 slugs a = F/m = 2400 lb/125 slugs = 19.2 ft/sec^2. With v_final = 0 and taking up as positive, the minimum stopping distance can be found using the kinematics equation v2_f = 2as + v^2 s = -v^2/2a = [-(-4.0 ft/sec)^2]/2 × 19.2 ft/sec^2 = -0.42 ft.

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Question:

Homeostatic regulation of many body functions is not achieved by the central nervous system but by the autonomic nervous system. Explain how the autonomic nervous system can bring about such regulation.

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Solution:

The autonomic nervous system is divided in two parts, both structurally and functionally. One part is called the sympathetic nervous system and the other is known as the parasympathetic nervous system. These two branches act antagonistically to each other. If one system stimulates aneffector, the other would inhibit its action. The basis for homeostatic regulation by the autonomic nervous system lies in the fact that the sympathetic and parasympathetic systems each sends a branch to the same organ, causing the phenomenon of doubleinnervation. Doubleinnervationby the autonomic nervous system, together with the action of the endocrine system, is the basis for maintaining homeostasis inside the body. If through the stimulation by one system of an organ, a substance or action is produced excessively, then the other system will operate to inhibit the same organ, thus reducing the production of that substance or in-hibiting that action. This is basically how the internal condition of the body is kept constant. An example will illustrate the above. Both the sympathetic and parasympathetic systems innervate the heart. The action of the former strengthens and accel-erates the heart beat while the latter weakens and slows the same. When a person is in fright, his heart beat involuntarily quickens owing, in part, to stimula-tion by the sympathetic system. To regain the normal state, the sympathetic system is overridden by the pa-rasympathetic system which decelerates the heart beat. In the normal condition , the heart receives impulses from the two antagonizing systems, and it is through a balance of the two that the proper rate of heart beat is main-tained .

Question:

Convert (a) - (c) from Infix to Prefix and Postfix expressions and (d) - (f) from Prefix to Infix and Postfix expressions. (a) A + B \textasteriskcentered C/D (b) A - C & D + B \uparrow E (c) X \uparrow Y \uparrow Z (d) + \textasteriskcentered - ABCD (e) + mA \textasteriskcentered BmC (f) \uparrow X + \textasteriskcentered YZW

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Solution:

For conversion from Infix to Prefix and Postfix forms we need to recollect the hierarchy and associativity rule. The former says that + has highest priority, then \textasteriskcentered, / and then +, -. The latter says that at the same level of priority associativity applies from left to right. e.g. A - B - C = (A - B) - C (a) A + B \textasteriskcentered C/D Conversion to Prefix form: Scanning from left to right we observe the operators +, \textasteriskcentered and/ \textbullet As \textasteriskcentered, and/ have higher priority than + and \textasteriskcentered precedes/, BC becomes the operand fol-lowed by /D. Thus the whole second operand takes the form /\textasteriskcenteredBCD. A is the first operand and + is the operator between them. Postfix forms\textemdash we have BC\textasteriskcentered then BC\textasteriskcenteredD/ & finally ABC\textasteriskcenteredD/+ A + B\textasteriskcenteredC/D = ABC \textasteriskcentered D/+ (b) A - C \textasteriskcentered D + B \uparrow E scan left to right First stepB\uparrowE= \uparrow BE, BE \uparrow\uparrowhas highest priority Second stepC\textasteriskcenteredD= \textasteriskcentered CD, CD \textasteriskcentered\textasteriskcenteredhas 2nd highest priority Third StepA -C \textasteriskcentered D= - A \textasteriskcentered CD, ACD +--and + are on the same level Final stepA - C \textasteriskcentered D+B \uparrow E= + - A \textasteriskcentered CD \uparrow BE, ACD \textasteriskcentered - BE \uparrow + (c) X \uparrow Y \uparrow Z both \uparrow on same level InfixPrefixPostfix First stepX\uparrowY=\uparrow XY,XY \uparrow Final stepX \uparrow Y\uparrowZ=\uparrow\uparrow XYZ,XY \uparrow Z\uparrow Conversion from Prefix to Infix and Post fix expression. (d) + \textasteriskcentered - A B C D InfixPrefixPostfix 1st step-AB=(A - B),AB - 2nd step\textasteriskcentered-ABC=(A - B) \textasteriskcentered C,AB - C\textasteriskcentered Final step+ \textasteriskcentered -ABCD=(A - B)\textasteriskcenteredC+D,AB - C\textasteriskcentered+ (e) + mA \textasteriskcentered B m Cm is unary operator acting on immediate operand with - sign & parenthesis. InfixPrefixPostfix 1st stepmC=(- C),Cm 2nd step&BmC=B & - C,BCm\textasteriskcentered Final step+ mA &BmC=(-A) + B&(-C),AB - C\textasteriskcentered+ (f) \uparrow X + \textasteriskcentered YZW/ InfixPrefixPostfix 1st step\textasteriskcenteredYZ=Y \textasteriskcentered Z,YZ \textasteriskcentered 2nd step+\textasteriskcenteredYZW=Y \textasteriskcentered Z + W,YZ \textasteriskcentered W + Final step\uparrow X +\textasteriskcenteredYZW= X \uparrowY \textasteriskcentered Z + W,XYZ \textasteriskcentered W + \uparrow

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Question:

During the course of World War I, 1.1 × 10^8 kg of poison gas was fired on Allied soldiers by German troops. If the gas is assumed to be phosgene (C0C1_2), how many molecules of gas does this correspond to?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E02-0046.htm

Solution:

To solve this problem we must convert mass to number of moles and then multiply by Avogadro's number to obtain the corresponding number of molecules. The number of moles of gas is given by moles = [mass (grams)/(molecular weight of C0C1_2 (g/mole)] The mass of gas is 1.1 × 10^8 kg = 1.1 × 10^8 kg x 10^3 g/kg = 1.1 × 10^11 g. The molecular weight of COC1_2 is obtained by adding the atomic weights (atm.wgt.) of its consti-tuents . Thus, molecular weight (COC1_2) =atmwgt(C) +atmwgt(O) + 2 ×atmwgt(C1) = 12.0 g/mole + 16.0 g/mole + 2 × 35.5 g/mole = 99 g/mole. Hence, the number of moles of gas is moles = [(mass)/(molecular weight)] = [(1.1 × 10^11 g)/(99 g/mole)] \cong 1.1 × 109 moles. Multiplying the number of moles by Avogadro's number, we obtain the number of molecules of gas: number of molecules = moles × Avogadro's number = 1.1 × 10^9 moles × 6 × 10^23 molecules/mole = 6.6 × 10^32 molecules.

Question:

Apply Hund's rules to obtain the electron configuration for Si, P, S, Cl, and Ar.

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Solution:

The ground state of an atom is that in which the electrons are in the lowest possible energy level. Each level may contain two electrons of opposite spin. When there are several equivalent orbitals of the same energy, Hund's rules are used to decide how the electrons are to be distributed between the orbitals: (1)If the number of electrons is equal to or less than the number of equivalent orbitals, then the electrons are assigned to different orbitals. (2)If two electrons occupy two different orbitals, their spins will be parallel in the ground state. Hund's rules states that the electrons attain positions as far apart as possible which minimizes the repulsion obtained from interelectronic forces. To solve this problem one must: (1)Find the total number of electrons within the atom (2)Determine the number of valence electrons (3)Find the number of electrons in the highest equivalent energy orbital. The total number of electrons in an atom is equal to that atom's atomic number. Thus,Sihas14electrons Phas15electrons Shas16electrons Clhas17electrons Arhas18electrons Next, from the orbital configuration: 14 Si1s_2 2s^2 2p^6 3s^2 \mid 3p^2 15 P1s^2 2s^2 2p^6 3s^2 \mid 3p^3 16 S1s^2 2s^2 2p^6 3s^2 \mid 3p^4 17 Cl1s^2 2s^2 2p^6 3s^2 \mid 3p^5 18 Ar1s^2 2s^2 2p^6 3s^2 \mid 3p^6 One knows that the highest equivalent energy orbital is the 3p orbital. The number of electrons in this orbital increases by 1 starting with 2 for Si, then, 3 for P, 4 for S, 5 for Cl, and 6 Ar, Thus, Ar closes this orbital. Using Hund's rules, the electron configurations can be written as shown in the accompanying figure.

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Question:

Explain excretion in the earthworm.

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/Users/wenhuchen/Documents/Crawler/Biology/F12-0296.htm

Solution:

The earthworm's body is composed of a series of segments internally partitioned from each other by membranes. In each segment of its body, there are a pair of specialized excretory organs, called nephridia. They open independently from the body cavity to the outside. The various nephridia are not connected to each other. A nephridium consists of an open ciliated funnel or nephrostome (corresponding to the flame cell in planaria) which opens into the next anterior coelomic cavity. A coiled tubule running from the nephrostome empties into a large bladder, which in turn empties to the outside by way of the nephridiopore (see Figure). Around the coiled tube is a network of blood capillaries. Materials from the coelomic cavity move into the nephridium through the open nephrostome partly by the beating of the cilia of the nephrostome, and partly by currents created by the contraction of muscles in the body wall. Some materials are also picked up by the coiled tubule directly from the blood capillaries. Substances such as glucose and water are reabsorbed from the tubule into the blood capillaries, while the wastes are concentrated and passed out of the body through the nephridiopore. The earthworm daily excretes a very dilute, copious urine, which amounts to 60 percent of its total body weight. The principal advantage of this type of excretory system over the flame cell is the association of blood vessels with the coiled tubule, where absorption and reabsorption of materials can occur.

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Question:

An airplane is flying at Mach 0.5 and carries a sound source that emits a 1000-Hz signal. What frequency sound does a listener hear if he is in the path of the air-plane after the airplane has passed?

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Solution:

The frequency of a wave disturbance depends on the relative motion of the source and observer. This phenomenon is called the Doppler effect and can be determined. The wavelength \lambda of a wave can be defined as \lambda = distance / no. of waves The no. of waves that a source of frequencyv_Semits in time t is justv_St. If the medium permits the waves to travel at velocity v then the distance they cover in time t due to their own motion isvt. Since the source is also moving toward the listener at velocityv_Sand covers a distancev_St, this means that the waves have a distancevt-v_Stto be spread out in. Therefore \lambda = (vt-v_St)/(v_St) = (v -v_S)/v_S The frequencyv_Lof the wave as observed by the listener as the source moves toward him is v_L= v/\lambda_L= v × [v_S/(v -v_S)] =v_S[v/(v -v_S)] = [v_S/{1 - (v_S/v)}] If the source moves away from the listener, thev_Sis considered negative in the above expression. A speed of Mach 0.5 means that the airplane moves at half the speed of sound in air. Thereforev_S/v = 0.5. As the airplane moves toward the listener v_L =v_S/(1 - 0.5) = 2v_S so that the listener hears a 2000-Hz sound. After the airplane has passed the listener, the frequency he hears is v_L =v_S/[1 + (v_S/v)] =v_S/[1+0.5] = (2/3)v_S so that the listener hears a 667-Hz sound.

Question:

The following observations were made for the reaction NH_4CNO \rightleftharpoons NH_4^+ + CNO^- \rightarrow(NH_2)_2CO : 1) Addition on an equimolar concentration of KCNO with respect to NH_4CNO doubled the initial rate of the reaction, 2) Addition of an equimolar concentration of NH_4Cl doubled the initial rate of reaction. From these observations, (a)Determine the order of the reaction, (b) Write the kinetic rate law expression for the reaction. (.c) Discuss the influence,, if any, of the following equilibrium upon the reaction kinetics: NH_4 + CNO^- \rightleftharpoons NH3 + HCNO with K_equil = [{(NH_3)(HCNO)} / {(CNO^-)(NH_4^+)} ] = 10^-4 .

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Solution:

(a) The order of a chemical reaction is the number of distinct particles that must come together to form the activated complex. This factor can also be called the molecularity of the reaction. To see which particles directly determine the reaction rate, note the effect on the rate when more of a certain type of particle is added. You are told that when KCNO is added, that is, the KCNO concentration is doubled, the rate doubles. This means the reaction must be first order with respect to CNO^- (since, in solution, KCNO dissociates to CNO^-) . When [NH_4Cl] is doubled, the reaction rate is again doubled. Thus, the re-action is first order with respect to NH_4^+ , since, in solution, NH_4Cl dissociates into NH_4^+. Since doubling the amount of either NH_4^+or CNO^- doubles the reaction rate, the he reaction must be second-order overall. (b)By definition, a second order reaction with two different molecules has rate = {-d[A]} /dt = {-d[B]}/dt = k[A][B], where k = rate constant and A and B are the two different molecules in A + B \rightarrow products. Thus, for this reaction, the rate law becomes rate = {-d [NH_4 CNO]} / [dt] = {d[NH_2]_2co} / (dt) = k[NH_4^+ ][CNO^-] (c)To answer this part, notice the following: The equilibrium constant for the formation of NH_3 and HCNO is extremely small, which means that, in solution, the con-centration of these species in pure NH_4CNO solution should also be small, i.e., [NH_3] / [NH_4^+] = [HCNO] / [CNO^-] = (10^-4)^1/2 = .01. If either NH_3 or HCNO is added to the solution, the reaction is driven to the left. But, even if all of the other re-actants available were consumed, the concentration of NH_4^+ and CNO^- would increase by only 1%, which means that the reaction rate would increase by 2%, which is not easily detected.

Question:

In a competitive market (one in which noindividual supplier is powerful enough to influence prices),the relation be-tween the supply and demand of a good is assumed to be the followings If the supply, S, exceeds the demand, D (surplus), the demand falls. If D exceeds S (shortage), the price rises. Assuming there is no delay between the occurrence of surplus or shortage and the price change it causes, write a FORTRAN program which uses the modified Euler method (the method is fully explained, and the subroutine for it is given in the SIMULATION chapter) to simulate this system from time t=o to t=t_F, if the price of the good (in dollars per unit of good) at time t is p(t), p(o) = Po>o, and it is observed that: Case(1): D = f(p(t)) and S = h(p(t)) where f and g are linear quantities. Case (2): D = f(p(t)) is inversely proportional to[p(t)]^2 with proportionality factor d_1>2po and S is directly proportional to [p(t)]^0.11 with proportionality factor s_1. D and S are expressed as units of good per time period. Compare the modified Euler approximations with exact values, where feasible.

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Solution:

First note that exogenous factors, such as government price control, will be neglected in this model. Negative demand and supply can be interpreted as return of goods, as the diagram below shows (the arrows represent flow of goods): Also note that D = f(p), S = g(P) implies D, S vary only with price: all other influences (e.g. tastes, income, technological change, etc.) are held fixed in this model. Since both demand and supply depend on price and both determine price, we recognize that this system has feed-back. Therefore, the block diagram of the system will contain a linkage between output and input: The assumed relation between supply, demand and price can be expressed mathematically: Ṗ = K (D - S)(1) where ṗ is the time rate of change of price (in dollars per unit time period) , and K is a constant of propor-tionality, 0

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Question:

Two 250-turn circular Helmholtz coils are placed parallel to one another and separated by a distance equal to their common radius. Find the value of the magnetic induction at a point on the axis between them when current flows through both coils in the same sense, and show that the field is almost uniform about the midpoint.

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Solution:

The magnetic induction due to a single coil at a point along the axis a distance y from the plane of the coil can be found, using the Biot - Savart law dB^\ding{217} = (\mu_0 / 4\pi) [(I dl^\ding{217} × r^\ding{217})/(r^3)] (see figures A and B). Due to the symmetry of the loop, the vertical components of the dB^\ding{217} contributions by all the elements of current carrying wire dl^\ding{217}, cancel. The horizontal components add, however (see figure B). The magnitude of dB is dB = (\mu_0 /4\pi) [(I dlr sin \alpha)/(r^3)] where \alpha is the angle between dl^\ding{217} and r^\ding{217}. B_1 = (\mu_0 /2) [(Ia sin \alpha)/(r^2)] = (\mu_0 /2) [(ia^2)/r^3] = (\mu_0 /2) [Ia^2/(a^2 + y^2 )^3/2] Similarly, at the same point the magnetic induction due to a single turn of the second coil is (see figure A) B_2 = (\mu_0 /2) [Ia^2 /{a^2 + (a - y)^2 }^3/2] These act in the same direction, (for the direction is determined by the direction of the vector dl^\ding{217} × dr^\ding{217}, which is the same for both coils) and thus the total effect at 0 due to the n (= 250) turns of both coils is B = n (B_1 + B_2 ) = [(250 \mu_0 Ia^2)/2] [{1/(a^2 + y^2 )^3/2} + {1/[a^2 + (a - y)^2]^3/2} If y = a/2 , then B = [(8 × 250 \mu_0I)/5^3/2] a. Further, dB/dy = [(250 \mu_0Ia^2)/2] [{- 3y/(a^2 + y^2)^5/2} + {3(a - y)/[a^2 + (a - y)^2]^5/2}] = 0 if y = a/2. Also (d^2 B/dy^2) = [(250 \mu_0Ia^2) / 2] [{- 3 / (a^2 + y^2 )^5/2} - {3 / [a^2 + (a - y)^2]^5/2} + {15y^2 / (a^2 + y^2 )^7/2} + {[15(a - y)^2] / [a^2 + (a - y)^2]^7/2}] = [(250 \mu_0Ia^2) / 2] [{[15y^2 - 3(a^2 + y^2 )]/[(a^2 + y^2 )^7/2]} + {(15(a - y)^2 - 3[a^2 + (a - y)^2]) / ([a^2 + (a - y)^2]^7/2)] Thus dB/dy and d^2 B/dy^2 are each equal to zero at the point y = a/2, midway between the coils. Hence B hardly varies around that point, giving a large region of uni-form field midway between the coils. With this particular spacing of the coils, the dropping off in the value of B due to one coil as we move away from it is compensated for by the increase in B due to the other coil for much of the region be-tween them. The situation is illustrated in figure C. The solid lines give the magnitude of B due to each coil separately at various distances along the axis. The dashed line shows the combined effect of the two coils, and the region of uniform field around the midpoint of the system is clearly seen.

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Question:

In Drosophilia, the dominant gene G codes for gray body color and the dominant gene N codes for normal wings. The recessive alleles of these two genes result in black body color (g) and vestigial wings (n) respectively. Flies homozygous for gray body and normal wings were crossed with flies that had black bodies and vestigial wings. The F_1 progeny were then test-crossed, with the following results: Gray body, normal wings236 Black body, vestigial wings253 Gray body, vestigial wings50 Black body, normal wing61 Would you say that these two genes are linked?If so, how many map units apart are they on the chromosomes?

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Solution:

To determine if the two genes are linked, we look at the F_2 progeny. We notice that there are, two large groups (236, 253) which are approximately equal, and two small groups (50, 61) which are also nearly equal. We can reduce these numbers to small whole number ratios: gray body,normal wings236/50 roughly equals5 black body,vestigial wings253/50 roughly equals5 gray body,vestigial wings61/50 roughly equals1 black body,normal wings50/50 roughly equals1 Since the heterozygous F_2 parent (the result of a cross from two homozygous parents) is crossed with a double recessive, we would expect a 1 : 1 : 1 : 1 phenotypic ratio among the offspring (see previous problem). However, the ratio here is 5 : 5 : 1 : 1. This significant departure from the 1 : 1 : 1 : 1 ratio indicates that linkage is indeed likely. The fact that the ratio divides the progeny into four groups, two large and of equal size and two small and of equal size, is also a typical result of a cross involving linked genes. To calculate the distance between the genes, we will have to rely on the fact that the frequency of crossing over depends on this distance. It is logical to assume that the farther apart two genes are on a chromosome, the more likely it is that a crossover will occur, because there is a larger possible region in which it can occur. The process of locating genes on chromosomes is called mapping. The percentage of crossing over or recombination gives us no information about the absolute distances between the genes but it does give us relative distances between them. Suppose, for example, that genes A and B have a recombination frequency of 20% and genes A and C have a frequency of 40%. From this data, we cannot tell the exact distance of B from A or C from A but we can say that C must be twice as far from A as B is from A, since twice as much crossing over occurred in the distance between C and A. Instead of having to compare two distances every time we want to refer to the separation of two genes, we have, by convention, established a standard measurement known as a map unit. A map unit is defined as the distance on the chromosome within which a crossover occurs one percent of the time. When genes on a chromosome are allocated their respective positions on that chromosome, we can obtain what we call a genetic map. To obtain the map distance between two genes, we have to know the frequency of crossing over between them. Recombination frequency is defined as the ratio of the number of recombinants to the whole progeny, that is, Because recombinants can only result from a cross-over event, their numbers in the population will be small for closely linked genes. The recombinants in this problem are found in the two smaller groups, and their total is 50 + 61 or 111. The total number of progeny is 236 + 253 + 50 + 61 or 600. Therefore: RF= [(number of recombinants) / (total number of progeny)] = [(111) / (600)] = 0.185. Since the map units are expressed as a percentage of recombination, a RF of 0.185 is the same as 0.185 × 100 or 18.5 map units. Hence a map distance of 18.5 map units separates the gene for body color from the gene for wing size (See Fig.)

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Question:

A sugar solution was prepared by dissolving 9.0 g of sugar in 500 g of water. At 27\textdegreeC, the osmotic pressure was measured as 2.46 atm. Determine the molecular weight of the sugar.

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Solution:

The molecular weight of the sugar is found by determining the concentration, C, of sugar from the equation for osmotic pressure, \pi = CRT, where \pi is the osmotic pressure, R = universal gas constant = 0.08206 (liter-atm / mole) -\textdegreeK , and T is the absolute tem-perature . The osmotic pressure is measured as \pi = 2.46 atm and the absolute temperature is T = 27\textdegreeC + 273 = 300\textdegreeK, hence \pi = CRT, 2.46 atm = C × 0.08206 (liter-atm / mole-\textdegreeK) × 300\textdegreeK, or C = (2.46atm) / [(0.08206(liter-atm / mole -\textdegreeK) × (300\textdegreeK)] = 0.10 (mole / liter). If we assume that the volume occupied by the sugar molecules in so small a concentration can be neglected, then in 1 liter of solution there is approximately 1 liter of water, or 1000 g of water, and C = 0.10 (mole / 1000 g). Therefore, there is 0.10 mole of sugar dissolved in 1000 g of water. g of sugar dissolved in 500 g of water is equi-valent to 18.0 g of sugar dissolved in 1000 g of water (9.0 g / 500 g) = (18.0 g / 1000 g). But since C = (0.10 mole / 1000 g), the 18.0 g of sugar must correspond to 0.10 mole of sugar. Therefore, the molecular weight of the sugar is (18 g / 0.1 mole) = (180 g / mole).

Question:

Acompressionalwave of frequency 250/sec is set up in an iron rod and passes from the rod into air. The speed of the wave is 1.6 x 10^4 ft/sec in iron and 1.1 x 10^3 ft/sec in air. Find the wavelength in each material.

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Solution:

The frequency f of a wave remains constant as it passes from one medium to another. The velocity v and wavelength \lambda of the wave vary in accordance with the relationship, v =\lambdaf In iron \lambda = (v/f) = {(1.6 × 10^4 ft/sec)/(250/sec)} = 64 ft In air \lambda = (v/f) = {(1.1 × 10^3 ft/sec)/(250/sec)} = 4.4 ft

Question:

In the electroplating of nickel from a solution containing Ni^2+ ion, what will be the weight of the metal deposited on the cathode by a current of 4.02 amperes flowing for 1000 minutes?

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Solution:

When Ni^2+ plates out of solution, the reaction can be stated as Ni^2+ + 2e^- \rightarrow Ni . This means that for every Ni^2+ ion that deposits on the cathode, 2 electrons must be used. These electrons will be obtained from the current. One mole of electrons is commonly called one Faraday. One can see from the reaction that 2 Faradays are needed to convert one mole of Ni^2+ to Ni. (1 Faraday = 96,500 coulombs). The MW. of Ni is 58.7, therefore, 2 Faradays will plate out 58.7g of Ni. Since the number of coulombs determines the amount of Ni that plates onto the cathode, the number of coulombs that flow in this current during 1000 minutes needs to be calculated. It is stated in the problem that a current of 4.02 amperes flows for 1000 minutes. One ampere is de-fined as one coulomb per second ampere = coulomb/second. To find the total number of coulombs that were transmitted to the Ni^2+ ions, the amperage of the current must be multiplied by the time the current was flowing. The time must be converted from minutes to seconds by use of the conversion factor, 60 secs/1 minute. no. of coulombs = 4.02 coulombs/sec. × 1000 min. × 60 sec/minute = 24,120 coulombs. Since one knows that 2 Faradays or 193,000 coulombs (2 × 96,500 coul) plate out 1 mole, or 58.7g, of Ni, one can set up the following ratio to determine the number of grams that 24,120 coulombs plate out. Let x = the number of grams of Ni^2+ that 24,120 coulombs will convert to Ni . (58.7 grams) / (193,000 coulombs) = (x) / (24,120 coulombs) x = (24,120 × 58.7 g)] / (193,000) = 7.34g.. 7.34g of Ni will be deposited.

Question:

Describe the structure and function of the human placenta.

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Solution:

The placenta is an organ composed of both maternal and fetal tissues that supplies the developing embryo with nutrients and oxygen and enables the embryo to excrete carbon dioxide and other metabolic wastes. It also functions as an endrocrine gland, secreting estrogens, progesterone, human chorionic gonadotropin (HCG), and human placental lactogen (HPL). The former two are steroids, while the latter two are protein hormones. HCG functions to preven regression of the corpus luteum, while HPL stimulates secre-tion of the steroid hormones from the corpus luteum. The placenta forms following the implantation of the embryo within the uterine wall. Lytic enzymes secreted by the trophoblasts break down the capillaries of the uterine endometrium causing the maternal blood to pool in that area. Subsequently, the trophoblast layer surrounds and sends out villi into this region (see Figure). These villi become integrated by the umbilical arteries and veins which are extensions of the fetal circulatory system. The fetal blood in the capillaries of the villi come in close contact with the maternal blood in the tissue sinuses between the villi. Nutrients diffuse from the maternal blood through the trophoblast layer into the fetal capillaries, and waste products from the fetus is eliminated into the maternal blood. Note, how-ever, that at no time or place does the blood of the fetus mix with that of the mother. Blood from the mother enters the placenta via the uterine artery and leaves the region via the uterine veins. Similarly, the fetal blood never leaves the fetal vessels, which consist of umbilical arteries, the capillary network in the placentalvilli, and the umbilical vein.

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Question:

What is the weight of 1,000 cubic feet of air at STP?

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Solution:

To solve this problem, one must first recognize that air is a mixture of 80% N_2 and 20% O_2. One should also know that 22.4 cubic feet of any gas at STP weighs its ounce molecular weight. For example, in this problem, 22.4 ft.^3 of O_2 would weigh 32 oz. 22.4 ft.^3 of N_2 would weigh 28 oz. Using this information, if one knows what 22.4 ft.^3 of airweighs, then one can determine how much 1000 ft. of airweighs. Since air is a mixture, one can find its ounce molecular weight by multiplying the per cent composition of each gas times each gas' ounce molecular weight and adding these two values: for N_2 : (0.8) (MW N_2) = (0.8) (28 oz) = 22.4 oz for O_2 : (0.2) (MW O_2) = (0.2) (32 oz) = 6.4 oz. The ounce molecular weight for air is 22.4 + 6.4 = 28.8 oz., and has a volume of 22.4 ft.^3 If 22.4 ft.^3 of air has a weight of 28.8 oz then, the weight of 1000 ft.^3 can be found through the following proportion: [(22.4 ft.^3) / (28.8 oz)] = [(1000 ft.^3 ) / (unknown weight)] Solving for the unknown weight of air weight of air = [{(1000 ft.^3) (28.8 oz.)} / {22.4 ft.^3}] = 1.29 × 10^3 oz.

Question:

Nitroglycerin (C_3 H_5 ( NO_3) _3) explodes according to the follow-ing reaction: 4C_3 H_5 (NO_3) _3 (l) \rightarrow 12CO_2 (g) + 6N_2 (g) + O_2 (g) + 10H_2 O (g), producing only gaseous products. What is the total volume of gaseous products produced at standard temperature and pressure (STP) when 454 g of nitroglycerin explodes? The molecular weight of nitroglycerin is 227 g/mole.

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Solution:

This problem is an application of the ideal gas equation, PV =nRT, where P = pressure, V = volume, n = number of moles, R = gas constant, and T = absolute temperature. Solving for V, V = [(nRT) / (P)] STP is, by definition, 0\textdegreeC(= 273\textdegreeK) and 1atmpressure. Hence, T = 273\textdegreeK and P = 1 atm. Also, R = 0.082 liter -atm/mole-deg. We must find the number of moles, n, of gaseous products. The number of moles of nitroglycerin we started with is equal to its mass divided by the molecular weight, or 454 g/227 g/mole = 2 moles of nitroglycerin. Dividing the equation for the reaction of nitroglycerin by 2 (so that 4C_3 H_5 (NO_3)_3 becomes 2C_3 H_5 (NO_3) _3), we obtain: 2C_3 H_5 (NO_3)_3 (l) \rightarrow 6CO_2 (g) + 3N_2 (g) + (1/2)O_2 (g) + 5H_2 O (g). Thus, our 2 moles of nitroglycerin will produce a total of 6 + 3 + (1/2) + 5 = 14.5 moles of gaseous products, so that n = 14.5 moles. Substituting the values of n, R, T, and P into the equation for V gives: V = (nRT/P) = (14.5 moles × 0.082 liter -atm/mole - deg × 273\textdegreeK) /(1atm) = 325 liters.

Question:

Write a PL/I program that will simulate the moves of disks in the Towers of Hanoi Game.

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Solution:

This is an ancient puzzle, in which there are three pegs and 64 disks with holes through the center. The object is to move the entire tower of disks from one peg to another in the fewest possible numbers of moves. Disks must always be stacked so that a larger disk is never on top of a smaller disk. We have chosen a stack of eight disks, which requires 255 moves, because to move 64 disks, you have to make 18, 446, 744, 073, 709, 551, 615 moves - a formidable task even for a computer. In general, 2^n - 1 moves are requir-ed to transfer the tower to another peg (where n is the num-ber of disks in the tower). The recursively defined solution in PL/I follows: /\textasteriskcentered MOVE N DISKS FROM POLE X TO POLE Y. USE POLE Z TO \textasteriskcentered/ /\textasteriskcentered STORE DISKS IF NECESSARY \textasteriskcentered/ HANOI: PROCEDURE (X, Y, Z, N) RECURSIVE; DECLARE (X, Y, Z, N) FIXED; IF N = 1 THEN PUT SKIP LIST ('MOVE DISK FROM',X,'TO',Y); ELSE DO; CALL HANOI(X, Z, Y, N - 1) ; PUT SKIP LIST ('MOVE DISK FROM', X, 'TO', Y); CALL HANOI(Z, Y, X, N - 1); END; END HANOI;

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Question:

Why does milk \textquotedblleftspoil\textquotedblright when kept in a refrigerator?

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Solution:

Even though bacteria may be provided with the proper nutrients for cultivation , it is necessary to determine the physical environment in which they will grow best. Bacteria exhibit diverse reactions to the temperature of their environment. The process of growth is dependent on chemical reactions , and the rates of these reactions are influenced by temperature. Temperature therefore affects the rate of growth of bacteria. Most bacteria grow optimally within a temperature range of 25 to 40\textdegreeC. (The normal temperature of the human body is 37\textdegreeC). These bacteria are termedmesophiles. There are bacteria that grow best at temperatures between 45 to 60\textdegreeC. These bacteria are termed thermophiles . Somethermophileswill not grow at temperatures in the mesophilic range. At the opposite end of this thermal spectrum are the psychrophiles , bacteria which are able to grow at 0\textdegreeC or lower. Most psychrophiles grow optimally at higher tempera-tures of about 15-20\textdegree C. Thepsychrophilesare responsible for the spoilage of milk in the cool temperatures of a refrigerator (about 5\textdegreeC). After a week or so, pasteurized milk will begin to "spoil." The accumulation of metabolic products of psychrophilic bacteria will impart an abnormal flavor or odor to the milk. The milk might become viscous, which is a condition referred to as "ropy" fermentation . The viscosity is caused by the accumulation of agumlike material which normally forms a capsule around each bac-terium. Sweet curdlingmay also occur, caused by the coagulation of casein, a milk protein .

Question:

The biomass of each trophic level is usually much less than that of the preceding lower trophic level. Define the term biomass and explain the factors that determine the biomass of trophic levels in ecosystems.

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Solution:

Biomass is the total mass of the living material present in a certain category, whether it is a trophic level or an ecosystem. The distribution of biomass within some ecosystems can be represented by a pyramid, with the first trophic level (producers) at the base and the last consumer level at the apex. In general, the decrease of energy at each successive trophic level means that less of the biomass can be supported at each level. If the animals of each trophic level were of the same size, they would have to be rarer and rarer toward the top of the pyramid where energy is in shortest supply. Since the animals at the top are usually larger, they are sharing an even smaller supply of energy. The animals high in food chains therefore, must be few, and the pyramid of biomass is a direct consequence of this. Thus the total mass of carnivores in a given community is almost always less than the total mass of herbivores. The size, growth rate and longevity of the species at the various trophic levels of a community are important in determining whether or not the pyramidal model will hold for the biomass of the community. In fact, biomass pyramids of different communities vary greatly. The variability of biomass pyramids exist because the plant producer organisms exhibit extreme variability in their ability to undergo photosynthesis. The small algae of some ocean communities can greatly outproduce most land plants on a per gram basis. This is also because of the high metabolic and reproductive rates of algae. Consequently, they are able to support a pro-portionately much larger biomass of herbivores. The production of ocean herbivores is still only about ten percent (due to ecological efficiency) that of algae. But the biomass of the herbivores is increased because the turnover rate (the rate at which they are consumed and replaced) of the algae is very high. Biomass normally tends to decrease with each successive trophic level, forming the shape of a pyramid. Soma communities show an inverse pyramid biomass relationship, such as the case of open ocean algae primary producers.

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Question:

It is known that DNA (Deoxyribonucleic acid) exists as a two stranded molecule. Evidence for this stems from the fact that when ^15N isotope is added to bacteria that are reproducing, 50% of the nitrogen in the DNA of the first new generation of bacteria is ^15N. (a) How does this observation support the double helix (2 strands) structure of DNA; (b) If the first new gener-ation of bacteria were grown and reproduced on only ^14N, what isotopic makeup would you expect to find in the DNA of the second generation?

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Solution:

(a) You are told that the new generation had DNA of 50% ^15N isotope. The parents of this generation possess DNA with only ^14N. For their progeny, the new generation, to have 50% ^15N, DNA had to be synthesized. If all the DNA of the progeny were synthesized, its DNA content would be 100%^15N. But only 50% of the nitrogen is ^15N. This means, therefore, that half of the DNA from the parents, which is not newly synthesized, is used in the new generation. Half of the DNA is newly synthesized (^15N) and half already present (^14N). This implies the exist-ence of two strands. This can be better seen if you write of that type of nitrogen, and there are 2 strands, each type makes up only 50% of the total nitrogen. (b) The new generation is reproducing on ^14N. The progeny will have one half of its DNA newly synthesized on ^14N and half of its DNA already present. You can picture it as the following: The starred nitrogen 14 is that already present. The unstarred was that newly synthesized. From this, you see that half of the second new DNA generation would be ex-pected to contain only ^14N strands (starred and unstarred) , and the other half would contain 50% ^14N and 50% ^15N.

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Question:

(a) What will be the frequencies of the first and the second (a) What will be the frequencies of the first and the second overtones of a pipe closed at one end of length 2 ft? (b) What will be the frequencies of the first and second overtones of an open pipe 2.5 ft long? (c) Will there be any common beat frequency between these overtones?

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Solution:

(a) Not all wave shapes can be fitted into a closed (or semi-closed) pipe. Only those waves which satisfy certain boundary conditions can exist in the enclosure. Only those waves which have a node (a point of zero amplitude) at a closed end of the pipe and which have an antinode (a point of maximum amplitude) at an open end, can exist in the pipe. The wave with the lowest frequency which can exist in a pipe closed at one end is one which has only one node (at the closed end) and one antinode (at the open end). (See figure A). Its wavelength is \lambda = 4l and its frequency n_f (the fundamental fre-quency) is, using the relation \lambdan_f = v, where v is the velocity of sound, n_f = v/\lambda = v/4l = [(1100 ft/sec) / (4 X 2 ft)] = 137.5 sec^-1 The first overtone contains two nodes and two antinodes. The second overtone contains three nodes and three antinodes. In general, the nth overtone will contain n + 1 nodes and n + 1 antinodes. The first and second overtones of the pipe will then be the third and the fifth harmonic respectively, i.e., frequencies 3 times and 5 times that of the fundament-al (see figure A). \therefore n_1st = 3(137.5) = 412.5 sec^-1 n_2nd = 5(137.5) = 687.5 sec^-1 The only waves which can exist in an open pipe are those which have antinodes at both ends, of the pipe. The enclosed wave with the lowest possible frequency, n_f (the fundamental) space will contain two antinodes and one node (see figure B). The first overtone contains three antinodes and two nodes. The second overtone contains four antinodes and three nodes. In general, the nth overtone contains n + 2 antinodes and n + 1 nodes. The first and second overtones will then be the second and third harmonics, i.e., frequencies 2 times and 3 times that of the fundamental. The wavelength of the fundamental is \lambda = 2l and its frequency n_f' is n_f' = v/\lambda = v/2l = [(1100 ft/sec) / (2 X 2.5 ft)] = 220 sec_-1 . Hencen'_1st = 2(220) = 440 sec^-1 n'_2nd = 3(220) = 660 sec^-1 The frequency difference for the first overtones is 440 - 412.5 = 27.5. The frequency difference for the second overtones is 687.5 - 660 = 27.5. Thus a common beat frequency is 27.5 cycles per sec.

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Question:

A parallel beam of light falls normally on the first face of a prism of small angle \alpha. At the second face it is partly transmitted and partly reflected, the reflected beam striking the first face again and emerging from it in a direction making an angle of 6\textdegree30' with the reversed direction of the incident beam. The refracted beam is found to have undergone a deviation of 1\textdegree15' from the original direction. Calculate the refractive index of the glass and the angle of the prism.

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Solution:

We must first solve for the refractive index, n, of the prism glass. Applying Snell's Law to the refraction at point c (see figure), we obtain n sin \textphi_1 = n' sin \textphi_2 But n' is the refractive index of air, which is 1. Hence, n sin \textphi_1 = sin \textphi_2(1) Similarly, applying Snell's Law to the refraction at point B n sini= sin \delta_2(2) Now, we must relate \textphi_1, \textphi_2 andito known Quantities. Note that \angle HCP = \angle DCE, since they are vertical angles. Therefore, \angle HCP = \angle DCE \textphi_2 - \delta_1 = \textphi_1 or \delta_1 = \textphi_2 - \textphi_1(3) Noting that DC and BS are parallel, i= 2\textphi_1(4) We need one more equation relating any of O_1,O_2 andito \alpha. \angle ACD = 90\textdegree - \alpha . But \angle ACD + \textphi_1 = 90\textdegree Hence 90\textdegree - \alpha + \textphi_1 = 90\textdegree and \textphi_1 = \alpha(5) If all the angles (\textphi_1, \textphi_2,i, \delta_1, \delta_2) are small, we may approximate the sine of an angle by the angle itself. Using (1) and (2) n \textphi_1 \approx \textphi_2 ni\approx \delta_2(6) Taking the ratio of each equation in (6) (\varphi_1/i) \approx (\varphi_2/\delta_2) Hence,\varphi_1 \approx (i\varphi_2/\delta_2) Using (4)\varphi_1 \approx [(2\varphi_1 \varphi_2)/\delta_2] or \delta_2 = 2\varphi_2(7) Hence, solving (3) for \varphi_1 \varphi_1 = \varphi_2 - \delta_1 Since \varphi_2 = (\delta_2/2) \varphi_1 = (\delta_2/2) - \delta_1(8) Solving the first equation of (6) for n n \approx (\varphi_2/\varphi_1) with (7) and (8) n = [(\delta_2/2)/{(\delta_2/2) - \delta_1}] = [(1)/{1 - (2\delta_1/\delta_2)}] Using the given data n = (1)/{1 - [2(1\textdegree15')/(6\textdegree30')]} n = (1)/{1 - [2(1 1/4\textdegree)/(6 1/2\textdegree)} n = (1)/{1 - [(5/2)/(13/2)]} = [(1)/{1 - (5/13)}] n = [13/(13 - 5)] = (13/8) = 1.625 Furthermore, from (5) \alpha = \varphi_1 But, equation (8) tells us that \alpha = \varphi_1 = (\delta_2/2) - \delta1 = (6\textdegree30'/2) - 1\textdegree15' Hence\alpha = 3\textdegree15' - 1\textdegree15' = 2\textdegree

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Question:

A 100 g block of copper (s_cu= .095) is heated to 95\textdegreeC and is then plunged quickly into 1000 g of water at 20\textdegree C in a copper container whose mass is 700 g. It is stirred with a copper paddle of mass 50 g until the temperature of the water rises to a steady final value. What is the final temperature?

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Solution:

The heat lost by the hot copper block as it cools to temperaturet_f m_blocks_cu(t_95 -t_f) = (100 g) ( .095 cal/g) (950C -t_f) wheres_cuis the specific heat of copper. The heat gained by the water, the container, and the paddle is (m_water_ S(H)2 O+m_container_S_cu+m_paddle_S_cu) (t_f- t_20) Here S(H)2 Ois the specific heat of water. Then [(1000 g) (1 cal/g) + (700 g) (.095 cal/g) + (50g) (.095 cal/g)] (t_f- 20\textdegree C)] Equating the heat lost to the heat gained: (100 g) (.095 cal/g) (95\textdegree C -t_f) = [(1000 g) (1 cal/g) + (700 g) (0.95 cal/g) + (50 g) (.095 cal/g)] (t_f- 200C) Regrouping and solving fort_f: (9.5) (95\textdegreeC -t_f) = (1000 + 66.5 + 4.75) (t_f- 20\textdegreeC) 902.5\textdegreeC - 9.5t_f= 1071.3 (t_f- 20) = 1071.3t_f - 21430\textdegreeC 22330\textdegreeC = 1081t_f t_f= (22330/1081)0C = 20.6\textdegreeC.

Question:

The equation for cream of tartar and baking powder, in action with hot water is: NaHCO_3 + KHC_4H_6O_6\ding{217} NaKC_4H_4O_6 + H_2O + CO_2 (g) From the use of 4.2 g of NaHCO_3 (baking powder), what volume of CO_2 should be liberated through the dough at 273\textdegreeC?

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Solution:

From thestoichiometryof the equation, 1 mole of NaHCO3 produces 1 mole of CO_2. The molecular weight of NaHCO_3 is 84 g / mole. Thus, 4.2 g is (4.2 g / 84 g / mole) 0.05 moles. 1 mole of any ideal gas at STP (Standard Temperature and Pressure) occupies 22.4 l; thus the volume of CO_2 gas liberated at STP would be (22.4 l / mole) (0.05mole) = 1.12 l. However, this must be converted from 1.12 l at STP (0\textdegreeC = 273\textdegreeK, 1atm) to a temperature of 273\textdegreeC = 546\textdegreeK. To this conversion, Charles' Law is used (T_1 / T_2) = (V_1 / V_2), where T is temperature in \textdegreeK, and V is the volume at that temperature. Thus, (546\textdegreeK / 273\textdegreeK) = (V_1 / 1.12 l) Volume of CO_2 = V_1 = 2.24 l.

Question:

How are tracers useful as a debugging aid?

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Solution:

Suppose a computer runs and executes a program but produces incorrect results. If the programmer is unable to locate the bugs by the usual methods (running test data, checking size dimensions, etc.), an emergency technique called tracing can be applied. Tracing is usually done with an interpreter, whose routine dissects, analyzes and executes pseudo-instructions. (Pseudo-instructions do not form part of the program but are neces-sary for its execution). The tracer is an interpretive routine that analyzes and executes real machine instructions one at a time. To understand its opera-tion consider the following problem: The trouble with the program being debugged is suspected to be in the action of a loop. While the loop is being executed, the values of some variables go out of bounds. Using the tracer routine, the contents of the accumulator are printed out before each instruc-tion in the loop is executed. The point at which the variables misbehave can thus be located. The advantage of the trace routine is that the complete internal program is available for external inspection. This is, at the same time, its disadvantage. A series of nested loops may require a few thousand iterations. All these iterations will be executed and printed at printer speed. Clearly, the computer time involved is prohibitive. Hence, tracing functions have been modified to execute only non-branch commands, attached to high speed printers. If a branch is met, selected registers are printed before execution of the branch.

Question:

What are the differences between the queenbee, the worker bee , and the drone?

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Solution:

A honey bee colony consists of a single re-productive female, the queen , a few hundred males, called drones, and thousands of workers, which are sterile females. Drones hatch from unfertilized eggs, and are haploid , while females hatch from fertilized eggs, and are thus diploid. The type of food fed to the female larva determines whether it will develop into a worker or queen. A diet of "royal jelly" causes her to become a fertile queen . The work performed by a worker bee is determined by its age. For two weeks after metamorphosis, workers function as nurse bees, first incubating the brood and preparing brood cells, later feeding the larvae. Then the sterile females become house bees for one or two weeks, working as storekeepers, waxsecreters, or guards and cleaning the hive, or ventilating it by fanning their wings. The worker finally becomes a food gatherer for four or five .weeks, and collects nectar, pollen, and water. A queen bee soon after hatching mates several times with male drone bees. The queen accumulates enough sperm to last for her lifetime, and stores the sperm in-ternally in aspermatotheca. Thereafter shelays fertilized and unfertilized eggs, as many as a thousand a day. The drone dies after copulation; its reproductive organs literally explode into the female . When the number of bees in a hive becomes too large, about half the drones and workers, a new queen bee, and the old queen bee migrate to a new location and begin a new society. The old colony is left with developing queens.

Question:

Describe feeding in planarians.

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Solution:

Planarians belong to the group of fresh water flatworms and have a digestive system which lacks an anal opening. The mouth of the planarian opens into a cavity that contains the muscular pharynx, or proboscus, which can be protruded through the mouth directly onto the prey (see Figure) . In feeding, the planarian moves over its food object, which may be a small worm, crustacean or insect larva, and traps it with its body. The pharynx is then extended and attached to the food material, and by sucking movements produced by muscles in the pharynx, the food is torn into bits and ingested. Digestive enzymes are released in order to assist the planarian in breaking down the food prior to ingestion. The pharynx delivers the food into the three-branched gastrovascular cavity. The branching of the planaria's gastrovascular cavity into one anterior and two posterior branches provides for the distribution of the end products of digestion to all parts of the body. Flatworms with three branched gut cavities are called triclads, in contrast to those with many branches, called polyclads. Most of the digestion in planaria is intracellular, which means that it occurs in food vacuoles in cells lining the digestive cavity. The end products of digestion diffuse from these cells throughout the tissues of the body. Undigested materials are eliminated by the planaria through its mouth. The mouth, then serves as both the point of ingestion and the point of egestion. It is interesting to note that planarians can sur-vive without food for It is interesting to note that planarians can sur-vive without food for months, gradually digesting their own tissues, and growing smaller as time passes.

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Question:

What is the impedance of a 1-\muf capacitor at angular frequencies of 100, 1000 and 10,000rad/sec?

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Solution:

The circuit element is a capacitor, therefore the Impedance is purely reactive. At an angular frequency of 1000rad/sec, the reactance of a 1-\muf cap-acitor is X_C = (1/wC) = [1/{10^3 (rad/sec)} × 10^-6f] = 1000 ohms. At a frequency of 10,000rad/sec the reactance of the same capacitor is only 100 ohms, and at a frequency of 100rad/sec it is 10,000 ohms.

Question:

What is the average kinetic energy of air molecules at a temperature of 300\textdegree K? What would be a typical speed for nitrogen molecules, which are the major constituent of air? The mass of a nitrogen molecule is approximately 4.68 × 10^-26 kg.

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Solution:

The kinetic energy of a particle can have more than one term which is quadratic in velocities. If the molecules of an ideal gas display only three dimensional translational motion, then the kinetic energy of a molecule is k = [(1/2)mv_x^2] + [(1/2)mv_y^2] + [(1/2)mv_z^2] wherev_x,v_y,v_zare the velocity components respectively along the x, y, and z directions. There are three indepen-dent quadratic terms and we say that the system has three degrees of freedom. Theequipartitiontheorem states that the average kinetic energyE_kof a particle when the system is in thermal equilibrium is E_k = (1/2)nkT where n is the number of degrees of freedom of the system. For T= 300\textdegree K, we have E_k= (3/2)kT = (3/2)(1.38 × 10^-23 J/\textdegreek)(300\textdegree K) = 6.21 × 10^-21 J The average speed of a nitrogen gas molecule may be found using the kinetic-energy relation,E_k= (1/2)mv^2 . For the nitrogen molecule, then, the average speed v is v^2 =E_k/2m E_k= [(2)(2.61 × 10^-21 J)]/[4.68 × 10^-26 kg] = 0.265 × 10^6 m^2/s^2 = 26.5 × 10^4 m^2/s^2 orv = 5.15 × 10^2 m/s.

Question:

The ameba has no mouth for ingestion of food, but can take food in at any part of the cell. Explain this method of ingestion. What occurs following ingestion?

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Solution:

In the amoeba, pseudopodia (false feet) are used for nutrient procurement. These are temporary projections of cytoplasm which extend around the prey in a cup-like fashion eventually enveloping it completely. These pseudopodia can form anywhere on the surface of the amoeba. The enclosing of the captured organism by cyto-plasm results in the formation of a food vacuole within the amoeba. Usually, the pseudopodia are not in intimate contact with the prey during engulfment and a considerable amount of water is enclosed within the food vacuole along with the captured organism. Engulfment may also involve complete contact with the surface of the prey, and the resulting vacuole is then completely filled by food. Death of the prey takes from 3 to 60 minutes and results primari-ly from a lack of oxygen. Digestion occurs within the food vacuole. In amoeba, as in man, digestion is controlled by enzymes, and dif-ferent enzymes act at definite hydrogen-ion concentrations. The enzymes that function in the vacuoles of an amoeba enter by fusion of the vacuoles withlysosomes. The enzymes hydrolyze the proteins into amino acids, fats into fatty acids and glycerol, and carbohydrates into simple sugars. These end products of digestion are absorbed by the rest of the organism through the vacuolar membrane. The food vacuole is not stationary within the organism but circul-ates in the fluid cytoplasm. Undigested material is egested from the cytoplasm. Like that of ingestion, the point ofegestionis not fixed; food vacuoles containing undigested material may break through the surface at any point.

Question:

500 g of ice at 0\textdegreeC is added to 500 g of water at 64\textdegreeC. When the temperature of the mixture is 0\textdegreeC, what weight of ice is still present? Heat of fusion of H_2 O = 80cal/g.

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Solution:

The amount of heat used to melt the ice is equal to the amount of heat lost by the water. The amount of heat lost by the water is determined by using the specific heat of water. The specific heat of water is 1 cal/g degree. This means that for each gram of water cooled 1 degree, 1 calorie of heat will be evolved. Here, 500 g of H_2 O is lowered 64\textdegree - 0\textdegree or 64\textdegree. The heat evolved by the water can now be found. no. of calories evolved= 1 cal/g-degree × 500 g × 64\textdegree = 32000 cal. The heat of fusion is the quantity of heat necessary to liquefy 1 g of a solid substance at constant temperature at its melting point. Therefore, if the heat of fusion of water is 80 cal/g, then it takes 80 cal to melt one gram of ice. You found that 32000 calories are absorbed by the ice. The weight of ice melted by 32000 cal is found by dividing 32000 cal by 80 cal/g. no. of grams of ice= (32000 cal) / (80 cal/g) = 400 g Thus 400 g of the ice is melted. Originally, there was 500 g of ice, therefore 100 g of ice is left.

Question:

A box weighing 100 pounds is pushed up an inclined plane 10 feet long with its upper end 4 feet above the ground. If the plane is 80% efficient, what is the force of friction?

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Solution:

In approaching this problem, a careful plan of attack must be laid out. We are asked to find the force of friction in an inclined plane, given the di-mensions and efficiency of the plane. Reasoning back-wards, we begin by noticing that the effort consists of the force gravity exerts down the plane, plus the friction. The force exerted down the plane by gravity is calculable from the dimensions of the inclined plane and the weight of the box by constructing a proportion between the force triangle and the inclined plane (see diagram). 4'/10' = x/100 lb x = 40 lb Efficiency is calculated from AMA over IMA. IMA is calculated from the dimensions of the inclined plane (length over height, in this case 10 ft/4 ft = 2.5). The AMA is resistance over effort. The resistance is simply the weight of the box. We are only left with one unknown - the frictional force, which we now solve for efficiency= AMA/IMA = (resistance/effort) / (2.5) = [resistance/(x + F_friction)] / 2.5 = [100 lb/(40 lb + F_friction)] / 2.5 = 80 % = .80 Hence,F_friction= 10 lb.

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Question:

You have two half-cells coupled together to form a concentration cell: H_2(1 atm) \rightarrow 2H^+(.1M) + 2e^- and H_2 (1 atm) \rightarrow 2H^+ (1.0M) + 2e^-. Calculate the free energy and electrical work obtainable. F = 96,500 coulombs.

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Solution:

To find the free energy and electrical work obtainable, calculate the maximum cell voltage in this cell, since ∆G = free energy = -n F∆E, where n = number of electrons transferred, F = Faraday constant, and ∆E = cell voltage. To find ∆E, use the Nernst equation, which states ∆E = ∆E\textdegree-.059/N log K, where ∆E\textdegree = voltage at standard conditions, and K = equilibrium constant of reaction. To find ∆E\textdegree, note that the half-reactions are given as oxidation processes (loss of electrons). In a cell, however, the overall reaction is the sum of the oxidation and reduction process (gain of electrons). Therefore, you reverse one of these half-reactions to obtain a reduction. If you reverse H_2 (1 atm) \rightarrow 2H^+(1.0M) + 2e^-,you obtain 2H^+(1.0M) + 2e^- \rightarrow H_2 (1 atm) :reduction oxid:H_2 (1 atm) \rightarrow2H^+(.1M) + 2e^- Total (overall) H^+(1.0M) \rightarrow H^+(.1M) From this, you see that only one electron is being transferred per H^+, which means N = 1. ∆E\textdegree = standard potential = E\textdegree_products - E\textdegree_reactants . But the product and reactant are the same species, which means the difference is zero. Thus, ∆E\textdegree = 0. To find K, equilibrium constant, remember that the ratio of concentrations of products to reactants, each raised to the power of its coefficient in the equation. Thus K = {[H^+ ]_product } /{[H^+ ]_reactant} . for this reation. Thus, K = (.1M) / (1.0M) Recalling, ∆E = ∆E\textdegree = -(.059 / N )log K, you can sub-stitute to obtain the Following: ∆E = 0 -(.059 / 1) log [(.1)/(1.0)] = 0 - (.059) (-1) = + 0.59 V. Recalling ∆G = - n F∆E, you have all the values to find ∆G. Substituting and solving, ∆G = -(1) (96500) (.059) = 5693.5 (coul-volt / mole) or∆G = 5693.5 [(1cal) / (4.184 coul-volt)] [(Kcal) / (1000cal)] = -1.36 (Kcal / mole) Thus, the maximum work obtainable, the free energy of the system, equals -1.36 (Kcal/mole).

Question:

What are the main parts of the CPU?What tasks must the CPU perform during each instruction cycle?

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Solution:

Figure 1 shows the main parts of the CPU. The CPU performs the following tasks during each instruc-tion cycle. 1.It places the address of the instruction on the ad-dress bus. 2.It takes the instruction from the input data bus and decodes it. 3.It fetches the address and data required by the in-struction. 4.It performs the operation specified by the operation code. The instruction may be a logical or arithmetic function, a data transfer, or a management function. 5.It provides appropriate responses to incoming control signals, such as interrupts. 6.It provides status, timing, and control signals to the memory and I/O sections.

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Question:

Although some adult amphibia are quite successful as land animals and can live in comparatively dry places, they must return to water to reproduce. Describe the process of reproduction and development in the amphibians.

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/Users/wenhuchen/Documents/Crawler/Biology/F13-0335.htm

Solution:

The frog (order Anura) will be used as a representative amphibian in this problem. Reproduction in the frog takes place in the water. The male siezes the female from above and both discharge their gametes simultaneously into the water. This process is known as amplexus. Fertilization then takes place externally, forming zygotes. A zygote develops into a larva, or tadpole. The tadpole breathes by means of gills and feeds on aquatic plants. After a time, the larva under-goes metamorphosis and becomes a young adult frog, with lungs and legs. The same type of system is seen in the salamanders (order Urodela). Like the metamorphosis of insects and other arth-ropods, that of the amphibia is under hormonal control. Amphibia undergo a single change from larva to adult in contrast to the four or more molts involved in the development of arthropods to the adult form. Amphibian metamorphosis is regulated by thyroxin, the hormone secreted by the thyroid gland, and can be prevented by removing the thyroid, or the pituitary which secretes a thyroid-stimulating hormone modulating the secretion of thyroxin.

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Question:

Water flows at a rate of 2.5 m^3 \textbullet s^-1 over a waterfall of height 15 m. What is the maximum difference in tem-perature between the water at the top and at the bottom of the waterfall and what usable power is going to waste? The density of water is 10^3 kg \textbullet m^-3 and its specific heat capacity is 10^3 cal \textbullet kg^-1 \textbullet C deg^-1.

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Solution:

The water loses potential energy and gains kinetic energy in falling over the waterfall. The maximum possible temperature difference between the water at the top and at the bottom of the falls occurs if all this kinetic energy is converted to heat. The potential energy lost,mgh, is completely converted to heat in this case. The power available is the potential energy lost in a time \cyrchar\cyrt, or P = (mgh) / \cyrchar\cyrt = (\rhoVgh) / \cyrchar\cyrt(1) where m is the mass contained in a volume of water V, and \rho is the mass density of water. Note that V/\cyrchar\cyrt is the volume of water passing over the waterfall per unit time. Hence, P = 10^3 kg \textbullet m^-3 × 2.5 m^3 \textbullet s^-1 × 9.8 m \textbullet s^-2 × 15 m = 3.675 × 10^5 W = 367.5 kW. Under the conditions we have assumed, all this power goes into heat. The rise in temperature, ∆t, of a mass of material m, caused by an amount of heat Q is given by the relation amount of heat Q is given by the relation Q = mc ∆t where c is the specific heat capacity of the material. Hence ∆t = Q/mc = (Q/\cyrchar\cyrt) / (cm/\cyrchar\cyrt) In our case, P = Q/\cyrchar\cyrt and ∆t = P/(cm/\cyrchar\cyrt) = P\cyrchar\cyrt/mc Furthermore, m is the mass of water in a volume V, or m =\rhov whence ∆t = (\rhoVgh)/(\rhoVc) =gh/c ∆t = (9.8 m \textbullet s^-2 × 15 m) /(4.186 × 10^3 J \textbullet kg^-1 \textbullet C deg^-1) = 0.035\textdegree C. This is the temperature change experienced by a mass m of water in falling through a distance h.

Question:

What is the maximum number of emission lines one would expect to see in a molecule containing only the six electronic energy levels depicted in diagram A?

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Solution:

Emission is observed when an electron falls from anelectronic energy level of higher energy to one of lower energy. From the energy level n = 6, electrons can fall to energy levels of n = 5, 4, 3, 2, and 1. Thus, five emission lines are drawn from the energy level n = 6. By using a similar process, one determines that the maximum number of emission lines from energy levels n = 5, 4, 3, 2, and 1 are four, three, two, one, and zero, respectively. Because there are no energy levels lower than n = 1, no emission lines are obtained from it. The total number of emission lines is found by taking the sum of all the emission lines from the various energy levels. Thus, 5 + 4 + 3 + 2 + 1 + 0 = 15 mission lines can be expected. The possible emission processes are depicted in diagram B.

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Question:

A triangular plate is immersed in water (of density \rho) with one vertex at the surface and the others at depths of 6 in. and 12 in. What is the thrust on the plate due to the pressure of the water only? Its areas is 63 in^2 (See figure (a))

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Solution:

At depth z from the water's surface, the press-ure is g\rhoz, the symbols having their usual significance. The total thrust on the small element, parallel to the surface, shown in figure (a) is dF = g\rhozy dz. The total thrust on the plate is found by summing this differential element of force over the entire tri-angle. F = ^z0\int_0 g\rhozy dz = g\rho[( ^z0\int_0 \rho'zy dz)/(\int^z0_0 \rho'y dz)] × \int^z0_0 y dz ' ' wherep' is the density of the plate material. p' is the density of the plate material. Here,^z0\int_0 \rho'y z dz/^z0\int_0 \rho'y dzis the location of the plate'scenter of mass relative to the water's surface. ^z0\int_0 y dz is the area of the plate. The total thrust on the plate is thus seen to be the pressure at the center of mass of the plate multi-plied by the area of the plate. This is a general result for all plates, as can be seen from the general nature of the derivation. In the particular case of the triangular plate, the area of 63 in^2 = 63/144 ft^2 is given. The center of mass of a triangular plate is two- thirds of the way from a vertex to the middle of the opposite side. From figure (b), we see that triangles BEC and DFC are similar. Hence CF/CE = CD/CB CF = [CD/CB] CE Since D is the mid-point of BC CF = (1/2)CE = (1/2) (6") = 3" Hence, D is 9" below the water's surface. But the center of mass is two- thirds of AD from A and must be 6 in. = (1/2) ft from the surface. In this case, therefore, F = g\rho {location of C.M.} × {area of plate) = 32 ft/s^2 × 1.94 sl/ft^3 × (1/2) ft × 63/144 ft^2 = 15.58 lb where we used the fact that \rho = 1.94 sl/ft^3 for water.

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Question:

Write a BASIC program which compares two strings according to their alphabetic ordering. Output both strings and their relationship to each other.

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Solution:

In this problem string variables are denoted by A$ and B$. They can be compared to each other using the relational operators <, =, or > . 1\O\O REM COMPARES STRINGS FOR ORDER 11\O PRING "A$"; 12\O INPUT A$ 13\O IF A$ = "STOP" THEN 24\O 14\O PRINT "B$"; 15\O INPUT B$ 16\O IF A$ < B$ THEN 22\O 17\O IF A$ = B$ THEN 2\O\O 18\O PRINT A$; "IS GREATER THAN"; B$ 19\O GO TO 1\O\O 2\O\O PRINT A$; "IS EQUAL TO"; B$ 21\O GO TO 1\O\O 22\O PRINT A$; "IS LESS THAN"; B$ 23\O GO TO 1\O\O 24\O END

Question:

Suppose you were given a cage full of small, rat-like animals from the Rocky Mountain region and another cage full of similar animals from the Appalachian Mountains. How would you determine if the two groups were of the same or different species?

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Solution:

The species is the fundamental unit of biological classification. A species is defined as a group of organisms that is closely related structurally and functionally, which in a natural environment inter-breeds and produces fertile offspring, but which seldom, if ever, breeds with organisms of another species. A species is reproductively isolated. The dog and the cat are examples of two different species. Dogs and cats cannot breed with one another. Closely related species may interbreed, however, the offspring are rarely fertile. The horse and the donkey are examples of closely related species which can mate and produce viable offspring. This offspring is the mule, which is unable to reproduce. A mule can only result from the union of a horse and a donkey. The fact that mules are sterile helps to establish the fact that the horse and the donkey are of different species. Species may be subdivided into subspecies, varieties, or races. The varieties are kept distinct either through geographic isolation, seasonal differences in reproductive patterns, or by the intervention of man. Dogs offer an example of controlled breeding. If unres-trained by man, the different varieties interbreed to produce mongrels. In certain cases, interbreeding between different subspecies is very difficult. The greatest problem in the case of dogs would be size; however, sexual attraction between different varieties exists and it can be proven that both varieties belong to the same species because they are structurally and functionally similar. Both the small and large varieties could breed with a medium sized variety, proving that the varieties are not reproductively isolated. If interbreeding between varieties is prevented by seasonal variation of reproduc-tive patterns, careful regulation of laboratory conditions could enable two different subspecies to be ready for reproduction at the same time. In the case of the rat-like animals, the two groups should be studied to see if they are similar structurally and functionally. If so, it must be determined whether the animals could breed with one another. If they could and if they produce fertile offspring, the two groups would be classified as members of the same species. If the off-spring were all sterile, it would mean that the two groups of animals belonged to two different, closely related species. If the two groups seemed similar after studying their structural and functional characteristics, but did not breed with one another, an examination of their natural reproductive patterns might offer information as to why breeding, did not occur under laboratory conditions. If conditions were altered and fertile offspring were pro-duced, the two groups of animals could be members of the same species. If breeding did not occur under any condi-tions, or through an intermediate species, one would con-clude that the two were members of different species.

Question:

Explain how the BSA and BUN instructions can be used for subroutine call and return.

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Solution:

Figure 1 demonstrates the use of the BSA and BUN instructions in writing subroutines. The actual sub-routine in Figure 1 starts at location m + 1 and continues to location m + z. After fetching the BSA instruction from memory the con-tents of the PC are 026. During the execute cycle the contents of the PC are stored in location m and the ad-dress m+1 is stored in the PC. Thus, the computer jumps to the subroutine and saves, the return address in loca-tion m. The computer then executes all of the instruc-tions in the subroutine. Finally the computer reads the BUN instruction at the end of the subroutine. Notice, however, that BUN is an in-direct instruction (specified by the 1 in bit I of the instruction). During this instruction the address specified in location m is stored in the PC; hence the PC contains the return address and returns to location 026 after executing the BUN instruction.

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Question:

Analyze the synchronous sequential circuit given below. Assume the inputs are binary levels and that the following state assignment is used. y = 0 \equiv A y = 1 \equiv B Use Karnaugh maps to find a) the state tableb) the state diagram

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Solution:

The given sequential circuit is built of AND, OR, and NOT gates and a memory device called a D Flip-Flop. Having an external clock connected to the circuit means that the memory changes state only during a clock pulse. The operation of a D Flip-Flop goes like this: When the clock pulse goes from 0 to 1, data enters the flip-flop. When the clock pulse returns to 0 from 1 the output state changes may occur. Flip-flops which work under these con-ditions are called EDGE-TRIGGERED. This sequential circuit has only one Flip-flop and hence only two states, A and B. The best way to examine the behavior of the given sequential circuit is by drawing the timing diagram. The timing diagram for the memory circuit is given in Figure 1; In general, the operation of the sequential circuit given may be completely defined by a state-table which lists all possible operating conditions, as shown in figure 2. The logic followed in the construction of the state-table is as follows: To fill in the upper left hand corner assume a present state y^k = 0 and the input x^k = 0. Following these signals through the combinational circuit it is found that the next state is y^k = 0 and that the output is j = 0. Hence the entry in the upper left; block is 0/0 . The initial conditions for the upper right block is y^k = 0, and x^k = 1. Applying these signals to the combina-tional circuit given yields the output j = 0. So that the entry in this block is 1/0. To obtain the completed state diagram the states which the circuit assumes must be identified. Identification may be done as follows: y = (0) \equiv A y = (1) \equiv B The completed state table is shown in Fig. 3: b) To construct the state diagram, the state table is used. Examination of the state table reveals that these are the only two states, A and B, which the circuit may assume. These states are represented in a state diagram as shown in figure 4. Note that assuming an initial state, y^k = 0 and x^k = 0, the output is zero and the next state is still A. When y^k = 0 and x^k = 1 the next state is the state B and the output is y = 0. When x^k = 0 is applied to state B the next state is again state B and output is y = 0. When x^k = 1 state B changes to state A with an output of j = 1.

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Question:

A delicate machine weighing 350 lb is lowered gently at constant speed down planks 8 ft long from the tailboard of a truck 4 ft above the ground. The relevant co-efficient of sliding friction is 0.5. Must the machine be pulled down or held back? If the required force is applied parallel to the planks, what is its magnitude? The machine is reloaded in the same manner, a force of 330 lb being applied. With what velocity does it reach the tailboard? What kinetic energy and what potential energy has it then acquired and how much work has been performed in overcoming friction? What relation-ship exists between these quantities?

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Solution:

The forces acting on the machine are four in number. (1) The weight, mg^\ding{217}, acting vertically downward. (2) The normal force exerted by the plane, N^\ding{217}. (3) The frictional force F^\ding{217}, acting up the plane opposing the motion down it. If \mu is the coefficient of sliding friction, this force has magnitude \muN. (4) The force P^\ding{217} necessary to keep the machine moving with constant speed. In the diagram this is drawn acting up the plane. If the machine has to be pulled down, P^\ding{217} will be negative. Resolve the force mg^\ding{217} into its components along the plane and at right angles to the plane. The forces are in equilibrium since there is no acceleration taking place. Hence, by Newton's Second Law, N = mg cos \textthetaand mg sin \texttheta = P + \muN = P + \mu mg cos \texttheta. P = mg (sin \texttheta - \mu cos \texttheta). Therefore P = mg(sin \texttheta - \mu cos \texttheta). But sin \texttheta = 4/8 = (1/2), as shown in figure (a) \therefore\texttheta = 30\textdegree. \thereforeP = 350 lb [(1/2) - 0.5 (\surd3/2)] = 350 × 0.067 lb = 23.45 lb. The machine must be held back with a force of this magnitude. During the loading process, the forces acting are those shown in figure (b). Compared with the previous case, (1) and (2) are the same as before, (3) is of the same magnitude but, since it still acts against the motion, its direction is now reversed; (4) is replaced by the force P'^\ding{217} supplied by the loaders. There is no tendency to move at right angles to the plane. Thus N = mg cos \texttheta. The net force up the plane is P' - mg sin \texttheta - \muN = P' - mg(sin \texttheta + \mu cos \texttheta) = 330 lb - 350 [(1/2) + 0.5 (\surd3/2)] lb = (330 - 326.55)lb = 3.45 lb. This force is acting on a mass of 350/32 slugs, and will produce an acceleration a = 3.45/(350/32) ft/s^2 as a result of Newton's Second Law. The velocity after the machine has traveled 8 ft. from rest is thus given by the kinematics equations for constant acceleration. In this case, we use the equation v^2 = v^2_0 + 2a(x - x_0) where x - x_0 is the distance travelled along the plane, a is the machine's acceleration parallel to the plane, and v_0 is its initial velocity. Since v_0 = 0, v^2 = 2 × [(32 × 3.45 ft/s^2)/(350)] × 8 ft orv = 2.25 ft/s. The kinetic energy at that time is (1/2) mv^2 = (1/2) × (350/32) slugs × 2 × [(32 × 3.45)/(350)] × 8 ft^2 /s^2 = 27.6 ft-lb. Or, alternatively, the work-energy theorem tells us the kinetic energy is the net force up the plane times 8 ft = 3.45 lb × 8 ft = 27.6 ft\bulletlb. The potential energy is Wh = 350 lb × 4 ft = 1400 ft-lb. The work done in overcoming friction is F^\ding{217} \textbullet s^\ding{217} = \muN × 8 ft = 8 ft × \mumg cos \texttheta = 1212.4 ft-lb. The work done by the applied force P' is P'^\ding{217} \textbullet s^\ding{217} = 330 lb × 8 ft = 2640 ft-lb. But 27.6 + 1400 + 1212.4 = 2640. Thus the work done by the applied force equals the kinetic energy plus the potential energy gained by the machine added to the work done to overcome friction. This is merely a statement of the conservation of energy applied to this problem.

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Question:

Find thehydroniumion concentration of .1M HOAC (acetic acid) solution. Assume k_a = 1.75 x 10^-5 for acetic acid.

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Solution:

You want to represent the equilibrium constant expression for the reaction, which necessitates a balanced equation. After writing the expression, you want to express the concentrations in terms of the same variables and solve for it. Begin by writing the balanced equation for the reaction of acetic acid in water. The acid will donate a proton (H^+) to the only available base, H_2O. Thus, HOAC + H_2O \rightarrow H_3O^+ + OAC^-. [H_3O^+], the hydroniumconcentration, is the quantity you are looking for. The equilibrium constant expression measures the ratio of the concentrations of the products to the reactants, each raised to the power of their respective coefficients in the chemical equation. Thus, the constant, Ka, = {[OAC-] [H_3O^+]}/[ HOAC]. Notes H_2O is omitted, since it is considered a constant. Ka = 1.75 × 10^-5 . Equating, {[OAC-] [H_3O^+]}/[HOAC] = 1.75 × 10^-5. Let x = concentration of H_3O^+. According to the reaction, [H_3O^+] = [OAC^-], thus, x = concentration of [OAC^-], also. If the initial concentration of HOAC is .1 and X (moles/liter) of [H_3O^+] are formed, then you have (.1-x) moles/liter of HOAC left. Substituting these variables into the equilibrium constant expression, you have [x^2/(.1 - x)] = 1.75 × 10^-5 . Solving, x = [H_3O^+] = 0.0013M

Question:

A man is exposed to a pathogenic bacterium. If inflammation occurs, has the body's first line of defense been penet-rated?

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Solution:

This answer will describe the body's first two lines of defense against invading pathogenic micro-organisms. The first line of defense is basically me-chanical, involving physical barriers such as the skin and mucous membranes. The tough layer of skin forms a pro-tective barrier as long as it is intact. The sticky mucous secretions of the respiratory and digestive tract, along with other tissues, collect and hold many pathogens until they can be disposed of. Cilia, small hairlike appendages lining the epithelial cells of the body cavities, help sweep bacteria way from susceptible areas. The mechanical processes of coughing, sneezing, tear shedding, and sali-vating provide a clearing mechanism that flushes away bacteria. Both enzymes, such as lysozyme in tears, and the acidity or alkalinity of body secretions prevent bacteria from entering further into the body. The second line of defense is cellular. If pathogens succeed in penetrating the first line of defense and enter deeper tissue, they may be engulfed and destroyed by cells in the body named phagocytes. This process is called phagocytosis. Phagocytes include macrophages and leukocytes (white blood cells). Inflammation results from damage to body tissue caused by the invading pathogen. There is an accumulation of serum and leukocytes in the area, and this produces the signs of inflammation: swelling, heat, pain and redness. Although the phagocytes destroy the bacteria by ameboid engulfment, some phagocytes are themselves also destroyed after the engulfment occurs. Dead phagocytes, bacteria and serum that result from an inflammation are together termed pus, which may be colored according to the pigment of the bacteria involved in the infection. We thus see that inflammation occurs when the body's first line of defense is broken and the second line of defense, phagocytosis, takes over.

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Question:

Most bread is made from flour, water, fat, sugar and salt. Yeast is added as a leavening or "raising agent". Besides providing flavor, explain the function of salt in bread.

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Solution:

The small portion of salt in bread (0.88 - 1.18 % by weight) helps to regulate the rate of fermen-tation of the yeast. Fermentation is the breakdown of complex molecules in organic compounds. Without salt, fermentation proceeds too rapidly and the bread becomes too porous or "light". With excess amounts of salt, fermentation is too slow, which results in the bread becoming too compact or heavy. Thus, the amount of salt present is crucial to the formation of quality bread.

Question:

A cattle breeder wants to establish a pure-breeding herd of roan short-horned cattle. What could you tell him about his chances for success in such a venture?

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/Users/wenhuchen/Documents/Crawler/Biology/F25-0658.htm

Solution:

Using information obtained in the previous problem we can determine if it is possible to establish a pure-breeding line of roan cattle. Roan short-horned cattle are heterozygotes. Since they have both codominant genes (the gene for red color and the gene for white color) it is impossible for them to ever breed pure. To illustrate this, let us look at the results of a mating between a roan cow and bull. Let R represent the gene for red color and W represent the gene for white color. Red cattle have the genotype RR. White cattle have the genotype WW. Roan cattle, because their color results from equal contribution of both red and white, have the genotype RW. If we mate two roan cattle: We see that one red and one white individual will be produced for every two roan individuals. Therefore, a cattle breeder might get twice as many roan cattle than red or white, but he can never be assured of having a pure-breeding herd of roan cattle. If the breeder would settle for having large numbers of roan cattle within his herd, he could continue to mate roan cattle together, but now, whenever red or white cattle are produced, mate them to each other. Such a cross would result only in roan individuals.

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Question:

Charge is uniformly distributed over a spherical volume. Show that the electric intensity at any point inside the volume is the same as if all the charge closer to the center were concentrated there, and the rest of the charge removed.

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Solution:

Consider any sphere concentric with the charged volume and of lesser radius r than the charged volume (see figure). By symmetry, the electric intensity must be the same at all points on the surface of a sphere of radius r since the charge is uniformly distributed within the sphere. Thus, the electric field intensity must be purely radial in direction. Hence the flux of intensity from such a sphere is, by definition, \varphi_e= \int_S E^\ding{217} \textbullet dS^\ding{217} In this surface integral, E^\ding{217} is the electric field in-tensity, and dS^\ding{217} is a vector element of area for the surface S. In this case, S will be the above- mentioned sphere of radius r. Since E^\ding{217} is a function of r only, and radial in direction, we may write E^\ding{217} = E (r) ȓ where ȓ is a unit vector in the radial direction, originating at the center of the sphere shown in the figure. But dS^\ding{217} is also in the radial direction, since it is an element of surface area of the sphere of radius r. Then dS^\ding{217} = dS ȓ Using the last 2 equations, we evaluate \varphi_e: \varphi_e = \int_S E(r)ȓ \textbullet ȓ dS = \int_S E(r) dS Since E(r) is constant over any sphere of radius r, \varphi_e = E(r) \int_S dS = E(r) × 4\pir^2(1) where 4\pir^2 is the surface area of S. But by Gauss' Law, \varphi_e = Q/\epsilon_0 where Q is the net charge enclosed by the surface over which \varphi_e is evaluated (S). Combining (1) and (2) E(r) = Q/ 4\pi\epsilon_0 r^2 where Q is the total charge inside the sphere of radius r. But the same equation would have been obtained if all the charge inside a radius r were concentrated at the center and the rest removed, which proves the premise stated in the problem.

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Question:

Calculate the percentage of the total wet weight of a hepatocyte that is contributed by the cell membrane. Assume the hepatocyte to be a cube 20\mu on an edge and that the membrane is 90 \AA thick and that its specific gravity, as well as that of the entire cell,is 1.0.

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Solution:

Because the specific gravity is the same for the membrane and the rest of the cell, the weight percentage is equal to the volume percentage. If the hepatocyte is assumed to be a cube, its volume is equal to s^3, where s is length of the side. The membrane is composed of six rectangular solids. Each solid is 90 \AA by 20\mu × 20\mu and one is found on each face of the cube. The volume of the membrane is equal to the length times the product of the height and width times 6. The percentage of the total weight of the cell that is taken up by the membrane is the quotient of the volume of the membrane divided by the total volume of the cell multiplied by 100. Since 1 \AA = 10^-4 \mu Volume of cell = (20\mu)^3 = 8.0 × 10^3 \mu^3 Volume of the membrane = (9.0 × 10^-3\mu) (20\mu) (20\mu) × 6 = 21.6 \mu^3 Percent of weight taken up by the membrane = = [(21.6 \mu^3)/ (8.0 x 10^3 \mu^3)] × 100 = .27%.

Question:

A hungry chimpanzee is able to solve the problem of getting a banana that is out of reach by stacking boxes on top of one another. Explain.

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Solution:

This type of behavior is called insight learning. Considered to be the highest form of learning, it is most prevalent in primates and man. Insight is the ability to respond correctly on the first and sub-sequent trials to problems not encountered before. The animal is able to apply its prior learning experience in other situations to a new problem and solve the problem without having to use a trial-and-error process. In the situation given, the chimpanzee used insight to solve a problem it had never encountered before. Insight is the ability to incorporate concepts and prin-ciples with past experience in order to solve problems. During his survey of the problem, the chimp might mentally or visually internalize a trial-and-error process. By mentally integrating several elements ori-ginally learned separately, the chimpanzee can figure out the solution without doing the actual trial-and-error process and learning by mistake. In humans, insight learning is often referred to as reasoning. Dogs and cats demonstrate lack of insight or reasoning by being unable to perform well on detour tests. Detour tests are the simplest techniques used to investigate problem- solution capacities in animals. In these tests, an animal is blocked from a direct approach to food which he can see and smell by a barrier. In order to get to the food he must first walk away from the food and detour around the barrier (Fig. 1). Monkeys and chimpanzees are usually successful initially without having to rely on a trial and error process.

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Question:

What are plant hormones?Discuss the important effects of theauxins.

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Solution:

Plant hormones, like animal hormones, are organic substances which can produce striking effects on cell metabolism and growth even though present in extremely small amounts. They are produced primarily in actively growing tissues, especially in the apicalmeristemsof the plant. Like animal hormones, plant hormones usually exert their effects on parts of the plant body somewhat removed from the site of their production. Movement of plant hormones to target regions of the plant is made possible by the presence of phloem. Auxins may be regarded as the most important of the plant hormones, since they have the most marked effects in correlating growth and differentiation to result in the normal pattern of development. The differential distribution ofauxinin the stem of the plant as it moves down from the apex (where it is produced) causes the plant to elongate and bend toward light.Auxinfrom seeds induces the maturation of the fruit. Auxin from the tip of the stem passes down into the vascular cambium below and directs the tissue toward differentiating into secondary phloem and xylem.Auxinalso stimulates the differentiation of roots-by placing a cut stem in a dilute solution ofauxin, roots can be readily produced. The auxins , in addition, deter-mine the growth correlations of the several parts of the plant. They inhibit development of the lateral buds and promote growth of the terminal bud. Finally,auxinscontrol the shedding of leaves, flowers, fruits, and branches from the parent plant. By inhibiting the forma-tion of the abscission layer between leaf petioles and the stem or branch, theauxinsprevent the leaves from being shed.

Question:

An inductor of inductance 3 henrys and resistance 6 ohms is connected to the terminals of a battery of emf 12 volts and of negligible internal resistance, (a)Find the initial rate of increase of current in the circuit, (b) Find the rate of increase of current at the instant when the current is 1 ampere. (c) What is the instantaneous current 0.2 sec after the circuit is closed?

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Solution:

(a) Using Kirchoffs voltage law around the cir-cuit yields V = Ri(t) + L[{di (t)} / dt] or[{di (t)} / dt] = (V/L) - (R/L) i (t).(1) or[{di (t)} / dt] = (V/L) - (R/L) i (t).(1) Here, L (di/dt) is the EMF induced in the inductance due to the changing current. Equation (1) relates the rate of change in current at a time t, to the current at time t. At t = 0, the initial current is zero. Hence the initial rate of in-crease of current is (di/dt) = (V/L) = (12 V/3h) = 4 amp/sec. (b)When i = 1 amp, (di/dt) = (12 V/3h) - (6\Omega/3h) × 1 = 2 amp/sec. , (c) The current at any time can be found by solving equation (1) for i. The function i = (V/R) [1 - \epsilon^-(R/L)t ] can be shown by substitution into the differential equa-tion to satisfy both equation (1) and the initial condition i = 0 at t = 0. At t = 0.2 sec, we then have i = (V/R) [1 - \epsilon^-(Rt/L) ] = (12 V/6) [1 - \epsilon^-6×.2/3 ] = 2 [1 - \epsilon^-.4 ] amps = 2(1 - .672) amps = 0.65 amps.

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Question:

For the reaction I_2 (g) \rightleftarrows 2I(g), K = 3.76 × 10\Elzbar5at 1000\textdegreeK . Suppose you inject 1 mole of I_2 into a 2.00 liter-box at 1000\textdegree K . What will be the final equilibrium concentrations of I_2 and of I?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E09-0318.htm

Solution:

Final equilibrium concentrations can be determined from the equilibrium constant expression. For this reaction, K, the equilibrium constant, equals [ I]^2 / [I_2] . You are given this value as 3.76 × 10\Elzbar5at 1000\textdegreeK. Equating, there-fore, you obtain . 3.76 × 10\Elzbar5= [ I ]^2 / [I_2] . You inject 1 mole of I_2in a 2 liter box, which means the initial concentration is 1 mole/2 liter = 0.5M. Let x = the amount of I_2 that decomposes. Then, at equilibrium [I_2 ] = .5\Elzbar x, i.e., the initial amount minus the decomposed amount. Whatever decomposes yields I. For every mole of I_2 that decomposes, two moles of I are generated. This can be seen from the coefficients in thechecmicalequation. Thus, at equilibrium, [I] = 2x. Substituting these values in the equilibrium constant expression, you obtain (2x)^2 / (.5\Elzbar x) = 3.76 × 10\Elzbar5. Solving for x you obtain x = 2.17 × 10\Elzbar5, using the quadratic equation. Thus, the concentrations become I = 2x = 4.34 × 10\Elzbar5M and I_2 = .5\Elzbar x = .498M.

Question:

Write out complete equations for the molar free energy of formation of nitric oxide, ammonia, and water vapor. 4NH_3 (g) + 5O_2 (g) \rightarrow 4NO(g) + 6H_2O (g). Using the data from the accompanying table, determine the ∆G\textdegree for this reaction. If ∆S\textdegree = 42.94cal, what is ∆H\textdegree for the reaction? Cal-culate the value of its equilibrium constant at 298\textdegreek; at 798\textdegreek. Calculate ∆G\textdegree at 798\textdegreek. Refer to the following table Standard Heats of Formation, ∆H\textdegree, and Standard Free Energies of Formation, ∆G\textdegree, in Kcal/mole at 25\textdegreeC Energies of Formation, ∆G\textdegree, in Kcal/mole at 25\textdegreeC Substance ∆H\textdegree ∆G\textdegree NH_3 (g) NO (g) H_2O (g) - 11.0 + 21.6 - 57.8 -4.0 + 20.7 - 54.6

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Solution:

∆G may be defined as net energy (free energy) available for the production of useful work. It can predict spontaneity, maximum work obtainable, and max-imum yields in an equilibrium reaction. There exist a number of equations that allow measurement of ∆G quantitatively. You will employ these in answering some parts of this question. You commence as follows: To write the equations for molar free energy of formation of the given substances, consider their components and, with the appropriate coefficients, write a balanced equation so that only 1 mole is formed. You have,then, Ammonia:1/2 N_2 (g) + 3/2 H_2 (g) \rightarrow NH_3 (g) Nitric oxide:1/2 N_2 (g) + 1/2 O_2 (g) \rightarrow NO (g) Water vapor:H_2 (g) + 1/2 O_2 (g) \rightarrow H_2O (g) The ∆G\textdegree ' s for these are, respectively, - 4.0, + 20.7, and -54.6 kcal/mole. The ∆G\textdegree for the overall reaction = ∆G\textdegree _products - ∆G\textdegree _reactants, each multipled by its coefficients in the chemical equation. Therefore, ∆G\textdegree _overall = 4∆G\textdegree_NO (g) + 6∆G\textdegreeH2O (g)- 4∆G\textdegreeNH3 (g)- 5∆G\textdegreeO2 (g)= 4 (20.7) + 6 (- 54.6) - 4 (- 4.0) - 5 (0) = - 228.8 kcal/mole. [Note: by definition the ∆G\textdegree's of all elements are zero, which explains why ∆ G\textdegreeO2 (g)= 0. ] To find ∆H\textdegree, remember that ∆H\textdegree = ∆G\textdegree + T∆S\textdegree, where ∆H\textdegree = change in enthalpy, T = tempera-ture in kelvin (Celsius plus 273\textdegree) and ∆S\textdegree = change in entropy; T = 25\textdegreeC or 298\textdegreeK since all ∆G\textdegree's are measured at 25\textdegreeC. By substituting, you find ∆H\textdegree = -228.8 + (298) × (42.94) × (1kcal/1000cal) = - 228.8 + 12.8 = 216.0 kcal/mole. (The 1 kcal/1000 cal is a conversion factor.) To find the equilibrium constant at a given temperature, remember that ∆G\textdegree = -2.303 RT log k, where R = universal gas constant, T = temperature in kelvin ( Celsius plus 273\textdegree) and k = the equilibrium constant. Thus, at 298\textdegreek, you have -228.8 = -2.303(1.987) × (298) log k. Solving, k = antilog [(- ∆G\textdegree)/(2.303 RT)] = antilog [{(228.8) (1000)} / {(2.303) (1.987) (298)}] = antilog 167.8 = 6 101 6 7. To find k at 798\textdegreek, use the equation log (k_2 /k_1) = [(- ∆H\textdegree)/(2.303 R)][(1/T_2 ) - (1/T_1 )] Substituting, log [(k7 9 8)/(k2 9 8)] = [(- ∆H\textdegree)/(2.303 R)] [(1/798) - (1/298 )] = - [{(-216.0) (1000 cal/kcal)}/{(2.303) (1.987)}]. Rewriting and solving, k7 9 8= (k2 9 8) antilog (-99.2) = (6 × 101 6 7) (6 × 10-1 0 0) = 4 × 106 8 To find ∆G\textdegree at 798\textdegreek, use the equation ∆G\textdegree = -2.303RT log k: ∆G\textdegree = -2.303RT log k78 9. ∆G\textdegree = (-2.303) (1.987) (798) log 4 × 106 8= -2.5 × 10^5 cal/mole = - 250 kcal/mole.

Question:

a) Liquid HCI is anonelectrolyte, but aqueous HCI is a strong electrolyteb) Liquid HCN is anonelectrolyteas is aqueous HCN.

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Solution:

An electrolyte is a substance which exists as ions. Both liquid HCN and HCI are not electrolytes because they consist of neutral atoms. No ions exist. The case with an aqueous solution requires a little more investigation. In water, HCI is ionized via the reaction HCI + H_2O \rightleftarrows H_3O^+ + Cl^-. Thus, you havehydroniumions and chloride ions in solution. Thus, aqueous HCI is an electrolyte. HCN also dissociate when placed in water. But, to what extent? H_3O^+ and CN^- ions are only to a very slight extent produced, since HCN's dissociation constant is so small. Thus, while HCN may be called an electrolyte, it is an extremely weak one.

Question:

During a moon landing, one of the experiments performed was the measurement of the intensity of solar wind. As a collector, an aluminum strip of about 3000 cm^2 area was used. It was found that in 100 min, a mass of 3.0 × 10-^10 g of H atoms was collected (by the sticking of H atoms to the strip) . What was the intensity of the solar wind (in numbers of atoms per cm^2 per second)?

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Solution:

The intensity of the solar wind is the number of H atoms striking the unit area in unit time, intensity = (number of atoms intensity) / (area × time) . We must determine the number of atoms. This is accomplished by converting mass to moles and then multiplying by Avogadro's number. The number of moles is given by moles = mass of H atoms/ atomic weight of H = (3.0 × 10^-10 g) / (1.0 g/mole) =3.0 × 10^-10 mole. Since there are an Avogadro's number of atoms in one mole, the number of atoms in 3.0 × 10^-10 mole is number of atoms = moles × Avogadro's number = 3.0 × 10^-10 mole × 6 × 10^23atoms/mole = 18 × 10^13 atoms. The intensity of the solar wind is then intensity = (number of atoms) / (area × time) = (18 × 10^13 atoms) / (3000cm^2 × 100 min) = (18 × 10^13 atoms) / (3000cm^2 × 100 min × 60 sec/min) = 1 × 10^7 atoms / cm^2 - sec.

Question:

You are handed a deck of punched cards with each card bearing two dates in the twentieth century. The format for June 13, 1979, for example, is 061379. Design a generalized flowchart with instructions to output the two dates along with the number of days elapsed between them. It should not matter which of the puncbed dates comes earlier. Also, remember to account for leap years and errors in the data. A card containing all zeros signifies the end.

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Solution:

The best way to attack this problem is by writing a series of flowcharts, each being more detailed than the previous one. The concept of refinement is used here, because there are many subtasks that must be accomplished throughout. The first flowchart to be written is only a skeleton of the entire algorithm and is shown in figure 1. We Include annotation nodes where the subsequent flowcharts should fit in. Let us call the six input items A, B, C, D, E, and F. Assume that the two dates are June 1, 1978 and May 15, 1979. Then, A, B, and C are 06, 01 and 78 respectively and D, E, and F are 05,15 and 79 respectively. To find intervening days in a single month, control passes to the box containing T \leftarrow T + (E - B). To find longer periods of time, you must write more complicated instructions. First, let us examine node 7. This section evaluates the number of days in the month given. February is set aside until we have to adjust for leap years. The flowchart for node 7 is given in figure 2. This section checks the data for all types of errors. The variable KMT is used in the error messages to indicate that the KNTth card is wrong. FLAG remains at zero until both pairs of dates, (A and B, D and E) are checked on each card. X and Y store the dates temporarily while error testing is done. The following relations are checked; if there is an error, a message is output. 1) No month code can be greater than 12. 2) If the month code is 4, 6, 9, or 11 (April, June, September, or November), a maximum of 30 days is allowed. 3) If the month is neither February nor one of the above, it must have a maximum of 31 days. The flag must be raised to 1 before control passes to the next card. When the next pair of dates is examined through the flowchart, KNT is increased by 1, and the main program continues. Node 3 rearranges the dates if the first group ABC follows DEF on the calendar. Year, month, and day are evaluated in that order, and appropriate action is taken. The flowchart for node 3 is given in figure 3. Node 5 calculates the number of days if the dates are not in the same year. Leap years are taken into account by adding 366 days to the elapsed time. Further refinements for the elapsed months and days will come in node 6. The flowchart for node 5 is shown in figure 4. To explain the next node, we must first introduce the following data table. It gives the month codes, the elapsed days in a year starting with the first of each month(s), the days remaining in a year starting with the last day of each month (W), and the number of days in each month (Z). For leap years, the entries in the table are adjusted in the program. The table is shown in figure 5 and the flowchart in figure 6. Month S W Z 1 0 334 31 2 31 306 28 3 59 275 31 4 90 245 30 5 120 214 31 6 151 184 30 7 181 153 31 8 212 122 31 9 243 92 30 10 273 61 31 11 304 31 30 12 334 0 31 fig. 5 Finally, if the dates are in the same year but not in the same month, we turn to node 4. The data table is used in this section to adjust the dates accordingly. The flowchart of node 4 is given in figure 7.

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Question:

Differentiate between direct and indirect respiration, and between external and internal respiration.

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/Users/wenhuchen/Documents/Crawler/Biology/F16-0392.htm

Solution:

The phenomenon in which the cells of an organism exchange oxygen and carbon dioxide directly with the surrounding environment is termed direct respiration. This form of respiration is a fairly simple process and occurs in small, aquatic animals such as paramecia or hydras. In these animals, dissolved oxygen from the surrounding water diffuses into the cells, while carbon dioxide within the cell diffuses out; no special respiratory system is needed. With the evolution of animals into larger and more complex forms, it became impossible for each cell to exchange gases directly with the external environment. Consequently, it became necessary for these organisms to have a specialized organ system that would function in gas exchange with the environment. This structure must be thin-walled, and its membrane must be differen-tially permeable. In addition, the membrane must be kept moist so that oxygen and carbon dioxide could dissolve in it, and it must have a good blood supply. The process of respiration employing this organ system is called indirect respiration. For indirect respiration, the lower vertebrates developed gills and the higher vertebrates developed lungs. During indirect respiration, gas exchange between the body cells and the environment may be categorized into two phases: an external and an internal phase. External respiration is the exchange of gases by dif fusion that occurs between the lungs and the bloodstream. Oxygen passes from the lungs to the blood and carbon di-oxide passes from the blood to the lungs. Internal respi-ration takes place throughout the body. In the latter, there is an exchange of gases between the blood and other tissues of the body, with oxygen passing from the blood to the tissue cells and carbon dioxide passing from the cells to the blood. This phase, along with the external phase, relies on the movement of gases from a region of higher concentration to one of lower concentration.

Question:

What will be the maximum kinetic energy of the photoelectrons ejected from magnesium (for which the work function \varphi = 3.7eV) when irradiated by ultraviolet light of frequency 1.5 × 10^15 sec^-1 ?

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/Users/wenhuchen/Documents/Crawler/Physics/D33-0973.htm

Solution:

The energy of a photon with frequency 1.5 × 10^15 sec^-1 is \epsilon =hv= (6.6 × 10^-27 erg-sec) × (1.5 × 10^15 sec^-1) = 9.9 × 10^-2 erg × [1eV / (1.6 × 10^-12 erg)] = 6.2eV The maximum kinetic energy of a photoelectron is obtained from what-ever energy is left over after the collision of a photon and surface electron has occurred. Since the electron loses energy equal in amount to \varphi while leaving the metal, we may write KE =hv- \varphi = 6.2eV- 3.7eV = 2.5eV

Question:

Develop a FORTRAN subprogram to evaluate the integrationof f(x)dxbetween the limits of A and B using GAUSSIANquadrature, which is expressible as ^B\int_Af(x)dx= [(B - A) / 2]^N\sum_i_=1w_if [{(B - A)t_i+ (B + A)} / 2] wherew1, w2...w_Nare the weighting coefficients and t1, t2 ,...,t_Nare the roots of the Legendre polynomial P_N(t) = 0

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Solution:

We need the result from elementary numerical analysis that the weightsw_Kcan be expressed in the form w_K= 1\int_-1 L_K(x)dx= [1 / {(P_n+1)(x_K)}] 1\int_-1 [{(P_n+1)(x)dx} / (x -x_K)] whereL_K (x) is a Lagrange polynomial andP_n(x) is a Legendre polynomial. Our program will let the value of N range from 3 to 6. The compu-tation startswith N = 3. The program will compare the result based on N = 3 with thatbased on N = 4. The results must satisfy the criterion \epsilon\geq [{(A_n+1) -A_n} / A_n] whereA_n+1 is the answer based on N + 1 points, and A_n is the answer basedon N points, and \epsilon is taken to be 10-4 . If the re-sult fails to pass the abovetest, the value of N will be increased by one. The maximum value of N is set to be 6. SUBROUTINE GAUSS (A, B, X, F, KOUNT) CINTEGRATION OF F(X).DX BY GAUSSIAN QUADRATURE CBETWEEN LIMITS OF A AND B CNOMENCLATURE CF = F(X) UNDER INTEGRAL SIGN CKOUNT = INTEGER USED TO CONTROL METHOD COF EXECUTION CANS = ANSWER TO INTEGRATION DIMENSIONW(4,6), T(4,6) IF (KOUNT) 8, 10, 8 8GO TO 30 10ANS = 1.0 IP0INT = 3 EPS = 10.E - 05 CSTORE WEIGHT COEFF AND LEGENDRE ROOTS DO 2 I = 1,4 IP2 =1 + 2 2READ 1, (W(I,K), K = 1,IP2), (T(I,K), K = I,IP2) 1FORMAT (5F15.10) CCHANGE INTEGRATION LIMITS TO (-1 TO +1). CEVALUATE COEFF OF NEW FUNCTION C = (B - A)/2. 18IP0INT = IPOINT + 1 TEMP = ANS ANS = 0 IPM2 = IPOINT - 2 KOUNT = 1 CEVALUATE NEW VARIABLES WHICH ARE EXPRESSED CIN TERMS OF LEGENDRE ROOT. 20X = C\textasteriskcenteredT(IPM2,KOUNT) + (B + A)/2 . RETURN CCARRY OUT INTEGRATION BY CALCULATING CANS = C\textasteriskcentered(W1\textasteriskcenteredF(X1) + W2\textasteriskcenteredF(X2)+...). 30ANS = ANS + C\textasteriskcenteredW(IPM2,KOUNT)\textasteriskcentered F KOUNT = KOUNT + 1 IF (KOUNT - IPOINT) 20,20,40 40IF (IPOINT - 3) 18, 18, 50 CNEXT DETERMINE WHETHER THE DEVIATION COF ANSWER IS WITHIN THE LIMIT CIF NOT,TAKE1 MORE INTEGRATION POINT 50DELT =ABSF(ANS - TEMP) RATIO = DELT/TEMP 7IF (RATIO-EPS) 70, 70, 80 CPUNCH OUT ANS IF DEVIATION WITHIN LIMIT 70PUNCH 72, IPOINT, ANS 72FORMAT (///5X, 16H BY CONVERGENCE, ,I2, 24H 1 POINT GAUSS.QUADRATURE, 15H GIVES ANSWER= , 2 E14.8//) KOUNT = 7 RETURN 80IF (IPOINT - 6) 18, 100, 100 100PUNCH 102, IPOINT, ANS CANSWER PUNCHED OUT AFTER 6 POINT CGAUSSIAN INTEGRATION IS STILL NOT CWITHIN THE LIMIT. 102FORMAT (///5X, 22H BY LIMITS OF PROGRAM, ,I2, 1 14H POINT GAUSS., 23H QUADRATURE GIVES ANS. =, 2 E14.8//) KOUNT = 7 RETURN CDATA FOR WEIGHTING COEFFICIENTS CAND LEGENDRE ROOTS END

Question:

What contributions did Darwin make to the theoryof evolution?

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Solution:

Darwin made a twofold contribution to the study of evolution. He presenteda mass of detailed evidence and convincing argument to prove thatorganic evolution actually occurred, and he devised the theory of naturalselection, to explain how organic evolution works. Darwin spent a major part of his life studying the animals, plants, andgeologic formations of coasts and islands, at the same time making extensivecollections and notes. When he was studying the native inhabitantsof the Galapagos Islands, he was fascinated by the diversity of thegiant tortoises and the finches that lived on each of the islands. The diversitywas gradual and continual, and could not be explained by the theoryof special creation, which stated that all living things are periodically destroyedand recreated anew by special, unknown acts of creation. As Darwin mused over hisob-servations,he was led to reject this commonly heldtheory and seek an alternative explanation for his ob-servations. Then Darwin came up with the idea of natural selection. He believedthat the process of evolution occurred largely as a result of the selectionof traits by constantly operating natural forces, such as wind, flood; heat, cold and so forth. Since a larger number of individuals are bornthan can survive, there is a struggle for survival, necessitating a competitionfor food and space. Those individuals with characteristics that betterequip them to survive in a given environment will be favored over othersthat are not as well adapted. The surviving individuals will give rise tothe next generation, trans-mitting the environmentally favored traits to thedescendents. Since the environment is continually changing, the traits tobe selected also change. Therefore, the operation of natural selection overmany years could lead ultimately to the development of descendants thatare quite different from their ancestors - different enough to emerge asa new species. The formation, extinction, and modification of a species, therefore, are regulated by the process of natural selection. This principle, accordingto Darwin, is the major governing force of evo-lution.

Question:

Compare photosyntheticphosphorylationand oxidative phosphorylation. What are the roles offerredoxinand plastoquinone?

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Solution:

Photosynthetic and oxidativephosphorylationare both ATP - synthesizing processes. They differ, however, in their source of energy, site of reaction, and in the nature of their electron donor and acceptor. Photosyntheticphosphorylation, as its name suggests, utilizes light energy to synthesize ATP from ADP and inorganic phosphate, light energy striking a green plant is absorbed by the chlorophyll. The excited chlorophyll molecule ejects a high energy electron which passes down an electron transport chain. During this passage, the electron returns to its original energy level, producing ATP molecules in the process. Photosyntheticphosphorylationuses neither oxygen nor organic substrates. The chlorophyll molecule, serves as both the electron donor and acceptor (though in non - cyclicphotophosphorylation, additional electrons are supplied by water to reduce NADP^+). The excited chlorophyll molecule passes an energetic electron toferrodoxin, and having lost an electron, is now able to accept another. The end result is that the energy from sunlight is converted to chemical energy stored in ATP. The entire process occurs in the chloroplasts. Oxidativephosphorylationtakes place in the mito-chondria. The process is incapable of utilizing sunlight as its energy source. Rather, it uses the energy released when certain reduced substrates are oxidized. The electrons released are passed in a series ofredox(oxidation / reduction) reactions down the electron transport chain, a system of electron acceptors of decreasing re-duction potential. The energy of oxidation is captured in the form of an energy - rich phosphate - ester bond in ATP. In oxidativephosphorylation, the ultimate electron ac-ceptor is oxygen and the primary electron donor is a sug-ar or some other organic substance. Ferredoxinis a component of the electron carrier chain in the chloroplasts of plant cells. It is an iron - containing protein that functions in capturing energy - rich electrons from excited chlorophyll molecules and in passing them to the next electron acceptor of the chain.Plastoquinone is another electron carrier in the chloroplast; it can receive electrons from eitherferredoxinorcytochromeand transfer them to thecytochromes.

Question:

What is the frequency of a 2-cm sound wave in sea water? What is the frequency of a 2-cm sound wave in sea water? The velocity of sound in sea water is \upsilon = 1.53 × 10^5 cm/sec.

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Solution:

The 2-cm is the wavelength \lambda of the sound wave. The frequency is given by v = \upsilon/\lambda = (1.53 × 10^5 cm/sec)/(2 cm) 1.53 × 10^5 cm/sec)/(2 cm) = 7.6 × 10^4 Hz = 76 kHz which is an ultrasonic wave (that is, above the human audible range). A 2-cm sound wave in air would be audible.

Question:

What is meant by the equilibrium of a chemical reaction? What factors influence the position of the equilibrium?

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/Users/wenhuchen/Documents/Crawler/Biology/F01-0009.htm

Solution:

Chemical equilibrium is the state in which two opposing reactions (called forward and reverse) are proceeding at the same rate. Not all chemical reactions are reversible. Only if the products of a chemical reaction (the forward one) can recombine to form the original substances will it be possible to have the two opposing reactions necessary to reach a state of equilib-rium. This state is reached when the rate of the forward reaction equals the rate of the reverse reaction. For example, the union of hydrogen and nitrogen to form ammonia (equation a) and the reverse or opposing reaction, which is the decomposition of ammonia to nitrogen and hydrogen (equation b) can be rewritten as a reaction at chemical equilibrium (equation c): (a) 3H_2 + N_2R_a\rightarrow2NH_3where R_a = rate of reaction a (b) 2NH_3R_b\rightarrow3H_2 + N_2whereR_b= rate of reaction b (c) 3H + N_2(R_b\rightarrowR_a\rightarrow)2NH_3where Ra=R_b. It should be noted that the concentrations of reactants and products in reaction (c) or in any reaction at equilibrium may vary greatly. The amounts of substances present at equilibrium is dependent on the nature of the reactants, and on the pressure and the temperature at which equilibrium is established. The presence or absence of a catalyst has no influence on the relative amounts of products or reactamts present at equilibrium. The nature of the reactants affects the equilibrium concentrations. This is due to the difference in strengths of bonding forces. Consider the general equations: (d) A_2 + B_2 \rightarrow 2AB (e) 2AB \rightarrow A_2 + B_2 If the energy needed to bond A_2 and B_2 in reaction (d) is greater that the energy needed to break the A-B bond in reaction (e), then reaction (e) has a greater tendency to occur than reaction (d). In this case, the equilibrium would favor one side: A_2 + B_2 \rightleftharpoons 2AB The effect of pressure on an equilibrium mixture is summed up best by LeChatelier'sprinciple which states that if the pressure on a system at equilibrium is changed, there is a tendency to diminish the change in pressure. The reaction that produces the fewer number of molecules and tends to lower the pressure is the reaction favored by the application of high pressure to the system. Thus if the equilibrium mixture: N_2 + 3H_2 \rightleftarrows 2NH_3 (1 molecule + 3 molecules) = 4 molecules2 molecules is established at a relatively low pressure, the equilibrium mixture contains a smaller proportion of ammonia and a larger proportion of nitrogen and hydrogen than it would at a high pressure. It should be realized that this only applies to a gaseous system. Increasing the pressure of a liquid or a solid would not appreciably increase the concentrations, and thus would have little effect on the equilibrium concentrations. A change in pressure has no effect on a gaseous mixture at equilibrium if neither the forward nor the reverse reaction diminishes the change in pressure; for example: H2+ Cl_2 \rightleftarrows 2HCl (1 molecule + 1 molecule) = 2 molecules2 molecules The influence of temperature is similar to that of pressure. A rise in temperature increases the rate of all reactions. Thus, both the rates of the forward and the backward reactions would be increased. But, one will probably be increased more than the other. To determine which reaction's rate will be increased and which side will be favored in equilibrium we can use a modified version of LeChatelier' s principle using temperature instead of pressure as the stress. If the temperature is increased, an endothermic reaction will be favored over an exothermic reaction because it will take in heat and thus, lessen the amount of heat present. An exothermic reaction would be favored by a decrease in temperature, because this reaction liberates heat to the surroundings and does so more readily if the surroundings are at a relatively low temperature.

Question:

Describe the life cycle of Selaginella.

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Solution:

Selaginella is a member of the subphylum Lycophyta. Its dominant sporophyte generation generally consists of a branched, prostrate stem with short upright branches, normally only a few inches high. Both horizon-tal and upright stems are sheathed with small leaves in four longitudinal rows or ranks. At the ends of the upright branches, reproductive leaves, called sporophylls group to form cones or strobili. Sporangia grow in or near the axils of the sporophylls. Two types of sporangia are present in Selaginella, the microsporangia, which produce small microspores; and the megasporangia, which produce large megaspores. A single strobilus usually contains both types of sporangia and thus produces both types of spores. In a developing megasporangium, all but one of the spore mother cells degenerate. This remaining cell undergoes meiosis and forms four megaspores. Each megaspore is capable of giving rise to a female game-tophyte. In the microsporangium, only a few spore mother cells degenerate. Each of the 250 or so cells that remain gives rise to four microspores by meiosis. All microspores are capable of developing into the male microgametophyte. Repeated mitotic cell divisions of the megaspore result in the formation of the female gametophyte within the megaspore cell wall. The megaspore may be shed from the strobilus at any stage in the development of the megagametophyte, or may remain in the strobilus well after fertilization until the completion of early embry-onic development. The material stored in the megaspore is the major source of food for the female gametophyte and the developing embryo. The gametophyte increases in size, eventually rupturing the megaspore wall, and the small, colorless megagametophyte tissue protrudes, exposing the archegonia. The haploid microspore also undergoes mitosis and forms two cells. The smaller cell, the prothallial cell, remains vegetative and does not divide further. The other cell, by repeated divisions, develops into a gametophyte, or antheridium, composed of a jacket of sterile cells enclosing either 128 or 256 biflagellated sperm, the gametophyte remains in the microspore cell, which is released from the microsporangia and may drop near a megaspore either on the ground or on a strobilus. When wet by dew or rain, the microspore wall splits, and the sperm within are free to swim to the megagametophyte and fertilize the haploid egg, thus initiating the diploid sporophyte generation. Of the two cells formed by the first division of the zygote, only one develops into an embryo. The other grows into an elongated structure, the suspensor, which pushes the developing embryo into gametophytic tissue for food supply. The young Selaginella soon acquires its own food-making mechanism and becomes an independent sporophyte plant. In some species, as in the seed plants, the embryo is produced in the female gametophyte while it is still within the sporophyte.

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Question:

What evidence supports the theory that vascular plants evolved directly from green algae, and not from bryophytes?

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Solution:

At one time it was believed that the vascular plants evolved from a less advanced bryophyte, such as the liverworts or hornworts. But although there is fossil evidence of true vascular plants in the Silurian period some 360 million years ago, the first evidence of the bryophytes does not appear until the Pennsylvanian period, which began about 100 million years later. The fact that the bryophytes first appear in a later period than the vascular plants makes it unlikely that the vascular plants could have evolved from bryophytes. Since green algae contain pigments (chlorophylls), principal food reserves (starch) similar to those of the higher plants, and since fossil evidence shows they invaded the land some 140 million years earlier than the first group of vascular plants, botanists are now inclined to believe that ancient green algae were the common ancestors of both higher nonvascular and vascular plants. Hence bryophytes and vascular plants are now believed to have derived from green algae and to have diverged in their evolutionary processes long ago .

Question:

Why are the stem reptiles especially important in evolution?

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Solution:

The stem reptiles (cotylosaurs) were the first reptiles to arise. They arean important group of organisms because they were the ancestral formof a variety of important animals, including the mammals. The stem reptilesoriginated in the Paleozoic Era and gave rise directly to the many formsof reptiles that flourished in the Mesozoic Era - the turtles,tuatera, lizards, and snakes. These forms have survived a long evolutionary historyto become the modern reptiles. The stem reptiles also gave rise to severalother lineages, including thetherapsidsthat ultimately led to the mammals, and thethecodontsthat in turn gave rise to crocodiles, birds, flyingreptiles called pterosaurs, and the great assemblage of reptiles calleddinosaurs.

Question:

Show that a two input NOR gate can be achieved with two transistor inverters in parallel as shown in figure 1.

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Solution:

A transistor operates as a switch. When the base voltage is high, > .7V base to emitter, the collector-emitter junction forms a closed switch and when the base voltage is low, < .7V base to emitter, the collector-emitter junction forms an open switch. Hence, when Vin 1, and Vin 2are low, both switches are open, preventing current from flowing through T_1 or T_2, and the output voltage is high (\approx 5V). When V_in 1 (or Vin 2) is high and other input is low, then T1 (T2) is closed. In this case the output voltage is low, because current flows through one of the switches, connecting V_out- to ground. When Vin 1and Vin 2are high, switches T1 and T2 close, connecting V_out to ground. This circuit behavior can be tabulated as follows: Vin 1 Vin 2 V_out L L H L H L H L L H H L Using positive logic, this table is seen to be the truth table for a NOR gate.

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Question:

A metal disc is suspended from its center by a torsion bar, as shown in the figure. Find the period of oscilla-tion for small angular displacements from equilibrium. Torsion bars have the property of exerting a torque which is proportional to the angular displacement and oppositely directed.

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Solution:

The angular displacement \Phi is opposed by a torque \tau = -\cyrchar\cyrk \Phi where \cyrchar\cyrk is the torsion constant, and corresponds to the force constant of a tensile spring. The angular acceler-ation \alpha and the moment inertia of the disc are related to \tau by \tau = I \alpha = I[(d^2\Phi)/(dt^2)] henceI [(d^2\Phi)/(dt^2)] =\Elzbar\cyrchar\cyrk \Phi. The resulting differential equation for \Phi is [(d^2\Phi)/(dt^2)] + \Omega^2 \Phi = 0 where \Omega = \surd(\cyrchar\cyrk/I). This equation defines a simple harmonic motion, given by\Phi(t) = \Phi_0 sin \Omegat, \Phi_0 is themaximum angular displacement and we assumed that \Phi = 0 at t = 0. The period T is T = 2\pi/\Omega = 2\pi\surd(I/\cyrchar\cyrk).

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Question:

For principal P, a bank pays R interest compounded annually . Define a function in FORTRAN which calculates the deposit at the end of n years. Use the formula NEW DEPOSIT = P ((1+R)^n- 1)/R

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Solution:

Let P be the original principal. After 1 year, the new principal will be P_1 = P (1+R). After two years, it will be P_2 = P1(1+R) = P (1+R) (1+R) =P(1+R) ^2. After n years, it will be P_n = P (1+R)^n. By NEW DEPOSIT is understood the amount in the bank after n years less the original principal. In FORTRAN we can write FUNCTION ND (P, R, N) ND = P\textasteriskcentered(((1.+R) \textasteriskcentered\textasteriskcenteredN) - 1.)/R RETURN END [Note: Remember to declare variables as INTEGER or REAL in the main program.]

Question:

Based upon the followingthermochemicaldata, show that ozone, O_3, is considerably more stable than a cyclic structure would suggest. The enthalpy for the O-O bond is approximately 33 Kcal / mole. 1(1/2) O_2\ding{217} O_3,∆H_formation= + 34.5 kcal O_2\ding{217} 2O∆H_dissociation= + 119 kcal.

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Solution:

The bond energy is the amount of energy needed to separate atoms joined in a chemical bond. The greater the bond energy the more stable the bond. The stability of a molecule is proportional to the sum of the bond energies of the bonds in the molecule. If O_3 assumes a cyclic conformation, there are 3 O-O bonds present. Thus, the stability of this structure is proportional to 3 times the bond energy of an O-O bond, 33 kcal / mole. stability of the cyclic structure \alpha 3 × 33Kcal / mole or 99 kcal / mole One can calculate the actual stability of O_3 by using the data supplied in the problem. The overall re-action for the dissociation of O_3 is (i)O_3 \ding{217} 3O. One can derive this equation from the two given in the problem. The reverse of the equation and the ∆ dissociation for the formation of O_3 is (ii)O_3 \ding{217} 1(1/2) O_2∆H_diss= - 34.5 kcal / mole If one multiplies the second equation by 3 / 2 it and its ∆H becomes: (iii)3/2 O_2 \ding{217} 3O∆H_diss= 178.5 kcal / mole One obtains the overall equation (i) by adding to-gether equations (ii) and (iii). One finds the overall ∆H by adding together the ∆H's of these reactions. (ii)O_3\ding{217} 3/2 O_2∆H= - 34 kcal / mole (iii)+ 3/2 O_2 \ding{217} 3O+ ∆H= 178.5 kcal / mole O_3 + 3/2 O2\ding{217} (3/2)O_2 + 3O144.5 kcal / mole OrO_3\ding{217} 3O The actual stability of O_3 is, therefore, proportional to 144.5 Kcal / mole. This is greater than that which would be expected if O_3 had a cyclic structure.

Question:

The figure shows a current of 25 amp in a wire 30 cm long and at an angle of 60\textdegree to a magnetic field of flux density 8.0 × 10^-4 weber/m^2 . What are the magnitude and direction of the force on this wire?

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Solution:

We must find the magnetic force on a wire placed in a magnetic field. The force on a differential element of the current carrying wire, dl^\ding{217}, is dF^\ding{217} = I dl^\ding{217} × B^\ding{217}(1) where I is the current in the wire, dl^\ding{217} is a vector tangent to the wire in the direction of I, and B^\ding{217} is the magnetic induction. The net force on the wire is found by integrating dF^\ding{217} over the length of the wire (this amounts to adding the contributions of each current element dl^\ding{217} to the net force). Hence, F^\ding{217} = ^l\int_oI dl^\ding{217} × B^\ding{217}(2) where l is the length of the wire. Because I is constant and B^\ding{217} is uniform (independent of position) we may rewrite (2) as F^\ding{217} = I [ ^l\int_o dl^\ding{217} ] × B^\ding{217}or F^\ding{217} = I l^\ding{217} × B^\ding{217}(3) where l^\ding{217} is a vector whose magnitude is the length of the wire, and whose direction is the direction of current flow in the wire. Now, in general, the magnitude of a cross product (such as \vert l^\ding{217} × B\ding{217}\vert) is defined as: \vert l^\ding{217} × B\ding{217}\vert = lB sin \texttheta where l is the magnitude of l^\ding{217}, B is the magnitude of B^\ding{217}, and \texttheta is the angle between the directions of l^\ding{217} and B^\ding{217}. Because we are asked to find the magnitude of the force, we have: F = IlB sin \texttheta(4) Substituting the given values into (4), we obtain: F = (25 amp) (.3m) [8 × 10^-4 (weber/m^2)] (sin 60 o ) orF = 5.2 × 10^-3Newtons The wire is pushed away from the reader, from the stronger toward the weaker field.

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Question:

Describe, by means of block diagrams, how an S-C type flip-flop can be derived using NOR-gates and delay-lines, and then obtain an excitation equation for the S-C flip- flop.

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Solution:

The S-C (i.e., set-clear) type flip-flop can be obtained using NOR-gates and Delay-lines as in fig. 1. From figure 1: A = (y' + C)' and, B = (y + S)'. Also, after a time-delay ∆, when steady state is reached, the next states y\rightarrowand y'\rightarrowof the outputs y and y' respec-tively, are given by the equations: \rightarrow \rightarrow y\rightarrow= A = (y' + C)', \rightarrow and, y'\rightarrow= B = (y + S)'. \rightarrow To show that the circuit is indeed a flip-flop we substi-tute different combinations of values of S and C into the next state equations for y\rightarrowand \rightarrow y'\rightarrow. \rightarrow i) Put S = 0 and C = 0 \thereforey\rightarrow= (y' + 0)' = (y')' = y \thereforey\rightarrow= (y' + 0)' = (y')' = y \rightarrow and, y'\rightarrow= (y + 0)' = (y)' = y' \rightarrow Thus, there is no change in the circuit, ii) Put S = 1 and C = 0 \therefore y\rightarrow=(y' + 0)' = (y')' = y \rightarrow and, y'\rightarrow= (y + 1)' = (1)' = y'\rightarrow= (y + 1)' = (1)' = \rightarrow 0 And, as y'\rightarrow= 0 therefore y\rightarrowbecomes = 1. \rightarrow \rightarrow Thus, the flip-flop sets iii) Put S = 0 and C = 1 \therefore y\rightarrow= (y' + 1)' = (1)' = 0 \therefore y\rightarrow= (y' + 1)' = (1)' = 0 \rightarrow and, and, y'\rightarrow= (y + 0)' = y' = 1 \rightarrow Thus, the flip-flop clears, iv) The condition S = 1 and C = 1 is not allowed because it leads to ambiguity in the outputs y\rightarrowand y'\rightarrow: \rightarrow \rightarrow y\rightarrow= (y' + 1)' = (1)' = 0 y\rightarrow= (y' + 1)' = (1)' = 0 \rightarrow and, y'\rightarrow= (y + 1)' = (1)' = 0 \rightarrow But both y\rightarrowand y'\rightarrowcannot be equal to zero because y\rightarrowand y'\rightarrowmust be \rightarrow \rightarrow \rightarrow \rightarrow complements of each other. Hence, for an S-C type flip-flop, there must always be a constraint on the input, i.e., S\bulletC = 0, so that both S and C cannot be = 1. The above conditions can be plotted on a Karnaugh map. The constraint condition (S=1, C=1) gives rise to 'don't-care' states "d", because they will never occur in actual working of the flip-flop. Hence, the Karnaugh map is shown in figure 2. This Karnaugh Map now allows us to write down the excita-tion equation as follows: y\rightarrow= S + C'y \rightarrow = S × y' + C'y (equivalent statement) \thereforey\rightarrow= Sy' + C'y \rightarrow

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Question:

The nucleus of an atom has a charge +2e, where e is the electronic charge. Find the electric flux through a sphere of radius 1 \textdegreeA(10^-10 m).

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/Users/wenhuchen/Documents/Crawler/Physics/D17-0567.htm

Solution:

This problem is solved most directly by using Gauss's Law: \varphi_E = 4\pik_Eq where g is the total charge enclosed by the sphere. The orientation of the charge within the sphere does not matter. Gauss's Law yields: \varphi_E = 4\pik_E(2e) =8\pik_Ee We can also find the flux from its definition: \varphi_E = \int E^\ding{217} \textbullet dS \varphi_E = \int E^\ding{217} \textbullet dS \ding{217} where dS^\ding{217} is the differential surface area vector which is always perpendicular to the surface. Taking the nucleus to be at the center of the sphere, we see that the field, being radial, is perpendicular to the sphere surface. Since the nucleus is equidistant from all points on the sphere, the magnitude of the field is constant over the entire surface, and is, by definition: E = k_E (2e/R^2) where R is the radius of the sphere. Thus, we may take E^\ding{217} outside of the integral sign: \varphi_E = 2k_E (e/R^2) \int cos 0\textdegree dS = 2k_E(e/R2) \int ds = = 2k_E (e/R^2) S = 2k_E(e/R2) \bullet 4\piR^2 = 8\pik_Ee The E^\ding{217} vector makes an angle of 0\textdegree with the dS^\ding{217} vector, since both are perpendicular to the surface of the sphere (see diagram). The cosine is introduced since the integrand is the inner product of the two vectors E and dS^\ding{217}, and by definition, if A^\ding{217} and B^\ding{217} are vectors then: A^\ding{217} \textbullet B^\ding{217} = \vertA^\ding{217}\vert\vertB^\ding{217}\vert cos (A, B) where (A, B) is the angle between A^\ding{217} and B^\ding{217}. S = 4\piR^2 is the surface area of the sphere. Substituting k_E = 9 × 10^9 (N-m^2/coul.^2) and e = 1.6 × 10^-19 coulombs we obtain the magnitude of the flux. \varphi_E = (8) (3.14) [9×10^9 (N-m^2/coul.^2)] (1.6 × 10^-19 coul.) = 3.61 × 10^-8 (N-m^2/coul.) We note that this result is independent of the radius of the sphere. We could have seen this immediately from Gauss's Law, since the flux through the sphere is entirely determined by the amount of charge enclosed within it.

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Question:

A 1-gram mass falls through a height H = 1 cm. If all of the energyacquired in the fall were converted to yellow light (\lambda = 6 × 10^-5 cm), how many photons would be emitted?

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Solution:

The energy acquired in the fall is just the potential energy: PE =mgh = (1 g) × (980 cm/sec^2) × (1 cm) = 980 ergs The energy of each yellow photon is l =hv= (hc/\lambda) = [{(6.6 × 10^-27 erg) × (3 × 10^10 cm/sec)}/(6 × 10^-5 cm)] = 3 × 10^-12 ergs/photon whichrepresents one photon having a total energy of 3 × 10^-12 ergs. Since nphotons have a total energy of 980 ergs, we may set up the following proportion: (nphotons/1 photon) = [(980 erg)/(3 × 10^-12 ergs)] From which we haven = 3 × 10^14 photons.

Question:

The mean length of a Rowland ring Is 50 cm and Its cross section is 4 cm^2. Use the permeability curve given below to compute the mag-netomotive force needed to establish a flux of 4 × 10^-4 weber in the ring, (a) What current is required if the ring is wound with 200 turns of wire? (b) If an air gap one millimeter in length is cut in the ring, what current is required to maintain the same flux?

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Solution:

(a) As can be seen from Fig. 2, a Rowland ring is a torus of a given ferromagnetic material with two colls around. The first long coil is used to set up the circular magnetic field B Inside the ring. As the current I in the first coil changes, an induced e.m.f. will be set up in the second coil. Thus, by measuring the voltage at the terminals of the second coil, we can measure B. If the ferromagnetic material is not present, the magnetic field B_0 in the toroidal coil is given by (similar to a solenoid) B_0 = \mu_0 n I where n is the number of turns of the first coil per unit length. We can find B_0 by measuring I. As we insert the ferromagnetic core in-side the coil, B_0 will induce a magnetic field (called the magnetiza-tion M) in the direction of B_0, in the core. The resultant magnetic field B will be the sum of these fields, (for magnetic materials, M is often much larger than B^0). Defining the magnetic intensity H as B_0 = \mu_0 H , the expression for B is given by B = \mu_0 (H + M) = \muH where M is the magnetization and \mu the permeability of the core. From the magnetization curve, the H required to set up a magnetic field is 2 × 10^2 amp/m. At this point the permeability is \mu = (B/H) 5 × 10^-3 w/amp.m B = (\varphi/A) = [(4 × 10^-4 w)/(4 × 10^-4 m^2 )] = 1 w/m^2 . In analogy with electric circuits, one can represent a magnetic material subject to a magnetic flux as a magnetic circuit. The circuit for this problem is given in Fig. 3, where the loop variable is magnetic flux and the magnetomotive force that gives rise to \varphi_B in a coil is taken to be nIl. With these definitions, the reluctance of our ring with \varphi_B through it, is given by R = (\scorpio/\varphi_B ) = (nIl/BA) The magnetic field B inside the ring is B = \muH = \mu(B_0 /\mu_0 ) = \munl , Thus R = (1/\muA) = [(0.5 m)/{5 × 10^-3 (w/amp.m) × 4 × 10^-4 m^2}] = 2.5 × 10^5 amp/w, and since \scorpio = \varphiR, the required magnetomotive force is M = 4 × 10^-4 w × 2.5 × 10^5 (amp/w) = 100 amp-turns. Using the expression for the magnetomotive force \scorpio = NI where N is the total number of turns in the coil, the required cur-rent is obtained as I = (\scorpio/N) = [(100 amp turns)/(200turns)] = 0.5; amp. (b) The air gap will introduce a new reluctance R_a into the magnetic circuit, in series with the reluctance of the ring (see Fig. 4). Since the same flux \varphi_B goes through both of the reluctances, the loop equation for this magnetic circuit is \scorpio = (R + R_a ) \varphi_B where R_a = (l_a /\mu_0 A) = [(10^-3 m)/(12.57 × 10^-7 w/amp.m × 4 × 10^-4 m^2 )] = 20 × 10^5 amp/w . Hence, assuming that the change in length of the ring is negligible and \varphi_B in the ring is kept constant, the new magnetomotive force for this arrangement is \scorpio = (2.5 × 10^5 + 20 × 10^5 ) × 4 × 10^-4 = 900 amp-turns. Therefore the current in the first coil should be increased to I = (\scorpio/N) = [(900 amp-turns)/(200turns)] = 4.5 amp .

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Question:

An airplane lands on a carrier deck at 150 mi/hr and is brought to a stop uniformly, by an arresting device, in 500 ft. Find the acceleration and the time required to stop.

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Solution:

Converting units to ft-sec, v_0 = (150 mi/hr) × [(5280 ft/mi)/(3600sce/hr)] = 220 ft/sec. Since there is a constant deceleration, 2as = v^2_1 - v^2_0 2a(500 ft) = 0 - (220 ft/sec)^2 a = {-(220 ft/sec)^2 }/ {2(500 ft)} = -48.4 ft/sec^2 . Solving for t in the following formula, v_1 = v_0 + at t= (v_1 - v_0)/a = [0 - 220 ft/see]/[ - 48.4 ft/sec^2] = 4.55 sec.

Question:

If one atom of hydrogen (mass 1.673 × 10^-27 kilogram) unites with one atom of lithium (mass 11.648 × 10^-27 kilogram) to form two helium atoms (mass 6.646 × 10^-27 kilogram each), how much kinetic energy in joules is released in this nuclear reaction? How many atoms of hydrogen would have to be transformed to generate 9.8 joules (9.8 joules of work are done when 1 kilogram of mass is raised 1 meter)?

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Solution:

The energy released in this reaction manifests itself as a loss of mass in the final product. This means that there must be a difference between the mass of the reactants and products, and this mass difference appears as energy. The mass of the products, one atom of hydrogen and one atom of lithium, is equal to the mass of two helium atoms plus the mass difference. 1 hydrogen atom + 1 lithium atom = 2 helium atoms + \DeltaM 1.673 × 10^-27 kg + 11.648 × 10^-27 kg = 2(6.646 × 10^-27)kg + \DeltaM Whence \DeltaM = 0.029 × 10^-27 kg From Einstein's relation \DeltaE = \DeltaMc^2 we find \DeltaE = 0.029 × 10^-27 kg × (3 × 10^8 m/sec)^2 = 0.26 × 10^-9 joules The number of hydrogen atoms required for 9.8 joules is N = 9.8 joules / (0.26 × 10^-9 joules) = 38 × 10^9 atoms approximately.

Question:

Distinguish between the terms "in situ,\textquotedblright "in vitro," and "invivo."

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Solution:

These terms are all used to refer to where a biochemical reaction or process takes place. "In situ" is Latin for "in place"; it refers to a reaction that occurs in its natural or original position. "In vitro" is Latin for "in glass"; it refers to a reaction that does not occur in the living organism, but instead occurs in a laboratory such as a test tube. "In vivo" is Latin for "in the living" and refers to a reaction that occurs in the living organism.

Question:

What led to the demise of the western Neanderthal population? How and in which area or areas of the world did modern man, represented first by Cro-Magnon man, arise?

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Solution:

The Neanderthal race can be differentiated into two forms: the classical Neanderthals in western Europe, and progressive Neanderthals, ortransitionals, in central and eastern Europe and the Middle East. There are conflicting points of view as to the taxonomy of the western group. Some would designate it as a separate species with the name Homo neanderthalensis . Others include it in the species sapiens and simply recognize a subspecies designation: Homo sapiensneaderthalensis. One of the major reasons for this taxonomic discrepancy is the recognition of transitional forms that cannot be assigned to either the Neanderthal or modern sapiens group. An important point is that the western European group was most likely genetically isolated during theWurmI glacial period (40,000 to 75,000 years ago), while groups to the east and south presumably had contact with each other. The populations at this point in time were geographically widespread. A great deal of morphological variability existed among them and the gene pool as a whole evolved from either Homo erectus or some undiscovered group very similar to it. Cro-Magnon man, the early representative of Homo sapiens (or modern man) appeared around 40,000 years ago (the date is disputed). There are various viewpoints concerning the origin of Homo sapiens. Some believe that the western Europeanneanderthalpopulations died out and lent little of their genes to modern man, while the tran-sitional forms in eastern and central Europe and in the Middle East evolved further into modern man. Two other points of view \rule{1em}{1pt} thePresapienand the Preneanderthal \rule{1em}{1pt} have much in common. Both agree with the first in consi-dering Neanderthals of western Europe as dead ends of hominid evolution, becoming extinct as the climatic conditions to which they were adapted changed. ThePresapientheory suggests that Homo sapiens originated as a distinct lineage, completely separate from that line leading to theNeaderthals. ThePreneanderthaltheory explains that there was a broad and varying population, one segment of which was an isolated, European, cold-adapted group (that is, the western Neanderthals) the other segment inhabited the Near East and became modern man. Another general theory views that modern man arose from a stock that existed about 250,000 years ago during an interglacial period, through a number of intermediate populations. While the western European Neanderthal group was slowly evolving in isolation in a cold climate, more favorable climatic conditions farther to the east and south and continual migration of hunting groups produced modernsapienpopulations. This does not exclude the possibility of some western European Neanderthal genes being incorpo-rated into modern populations. It is very likely that some of their genes did find their way into modern populations. The western Neanderthals became extinct, but how this occurred is not known for sure. Some anthropologists believe that they did not become extinct in the strict sense of the term. Rather, they were absorbed by interbreeding with emigrating, anatomically and culturally more modern populations, probably Cro-Magnon man. They reasoned that climatic amelioration probably allowed greater and greater population contact. If this was the case, thenneandertha-lensisand sapiens should not be considered as separate species (due to their ability to interbreed). For this and the reason mentioned earlier, some scientists classify neanderthalensis as a sub-species of sapiens. Some scientists on the other hand, believe that these two hominids belong to separate species, and Neanderthal man was eliminated by Homo sapiens by either combat or competition. Some even postulated a war, in whichCro-Macmon populations flooded western Europe and killed off the Neanderthals. We have yet to wait for further archaeological evidence to settle this conflict. But we can be certain at this stage that the process of cultural exploitation did occur. Old tools and methods gave way to newer, finer and better adapted methods. Thus, less technologically advanced cultures were slowly forced out of existence. Where then did modern man arise? The center of origin of modern man appears to have been in Asia, specifically in the general region of the Caspian Sea, but this is being disputed. It is important to point out as a concluding comment that there is at present a great deal of differing opinions concerning the origin and evolution of Homo sapiens, as seen in the many disagreements among various textbooks on this issue.

Question:

In studying the kinetics of the reaction X(g) + Y(g) \rightarrow Z(G) at 800\textdegreeK, the data in the following table were observed. (a)Write the rate law for this reaction, (b) What is the numerical value of the specific rate constant? (c) What would be the initial rate of Z formation starting with 0.15 M X and 0.15 M Y? (d) How would the rate in (c) be changed if, after the reaction had just begun, the volume of the container were abruptly doubled? Initial concentration ofX.M Initial concentration of Y.M Initial rate of formation of Z.M/min 0.10 0.10 0.030 0.20 0.20 0.240 0.20 0.10 0.120

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Solution:

(a) All of the components in this system are in the gaseous phase, thus the system is homogeneous. The rate of a reaction in a gaseous homogeneous system is dependent upon the concentration of the reactants. In a gaseous system, concentration is a function of the press-ure of the system. From the data given in the table, one can see that when the concentration of Y is held constant and the concentration of X is doubled, the rate is quadrupled. Therefore, the rate is proportional to [X]^2 . When the concentration of X is held constant and the concentration of Y is doubled, the rate is doubled and is thus proportional to [Y]. The rate law for this reaction can thus be written Rate = k [X]^2[Y] where k is the rate constant. (b) The numerical value of k is found by substituting in values for [X], [Y] and the rate into the rate law. Con-centration values for X and Y may be obtained from any one line of the table. The values obtained from the first line of the table are (0.030 M/min ) = k (0.10 M)^2(0.10 M) (0.030 M/min ) / [(0.10 M)^2(0.10 M)] = k = 30.0 M^-2 min^-1 = k (c) One can solve for the initial rate of the formation of Z by using the rate law and k when [X] and [Y] are given. Thus, Rate = k[X]^2[Y] Rate = (30.0 M^-2 min^-1)(0.15 M)^2 (0.15 M) =.101 M/min. (d) According to Boyle's Law the pressure is in-versely proportional to the volume, Therefore, if the volume is doubled the pressure is halved. Because the re-actants in this system are gases . when the pressure is halved, the concentration is halved, One can find the rate of the reaction described in (c) by using the rate law with the concentrations of the reactants halved Rate = 30.0 M^-2 min^-1 (1/2(0.15 M))^2(1/2(0.15 M)) = 30.0 M^-2 min^-1 (.075 M)^2(.075 M) = 1.27 × 10^-2 M/min The percent of the first rate that the second rate is equal to is found by dividing the second rate by the first and then multiplying the quotient by 100. (Rate_2 / Rate_1) × 100 = {(1.27 × 10^-2) / (1.01 × 10^-1)} × 100 \textasciigrave= 12.5% or 1/8 . The second rate is, therefore, 1/8 of the first.

Question:

Write a BASIC program which multiplies a polynomial of degree 2 by a polynomial of degree 1.

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Solution:

It is instructive to study a concrete example. Consider 3X^2 + 11X - 5 2X + 7 21X^2 + 77X - 35 6X^3 + 22X^2 - 10X 6X^3 + 43x^2 + 67X - 35 or simply 3 + 11 - 5 2 + 7 21 + 77 - 35 6 + 22 - 10 6 + 43 + 67 - 35 The program can be set up by putting 3, 11, and -5 in one computer list, and 2 and 7 in a second list, while leaving room for the product coefficients 6, 43, 67, and -35 in a third list. The columns are labeled as below 3210Col. no if zero subscript allowed 4321Col. no. with no zero subscript 311-5 2+7 21+ 77-35 6 + 22- 10 6 + 22+ 67-35 In general, multiplying a number in column I by a number in column J results in a productin column I + J - 1 (unless zero subscript or a zero column is allowed, in which case the product column is I + J). Thus, product totals can be accumulated with a statement of the form XXX LET P (I+J-1) = P(I+J-1) + F(I)\textasteriskcenteredS(J) where the P list is initialized to 0. 8REM LINES 10 THRU 40 READ AND 9REM PRINT THE FIRST POLYNOMIAL 10FOR X = 3 TO 1 STEP - 1 20READ F[X] 30PRINT F[X]; 40NEXT X 50PRINT "TIMES"; 51REM LINES 60 THRU 90 READ AND 59REM PRINT THE SECOND POLYNOMIAL 60FOR Y = 2 TO 1 STEP - 1 70READ S[Y] 80PRINT S[Y]; 90NEXT Y 98REM 100 THRU 120 SET THE RESULT LIST 99REM TO ALL ZEROS 100FOR W = 1 TO 4 110LET P (W) = 0 120NEXT W 128REM LINES 130 THRU 170 DO THE 129REM ACTUAL MULTIPLYING 130FOR I = 1 TO 3 140FOR J = 1 TO 2 150LET P(I+J-1) = P(I+J-1) + F(I)\textasteriskcenteredS(J) 160NEXT J 170NEXT I 180PRINT "YIELDS"; 188REM AND NOW THE 189REM 'ANSWER LIST' IS PAINTED 190FORZ = 4TO1STEP - 1 200PRINT P (Z); 210NEXTZ 218REM THE FIRST 3 NOS. REP. 3X\uparrow2 + 11X - 5 219REM THE NEXT 2 NOS . REP. 2X + 7 220DATA 3,11, -5, 2, 7 230END

Question:

Write a program, using IBM/360 Assembly language, to check if a number, greater than 2, is a prime number. Make use of the flowchart of Fig. 1.

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Solution:

On examining the flow chart, we see that we shall need the following registers: A register for storing the variable N:say,Register2. A register for the NUMBER to be checked:say,Register3. A register for storing the constant '1':say,Register4. A register for storing the constant 'O':say,Register5. Registers for multiplication and division:say,Registers , 6 and 7. A utility register:say,Register8. With these registers alloted, we can now write the program as follows. MAINSTART0 PRINTNOGEN INITIAL L2, = F '2'SET REGISTER 2 TO '2',FOR AN INITIAL VALUE OF N = 2 L4, = F '1'SET REGISTER 4 TO '1' L5, = F'0'SET REGISTER 5 TO 'O' RWD3READ THE VALUE OF THE NUMBER \textasteriskcenteredTO BE CHECKED INTO REGISTER 3 LOOPL7, 4LOAD REG. 7 WITH CONTENTS OF REG. 4 MR6, 3MULTIPLY CONTENTS OF REGS. 6 & 7 BY REG. 3 (i.e., MULTIPLY 1 BY NO. TO BE CHECKED) & STORE THE RESULT IN REG. 6 AND REG.7, AS A COUBLE-WORD. DR6, 2DIVIDE CONTENTS OF REG. 6 \textasteriskcentered& 7, i.e., THE NUMBER, BY \textasteriskcenteredCONTENTS OF REG. 2, i.e., N. \textasteriskcentered(THE QUO-TIENT GETS \textasteriskcenteredSTORED IN REG. 7 & THE REMAINDER IN REG. 6) CR5, 6COMPARE REMAINDER WITH `0' BMCHKIF NEGATIVE, BRANCH TO LABEL CHK WWD4IF NOT NEGATIVE, \textasteriskcenteredREMAINDER IS ZERO, HENCE, \textasteriskcenteredNUMBER IS NOT A PRIME, HENCE, WRITE '1' AS AN INDICATION BOUTBRANCH TO END THE PROGRAM CHKAR2, 4ADD CONTENTS OF REGISTERS 2 & 4, i.e., INCREMENT N TO N + 1. LR8, 2LOAD INCREMENTED 'N' IN REG. 8 SR8 , 3SUBTRACT 'NUMBER' FROM 'N' BMLOOPIF NEGATIVE, IT MEANS \textasteriskcentered'NUMBER' IS GREATER THAN \textasteriskcentered'N'. HENCE, REPEAT THE LOOP. WWD3IF NOT NEGATIVE, IT MEANS 'NUM-BER' = 'N' HENCE, \textasteriskcentered'NUMBER' IS A PRIME \textasteriskcenteredNUMBER, AND HENCE, PRINT \textasteriskcenteredTHE 'NUMBER' ITSELF AS AN IN-DICATION. OUTEXIT ENDMAIN

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Question:

What is the difference in the levels of a liquid in two connecting capillaries of diameters D_1 and D_2? The surface tension of the liquid is \sigma. The edge angles of the surface films are zero.

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Solution:

The surface forces, S_1 add S_2, acting on the water surfaces in capillaries 1 and 2 (as shown in the figure), respectively are S_1 = \piD_1\sigma,S_2 = \piD_2\sigma. If W_1^\ding{217} and W_2^\ding{217} are the weights of the water columns in tubes 1 and 2, the net force (neglecting the air pressure), at the bottom of each capillary is F_1^\ding{217} = W_1^\ding{217} + S_1^\ding{217}, F_2^\ding{217}= W_2^\ding{217} + S_2^\ding{217} Corresponding pressures are P_1 = F_1/A_1 = W_1/A_1 - S_1/A_1 = gh_1\sigma - (S_1/A_1) P_2 = F_2/A_2 = W_2/A_2 - A_2/A_2 = gh_2\sigma - (S_2/A_2) where g is the gravitational acceleration, \sigma is the density of the liquid, and A_1 and A_2 are the cross-sectional areas of the tubes. The hydrostatic pressures at the bottoms of the tubes should be equal, hence we have P_1 = P_2 giving gh_1\sigma - (S_1/A_1) = gh_2\sigma - (S_2/A_2), or\sigmag (h_2 - h_1) = (S_2/A_2) -(S_1/A_1), \sigmag (h_2 - h_1) = [(\piD_2\sigma) / {(1/4)\pi D^2_2 }] - [(\piD_1\sigma) / {(1/4)\pi D^2_1}] = 4\sigma[(1/D_2 ) - (1/D_1)]. The difference in the levels of the water columns is therefore, (h_2 - h_1) = [ {4\sigma(D_1 - D_2)} / (\sigmag D_1D_2) ].

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Question:

Discuss the physiological basic of phototropism.

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Solution:

A physiological explanation of phototropism that has been proposed involves a group of plant hormones called the auxins. The bending of a plant shoot toward light in phototropism takes place as a result of unequal rates of growth between the side facing the light and the side shielded from the light. Since growth of an organism involves cell division and enlargement, the difference in the rate of growth in the two sides of the plant reflects a difference in the rate of cell division or enlargement or both. Auxins are known to accelerate cell growth in actively growing regions, such as stem tips and vascular cambium. These hormones have been demonstrated to be produced by the apical meristem, and shown to be transported down the stem from the apex in an active process. In response to unidirectional light, it is found that there is a differential distribution of auxin in the stem, the side of the stem facing light receiving a lower concentration of auxin than the side away from light. Phototropism is proposed to be the effect of this uneven distribution of auxin as it moves down the stem from the apex. This mechanism suggests that as light strikes a plant unilaterally, the absorption of light energy by pigments such as carotenoids or flavins initiates the differential distribution of auxin. The exact mechanism by which this occurs is not known, but several hypotheses have been advanced. Some workers suggest that light causes greater destruction of auxin on the lighted side of the stem, or that light produces oxidation products of auxin which inhibit auxin transport down the lighted side. Others postulate that the plant redistributes the auxin that is being synthesized and transports it laterally to the shaded side. However it is accomplished, it is certain that the shaded side has a higher auxin concentration and elongates faster than the illuminated side with a lower auxin level. The effect is that the tip and the top part of the stem curve toward the source of the light.

Question:

A given sample of pure compound contains 9.81 g of zinc, 1.8 × 10^23 atoms of chromium and 0.60 mole of oxygen atoms. What is the simplest formula?

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Solution:

The simplest formula for this compound is found from the simplest ratioof moles Zn : moles Cr : moles O. One is given the number of moles ofO but must find the number of moles of Zn and Cr. One is given that thereis 9.81g of Zn. One can find the number of moles by dividing 9.81 g by the MW of Zn. (MW of Zn = 65.4) . no. of moles = (9.81g) / (65.4 g/mole) = 0.15 moles One is given that there are 1.8 × 10^23 atoms of Cr present. The number of molescan be found by dividing the number of atoms by the number of atomsin one mole, 6.02 × 10^23 . no. of moles = (1.8 × 10^23 atoms) / (6.02 × 10^23 atoms/mole) = 0.30 moles The ratio ofZn :Cr : O is .15 : .30 : .60 or 1 : 2 : 4. The simplest formulais ZnCr_2O_4.

Question:

Where is the respiratory center and what are its functions? Describe briefly the neural control of breathing.

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Solution:

Breathing requires the coordinated contrac-tion and relaxation of several muscles. This is achieved by the respiratory center, which is composed of special groups of cells in the medulla and pons of the brain. From this respiratory center, nerve impulses are rhythmically discharged to the intercostal (rib) muscles, resulting in their periodic contraction every 4 to 5 seconds. The breathing movements are automatic and occur without voluntary control under normal conditions. We can voluntarily hold our breath, but not indefinitely, since then the automatic center eventually takes over and forces us to exhale. By experimentation, neurons have been found in the medulla which propagate action potentials in perfect synchrony with inspiration. A smaller number of other neurons have been discovered which discharge synchronous-ly with expiration. These two types of neurons are called inspiratory and expiratory neurons, respectively. These may also be referred to as inspiratory and expiratory centers. Electrical stimulation of the inspiratory neurons can produce a maximal inspiration. Conversely, electrical stimulation of the expiratory neurons shuts off inspiration abruptly, and produces contraction of the expiratory muscles. The question arises as to what induces firing of the medullary inspiratory neurons. It appears that these neurons have inherent autorhythmicity - the capacity for periodic self-excitation. However, synaptic input from other neurons plays an essential role in regulating the rhythmicity of these inspiratory neurons. There are three vital inputs to the medullary inspiratory neurons, which play a role in modulating respiratory rhythm. They are (1) direct intracranial connections between the medulla and the pons (2) reciprocal connections with the medul-lary expiratory neurons, and (3) afferent input from stretch receptors in the lung (see Figure). The connections between the inspiratory and expiratory centers are inhibitory in nature. Thus at the beginning of inspiration, when the inspiratory neurons are firing, the expiratory center is prevented from firing so that expiration is inhibited. When inspiration ceases, ex-piratory inhibition also stops, and expiration is able to occur which then, in turn, inhibits inspiration. These reciprocal connections serve to synchronize inspiration and expiration. As was previously mentioned, the medullary inspiratory center is connected to the pons. This area of the pons is often called the pneumotaxic center, and destruc-tion of this center produces profound changes in respira-tion. The medullary neurons receive neural input from the pons, which exerts a tonic effect upon the inspiratory neurons. It is also likely that the pneumotaxic center serves as a central relay station for the respiratory in-hibition initiated by the lung stretch receptors. As the lungs expand during inspiration, these receptors are stimulated, and impulses travel up the afferent nerves to the brainstem, where they aid in terminating inspiration. To summarize, the medullary inspiratory neurons primarily control the cycle of ventilation. These inspiratory neurons innervate the inspiratory muscles. The spontaneous increase in the firing rate of these inspiratory neurons is a crucial factor for initiating ventilation. Due to the inhibitory impulses from medullary expiratory neurons and pulmonary stretch receptors which act through higher brain centers, these neurons will stop firing. The cessation of activity by the inspiratory neurons releases the inhibition on the expiratory neurons so that expiration can passively occur. Only in forced expiration are the expiratory muscles themselves used. This active expiratory movement is synchronized with the passive component of expiration. This is possible because of the reciprocal connections between the medullary inspiratory and expiratory centers.

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Question:

A closed loop of wire is placed in a magnetic field in such a way that flux points into the coil as shown in the figure where the x's indicate the tail ends of flux lines. If the flux density suddenly increases, will current be induced in the loop, and if so will it point clockwise or counterclockwise?

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Solution:

This problem is an application of Lenz's Law, by virtue of which a current will be induced in the loop. The direction of the current is found by noting that a change in flux induces a current which by its own electromagnetic action opposes the change re-sponsible for it. This suggests that the current, whatever its direction, develops a magnetic field out of the paper inside the loop, i.e., opposed to the original field. The right hand rule is used to find the current direction in the coil. If the thumb of the right hand points in the direction of the magnetic field developed by the loop, then the fingers will curl in the direction of the current in the loop. In this case the current must point in a counterclock-wise direction.

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Question:

Using the ideal gas law, show that for an ideal gas of density \rho, P/\rhoT is a constant.

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Solution:

The ideal gas law reads PV = nRT, where P = pressure, V = volume, n = number of moles, R = gas constant, and T = absolute temperature. The number of moles is equal to the mass m of a sample of gas divided by the molecular weight MW of that gas: n = m/MW. Then PV = nRT = (m/MW) RT. Dividing both sides by the volume,we obtain P = (m/MW) (RT/V). But mass per unit volume is P = (m/MW) (RT/V). But mass per unit volume is density, hence \rho = m/V and this equation becomes P = (\rho/MW) RT or p /(\rhoT) = R/MW. Since R is a universal constant and MW, the molecular weight, is constant for a particular gas, P/(\rhoT) = R/MW = constant.

Question:

(a) 3, 3-dimethylpentane, (b) n- heptane, (c) 2- methylheptane, (d) n-pentane, and (e) 2-methylhexane. Arrange these compounds in order of decreasing boiling points (without referring to tables).

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Solution:

To rank the boiling points of these substances, you must consider the rules which govern the property of boiling points. Except for very small alkanes, the boiling point rises 20 to 30 degrees for every carbon that is added to the chain. (Alkanes are saturated hydrocarbons of the general formula C_nH_2n+2,where n = number carbon atoms.) Alkanes of the same carbon number but different structures (isomers) will have different boiling points. A branched- chain isomer has a lower boiling point than a straight- chain isomer. This is because the shape of a branched- chain molecule tends to approach a sphere, decreasing the surface area, so that the intermolecular forces are easily overcome at a lower temperature. With this in mind, you can proceed to solve the problem. Because these rules pertain only to structures, you must write them out. c. has the greatest carbon content, 8 carbon atoms, which means it has the highest boiling point, a, b and e all have 7 carbon atoms; thus carbon content is equal. But only b is not branched, which means it has the highest boiling point out of this group, e is the next highest since it is less branched than a. d has the lowest boiling point, since it has the lowest carbon content (5 atoms). Thus, you rank them in order of decreasing boiling point as c, b, e, a, d.

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Question:

Assume that an A molecule reacts with two B molecules in a one-step process to give AB_2. (a) Write a rate law for this reaction. (b) If the initial rate of formation of AB_2 is 2.0 × 10^-5 M/sec and the initial concentrations of A and B are 0.30 M, what is the value of the specific rate constant?

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Solution:

(a) The overall equation for this reaction is A + 2B = AB_2 Since no other information is provided about the reaction, the rate law for the reaction is assumed to be written Rate = k [A][B]^2, where k is the rate constant and [ ] indicates concentration. (b) One can solve for k using the rate law when the rate, [A] and [B] are given as they are in this problem. Rate = k [A][B]^2 2.0× 10^-5 M/sec = k (0.30 M)(0.30 M)^2 (2.0× 10^-5 M/sec) / {(0.30 M)(0.30 M)^2} = k k = 7.41 × 10^-4 M^-2 sec^-1.

Question:

How is a human's perception of a flower different from that of a bee?

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Solution:

It is known that pigeons can sense the earth's magnetic field and bats can detect ultrasonic signals, both of which humans cannot detect. It is therefore obvious that humans do not have all the sensory skill possessed by other animals. Another fascinating example of animal sensory perception, is the detection of ultravi-olet light by the honeybee. Certain flowers which appear to us to be plainly colored, actually contain highly contrasting patterns in the ultraviolet range. If one should look at these flowers through a transformer that shifts the ultraviolet energies to the visible spectrum, one could see how a bee perceives these flowers (See photograph). The patterns, completely invisible to the human eye, serve as guides to the bee in reaching the location of nectar and pollen. The two photographs show the different views of the same flower as seen by a human eye (A) and a bee (B). The bee is sensitive to ultraviolet light which is invi-sible to human beings.

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Question:

The pressure in an automobile tire is measured by a tire gauge in winter at 0\textdegreeC as 30 pounds per square inch (psi). The gauge measures the difference between the actual tire pressure and atmospheric pressure (15 psi). Assume that the tire does not leak air and does not change its volume. What will be the new gauge pressure if the same tire is measured during the summer at 50\textdegreeC?

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Solution:

We will solve this problem by employing a modified form of the ideal gas law. The ideal gas equation is PV =nRT, where P is the total pressure (P = atmospheric pressure + gauge pressure), V is the tire volume, n the number of moles of air, R the gas constant, and T the absolute temperature. Bearing in mind the fact that n, V, and R are constant, we may write this equation for the two separate sets of conditions, as P_summerV = nRT_summerandP_winterV = nRT_winter. Dividing the first by the second gives (P_summerV) / (P_winterV) = [(nRT_summer) / (nRT_winter)] or(P_summer) / (P_winter) = (T_summer) / (T_winter) Solving forP_summer, we have P_summer= [(T_summer)_ / (T_winter)] ×P_winter. P_winter= atmosphericpressure + gauge pressure = 15 psi + 30 psi = 45 psi, T_summer= 50\textdegreeC = 323\textdegreeK, andT_winter= 0\textdegreeC = 273\textdegreeK. Hence, P_summer= [(T_summer) / (T_winter)] ×P_winter = [(323\textdegreeK) / (273\textdegreeK)] × 45 psi = 53 psi. The gauge pressure will beP_summer- atmospheric pressure, or 53 psi - 15 psi = 38 psi .

Question:

In what ways do ferns resemble seed plants? In what respects do they differ from them?

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Solution:

ThePterophyte(ferns) and theSpermophyte(seed plants- gymnosperms and angiosperms) are alike in a number of respects. They are both terrestrial plants and as such have adapted certain similar anatomical structures. The roots, of both plants aredefferentiatedinto root cap, an apicalmeristem, a zone of elongation and a zone of maturation. Their stems have a protective epidermis, supporting, and vascular tissues: and their leaves have veins,chlorenchymawith chlorophylls, a protective epidermis andstomatae. The ferns also resemble the seed plants in that thesporophyteis the dominant generation. The characteristics that distinguish ferns from the seed plants include the structure of the vascular system, the location of the sporangia, the absence of seeds, the structure and transport of sperm, and the patterns of reproduction and development. Unlike the seed plants, ferns have onlytracheidsin their xylem and no vessels. They bear their sporangia in clusters on their leaves (fronds), in contrast to the seed plants which carry their sporangia on specialized, non-photo-synthetic organs, such as the cone scales of a gymnosperm. The ferns produce no seeds and the embryo develops directly into the newsporophytewithout passing through any protected dormant stage as seen in the seed plants. The ferns retain flagellated sperm and require moisture for their transport and subsequent fertilization. The seed plants, on the other hand, have evolved a mechanism ofgameticfusion by pollination, i.e., the growth of a pollen tube. The pollen tube eliminates the need for moisture and provides a means for the direct union of sex cells. The ferns also differ from the seed plants in their life cycle. While in both, thesporophyteis the dominant generation, the fern gametophyte is an inde-pendent photosynthetic organism whereas the seed plant's gametophyte, bearing no chlorophyll, is parasitic upon thesporophyte. Also, the gametophyte of the seed plant is highly reduced in structure. The cycad (a gymnosperm) male gametophyte, for instance, consists of only three cells. The ferns, furthermore, are unlike the seed plants in being homosporous , that is, producing only one kind of spore. The seed plants, on the contrary, have two types of spores, the larger, female spores and the smaller, male spores.

Question:

What are binary trees? Write a program that reads in 50 values and places them in a binary tree. After reading these values the program should print the values in the tree in ascending order.

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Solution:

Binary trees are very similar to a linked list. Consider the following structure struct binary _ tree { int number; struct binary _ tree \textasteriskcenteredright _ node; struct binary _ tree \textasteriskcenteredleft _ node; }; }; A single element in a binary tree is visualized as follows Binary trees are commonly used to sort information using the following methods them in ascending order is as follows: #define TRUE 1 #define FALSE 0 #define NULL '\textbackslash0' structbinary _ tree { intnumber; structbinary _ tree \textasteriskcenteredright _ node; structbinary _ tree \textasteriskcenteredleft _ node; } ; main ( ) { intnumber _ entered; /\textasteriskcenterednumber entered or \textasteriskcentered/ /\textasteriskcenteredread in\textasteriskcentered/ intnumber _ of _ values; /\textasteriskcenterednumber of values to be read in\textasteriskcentered/ inti; intlocation _ not _ found; /\textasteriskcenteredto track the location of the value \textasteriskcentered/ char \textasteriskcenteredcalloc ( ); struct binary _ tree \textasteriskcenteredroot, \textasteriskcenteredcurrent _ node, \textasteriskcenterednew _ node; root = (struct binary _ tree \textasteriskcentered) calloc (1, sizeof (structbinary _ tree)); number _ of _ values = 50; /\textasteriskcenterednow get the 1st value \textasteriskcentered/ current _ node = root; current _ node \rightarrow number = get _ integer ( ); current _ node \rightarrow right _ node = NULL; current _ node \rightarrow left _ node = NULL; /\textasteriskcenteredsince the 1st value has been read in \textasteriskcentered/ /\textasteriskcenteredwe are left with 49 more values\textasteriskcentered/ for (i = 1; i < 50; i++) { /\textasteriskcenteredtake the next value\textasteriskcentered/ number _ entered = get integer ( ); /\textasteriskcenteredstart at the top i.e. the root \textasteriskcentered/ current _ node = root; location _ not _ found = TRUE; new _ node = (structbinary _ tree \textasteriskcentered) calloc (1,sizeof (structbinary _ tree)); new _ node \rightarrow number = number _ entered; new _ node \rightarrow right _ node = NULL; new _ node \rightarrow left _ node = NULL; /\textasteriskcenterednow we have to find the location \textasteriskcentered/ /\textasteriskcenteredin the tree for new value\textasteriskcentered/ while (loation _ not _ found) { if (number _ entered < current _ node \rightarrow number) if (current _ node \rightarrowleft _ node == NULL) location _ not _ found = FALSE; else current _ node = current _ node \rightarrow left _ node; else if (current _ node \rightarrow right _ node == NULL) location _ not _ found = FALSE; else current _ node = current _ node \rightarrow right _ node; } /\textasteriskcenteredend while \textasteriskcentered/ /\textasteriskcenterednow to place the new node in the tree\textasteriskcentered/ /\textasteriskcenteredat the correct location\textasteriskcentered/ if (number _ entered < current _ node \rightarrow number) current _ node \rightarrow left _ node = new _ node; else current _ node \rightarrow right _ node = new _ node } print _ tree (root); /\textasteriskcenteredprint _ tree prints out the sorted values\textasteriskcentered/ } print _ tree (node) structbinary _ tree \textasteriskcenterednode; { if (node \rightarrow Left _ node ! = NULL) print _ tree (node \rightarrow left _ node); printf ("%d", node \rightarrow number); if (node \rightarrow right _ node ! = NULL) print _ tree (node \rightarrow right _ node); } Explanation of Routines Used: 1) get _ integer ( ) This is an assumed routine (i.e. not present in the li-brary) which gets next integer from the input. As an excercise write this routine. 2) print _ tree (node) The function print _ tree ( ) invokes itself recursively to print the values contained in the tree. This routine searches each left node until node \rightarrowleft equals NULL. It then prints the value contained at the node and returns up one level in the tree and prints the value at that node. The routine further checks whether the node has a right node and if so starts the same process at that node.

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Question:

Design an 8 × 3 encoder.

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Solution:

An encoder is a device with 2^n input lines (in this case eight) and n output lines (in this case 3). Only one of the input lines can be excited at any one time. The encoder generates on its output lines a code corresponding to the input line that was excited. The truth table for the 8 × 3 encoder is shown in figure 1. Bits I_0 through I_7 are the input lines and bits A_0 through A_2 are the output lines. INPUTS I_0I_1I_2I_3I_4I_5I_6I_7 OUTPUTES A_0A_1A_2 10000000 000 01000000 001 00100000 010 00010000 011 00001000 100 00000100 101 00000010 110 00000001 111 Fig. 1 \rule{1em}{1pt} truth table output lines. A_0 is most significant output bit; A_2 is least significant output bit. From the truth table it is seen that A_0 = I_4 + I_5 + I_6 + I_7 A_1 = I_2 + I_3 + I_6 + I_7 A_2 = I_1 + I_3 + I_5 + I_7 The logic diagram of the encoder is obtained from these equations. Figure 2 shows the encoder.

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Question:

A sample of a gas exhibiting ideal behavior occupies a volume of 350 ml at 25.0\textdegreeC and 600 torr. Determine its volume at 600\textdegreeC and 25.0 torr.

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Solution:

We can solve this problem by determining the number of moles, n, of gas from the ideal gas equation, PV = nRT (where P = pressure, V = volume, n = number of moles, R = gas constant, and T = absolute temperature) by using the first set of conditions and then substituting this value of n along with P and T from the second set of conditions into the ideal gas equation and solving for V. However, we can save useless calculation by denoting the first and second sets of conditions by subscripts \textquotedblleft1\textquotedblright and \textquotedblleft2\textquotedblright, respectively, to obtain P_1 V_1 = nRT_1,andP_2 V_2 = nRT_2, where n is the same in both cases. Dividing the second of these equations by the first we obtain [(P_2 V_2 )/(P_1 V_1 )] = [(nRT_2 )/(nRT_1 )]or(P_2 V_2 / P_1 V_1 ) = (T_2 /T_1), where we have cancelled n and R (both constant). Solving for V_2, we obtain V = [(P_1 V1 )/(P_2 )] × (T_2 /T_1 ) = [(600 torr × 350 ml)/(25.0 torr)] × [(873.15\textdegreeK)/(298.15\textdegreeK)] = 24.6 × 10^3 ml = 24.6 l.

Question:

An electron is projected into a magnetic field of flux density B = 10 (w/m^2 ) with a velocity of 3 × 10^7 m/sec in a direction at right angles to the field. Compute the magne-tic force on the electron and compare with the weight of the electron.

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Solution:

The force on the electron in a magnetic field is given by F^\ding{217} =qv^\ding{217} × B^\ding{217} where v^\ding{217} is the velocity of the electron and B^\ding{217} is the flux density of the magnetic field. In this case the velocity is perpendicular to the magnetic field so the force may be computed by a straightforward multiplication instead of taking the cross product. The magnetic force is F =qvB= 1.6 × 10^-19 × 3 × 10^7 × 10 = 4.8 × 10^-11newtonThe gravitational force, or the weight of the electron, is F = mg = 9 × 10^-31 × 9.8 = 8.8 × 10^-30newton. The gravitational force is therefore negligible in comparison with the magnetic force.

Question:

Given that k = 8.85 at 298\textdegreeK and k = .0792 at 373\textdegreeK, calculate the ∆H\textdegree for the reaction of thedimerizationof NO_2 to N_2 O_4. Namely, 2NO_2 (g) \rightleftarrows N_2 O_4 (g).

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Solution:

∆H\textdegree is the standard enthalpy change, a measure of the heat content of the system. It is quantitatively related to the equilibrium constants of a system at dif-ferenttemperationsby thevan'tHoff equation. log (k_1/k_2 ) = [(∆H\textdegree)/(19.15)] [(1/T_1) - (1/T_2)], where k_2 and k_1 are equilibrium constants, and T_1 and T_2 are temperatures in Kelvin. Given that k_1 = 8.85 at T_1 - 298\textdegreeK and k_2 = 0.0792 at 373\textdegreeK, one can find ∆H\textdegree by the substitution of these values in the equation. Thus, log [(.0792)/(8.85)] = [(∆H\textdegree)/(19.15)] [(1/298) - (1/373)]. Solving for ∆H\textdegree, one obtains ∆H\textdegree = - 58, 200 J/mole for the reaction.

Question:

A tank of water is placed on a scale, which registers its weight as W = Mg. What is the change in the scale reading if a block of steel, of weight w = mg, is lowered into the tank as shown in figure (B)?

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Solution:

The figure shows the situation. The steel is held in the water by a string with tension T. We wish to find the force exerted by the system on the scale. By Newton's Third Law, this is equal in magni-tude to the force exerted by the scale on the system, shown in figure (A). Since this system is in equi-librium, the net force acting on it is zero and T + N - (M + m)g = 0 N = (M + m)g - T(1) However, in order to solve for N, we must know T. We can obtain T by applying Newton's Second Law to the system shown in figure (C). Since this system is in equilibrium, we may write T + F_B - mg = 0 where F_B is the buoyant force on the block. By Archimedes' Principle, F_B is equal to the weight of water displaced by the steel. Hence, F_B = \rho_wgv F_B = \rho_wgv where V is the volume of the cube and \rho_w is the density of water. Then, T = mg - F_B = mg - \rho_wgV(2) Using (2) in (1), N = (M + m)g + \rho_wgV - mg orN = Mg + \rho_wgV This is the "weight" registered by the scale. If the tank were weighed without the steel, we'd find N' = Mg The difference in these 2 readings is N - N' = Mg + \rho_wgV - Mg = \rho_wgV or the buoyant force.

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Question:

Discuss the important effects of gibberellins, cytokinins, and abscisic acid on plant growth.

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Solution:

Gibberellins, cytokinins and abscisic acid are the three major types of plant hormones that have been identified in addition to auxins. These three hormones interact with auxins and with each other to regulate biological activities in plants. Gibberellins and cyto-kinins have dominant roles in controlling the early phases of growth and development; auxins become dominant later in controlling cell elongation, and abscisic acid opposes the functions of all these three growth-promoting factors to bring about a well-regulated, balanced pattern of plant growth. Gibberellins function to lengthen stems, stimulate seed germination, induce flower formation, and increase the size of fruits in some species of plants. The for-mer two are their most vital functions. Seed germina-tion is examined here: Just before germination, the embryo of a seed secretes a gibberellin which induces the production of \alpha-amylase, an enzyme which hydrolyzes the stored starch for energy. The hormone also acti-vates other enzymes of the seed which breaks down the materials of the seed coat, weakening the coat to facilitate the breakthrough of the germinating embryo. It has been shown that actinomycin D inhibits the synthesis of \alpha-amytlase in response to gibberellin. Actinomycin-D is a known inhibi-tor of DNA-dependent RNA synthesis; this suggests that gibberellins cause the activation of certain enzymes by regulating, in some fashion, the expression of the genetic information contained in the DNA, perhaps by uncovering a portion of the DNA so that it can be transcribed to produce a specific RNA segment. This segment can then be translated to form an enzyme (which is a protein). Cytokinins function in stimulating growth of cells and accelerating their rate of division. Cytokinins can also change the structure of plant tissue: at certain concentrations, cytokinins are shown to cause root and shoot formation in plant tissue cultures. Cytokinins are also found to oppose auxins by causing lateral buds to develop and thus modifying apical dominance. In addition, cytokinins have been demonstrated to prevent leaves from yellowing and hence play a role in delaying senescence. A possible explanation for the mechanism of this function is that cytokinins prevent the progressive "turning off" of segments of DNA (associated with aging) that are responsible for the synthesis of enzymes and protein and the production of other compounds such as chlorophyll. Abscisic acid was first reported in the early 1960's, and is now known as a growth-inhibitor in plants. Abscisic acid is collected largely from the ovary bases of fruits, and its concentration here is highest at the time of fruit drop. The hormone acts antagonistically to gibberellins by inhibiting the production of \alpha-amylase in seeds and triggering seed dormancy. It opposes cyto-kinins by inactivating vegetative buds, causing the yel-lowing of leaves, and reducing the growth rate of plants. Furthermore, it opposes auxins by accelerating abscission (hence its name) and promoting leaf, fruit, flower and branch shedding. It also promotes stomatal closure. In short, by checking growth and reproductive activities of the plant, abscisic acid interacts with the three growth - promoting hormones to bring about an optimal rate of growth and differentiation.

Question:

Explain what is known as the 'Count and/or Compare' methodof stopping a computer job.

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Solution:

Sometimes, computer programs may be such that the computer will enterinto a segment of the program which makes the computer do an endlessnumber of loops. This will happen because the programmer has notprovided the instructions which control the exit from the loop. In such a casethe program will be terminated by the operating system when the allotedtime for the job has been exhausted. The careful programmer always watches out for these pit-falls in his program, and if there is a possibility of end-lessloopings(e.g., in a programthat calculates the sum of all even numbers greater than one), he normallyplaces a maxi-mum limit to the amount times the loop is carried out, or, a maximum limit to the value of the result of his program. When this maximum limit is reached, the computer is in-structed to exitthe loop and carry out the remainder of the program. This method is illustrated in the program below: SUM: PROCOPTIONS(MAIN); DCL (X,SUM)FIXED(4)INIT(0); LOOP: GET LIST(X); /\textasteriskcentered A VALUE OF X IS READ OFF A CARD \textasteriskcentered/ SUM=SUM+X ; IF SUM>100 THEN GOTO FINISH; /\textasteriskcentered HERE, WE SET A MAXIMUM LIMIT OF 100 TO THE VALUE THAT THE SUM OF ALL THE DIFFERENT NUMBERS 'X' WHICH ARE READ FROM THE CARD/CARDS, CAN REACH \textasteriskcentered/ GOTO LOOP; FINISH: PUTLIST(SUM); END SUM;

Question:

A 1-kg block slides down a rough inclined plane whose height is 1 m. At the bottom, the block has a velocity of 4 m/sec. Is energy conserved?

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Solution:

Energy will be conserved if the kinetic energy gained by the block is equal to the potential energy lost. At top:PE = mgh = (1 kg) × (9.8 m/sec^2) × (1 m) = 9.8 J. At bottom:KE = (1/2) mv^2 = (1/2) × (1 kg) × (4 m/sec)^2 = 8 J. Apparently, energy is not conserved. But we know that friction is present between the block and the rough plane. A certain amount of energy (1.8 J) has evidently been ex-pended in overcoming this friction. This amount of energy appears as thermal energy and could be detected by measuring the temperature rise in the block and the plane after the slide is completed.

Question:

1) E^\ding{217}_T = \^{\i} E_o sin [2\pi(z / \lambda) - (vt)] + \^{\j} E_o cos [2\pi(z / \lambda) - (vt)]. 2) E^\ding{217}_T = \^{\i} E_o sin [2\pi(z / \lambda) + (vt)] + \^{\j} E_o sin [2\pi(z / \lambda) + (vt) - (1 / 8)]. 3) E^\ding{217}_T = \^{\i} E_o sin [2\pi(z / \lambda) - (vt)] - \^{\j} E_o sin [2\pi(z / \lambda) - (vt)].

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Solution:

1) The two components of E^\ding{217}_T are E_x = E_o sin [2\pi{(z/\lambda) - vt}],E_y = E_o cos [2\pi{(z/\lambda) - vt}]. These are actual components of a real vector, not phasors, so we want to construct the resultant at various times and also some convenient place like z = 0. t0\tau / 4\tau / 23\tau / 4\tau / 8(\tau = 1 / v ) E_x0-Eo0E_o(-E_O / \surd2) E_yE_o0-E_o0(+E_O / \surd2) The resultant vector is simply a vector of length E, rotating counterclockwise. The polarization is circular. 2) The components of the given vector are E_x = E_o sin [2\pi{(z / \lambda) + (vt)}] E_y = E_o sin [2\pi{(z / \lambda) + vt - 1 / 8}] Again we consider the point z = 0 and follow the resultant as time progresses. t 0 \tau / 8 2\tau / 8 3\tau / 8 4\tau / 8 5\tau / 8 6\tau E_x 0 +E_osin(\pi/4) E_o E_osin(\pi/4) 0 -E_osin(\pi/4) -E_o E_y -E_osin(\pi/4) 0 +E_osin(\pi/4) E_o E_osin(\pi/4) 0 -E_osin(\pi/4) The locus of the end of E_Resultant is an ellipse at 45 de-grees to the axes, as shown in figure 2b. To find the semi-major axis, look at the time when E_x = E_y this is halfway between t = 0 and t = \tau / 8: t = \tau / 16. Hence E_x = E_o sin (\pi / 8) , E_y = E_o sin (-\pi / 8) : E_R = E_o sin (\pi / 8) \surd2 = 0.542E_o To find the semi-minor axis, which is \tau / 4 later, t = \tau / 16 + \tau / 4 = 5\tau / 16. So we have: E_x = E_o sin (5\pi / 8) , E_y = E_o sin (3\pi / 8), E_R = E_o sin (3\pi / 8)\surd2 = 1.31E_o. 3) The components of E^\ding{217}_T are E_x = E_o sin [2\pi (x / \lambda -vt)]. E_y = -E_o sin [2\pi(z / \lambda - vt)]. As before, z = 0 is most convenient point. t0\tau / 8\tau / 43\tau / 84\tau / 85\tau / 8 E_x0E_o sin(\pi/ 4)Eo+E_osin (\pi / 4)0-E_o sin(\pi / 4) E_y0-E_o sin(\pi/ 4)-E_o-E_o sin(\pi/ 4)0E_0sin(\pi/ 4) Thus the polarization is linear, along a line at 45 degrees to the axes. It has amplitude E_o / \surd2.

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Question:

Gills are very important for the proper exchange of gases between an aquatic organism and its environment. Using the fish as an example, describe the structure of a gill.

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/Users/wenhuchen/Documents/Crawler/Biology/F16-0408.htm

Solution:

The gills of a fish are located on both sides of: the animal, slightly posterior to the eyes. A bony plate known as the operculum completely covers each gill except at the posterior margin (see figure 1). This opening allows water, which has entered through the mouth and passed over the gill, to exit from the body. The gill is composed of many filaments, each subdivided into numerous lamellae which serve to increase the surface area for gas exchange (see Fig. 2). It is across these lamellae that gas exchange occurs. Running through the filaments are two blood vessels called the afferent and efferent blood vessels. The former carries deoxygenated blood to the gill, while the latter carries freshly oxygenated blood to the rest of the body. The gill filaments are supported at their base by bony extensions of the ventral skeletal column, called gill arches. The filaments themselves do not have any other means of support, because they receive sufficient support from the water around them. The blood vessels in the filaments are separated from the water by only two layers of cells: the single cell layer of the wall of the blood vessel and the single layer of cells making up the filament surface. Oxygen moves by diffusion from the water, across the intervening cells and into the blood. This blood then leaves the gill through the efferent vessel and transports the oxygen to the rest of the body.

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Question:

Suppose that a small, electrically charged metal ball is lowered into a metal can as illustrated in the Figure. Show how the charge is distributed when (a) the ball is inside the can but not touching and (b) after the ball touches the inside of the can.

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/Users/wenhuchen/Documents/Crawler/Physics/D17-0554.htm

Solution:

When the charged sphere is lowered to the position as in Figure (b), free electrons from atoms in the metal migrate to the inner surface because the positive charge of the ball exerts an attractive force. Since the net charge of the isolated can was originally zero, there is now a charge imbalance within the conductor. An excess of positive charge results. In a fraction of a second, due to transient currents within the conductor, the positive charges can be thought to mutually repel, spreading to the outside surface of the can. The outside surface then has a positive charge equal to the negative charge on the inside surface. As the ball and can touch, they form a single conductor and the electrons on the inner surface of the can move onto the metal ball and neutralize the positive charge carried by the ball. The final result is that the excess charge of the metallic sphere, placed in contact with an insulated metal can, resides entirely on its outside surface (see Figure c). This experiment provides a verification of Gauss's law. If a Gaussian surface is constructed inside the outer surface of the metal can, then there is no net charge within the surface. Then, according to Gauss's law, \varphi = Q/\epsilon_0 where \varphi is the electric flux through the Gaussian surface due to the net charge Q within the surface. This then becomes \varphi = 0. Any excess charge must therefore reside on the outer surface of the conductor, outside the Gaussian surface.

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Question:

The pituitary gland has been called the master gland. Is this term justified? Where is the gland located and what does it secrete?

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/Users/wenhuchen/Documents/Crawler/Biology/F21-0533.htm

Solution:

The pituitary gland, also known as the hypophysis, lies in a pocket of bone just below the hypothalamus. The pituitary gland is composed of three lobes, each of which is a functionally distinct gland. They are the anterior, intermediate and posterior lobes. The anterior and posterior lobes are also known as the adenohypophysis and neurohypophysis, respectively. In man, the intermediate lobe is only rudimentary and its function remains unclear, it secretes melanocyte stimulating hormones (MSH) which is known to cause skin darkening in the lower vertebrates. The anterior lobe is made up of glandular tissue which produces at least six different protein hormones. Evidence suggests that each hormone is secreted by a different cell type. Secretion of each of the six hormones occurs independently of the others. One of the hormones secreted by the anterior pituitary is known as TSH (thyroid stimulating hormone), which induces secretion of the thyroid hormones from the thyroid gland. Thyroid hormones refers to two closely related hormones,thyroxine(T_4) and triiodothyronine (T_3) . Another hormone secreted by the anterior pituitary is ACTH (adrenocorti- cotrophic hormone), which stimulates the adrenal cortex to secrete cortisol. The anterior pituitary is also responsible for the release of the gonadotropic hormones, FSH (follicle stimulating hormone) and LH (luteinizing hormone) . These hormones primarily control the secretion of the sex hormones estrogen, progesterone and testosterone by the gonads. FSH and LH also regulate the growth and development of the re-productive cells (sperm and ovum) . TSH, ACTH, FSH and LH are all trophic hormones in that they stimulate other endocrine glands to secrete hormones. There are two hormones secreted by the anterior pituitary that do not affect other hormonal secretions, but rather act directly on target tissues. One of these is called prolactin, which stimulates milk produc-tion by the mammary glands of the female shortly after giv-ing birth. The other hormone is called growth hormone, which plays a critical role in the normal processes of growth. The posterior lobe of the pituitary gland is actually an outgrowth of the hypothalamus and is true neural tissue. The posterior pituitary differs from the anterior pituitary with respect to embryological origin as well as types of hormones secreted. It releases two hormones called oxytocin and vasopressin. Oxytocin principally acts to stimulate contraction of the uterine muscles as an aid to parturition. Emotional stress may also cause the release of this hormone, and is frequently the cause of a miscarriage. Oxytocin is also responsible for the milk let-down reflex. It stimulates smooth muscle cells of the mammary glands which causes milk ejection. Antidiuretichormone (ADH) stimulates the kidney tu-bules to reabsorb water and thus plays an important role in the control of plasma volume. In addition, ADH can increase blood pressure by causing arteriolar constriction. Thus ADH is also called vasopressin. It should now be clear why the pituitary is called the master gland; it secretes at least nine hormones, some of which directly regulate life processes while others control the secretion of other glands important in development, behavior and reproduction.

Question:

Gland secretions and smooth muscle contraction in the gastrointestinal region are regulated by both nerves and hormones. Describe these hormones and their effects.

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/Users/wenhuchen/Documents/Crawler/Biology/F21-0535.htm

Solution:

The gastrointestinal region consists of the stomach, the small and large intestines and parts of the liver and pancreas. It is the site where complex molecules are catabolized into simpler forms which are absorbed into the circulation. These simpler molecules are transported to different synthetic sites of the body. The deposition of proteins into the stomach provides the major chemical stimulus for the release of the hormone gastrin. This hormone may also be released as a result of stimulation of the parasympathetic fibers of the vagus nerves to the stomach. Gastrin is secreted by cells in the mucosa of the pyloric region of the stomach.It has mutiple sites of action throughout the gastroin-testinal system. Gastrin stimulates secretion of hydro-chloric acid into the stomach as well as the contraction of the gastroesophageal sphincter, a smooth muscle. Gastrin also increases gastric motility. Stimulation of the wall of the duodenum causes release of the hormone enterogastrone. This hormone inhibits the secretion of gastrin. Under normal conditions, the amount of gastric juices secreted by the stomach is determined by the balance between levels of enterogastrone and gastrin. The pancreas can be divided into two portions. The endocrine portion of the pancreas secretes the hormones insulin and glucagon. The exocrine portion of the pancreas secretes solutions containing high concentrations of sodium bicarbonate and large numbers of digestive enzymes. Secretin is a hormone that is released by the mucosal cells of the duodenum. The stimulus for secretin release is the presence of acid in the duodenum. Secretin will elicit an increase in the amount of bicarbonate and the volume of fluid released by the pancreas, but stimulates little enzymatic secretion. The action of secretin is predominantly upon the acinar cells of the pancreas, which secrete bicarbonate. This bicarbonate will neutralize the acid secretions entering the small intestine from the stomach. Another hormone secreted by the duodenum is cholecystokinin (CCK) . The presence of amino acids and fatty acids in the duodenum will stimulate its release. CCK will increase pancreatic enzyme secretion, which leads to additional fat and protein digestion. It may also cause contraction of the gallbladder, especially after a fatty meal. This contraction will release bile into the intestinal tract. Bile is essential for digestion of fat in the small intestine. It is produced in the liver and stored in the gallbladder until the stimulatory action of CCK causes its release. As chyme leaves the stomach to enter the duodenum, its presence stimulates the release of CCK and secretin from the wall of the duodenum. These hormones then stimulate the secretion of enzymes and bicarbonate by the pancreas. These, pancreatic enzymes will act upon large molecules in the chyme to produce simple sugars, amino acids and fatty acids which can readily be absorbed by the intestinal cells. These enzymes also, serve as a chemical stimulus for hormonal release until the nutrients are absorbed. Ulcerative damage to the intestinal walls by acids released from the stomach is prevented by pancreatic secretion of bicarbonate. In addition, bicarbonate creates an alkaline environment, which favors the action of pancreatic enzymes. The release of CCK and secretin also inhibits gastric secretion and motility, thus slowing the movement of digested nutrients and allowing neutralization, digestion and absorption to occur.

Question:

a) The maximum pressure variation P that the ear can tolerate in loud sounds is about 28 N / m^2 (=28 Pa). Normal atmospheric pressure is about 100,000 Pa. Find the corresponding maximum displacement for a sound wave in air having a frequency of 1000 Hz. b) In the faintest sound that can be heard at 1000 Hz the pressure amplitude is about 2.0 × 10^-5 Pa. Find the correspond-ing displacement amplitude.

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/Users/wenhuchen/Documents/Crawler/Physics/D37-1090.htm

Solution:

(a) The pressure amplitude is given by: y_m = P / k\rho_0v^2. The speed of sound in air is, v = 331 m/s so that k = (2\pi / \lambda) = (2\piѵ / v) = (2\pi × 10^3 / 331)m^-1 = 19m^-1. The density of air \rho_0 is 1.22 kg/m^3. Hence, for P = 28 Pa y_m = [28 / {(19)(1.22)(331)^2}]m = 1.1 × 10^-5m. The displacement amplitudes for the loudest sounds are about 10^-5m, a very small value. b) From y_m = P / k\rho_0v^2, using these values for k, v, and \rho_0, with P = 2.0 × 10^-5 N/m^2, y_m \cong 8 × 10-12m \cong 10^-11m.

Question:

In a bubble chamber, a pion collides with a proton and three particles are produced, as shown in the Figure. One of the particles is the neutral K^0, which after traveling a distance of about 1 × 10^-1 m decays into two pions of opposite charge. If the speed of the K^0 is 2.24 × 10^8 m/s, determine the rest lifetime.

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/Users/wenhuchen/Documents/Crawler/Physics/D32-0952.htm

Solution:

The Lorentz transformation, of the special theory of relativity, may be used to relate the rest lifetime of the neutral particle, t_0 to the relative lifetime, t, as measured in the laboratory frame of reference. The laboratorylifetime t is given by. t = (distance traveled between creation and annihilation , d) / (speed of the particle, v) d / v = (1 × 10^-1 m) / (2.24 × 10^8 m/s) = 4.5 × 10^-10 s Using the relativistic expression for time dilation, the neutral particle's lifetime in its rest frame is t_0 = t \surd{1 - (v^2c^2)} = 4.5 × 10-10\surd [1 - {(5 × 10^16)/(9 × 10^16)s}] = 4.5 × 10-10\surd(4/9)s = (2/3)(4.5 ×10^-10) = 3.0 × 10^-10 s

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Question:

Gyroradius :- What is the radius of the cyclotron orbit in a 10-kiloguass field for an electron of velocity 10^8 cm/sec normal to B?

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/Users/wenhuchen/Documents/Crawler/Physics/D21-0715.htm

Solution:

In the cyclotron , a particle is launched from pointS . A voltage v is applied across the \textquotedblleftgap\textquotedblright (G) . A magnetic field B exerts a centripetal force towards the centre . The particle is accelerated by the voltage v each time it crosses the gap , which increases the radius of the orbit . Thus the particle describes a spiral . The centripetal force which keeps the particle in orbit is given by F \ding{217} = (q/c) v = (q/c) v \ding{217} × B \ding{217} The magnetic field is perpendicular to the linearvelocity of the particle so that the force exerted on the particle is towards the center of the cyclotron. The magnitude of the force is (q/c) v B sin \texttheta, and since sin 90 o = 1 , we have F = q/c vB . Since within each orbit we can think of the particle as describing a circle, its acceleration is simply \omega2R or (v^2 /R) where v is the linear velocity. 2 R or (v^2 /R) where v is the linear velocity. Since F = ma (q/c) vB = m (v^2 /R) R = (mvc)/qB Substituting our values we have: R = [(0.911 × 10^-27 gm 10^8 cm/sec 3 × 10^10 cm/sec)/(4.8 × 10^-10 esu 10^4 guass)] \approx5 . 7×10-4cm 5 . 7 10 4 cm

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Question:

A chemist is given an unknown element X. He finds that the element X has an atomic weight of 210.197amu and consists of only two isotopes, ^210X and ^212X. If the masses of these isotopes are, respectively, 209.64 and 211.66amu, what is the relative abundance of the two isotopes?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E19-0707.htm

Solution:

The relative abundance of the two isotopes is equal to their fraction in their element. The sum of the fractions of the isotopes times their respective masses is equal to the total atomic weight of X. The element X is composed of only two isotopes. The sum of their fractions must be equal to one. Solving: Let y = the fraction of the 210 isotopes. Since, the sum of The fraction of the fraction is one, 1 - y = the fractions of the 212 isotope. The sum of the fractions times their masses, equals the atomic weight of X. 209.64 y + 211.66(1 - y) = 210.197; y = 0.7242 = fraction of ^210X. 1 - y = 0.2758 = fraction of ^212X. By percentage the relative abundance of ^210X is 72.42% and ^212X is 27.58%.

Question:

Both food and air pass through the pharynx. Explain how these are channeled into the proper tubes so that the food goes into the stomach and the air goes to the lungs.

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/Users/wenhuchen/Documents/Crawler/Biology/F16-0400.htm

Solution:

Under normal conditions the glottis (the opening to the larynx) is open and air passes freely into the larynx or voice box, and ultimately into the lungs. When we swallow, however, the larynx moves up so that the glottis is closed by the epiglottis (a flap of tissue) and the food or liquid passes into the esophagus behind the trachea. Swallowing is a complex reflex initiated when pressure receptors in the wall of the pharynx are stimulated. These receptors send impulses to the swallow-ing center in the medulla, which coordinates the swallowing process via nerves to the skeletal muscles in the pharynx, larynx, and upper esophagus, as well as the smooth muscles of the lower esophagus. Once swallowing has been initi-ated, it cannot be stopped. Following the swallowing reflex, the glottis once again opens, and remains open until the next swallowing reflex is initiated. Because of the placement of the epiglottis over the glottis, breathing ceases momentarily during swallowing. After the completion of the reflex, which takes about one second, breathing resumes.

Question:

An inexperienced nurse measures the blood pressure in the artery of the upper arm of a man. She then measures the blood pressure in the artery of the man's leg. The nurse obtains a different value. Why?

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/Users/wenhuchen/Documents/Crawler/Biology/F15-0386.htm

Solution:

During systole, the left ventricle contracts, forcing blood under pressure into the aorta and the arteries. Since the arterial walls are elastic (due to the elastic fibers in the outer layer of arteries), the rush of blood stretches the arteries and exerts a pressure on them. Blood pressure is the actual force exerted by the blood against any unit area of a vessel wall. The maximal pressure caused by the contraction of the heart is called the systolic pressure. In a normal adult at rest, the systolic pressure averages about 120 mm of mercury. This value means that the force exerted would be sufficient to push a column of mercury up to a level of 120 millimeters. During diastole, the heart relaxes and does exert a pressure on the arterial blood and the arterial pressure decreases. The stretched arterial walls recoil passively, maintaining some pressure which drives the blood. The pressure does not fall to zero during diastole because the next ventricular contraction occurs quickly enough to restore the pressure. The minimal pressure during the relaxation phase of the heart is called the diastolic pressure. The normal adult value is about 80 mm of mercury. The systolic and diastolic pressures are usually written as a fraction: 120/80. These values of blood pressure are only specific for the arteries in the upper arms. As the nurse should know, the blood pressure values are different in other parts of the body. The blood pressure decreases as the blood moves further away from the heart, with the lowest value occurring in the vena cava. The arterial pressure in the systemic circulation is inversely proportional to vascular resistance. That is, as resistance to blood flow increases, pressure decreases. Vascular resistance is directly proportional to the viscosity of the blood and the length of the vessel, and inversely proportional to the fourth power of the radius of the vessel. This is Poiseuille'sLaw: R = [(8 n l) / (\pi r 4)]where n = viscosity l = length r = radius All these factors tend to produce friction between the blood flowing in the vessel and the vessel walls, decreasing the blood pressure. Note that the single most important determinant of resistance is the radius of the vessel. In the aorta, the resistance is very low since the blood has not traveled far and the radius is large. Resistance in the large arteries is also low, so that the pressure is still about 100 mm Hg (Hg means mercury). The resistance begins to increase rapidly in the arterioles, as the radii of these vessels decrease. The pressure of the blood as it leaves the arterioles to enter the capillaries decreases to about 30 mm Hg. In the capillaries, the velocity of the blood flow is very low. The pressure at the beginning of the veins is about 10 mm Hg and decreases to almost 0 mm Hg at the right atrium. One would not expect the pressure to decrease so much in the veins since they are wider in width than the capillaries. This larger radius should cause a decrease in resistance. However, pressure in the veins does not increase because the walls of veins are easily distensible.Distensibilityarises from the fact that veins are not rigid structures; they have only a thin muscle coat. We know that very little pressure is generated in filling an easily distensible object, hence the pressure of the veins is negligible.

Question:

"The ovule of a rose is analogous to the capsule of a moss." Explain what is wrong with this statement.

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/Users/wenhuchen/Documents/Crawler/Biology/F08-0206.htm

Solution:

The ovule of a rose is analogous to the capsule of a moss only in that both are sporangia, meaning that they produce spores. The ovule is amegasporangiumand produces female haploid spores by meiotic division of the megaspore mother cell. One of these spores will develop into the female gametophyte. Necessarily the rose plant must also produce male spores, and indeed themicrosporangiaor pollen sacs are the site of male haploid spore production. The male spores will give rise to male gametophytes. The rose plant, then, produces two types of spores which differ in size and function, and is said to beheterosporous. The moss, like other lower plants, ishomosporous. Its sporangium is the capsule, and it is the only type of sporangia that the moss possesses. The capsule gives rise to diploid mother spore cells which undergo meiotic divisions to form haploid spores. These spores are uniform in appearance, and all germinate alike into bisexual gametophytes bearing both male and female sex organs. The ovule of a rose and the capsule of a moss differ then, in that the ovule produces only those spores which will develop into female gametophytes, while the capsule produces spores which will give rise to bisexual gametophytes.

Question:

A metal has an atomic weight of 24. When it reacts with a non-metal of atomic weight 80, it does so in a ratio of 1 atom to 2 atoms, respectively. With this information, how many grams of non-metal will combine with 33.3 g of metal. If 1 g of metal is reacted with 5 g of non-metal, find the amount of product produced.

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Solution:

To answer this problem, write out the reaction between the metal and non-metal, so that the relative number of moles that react can be determined. You can calculate the number of grams of material that react or are produced. You are told that 1 atom of metal reacts with 2 atoms of non-metal. Let X = non-metal and M = metal. The compound is MX_2. The reaction is M \rightarrow 2X + MX_2. Determine the number of moles that react. You have 33 g of M with an atomic weight of 24. Therefore, the number of moles = [(33 g) / (24 g/mole) = 1.375 moles. The above reaction states that for every 1 mole of M, 2molesof X must be present. This means, therefore, that 2 × 1.375 = 2.75 moles of X must be present. The non-metal has an atomic weight of 80. Thus, recalling the definition a mole, 2.75 mole = [(grams) / (80 g/mole)]. Solving for grams of X, you obtain 2.75 (80) = 220 grams. Let us consider the reaction with 1 g of M with 5 g of X to produce an unknown amount of MX_2. The solution is similar to the other, except that here you consider the concept of a limiting reagent. The amount of MX_2 produced from a combination will depend on the substance that exists in the smallest quantity. Thus, to solve this problem you compute the number of moles of M and X present. The smaller number, (based on reaction equation) is the one you employ in calculating the number of moles of MX_2 that will be generated. You have, therefore, M_moles = (1/24) = .04166 moles M X_moles= (5/80) =.0625 moles X. Using 0.0625 moles X, only .03125molesof MX_2 will be produced, since the equation informs you that 1 mole of MX_2 is produced for every two moles of X. For M, it is a 1 : 1 ratio, so that .04166 moles of MX_2 would be generated. Therefore, X is the limiting reagent. The atomic weight of MX_2 is 184. Thus, the amount produced is 184 g/mole (.03125moles) = 5.75 g.

Question:

A uniform cylinder rolls from rest down the side of a trough whose vertical dimension y is given by the equation y = Kx^2. The cylinder does not slip from A to B, but the surface of the trough is frictionless from B on toward C. (See figure). How far will the cylinder ascend toward C? Under the same conditions, will a uniform sphere of the same radius go farther or less far toward C than the cylinder?

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Solution:

Since we do not know the actual frictional force acting on the cylinder, we cannot use dynamical methods to solve for the final position of the cylinder. Our only other recourse is to use the prin-ciple of conservation of energy to relate the energy of the cylinder at point A to its energy at point C. By doing this, we will find an expression for the final position of the cylinder, and the problem will be solved. Since friction acts along path AB, but not along path BC, we ap-ply the principle of energy conservation in 2 steps. First, we relate the cylinder's energy at points A and B. Then, using the data ob-tained from the first step, we relate the cylinder's energy at points B and C. The cylinder begins from rest at point A, and therefore has only potential energy. Taking the reference level for potential energy at y = r (see figure), we obtain E_A = mgy_1(1) for the cylinder's energy at A. In traveling from A to B, friction is present. However, this force does no work because we are given the fact that the cylinder is not slipping. By definition, this means that the velocity of the con-tact point of cylinder and surface is zero. Under these conditions the velocity of the cylinder's center of mass, v, is related to the angular velocity by v = \omegar(2) where r is the cylinder radius. Hence, the energy of the cylinder at B is E_B = (1/2) mv^2 + (1/2) I\omega^2(3) where I is the cylinder's moment of inertia. The first term repre-sents the cylinder's translational motion, and the last term represents its rotational motion. Now, in going from B to C, no friction acts. As a result, the cylinder cannot roll, and the rotational motion it had at B is pre-served throughout the trip to C. At C, the cylinder's center of mass has no velocity, but it is still spinning, with angular velocity \omega , about its center of mass. The energy at C is then E_C = mgy_2 + (1/2)l\omega^2(4) By the principle of conservation of energy E_A = E_B and E_B = E_C. Therefore, using (1), (3) and (4) mgy_1 = (1/2) mv^2 + (1/2) I\omega^2(5) mgy_2 + (1/2) I\omega^2 = (1/2) mv^2 + (1/2) I\omega^2(6) form (2) \omega = v/r Substituting this in (5) mgy_1 = (1/2) mv^2 + (1/2) I(v^2/r^2)(7) Solving for v^2 mgy_1 = (v^2/2) [m + (I/r^2)] or v^2= (2mgy_1)/(m + I/r^2)(8) From (6) mgy_2 = 1/2 mv^2(9) We may eliminate v^2 from (9) by substituting (8) in (9), whence mgy_2 = 1/2 m[(2mgy_1)/(m + I/r^2)] then y_2 = (my_1)/(m + I/r^2) for a cylinder, I = (1/2) mr^2 and y_2 = (my_1)/[m + (1/2)mr^2 /r^2] = (my_1)/[(3/2) m] y_2 = (2/3)y_1. For a sphere, the above analysis still holds. Since I = 2/5 mr^2 y_2 = (my_1)/[m + (2/5) mr^2/r^2] = (my_1)/[(7/5) m] y_2 = (5/7) y_1. Hence, the sphere travels further.

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Question:

Compute and print out a table of the function Y = 1/(4 - X^2) inthe range 0 \leq X < 2 using a variable step interval. Start out witha step width of 0.1 (i.e. X = 0,0.1, 0.2...1.9). Reduce the stepwidth by a factor of 10 whenever X would reach 2 duringnext iteration, or whenever two successive values differby more than 10. Stop the tabulation when Y reaches 100. Details of printout: The x and Y values are to be printedin two parallel columns, in the 2nd and 4th field. Headings are to be printed as shown below. An extra blank lineis to be inserted whenever the step interval is changed. FUNCTION TABLE OF F (X) = 1/(4 - X\textasteriskcentered\textasteriskcentered2) 2 blank lines \rightarrow { VALUE OF XVALUE OF F(X) 2 blank lines{ 0\textbullet00000E + 002\textbullet50000E - 01 \textbullet\textbullet \textbullet\textbullet \textbullet\textbullet \textbullet\textbullet

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G13-0348.htm

Solution:

This problem deals with two aspects. (1) Formatting the output in desired form; Using LIST-directed input-outputmethods, assuming that at this point we are unaware of the EDIT form of input output. (2) Computation of Y. 1) Formatting: The first heading FUNCTION TABLE OF F (X) = 1/(4 - X\textasteriskcentered\textasteriskcentered2) has to be placed in the center of page. We can have 132 letters printed in a single line. 132 characters are dividedin five or six fields. Assuming a division of five fields each field has 26 characters; the center of the page would be at the 132/2 = 66th character. Now, the heading is to be centered and it has 37 characters includingblanks. Hence the 19th character i.e. letter F of F (x) must be at thecenter. The heading will be between the 2nd and 3rd fields. Leaving thefirst field, we are left with 40 characters to the center i.e. 66 - 26 = 40. Out of 40, 18 characters are towards the left i.e. the se-cond field as used bythe heading. So we have to insert 40 - 28 = 22 blanks in the PUT statement. UsingSKIP(0) and the same number of blanks followed by (37) - will underline the heading. As two blanks are required between heading & subheading, useSKIP(2) statement. The value of X and value of F(X) areto be in the 2nd and 4th fields. Hence, in the PUT statement we SKIP the1st & 3rd field by using the null string i.e. '' 2) To compute Y, we are given Y = 1/(4 - X\textasteriskcentered\textasteriskcentered2). Use the DO loop to havedifferent values of X between 0 \leq X < 2. First we increment X in steps of 0.1. At every step we store the calculatedvalue of Y in Y_OLD. Increment X and compute Y. Take the differenceof (Y-Y_OLD). If this difference is \leq10 we store this Y as computedY in Y_OLD, increment X and proceed. At the same time we checkfor X<2. When X = 2 or (Y-Y_OLD) >10, reduce the step width by a factorof 10. Also we check if Y<100. When this fails we exit the loop and endthe program. The program is as follows. TABLE:PROCEDUREOPTIONS(MAIN); DCL( (X,Y,XINT,Y_OLD) INIT (0) , (X1,XFIN,S)) FIXED(9,5); PUTLIST('' , 'FUNCTION TABLE OF F(X) = 1/(4 - X\textasteriskcentered\textasteriskcentered2) ') ; PUT LIST ('' ,')-----------------------') SKIP (0); PUTSKIP(2); PUTLIST('' , 'VALUE OF X' , '', ' VALUE OF F (X) ') ; PUT LIST ('' , '-------' , '', ' ----- ----' ); SKIP(0); PUT SKIP (2);/\textasteriskcentered WITH ABOVE STATEMENTS WE FORM THE TABLE \textasteriskcentered/ XFIN = 2; S = 0.1;/\textasteriskcentered S IS INCREMENT STEP OF X \textasteriskcentered/ LOOP1:DO WHILE (Y<100); LOOP2:DO X=XINT TO (XFIN-S) BY SWHILE((Y-Y_OLD)\leq10) ; Y_OLD =Y ;/\textasteriskcentered STORE PREVIOUS VALUE OF Y \textasteriskcentered/ Y = 1/(4-X\textasteriskcenteredX); /\textasteriskcentered COMPUTE Y \textasteriskcentered/ IF Y>100 THEN STOP; PUTLIST(' ' , X,' ',Y) SKIP; X1 = X; END LOOP2; S = S/10;/\textasteriskcentered REDUCE S BY FACTOR OF 10) XINT = X1 + S /\textasteriskcentered REINITIALIZE XINT \textasteriskcentered/ Y_OLD=Y; PUT SKIP;/\textasteriskcentered EXTRA BLANK INSERTED WHEN STEP INTERVAL IS CHANGED \textasteriskcentered/ END LOOP1; END TABLE; During the first run XINT = 0, S = 0.1/ XFIN-S = 1.9 there-fore LOOP2: DO X = 0 TO 1.9 BY 0.1 WHILE ((Y-Y_OLD)>10). We go out of this loop either when X = 2 or Y-Y_OLD) >10. Now X1 holds the value of X, when we exit from the loop. Say for the first runX1 = 1.9, X became = 2 & we exit out of LOOP2. Now S = S/10 = .01. XINT = X1 + S = 1.91 ENTER LOOP2 again whence LOOP2: DO X = 1.91 TO 2 BY .01 WHILE ((Y-Y_OLD)<10); andgo on until Y becomes \geq100. We go out of LOOP1 and proceed to END;

Question:

The central dogma of biochemical genetics is the basic relationship between DNA, RNA, and protein. DNA serves as a template for both its own replication and synthesis of RNA, and RNA serves as a template for protein syn-thesis. How do viruses provide an exception to this flow scheme for genetic information?

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Solution:

The central dogma of biochemical genetics can be summarized in the following diagram: Arrow 1 signifies that DNA is the template for its own replication. Arrow 2 signifies that all cellular RNA molecules are made on DNA templates. All amino acid sequences in proteins are determined by RNA templates (arrow 3). However, in the viral infection of a cell, RNA sometimes acts as a template for DNA, providing an excep-tion to the unidirectionality of the scheme involving replication, transcription and translation. The agents involved are certain RNA viruses. For example, an RNA tumor virus undergoes a life cycle in which it becomes a prophage integrated into the DNA of a host chromosome. But how can a single- stranded RNA molecule become in-corporated into the double-helical structure of the host DNA? It actually does not. An RNA tumor virus first adsorb to the surface of a susceptible host cell, then penetrates the cell by an endocytotic (engulfment) process, so that the whole virion (viral particle) is within the host cell. There, the particle loses its protein coat (probably by the action of cellular proteolytic enzymes). In the cytoplasm, the RNA molecule becomes transcribed into a complementary DNA strand. The enzyme mediating this reaction is called reverse tran-scriptase and is only found in viruses classified as retro- viruses. (Thus, cells do not have the capacity to transcribe DNA fron an RNA template.) Cellular DNA polymerase then converts the virally produced single stranded DNA into a double-stranded molecule. This viral DNA forms a circle and then integrates into the host chromosome, where it is transcribed into RNA needed for new viral RNA and also for viral- specific protein synthesis. The RNA molecules and protein coats assemble, and these newly made RNA tumor viruses are enveloped by sections of the cell's outer membrane and detach from the cell surface to infect other cells. The release of the new virions does not require the lysis of the host cell and is accomplished by an evagination of the outer membrane. Other cells are in-fected through the fusion of the envelope and the potential host cell's outer membrane, thereby releasing the virion into the cytoplasm. Unlike the \lambda phage infection of anE.colicell, the RNA tumor virus does not necessari-ly interfere with normal cellular processes and cause death. Thus, the enzyme reverse transcriptase of RNA viruses provides the basis for the single known exception to the standard relationship between DNA, RNA and proteins by catalyzing the transcription of DNA from RNA. The virus causing AIDS (Acquired Immune Deficiency Syndrome) is a retrovirus and thus utilizes the reverse transcriptase enzyme.

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Question:

(1) What is the electric intensity in a copper conductor of resistivity \varphi = 1.72 × 10^-8 ohm meter having a current density J = 2.54 × 10^6 amp/m^2? (2) What is the potential difference between two points of a copper wire 100 m apart?

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Solution:

(1) By definition, E^\ding{217}, the electric field, is related to the current density, J^\ding{217}, through the relationship E^\ding{217} = J^\ding{217}/\sigma But \sigma = 1/ \varphi, and therefore, \vertE^\ding{217}\vert = \vertE^\ding{217}\vert = \varphi\vertJ^\ding{217}\vert = (1.72 × 10^-8 ohm\bulletm) (2.54 × 10^6 amp/m^2) = 4.37 × 10^-2 volt/m. (2) E is related to V by V_v - V_a = \rule{1em}{1pt} ^b\int_a E^\ding{217} \bullet dl^\ding{217}. From the figure, E^\ding{217} is parallel to the axis of the cylindri-cal wire. If we evaluate (1) along a line in the direction of E and parallel to the cylinder axis, we obtain V_b - V_a = E(a - b) V_b - V_a = [4.37 × 10^-2 (volt/m)] (0 - 100m) V_b - V_a = \rule{1em}{1pt} 4.37 volts. Therefore, V_b is at a lower potential than V_a. A second method of solution is to calculate the re-sistance (R) of a length l of the wire from R = \varphil/A R = \varphil/A where A is the cross sectional area of the wire. Using Ohm's Law, V_a - V_b = [\varphil/A] I If A and the current I are given, V_a - V_b may be found.

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Question:

A 5-gm block of aluminum with at 250\textdegreeK is placed in contact with a 15-gm copper block at 375\textdegreeK. The equilibrium temperature of the system is 321\textdegreeK. The specific heat of aluminum is c_v = 0.91 joules/gm-\textdegreeK, and that of copper c_v = 0.39 joules/gm-K. What is the change in entropy of the system when the two blocks of metal are placed in contact?

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/Users/wenhuchen/Documents/Crawler/Physics/D15-0526.htm

Solution:

The entropy is defined as being the area under the mc_v /T vs T curve. We can always find this area by drawing the curve and counting the squares. However, if c_v is a constant, we can obtain the answer from a formula. The area is just ∆S = ^T2\int_T1 m (c_v/T) dt = mc_v 1n(T_2/T_1) where T_2 is the final temperature, T_1 the initial and m is the mass of material. The above expression is correct when v = const., which is approximately true for a solid. Now, the aluminum warmed up from 250\textdegreeK.; thus, ∆S_AL = 5 gm × 0.91 (joules/gm-\textdegreeK) × 1n (321\textdegreeK/250\textdegreeK) = 1.14 joules/\textdegreeK The copper was cooled 375\textdegreeK to 321\textdegreeK. ∆S_CU = 15 × 0.39 (joules/gm-\textdegreeK) × 1n (321\textdegreeK/375\textdegreeK) = - 15 × 0.39 1n (375/321) joules/\textdegreeK = - 0.91 joules/\textdegreeK We can then determine that ∆S =∆SAL+ ∆S_Cu = (1.14 - 0.91) joules/\textdegreeK = 0.23 joules/\textdegreeK There was net increase in entropy of the system by 0.23 joules/\textdegreeK. even though the internal energy remained constant.

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Question:

A 1.4447-gm sample of coal is burned in an oxygen-bomb calorimeter. The rise in temperature of the bomb (mass m_c) and the water surrounding it (mass m_w) is 7.795 F0 = 4.331C0. The water equivalent of the calo-rimeter [= m_w + m_c(c_c/c_w)] is 2511 gm. What is the heating value of the coal sample?

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/Users/wenhuchen/Documents/Crawler/Physics/D13-0481.htm

Solution:

The heat flowing in or out of a body of mass m with specific heat c is given by Q =mc∆T where ∆T is the change in temperature of the body. Therefore, Heat liberated = (2511 gm) (1 cal/gm C0) (4.331 C0) = 10,875 cal Heating value = (heat liberated / mass of heating agent) = (10.875 cal) / (1.4447gm) = 7525 (cal/gm) = 13,520 [Btu/Ib]

Question:

Distinguish between cell-mediated immunity and humoral Distinguish between cell-mediated immunity and humoral immunity. What structures are responsible for each type of immunity?

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/Users/wenhuchen/Documents/Crawler/Biology/F14-0364.htm

Solution:

It was originally thought that all antibodies were secreted into the blood or other body fluids, such as saliva, mucous and tears. Such antibodies, which function only after they are secreted into body fluids, are called humoral antibodies, and the immunity mediated by them is called humoral immunity. It eventually became clear that other types of immunological responses were due to anti-bodies which remain bound to their parent lymphocyte. The immunity mediated by these special lymphocytes is called cell-mediated immunity. A variety of antigens evoke the cell-mediated immune response for example, the bacterium Mycobacterium tuberculosis (the causative agent of tubercu-losis). We can now distinguish two different types of lymphocytes involved in the immunological response. The T, or thymus lymphocytes are responsible for cell-mediated immunity, whereas the B, or bursal lymphocytes are re-sponsible for humoral immunity. Both types of lymphocytes originate from bone marrow stem cells. Those stem cells which migrate to the thymus differentiate into T cells while those which migrate into the bursa differentiate into B cells. (The bursa is a gland in birds, although we do not yet know its mammalian counterpart.) Both B and T lymphocytes enter the circulation and colonize in lymphoid tissue such as the spleen and lymph nodes. In the presence of antigens, T lymphocytes transform into lymphoblasts, cells which do not secrete their anti-bodies, but which are involved in cell-mediated immunity. When B lymphocytes are antigenically stimulated, they transform into plasma cells which produce and secrete the immunoglobulins involved in humoral immunity. A given antigen predominately leads to either a cell-mediated or humoral immunological response. As a general rule, the humoral response is seen as a defense against bacteria and viruses in the body's fluids, while the cellular response is effective against bacteria and viruses in infected host cells as well as against fungal and protozoan infections.

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Question:

How much energy would be released if deuterium could be made to form helium in a fusion reaction?

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/Users/wenhuchen/Documents/Crawler/Physics/D36-1048.htm

Solution:

The difference in the mass of the helium nucleus, and the two deuterium nuclei, is the mass defect. \Deltam. ^mHe - 2m_2H = \Deltam Noting that 1 atomic mass unit (a.m.u.) equals 1.66 × 10^-27 kg, we may express all masses in terms of a.m.u.'s. m_2H - 2.01419 amu m_He = 4.00278 amu thus \Deltam = 0.0256 amu. This mass loss resulting from the fusion of the two deuterium nuclei into a helium nucleus, is converted into energy and released during the fusion process. E = \Deltamc^2 = (.0256 amu) × [3 × 10^8 (meter/sec)]^2 = (.0256 amu) [1.66 × 10-27(kg/amu)] (9 × 10^16 m^2/s^2) (The unit of ∆m was changed to kilograms to be compatible with the MKS system being used.) Hence E = \Deltamc^2 = .383 × 10^-11 Joules

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Question:

Determine the entropy difference between solid and liquid states for a substance melting at 100\textdegreeC and having a heat of fusion of 10,000 J/mol.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E15-0534.htm

Solution:

At the melting point, the liquid and solid are in equilibrium, which means the change in free energy (∆G), i.e., the energy available for useful work, is zero. This is due to the fact that ∆G is defined as the differ-ence of ∆G of the liquid and the ∆G of the solid. Because the system is in equilibrium, ∆G_liq= ∆G_solid. Therefore, ∆G for the whole system is ∆G_solid- ∆G_liq= 0. With this in mind, you can compute the entropy change (∆S), i.e., the change in randomness of the system from the equation ∆G = ∆H - T∆S, where ∆H = enthalpy or heat content and T = temperature in degrees Kelvin (Celsius plus 273\textdegree). For this problem you are given ∆H in the form of heat of fusion, which is the amount of heat necessary to melt one mole of solid. You are also told T. ∆G is equal to zero. Substituting: ∆G = 0 = ∆H - T∆Sor∆S = (∆H/T) ∆S = [(10,000 J/mole)/(373\textdegreeK)] = 26.8 [(J)/(mole \textdegreeK)]

Question:

A roller coaster starts from rest at the highest point of the track 30 m above the ground. What speed will it have at ground level if the effect of friction is neglected?

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/Users/wenhuchen/Documents/Crawler/Physics/D05-0235.htm

Solution:

The given observables are v_0 = 0 m/s, h_0 = 30 m, and h_f = 0 m. When the cart is at the top of the track, the total energy is all potential energy. E_0 = mgh_0 When the car is moving with speed v at the bottom, the total energy is E_T = (1/2) mv^2 + mgh_f = (1/2)mv^2since h_f = 0 Since the total energy is constant, we have, by the principle of conservation of energy, m gh_0 = (1/2)mv^2 v^2 = 2gh_0 = (2) (9.8 m/s^2) (3 × 10^1 m) = 5.88 × 10^2 m^2/s^2 v = 2.4 × 10^1 m/s Notice that the mass of the roller coaster was not required to solve the problem.

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Question:

Which tissue in the body is responsible for the rapid transmission of information?

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/Users/wenhuchen/Documents/Crawler/Biology/F02-0053.htm

Solution:

To some extent, all cells have the property of irritability, the ability to respond to stimuli. Nervous tissue, however, is highly specialized not only for receiving and responding to such stimuli, but also for the transmission of stimuli. Nerve cells are easily stimulated and can transmit impulses very rapidly. Each nerve cell is specific for the type of information it transmits and the impulse is directed and coordinated to specific areas in the body. Nervous tissue consists of neurons, cells that conduct electrochemical nerve impulses. Each neuron has an enlarged cell body, which contains the nucleus and two or more thin, hairlike processes extending from the cell body. There are two distinct types of these processes; they differ in the direction they normally conduct a nerve impulse. Axons conduct nerve impulses away from the cell body, while dendrites conduct impulses toward the cell body. The neurons are connected together in chains or networks in order to relay impulses for long distances to different parts of the body. The junction between the terminals of the axon of one neuron and the dendrite of the next neuron in line is called a synapse. The axon and dendrite do not actually touch at the synapse. There is a small gap between the two processes. An impulse can travel across the synapse only in one direction, from an axon to a dendrite. In this way the synapse functions in preventing impulses from backflowing in the wrong direction. A group of axons bound together by connective tissue consti-tutes a nerve. The functional combination of nerve and muscle tissue is fundamental to all multicellular animals except sponges. These tissues give animals their charac-teristic ability to move rapidly in response to stimuli. In other words, muscle contraction and thus movement are initiated and controlled by nervous tissue.

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Question:

Describe in words the function and the output of the following COBOL program. IDENTIFICATION DIVISION PROGRAM-ID.ID-DISPLAY. ENVIRONMENT DIVISION. CONFIGURATION SECTION. SOURCE-COMPUTER.H-200. OBJECT-COMPUTER.IBM-360-40. DATA DIVISION WORKING-STORAGE SECTION. 77IDPICX(20). PROCEDURE DIVISION. FIRST-PARA. DISPLAY "ENTER OPERATORS ID" UPON CONSOLE. ACCEPT ID FROM CONSOLE. DISPLAY "THANK YOU," ID UPON CONSOLE. DISPLAY ID. STOP RUN.

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Solution:

This is an introductory computer program written with COBOL language (CommonBusiness-OrientedLanguage) . COBOL was specifically created for handling the masses of data that are necessary in business data processing. Specific features of COBOL language are as follows: 1) Identification Division entries IDENTIFICATION DIVISION. (header) This part is used to identify the program to the computer. PROGRAM-ID. This is a special paragraph which belongs to IDENTIFICATION DIVISION. It appears in the format as follows; PROGRAM- ID. program-name The program-name identifies the program, in this case it is called ID-DISPLAY. 2)Environment Division entries: ENVIRONMENT DIVISION, header CONFIGURATION SECTION, header SOURCE-COMPUTER, computer-name entry. OBJECT-COMPUTER. computer-name entry. Environment division describes the environment, or equipment, on which the program will be run. Entries in this division will vary depending on the computer available. As specified in line 5 of the example program, the source computer is a Honeywell 200. (program compiler), and the object computer is an IBM-360-40 (program executer.). 3) Data Division entries: DATA DIVISION header WORKING-STORAGE SECTION header Level number 77 PICTURE clause Picture character X The third division of the COBOL program describes the data to be used in the procedure division. Input data, output data, and working data must all be described. In the actual format, the level number and Picture appear as fol-lows: 77 data-name PIC picture. In this format, the level number and PIC are required parts and must be used as specified. The programmer fur-nishes the data-name, in this case ID, and the actual pic-ture. The PICTURE clause tells the computer how many char-acters are in the data item, and what kind of characters they are. The most useful all purpose Picture character is X, which is used to tell the computer that the corresponding character in the data item might be anything--at least any character in the standard character set. Thus when PIC X(20) is written it tells the computer the data item is 20 characters long and they are all standard characters. 4)Procedure Division entries: PROCEDURE DIVISION, (header) DISPLAY statement ACCEPT statement DISPLAY statement DISPLAY statement STOP RUN. statement In the procedural part of the program the programmer tells the computer exactly how to solve a problem. For the PROCEDURE DIVISION there should be a paragraph- name. This paragraph-name may be referred to during the execution of the program, or it may be used for documentation, to explain the program to a reader. In the example given the paragraph-name is FIRST-PARA. Programs usually require some kind of input and output data. In COBOL, the ACCEPT statement is used to handle low volume input from either the card reader or the console key-board, and the DISPLAY statement is used to handle low-volume output to either the printer or the console. With the UPON CONSOLE option, as shown in the example program, DISPLAY will cause the message to be printed on the console typewriter or be displayed on the cathode ray tube, depending on the system hardware. For-mats for both statements are given below. ACCEPTdata-name [FROM CONSOLE.] Restrictions: 1. Brackets [ ] indicate optional clause. 2. Braces { } indicate choose one. 3. Literal must be exactly what you want displayed, enclosed in quotes. 4. Data-name must have been described in Data Division. STOP RON. This statement is used to end every program. The given example program can now be explained with more detail. After identifying the program as ID-DISPLAY, the programmer defines the machinery to be used in handling the problem, namely a Honeywell-200 to compile the program, and an IBM 360 Model 40 to solve, or rather, execute it. In DATA DIVISION, the variable ID is identified as an independent variable with the use of level number 77. Therefore the identified variable ID has no subdivisions in itself, and has 20 characters from the standard character set. Procedure division header is followed immediately by the paragraph name FIRST-PARA, then the computer is instructed to ask for the operators ID by displaying "Enter Operators ID" on either the cathode ray tube or console typewriter. The computer places the operators ID in the variable ID. Then the sentence "THANK YOU" followed by the opera-tors ID will appear on the screen or on the typewriter (de-pending on the setup) . The computer will then print the ID on the standard output device, usually a printer, because the DISPLAY ID statement did not specify an output device. The machine then stops execution with the STOP RON statement.

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Question:

An automobile accelerates at a constant rate from 15 mi/hr to 45 mi/hr in 10 sec while traveling in a straight line. What is the average acceleration?

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Solution:

The magnitude of the average acceleration, or the rate of change of speed in this case, is the change in speed divided by the time in which it took place, or a= {(45 mi/hr) - (15 mi/hr)} / (10 sec - 0) = (30 mi/hr)/10 sec. Changing units so as to be consistent, a= [(30 mi/hr) × (5280 ft/mi) × (hr/3600 sec)]/ 10 sec = (44 ft/sec)/(10 sec) = 4.4 ft sec^2 This statement means simply that the speed increases 4.4 ft/sec during each second, or 4.4 ft/sec^2 .

Question:

Why are the elements carbon, hydrogen, oxygen, and nitrogen (C, H, O, N) of immense importance in living matter ?

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Solution:

Carbon, hydrogen, oxygen and nitrogen are the four most abundant elements in living organisms. In fact, they make up about 99 percent of the mass of most cells. These four elements must possess some unique molecular ability which caused them to be se-lected to be the major constituents of life. This can be seen by comparing the relative abundance of the major chemical elements in the earth's crust to those in the human bo dy . Examining the table shows that the only close correspondence is for oxygen. Thus, we can assume that there is some reason why these 4 elements have been selected for their role in life. Earth's crust element % human body element % O 47.0 H 63.0 Si 28.0 O 25.5 Al 7.9 C 9.5 Fe 4.5 N 1.4 Ca 3.5 Ca 0.31 Na 2.5 P 0.22 K 2.5 Cl 0.08 Mg 2.2 K 0.06 Ti 0.46 S 0.05 H 0.22 Na 0.03 C 0.19 Mg 0.01 Relative abundance of the major chemical elements in earth' s crust and in the human body as the percent of total number of atoms. One property that makes these elements special is that they can readily form covalent bonds by electron pair sharing. To complete their outer shell, hydrogen needs one electron, oxygen needs two, nitrogen needs three and carbon needs four. Thus, these four elements react with themselves and each other to form a large number of covalent compounds. Furthermore, carbon, oxygen, and nitrogen can also form double bonds by sharing two electron pairs. Even further, carbon can form triple bonds. This gives these elements a lot of versatility in forming chemical bonds. Another property that makes carbon, oxygen, hydrogen and nitrogen uniquely fit for their role in living matter is that they are the lightest elements that can form covalent bonds. The strength of a covalent bond is inversely related to the atomic weights of the bonded atoms. Thus these four elements are capable of forming very strong covalent bonds.

Question:

Write a program using APL that evaluates the focal length of a lens.

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Solution:

The formula for the focal length of a lens is F = [(N) (R1) (R2)] / [(N - 1) ((N(R1 + R2) - T (N - 1))] whereF is the focal length N is the index of refraction of the glass T is the thickness of the lens R1 and R2 are the radii of curvatures. Let us first give the program a name: LENGTH. Its procedure looks as follows: \nablaLENGTH [1]NUMBER \leftarrow N × R1 × R2 [2]DENOM\leftarrow (N - 1) × (N × R1 \div R2) - T × N - 1 [3]FOCAL\leftarrow NUMBER \div DENOM [4]FOCAL [5]\nabla When writing an APL program, one must first enter the definition mode. This is achieved by typing the upside- down triangle, called "DEL", followed by a program name. The next DEL met in the procedure of the program indicates its end. The value of the numerator is stored under NUMER, and the value obtained in the denominator-under DENOM. The so-lution is stored under the variable FOCAL. Line [4] tells the computer to print FOCAL when execution begins. Line [5] which contains a DEL, indicates that the definition mode is complete. The following is a sample output: N \leftarrow 1.3275 T \leftarrow 0.375 R1 \leftarrow 8.0 R2 \leftarrow 7.85 LENGTH 12.16918 [Note that for special types of lenses, the radii of curvature can take on negative values as well. This illus-tration is the most general case to demonstrate the useful-ness of APL.]

Question:

A biology student from Nebraska, vacationing at a Virginia beach, observes that the level of activity of a fiddler crab changes with the cycle of the tides. When he brings some crabs back with him to Nebraska, he notices that a similar activity cycle is shown, yet there are no ocean beaches for miles. Explain.

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Solution:

The fiddler crab lives on the beaches of the eastern coast of the United States. These crabs are very active at low tide. At high tide, they are inactive and usually burrow in the sand so they will not be swept away to sea. Since there are approximately two low tides and two high tides every 24 hours, the crabs undergo a cycle of activity. The biology student might expect this cycle of activity to stop when the crabs are in Nebraska. The fact that it does not suggests that the actual presence of the tides is inconsequential to the crab's level of activity. That is, the sensing of the water is not criti-cal. The crucial determinant of the level of activity of the crab is not the presence of the tides, but the gravitational pull of the moon which brings about the tides. The fiddler crab has a sensory mechanism which can detect the moon's gravitational pull, and its activity cycle varies according to this pull. However, the fiddler crab in Nebraska does not have the identical cycles of activity that it showed in Virginia. Since the time of the "rising" of the moon is slightly different in Nebraska, the crab will gradually reset its activity cycle. This cycle would correspond to the times of the tides were Nebraska located on an ocean.

Question:

A car on a country road in Maryland passes over an old- fashioned hump-backed bridge. The center of gravity of the car follows the arc of a circle of radius 88 ft. Assuming that the car has a weight of 2 tons, find the force exerted by the car on the road at the highest point of the bridge if the car is traveling at 30 mph. At what speed will the car lose contact with the road?

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Solution:

The forces acting on the car at the highest point of the bridge are its weight W^\ding{217} = mg^\ding{217} downward and the normal force N^\ding{217} exerted by the bridge upward. These cannot be equal, since there must be a net downward force to provide the acceleration necessary to keep the car traveling in a circle. Thus mg - N = mv^2/r, by Newton's Second Law. N= m [g - (v^2/r)] = W/g [g - (v^2/r)] = W [1 - (v^2/rg)]. where W is the weight of the car. Here v = 30 mph = 44 ft/s. \thereforeN= 2 tons[1 - {(44^2 ft^2/s^2)/(88 × 32 ft/s^2 ) } ] = 2 [ 1 - (11/16)] ton = 5/8 ton. But action and reaction are equal and opposite. Thus, if the road exerts a force of 5/8 ton on the car, the car exerts the same force on the road. The car loses contact with the road when N = 0, that is, when v^2 = rg. Thus the speed required is v= \surdrg = \surd(88 ft × 32 ft/s^2) = 16 \surd11 ft/s = 53.1 ft/s = 36 mph.

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Question:

Bubbles of air escaping from a cylinder of compressed air at the bottom of a pond 34 ft deep are observed to expand as they rise to the surface. Approximately how much do they increase in volume if it can be assumed that there is no change in temperature?

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Solution:

Recognizing this as a situation involving Boyle's Law, it follows that V_t/V_b=p_b/pt where the subscripts b and t refer to bottom and top respectively. Furthermore, p and V represent the pressure and volume, respectively. But p_t is atmos-pheric pressure (1 atmosphere = 14.7 lb/in.^2). The pressure at the bottom of the pond is P_b = P_t + (\rhog)h where \rho is the density of water, g = 32 ft/s^2, and h is the depth of point b relative to the surface of the pond (point t). Therefore, p_b = 1atm+ (62.4 lb/ft^3) (34 ft) p_b = 1atm+ 2121.6 lb/ft^2 Since 1 lb/ft^2 = 1/144 lb/in^2 2121.6 lb/ft^2 = (2121.6/144) lb/in^2 = 14.73 lb/in^2 Because 14.7 lb/in^2 = 1atm 2121.6 lb/ft^2 \approx 1 atm. Hencep_b= 2 atmospheres (29.4 lb/in.^2) It follows thatV_t/V_b= 2/1

Question:

A uniform eccentric drive wheel is circular and of radius 4 in. It has a circular hole cut in it, of radius 1/2 in., for the drive shaft. The center of the hole is 1/2 in. from the center of the wheel. What is the location of the center of gravity of the drive wheel?

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Solution:

By symmetry the center of gravity must lie on the diameter AB which passes through O and X, the centers of the circular wheel and circular hole. Set AB horizontal and let Y, which is a distance x from O, be the location of the center of gravity of the drive wheel. Weights acting vertically are now at right angles to AB. If the circular piece removed to form the hole were replaced, the resultant of the weight W\ding{217} - w\ding{217} of the drive wheel plus the weight w\ding{217} of the piece replaced would have to be the weight W\ding{217} of the whole circle which acts at O. But the moment of the re-sultant weight about any point in the plane of the wheel must be equal to the sum of the moments about the same point of the individual forces making up the resultant. For simplicity let this point chosen be O. The weight of the circle replaced must act at X (see figure). The moment of the resultant weight about O is zero. Hence (W - w) x - w × (1/2) in. = 0. \therefore x = (1/2) in. × [w/(W - w)](1) Butw = mg W - w = (M - m)g where m is the mass of the piece originally occupying the space of the hole, and M - m is the mass of the drive wheel minus the hole. Then x = (1/2) in. × [m/(M - m)](2) If \sigma is the surface density of the material of which the drive wheel is composed m = \sigma \pi (1/2 in.)^2 M - m = \sigma \pi [(4 in.)^2 - (1/2 in.)^2] Using this in (2) x = 1/2 in. × [{\sigma \pi (1/4 in^2.)}/{ \sigma \pi (16 - 1/4) in^2.}] x = 1/2 in. × [(1/4)/(63/4)] = 1/126 in. The center of gravity of the drive-wheel is thus (4 - 1/126)in.= 3.992 in. from point A in the figure.

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Question:

Bone, like other connective tissues, consists of cells and fibers, but unlike the others its extracellular components are calcified, making it a hard, unyielding substance ideally suited for its supportive and protective function in the skeleton. Describe the macroscopic and microscopic structure of bone.

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/Users/wenhuchen/Documents/Crawler/Biology/F19-0477.htm

Solution:

Upon inspection of a long bone with the naked eye, two forms of bone are distinguishable: cancellous (spongy) and compact. Spongy bone consists of a network of hardened bars having spaces between them filled with marrow. Compact bone appears as a solid, continuous mass, in which spaces can be seen only with the aid of a microscope. The two forms of bone grade into one another without a sharp boundary. (See Fig.1.) In typical long bones, such as the femur or humerus, the shaft (diaphysis) consists of compact bone surrounding a large central marrow cavity composed of spongy bone. In adults, the marrow in the long bones is primarily of the yellow, fatty variety, while the marrow in the flat bones of the ribs and at the ends of long bones is primarily of the red variety and is active in the produc-tion of red blood cells. Even this red marrow contains about 70 percent fat. The ends (epiphyses) of long bones consist mainly of spongy bone covered by a thin layer of compact bone. This region of the long bones contains a cartilaginous region known as an epiphyseal plate. The epiphyseal cartilage and the adjacent spongy bone constitute a growth zone, in which all growth in length of the bone occurs. The surfaces at the ends of long bones, where one bone articulates with another are covered by a layer of cartilage, called the articular cartilage. It is this cartilage which allows for easy movement of the bones over each other at a joint. Compact bone is composed of structural units called Haversian systems. Each system is irregularly cylindrical and is composed of concentrically arranged layers of hard, inorganic matrix surrounding a microscopic central Haversian canal. Blood vessels and nerves pass through this canal, supplying and controlling the metabolism of the bone cells. The bone matrix itself is laid down by bone cells called osteoblasts. Osteoblasts produce a substance, osteoid, which is hardened by calcium, causing calcification. Some osteoblasts are trapped in the hardening osteoid and are converted into osteocytes which continue to live within the bone. These osteocytes lie in small cavities called lacunae', located along the interfaces between adjoining concentric layers of 'the hard matrix. Exchange of materials between the bone cells and the blood vessels in the Haversian canals is by way of radiating canals. Other canals, known as Volkmann's canals, penetrate and cross the layers of hard matrix, connecting the different Haversian canals to one another. (See Fig. 2.) With few exceptions, bones are invested by the periosteum, a layer of specialized connective tissue. The periosteum has the ability to form bone, and contri-butes to the healing of fractures. Periosteum is lacking on those ends of long bones surrounded by articular cartilage. The marrow cavity of the diaphysis and the cavities of spongy bone are lined by the endosteum, a thin cellular layer which also has the ability to form bone (osteogenic potencies). Haversian type systems are present in most compact bone. However, certain compact flat bones of the skull (the frontal, parietal, occipital, and temporal bones, and part of the mandible) do not have Haversian systems. These bones, termed membrane bones, have a different architecture and are formed differently than bones with Haversian systems.

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Question:

If 1 mole of HCI and 1 mole of NaC_2H_3O_2 are mixed in enoughwater to make one liter of solution, what will be the concentrationsof the species in the final equi-librium? K_diss=1.8 × 10^-5 for NaC_2H_3O_2.

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Solution:

To answer this question, you must consider what is happening at equilibrium. This necessitates de-fining K _dissociation' which is an equilibrium constant HCI and NaC_2H_3O_2 are strong electrolytes, which means that, in solution, they are completely dissociated. You have, therefore, H^+, CI^-, Na^+,and C_2H_3O_2^- ions present in the solution. The Na^+ and CI^- do not associate, and need not be considered. Thus, you must only consider the formationof HC_2H_3O_2 from H^+ and C_2H_3O^- . The equation for this reaction canbe written H^+ + C_2H_3O_2^- \rightleftarrows HC_2H_3O_2 This reaction can proceed in both directions, an equi-librium exists, as the doublearrow indicates. The equi-librium constant (K_eq) for this reaction is equalto [(HC_2H_3O_2)/{(H^+)(C_2H_3O_2^-)}] . K_dissociationmeasures the equilibrium quantitatively. The dissociation reactionfor HC_2H_3O_2 can be written HC_2H_3O_2 \rightleftarrows H^+ + C_2H_3O_2^-. The dissociation constant, K_diss= [{(H^+)(C_2H_3O_2^-)}/(HC_2H_3O_2)] = 1.8 × 10^-5 . By examination, you can see thatK_eqfor the association reaction is equalto l/K_diss. Thus, K_eq= (1/K_diss) = [(1)/(1.8 × 10^-5)] = [(HC_2H_3O_2)/{(H^+)(C_2H_3O_2^-)}] To rewrite into a more convenient form for solving, take the reciprocal of eachside. 1.8 × 10^-5 = [{(H^+)(C_2H_3O_2^-)}/(HC_2H_3O_2)] The final concentrations of the species, the unknowns, will be those atthe equilibrium. Let y be the concentration of HC_2H_3O_2 at equilibrium. The concentrations of both H^+ and C_2H_3O_2^- can be represented by 1 - y, Initially, you started with 1 mole/liter of each, therefore, each y mole/liter thatassociates to form HC_2H_3O_2 must be sub-tracted from the initial concentration. You can now sub-stitute these variables into the expression forK_dissto obtain [{(1 - y) (1 - y)}/(y)] = 1.8 × 10^-5 Solving for y, using the quadratic formula, you obtain y = .996. Therefore, the concentrations of the species are [H^+] = 1 - y = .004 M [C_2H_3O_2^-] = .004 M [HC_2H_3O_2] = .996 M.

Question:

Explain what is Computer - assisted numerical machine control . Explain the advantages, and describe how a program may be written.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G09-0218.htm

Solution:

Numerical machine control is the automatic control of machines by numerical data, on punched cards or tapes. The advantage of this method is that when an item is to be produced, all the minute details and operations are planned out in the office. Then the cards or tapes arepunched out, and from the moment the tape is taken to the shop floor and installed in the machine, production starts as pre-planned. Thus, management flexibility, and efficiency are both achiev-ed. Computers come in to assist when the problems become too complex to be handled manually. For example, when the tools have to move along continuous paths which form arcs of circles, parabolas, etc., the paths have to be calculated, given only a few individual points. The computer also has a number of built-in functions and subroutines, so that complicated operations just have to be called by their name, and the computer carries out all the details. In fact, some computer languages have been developed (e.g., APT) which use English-like words and abbreviations to write programs to carry out different machining operations . For simpler position control machining, the program must specify the starting and ending positions of various tools, the spindles to be used, the speeds and feed rates, various auxiliary functions such as whether coolant fluid should be on, etc. For example, the Avery Machine Tool Com-pany programming system uses commands such as; Command Description G8 0 CANCEL ALL PREVIOUS INSTRUCTIONS. G8 1 MOVES SPINDLE DOWN FOR DRILLING AND THENRAISESUP AGAIN. G8 2 DRILLS AS ABOVE, BUT DWELLS FOR A WHILE BEFORERAISINGUP. G8 4 SPINDLE DESCENDS, DOES A TAPPING OPERATION, RISES UP AGAIN. F SPECIFIES SPINDLE FEED RATE, e.g., F02 (IT COULD, e.g., MEAN 3"/MIN.). S SPECIFIES SPINDLE SPEED, e.g., S07 (IT COULD, e.g., MEAN 300 RPM). T SPINDLE SELECTION, e.g., T01 (i.e., SPINDLE NO. 1 TO BE OPERATIVE). X.... MOVE IN THE DIRECTION OF X-AXIS BY AMOUNT SPECIFIED. Y.... MOVE IN THE DIRECTION OF Y-AXIS BY AMOUNT SPECIFIED.. Z.... MOVE IN THE DIRECTION OF Z-AXIS BY AMOUNT SPECIFIED. R.... MOVE IN THE DIRECTION OF THE Z-AXIS BY THE AMOUNT SPECIFIED, BUT AT A RAPID TRAVERSING RATE. THE REST OF THE DISTANCE (Z-R) IS TO BE TRAVERSED AT THE RATE SPECIFIED BY F. M06 COOLANT ON. M30 END OF TAPE REWIND TAPE TO STARTING POINT. ETC. ETC. The above commands can be used to write out programs for the machining of variousworkpieces.

Question:

500 g of lead shot at a temperature of 100\textdegree C are poured into a hole in a large block of ice. If the specific heat of lead is .03, how much ice is melted?

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Solution:

As the lead is poured into the hole in the ice, the latter will melt (gain heat energy) and the former will cool off (lose heat energy). By the principle of conservation of energy, we may write heat lost by lead = heat gained by ice(1) Now assuming that the lead doesn't undergo a phase change during the process, we have heat lost by lead =m_lc_l\vert∆T\vert(2) wherem_lis the mass of the lead,c_lis the specific heat, and \vert∆T\vert is the magnitude of the temperature change ex-perienced by the lead. Unlike the lead, the ice changes phase during the process. Assuming that not all of the ice is melted, the portion that is melted will be in equilibrium with the remaining ice. Hence, heat gained by ice = m_i L(3) where m_i is the mass of ice melted, and L is the heat of fusion of ice. Then using (3) and (2) in (1) m_lc_l∆T_l= m_i L orm_i = (m_lc_l│∆T│) / L m_i = {(500 g) (.03 cal/g0c) │(00c - 1000c)│} / {80 cal/g} m_i = 18.75 g Note that the final temperature of the lead is 0\textdegreeC. Since not all the ice is melted, the lead comes into equilibrium with the ice and water at 0\textdegreeC.

Question:

During exercise, certain changes occur within the body of the active animal. Explain in what ways these changes affect the activity of the heart.

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Solution:

During exercise, the tissues become increasingly metabolically active and require much more oxygen and nutrients than are needed when the tissues are at rest. The heart is stimulated by this increased activity and adapts to it by increasing cardiac output- the volume of blood pumped per minute. Cardiac out-put is the product of stroke volume (blood volume pumped per beat) and heart rate. Since stroke volume and heart rate increase with exercise, the cardiac output increases: Cardiac output = Stroke volume × heart rate ml/min = ml/beat × beat/min The heart normally pumps about 75 milliliters of blood per beat, but can pump as much as 200mIsduring strenuous exercise. The normal heart rate of 70 beats per minute can increase to as much as 200 beats per minute during exercise. One change that occurs during exercise is an increase in the carbon dioxide content of the blood. As the tissues become more metabolically active, more ATP is made with a concomitant increase in carbon dioxide as a waste product. The CO_2 diffuses from the tissues into the bloodstream. This blood stimulates certain receptors in the vessels, which then stimulate the heart to in-crease its volume per beat. Increased muscle activity also causes other changes. The movement of the muscle during exercise exerts a greater than normal pressure on the veins, causing more blood to flow into the heart. The increased blood volume stretches the heart muscle. Since the contractile power of any muscle is increased (up to a point) by the tension it experiences when it begins to contract, the increased blood volume causes a greater quantity of blood to be pumped per heartbeat. The third change which occurs during exercise does not affect the blood volume per beat, but affects the heart rate. During exercise, excess heat is produced and raises the body temperature a few degrees. The extra heat causes thesino-atrialnode to increase its rate of stimulation, thus increasing the number of beats per minute. It is thought that the increased temperature increases the permeability of the muscle membrane of the S-A node to various ions (Na^+ and K^+), thus accelerating the stimulatory process. (It is the change in membrane permeability to Na^+ ions which causes an impulse to arise). This temperature sensitivity of the S-A node explains the increased heart rate accompanying a fever. The heart thus helps the tissues of the body acquire more oxygen and remove more waste by increasing both the volume of blood pumped per beat and the number of beats per minute. There are also autonomic nervous changes that occur during exercise that work to increase the cardiac output (see next question). Summary of cardiovascular changes during exercise (+ increase;- decrease) skeletal muscle blood flow + 175% systolic arterial pressure + 50% diastolic arterial pressure no change stroke volume + 20% heart rate + 100% mean arterial pressure + 15% cardiac output + 120% blood flow to kidneys - 30% total peripheral resistance - 50%

Question:

Write a BASIC program to round numbers of the form XX.X to the nearest integer. Your program should read in the numbers from a DATA statement and print out the original number and the rounded number.

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Solution:

This program introduces the library function INT(X), which returns the integer part of the input number X. We need one elementary result from numerical analysis, namely that the rounded value of X is given by INT(X + 0.5) to the nearest integer. 1\O REM ROUNDS NUMBERS 2\O PRINT "NUMBER", "ROUNDED 3\O FOR A = 1 TO 8 4\O READ X 5\O LET Y = INT(X + \O.5) 6\O PRINT X,Y 7\O NEXT A 8\O DATA 16.6, 14.8, 25.6, 4.8, 87.8, 67.2, 35.1, 93.2 9\O END It is not needed to add \O.5 to X (statement 5\O) for those systems where command INT(X) rounds off, instead of just cutting the fractional part of X.

Question:

(1) concentrated H_2SO_4 , on ethyl alcohol (2) citric acid plus KOH.

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Solution:

(1) Here, you have an acid, H_2SO_4 , reacting with the alcohol, ethanol. There exist two reactions that can occur. Which ever one predominates depends on the conditions. If the temperature is raised to 140\textdegreeC, you have the formation of ethyl ether, C_2H_5OC_2H_5. Let us investigate why. When the H_2SO_4 is added to ethanol, it protonates the OH, i.e., The protonated alcohol undergoes the following reaction: A water molecule is expelled. In the solution you have other ethanol molecules, thus, the following can occur; The C_2H_5+ needs electrons, which can be supplied by the oxygen atom in another alcohol molecule. Once this occurs, a proton is expelled to obtain the ether. The proton reacts with the existing HSO_4-, so that H_2SO_4 is not diminished. You have the following: If the temperature is raised above 160\textdegreeC, a different set of reactions occur. Again, the acid protonates the alcohol to form Water departs to yield C_2H_5+. Let us write this as This time, however, it does not react with another alcohol molecule but expels a proton immediately to form the alkene, ethylene or ethene. This is the final product, ethylene, at a temperature greater than 160\textdegreeC. (2) Here you have citric acid reacting with a base, KOH. This suggests the formation of a salt and water. To predict which salt, write out the structure of citric acid. The structure is, To understand how it reacts, you need to know the origin of its acidic nature. The acidic nature is determined by the H of the carboxylic group, COOH. Thus, you can anticipate the following reaction:

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Question:

A ball of mass 100 g is attached to the end of a string and is swung in a circle of radius 100 cm with a constant linear velocity of 200 cm/sec. While the ball is in motion, the string is shortened to 50 cm. What is the change in the velocity and in the period of the motion?

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Solution:

We define the initial state of the ball to be the circular path of 100 cm radius and the final state to be the path of 50 cm radius. The initial angular momentum of the object is L = mvr = (100 g) × (200 cm/sec) × (100 cm) = 2 × 10^6 g-cm^2 /sec The initial period is T = 1/f where f is the frequency of rotation of the ball in its initial motion. But 2\pif = \omega where \omega is the initial angular velocity of the ball. Hence T = 1/f = 2\pi/\omega Also, \omega = v/r where v is the velocity of the ball and r is its distance from the axis of rotation. Therefore T = 2\pi/(v/r) = 2\pir/v = [{2\pi × (100 cm)}/{200 cm/sec}] = \pi sec An alternate derivation is as follows. The velocity v of the object about the initial circular path is constant. Hence, v = (distance)/(time) In one period, however, the object moves a distance of one circumf-erence length. Therefore v = (2\pir)/T orT = (2\pir)/v Shortening the string does not apply any torque to the ball because the applied force lies along the line connecting the ball with the center of rotation. Therefore, the torque is zero. However, \tau= dL\ding{217}/dt \ding{217} Then O = dL\ding{217}/dt \ding{217} orL\ding{217}= constant. \ding{217} Hence the final angular momentum, L', is equal to the initial angular momentum L: L' = mv'r' = L Thus, the final velocity is v' = L/mr' = [2 × 10^6 g-cm^2/sec]/[(100 g) × (50 cm)] = 400 cm/sec . The new period is T' = 2\pir'/v' = [2\pi × (50 cm)]/(400 cm/sec) = \pi/4 sec. Therefore, decreasing the radius by a factor of 2 has increased the linear velocity by the same factor but has decreased the period by a factor of 4.

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Question:

Find all integers less than 50,000 that equal to the sum of the factorials of their digits. Write a computer program to solve this problem.

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Solution:

To gain a better understanding of the problem, consider the following examples: 1) The number 7666 is not equal to the sum of its factorial digits since 7! + 6! + 6! + 6! = 5040 + 720 + 720 + 720 = 7200 \not = 7666. 2) Let n = 145. Then 1! + 4! + 5! = 1 + 24 + 120 = 145. Hence 145 is equal to the sum of the factorials of its digits. In the writing of the program, first store the value of n! up to 9! in the memory. Then store each digit of the given number in a location cell. Actually there are only 4 numbers that satisfy the above property: 1,2,145 and 40585. But this is a problem in programming as well as number theory; so, on with the program. 5DIM D(4), F(9) 10MAT D = ZER REMEACH DIGIT OF N (5 DIGITS FOR EVERY NUMBER) REMIS STORED IN D(J). BEGIN WITH 00001. 15LET D(4) = 1 REMNOW STORE FACTORIAL VALUES. 20LET F(0) = 1 25LET F(1) = 1 30LET F(2) = 2 35LET F(3) = 6 40LET F(4) = 24 45LET F(5) = 120 50LET F(6) = 720 55LET F(7) = 5040 60LET F(8) = 40320 65LET F(9) = 362,880 70PRINT "NUMBER IS", " " , " SUM OF FACTORIALS OF ITS DIGITS IS" 75PRINT REMNOW ADD UP THE DIGITS IN EACH REMD(J), MULTIPLYING BY CORRESPONDING POWERS OF TEN FOR REMINCREASING J. 80LET S = D(4) + 10\textasteriskcentered D(3) + 100\textasteriskcentered D(2) + 1000\textasteriskcentered D(1) + 10000\textasteriskcentered D(0) 85LET S1 = 0 90LET W = 0 95FOR J = 0 TO 4 100IF W < > 0 THEN 115 105IF D(J) = 0 THEN 140 110LET W = 1 115LET V = D(J) 120IF V = 9 THEN 160 125IF V < > 8 THEN 135 130IF S < 40000 THEN 160 REM NOW ADD THE FACTORIAL OF V TO S1 135LET S1 = S1 + F(V) 140NEXT J REM NOW WE TEST FOR EQUALITY OF S AND S1 145IF S < > S1 THEN 160 150PRINT 155PRINT S, " ", " ", S1 160LET D(4) = D(4) + 1 165FOR J = 0 TO 4 170IF D(J) = 0 THEN 200 175IF D(J) < 10 THEN 200 180LET D(J) = D(J) - 10 190LET D(J-1) = D(J-1) + 1 195GO TO 165 200NEXT J 205IF D(0) < 5 THEN 80 210STOP 215END

Question:

Write a BASIC program to express current as a function of time for the circuit below. Assume that the switch is closed at time t = 0 and that the resistance R is given as a function of current byR = a + bi2 . Let E = 100 volts; L = 2 henries; a = 50 ohms, b =25 ohms/amp2Use the 4th order Runge-Kutta method. 25 ohms/amp2

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Solution:

According to Kirchhoff's voltage law, the total number of voltage gains and drops around a circuit must be zero. Hence: E - L (di/dt) -Ri = E - L (di/dt) - (a + bi2)i = 0. Isolating the rate of change of current, we get (di/dt) \equiv (E/L) - (b/L) i3 - (a/L) i. A straightforward application of the fourth order Runge-Kutta method yields i_new = i_old + (∆t/6) (k1 + 2k2 + 2k3 + k4) where k1 = (E/L) - (b/L) i3_old -(a/L) i_old k2 = (E/L) - (b/L) (i_old + (1/2) k1∆t)3 - (a/L) (i_old + (1/2) k1∆t), k3 = (E/L) - (b/L) (i_old + (1/2) k2∆t)3 - (a/L) (i_old + (1/2) k2∆t) k4 = (E/L) - (b/L) (i_old + k3∆t)3 - (a/L) (i_old +k3∆t) In the program below D is the increment on the time value. 1\O\OREM CIRCUIT ANALYSIS 1\O1REM RUNGE KUTTA METHOD 1\O5PRINT ''TIME", "CURRENT" 1\O6PRINT "(SEC)", "(AMPERES)" 1\O7PRINT 11\OREAD E,L,A,B 12\OREAD D, T2 13\OLET I1 = \O 14\OFOR T = \O TO T2 STEP D 15\OPRINT T, I1 16\OLET K1 = E/L - B\textasteriskcenteredI1\uparrow3/L - A\textasteriskcenteredI1/L 161LET K2 = E/L - B\textasteriskcentered(I1+.5\textasteriskcenteredK1\textasteriskcenteredD)\uparrow3/L - A\textasteriskcentered(I1+.5\textasteriskcenteredK1\textasteriskcenteredD)/L 162LET K3 = E/L - B\textasteriskcentered(I1+.5\textasteriskcenteredK2\textasteriskcenteredD)\uparrow3/L - A\textasteriskcentered(I1+.5\textasteriskcenteredK2\textasteriskcenteredD)/L 163LET K4 = E/L - B\textasteriskcentered(I1+K3\textasteriskcenteredD)\uparrow3/L - A\textasteriskcentered(I1+K3\textasteriskcenteredD)/L 17\OLET I2 = I1 + D\textasteriskcentered(K1+2\textasteriskcenteredK2+2\textasteriskcenteredK3+K4)/6.0 18\OLET I1 = I2 181NEXT T 8\O\ODATA 1\O\O, 2, 5\O, 25 8\O1DATA 5.0E-3, 0.1 999END Computer output reveals that a steady state current of about 1.18 amps is reached in 0.1 sec. Note that with no current dependence in the resistance this steady state current would be somewhat higher, namely 2 amps = E/a.

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Question:

In humans and some other mammalian species, offspring are usually produced singularly. However, sometimes two or more offspring are born at the same time. What are the different methods of producing twins in humans?

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Solution:

It is not unusual for human mothers to give birth to more than one offspring at the same time. Basically, there are two kinds of twinning that can occur. One results from the simultaneous release of two eggs (one from each ovary). Both of these eggs can be ferti-lized and develop. Such fraternal twins or heterozygous twins may be of the same or different sex and have the same degree of resemblance that normal brothers and sisters have. They are entirely independent individuals who have about 25% of their genetic information in common, as do normal siblings. Although they may be situated close together in the uterus, fraternal twins have separate fetal membranes. True or identical twins, also termed homozygous twins, are the products of a single egg fertilized by a single sperm. At some early stage of development, the egg divides into two (or more) independent parts, with each developing into a separate fetus. Since they come from the same fertilized egg, they are genetically identical and, therefore, are of the same sex. Such twins usually arise from the same blastocyst. Two sepa-rate inner cell masses may arise from a single blastocyst or a single cell mass may divide into two. In these cases, the twins will have separate amnions and umbilical cords, but share the chorion and placenta. If, however, a single cell mass develops into two embryos, the twins will share amnion, chorion, and placenta (see Figure). Occasionally, identical twins develop without separating completely and are born joined together, and are termed Siamese twins. All grades of union have been known to occur, from almost complete separation to fusion throughout most of the body so that only the head or legs are double. Sometimes the two twins are of different sizes and degrees of development. One might be quite normal, while the other might be an incompletely formed parasite of the first. Such errors of development usually cause death during or shortly after birth.

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Question:

The cross-sectional area of the copper wire used in a household wiring system is approximately 3 mm^2 = 3 × 10^-6 m^2. Find the resistance of a copper wire 10 m long, at 20\textdegreeC, if the resistivity of copper is 1.72 × 10^-8 ohm. m.

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Solution:

The resistivity \rho of an isotropic material is defined as the ratio of the electric field E it is placed in and the current density J flowing through it as electric field E it is placed in and the current density J flowing through it as a result of this electric field: a result of this electric field: \rho = [{E (volts/m)}/{J (amp\bullet/m^2 )}] = (E/J) (ohms/m)(1) \rho = [{E (volts/m)}/{J (amp\bullet/m^2 )}] = (E/J) (ohms/m)(1) As in the figure, we take the wire to be a cylindrical conductor and As in the figure, we take the wire to be a cylindrical conductor and apply a voltage across the ends. A finite current I and a non-zero electric apply a voltage across the ends. A finite current I and a non-zero electric field will be set up in the wire since we have an imperfect conductor field will be set up in the wire since we have an imperfect conductor (\rho \not = 0). The cross-sections at each end are small enough to be assumed (\rho \not = 0). The cross-sections at each end are small enough to be assumed to be equipotential surfaces. By definition, to be equipotential surfaces. By definition, V_b - V_a =- ^b \int _a E^\ding{217} \bullet dl^\ding{217}(2) V_b - V_a =- ^b \int _a E^\ding{217} \bullet dl^\ding{217}(2) andI = \intJ^\ding{217} \bullet ds^\ding{217}(3) andI = \intJ^\ding{217} \bullet ds^\ding{217}(3) where E^\ding{217} and J^\ding{217} are the electric field intensity and current density, where E^\ding{217} and J^\ding{217} are the electric field intensity and current density, respectively. In this case, E^\ding{217} is parallel to the axis of the wire, and if we respectively. In this case, E^\ding{217} is parallel to the axis of the wire, and if we evaluate (1) along a path parallel to the axis of the wire, we obtain evaluate (1) along a path parallel to the axis of the wire, we obtain V_b - V_a = El(4) V_b - V_a = El(4) Similarly, J^\ding{217} is parallel to the axis of the wire, and (3) may be Similarly, J^\ding{217} is parallel to the axis of the wire, and (3) may be evaluated. Hence, evaluated. Hence, I = JA(5) I = JA(5) Using (4) and (5) in (1), we obtain Using (4) and (5) in (1), we obtain \rho = (E/J) = [{(V_b - V_a )/ l} / (I/A)] \rho = (E/J) = [{(V_b - V_a )/ l} / (I/A)] \rho = [{(V_b - V_a )A } / (Il)] \rho = [{(V_b - V_a )A } / (Il)] But [(V_b - V_a )/(I)] = R, by Ohm's Law, where R is the resistance of the wire. Hence \rho = (RA/l) Solving for R, R = (\rhol/A) . Using the data provided R = [{(1.72 × 10^-8 ohm \bullet m) (10 m)}/(3 × 10^-6 m^2 )] R = .057 ohm.

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Question:

Distinguish between structural genes, regulatory genes, promotor genes, and operator genes.

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Solution:

Structural genes are those genes which code directly for the synthesis of proteins required either as enzymes for specific metabolic processes or as structural units of the cell or organism. These genes specify the primary amino acid sequence of polypeptides. Oftentimes, related structural genes, such as those whose proteins catalyze sequential reactions in a metabolic pathway, will be positioned in a linear sequence along a region of the chromosome. When these genes are transcribed, a single continuous mRNA molecule is formed. Regulatory genes code for inhibitory protein molecules known as repressors. These repressors act to prevent the activity of one or more structural genes by blocking the synthesis of their gene products. They do this by binding to operator genes. An operator gene is a specific region of the DNA molecule that exerts control over a specific group of structural genes by serving as a binding site for a given repressor molecule. When not bound to the repressor, the operator is transcribed along with the structural genes with which it is associated, though it is probably not translated. Promotor genes are also associated with a given gene or group of genes. It serves as a binding site for RNA polymerase, the enzyme responsible for transcription of DNA. The operator gene is located between the promotor and structural genes on the chromosome. When the repressor binds to the operator, it prevents the movement of RNA polymerase along the DNA molecule, thereby inhibiting trans-cription of the structural genes. When the repressor is not bound to the operator, transcription is free to occur. The term operon is used to designate a given system of structural genes, along with their associated promotor, operator, and regulatory genes. Unlike the promotor and operator, a regulatory gene is not necessarily located in the proximity of the operon with which it is associated. In addition, there is evidence that a given repressor molecule may control more than one group of structural genes. Although the specific mechanistic and physical rela-tionships of repressor, operator, promotor, and structural genes are relatively clear for microorganisms, they have not as yet been clearly defined in higher organisms. This is because crossing over, translocations, and inversions that occur in diploid organisms can act to disrupt clusters of linked genes. The overall regulation model, however, probably does operate in higher organisms.

Question:

Explain what difficulties would arise if messenger RNA molecules were not destroyed after they had produced some polypeptide chains.

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Solution:

Upon translation, mRNA molecules give rise to polypeptide chains. Normally, mRNA molecules are short- lived and are broken down by RNAase after a few transla-tions, thereby allowing a cell to control its metabolic activity. If an mRNA molecule were not destroyed, it would continue to synthesize its protein, and soon there would be an excess of this protein in the cell. This condition leads to some important implications for the cell. The continual translation of mRNA into proteins would entail a serious depletion of the energy store in the cell. For example, before an amino acid can attach to a tRNA molecule, it has to be activated. Activation is brought about by the hydrolysis of one molecule of ATP. So for each amino acid in the polypeptide chain, one molecule of ATP must be used. If translation were to proceed ceaselessly because the mRNA is not destroyed, large amounts of ATP molecules would be consumed and the energy supply in the cell depleted. In addition, the cell would accumulate proteins that it may not need. This use of large amounts of energy to produce unneeded protein would be a wasteful process. Indeed, excessive accumulation of a protein may even be harmful to the cell or organism. For example, given the right environ-ment, a protease such as pepsin, if present in more than sufficient amounts, will eat away the wall of the stomach, forming an ulcer.

Question:

Calculate the self-inductance of a coil if a rate of change of current of 5 amperes per second produces a backemfof 2 volts.

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Solution:

A change in current through the coil produces a change in magnetic flux through the coil. By Faraday's law, this changing flux induces an EMF in the coil, in such a direction as to oppose the flux change which produced it. This EMF is termed the back EMF of the coil. The magnitude of this back EMF, E, is E = L [(∆I)/(∆t)] where L is a constant of proportionality which depends on the geometry of the coil. It is termed the self- inductance of the coil. We are given that E = 2 volts and [(∆I)/(∆t)] = 5 amp/sec. Therefore L = [(E)/(∆I/∆t)] = (2/5) = 0.4henry.

Question:

Kodel is a polyester fiber. The monomers are terephthalic acid and 1,4-cyclohexanedimethanol. Write the structure of a segment of Kodel containing at least one of each monomer unit.

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Solution:

Kodel is a polymer of significantly different properties than its constituent monomers. This particular polymer segment is made by the combination of a carboxylic acid and an alcohol? this addition reaction is called esterification. Structures of the monomers are shown above in Figure A and B. The esterification reaction occurs when the - OH group on the acid breaks off and forms a water molecule with one of the hydrogen atoms of the alcohol. Thus, a monomer of Kodel looks like Figure C (above) and forms a continuous chain.

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Question:

A solenoid switch is activated when the magnetic induction on the axis is 5 × 10^-4 Wb\bulletm^-2. The solenoid has 50 turns per cm and an inductance of 180 mH, and is operated by a 12V battery. Find the time lag when it is employed in a circuit of resistance 90 \Omega.

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Solution:

The situation is shown in the figure. When the magnetic induction in the solenoid has the value B, the current i passing through the coil is given by the equation B = \mu_0 ni , where n is the number of turns per unit length of the solenoid. Further, the current at any time t after the solenoid circuit is switched on is given by the relation i' = (V/R) [1 - e^-Rt/L ] where L is the solenoid inductance.(See figure).Hence, B = \mu_0 ni' = [(\mu_0 nV)/R] [1 - e^-Rt/L ] Solving for t as a function of B, [(BR)/(\mu_0 nV)] = 1 - e^-Rt/L e^-Rt/L= 1 - [(BR)/(\mu_0 nV)] Taking the logarithm of both sides of the equation - (Rt/L) = ln [1 - {BR/(\mu_0 nV)}] Then t = - (L/R) ln[1 - {BR/(\mu_0 nV)}] Hence, when B = 5 × 10^-4 W\bulletm^-2 , t is t = - [(180 × 10^-3 H)/(90 \Omega)] ln 1- [(90\Omega × 5 × 10^-4 Wb\bulletm^-2 )/ (4\pi × 10^-7 N\bulletA^-2 × 5 × 10^3 m^-1 × 12V)] = -2 × 10^-3 × 2.303 log (1 - 0.597)s = + 4.606 × 10^-3 log(1/0.403)s = 1.818 × 10^-3 s. The time lag is thus 1.82 × 10^-3 s.

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Question:

An \alpha-particle (mass 6.64 × 10^-27 kg, charge 3.20 × 10^-19 coul) is scattered through 120\textdegree by a gold nucleus (mass 3.28 × 10^-25 kg, charge 1.26 × 10^-17 coul). The initial kinetic energy of the \alpha-particle was 1.5 × 10^-12 J. Cal-culate the impact parameter, and the distance of nearest approach of the \alpha-particle to the nucleus.

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Solution:

When a charged particle q_1 is scattered by another stationary charge q_2, the scattering angle \texttheta is determined by the Coulomb force between the two charges, the speed of the incident particle, V_o, and the distance b between the initial trajectory of the incoming particle and the line parallel to this trajectory passing through q_2. The distance b is called the impact parameter, as shown in the figure. Rutherford's formula for the scattering angle is tan \texttheta/2 = (k_Eq_1q_2)/mv^2_0b) Therefore, the impact parameter b is b = [k_Eq_1q_2)/(mv^2_0 tan(\texttheta/2)] = [(9 × 10^9 × 3.20 × 10^-19 × 1.26 × 10^-17) / (2 × 1.5 × 10-12 × 1.732)] = 7.0 × 10^-15 m

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Question:

Considering air to be an ideal gas to a first approximation, calculate the ratio of the specific heats of air, given that at sea level and STP the velocity of sound in air is 334 m\bullets^-1, and that the molecular weight of air is 28.8 g-mole^-1.

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Solution:

The speed of sound in air is c = \surd(\beta/\rho)(1) where \rho is the density of air, and \beta is its bulk modulus. The latter is given by \beta = -∆p/(∆V/V) where ∆V/V is the fractional change in volume of a volume element of air when it is exposed to a change in pressure ∆p. For infinitesimal increments, we may write \beta = - [dp/(dV/V)]or-[(Vdp)/(dV)](2) Now, the compressions and rarefactions of the sound waves travelling through air are adiabatic. Hence, the pressure experienced by a volume of air, V, must satisfy pV^\Upsilon = constant = \alpha(3) where \gamma = Cp/Cv, the ratio of the molar specific heat at constant pressure and the molar specific heat at constant volume. Then, using (3) dp/dV= (d/dV)(\alpha/V^\Upsilon) = (d/dV)(\alphaV^-\Upsilon) dp/dV= - \Upsilon \alphaV-\Upsilon-1 Since \alpha =pV^\gamma, this becomes dp/dV= -\Upsilon pV^\gammaV^-\Upsilon-1 =-\UpsilonpV^-1 Using this relation in (2) \beta = - V(-\Upsilon p V^-1) =\gammap Inserting this in (1) c = \surd(\Upsilonp/\rho)(4) But, if air is assumed to be an ideal gas, it must follow the ideal gas law, or pV =\muRT where T is the temperature of air in degrees Kelvin, and \mu is the number of moles of air in a given volume of the gas, V. Then p = (\muRT)/V Now\mu = M/M0 where M is the mass of air in a volume V, and M_0 is the mass of one mole of air. Hence p = MRT/M_0V =\rhoRT/M0 by definition of \rho. Using this in (4) c = \surd(\gammaRT/M_0) whence\Upsilon = M_0c^2/RT = (28.8 \bullet mole^-1 × 33,400^2 cm2\bullet s^-2)/(8.31 × 10^7 ergs\bulletmole^-1\bulletKdeg^-1 × 273 K deg) = 1.415.

Question:

What is the volume, in cubic centimeters, of a cube whichis 150.0 mm along each edge?

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Solution:

There are 10 mm in 1 cm; therefore, millimeters can be converted tocentimeters by multiplying the number of millimeters by 1 cm/10 mm. lengthof edge in cm = 150 mm × 1 cm/10 mm = 15 cm. The volume of a cube is equal to the length of the side cubed. volume= (15 cm)^3 = 3375 cc.

Question:

What social and behavioral characteristics are paralleled in the primates, in particular chimpanzees, gorillas and man?

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Solution:

Chimpanzees, gorillas and man, as well as some other primates, have evolved grouping lifestyles that are composed of members of their family and relatives. These social groups, as they are referred to by anthropologists, are highly organized and structured. These primates grow up with similar life cycles: (1) a prenatal stage, (2) birth, (3) infancy, (4) childhood, (5) adolescence, and (6) adult-hood. All higher primates usually space their births from 2 to 5 years apart for adequate nutrition and proper maternal care. As one ascends the evolutionary ladder of primates, there is observed a lengthening of period in which the offspring remains dependent on the mother. The lengthened prenatal and infancy periods might be necessary in the higher primates for proper development of the nervous system, which is characteristically highly complex, especially in certain regions of the brain. After birth the young stay with their mothers for a relatively long time compared to other animals. During infancy and childhood, chimpanzees, gorillas and man are relatively devoid of responsibilities. They spend their days in a play situation with their siblings and/or other young, where they learn how to conduct themselves in a social atmosphere. These peer play groups are very important in that discrete behaviors learned in this en-vironment will help to ensure the groups continued existence. As these primates enter adulthood, hunting for food becomes a primary responsibility. Food procurement by the primates highlights the great degree of sexual dimorphism in the primates, as in many other animals as well. For example, male chimpanzees possess greater strength and longer canine teeth than the females. Some sexually dimorphic characteristics have been documented to correlate to a division of labor between the males and females. Male chimpanzees hunt and are the protectors of the group, while female chimpanzees bear and care for the young and gather vegetation. Although gorillas are herbivores, both chimpanzees and humans are omnivores. The chimpanzees' diet consists of approximately 25% meat and 75% gathered vegetables. The human's diet is more variable. Hunting for meat by primates used to be regarded only as a human feature, but it has also been observed in chimpanzees and baboons. The social organization involved in hunting requires a form of group communication for trapping the prey, even when only small animals are being hunted. Chimpanzees and man bring the hunted food back to the social group where it is shared. The organization for sharing food between the young, adult males and females is a rather complex feature when compared to the lifestyles of other carnivorous animals. The social interaction arising from the hunting and sharing of food may have been the step that bound communication and group unity in the beginnings of culture. All primates communicate with each other either by vocal or nonvocal signals. Finally, primate groups occupy a home range: a specific geographic area in which they remain and protect.

Question:

How many grams of Ca(C_2 O_4 ) (calcium oxalate) will dissolve in water to form 1.0 liter of saturated solution? The K_sp of Ca(C_2 O_4 ) is 2.5 × 10-^9 mole^2 / liter^2 .

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Solution:

This problem is an application of the solubility product constant (K_sp ) expression for dissociation of a solid. In general, if a solid, A, dissociates into ions B,C,D,... according to the equation A \rightarrow bB + cC + dD + ... , the ion product constant (I.P.) is given by I.P. = [B]^b [C]^c [D]^d ... . At the point where addition of solid will result in precipitation, the ion product constant is equal to the solubility product constant. (When I.P. > K_sp , precipitation will occur.) Thus, for a solution just on the verge of precipitation, K_sp = [B]^b [C]^c [D]^d ... . The dissociation of calcium oxalate is given by the equation Ca(C_2 O_4 ) \rightleftarrows Ca^2+ + (C_2 O_4 ^2-) and the K_sp is K_sp = 2.5 × 10-^9 mole^2 /liter^2 = [Ca^2+] [C_2 O_4 ^2-] . Let the concentration of the calcium ion be x. Then, since one C_2 O_4 ^2- ion is produced for every Ca^2+ ion produced, [Ca^2+] = [C_2 O_4 ^2-] = x . Substituting into the expression for K_sp gives [Ca^2+] [C_2 O_4 ^2-] = 2.5 × 10-^9 mole ^2/liter^2 (x)(x) = 2.5 × 10-^9 mole ^2/liter^2 x = (2.5 × 10-^9 mole ^2/liter^2 )^(1/2) x = 5.0 × 10-^5 mole/liter = 5.0 × 10-^5 M . Thus, the concentration of Ca(C_2 O_4 ) will be 5.0 × 10-^5 M . To con-vert to mass, we use the relationship (mass = concentration × volume × molecular weight). The molecular weight of Ca(C_2 O_4 ) is 128g/mole. Hence, the required mass of Ca(C_2 O_4 ) is, for one liter of solution, mass = concentration × volume × molecular weight = 5.0 × 10-^5 M × 1l × 128g/mole = 0.0064g .

Question:

Explain how speciation by hybridization occurs. Contrast it withspeciation by geographic isolation.

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Solution:

Occasionally members of two different but closely related species maycross and successfully produce viable hybrid offspring. Such hybrids oftwo different species, however, are usually not fertile. Their unlike chromosomescannot pair in meiosis, and the resulting eggs and sperm donot receive the proper assortment of chromosomes. The hybrids, therefore, cannot continue their genetic line unless their chromosomes undergosome change that will allow them to synapse. One such change, whichhas been frequently observed in some species of plants, is the doublingof chromosomes in the gametes. This results in the production of tetraploidsfrom the normal diploids. In thetetraploidprogenysynapsiscan occurbetween dup-licated chromosomes, meiosis can take place in a normalfashion, and normal fertile eggs and sperm will be produced. Thereafter, thetetraploidhybrids will breed true, but will be unable to crosswith either of the pa-rental types. This is because a cross between a tetraploidhybrid and a diploid parent will produce triploid off-spring, which, becauseof the highly irregular distribution of their chromosomes at meiosis, are sterile. Thus, thetetraploidhybrids (or any otherpolyploids withan even chromosome number) become a new, distinct species. Through hybridization, the best characters of each of the original speciesmay be combined in descendents better able to survive than the originalspecies, and the worst combined in descendents that are doomed tobe eli-minated by natural selection. Hence, hybridization tends to producenew species that are better adapted than the parental species. As was previously seen, hybridization with doubling of chromosomesproduces new species with characteristics bridging the two parentalspecies. In such cases, there is a convergence of characteristics. Geographic isolation, in contrast tends to produce new species with divergingcharacteristics. Where hybridization results in formation of one speciesfrom two parental forms, geographic isolation results in two or more distinct species from one common parental form.

Question:

(a)Write a FORTRAN program to find approximate solutions to the first order differential equationdy/dx= f(x,y) with initial condition y(X_0) = Y_0 using Euler's method. (b)Write a FORTRAN program to find approximate solutions to the second order differential equation d^2y/dx^2 = g(x,y,y') with initial conditions y(x_0) = Y_0 and y'(x_0) = y'_0 using Euler's method.

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Solution:

(a) Euler's method assumes that during the interval (x,x+\Deltax),dy/dx remains constant and that \Deltay/\Deltax=dy/dx= f(x,y) \Rightarrow\Deltay=\Deltaxf(x,y). Thus, y_new=y_old+\Deltay=y_old+\Deltaxf(x_old_ ,y_old). Or, stating this as a difference equation: yn+1=y_n+\Deltaxf(x_n,y_n)(1) forn = 0,1,...,N. The FORTRAN main program is: READ N, DELTAX,X,Y PRINT, T,X,Y DO 1\O I = 1,N X = X + DELTAX CUSE EQUATION (1) Y= Y + DELTAX \textasteriskcentered F(X,Y) PRINT, X,Y 1\OCONTINUE STOP END Note: F(X,Y) must be defined in a FUNCTION subprogram (b) Rewrite the equation as a system of simultaneous first-order equations:y' =dy/dx= z z' =dz/dx= g(x,y,z) with initial conditions y(x_0) = y_0 and z(x_0) = y'(x_0) = y'_0 The FORTRAN main program is: READ, N, DELTAX,X,X,Z PRINT, X,Y,Z DO 1\O I = 1,NZPRIME = G(X,Y,Z) X = X + DELTAX Y = Y + DELTAX \textasteriskcentered Z Z = Z + DELTAX \textasteriskcentered ZPRIME PRINT, X,Y,Z 10CONTINUE STOP END Note:Again, G(X,Y,Z) must be defined. Also, for a fixed value of Ax, the error increases as \vertx - x_0\vert increases. In addition, for a fixed value ofx_n, the error increases as\Deltaxincreases and is of order (\Deltax)^2 .

Question:

Write a BASIC program to calculate the principal after N years on a balance (original principal) of B at time 0. Assume that interest is compounded yearly at 5% effective yield. Print the results for N = 1, 3, 5.

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Solution:

The formula we need is P = P_O (Hi)^Nwhere P_O = B,i= 0.05, and N takes on values 1, 3, and 5. This program introduces the STEP option in a FOR-NEXT loop. Data of P_O = B = 200 is used. 1\OREM CALCULATES PRINCIPAL ON 2\OREM A BALANCE AT 5% INTEREST 3\OREM B = ORIGINAL BALANCE, P = PRINCIPAL 4\OREAD B 5\OPRINT "PRINCIPAL", "BALANCE", "YEARS" 6\OFOR N = 1 TO 5 STEP 2 7\OLET P = B\textasteriskcentered1.05 \uparrow N 8\OPRINT P, B, N 9\ONEXT N 1\O\ODATA 2\O\O 11\OEND

Question:

Consider the formation of an ionic molecule, AB, by the donation of an electron from atom A to atom B. If the ionization potential of A is 5.5eVand the electron affinity of B is 1.5eV, and the ion B^- has a radius of 0.20 nm, what is the maximum size of A^+ that would lend itself to the formation of an energetically stable bond?

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Solution:

When two atoms form an ionic bond, the energy difference between the ionization potential of one and the electron affinity of the other goes into the electrostatic attraction between the ions of the two atoms. This energy difference is related to theinternuclearseparation between ionized atoms by the following relationship: E = 1.44 × [(q_1q_2) / r] where E is the energy difference in electron-volts (eV) , q_1 and q_2 are the charges on the ions in electrostatic units (esu), and r is theinternuclear separation in nanometers (nm) . In this problem we can determine E and use this to find r. When A ionizes to A+, it absorbs 5.5eVof energy. When B accepts an electron to form B^-, it gives up 1.5eVof energy. Hence, a total energy of E = 1.5eV- 5.5eV= - 4.0eVgoes into the formation of the AB ionic bond. The charge on A^+ is q_1 = + 1esuand that on B^- is q_2 = -1esu. One can now determine theinternuclearseparation of this bond by solving the equation E= 1.44[(q_1q_2) / r] for r to obtain: r= (1.44 q_1q_2) / E = [(1.44) (1esu) (-1esu)] / (-4.0eV) =0.36 nm. Hence, the nucleus of A is separated from the nucleus of B by 0.36 nm. Since the radius of B^- is 0.20 nm, the maximum radius of A^+(at the point where it just "touches" B^-) is 0.36 nm - 0.20 nm = 0.16 nm.

Question:

Show that momentum is conserved in a chemical reaction in which the atoms of the reactants are rearranged or ex- changed while conserving the total mass. We assume that there are no external forces (see Figs. A,B).

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Solution:

Let the reaction be represented by A + BC \rightarrow B + AC where BC means a molecule consisting of atoms B and C. In the reaction, atom C attaches itself to atom A to form AC. In the inertial frame of Figure A, let us write the equation of conservation of energy before and after the collision (here, the final total energy should also include the energy lost during collision) E_i = E_for (1/2)M_Av_A^2 + (1/2)(M_B+ M_C)v_BC^2 = (1/2)M_Bw_B^2 + 1/2(M_A + M_C)w_AC^2 + \Delta\epsilon Here \Delta\epsilon represents changes in the binding energy of the molecules taking part in the reaction. In a second inertial frame moving with velocity V with respect to the first the new velocities of the particles will be their velocities in the old frame minus the velocity of the old frame relative to the new one. With this re-placement of v_A \ding{217} by v_A \ding{217} - V \ding{217} , etc. the law of conservation of energy in the new frame is (Fig. B). (1/2) M_A(v_A - V)^2 + (1/2) (M_B + M_C) (v_BC - V)^2 = (1/2) M_B(w_B - V)^2 + (1/2)(M_A + M_C)(w_AC - V)^2 + \Delta\epsilon On writing out the squares of the quantities in parentheses and comparing this equation with the previous one, we see that the two equations are con-sistent if M_Av_A + (M_B + M_C)v_BC = M_Bw_B + (M_A + M_C)w_AC which is exactly a statement of the law of conservation of linear momentum.

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Question:

What is bile and what is its role in digestion? Where is bile manufactured and how does it reach the food undergoing digestion?

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/Users/wenhuchen/Documents/Crawler/Biology/F17-0436.htm

Solution:

Bile is very important for proper digestion, although it contains no enzymes. It is highly alkaline and helps neutralize the acid in the chyme as it leaves the stomach and enters the small intestine. This is necessary in order for the intestinal enzymes to function properly. Bile is composed of bile salts, lecithin, cholesterol and bile pigments. The first three are involved in the emulsification of fat in the small intestine. The bile pigments give bile its color. The major bile pigment is bilirubin. Bilirubin is actually a breakdown product of hemoglobin, the oxygen- carrying protein in red blood cells. In the large intestine, bilirubin and other bile pigments are further converted by bacteria into brown pigments, which give rise to the color of feces. If the bile duct is blocked so that the pigments cannot be excreted in the bile, they will be reabsorbed by the liver. Eventually, the pigments will accumulate in the blood and tissues, giving the skin a yellow color; this condition is called jaundice. Cholesterol is a large fat-like molecule that has a very low solubility in body fluids. This may lead to deposits of cholesterol in the heart and arteries, which could result in heart disease or arteriosclerosis. The liver excretes excess cholesterol in the bile. Gallstones result from the accumulation of excess insoluble cholesterol in the gallbladder. Bile salts are the most active part of bile. They are salts of glycocholic acid, which is made from cholesterol. Unlike cholesterol, bile salts are very soluble. These salts are essential for digestion of fats. Butter and oil are fats which constitute part of a group of molecules called lipids. Lipid molecules are insoluble in water and tend to coalesce to form globules. The enzymes that digest lipids, called lipases, can only work on the surface of these globules. Alone, it would take weeks for lipases to complete fat digestion in this manner. Bile salts solve this problem by having detergent-like properties - they coat the globules and break them up into millions of tiny droplets called micelles. This process, called emulsification, greatly increases the surface area exposed to attack by lipases, speeding up lipid digestion. Bile salts are conserved by the body, and are reabsorbed in the lower part of the intestine, carried back to the liver through the bloodstream, and secreted again. The liver, one of the body's largest organs, constantly secretes bile. (600 - 800 ml. a day) A network of ducts collects the bile and passes it into the gallbladder, where it is stored until needed. The gallbladder is a small muscular sac that lies on the surface of the liver. When food enters the duodenum, certain receptor cells in the wall of the intestine sense the presence of fats in the chyme. Stimulated by the fats, the receptor cells in the duodenum secrete a hormone, called cholecystokinin (CCK) into the bloodstream. This hormone causes inhibiton of gastric motility and contraction of the gallbladder, forcing bile out through the bile duct and into the small intestine.

Question:

How many moles of Al_2 O_3 can be formed when a mixture of 0.36 moles of aluminum and 0.36 moles of oxygen is ignited? Which substance and how much of it is in excess of that required? 4Al + 3O_2 \rightarrow 2Al_2 O_3

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/Users/wenhuchen/Documents/Crawler/Chemistry/E05-0193.htm

Solution:

In this reaction 4 moles of Al and 3 moles of form 2 moles of Al_2 O_3. One is told that 0.36 moles of Al and of O_2 are available for the reaction. One can see from the equation that a greater number of moles of Al are needed for the reaction than O_2. Thus, one should assume that all 0.36 moles of Al will be used, but not all 0.36 moles of the O_2. Since 4 moles of Al are needed for every moles of O_2, the following ratio holds: (4/3) = [(number of moles of Al) / (number of moles of O_2 )] If 0.36 moles of Al are used, one can solve for the number of moles of O_2 that are needed to react with them. (4/3) = [(0.36moles) / (number of moles of O_2 )] number of moles of O_2 = [(3 × 0.36) / (4)] = 0.27 moles Since only 0.27molesof O_2 are needed, and there are 36 moles present, there is an excess of 0.09 moles of O_2. From the reaction, one knows that there are 2 moles of Al_2 O_3 formed for every 4molesof Al reacted. Therefore, a ratio can be set up to determine the number of moles of Al_2 O_3 formed from 0.36 moles of Al. (4/2) = [(number of moles of Al) / (number of moles of Al_2 O_3) If 0.36 moles of Al are reacted, one can determine the number of moles of Al_2 O_3 formed. (4/2) = [(0.36moles) / (number of moles of Al_2 O_3)] number of moles of Al_2 O_3 = [(0.36 × 2) / (4)] = 0.18 moles Note: One does not determine the moles of Al_2 O_3 from O_2, since the latter is in excess.

Question:

A cylinder is designed to hold one quart of liquid (or 58 cubic inches). What should the dimensions of the cylinder be in order to minimize the amount of material necessary to construct it?

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G21-0513.htm

Solution:

We use an approximation method known as bracketing. Of course, calculus could give us a precise solution, but we will demonstrate the usefulness of this method. First we know that the surface area S of the cylinder is given by S = 2\piR2 + 2\piRH, where R is the radius and H is the height. We also know that volume V = 58 in.3 = \pi R2H. We can solve for H by dividing both sides by \piR2 : H = V/\piR2 = 58 in.3/\piR2 Substituting into the surface area formula, we have S = 2\piR2 +2\piR(58/\pitR2) After combining terms, S = (2\piR2) + (116/R) We seek to minimize S by trying various values for R. We begin with very small and very large R's, and we gradually approach the solution from both sides. A flow chart illustrates the plan of the program: K is a counter; D is the increment on radius R. We begin by initializing the radius to 1, and then we calculate the surface area. If the calculated S is less than OLDS, an arbitrary initial guess at the surface area, OLDS is replaced by the calculated S, and the radius is incremented again. If S is greater than or equal to OLDS, the radius is too big. In this case, the increment D must be reduced to one-tenth its previous size. The variable K serves as a counter of precision. When 8 digits of precision have been reached, the solution of R is printed. Below is the program segment: R = 0.0 D = 1.0 OLDS = 1000.0 K = 0 2\OR = R + D S = (2.0\textasteriskcentered3.14159\textasteriskcenteredR\textasteriskcentered\textasteriskcentered2) + (116.0/R) IF (S.GE.OLDS) GO TO 3\O OLDS = S GO TO 2\O 3\OR = R - D D = D/10.0 K = K + 1 IF (K.GE.8) GO TO 4\O GO TO 2\O 40WRITE (5,\textasteriskcentered) R [Note use of FORMAT-free WRITE statement.]

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Question:

Fiddler crabs are a dark brown color during the day and change to a pale brown color at night. Is it possible to reverse this pattern and make the crabs turn dark at night and light during the day? How?

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/Users/wenhuchen/Documents/Crawler/Biology/F31-0816.htm

Solution:

Fiddler crabs undergo a daily cycle of color change. However, this is not just a simple adjustment to conditions of light and dark. When the crabs are brought into a room, which is kept dark continuously, they still turn dark brown at dawn and light brown at sunset. These crabs have an internal sense of time, called a "biological clock." It is possible to reset this biological clock by artificially producing periods of light and darkness. For example, we can keep the crabs in a room in which a bright light is turned on at sunset and shut off at sunrise. After several days of this treatment, the crabs begin to turn dark brown at sunset and pale brown at sunrise. Even when the crabs are exposed to constant light or constant darkness, the 24 hour rhythm continues. But this cycle is 12 hours out of phase with the true day, showing that the crabs' internal clock has been reset. Rhythms of this sort, with approximately 24- hour periods, are called circadian rhythms. The biological clock of the fiddler crab can also be reset by exposing it to ice water. This type of ex-perimental manipulation seems to stop the clock. When it is taken out of the ice water, the clock restarts, but it has lost time. This manipulation seems to support the belief that biological clocks are controlled by metabolic rhythms, since metabolism slows down under very cold conditions.

Question:

A young's experiment is set up with the following characteristics: Monochromatic source (\lambda = 0.55\mu), slit separation d = 3.3mm, distance from slits to screen D = 3m (see figure). 1) Calculate the fringe spacing. 2) Place a sheet of glass with plane parallel faces and thickness e = 0.01 mm in front of slit F_1. a) Determine the direction of the displacement of the fringes and the formula giving the relationship for their displacement. b) Knowing that the fringes are displaced by 4.73 \pm .01 mm find the index of refraction of the glass and its error.

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/Users/wenhuchen/Documents/Crawler/Physics/D37-1086.htm

Solution:

Here we need to concentrate on interference between two slits. From a standard optics text we find that the intensity of the interference pattern on N slits is, I = (I_O / N^2) (sin^2N\delta / sin^2\delta)(1) where I_O is the initial intensity; I is the intensity at the screen; N is the number of slits; and \delta is given by, \delta = (\pid / \lambda )sin \texttheta(2) where \lambda is the wavelength of coherent monochromatic light from the source; d is the slit separation, and \texttheta is the angle between the normal to the slit system and the point on the screen. For small angles, sin\texttheta = (x / D) where x is the distance from the center of the screen to a point at which we wish to measure the intensity and D is the distance from slits to screen. Substituting into equation (2) gives the result \delta = (\pid / \lambda )(x / D)(3) For two slits N = 2. Substituting this into equation (1) we realize we can expand the sin^2 2\delta term because, sin2\delta = 2 sin\delta cos\delta.(4) Squaring both sides results in , sin^2 2\delta = 4 sin^2\delta cos^2\delta .(5) Substituting this into equation (1) and cancelling terms, I = I_O cos^2\delta .(6) The fringe spacing i is the distance between two maxima or two minima . We therefore let i = x_2 - x_1 and \delta = \pi . Applying equation (3), \pi = (\pidi) / (\lambdaD).(7) Solving for i, i = (\lambdaD) / (D)(8) Substituting the given values for \lambda, D, and d into equation (8), i= [{(.55 × 10^-6)(3)} / (3.3 \bullet 10^-3)] = 5 × 10^-4m .(9) Now we insert a sheet of glass of thickness 0.01 mm and index of refraction n in front of one of the slits. The light of this slit will now travel a different optical distance to a position x on the screen. Initially before the glass plate was inserted in front of the slit, the path difference was, \delta_1 = (F_1M) - (F_2M) = x(\lambda / i).(10) The glass plate will add an additional path difference (n-1)e 30 the new path difference is, \delta = x (\lambda / i) + (n - 1)e(11) so, the new \delta for a maximum will be p\lambda (p being an integer) or, p\lambda = x (\lambda / i) + (n - 1)e .(12) Solving for x, x = (i / \lambda) [(p\lambda) - (n - 1)e](13) so the shift in fringes will be the change in x: ∆x =-(i / \lambda)(n - 1)e .(14) To measure the index of refraction we solve for n: n = 1-(\lambda / i)(∆x / e) .(15) So we do the measurement as follows. Observe the fringe system with no glass plate, record the positions x of the maxima, then insert the glass plate, record the new positions of the maxima and determine the shift ∆x of the maxima. In our problem ∆x = -4.73 mm, substituting values into equation (15), n = 1 - [(0.55 x 10^-3) / (0.5)] [(-4.73) / (10^-2)]= 1.5203.(16) To determine the error in n we want to look at the relationship d(∆x) / ∆x. Differentiating equation (15) we have dn = -(\lambda / ie)d(∆x).(17) Solving equation (14) for -(\lambda / ie) , -(\lambda / ie)= (n - 1) / (∆x).(18) Substituting for -(\lambda / ie)in equation (17), dn = (n-1)[d(∆x) / (∆x)] . Substituting values dn = 0.5 [(2 × 10^-2) / (4.73)] = 2 × 10^-3 , so n = 1.520\pm0.002.

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Question:

Write a FORTRAN program that reads in two positive integers of at most five digits, and outputs allpalindromic numbers lying between the integers.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G22-0540.htm

Solution:

A famous palindrome in English is the line spoken by Napoleon: 'Able was I ere I saw Elba'. The letters occur from right to left in the same order as they appear from left to right.Palindromicnumbers are positive integers having this "mirror property", such as 44, 2992, and 75757. The integers 1,2,...,9 are also included aspalindromicnumbers. We want to search for allpalindromicnumbers between two positive integers which lengths are at most 5 digits. The approach used in solving this problem is to store the first integer in array K, digit-by-digit in reverse order. Then a comparison is made to determine whether the integer is equal to its "mirror image", the reversed-order integer. If so, the palindromicnumber is printed out. This process continues until the upper parameter is exceeded. If, after all the integers between the parameters are checked, there exist nopalindromicnumbers, a message is printed out indicating this fact. The program terminates when a number less or equal to zero appears on the data card as one of the two parameters. DIMENSION K(5), L(500) 10READ (5,100) KONE, KIWO 100FORMAT (215) CDO WHILE KONE OR KIWO IS GREATER THAN ZERO IF (KONE.LE.O.OR.KTWO.LE.O) GO TO 99 J = 0 CTEST THE INTEGERS KONE, KONE + 1,... KTWO DO 60 NTEST = KONE, KTWO NTEMP = NTEST CSTORE NTEMP DIGIT-BY-DIGIT IN REVERSE ORDER CIN ARRAY K DO 40 I = 1,5 M = 6 - I KTEMP = NTEMP/10 K(M) = NTEMP-(KTEMP\textasteriskcentered10) IF (KTEMP.LE.O) GO TO 20 NTEMP = KTEMP 40CONTINUE 20M2 = (M + 5)/2 DO 30 JTEMP = M,M2 LTEMP = 5 + M - JTEMP IF (K(JTEMP).NE.K(LTEMP)) GO TO 60 30CONTINUE J = J + 1 L(J) = NTEST 60CONTINUE IF(J.GT.O) GO TO 70 WRITE (5,101) KONE, KTWO 101FORMAT (1X,'NO PALINDROMES BETWEEN', I5 , 'AND', 15) GO TO 10 70WRITE (5,102)(L(N), N = 1,J) 102FORMAT (10(1X,I6)/) GO TO 10 99STOP END

Question:

2 moles of hydrogen chloride are to be made from 1 mole each of hydrogen (H_2) and chlorine (Cl_2) at 25\textdegree and 1 atm. Calculate the G for this chemical reaction. STANDARD ENTHALPIES OFENTROPIES OF CERTAIN FORMATIONSUBSTANCES ∆H;AT 25\textdegreeC AT 25\textdegreeC AND 1 ATM Compound∆H; Substance S(cal deg^-1 mole^-1) NO (g)21.600 kcal/moleH_231.21e.u. NO_2 (g)8.091O_249.00 HF (g)-64.2Cl_253.31 HCl(g)-22.063N_245.77 HBr(g)-8.66HCl44.61 HI (g)6.20

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/Users/wenhuchen/Documents/Crawler/Chemistry/E15-0533.htm

Solution:

Free energy (∆G) may be expressed quantitatively as ∆G = ∆H - T∆S, where ∆H is the change in enthalpy, ∆S is the change in entropy, and T is temperature in degreeskelvin. Entropy measures the randomness of the system, while enthalpy is the heat content. The reaction may be written H_2 (g) + Cl_2 (g) \rightarrow 2HCl (g) Because free energy is defined in terms of ∆H, ∆S, and T, you can see these values to solve for ∆G. The table of standard enthalpies of formation, states that ∆H\textdegree_f,_HCl= - 22.063 kcal/mole. Two moles ofHCl are formed, ∆H for the reaction is 2 moles × (-22.063) ^cal/_mole .The ∆S is the final minus theinitalentropy; S_f S_f - S_i = S_2HCl - S_(H)2 - S(Cl)2 - S_i = S_2HCl - S_(H)2 - S(Cl)2 = 2 (44.61) - 31.21 - 53.31 = 4.70 cal deg^-1 Substituting into ∆G = ∆H - T∆S; G = -44, 123 cal - (298) (4.70eu) = - 45, 523 cal.

Question:

The current i in a long straight wire is increasing at a steady rate i = 3.36 (1 + 2t) × 10-2amps. A small circular loop of wire of radius a = 0.1 cm is in a plane through the wire and its center is a distance r = 100 cm from the wire (figure). If the resistance of the loop is R = 8.99 × 10^-4 ohms what is the induced current i flowing round it and in which direction does it flow?

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/Users/wenhuchen/Documents/Crawler/Physics/D22-0761.htm

Solution:

The current i in the straight wire produces cir-cular magnetic lines of force. In the vicinity of the loop these lines are perpendicular to the plane of the loop. Be-cause the current in the long wire is time dependent, the magnetic field which it sets up at the site of the circu-lar coil will change with time. Furthermore, being that magnetic flux is defined as \varphi = \int B^\ding{217} \textbullet ds^\ding{217},(1) (where B^\ding{217} is the magnetic induction and the integral is a surface integral), \varphi will also be time dependent. Then, by Faraday's Law E.M.F. = [(d\varphi)/(dt)](2) an E.M.F. (electromotive force) will be induced in the coil of wire. If the resistance of this loop is R, then by Ohm's Law, the current in the loop is i_l= (E.M.F.)/R = -(1/R) (d\varphi/dt)(3) To evaluate this current, we must find \varphi , and, hence, B^\ding{217}. To calculate the latter, we use Ampere's Law, or \int B^\ding{217} \bullet dl^\ding{217} = (\mu_0 i enc(4) This integral must be evaluated over a closed path, i_enc is the net current passing through this path and \mu_0 = 4\pi × 10^-7 Weber/amp. m. The lines of magnetic induc-tion, B^\ding{217}, for a long wire are circles centered on the wire. Hence, to evaluate (4) , we use a circular path of radius r centered on the wire. In this way, B and di will be paral-lel, and, from (4) B2\pir = \mu_0 i_enc andB = (\mu_0 i_enc /2\pir) We may next find \varphi. Because the area elements ds^\ding{217} of the circular loop are all parallel to the lines of B^\ding{217} set up by the long wire, f reduces to \varphi = \int B^\ding{217} \textbullet ds^\ding{217} = \int Bds. Since the radius a of the loop is very small compared with its distance r from the wire, it is reasonable to assume that the magnetic field has this value at all points inside the loop. The flux through the loop is therefore \varphi = BA = [(\mu_0 i_enc )/(2\pir)] (\pia^2 ) But i_enc = 3.36 (1 + 2t) × 10^-2 amp/s and \varphi = [{(\mu_0 (3.36) (1 + 2t)10^-2 amp/s a^2 )}/(2r)] Using(3) i = - (1/R) (d/dt) [{\mu_0 (3.36) (1 + 2t)10^-2 amp/s) a^2 }/(2r)] = - (1/R) [\mu_0 (3.36) (2) {10^-2 (amp/s)} a^2 ]/(2r) = {[4\pi × 10^-7 (Weber/amp\bulletm)] (6.72) [10^-2 (amp/s)] (10^-6 m^2 )}/{(2) (1m) (8.99 10^-4 ohms)} = [(4.22 × 10^-14 Weber)/(8.99 × 10^-4 ohms \bullet s)] = 4.70 × 10^-11 (Weber/ohms \bullet s) . But 1 Weber = 1 (Newton \bullet m/amp) and 1 ohm = (1 volt/amp) = 1 joule/amp coul) . i_l = 4.70 × 10^-11 (Newton \textbullet m/amp s) \bullet (amp \bullet coul/joule) = 4.70 × 10^-11 [(Newton \textbullet m \textbullet coul)/(Newton \bullet m \bullet s)] = 4.69 × 10^-11 amp The magnetic field produced by the straight wire goes through the loop into the page in the figure and it is increasing with time. The current i_1 in the loop tries to counteract this by producing a magnetic field coming out of the page. The induced current must therefore flow counterclockwise round the loop.

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Question:

What is the minimum length L of a wall mirror so that a person of height h can view herself from head to shoes?

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/Users/wenhuchen/Documents/Crawler/Physics/D27-0844.htm

Solution:

This is not easily solved by a diagram. We suppose that the person stands a distance x from the wall and that her eyes (E) are a distance y from the top of her head (H). To look at her toes (F) she looks at point A which is the point of reflection of a light ray from her foot. A must be at a height halfway between her eyes and feet (so that the angle of incidence equals the angle of reflection). Similarly to look at the top of her head she looks at point B. If OP = h and BP = y/2, then the length of the mirror is AB = PO - BP - OA = h - y/2 - 1/2 (h - y) = h/2. Thus the minimum length of the mirror is h/2, and this does not depend on the distance x that the person is standing away from the mirror.

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Question:

a) Calculate the ratio of the volume of a hepatocyte to that of an E. coli cell. Assume the hepatocyte to be a cube 20\mu on an edge, b) Calculate the ratio of their surface areas, c) Calculate the (surface / volume) ratios for each.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E22-0786.htm

Solution:

A hepatocyte is an eucaryote, a large complex cell, and an E. coli cell is a procaryote, a small simple cell. An E. coli cell is cylindrical with a length of 2.0\mu and a cross-sectional diameter of 1.0\mu. The volume of a cylinder is equal to the product of its length and the area of its cross- section. Volume of an E. coli cell = 2.0\mu × \pi × (0.5\mu)^2 = 1.57\mu^3 Volume of a hepatocyte = (20\mu)^3 = 8.0 × 10^3\mu^3 The volume ratio is (hepatocyte / E. coil) = 8.0 × 10^3\mu^3 = 5.1 × 10^3 : 1 b) The surface area of a solid cylinder is equal to 2\pirl + 2\pir^2, where r is the cross-sectional radius and l is the length. Surface area of an E. coli = 2\pi (0.5\mu) (2.0\mu) + 2\pi(0.5\mu)^2 = (6.28\mu)^2 + (1.57\mu)^2 = 7.85\mu^2 The surface area of a cube is 6s^2, where s is the length of the edge. Surface area of a hepatocyte = 6(20\mu)^2 = 2.40 × 10^3\mu^2 The ratio of their surface areas is (hepatocyte) / (E.coil) = (2.40 × 10^3 u^2) /(7.85 \mu^2) = 306 : 1. (c) Surface to volume ratiost E. coli; (suface) / (volume) = (7.85\mu^2) / (1.57\mu^3) = 5\mu^-1 Hepatocyte; (suface / volume) = (2.40 × 10^3\mu^2) / (8.0 × 10^3\mu^3) = 0.3\mu^-1.

Question:

Two devices, whose resistances are 2.8 and 3.5\Omega, respective-ly, are connected in series to a 12-V battery. Compute the current in either device and the potential applied to each.

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/Users/wenhuchen/Documents/Crawler/Physics/D20-0675.htm

Solution:

Resistances in series add. The equivalent re-sistance of the two devices is given by R = 2.8\Omega + 3.5\Omega = 6.3\Omega . The current supplied by the battery can be computed using Ohm's law, I = (V/R) = [(12 V)/(6.3\Omega)] = 1.9 A . This current I flows through both devices since they are in series. Ohm's law may be applied to calculate the potential applied to each device: (see figure) V_1 = IR_1 = (1.9 A) (2.8\Omega) = 5.3V V_2 = IR_2 = (1.9 A) (3.5\Omega) = 6.7V. Note that the sum of these potentials is equal to the battery potential, 12V.

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Question:

What is meant by embryonic development? Describe the various stages of embryonic development.

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/Users/wenhuchen/Documents/Crawler/Biology/F23-0589.htm

Solution:

Embryonic development begins when an ovum is fertilized by a sperm and ends at parturition (birth). It is a process of change and growth which transforms a single cell zygote into a multicellular organism. The earliest stage of embryonic development is the one-celled, diploid zygote which results from the fer-tilization of an ovum by a sperm. Next is a period called cleavage, in which mitotic division of the zygote results in the formation of daughter cells called blastomeres. At each succeeding division, the blastomeres become smaller and smaller. When 16 or so blastomeres have formed, the solid ball of cells is called a morula. As the morula divides further, a fluid-filled cavity is formed in the center of the sphere, converting the morula into a hollow ball of cells called a biastula. The fluid filled cavity is called the blastocoel. When cells of the blastula differentiate into two, and later three, embryonic germ layers, the blastula is called a gastrula. The gastrula period generally extends until the early forms of all major structures (for example, the heart) are laid down. After this period, the de-veloping organism is called a fetus. During the fetal period (the duration of which varies with different spec-ies) , the various systems develop further. Though de-velopmental changes in the fetal period are not as dramatic as those occurring during the earlier embryonic periods, they are extremely important. Congenital defects may result from abnormal de-velopment during this period.

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Question:

A small boy pumps up his bicycle tires on a day when the temperature is 300\textdegreeK. Find the temperature of the air in the bicycle pump if the tire pressures are to be 24.5 lb/in^2 and the air in the pump is assumed to be compressed adiabatically. For air, \gamma = 1.40.

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/Users/wenhuchen/Documents/Crawler/Physics/D15-0527.htm

Solution:

During an adiabatic process, the quantity p(1 - \gamma)/\gammaTremains T constant, when p and T are the press-ure and the temperature of the gas. In the final stages of the pumping, air at T_1 = 300\textdegreeK and atmospheric press-ure p_1 = 14.7 lb/in^2 is drawn into the bicycle pump and compressed adiabatically to a pressure of p_2 = 24.5 lb/in^2 and a temperature T_2. Therefore p_1(1 - \gamma)/\gammaT_1 = p_2(1 - \gamma)/\gammaT_2 orT_2/T_1 = (p_1/p_2)(1 - \gamma)/\gamma Taking the logarithm of both sides log (T_2/T_1) = [(1 - \gamma)/\gamma] log (p_1/p_2) = - (0.40/1.40) log (14.7/24.5) = 0.0634. Using the logarithmic table, we get T_2/T_1 = 1.157 andT_2 = 1.157 T_1 = 1.157 × 300\textdegreeK = 347\textdegreeK = 74.1\textdegreeC.

Question:

DIMENSION A(10,15,7) Assuming that elements of arrays are stored in lexicographic order of their indices, find the displacement of the location of the element A(5,7,4) relative to the location of A(1,1,1). (b)If A is a three-dimensional FORTRAN array of dimension n_1 × n_2 × n_3, find the location of the element A(I,J,K) in terms of the location of A(1,1,1) and I, J and K, where I,J and K are simple integer variables. Assume that elements are stored in lexicographic order of their indices.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G12-0285.htm

Solution:

(a) All conventional computers require a multidimensional array to be stored as a linear sequence of elements. The compiler must, therefore, contain a procedure that computes the actual location of an element in an array from its specification in terms of indices. The map-ping from the multidimensional array to the linear array is not unique. However, if we assume that the elements of the multidimensional array are stored in the lexicographic order of their indices, wecaysay that A(5,7,4) comes in the fourth column of the seventh row of the fifth plane of the array; i.e., it is preceded by four planes, six rows and three elements. But the FORTRAN statement DIMENSION A(10,15,7) allocates a storage area equivalent to 10 planes of 15 rows with 7 elements in each row. Each of the 10 planes, therefore; contains 15 × 7 elements. If loc(X) is the location of element A(X), then loc(5,7,4) = loc(1,1,1) + 4(15\bullet7) + 6(7) + 3 = loc(1,1,1)+ 420 + 42 + 3 = loc(1,1,1) + 465. Hence, the displacement of the location of the element A(5,7,4) relative to the location of A(1,1,1) is 465. (b) Generalizing the reasoning of part (a), we can say that A(I,J,K) comes in the K-thcolumn of the J-throw of the I-thplane of the array, i.e., it is preceded by (I-1) planes, (J-1) rows, and (K-1) elements. But A's dimensions are n1× n_2 × n_3, which means that each plane contains n_2 rows and n_3 columns, or n_2n_3 elements. Therefore, loc(I,J,K) = loc(1,1,1) + (I-1)n_2n_3 + (J-1)n_3 + (K-1) The part played by array dimensions n_1, n_2, n_3 in storage mapping is one reason why FORTRAN (and most other programming languages) require the actual dimensions of arrays to be declared. Some compilers, e.g., the WATFIV FORTRAN compiler, generate a code that determines during execution of the program whether a reference to a subscripted variable lies out-side the declared array bounds. If it does, and were to remain undetected, the information might get recorded in locations that house other data, or even in the program itself, altering the program and, hence, its behavior Since such compilers as WATFIV are used for program testing, the increase in execution time due to these tests is well worth it.

Question:

How many calories are developed in 1.0 min in an electric heater which draws 5.0 amp when connected to a 110-volt line?

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Solution:

A resistance (the electric heater) which draws 5.0 amp when connected to a 110 volt line develops power (or energy per unit time) given by P = energy/time = I^2R But, by Ohm's Law, V = IR, hence P = I(IR) = IV The energy developed in 1.0 min is then E = IVt = 5 amp × 110 volts × 1 min × 60 sec/min = 33,000 Joules (The unit of time was converted to seconds to make it compatible with the MKS system being used.) since 1 calorie = 4.19 Joules E = 33,000 Joules × (1 cal / 4.19 joules)= 7.9 × 10^3 cal.

Question:

A chemist expands an ideal gas against a constant ex-ternal pressure of 700 mmHg, and finds its volume changes from 50 to 150 liters. He finds that 1.55 Kcal of heat have been absorbed in the process. Determine the internal energy change that took place. 24.217 cal = 1 liter-atm.

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Solution:

The internal energy change of a system at constant pressure is given by the formula ∆E = ∆H - P∆V, where ∆E = change in internal energy, ∆H = enthalpy or heat absorbed or released, P = pressure and ∆V = change in volume. In this problem, ∆V = 150 - 50 = 100 liters. The pressure, in atms, = (700 mm)(1 atm)/(760 mm) = .921 atm. To convert, liter-atm to kilocalories, multiply by 24.217 cal./liter-atm and then multiply by Kcal/1000 cal. Therefore, ∆E = 1.55 Kcal - (.921 atm)(100 l) (24.217 cal/liter-atm) (Kcal/1000 cal) = 1.55 - 2.23 = - .68 Kcal.

Question:

If it were not for Arithmetic Registers like the accumula-tor (AC), we would find it almost impossible to do effective programming. Using illustrations, indicate the basic proper-ties of the Accumulator.

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Solution:

Memory units, such as fullwords, are quite valu-able, but if we are unable to alter or operate on them, they become quite useless. The accumulator facilitates this. Let's for the moment assume that each unit of memory can hold four decimal digits. The Accumulator may be regarded as a special unit of memory in the computer. For simplic-ity, we will also assume that it also holds four decimal digits. 1) The AC has the power to copy data from any word in memory without destroying the word's contents. This process is analogous to making a photostat copy. This copy can then be used in other operations. The CLA (clear and add) instruction will do this copying. It first clears the ac-cumulator and then adds the contents of the specified loca-tion to it, as shown in Example 1. 2) The AC can also reverse the above process; ;it can store (STO) its contents in a specified word, with the use of the STO instruction. 3) The AC also performs arithmetic instructions. Thus, an ADD (add) instruction adds the contents of any specified word to the contents of the AC and retains the accumulated sum in the AC. To enable the AC to perform some other task, the new sum is then stored in a convenient location. Storage may be done as shown in Example 2. Since subtraction is merely another form of addition, the AC can also be used in subtractions. Using the SUB (subtract) instruction, this calculation may be done as in Once again, the sum will be stored in a convenient location of memory. The above are the basic properties of the accumulator. But the great flexibility that results from the AC can be seen in much more depth by studying its use in conjunction with another arithmetic register referred to as an MQ. The Multiply Quotient (MQ) register is generally combined with the AC when multiplications and divisions are to be done. If we assume that each register has a length of 17 bits, then in a multiplication or division, they merge to form a simple 34-bit field. In this 34-bit field, the MQ occupies the low order 17 bits. After a multiplication or division, the result is stored starting with the MQ register. In other words, the results are right-justified. In the case of division, the remainder is stored in the AC field.

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Question:

In a simple ecosystem there exists a field of clover. Mice eat theclover and skunks eat the mice. The skunks are in turn eatenby wolves. According to the ten percent rule of ecologicalefficiency, how much energy would a wolf receive fromthe original clover plants?

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Solution:

The average ecological efficiency pertrophiclevel is 10 percent. Therefore we expect that for every 10,000 calories available from clover plants, 1,000 calories will be obtained for use by a mouse. When a skunk eatsthis mouse, only 100 of the 1,000 calories will be available to the skunk. In the lasttrophiclevel of this food chain, the wolf will obtain 10 percentof the 100 calories transferred to the skunk. Thus, a mere 10 caloriesof the original 10,000 calories can be used for the metabolic processesof the wolf. A toptrophiclevel carnivore that is receiving only onethousandth of the original calories of the plants must be sparsely distributedand far ranging in its activities because of its high food consumptionrequire-ments. Wolves must travel as much as twenty miles a dayto acquire enough food. The territories of individual tigers and other greatcats often cover hundreds of square miles. If there were predators of wolves, these predators would only be able to make use of 10 percent of the10 calories that the wolf obtained or, one calorie. Hence, it is hardly worthwhilepreying on animals in the uppertrophiclevels.

Question:

Find the force on an electrically charged oil drop when it moves at a speed of 1 × 10^2 (m/s) across a magnetic field whose strength is 2 T. Assume that the charge on theoildrop is 2 × 10^-17 C.

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Solution:

The speed of the particle, v = 1 × 10^2 (m/s), the field strength, B = 2 T, and the charge, q = 2 × 10^-17 C, are given. The force on the drop due to its motion through the magnetic field is given by F^\ding{217} = q^\ding{217} v^\ding{217} × b^\ding{217}(1) Since the drop moves across the magnetic field, the angle between v^\ding{217} and B^\ding{217} is 90\textdegree. Equation (1) then reduces to \vertF^\ding{217}\vert =qvB= (2 × 10^-17 c) (1 × 10^2 m/s) (2 T) = 4 × 10^-15 N The force acts in a direction perpendicular to both B^\ding{217} and to v^\ding{217}. This force is very small compared with the weight of even a very small oil drop.

Question:

Discuss the role of a stimulus that elicits a behavior pattern.

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Solution:

Innate behavior patterns occur in response to certain stimuli detected by an animal's sensory re-ceptors. Of the myriad of stimuli encountered in any situation, an animal responds only to a limited number of stimuli. These stimuli, which elicit specific res-ponses from an animal, are referred to as sign stimuli. The sign stimuli involved in intra-specific communication are called releasers. The intensity of the stimuli necessary to evoke a behavior pattern is inversely proportional to the animal's motivation to perform that behavior. Motivation is the internal state of an animal which is the immediate cause of its behavior. The motivational aspects of behavior can be characterized by terms such as "drives", "goal- oriented behavior", and "satiation". Motivation is partly determined by hormones, influences from the brain, and previous ex-periences. The animal must possess neural mechanisms selectively sensitive to particular releasing stimuli (or sign stimuli). When the releasing or sign stimuli are appropriate to stimulate an animal and the animal is sufficiently motivated, a behavior pattern is initiated. The intensity of the stimuli necessary to evoke a behavior pattern is inversely proportional to the animal's motivation to perform that behavior. Thus, an animal highly motivated to perform a given behavior will require a less intense stimuli to invoke that behavioral response than an animal which is not as highly motivated.

Question:

Skeletal muscle contains about 5 × 10^-6 mole of ATP and (30 × 10^-6) mole of phosphocreatine per gram wet weight. Calculate how much work (in calories) a 400\rule{1em}{1pt}gram muscle can theoretically carry out at the expense of its high- energy phosphate bonds alone, assuming that both glycolysis and respiration are inhibited.

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Solution:

When they are hydrolyzed, the high-energy phosphate bonds of ATP and phosphocreatine contribute 7.3 × 10^3 and 10.3 × 10^3 cal/mole, respectively. With this information, one can calculate the amount of work that can be carried out by this muscle. total work = work done by ATP + work done by phosphocreatine The number of moles of each compound is found by multiplying the number of moles per gram by the weight of the muscle. This product is then multiplied by the number of cal / mole that are produced when the phosphate bonds of the particular compound are broken. No. of cal produced by ATP = 5.0 × 10^-6 moles/g × (400 g) × (7.3) × 10^3 (cal / mole) = 14.6 cal. No. of cal produced by phosphocreatine = 30 × 10^-6 mole/g × 400 g × 10.3 × 10^3 cal/g = 123.6. Total no. of cal produced = 14.6 + 123.6 = 138.2 cal.

Question:

A doctor is testing the potency of a new drug. He compares two groups of samples, an experimental group and a control group. Each group contained 10 samples and yielded 10 results. Every result was assigned a number from 1 to 10 according to the degree of success. Do the two means of these groups differ significantly? Write a FORTRAN program to find out. Control Experimental 10 7 5 3 6 5 7 7 10 8 6 4 7 5 8 6 6 3 5 2

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Solution:

This is a problem in hypothesis testing for the difference between means of two sets of data. The test used is the t-test and the t-statistic is computed according to the formula t = [x_1 -x_2] / [\surd(\sumx^2_1 - {(\sumx_1)^2 / N_1} + \sumX^2_2 - {(\sumX_2)^2 / N_2}] {(1/N_1) + (1/N_2)})] wherex_i- mean ofi-thgroupi= 1,2 ; \sumx^2_1 - sum of squared score values ofi-thgroup ; (\sumx_1)^ 2 - square of the sum of scores ini-thgroup; N_i : the number of scores in thei-thgroup. The summary statistics for the given problem are: \sumx_1 = 70; \sumx_2 = 50; \sumx^2_1 = 520; \sumx^2_2 = 286 (\sumx_1)^ 2 = 4900; (\sumx_2)2= 2500; N_1 = 10 N_2 = 10 . Using these the calculated t-value is found to be t = (7 - 5) / 1.82 = 1.10. This calculated t-value is compared with the t-value for N_1 + N_2 - 2 degrees of freedom at the 5% level of significance (i.e., ift_calculated

Question:

Writepseudocodeto describe an algorithm to find a particularsubarrayin a larger array of characters.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G08-0181.htm

Solution:

Suppose that the main array is M(I), I = 1,2,3,...,N and thesubarray is S(I), I = 1,2,3,...,K, where K \leq N. The idea is to compare successive elements of M(I) with S(1) until a match is found. If so, then the subsequent values of M(I) must be compared with S(2), S(3),...,S(K) to find the entiresubarray. It is assumed that both the main array and thesubarrayhave already been read in. The program will output "yes" if thesubarrayis found, "no" if not found. A logical flag is used to indicate if a match has been found. The beginning and end of nests are indicated by a dotted line. do for I = 1 to N - K + 1 by 1 if M(I) = S(1) then do FLAG \leftarrow 'TRUE' do for J = I + 1 to I + K - 1 by 1 if M(J) \not = S(J - I + 1) then FLAG \leftarrow 'FALSE' exit do for end if end do for if FLAG = 'TRUE' exit do for end if end do for if FLAG = 'TRUE' then output "YES, SUBARRAY FOUND" else output "NO, SUBARRAY NOT FOUND" end if then else end program

Question:

4 liters of octane gasoline weigh 3.19 kg. Calculate what volume of air is required for its complete combustion at S.T.P.

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Solution:

To answer this problem, you need to write the balanced equation for the combustion of octane gasoline. This means knowing what the molecular formula of octane gasoline is and what is meant by combustion. Octane is a saturated hydrocarbon, i.e., it is analkane. Now, by saturated hydrocarbon, you mean a compound that contains only single bonds between the carbon to carbon and carbon to hydrogen bonds.Alkanes have the general formula C_NH2 N+2'where N = number of carbon atoms. Since the prefix "oct" means eight, you know there are 8 carbon atoms, which indicates that 18 hydrogen atoms are present. Thus, gasoline octane has the formula C_8H_18 . Now, by combustion you mean the reaction of an organic compound with oxygen to produce CO_2 and H_2O. With this in mind, you can write the balanced equation for the reaction as 2C_8H_18 + 25O_2 \rightarrow 16CO_2 + 18H_2O. To find out what volume of air is required for combustion, you need the volume of O_2 required, since 20% of air is oxygen (O_2). To find the amount of O_2 involved, use the fact that at S.T.P. (standard temperature and pressure) 1 mole of any gas occupies 22.4 liters. Thus, if you knew how many moles of O_2 were required, you would know its volume. You can find the number of moles by usingstoichiometry. You are told that there exists 3.19 kg or 3190 g (1000 g = 1 kg). The molecular weight (MW) of octane is 114 grams/mole. Thus, since mole = [grams (weight)] / M.W.] , you have (3190 / 114) = 27.98 moles of gasoline. From the equation's coefficients, you see that for every 2molesof gasoline, 25molesof O_2 are required. Thus, for this number of moles of gasoline, you need (27.98) (25 / 2) = 349.78 moles of O_2. Recalling that 1 mole of gas occupies 22.4 liters at S.T.P., 349.78 moles of O_2 occupies (349.78) (22.4) = 7835.08 liters. Oxygen is 20% of the air. Thus, the amount of air required is (5 liters air) / (1 liter O_2 ) × 7835.08 liters O_2 = 39,175.40 liters air.

Question:

The frequency of middle C is 256 sec^-1 What is its wave length in air?

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Solution:

The velocity of sound in air varies with temper-ature. At 180 C, which is a common room temperature it is 3.4 × 10^4 cm/sec. Hence, since Wave velocity = Frequency × Wavelength we see that Wavelength = Wave velocity / Frequency = (3.4 × 10^4 cm/sec.) / 256 sec-1 = 133 cm. This is approximately 4 feet. When the pianist plays middle ,C, the high pressure regions traveling toward a listener through the air are about 4 feet apart, but they are moving with the very high speed of 3.4 x 10^4 cm/sec (760 mph) and so 256 of them arrive at the listener's ear every second.

Question:

An automobile battery produces a potential difference (or "voltage") of 12 volts between its terminals. (It really consists of six 2 volt batteries following one after the other.) A headlightbulb is to be connected directly across the terminals of the battery and dissipate 40 watts of joule heat. What current will it draw and what must its resistance be?

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Solution:

To find the current, we use the formula P = IV where P is the power dissipated (40 watts), I is the current, and V is the voltage. Therefore, I = (P/V) = [(40 watts)/(12 volts)] = [(40 joules/sec)/(12joules/coulomb)] = 3.33 coulombs/sec = 3.33 amps . The bulb draws 3.33 amps from the battery . From Ohm' s law i= (V/R) R = (V/i) R ohms = [(12 volts)/(3.33 amps)] R = 3.6 ohms . The resistance of the bulb must be 3.6 ohms . We may also compute the resistance using the formula P = (V^2 /R) .Therefore R = (V^2 /P) = [(12)^2 /(40)] = 3.6 ohms . This second formula for power may be obtained from the first as follows: P =iV But from Ohm's Lawi= (V/R). Therefore P = [{(V) (V)}/(R)] = (V^2 /R) .

Question:

Find the speed and period of an earth satellite traveling at an altitude h = 135 mi above the surface of the earth where g' = 30 ft/sec. Take the radius of the earth R = 3960 mi.

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Solution:

The satellite experiences a force in the radial direction, towards the center of the earth. This force, the gravitational force on the satellite, provides the centripetal force that will keep the satellite in its circular orbit. Using Newton's Second Law, F = ma, we may write ma = mg' where g' is the acceleration due to gravity at a height of 135 miles above the earth's surface. Because the satellite is in a circular orbit, a = v^2/d (see figure). Then mv^2/d= mg' andv= \surd(g'd) v= \surd[(30 f/s^2)(3960 mi + 135 mi)] But 1 mi = 5280 ft v= \surd[(30 f/s^2)(4095)(5280 f) v= \surd(6.49×10^8 f^2/s^2) = 2.55×10^4 f/s To find the period of the satellite, we note that its speed is constant. Hence v = 2\pid/T where 2\pid is the distance travelled by the satellite in 1 revolution and T is the time required for this traversal, T = 2\pid/v = [{(2)(\pi)(4095 mi)(5280 f/mi)}/{(2.55×10^4 f/s)}] = (1.36×10^8 f)/(2.55×10^4 f/s) = 5.33×10^3 s

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Question:

A vibrating tuning fork of frequency 384 cycles \bullet s^-1 is held over the end of a vertical glass tube, the other end of which dips into water. Resonance occurs when the top of the tube is 21.9 cm and also 66.4 cm above the water surface. Calculate the speed of sound in air and the end correction of the tube. Assuming that the temperature is 13\textdegreeC, what is the velocity of sound at 0\textdegreeC (v_0)?

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Solution:

Figure (C) shows the tuning fork in vibra-tion. In order to send a compression wave down the tube, one tine of the fork is pulled back to position c, and then is let go. During one oscillation, the tine moves from c to b and then back to c. Now, the pipes with which we are dealing have water at one end. This end must then be a displacement node, for the air at this end of the column is always at rest. Alternatively, it is a pressure antinode. Similarly, the open end of the tube is a displacement maximum, or a pressure node. Now, suppose the fork starts at point c and sends a compression wave down the tube. When this wave reaches the bottom of the tube, it must be reflected as a compression if that end is to remain a pressure anti-node. The reflected compression travels back up the tube. If resonance is to occur, the compression must reach the tine of the tuning fork when the latter is ready to move from b to c. (In this situation, the tine is preparing to send a rarefaction down the tube). If this does not occur, the top of the tube will not remain a pressure node, for the reflected pressure wave and the new pressure wave will not cancel. Hence, in 1/2 oscillation of the tuning fork, the wave travels a distance 2L (twice the tube length). In general, resonance will occur if the wave travels a distance 2L in an odd number of half oscillations of the fork. If the wave velocity is v and the frequency of the fork is v_source, the resonance condition is 2L = v(n + 1/2)/ v_source(n = 0, 1, 2 ...)(1) Since the wave frequency equals the source frequency, we may write v_wave = v_source= v Butv_wave \lambda = v where \lambda is the wavelength of the wave. Hence, (1) becomes 2L = v(n + 1/2)/v Solving this for v v = (2Lv)/(n + 1/2)(n = 0, 1, 2 ...)(2) Therefore, we may measure v if we know the tube length, the source frequency, and the value of n. In practice, the tube is put in the water, and slowly drawn out as the fork vibrates. When the first reso-nance occurs, n = 0, and the length of tube extending out of the water is measured. Again, the tube is drawn out of the water. When the second resonance occurs, n = 1, and the tube length is again measured, etc. Each resonance will give us the same value for v, as equation (2) indicates. Using the relation v = v\lambda where \lambda is the wavelength of the wave, we may also write (2) as v = [2L/(n + 1/2)] (v/\lambda) orL = (n + 1/2) (v/\lambda)(n = 0, 1, 2, ...)(3) In practice, an antinode never occurs quite at the end of an open pipe. Its position is just beyond the end of the pipe, the maximum displacement slightly overshooting the end (see figs. (a) and (b)). Thus for the first reso-nance (n = 0 in (3)) the length of the tube will be almost a quarter of a wavelength, and \lambda/4 = L + E, where L is the length of the air column and E is the end correction. Similarly, for the second resonance, 3\lambda/4 = L' + E, where L' is the length of the air column when the second resonance occurs. Thus, using our data, \lambda/2 = L' - L = (66.4 - 21.9) cm = 44.5 cmor\lambda = 89 cm. Therefore the velocity of sound is v = f\lambda = 384 s^-1 × 89 cm = 34176 cm \bullet s^-1 = 341.8 m \bullet s^-1. Further, E = \lambda/4 - L = (22.25 - 21.9) cm = 0.35 cm. The velocity of sound in a gas is proportional to the square root of the absolute temperature. Hence v_0/v = \surd[T_0/T] = \surd(2730K / 286\textdegreeK) = 0.977. \thereforev_0 = 0.977 × 341.8 m \bullet s^-1 = 333.9 m \bullet s^-1.

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Question:

What prevents the stomach from being digested by itsown secretions?

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Solution:

The lining of the stomach is composed of cells that secrete hydrochloricacid, gastric juice, and mucus. Mucus is a polymer made up ofrepeating units of a protein- sugar complex. A coat of mucus, about 1 to 1(1/2) millimeters in thickness, lines the inner surface of the stomach. Mucus is slightly basic. This alkalinity provides a barrier to acids, keeping thearea next to the stomach lining nearly neutral. In addition, the membranesof the cells lining the stomach have a low permeability to hydrogenions, preventing acid from entering the underlying cells. The cells that make up the stomach (and duodenum) lining do not lastlong, even under this protection. Cell division and growth replace the entirestomach lining every 1 to 3 days. Thus the mucus layer, the permeabilityof the membranes and the continual replacement of the cells comprisingthe lining all help protect the underlying tissues from the action ofproteolyticenzymes of the stomach and duodenum. For many people, however, this does not provide enough protection. If too much acid is released, perhaps because of emotional strainor because theproteolyticenzymes have digested away the mucus, anulcer will result. Ulcers are usually treated by eating many small, bland mealsthroughout the day. This helps to keep the acid level down.

Question:

Consider the following input containing information about the players of a soccer team: LAST NAMEBIRTHDATENO. GAMES PLAYEDGOALS SCORED Write a program which punches cards with the same information, but in form: LAST NAME;BIRTH DATE: NO; GAMES PLAYED;GOALS SCORED

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Solution:

&ANCHOR = 1; &TRIM = 1 STRING = BREAK(' ') . CHARS SPAN(' ') NEXTCARD = INPUT: F(END) LOOPCARD STRING = CHARS ' ; ': S(LOOP) PUNCH = CARD: (NEXT) END The first line, &ANCHOR = 1; &TRIM = 1, introduces two new features of the SNOBOL IV language. First, it includes two separate statements. SNOBOL IV permits reordering of the several statements on one line, where each statement (except the last) ends with semicolon. Second, the first line introduces a new keyword &ANCH0R. If not otherwise specified, the initial value of &ANCH0R is zero, meaning that any pattern matching in that program will succeed whenever a pattern occurs anywhere in the subject. If the value of &ANCH0R is not zero, any pattern matching will succeed only if a pattern occurs in the beginning of the subject. Thus, in the previous problem, the two statements W0RD1LEN(1) . K K'K': F(START) could be replaced by &ANCH0R = 1 W0RD1'K': F(START) The second line includes BREAK and SPAN, the primitive functions whose values are pattern structures that match runs of characters. Patterns described by A RUN OF BLANKS, A STRING OF DIGITS, A WORD (RUN OF LETTERS) can be formed using SPAN as SPAN (' ') SPAN ('0123456789') SPAN ('ABCDEFGHIJKLMNOPQRSTUVWXYZ') respectively. Patterns described by i) Everything up to the next blank, ii) Everything up to the next punctuation mark, iii) Everything up to the next number, can be formed using BREAK as BREAK (' ') BREAK (',.;:.'?') BREAK ('+-0123456789') respectively. The pattern structure for SPAN matches the longest string that consists solely of characters, appearing in the argument. BREAK generates a pattern structure that matches the string up to, but not including, the break character in the argument. Thus, the second line of this program indicates that the variable STRING will be assigned a first matched string of characters up to a first blank (the same string will also be assigned to CHARS), concatenated with the run of blanks between it and the next string. The third line of the program reads in a first input card. The fourth line assigns the first string of the card up to a blank, concatenated with the run of blanks following it, to STRING and substitutes it by the same string, only now followed by a semicolon, rather than blanks. Then it takes the next string and run of blanks and does the same. The procedure repeats until all the strings on the card are substituted. Then the computer punches out the first card with the proper changes, goes back to the third line (labeled NEXT), reads in the next card, and performs the needed substitutions on it. The control exits from the loop and terminates the program when there are no more cards in the input. Considering the following input: PELE1938103530 the output will be: PELE; 1938; 10; 35; 30

Question:

Describe and give an example of a reflex arc.

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Solution:

To understand what a reflex arc is, we must know something about reflexes. A reflex is an innate, stereotyped, automatic response to a given stimulus. A popular example of a reflex is the knee jerk. No matter how many times we rap on the tendon of a person's knee cap, his leg will invariably straighten out. This ex-periment demonstrates one of the chief characteristics of a reflex: fidelity of repetition. Reflexes are important because responses to certain stimuli have to be made instantaneously. For example, when we step on something sharp or come into contact with something hot, we do not wait until the pain is experienced by the brain and then after deliberation decide what to do. Our responses are immediate and automatic. The part of the body involved is being withdrawn by reflex action before the sensation of pain is experienced. A reflex arc is the neural pathway that conducts the nerve impulses for a given reflex. It consists of a sensory neuron with a receptor to detect the stimulus, connected by a synapse to a motor neuron, which is attached to a muscle or some other tissue that brings about the appropriate response. Thus, the simplest type of reflex arc is termed monosynaptic because there is only one synapse between the sensory and motor neurons. Most reflex arcs include one or more interneurons between the sensory and motor neurons (see Figure 1). An example of a monosynaptic reflex arc is the knee jerk. When the tendon of the knee cap is tapped, and thereby stretched, receptors in the tendon are stimulated. An impulse travels along the sensory neuron to the spinal cord where it synapses directly with a motor neuron. This latter neuron transmits an impulse to the effector muscle in the leg, causing it to contract, resulting in a sudden straightening of the leg (see Figure 2).

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Question:

What is the energy of a photon of green light (frequency = 6 × 10^14vps)?

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Solution:

Planck's hypothesis states that E =hu, where u is the frequency of the radiation, and h is Planck's constant. Therefore, E = (6.63 x 10^-34 joule-sec) (6 x 10^14vps) E = 3.98 x 10^-19 joules.

Question:

Compare the teeth of mammals with those of lower vertebrates.

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Solution:

The chief differences between mammalian teeth and those of lower vertebrates is their respective embryonic origins. Of the three main types of embryonic tissues (ectoderm, mesoderm and endoderm), mammalian teeth are made up of two (ectoderm and mesoderm) while teeth of the lower vertebrates are derived solely from the ectoderm. Mammalian teeth result from outgrowths of the embryonic bone- producing tissue - the mesoderm. They are covered with a hard enamel derived from the embryonic ectoderm. Mammalian teeth are composed of living cells, supplied with blood vessels and nerves (see Figure 1). They are also highly evolved, consisting of different shapes adapted for different functions. The foremost teeth of most mammals are chisel-shaped incisors (see Figure 2). These are used for biting and cutting food. The incisors of rodents and horses grow continually and are worn down during eating. Meat eaters have sharp canine teeth that are used for tearing. All of a dog's teeth are sharp and pointed, and highly specialized for chewing raw meat. Herbivorous mammals have flat grinding teeth in the rear of their mouth that help crush and scrape the tough cellulose cell walls of plant cells. Figure 2. Structure and arrangement of teeth in different animals. (A) Snake: thin, sharp, backward-curved teeth that have no chewing function (the snake skull is here shown disproportionately large in relation to the other three). (3) Beaver (gnawing herbivore): few but very large incisors, no canines, premolars and molars with flat grinding surfaces. (C) Dog (carnivore): large canines, premolars and molars adapted for cutting and shearing. (D) Deer (grazing and browsing herbivore): six lower incisors (three on each side), but no upper incisors-these are functionally replaced by a horny gum; premolars and molars with very large grinding surfaces. Notice the large gap between the incisors and premolars. All the teeth of the lower vertebrates are sharp and pointed. The teeth of fish, reptiles, and amphibians have the same general shape, whether they are meat or plant eaters. The triangular, conical shape of these teeth in-dicates their evolutionary origin. The teeth of these vertebrates have evolved from scales. They are formed from the ectoderm (exterior tissue) of the embryo and are re-placed when lost. These teeth are not made up of living cells and do not contain blood vessels or nerve endings. Thus mammalian teeth are highly specialized, living organs while the teeth of the lower vertebrates are non-living, less evolved extensions of their skin tissue.

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Question:

You are given 3-bromohexane. Draw the stereoisomers of this compound; specify the R and the S enantiomers.

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Solution:

Non-superimposable mirror-image stereo- isomers are called enantiomers. It is not enough that a molecule has a mirror image, for everything has a mirror image; the mirror images must not be able to be superimposed on each other. The arrangement of atoms that characterize a particular stereoisomer is called its configuration. You want to specify the configuration in terms of R and S. Let us see what this means. In assigning configuration, you do the following after determining the chiral carbon: (1) Assign.a sequence of priority to the four atoms or groups of atoms attached to the chiral center. (A chiral center consists of a carbon atom to which 4 different groups are attached.) (2) Visualize the molecule so that the group of lowest priority is directed away from you. Now, observe the arrangement of the remaining groups. If, in going from the group of highest priority to the group of second priority your eye travels in a clockwise direction, the configuration is R; if counterclockwise, the configuration is S. Now, how do you determine priority? If the four atoms bonded to chiral center are all different, priority depends on atomic number. The atom of higher atomic number receives the higher priority. If they are isotopes, the higher mass number has priority. Now, if you can't determine priority from this, because you have 2 or more atoms that are the same attached to the chiral center, make a similar comparison of the next atoms (i.e., their atomic numbers) as you move out from the chiral center. With this in mind, you can proceed as follows: (a) 3-bromohexane. Recall, enantiomers are mirror image isomers so that you can say the enantiomers I and II, where the asterisk designates the chiral carbon. To specify R and S, the lowest priority group must be drawn away from you. H has the lowest priority since it has the smallest atomic number. It can be drawn away from you by placing it in a vertical position. By con-vention, groups in vertical positions are considered away from you. Thus, you can rewrite these enantiomers as shown in figure B. The hydrogens attached to the chiral center are now vertical, and, as such, away from us. The other groups surrounding the chiral center are Br, CH_2CH_3, and CH_2CH_2CH_3. Of the three, Br has highest priority because its atomic number is higher than carbons. The other two have carbons attached to the chiral center, so that no priority is possible here. You must move out from the chiral center. The next atoms reached are, again, carbon. No priority determination is possible. If you move out to the next atom, you find a carbon atom in CH_2CH_2CH3and a hydrogen atom in CH_2CH_3 (it doesn't have another carbon). Thus, because carbon has a larger atomic number than hydrogen, CH_2CH_2CH_3 has priority over CH_2CH_3. In summary, for the groups or atoms attached to the chiral center, the order of increasing priority can be shown by Recalling step (2), in going from highest to next lowest, then to next lowest, in the enantiomer on the left, you move in a clockwise direction. For enantiomer I, you move in a counterclockwise direction. Thus it is S and enantiomer II is R.

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Question:

(a) synthetic polymer, Thermosetting, (c) Thermoplastic, (d)Elastomers.

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Solution:

A synthetic polymer is a large molecule with a repeating structure. It may be synthesized by chemically combining many monomer or single units. Polymers which are used for molding are often referred to as resins or plastics. The polymers may be divided into three general types: Thermoplastic polymers - These substances will melt on being heated and can be molded easily into new shapes. Examples include polyvinylchloride, polyethylene, and nylon. Thermosetting polymers - These substances slowly char rather than melt. They are originally molded in their final shape or are machined into shape. They cannot be molded again after cooling. The best known example of this is Bakelite; it is used in the electrical industry as an insulating material. Elastomers- This type of material may be characterized as rubbery or elastic. Examples include natural rubber and the various synthetic rubbers such as neoprene.

Question:

SO_2CI_2 \rightarrow SO_2 + CI_2 is 8.0 minutes. In what period of time would the con-centration of SO_2CI_2 be reduced to 1.0% of the original?

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Solution:

If a reaction is first order, it means that the rate of the reaction is proportional to the concentration of a single reactant. For example, if substance A decomposes into the products B and C, A \rightarrow B + C, the rate is proportional to the concentration of A which is present at any time. Rate = k [A], where k is the rate constant for the reaction and [A] is the concentration of A. The rate at which the reaction proceeds is equal to the decrease in concentration of A with change of time. This can be written in the integrated form 2.303 log {[A] / [A_0]} = -kt, where [A] is the concentration of A after time t has elapsed, [A_0 ] is the original concentration of A and k is the rate constant for the reaction. In this problem, the final ratio of [A] to [A_0]is given as 1.0% or 0.010. Therefore, this equation can be used to find the time it takes for this process to occur, once one has calculated the rate constant k. The rate constant can be found if one remembers the relationship between the half-life of a reaction and k. Namely, t_1/2 =0.693 / k , where t, is the half life and k is the rate constant. The 1/2for this reaction is given as 8.0 minutes, therefore this relation can be used to obtain k. K = 0.693/ t_1/2 k = 0.693 / 8.0 minutes = 0.087 / minute Once k is known, it can be used in the original rate equation to determine the time it takes for this reaction to proceed until the concentration of A is 1.0% that of the original concentration. 2.303 log {[A] / [A_0]} = -kt.Substituting, 2.303log0.010 = - 0.087 / minute× t t = {(2.303log0.010) / (- 0.087 / min)} = {(2.303) (-2)} / (- 0.087 / min) = 53 minutes.

Question:

Why is it incorrect to say that man evolved from monkeys? What did he possibly evolve from?

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Solution:

The oldest fossils of Old World monkeys are ofParapithecusand several related genera that have been found in lower Oligocene rocks. Parapithecus was smaller than any of the modern monkeys or apes and was near the stem leading to humans (it had the same dental formula). Parapithecus may therefore represent a common ancestor of today's Old World monkeys, anthropoid apes and man. From that point, the evolution of the modern Old World monkeys diverged from that of the hominoids. The ancestry shared by New World monkeys, Old World monkeys, and the hominoids was separated even earlier in time. The New World monkeys diverged from the latter two at the beginning of the Oligocene; whileParapithecusfossils are dated to the middle Oligocene. By the Miocene (20 million years ago), the diver-gence of the Old World monkeys and the hominoid lineage is clearly evidenced by the fossil records. This can be verified by the examination of dental patterns. The hominoid fossils have 5-cusped molars rather than the 4-cusped molars of the monkeys. This indicates that man probably did not evolve directly from the monkeys. It is more correct to say that monkeys, man, and apes all evolved from a common ancestor independently. This ancestor had characteristics of the higher primates which evolved along different lines of specialization over millions of years. The Oligocene hominoid that gave rise to the lineage leading to man was a small, arboreal, monkey-like anthropoid equipped with binocular vision, living in a society of small troops which used visual and vocal communication. The oldest fossil found so far that shows features considered to be those of an ancestral hominoid is the skull ofDryopithecus.Dryopithecus is now known to have been a wide-ranging genus extending from western Europe to China that had some definite hominoid features. Dryopithecus is the most likely ancestor of the chimpanzee, gorilla, orangutan and man. The lineage of the gibbon is believed to have diverged into a distinct species before any of the other apes.

Question:

Suppose that in Millikan's experiment an oil drop weighing 1.6 × 10^\rule{1em}{1pt}13 N remains stationary when the electric field between the plates is adjusted to a value of 5 × 10^5 N/C. What is the charge on the oil drop in this case, and how many units of the charge quantum does this represent if the particle is moving with a constant velocity between the plates?

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Solution:

In the experiment, the charged particle is attracted upward by the electrostatic force. If the electric field in which the particle is moving is E, the electro-static force is F_e = qE where q is the charge. The gravitational force pulling the particle downward is its weight W = 1.6 × 10^\rule{1em}{1pt}13N. The particle is moving with a constant velocity (i.e., zero acceleration) therefore the net force acting on it must be zero. This means that F_e = qE = Wor q = W / E = (1.6 × 10^\rule{1em}{1pt}13N)/ (5 × 10^\rule{1em}{1pt}5N/ C) = 3.2 × 10\rule{1em}{1pt}19C. Electric charge in nature exists in integral multiples of 1.6 × 10^\rule{1em}{1pt}19 C, hence the particle of the problem has = (3.2 × 10^\rule{1em}{1pt}19C) / ( 1.6 × 10^\rule{1em}{1pt}19C/ quantum) = 2 quanta of charge.

Question:

A chemist has two reactants, P and Q, dissolved in one liter ofwater. The amounts of P and Q present are p grams and qgrams, respectively, and they are combining to form 2 apq gramsof the product R each second. It is also known that thereaction is reversible: in each second, 2br grams of R arebreaking up into br grams of each P and Q. In summary, thechange in one second is: - -amountof P increases by br - apq -amountof Q increases by br - apq -amountof R increases by 2apq - 2br . Design a program to simulate this reaction.

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Solution:

This reaction is actually continuous. One second divisions are used forconvenience. In general, at any time interval h, the amounts of P,Q, andR are as follows: p(t + h) = p(t) + hbr(t) - hap(t)q(t) q(t + h) = q(t) + hbr(t) - hap(t)q(t) r(t + h) = r(t) - 2hbr(t) + 2hap(t)q(t) As h becomes smaller, the numerical solutions converge to the continuous functions, yielding as close an approximation as desired. In fact , they converge to the solution of the following differential equations: dp/dt= br - apq dq/ dt = br - apq dr/ dt =- 2br + 2apq . Let us now declare all variables to avoid confusion: P,Q- number of grams of reactants. R - number of grams of product. A,B-stoichiometriccoefficients for P and Q, respectively in theequation AP +BQ \rightleftarrows R . H - reciprocal of the desired time interval, (in sec-1.) .) T - total time of simulated reaction (in sec.) C - control number.This variable allows the user to change desiredparts of the data. D - rate of change in 1/H seconds. The program looks as follows: 0PRINT "INPUT THE DESIRED CONTROL NUMBER" 11PRINT "0: STOP" 12PRINT "1: INITIAL GRAMS", "2: NEW COEFFICIENTS' 13PRINT "3: NEW TIME STEPS", "4: SIMULATE" 20INPUT C 30IF C = 0 THEN 280 40IF C = 1 THEN 120 50IF C = 2 THEN 140 60IF C = 3 THEN 160 70IF C = 4 THEN 180 110GO TO 10 120PRINT "INPUT INITIAL AMOUNTS OF P,Q, AND R" 125INPUT P,Q,R 130GO TO 10 140PRINT "INPUT MOLAR COEFFICIENTS FOR P AND Q" 145INPUT A,B 150GO TO 10 160PRINT "INPUT DESIRED AND TOTAL TIME INTERVALS" 165INPUT H,T 170GO TO 10 180PRINT "TIME", "P", "Q", "R" 190REM SIMULATION OF REVERSIBLE REACTION 200FOR I = 1 TO T 210FOR J = 1 TO H 220LET D = (B\textasteriskcenteredR - A\textasteriskcenteredP\textasteriskcenteredQ) / H 230LET P = P + D 240LET Q = Q + D 250LET R = R - (2.0\textasteriskcenteredD) 260NEXT J 270PRINT I,P,Q,R 275NEXT I 280END

Question:

A two-dimensional array named RST has twenty rows and twenty columns. Use FORTRAN to compute the product of the main diagonal elements of RST and store it in BPROD. A main diagonal element is the one that has the same row and column number, so that DPROD = 20\prod_i= 1RST(I, I).

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Solution:

The essential trick is to initialize DPROD =1.0 and then set up a loop which computes DPROD by multiplying its previous value by RST(I,I). DIMENSION RST (20, 20) DPROD =1.0 DO 30 I = 1,20 DPROD = DPROD\textasteriskcenteredRST(I,I) 30CONTINUE STOP END

Question:

Two particles of equal mass and equal but opposite velocities \pm v_i collide. What are the velocities after the collision?

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Solution:

Since no external forces act on the 2 particle system, we may use the principle of conservation of momentum. This principle will relate the velocities of the particles after the collision to their velocities before the collision. Therefore, m v_1i ^\ding{217} + m v_2i ^\ding{217} = m v_1f ^\ding{217} + m v_2f ^\ding{217}(1) where the subscript v_1i ^\ding{217} defines the initial velocity of particle 1, and similarly for v_2i ^\ding{217}, v_1f ^\ding{217}, and v_2f ^\ding{217}. Sub-stituting v_1i- ^\ding{217} = v_i ^\ding{217}and v_2i\ding{217}= - v_i ^\ding{217}, into (1), we obtain m v_i ^\ding{217} -mv_i^\ding{217}= mv_1f ^\ding{217}+ m v_2f\ding{217}= 0 Hence,v_1f\ding{217}= -v_2f ^\ding{217}. If the collision is elastic, then kinetic energy is conserved in the collision, and 1/2 m v1i^2 + 1/2 m v2i2= 1/2 m v_1f ^2 + 1/2 m v2f^2(2) But, v1i= v_2i = viand v1f= v2f=v_fbecause we are concerned only with magnitudes in (2). Therefore, 1/2 m vi^2 + 1/2 m vi2= 1/2 mv_f_2+ 1/2 mv_f^2 or,vi^2 =v_f^2 and the conservation of energy demands that the final speed v_f equal the initial speed v_i. If one or both particles are excited internally by the collision, thenv_f< v_i because some of the initial energy must go into the excitation energy. Hence, the final energy < initial energy. If one or both particles initially are in excited states of internal motion and on collision they give up their excitation energy into kinetic energy, thenv_fcan be larger than v_i.

Question:

Describe the organization of the PDP-8 computer, giving the various registers, memory, etc. Also explain the terms "address and "operation code."

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Solution:

The block diagram of Figure 1 shows the various elements of the PDP-8 computer. As shown in Figure 1, the PDP-8 memory has 4096 words or locations. Each word of the memory has 12 bits. Memory Address :Each word of the memory is numbered sequentially from 0 to 4095. This number associated with a word is called the "address" of the word. Thus, the "address" of a word shows the physical location of the word in memory. Actually, the PDP-8 uses the octal number system rather than the decimal number system. Hence, the address of a memory word (in octal) can be anything from 0 to 7777_8. To see this note that 4095 = 2^12 - 1 = 111111111111_2 = 111 111 111 111_2 = 7 7 7 7 _8 Thus, if the address of a word is 7777_8, the word is located at the last physical location in the memory. And, if the address of a word is 0, it is located at the first physical location in the memory. PC :The PDP-8 has a Program Counter (PC). The PC stores the address of the instruction to be executed (i.e., carried out) by the computer. At any given time the PC contains the address of the next instruction to be car- ried out, after the completion of the instruction which is presently being executed. The PC has 12 bits. MAR :The PDP-8 has a register called Memory Address Register (MAR). The MAR is used whenever any information is to be read from or written into the memory. The pro-cedures are as follows: If there is a memory location from which information is to be read from or written into, then the address of this location must be entered into the MAR. The MAR then communicates with the memory and the MBR to get the information transferred as de-sired. The MAR holds 12 bits. The number of bits in MAR depends on the size of the program counter. In this case the program counter has 12 bits. Therefore the number of bits in MAR is 12. MBR :The PDP-8 has another register called Memory Buffer Register (MBR) . Any information which is read from the memory is read into the MBR first. Then, from the MBR the information can be sent to wherever it is needed. Similarly, any information which is to be written into the memory must first be placed into the MBR. Then, from the MBR the information is written into the memory. In many instances, the main function of the MBR is to momentarily store data that is being transferred be-tween the general purpose registers and main memory. Registers manipulate data much faster than memory de-vices, and as such, memory buffer registers (MBR) were created to bridge the timing gap between registers and main memory. The outstanding feature of the MBR is that it can regulate its speed. When information is to be passed to memory, the MBR accepts this information from the high-speed registers and passes it at a slower speed to the memory device. The reverse is done when informa-tion is to be transferred from memory to a register. The MBR accepts this information from the low speed memory device and passes it at a higher speed to the register. Thus, MBR acts as a buffer between the memory and other registers in the computer. The bit size of the MBR depends on the bit size of one computer word. In the case of the PDP-8 one word is 12 bits long so MBR is also 12 bits long. IR :The PDP-8 has a register called Instruction Register (lR). At any given time the IR contains the Operation Code (OP) of the instruction which is presently being executed by the computer. The Operation Code present in the IR is coded information telling the computer what is to be done with the various data present in the different registers of the computer. The IR has 3 bits. AC :The PDP-8 has a general purpose register called the Accumulator (AC). The computer carries out addition, subtraction, multiplication, division, etc. by storing the intermediate and final operands and sums in the ac-cumulator. The AC has 12 bits. L :The PDP-8 has a 1-bit register called a Link Register (L). The Link register works in association with AC. Sometimes, the Link register and the AC are referred to by the combined mnemonic LAC, forming a combined 13-bit register. The Link register (normally referred to as the Link bit) can be set to 1 or cleared to 0 independently of the AC. But, while doing register shift-left and shift-right operations, the L bit works in association with the 12- bit AC, as LAC, to store the bit shifted into or out of the AC register.

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Question:

An electron in an oscilloscope tube is situated midway between two parallel metal plates 0.50 cm apart. One of the plates is maintained at a potential of 60 volts above the other. What is the potential gradient between the plates? What is the force on the electron?

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Solution:

The difference of potential between points a and b, is, by definition V_b \rule{1em}{1pt} V_a = \rule{1em}{1pt}b\int_a E^\ding{217} \bullet dl^\ding{217}(1) where the integral is evaluated over an arbitrary path between a and b and E^\ding{217} is the electric field intensity. In this example, E^\ding{217} is constant between the plates (see figure), and V_b \rule{1em}{1pt} V_a = \rule{1em}{1pt}E^\ding{217} \bullet^b\int_adl^\ding{217}(2) Also, we evaluate (2) over the path shown in the figure, since E^\ding{217} and dl^\ding{217} are in the same direction for this path. Then V_(x)2 \rule{1em}{1pt} V_(x)1 = \rule{1em}{1pt}E^\ding{217} \bullet ^(x)2\int_(x)1dl^\ding{217}= \rule{1em}{1pt}E(x_2 \rule{1em}{1pt} x_1) andE =(v_(x)2 v_(x)1) / (x_2 \rule{1em}{1pt} x_1) = \rule{1em}{1pt}(∆v / ∆x)(3) By definition, ∆v / ∆x is the potential gradient. Hence, for this case, by(3) \rule{1em}{1pt}E = ∆V / ∆x = \rule{1em}{1pt}(60 Volts / .5 × 10^\rule{1em}{1pt}2m) \rule{1em}{1pt}E = ∆V / ∆x = \rule{1em}{1pt}(60 Volts / 5 × 10^\rule{1em}{1pt}3m) = \rule{1em}{1pt}12 × 10^3 (V / m) Therefore, ∆V / ∆x = \rule{1em}{1pt}12 × 10^3 (V / m) E = 12 ×10^3 (V/m) By definition, the force on the electron is the product of E and the electron charge, or F = Eq = (1.2 × 10^4 nt / coul) × (\rule{1em}{1pt}1.60 × 10^\rule{1em}{1pt}19 coul) = -1.92 × 10^\rule{1em}{1pt}15 nt This force is toward the plate of higher potential.

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Question:

If one increases the angle away from the Z axis, how does the wave function for a 2P_Z electron change?

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Solution:

The wave function of a 2P_Z electron is \psi = [(1)/{4 \surd(2\pi)}] (Z/a_0)^5/2 e-Zr/2(a)0rcos\texttheta, where \psi is the wave function and \texttheta = the polar angle. The polar angle measures the angular deviation away from the vertical, or Z, axis. \psi^2 is proportional to the electron density at any given radius. One needs to consider only \psi and \texttheta to answer the question. To see how the wave function, \psi, changes when the angle from the Z axis is increased, take two values for \texttheta, and see what happens to \psi. For example, when \texttheta = 0 (that is, looking along the Z axis) ,cos\texttheta is one, the maximum value a cosine can assume. The wave function is dependent upon \texttheta. The cosine \texttheta is at a maximum, then \psi is at a maximum. Thus, \psi^2 is also at a maximum and the electron density is greatest when looking along the Z axis. When \texttheta = 900 (that is, when one is looking 900 away from the z axis, i.e. along the X or Y axis),cos\texttheta is zero, the minimum value a cosine can assume. This means, therefore, that \psi^2 is not at a minimum, which means that the probability of finding the electron at this location is zero. In summary, when the angle was increased from 0\textdegree to 90\textdegree, the wave function decreased. Thus, when one increases the angle away from the Z axis, the wave function and the electron density for a 2P_Z electron decreases.

Question:

Draw the logic diagram of a four-bit register using S-R flip flops.

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Solution:

Figure 1 shows the inputs and outputs of an S-R flip-flop. When clear is 0 the flip-flop is cleared sequentially (with-out the clock pulse requirement). The clear input must go to 1 during the normal clocked operation. Figure 2 shows the block diagram of the four-bit register. The data inputs are I_1, I2, I3,andI_4. The control inputs are CP (clock pulse), load, 2 3 , I_4. The control inputs are CP (clock pulse), load, and clear. The data outputs are A_1 , A_2 , A_3 , and A_4. The clock pulse is common to all four flip-flops and consists of continuous pulses. The circle under the tri-angle in each flip-flop indicates that flip-flop transi-tions occur during the falling edge of each clock pulse. The Load input is connected to all four flip-flops via the AND gates. When Load is 0 the inputs to S and R are 0, and Q remains in the previous state. When Load is 1 the in-puts to S and R are I andI, respectively; thus the output Q will show the value of I.

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Question:

In the figure, positive charge is distributed uniformly over the entries xy-plan, with a change per unit area, or surface density of change, \sigma. Find the electric intensity at the point P.

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/Users/wenhuchen/Documents/Crawler/Physics/D17-0566.htm

Solution:

Let the charge be subdivided into narrow strips parallel to the y-axis and of width dx. Each strip can be considered a line charge. The area of a portion of a strip of length L is L dx, and the charge dq on the strip is dq = \sigmaL dx. The charge per unit length, d\lambda, is therefore d\lambda = dq/L = \sigma dx. Considered as a line of charge the strip sets up at point P a field dE^\ding{217}, lying in the xz-plane and of magnitude dE = 2k\sigma(dx/r), which is the field due to a line of charge. The field can be resolved into components dE_x and dE_z. The components dE_x will sum to zero when the entire sheet of charge is considered. To see this consider the lines of charge at points x and - x. The x-components of each pair cancel each other. The resultant field at P is therefore in the z-direction, perpendicular to the sheet of charge. It will be seen from the diagram that dE_z = dE cos \texttheta and hence E = \int dE_z = 2k^+\infty\int_-\infty [(cos \texttheta dx)/r]. If we use \texttheta as the Integral variable [which varies between - (\pi/2) and (\pi/2) ], and note that r = a/cos \texttheta , x = a tan \texttheta, dx = a sec^2 \texttheta d\texttheta we obtain E = 2k\sigma^\infty\int_-\infty(cos \texttheta/r) dx = 2k\sigma^\pi/2\int_-\pi/2 [(cos^2 \texttheta) (a sec^2 \texttheta)} /a] d\texttheta = 2k \sigma\pi Note that the distance a from the plane to the point P does not appear in the final result. This means that the intensity of the field set up by an infinite plane sheet of charge is independent of the distance from the charge. In other words, the field is uniform and normal to the plane of charge. The same result would have been obtained if point P, had been taken below the xy-plane. That is, a field of the same magnitude but in the opposite sense is set up on the opposite side of the plane.

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Question:

Define 'Expressions' in COBOL language. What are arithmetic expressions and how are they evaluated?

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G11-0250.htm

Solution:

Expressions: An expression in the COBOL language may be defined as a combination of data-names, literals, and operators which may be reduced to a single value. Arithmetic Expressions: They are data names, identifiers, or numeric literals or, a series of data-names, identifiers and literals separated by arithmetic operators, which de-fine a single numeric value. The format of arithmetic expressions using arithmetic operators is as follows: data name or identifier or literal or {data name literal Identifier} operator { data name literal Identifier } The arithmetic operators are: OperationOperators Addition+ Subtraction- Multiplication{_\ast} Division/ Exponentiation\textasteriskcentered \textasteriskcentered The normal precedence rule (from high to low) is: 1) Exponentiation 2) Multiplication and Division 3) Addition and Subtraction When operators are all on the same hierarchical level, evaluation occurs in left to right order. Parentheses are used to alter the normal rules of precedence. In COBOL, parentheses are defined as having a higher order of prece-dence than any operator e.g. ,(A + B) / (C - D) cannot be expressed in COBOL as (A + B) / (C - D) as this repre-sents the formula A + (B/C)- D. Using parentheses it can be accurately expressed as (A + B) / (C - D) . Operations within parentheses have precedence over any other in a calculation: e.g., calculate A + B {_\ast} A - (C/2). with A = 4, B = 3 and C = 6. The expression will be evaluated as follows: Multiplication and division are on the same level, so first B{_\ast}A = 12 and then C/2 = 3 and thus it reduces to A + 12 - 3. Now + and - are on the same level. A + 12 = 16 is first evaluated and then 16 - 3 = 13 is the final evaluation. If we use parentheses instead: ((A + B) {_\ast} A - C) / 2 the sequence of evaluation is as follows 1.A + B = 7 2.7 {_\ast} A = 28 3.28 - C = 22 4.22/2 = 11 (RESULT).

Question:

Show that the moment of inertia of a body about any axis is equal to the moment of inertia about a parallel axis through the center of mass plus the product of the mass of the body and the square of the distance between the axes. This is called the parallel-axes theorem. Prove also that the moment of inertia of a thin plate about an axis at right angles to its plane is equal to the sum of the moments of inertia about two mutually perpendicular axes concurrent with the first and lying in the plane of the thin plate. This is called the perpendicular-axes theorem.

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/Users/wenhuchen/Documents/Crawler/Physics/D04-0182.htm

Solution:

Let I be the moment of inertia of the body about an arbitrary axis and 1_G the moment of inertia about the parallel axis through the center of mass G, the two axes being distance h apart. (See fig. (A)). By definition of the center of mass of a body relative to an arbitrary axis through a point P, we obtain I = \sum_i m_i r^2_i = \sum_i m_i(x_i^2 + y_i^2) = \sum_i m_i x^2_i + \sum_i m_i y_i^2 where the sum is carried out over all mass particles m_1 of the body, and r_i^2 is the distance form P to m_i. Now, x_i = xʹ_i + a y_i = yʹ_i + b as shown in figure (A). Here, (xʹ_i, yʹ_i) locates m_i relative to G, the center of mass. Then I = \sum_i m_i (xʹ_i + a)^2 + \sum_i m_i (yʹ_i + b)^2 I = \sum_i m_i (xʹ_i + a)^2 + \sum_i m_i (yʹ_i + b)^2 I = \sum_i m_i (xʹ_i + yʹ_i^2) + \sum_i m_i (a^2 + b^2) + 2a\sum_i m_ixʹ_i + 2b\sum_im_iyʹ_i But xʹ_i^2 + yʹ_i^2 = rʹ_i^2 and a^2 + b^2 = h^2, whence I = \sum_i m_i rʹ_i^2 + \sum_i m_i h^2 + 2a\sum_i m_ixʹ_i + 2b\sum_im_iyʹ_i By definition of the center of mass, however, \sum_i m_i xʹ_i = \sum_i m_iyʹ_i = 0,and I = \sum_i m_i rʹ_i^2 + \sum_i m_i h^2 = I_G + Mh^2 where M( = \sum_im_i) is the net mass of the body. This is the parallel-axes theorem. Although we derived this theorem in 2 dimensions, it is equally applicable in three dimensions. Take, in the case of the thin plate, the axes in the plane of the plate as the x- and y-axes, and the axis at right angles to the plane as the z-axis (see fig. (B) ). Then the moment of inertia of the plate about an axis perpendicular to the plate (the z-axis) is I_z = \sum_im_i r_i^2 where r_i locates m_i relative to 0. But r_i^2 = x_i^2 + y_i^2 where x_i and y_i are the x and y coordinates of m_i. Then I_z = \sum_im_i x_i^2 \sum_im_iy_i^2 But \sum_im_i x_i^2 = I_y and \sum_im_iy_i^2 = I_x, whence I_z = I_x+ I_y This is the perpendicular axes theorem.

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Question:

Electrons are located at definite energy levels in orbitals around the nucleus of an atom. Sometimes electrons move from one orbital to another. What happens to cause such a change?

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/Users/wenhuchen/Documents/Crawler/Biology/F01-0010.htm

Solution:

When an electron is in its ground state in the atom, it is occupying the orbital of lowest energy for that electron. There may be other electrons in the atom that are located in orbitals of even lower energy. The further the electron is from the nucleus, the greater the energy associated with that electron. An electron in the ground state may absorb energy from an external source (e.g. heat) , and be promoted to a higher-energy orbital. The electron would now be considered to be in an excited state. In a similar fashion, an excited electron may drop down to a lower-energy orbital (not necessarily the ground state), and in the process will decrease its energy content. This energy difference will be released in the form of a photon of light .

Question:

SINCALC: PROC; X = 2.5; N = 1; F1: FACT = 1; F2: POWER_X = X; SINX_2 = 0; REPEAT: SINX_2 = SINX_2 + POWER_X/FACT; F3: POWER_X = -POWER _X\textasteriskcenteredX\textasteriskcenteredX; F4: N = N + 2; F5: FACT = FACT \textasteriskcentered (N-1) \textasteriskcentered N; GO TO REPEAT; DONE: END;

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G08-0176.htm

Solution:

This program calculates the sine function as represented by a power series. The general formula is: sin (X) = (X/1!) - (X^3 / 3!) + (X5/5!) - \textbullet\textbullet\textbullet = ^\infty \sumn = 0(-1)^n [(X^2x+1 ) / (2x + 1)! ] The variable POWER_X is introduced to calculate the powers of x appearing in the series. Statement F2 initializes POWER_X and SINX_2. Statement F3 has two functions: 1) It increases the power of X by multiplying POWER_X twice by X, giving the odd exponents X3, X5, X7, etc. 2) The unary operator (-) alternates the sign of POWER_X. This is necessary to the definition of the function. As the program is written, we have an infinite loop. Execution of the statement REPEAT will continue indefinitely, since there is no way to reach the END statement. Let us insert control statements in the flowchart to amend this oversight. The programmer who designed this program neglected to choose an upper bound on the number of terms to be computed. We will write the flowchart so that only six terms are computed. To do this, we insert the variable KOUNT: when KOUNT is 6, control passes out of the loop.

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Question:

Consider a covalent bond between hydrogen and arsenic. It is known that the radii of hydrogen and arsenic atoms are respectively: 0.37 and 1.21 Angstroms. What is the approximate length of the hydrogen-arsenic bond?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E17-0621.htm

Solution:

One may assume that the two atoms are spheres. Thus, as in the figure shown, the bond length defined as the distance between the two nuclei is the sum of the lengths of the two radii. bond length = length of H radii + length of As radii. bond length = 0.37 \AA + 1.21 \AA = 1.58 \AA.

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Question:

Suppose the notion of a particle traveling along the x-axis is described by the equation x = a + bt^2 where a = 20 cm and b = 4 cm/sec^2. (a) Find the displacement of the particle In the time interval between t_1 = 2 sec and t_2 = 5 sec. (b) Find the average velocity in this time interval, (c) Find the instantaneous velocity at time t_1 = 2 sec.

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/Users/wenhuchen/Documents/Crawler/Physics/D03-0068.htm

Solution:

(a) The displacement of a particle moving from position r_1^\ding{217} to position r_2^\ding{217} is \Deltar \ding{217} = r_2 ^\ding{217} - r1^\ding{217}. In this problem, the motion of the particle is one-dimensional and we may neglect the vector nature of the displacement. Hence, \Deltax \Deltax = x(t_2) - x(t_1) = x(t_2) - x(t_1) \Deltax \Deltax = (a + bt^2_2) - (a + bt^2_1) = (a + bt^2_2) - (a + bt^2_1) \Deltax \Deltax = b(t^2_2 - t^2_1) = b(t^2_2 - t^2_1) \Deltax \Deltax = [(4 cm)/(sec^2)][25 - 4) sec^2 = [(4 cm)/(sec^2)][25 - 4) sec^2 \Deltax \Deltax = 84 cm = 84 cm The displacement is positive, so the particles position has increased in the positive direction along the x-axis.(x_2 > x_1) (b) Average velocity is given by the relation v_avg^\ding{217}= (\Deltar^\ding{217})/(\Deltat) =(r_2^\ding{217}_ - r_1^\ding{217})/(t2- t_1) Where r_2^\ding{217} and r_1^\ding{217} are the positions of the particle at times t_2 and t_1 respectively. This is a one-dimensional problem, hence v_avg= (\Deltax)/(\Deltat) = (84 cm)/(3 sec) = 28 cm/sec and it points in the positive x direction since\Deltax> 0. (c) The instantaneous velocity is v^\ding{217} = (dr^\ding{217})/(dt) Again, since we have a one-dimensional problem v= (dx)/(dt) = d/(dt)(a + b^2) = 2bt v(2 sec)= (2)(4 cm/sec^2)(2 sec) = 16 cm/sec.

Question:

Doubly charged \alpha-particles of energy 7.33 MeV are emitted from one isotope of thorium. What is the distance of closest approach of such an \alpha-particle to a gold nucleus? The mass of the \alpha-particle is 6.69 × 10^-27 kg, and the atomic number of gold is 79.

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/Users/wenhuchen/Documents/Crawler/Physics/D33-0997.htm

Solution:

The figure shows an energy diagram for the \alpha-particle gold nucleus interaction. Initially, we assume the \alpha-particle to be emit-ted from the thorium sample with a kinetic energy T_0. Since the thorium and gold nucleus are separated by an effectively infinite distance, the \alpha-particle gold nucleus system has no initial potential energy (V = 0). Hence, the initial energy of the system is totally kinetic and E_0 = T_0 = 7.33 MeV . Since energy is conserved, this is the value of E for all time t. Looking at the diagram, we see that for a given E, the closest the \alpha-particle can come to the gold nucleus is r_0 (distance of closest approach). This is true because, classically, the particle cannot penetrate the potential barrier to the left of r_0 , for, if it did, it would have a negative kinetic energy. This can be seen by noting that, at any time, the energy of the system can be written E = T + V or T = E - V . If V > E, as it is for values of r < r_0, then T < 0. Note that, at r_0, we may write V(r_0) = E or [(Zee')/(4\pi\epsilon_0 r_0)] = 7.33 MeV where Ze is the charge of the gold nucleus, and e' is the charge of the \alpha-particle. Since 1 eV = 1.602 × 10^-13 J (Zee')/(4\pi\epsilon_0 r_0) = 7.33 ×1.602 × 10^-13 J = 11.74310^-13 J Hence, r_0 = [(Zee')/{ 4\pi\epsilon_0 (11.743 × 10^-13 J)}] = [(79 × 1.602 × 10^-19 C × 2 × 1.602 × 10^-19 C)/(4\pi × 8.85 × 10^-12 C^2 \textbulletN^-1 \textbulletm^-2 × 11.743 × 10^-12 J)] = 3.102 × 10^-14 m.

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Question:

An isolated sphere 10 cm in radius is charged in air to 500 volts. How much charge is required? If this charge is then shared with another isolated sphere of 5 cm radius by connecting them together quickly with a fine wire, what is the final charge on each and what is the final potential of each?

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/Users/wenhuchen/Documents/Crawler/Physics/D17-0562.htm

Solution:

Figure A shows the first situation. (We assume both spheres are made of a conducting material.) The potential of a sphere is V = [1/(4\pi\epsilon_0)] (q/R) where \epsilon_0 is the permittivity of free space, and q is the charge on the sphere of radius R. Solving for q, and using the given data q = 4\pi\epsilon_0 RV q = {(10 × 10^-2 m) (500 V)} / {(9 × 10^9 N\bulletm^2/c^2)} = 5.6 ×10^-9 C(1) This is the charge needed to raise the sphere of 10 cm radius to 500 V. In figure B, we show the 2 spheres connected by a fine wire. In this situation, they have equal potentials and V_1 = V_2 or[1/(4\pi\epsilon_0)] [q_1/R_1] = [1/(4\pi\epsilon_0] [q_2/R_2] whenceq_1 = (R_1/R_2) q_2(2) Here, the subscript 1, refers to variables involved with the larger sphere, and similarly for the subscript 2 and the smaller sphere. Furthermore, since both spheres share the charge initially on the larger sphere, q_1 + q_2 = q (3) where q is given by (1). Using (2) in (3), we solve for q_2, [(R_1/R_2) + 1 ] q_2 = q q_2 = q/[(R_1/R_2) + 1 ] = q/(2 + 1) = q/3 whence q_2 = (5.6/3) × 10^-9 C \approx 1.86 × 10^-9 C Furthermore, q_1 = q - q_2 \approx (5.6 - 1.86) × 10^-9 C q_1 \approx 3.74 × 10^-9 C The final potential of the larger sphere is the same as the final potential of the smaller sphere. Both spheres then have a final potential of V = V_1 = V_2 = q_1/(4\pi\epsilon_0R_1) V \approx [(3.74 × 10^-9 C) (9 × 10^9 N \textbullet m^2/c^2)] / [10 × 10^-2 m] V \approx 336.6 Volts

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Question:

N^13 decays by \beta^+ emission. The maximum kinetic energy of the \beta^+ is 1.19MeV. What is thenuclidicmass of N^13?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E19-0718.htm

Solution:

In the emission of a \beta^+ particle, the daughter nuclide has a Z value one unit less than the parent with no change in A (mass). Thus, Z^(P)A\rightarrowZ - 1^(D)A + \beta^+, where P and D are the parent and daughter nuclides, respect-ively. The emitted positron (\beta^+) is unstable and is usually consumed after being slowed down by collisions. This is shown in the following reaction \beta^+ + \beta^-\rightarrow2\gamma In this reaction, 2 photons of light are produced. The reaction in this problem is _7N^13\rightarrow_6C^13 + \beta+ In this type of process, a simple difference of whole atom masses is not desired. Whole atom masses can be used for mass difference calculations in all nuclear reactions, except in \beta^+ processes where there is a resulting annihi-lation of two electron masses (one \beta^+ and one \beta^-). Thus, mass difference= (M_nfor _7N^13) - (M_nfor _6C^13) - Me = [(M for _7N^13) - 7M_e] - [(M for _6C^13) - 6 M_e] - Me = (M for _7N^13) - (M for _6C^13) - 2Me = (M for _7N^13) - 13.00335 - 2(0.00055) = (M for _7N^13) - 13.00445 Here,M_nis the nuclear mass (mass of neutrons and protons), M is the atomic mass, and M_e is the mass of an electron. This expression is equal to the mass equivalent of the maximum kinetic energy of the \beta^+. [(1.19MeV) / (931.5MeV/amu)] Then, 0.00128amu= (M for _7N^13) - 13.00445 or, M for _7N^13 = 13.00445 + 0.00128 = 13.00573amu.

Question:

On a long straight road a car accelerates uniformly from rest, reaching a speed of 45 mph in 11 s. It has to maintain that speed for 1(1/2) mi behind a truck until a suit-able opportunity for passing the truck arises. The car then accelerates uniformly to 75 mph in a further 11 s. After maintaining that speed for 3 min, the car is brought to a halt by a uniform deceleration of 11 ft/s^2. Illustrate the motion on a suitable diagram, and calculate (a) the total distance traveled, (b) the total time taken, (c) the average speed, and (d) the average acceleration in the first 142 s.

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/Users/wenhuchen/Documents/Crawler/Physics/D03-0077.htm

Solution:

A velocity-time diagram should be drawn. During the first 11 s the car accelerates uniformly to a speed of 45 mph = 66 ft/s. This part of the diagram is therefore a straight line OA inclined to the t-axis at an angle whose tangent is 66/11. The distance traveled, s_1, is the area under this portion of the graph. Thus s_1 = (1/2) × 11 s × 66 ft/s = 363 ft. In the second portion of the motion, the car travels for 1 (1/2) mi at a constant speed of 45 mph. This part of the graph, AB, is a straight line parallel to the t-axis, its length being t_2 = [(1(1/2) mi]/[45 mi/hr] × 60 min/hr × 60 s/min = 120 s. In the third portion of the motion, the car increases its speed by 30 mph = 44 ft \textbullet s^-1 at uniform acceleration in 11 s. This part of the graph is thus a straight line BC of slope 44/11. The distance traveled in this 11 s, s_3, is the area under this part of the graph, i.e., the shaded portion. s_3 = (1/2) × 11 s × 44 ft/s + 11 s ×66 ft/s = 968 ft. The next portion of the graph is again a straight line parallel to the t-axis. The time t, is 3 min = 180 s, and thus s_4 = 75 mi/hr × (3/60) hr = 3.75 mi. In the final part of the motion, the car is brought to rest from a speed of 110 ft \textbullet s^-1 by a uniform de-celeration of 11 ft/s^2. This portion of the graph, DE, is thus a straight line with a negative slope of 110/11. The time taken to come to rest, t_5, and the dist-ance traversed, s_5, are t_5= [(110 ft/s)/(11 ft/s)] = 10 s and s_5 = (1/2) × 10s × 110 ft/s = 550 ft. = (29,601/332) ft/s = 89.16 ft/s × (60 mph)/(88 ft/s) = 60.8 mph.

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Question:

Upon the addition of heat to an unknown amount of KCLO_3, .96 g of oxygen was liberated. How much KCLO_3 was present?

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Solution:

The key to answering this question is to write a balanced equation that illustrates this chemical re-action. From this, you can employ the mole concept to determine the weights of the substances involved. Given that oxygen is liberated and that oxygen gas exists as O^2 and not O, you can write the balanced equation 2KCLO_3 \rightarrow 2KCL + 3O_2. All atoms are accounted for. The coefficients indicate the relative number of moles that react. For example, every two moles of KCLO_3 yield 3 moles of oxygen. A mole is defined as weight in grams of a substance divided by its atomic or molecular weight. Therefore, you have [(.96 g) / (3 g/mole)] = .030 moles of oxygen. To calculate the weight of KCLO_3, you look at the balanced equation. You find that the number of moles of KCLO_3 is 2/3 the number of moles of oxygen. As such, the number of moles [(KCLO_3) / (3 moles O_2)] = .20 moles of KC1O_3. Recalling the definition of a mole, the number of grams of KC1O_3 (MW of KCLO_3 = 122.55) is .020 × (122.55 g/mole) = 2.5 g.

Question:

In the Bohr model of the hydrogen atom, the electron is considered to move in a circular orbit around the nuclear proton. The radius of the orbit is 0.53 × 10^-8 cm. What is the velocity of the electron in this orbit?

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/Users/wenhuchen/Documents/Crawler/Physics/D31-0910.htm

Solution:

If we assume that the only force acting is the electro-static force, F_E = -[(q_1q_2)/r^2] = (e^2/r^3) = [(4.80 × 10^-10 statC)^2 / (0.53 × 10^-8 cm)^2 = 8.2 × 10^-3 dyne In order to maintain a circular orbit, the electron must be subject to a centripetal acceleration: a_c = (v^2/r). The mass of the electron times this acceleration must equal the force on the electron: m_ea_c = [(m_ev^2) / r] = F_E or, solving for v, v = \surd[(rF_E) / (m_e)] = \surd[(0.53 ×10-8cm × 8.2 × 10^-3 dyne)] / (9.11 × 10^-28 g) = 2.18 × 10^-8 cm/sec Hence, the electron velocity is about 1 percent of the velocity of light.

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Question:

Under certain conditions, the rate equation for the formation ofwater 2H_2 + O_2 = 2H_2O isgiven by rate= k [H_2]^2 [H_2O] wherek is the rate constant. What is the overall order ofthis rateequation?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E13-0451.htm

Solution:

The overall order of a rate equation is equal to the sum of the exponentsto which the concentrations are raised. In the equation rate= k [H_2]^2 [H_2O], [H_2] is raised to the second power and [H_2O] is raised to the first power. Hence, the rate is second order in [H_2], first order in [H_2O], and 2 + 1 = 3, or third order overall.

Question:

Two atoms of scandium are to combine with three atoms of oxygen. If you start with 1 gram of scandium, how much oxygen is required? Scandium has an atomic weight of 44.96 g/mole. The at. wt. of oxygen is 15.999 g/mole.

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Solution:

The key to solving this problem is determining the number of moles that will be reacting. After this, one is able to calculate the mass by multiplying the atomic weight by the number of moles. Two atoms of scandium react with 3 atoms of oxygen, which is equivalent to saying 2 moles of scandium react with 3 moles of oxygen. A mole is defined as weight in grams of the substance divided by the atomic weight or molecular weight. If you start with 1 gram of scandium, then the number of moles of scandium is [(1.00 g) / (44.96 g/mole)] = .0222 moles. Since the number of moles of oxygen must be in a 3 : 2 ratio to scandium, multiply 3/2 by .0222 to determine the number of moles of oxygen that will react. .0222 = (3/2) .0333 moles of oxygen atoms. Recalling the definition of a mole, the mass of the oxygen that reacts is .0333 (15.999) = .533 g of oxygen.

Question:

A string is wrapped around a uniform homogeneous 3 lb cylinder with a 6 in. radius. The free end is attached to the ceiling from which the cylinder is then allowed to fall (as in the Figure), starting from rest. As the string unwraps, the cylinder revolves. (a) What is the linear acceleration of the center of mass? (b) What is the linear velocity, and (c) how fast is the cylinder revolving after a drop of 6 ft? (d) What is the tension in the cord?

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Solution:

(a) Isolate the cylinder, and indicate forces acting on it. There is no need to tabulate x- and y-components since they are all up or down forces. Set\sumF = ma Where \sumF is the net force acting on the cylinder, a is the acceleration of the cylinder's center of mass, and m is its mass. (Note that in problems such as this one, it is convenient to take the direction of motion as positive.) mg - T = ma\therefore a = g - (T.m)(1) Now, consider rotation. Consider torques about the center of mass 0, and set \sumL = l\alpha, where \sumL is the net torque about 0, I is the moment of inertia at the cylinder about 0, and \alpha is its angular acceleration. Hence \sumL = rT = l\alpha \thereforeT = I(\alpha/r) (Clockwise rotation corresponds to downward motion, already assumed to have the positive direction.) But I for a cylinder = 1/2 mr^2 , where r is the cylinder radius. In this problema = r\alpha T = (1/2)[(mr^2)/(r)] [a/r] = [ma/2] whereupon, using (1), a = [(-ma)/(2m)] + g = [(-a)/2] + g \therefore3a = 2g a = (2/3)g = 21.3 ft/sec^2(downward) (b) Now since a = constant (2/3 g), the linear motion is uniformly accelerated, such that the ve-locity of the center of mass is v^2 = v^2_0 + 2as,where s becomes the drop, h, that the cylinder experiences, and v_0 is its initial velocity. v^2 = 0 + (2) (21.3 ft/sec^2) (6 ft) v^2 = (12) (21.3 ft^2/sec^2) \thereforev = \surd256 = 16 ft/sec (c) The angular velocity \omega = v/r = (16 ft/sec)/[(1/2) ft] = 32 rad/sec (d) From (1) T = m(g - a) = m[g - (1/2) g] T = (1/3) mg = (1/3)(3 lb) = 1 lb

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Question:

If a living frog muscle is dissected out from a frog's body and placed in a closed container without oxygen, it can contract for a time when stimulated. Then it stops. But if the container is opened, it soon regains the ability to contract. Comment.

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/Users/wenhuchen/Documents/Crawler/Biology/F03-0078.htm

Solution:

The important clue in this question is the presence or absence of oxygen and how it may affect metabolism. When the muscle is still intact within the frog, the muscle cells should be undergoing aerobic respiration, since oxygen is available to each cell (by diffusion from the blood). However, once the muscle is removed and placed in an anaerobic environment , aerobic respiration is replaced by homolactic fermentation. Recall that this process involves the reduction of pyruvic acid to lactic acid due to the absence of any hydrogen acceptors. (see previous question) The build-up of lactic acid produces fatigue and muscle cramps. Since glycolysis produces only 2 net ATP per glucose molecule consumed, the ATP produced thus is not sufficient to replenish the ATP expended in muscle contraction. Eventually, the cell's ATP storage is depleted and conraction can no longer occur. However, when the container is opened, oxygen is reintroduced and is now available to accept hydrogen. Lactic acid is first converted back to pyruvic acid which eventually enters the TCA cycle and the respiratory system to produce more ATP. The presence of oxygen enables the muscle cells to undergo aerobic respiration with the concomitant increase in ATP production and oxidation of lactic acid back to pyruvic acid. This oxidation occurs in the liver. Under these conditions, the muscle regains its ability to contract. The same process occurs in humans after vigorous exercise. For example, once exercise is finished, heavy breathing continues. Such is the body's reaction in order to carry oxygen to the muscles as quickly as possible to repay the oxygen debt. Often this term is misleading, for oxygen debt does not mean the amount of oxygen the cells owe. Rather, it indicates the need of oxygen to oxidize the lactic acid accumulated when the cells are respiring anaerobically. The pyruvate formed as a result of the oxidation of lactate can enter into aerobic res-piration to generate a larger amount of ATP. However, some of the lactate, transported to the liver, gets oxid-ized to pyruvate and then converted back to glucose when oxygen becomes available.

Question:

If the density of ethylene is 1.25 g/liter at S.T.P. and the ratio ofcarbon to hydrogen atoms is 1 : 2, what is molecular weightand formula of ethylene?

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Solution:

This problem is solved once you know that at S.T.P. (Standard , Temperature ,Pressure, 0 \textdegreeC , and 1atm) one mole of any gas occupies 22.4 litters.Assume that one mole of ethylene gas ispresent .Density = mass/ volume. As such the mass of the gas = (1.25 g/l) (22.4l) =28.0g . Therefore one mole of ethylene weighs 28g. From the ratio given in thequestionyouknow the molecular formulacan be represented as (CH_2)_x . To obtain the actual molecular formula ,look for a compound that has a molecular weight of 28 g yet maintainsthe carbon: hydrogen ratioof 1:2 . By looking at the periodic tablefor atomicweights and throughsome arithmetic , you will find that theonly formula that meets these requirements is (CH_2)_2 or C_2H2 .This formulacan also be found by dividing the weight of 1 CH_2 into 28g . This solvedfor x in the expression (CH_2)_x. MW of CH_2 = 14. no. of CH_2= (28g) / (14g/CH_2) = 2 CH_2 = C_2H_4.

Question:

Describe the sex act.

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Solution:

The human sex act is generally divided into four phases. These are the excitement, plateau, orgasm and resolution phases. The excitement phase in the female is characterized by a rise in heart rate and blood pressure. Muscle tension also increases as the breasts swell and the genitalia become filled with blood. A lubricating fluid is secreted during this phase by the vaginal wall. In the male, the excitement phase is also characterized by increased blood pressure, heart rate and muscle tension. The penis becomes rigid as the erectile tissue becomes engorged with blood. The plateau phase is characterized by increased breathing rate, pulse rate and muscle tension in male and female. The orgasm phase is characterized by muscular contrac-tions of the vagina in the female and by the muscles in the genital area of the male. The orgasm phase is initiated by frictional force against the vaginal wall and clitoris in the female. In the male, orgasm is caused by pressure and friction against the nerve endings in the glans penis and in the skin of the shaft of the penis. The resolution phase is the final stage in the sex act. During this period body functions return to normal levels. Heart beat, pulse rate and breathing levels return to the precoital state.

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Question:

In a certain engine, fuel is burned and the resulting heat is used to produce steam which is then directed against the vanes of a turbine, causing it to rotate. What is the efficiency of the heat engine if the temperature of the steam striking the vanes is 400\textdegree K and the temperature of the steam as it leaves the engine is 373\textdegree K?

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Solution:

The efficiency E of a heat engine is the ratio of the net work W done by the engine in one cycle to the heat Q, absorbed from the high temperature source in one cycle. E = W/Q1 For the Carnot cycle, which describes the operation of a reversible heat engine, we know the efficiency to be, E = W/Q_1 = (Q_1 - Q_2)/Q_1 = (T_1 - T_2)/T1 where T_2 is the low temperature reservoir. Carnot stated that the efficiency of all Carnot engines operating be-tween the same two temperatures is the same and that no irreversible engine working between these two temperatures can have a greater efficiency. This means that the maxi-mum efficiency of this heat engine is given by, \varepsilon = (T_1 - T_2)/T_1 = (400\textdegreeK - 373\textdegreeK)/400\textdegreeK = 0.068 = 6.8%.

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Question:

Assume that one cubic foot of air near a thermonuclear explosion is heated from 0\textdegreeC to 546,000\textdegreeC. To what volume does the air expand?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E02-0033.htm

Solution:

Charles' Law (V_1 /T_1 = k, where V_1 is the initial volume, T_1 the initial absolute temperature, and k is a constant) states that the volume is directly proportional to the temperature. V \alpha T \alpha means ' is proportional to. ' The absolute temperature can be found by adding 273 to the temperature in \textdegreeC, T_1 = 0 + 273 = 273 T_2 = 546,000 + 273 = 546,273 Using the Charles' Law, one can set up the following ratio (V_1 /T_1 ) = (V_2 /T_2 ), where V_1 is the original volume, V_2 is the final volume, T_1 is the original temperature, and T_2 is the final tem-perature. Using this ratio, one can determine the final volume. (V_1 /T_1 ) = (V_2 /T_2 ) Substituting, one obtains (1 ft^3 /273) = (V^2 /546,273) V^2 = [(546,273)/(273)] = (1 ft^3) = 2001 ft^3 .

Question:

When the following reaction F_2 + Cl_2 \ding{217} 2ClF occurs, 26 Kcal / mole are released. The bond energies of F_2 and CI_2 are 37 and 58 Kcal / mole, respectively. Calculate the dissociation energy of theClFbond.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E17-0631.htm

Solution:

The dissociation energy is the amount of energy needed to break the chemical bonds of a molecule. The re-action for the dissociation of 2ClF bonds is 2ClF \ding{217} F_2 + Cl_2- This is the opposite of the reaction described in the problem, therefore, 26 Kcal / mole are consumed. The amount of energy consumed is equal to twice the dissociation energy ofClF(because 2 moles ofClFare present) sub-tracted from the sum of the dissociation energies of F_2 and Cl_2 ∆H = [∆H_F(2) + ∆H_CL(2)] - 2∆H_CL-F = - 26 kcal / mole 26 kcal / mole = 2∆H_Cl-F - 37 kcal / mole - 58 kcal / mole 2∆H_Cl-F = 26 kcal / mole + 37 kcal / mole + 58 kcal / mole = (121 / 2) kcal / mole = 60.5 kcal / mole. Thus, 60.5 Kcal are needed to break one mole ofCl-F bonds.

Question:

Using data from the accompanying table find the increase of enthalpy for 100 g of H_2 O going from ice at - 10\textdegreeC to liquid water at + 15\textdegreeC. Assume that the molar heat of fusion of ice is 6.02 kJ. Heat capacity in joules per mole-degree for various substances at the temperatures indicated H_2 O(s)at239\textdegreek33.30 H_2 O(s)at271\textdegreek37.78 H_2 O(l)at273\textdegreek75.86 H_2 O(l)at298\textdegreek75.23 H_2 O(l)at373\textdegreek75.90 H_2 O(g)at383\textdegreek36.28

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Solution:

Enthalpy is an indication of the changes in heat content of a system due to changes in the system. To solve this problem, one must be aware that in going from ice at - 10\textdegreeC to water at 15\textdegreeC, three things happen: (1) the ice is heated, (2) the ice is melted, (3) the liquid is heated. For the heating processes, (1) + (3), the change in enthalpy is quantitatively related to the heat capacity, C_p, and the change in temperature, ∆T by the relation is ∆H =C_p∆T. C_p, the heat capacity, is defined as the amount of heat required to raise one mole of material by one degree. The melting process, (2), is ex-pressed in terms of the molar heat of fusion, which is the amount of heat necessary to melt one mole of solid. The summation of the heat involved in all three processes will give the change in enthalpy. Because each step is defined in terms of moles, one must first calculate the number of moles involved. The molecular weight of H_2 O is 18. There-fore, there are 100/18 or 5.55molesof H_2 O in 100 g. Thus, (1) To heat ice, one needs, (5.55moles) (37.78 J mol^-1 deg-1) (10 deg) = 2,100 J (2) To melt ice, one requires (5.55moles)(6.02 kJ/mole)(1000 J/kJ) = 33,400 J (3) To heat liquid, one needs (5.55moles)(75.86 J mol^-1 deg^-1) (15 deg) = 6,300 J. The total increase in enthalpy is ∆H = 2,100 + 33,400 + 6,300 = 41,800 J.

Question:

A chemist dropped a 200 g object into a tank of water. It displaced 60 ml. of the water when it sunk to the bottom of the tank. In a similar experiment, it displaced only 50 g of an oil into which it was dropped. Calculate the density of the object and the oil.

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Solution:

The density (\rho) of a substance is defined as its mass divided by its volume. \rho = (mass - volume) Thus, to solve for the densities of the object and the oil, one must first calculate their respective masses and volumes The mass of the object is 200 g, but the volume is not given An object dropped in any liquid displaces a volume of liquid equal to the volume of the object. The object displaces 60 ml of water; therefore, the volume of the object is 60 ml Solving for the density of the object: \rho = (mass / volume) = (200 g / 60 ml) = 3.33 g / ml. One is given that the object displaces 50 g of the oil, and from the water experiment, it is known that the volume of the object is 60 ml. Since the object displaces the same volume of liquid as it occupies,60 ml of the oil weighs 50 g. Solving for the density of the oil: \rho = (mass / volume) (50 g / 60 ml) = 8.33 g / ml.

Question:

Make a diagram of the human brain. Label the principal parts and list the function(s) carried out by each.

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/Users/wenhuchen/Documents/Crawler/Biology/F20-0505.htm

Solution:

The human brain is the enlarged, anterior end of the spinal cord. This enlargement is so great that the resemblance to the spinal cord is obscured. The adult human brain has six major regions: the medulla, pons and cerebellum which constitute the hindbrain; the thalamus and cerebrum; both of which are in the forebrain; and the midbrain. The most inferior part of the brain, connected immediately to the spinal cord, is the medulla. Here the central canal of the spinal cord (spinal lumen) enlarges to form a fluid-filled cavity called the fourth ventricle. The medulla has numerous nerve tracts (bundles of nerves) which bring impulses to and from the brain. The medulla also contains a number of clusters of nerve cell bodies, known as nerve centers. These reflex centers control respiration, heart rate, the dilation and constriction of blood vessels, swallowing and vomiting. Above the medulla is the cerebellum, which is made up of a central part and two hemispheres extending side-ways. The size of the cerebellum in different animals is roughly correlated with the amount of their muscular activity. It regulates and coordinates muscle contraction and is relatively large in active animals such as birds. Removal or injury of the cerebellum is accompanied not by paralysis of the muscles but by impairment of muscle co-ordination. A bird, with its cerebellum surgically removed, is unable to fly and its wings are seen to thrash about without coordination. The pons is an area of the hindbrain containing a large number of nerve fibers which pass through it and make connections between the two hemispheres of the cerebellum, thus coordinating muscle movements on the two sides of the body. The pons also contains the nerve centers that aid in the regulation of breathing. In front of the cerebellum and pons lies the thick- walled midbrain. The midbrain is an important integrating region and contains the centers for certain visual and auditory reflexes. A cluster of nerve cells regulating muscle tone and posture is also present in the midbrain. A small canal runs through the midbrain and connects the fourth ventricle behind it to the third ventricle in front of it. The midbrain, pons and medulla are collectively called the brainstem. Ten of the twelve cranial nerves originate in the brainstem. The thalamus of the forebrain serves as a relay center for sensory impulses. Fibers from the spinal cord and parts of the brain synapse here with other neurons going to the various sensory areas of the cerebrum. The thalamus seems to regulate and coordinate the external signs of emotions. By stimulating the thalamus with an electrode, a sham rage can be elicited in a cat - the hair stands on end, the claws protrude, and the back becomes humped. However, as soon as the stimulation ceases, the rage responses disappear. The hypothalamus, located under the thalamus, is a collection of nuclei (a cluster of nerve cell bodies located within the central nervous system), concerned with many im-portant homeostatic regulations. Electrical stimulation of certain cells in the hypothalamus produces sensations of hunger, thirst, pain, pleasure, or sexual drive. The hypothalamus is also important for its influence on the pitui-tary gland, which is functionally under its control. Cells of the hypothalamus synthesize chemical factors that modu-late the release of hormones produced and stored in the an-terior pituitary. The hypothalamus has neural connections with the posterior pituitary. The hypothalamus and thalamus are collectively called the diencehpalon. The cerebrum, consisting of two hemispheres, is the largest and most anterior part of the human brain. In human beings, the cerebral hemispheres grow back over the rest of the brain, hiding it from view. Each hemisphere contains one cavity (one contains the first and the other the second ventricle, collectively known as the lateral ven-tricles), which is connected to the third ventricle. The outer portion of the cerebrum, the cortex, is made up of gray matter which comprises the nerve cell bodies. The grey matter folds greatly, producing many convolutions of the cerebral surface. These convolutions increase the surface area of the gray matter. The inner part of the brain is the white matter which is composed of masses of nerve fibers. Each cerebral hemisphere is divided into four lobes. The occipital or posterior lobe receives visual information. The temporal lobe, located on the side of the cerebrum just above the ears, receives auditory information. The frontal and parietal lobes are demarcated by the fissure of Rolando. Just in front of this fissure in the frontal lobe is the primary motor area, while the primary sensory area lies be- hind it in the parietal lobe. When all the areas of known functions are plotted, they cover only a small part of the total area of the human cortex. The rest, known as association, areas, are regions responsible for the higher intellectual faculties of memory, reasoning, learning and imagination, all of which help to make up one's personality. In some unknown way, the association regions integrate into a meaningful unit, all the diverse impulses constantly reaching the brain, so that proper response is made.

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Question:

Sulfur exists as S_2 vapor at temperatures between 700\textdegreeC and 1500\textdegreeC. At 1473 k it combines with hydrogen according to the equation H_2 (g) + (1/2)S_2 (g) \rightarrow H_2 S (g) . At 750\textdegreeC the equilibrium constant is 1.07 × 10^2 and at 1200\textdegreeC it is 4.39. Determine the heat of reaction in the temperature range 750\textdegreeC to 1200\textdegreeC, and the change in free energy at each of these temperatures.

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Solution:

The change in free energy ∆G is determined from the equation ∆G\textdegree = - 2.303 RT log k, where R is the gas constant, T the absolute temperature, and k the equilibrium constant. From this equation and the expression ∆G\textdegree = ∆H\textdegree - T∆S\textdegree, where ∆H\textdegree is the change in the heat of reaction and ∆S\textdegree the change in entropy, a relation between ∆H and T will be found. At T = 750\textdegreeC = 1023\textdegreeK,k = 1.07 × 10^2and ∆G\textdegree = - 2.303 RT log k = - 2.303 × 1.987 [(cal)/(mole-deg)] × 1023\textdegreeK × log (1.07 × 10^2) = - 9500 cal/mole = - 9.5 Kcal/mole. At T = 1200\textdegreeC = 1473\textdegreeK,k = 4.39and ∆G\textdegree = - 2.303 RT log k = - 2.303 × 1.987 [(cal)/(mole-deg)] × 1473\textdegreeK × log (4.93) = - 4330 cal/mole = - 4.33 Kcal/mole. To find an expression relating ∆H to T, consider the following two equations: ∆G\textdegree = - 2.303 RT log k,and∆G\textdegree = ∆H\textdegree - T∆S\textdegree. Setting these equal gives - 2.303 RT log k = ∆H\textdegree - T∆S\textdegree, or,log k = [(- ∆H\textdegree)/(2.303 RT)] + [(∆S\textdegree)/(2.303 R)]. For two different temperatures, T_1 and T_2 with equi-librium constants k_1 and k_2, respectively, log k_1 = [(- ∆H\textdegree)/(2.303 RT_1)] + [(∆S\textdegree)/(2.303 R)]and log k_2 = [(- ∆H\textdegree)/(2.303 RT_2)] + [(∆S\textdegree)/(2.303 R)]. Subtracting the first equation from the second, log k_2 - log k_1 = [{(- ∆H\textdegree)/(2.303 RT_2)} + {(∆S\textdegree)/(2.303 R)}] - [{(- ∆H\textdegree)/(2.303 RT_1)} + {(∆S\textdegree)/(2.303 R)}], or,log (k_2 /k_1) = [(- ∆H\textdegree)/(2.303 R)] [(1/T_2) - (1/T_1)]. Let T_1 = 1023\textdegreeK, k_1 = 1.07 × 10^2, T_2 = 1473\textdegreeK,andk_2 = 4.39. Then log (k_2 /k_1) = [(- ∆H\textdegree)/(2.303 R)] [(1/T_2) - (1/T_1)] log [(4.39)/(1.07 × 10^2)] = [(- ∆H\textdegree)/(2.303 × 1.987 cal/mole-deg)] [(1/1473\textdegreeK) - (1/1023\textdegreeK)] log [(4.39)/(1.07 × 10^2)] = [(- ∆H\textdegree)/(2.303 × 1.987 cal/mole-deg)] (0.000679\textdegreek^-1 - 0.000978\textdegreek^-1) or, ∆H\textdegree = [{log (4.39/107)}/(2.303 × 1.987 cal/mole-deg)] (0.000978\textdegreek^-1 -0.000679\textdegreek^-1) = - 21, 300 cal/mole = - 21.3 Kcal/mole. Thus, the heat of reaction in the temperature range 1023\textdegreeK to 1473\textdegreeK (750\textdegreeC to 1200\textdegreeC) is - 21.3 Kcal/mole.

Question:

Contrast the development of an amphibian, bird, and starfish egg up to the point of gastrulation.

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Solution:

In the isolecithal starfish egg, cleavage is radial (Figure 1). The first cleavage division passes through the long axis of the egg, splitting the egg into two equal daughter cells. The second cleavage plane is also vertical but at right angles to the first, separating the two cells into four. The third division is horizontal, at right angles to the other two, and splits the four cells into eight. Further divisions result in embryos containing 16, 32, 64, 128 cells, and so on, until a hollow ball of cells called the blastula, is formed. The wall of the blastula is a single layer of cells, the blastoderm, surrounding the blastocoel, the cavity in the center. It is this single-layered blastula which is converted later into a double-layered sphere, the gastrula. Amphibians also display radial cleavage but because of the unequal distribution of yolk in the telolecithal egg, the resulting blastomeres are not of equal size (Figure 2). The first cleavage begins as a shallow furrow at the animal pole and progresses gradually through to the vegetal pole. This first cleavage usually results in two equal blastomeres. The second cleavage is at right angles to the first. It begins at the animal pole even before the first cleavage has been completed at the vegetal pole. The result is four blastomeres of nearly equal size. The third cleavage is typically at right angles to the first two cleavages, and hence cuts horizontally across the egg's vertical axis. It passes well above the equator, so that the eight-celled stage commonly consists of four smaller animal blastomeres (micromeres), and four larger, yolk-laden vegetal blas-tomeres (macromeres). A cavity is present at the center of the group of blastomeres beginning at the eight-celled or 16-celled stage, which increases in size as cleavage progresses. The egg thud becomes a hollow ball of cells, namely, a blastula. Its blastocoel is roughly hemispherical. It has a dome of smaller animal cells and a floor of larger yolk-laden vege-tal cells. Telolecithal bird eggs also display radial cleav-age, but the cleavage is meroblastic. In bird eggs and other eggs which contain a large amount of yolk, cleavage occurs only in a small disk of cytoplasm in the animal pole (Figure 3). At first, all cleavage planes are vertical and all the blastomeres lie in a single plane. The cleavage furrows separate the blastomeres from each other but not from the yolk. The central blastomeres are continuous with the yolk at their lower ends, and the blastomeres at the circumference of the disk are contin-uous both with the yolk beneath them and with the uncleaved cytoplasm at their outer edge. As cleavage continues, more cells become cut off to join the ones in the center, but the new blastomeres are also continuous with the uncleaved underlying yolk. The central blastomeres eventually become separated from the underlying yolk either by cell divisions with horizontal cleavage planes or by the appearance of slits beneath the nucleated portions of the cells. Horizontal cleavages separate upper blastomeres and lower blastomeres. The upper blastomere is a cell with a complete plasma membrane, which is separated from its neighbors and from the yolk. The lower blastomere is a cell which remains connected with the yolk. The blastomeres at the margin of the disk, and the lower cells in contact with yolk, eventually lose the furrows that partially separated them and fuse into a continuous syncytium. This syncytium con-tains many nuclei and is termed the periblast. The periblast is believed to break down the yolk, thereby making its nutrients available for the growing embryo. The free blastomeres (with complete plasma membranes) become incorporated into two layers, an upper epiblast, and a lower hypoblast. Between these two layers is a cavity, the blastocoel. Below the hypoblast and above the yolk is the shallow subgerminal cavity, which appears only under the central portion of the blastoderm. The area of the blastoderm over the subgerminal space is relatively transparent (due to lack of yolk) and is called the area pellucida, whereas the more opaque part of the blastoderm (which contains some yolk) that rests directly on the yolk is called the area opaca.

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Question:

What results would you expect if the macronucleus were removed from a paramecium?

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Solution:

Ciliates differ from all other animals in possessing two distinct types of nuclei - a large macro-nucleus and one or more smaller micronuclei. The macro-nucleus is sometimes called the vegetative nucleus, since it is not critical in sexual reproduction, as are the micronuclei. The macronucleus is essential for normal metabolism, for mitotic division, and for the control of cellular differentiation. The macronucleus is con-sidered to participate actively in the synthesis of RNA, which is used in cell metabolism. Removal of the macronucleus from a ciliate causes cell death, even if a micronucleus is present.

Question:

Show that for large values of the principal quantum number, the frequencies of revolution of an electron in adjacent energy levels of a hydrogen atom and the radiated frequency for a transition between these levels all approach the same value.

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Solution:

We use the Bohr model of the hydrogen atom in the analysis of this problem. Assuming that the electron is orbiting about the proton in a circular orbit, we observe that its period is given by \tau = 2\pir/v(1) where r is the orbit radius, and v is the velocity of the electron. Bohr's condition on the angular momentum of the electron is L =nh/2\pin = 0, 1, 2,... In a circular orbit, L =mvr(2) Whence mvr=nh/2\pi andv =nh/2\pimr Inserting (2) in (1) \tau = 2\pir . (2\pimr)/nh \tau = [(4^2mr^2)/(nh)](3) To eliminate r in (3), we note that the Coulomb force acting on the electron is centripetal, and (1/4\pi\cyrchar\cyrie_0) (e^2/r^2) = (mv^2/r) where e is the electronic charge. Hence e^2r = mv^2r^24\pi\cyrchar\cyrie_0 r =[e^2/(4\pi\cyrchar\cyrie_0v^2m)](4) Substituting (2) in (4) r = [e^2/(4\pi\cyrchar\cyrie_0m)] [(4\pi^2m^2r^2)/(n^2h^2)] 1 = [(e^2\pim)r / (n^2h^2\cyrchar\cyrie_0)] andr = [(n^2h^2\cyrchar\cyrie_0)/ (e^2\pim)](5) Substituting (5) in (3) \tau = [(4\pi^2m)/(nh)] \bullet [(n^4h^4\cyrchar\cyrie^2_0)/(e4\pi^2m^2)] \tau = [(4n^3h^3\cyrchar\cyrie^2_0)/(e^4m)] Hence, the orbital frequency of the electron is v = [(me^4)/(4\cyrchar\cyrie^2_0n^3h^3)](6) We must now find how the energy of a hydrogen atom is quantized. The energy of the atom is potential and kinetic E = (1/2)mv^2 - [(e^2)/(4\pi\cyrchar\cyrie_0r)](7) To transfer E into a function of n, we eliminate v and r in (7) by using (2) and (5) v = (nh/2\pimr) = (nh/2\pim)[(e^2\pim)/(n^2h^2\cyrchar\cyrie_0)] v = [(e^2)/(2\cyrchar\cyrie_0nh)] r = [(n^2h^2\cyrchar\cyrie_0)/(e^2\pim)] andE = m/2[(e^4)/(4\cyrchar\cyrie^2_0n^2h^2)] - [e^2/(4\pi\cyrchar\cyrie_0)] [(e^2\pim)/(n^2h^2\cyrchar\cyrie_0-)] E = [(me^4)/(8\cyrchar\cyrie^2_0n^2h^2)] - [(me^4)/(4\cyrchar\cyrie^2_0n^2h^2)] E = [(-me^4)/(8\cyrchar\cyrie^2_0n^2h^2)] But, the energy is quantized and hv_n= [(-me^4)/(8\cyrchar\cyrie^2_0n^2h^2)] v_n= [(-me^4)/(8\cyrchar\cyrie^2_0n^2h^3)] The radiated frequency for a transition between 2 adjacent energy levels is v_n- v_n-1 = [(-me^4)/(8\cyrchar\cyrie^2_0h^3)] [{1/n^2} - {1/(n-1)^2}] v_n- vn-1= [(-me^4)/(8\cyrchar\cyrie^2_0h^3)] [{n^2-2n+1-n^2}/{n^2(n-1)^2}] v_n- vn-1= [(-me^4)/(8\cyrchar\cyrie^2_0h^3)] [{2n-1}/{n^2(n-1)^2}](7) We must show that (6) and (7) approach the same value for large n. If n>> 1, 2n - 1 \approx 2n and n -1 \approx n. Hence, v_n- vn-1\approx [(me^4)/(8\cyrchar\cyrie^2_0h^3)] [(2n)/(n^2 n^2)] v_n- vn-1\approx [(me^4) / (4n^3\cyrchar\cyrie^2_0h^3)] which is (6). At large quantum numbers, classical and quantum mechanical results agree for the energy levels lose their discrete characteristics. This is an example of Bohr's correspondence principle.

Question:

A factory produces bars whose exact length is unknown until after production. Its cutting shop receives orders for cut lengths of bar which must be met by cutting up the manufactured bars. For a given manu-factured bar of length 'L' a set 'C' of orders is chosen from the order list, which can be met by cutting up this bar, and with minimum wastage. Design a recursive back-tracking procedure which will construct 'C' for given ' L' and order list.Incorporate the procedure in a program which constructs 'C' from the following order list order1773 mm order2548 mm order365 mm order4929 mm order5548 mm order6163 mm order7421 mm order837 mm anda manufactured bar length L = 1848 mm.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G16-0419.htm

Solution:

PROGRAM SETOFORDERS (input, output); (\textasteriskcentered This program reads a bar length and a \textasteriskcentered) (\textasteriskcentered list of outstanding orders for lengths to be \textasteriskcentered) (\textasteriskcentered cut from such bars, and chooses by \textasteriskcentered) (\textasteriskcentered recursive trial and error the set of \textasteriskcentered) (\textasteriskcentered ordered lengths that should be cut to minimize \textasteriskcentered) (\textasteriskcentered bar wastage. \textasteriskcentered) CONSTMaxsizeoforderlist= 20; TYPE orders =1 ..Maxsizeoforderlist; orderset = set of orders; VAR size of orderlist :orders; Bestset :orderset; Bestlength , L:o ..Maxint; Length:Array [orders] of 1 . .Maxint; I:orders; (\textasteriskcentered sets are used to describe the trial \textasteriskcentered) (\textasteriskcentered solution, best solution, and potential components, \textasteriskcentered) (\textasteriskcentered The length of the orders are held in the \textasteriskcentered) (\textasteriskcentered array length.\textasteriskcentered) (\textasteriskcentered Best set contains the indices in length of \textasteriskcentered) (\textasteriskcentered the orders making up the best solution \textasteriskcentered) (\textasteriskcentered found so far.\textasteriskcentered) (\textasteriskcentered Best length denotes the total length of \textasteriskcentered) (\textasteriskcentered the orders in bestset.\textasteriskcentered) PROCEDURE CONSIDER (Triallength: integer; orderchosen ,ordersremaining:orderset); (\textasteriskcenteredTriallengthdenotes the total length of \textasteriskcentered) (\textasteriskcentered orders currently in the set orderchosen.\textasteriskcentered) (\textasteriskcenteredorderchosenis a set whose members are \textasteriskcentered) (\textasteriskcentered the indices in length of the orders in \textasteriskcentered) (\textasteriskcentered the current trial solution. \textasteriskcentered) (\textasteriskcenteredordersremainingis a set whose members \textasteriskcentered) (\textasteriskcentered are the indices in length of the orders \textasteriskcentered) (\textasteriskcentered available for incorporation into the trail \textasteriskcentered) (\textasteriskcentered solution. \textasteriskcentered) VAR x:0 ..Maxsizeoforderlist; BEGIN IFtriallength< = L Then begin Iftriallength>bestlength thenbegin bestlength : =Triallength; bestset : =orderchosen End; x : = o; while ordersremaining< > [ ] DO Begin RepeatX := X + 1 until x inordersremaining; ordersremaining : =ordersremaining- [x]; consider (Triallength+length[x],orderchosen+ [x],ordersremaining) End End End; (\textasteriskcenteredimprove\textasteriskcentered) Begin Read( L); write1n ('Manufactured bar length is', L : 5,'MM') write1n ; writeln('orderlistis'); Read (sizeoforderlist); For I:=1 tosizeoforderlistDO Begin read (length[I]); write1n(length[I],'MM')END; Bestlength := o;Bestset:= [ ]; (\textasteriskcentered Initially, trial set is empty and all \textasteriskcentered) (\textasteriskcentered orders are available for inclusion in the \textasteriskcentered) (\textasteriskcentered trial solution. \textasteriskcentered) consider (o,[ ] , [1 . .sizeoforderlist]); write1n; write1n ('optimal set of orders'); ForI: = 1 tosizeoforderlistDO If I inbestsetthenwriteln(length [I]:6 ,'MM'); write1n; write1n ('wastage=', L-Bestlength:4, 'MM') End.

Question:

There are two bacterial species growing in a particular nutrient broth. One species can use maltose as its carbon source , while the other cannot and is capable only of utilizing glucose . Each species has a characteristic appearance. How can one separate and identify the two species?

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/Users/wenhuchen/Documents/Crawler/Biology/F05-0128.htm

Solution:

Nutrient broth is a liquid medium which can support the growth of most heterotrophs. Nutrient broth is not a medium of any particular chemical composition but is simply a solution of complex raw materials which has been found to support the growth of many microorganisms. The point is to find a method to isolate each bacterial type. One way is to use a selective medium, which is a medium that either contains or lacks a particular substance which isnecesseryfor the growth of an organism. To achieve separation, one must make one kind of medium which contains only maltose as its carbon source, and another medium which contains only glucose. (Both media would contain other essential nutrients) Maltose is adissacharidewhich is composed of two glucose molecules. The maltose-utilizing bacteria have an enzyme that can break down maltose into glucose. They can therefore grow when presented with either maltose or glucose. But the glucose-utilizing bacteria lack this enzyme, and can use only glucose as a carbon source. The maltose plate thus serves as the selective medium. Agar (used as a solidification agent) is added to the media and poured into separate Petri dishes (shallow, circular plates used for growing microorganisms). By means of a transfer loop (also called an inoculum loop- a long metal instrument with a wire loop at one end), a portion of the bacterial suspension is then placed on the surface of the maltose solid medium and streaked over the surface. The same "inoculation" procedure is performed on the glucose plate. This streaking thins out the bacteria on the agar surface, so that individual bacterial colonies are separated. Each colony develops from a single bacterial cell and is therefore genetically pure (barring mutations). All the colonies on the maltose plate consist of maltose-utilizing bacteria, since only this species can grow. The glucose- utilizing bacteria will not be able to grow on this plate. Both types of bacteria should be present on the glucose plate , since maltose-utilizing bacteria also use glucose (the product of maltose breakdown). To select for glucose-utilizing bacteria, bacteria are transferred from any one colony on the glucose plate onto a maltose plate. If colonies form on the maltose plate, maltose-utilizing bacteria were selected. If no colony arises , it can be inferred that only glucose-requiring bacteria are present. Therefore, the colony on the glucose plate which corresponds to an area of no growth on the maltose plate will be the stock of glucose-requiring bacteria .

Question:

The heart does not extract oxygen and nutrients from the blood within the atria and ventricles. How is the heart sup-plied with its metabolic needs?

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/Users/wenhuchen/Documents/Crawler/Biology/F15-0381.htm

Solution:

The heart depends on its own blood supply for the extraction of necessary oxygen and nutrients. The blood ves-sels supplying the heart are known as the coronary vessels. The coronary artery originates from the aorta, just above the aortic valve, and leads to a branching network of small arteries, arterioles, capillaries,venules, and veins similar to those found in other organs. The rate of flow in the coronary artery depends primarily on the arterial blood pressure and the re-sistance offered by the coronary vessels. The arterioles in the heart can constrict or dilate, depending on the local meta-bolic requirements of the organ. There is little if any neural control. If the coronary vessels are blocked by fatty deposits, the heart muscle would become damaged because of decreased sup-ply of nutrients and oxygen. If the block is very severe andpresistsfor too long, death of heart muscle tissue may result; this condition is called heart attack. Low arterial pressure may also lead to a heart attack for the same reasons.

Question:

Find the period of a communications satellite in a circular orbit 22,300 mi above the earth's surface, given that the radius of the earth is 4000 mi, that the period of the moon is 27.3 days, and that the orbit of the moon is almost circular with a radius of 239,000 mi.

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/Users/wenhuchen/Documents/Crawler/Physics/D04-0145.htm

Solution:

In a circular orbit of radius r an earth satellite of mass m has a velocity v. The distance, d, the satellite moves in one revolution equals 2\pir. Its period T is the time it takes for the satellite to make one revolution. Therefore T = d/v = 2\pir/v(1) The centripetal force necessary to keep the satellite moving in a circle is supplied by the gravi-tational force exerted by the earth. By Newton's Second Law, G(M_Em/r^2) = mv^2/r where G is the gravitational constant and M_E is the mass of the earth. From equation (1), V^2/r^2 = (2\pi/T)^2 = 4\pi^2/T^2 andG(M_Em/r^2) = 4\pi2mr/T^2(2) The same arguments apply to the moon of mass M_M which moves in a circle of radius R with period T_M. G(M_EM_M/R^2) = 4\pi^2M_MR/T^2_M ;(3) Solving for GM_E in equations (2) and (3), we have respectively GM_E = 4\pi^2r^3/T^2 GM_E = 4\pi^2R^3/T^2_M Equating the above two expressions, 4\pi^2r^3/T^2 = 4\pi^2R^3T_M^2 orT^2/T^2_M = r^3R^3 Therefore, the period of the satellite can be found by substitution of the numerical values given. Using r = 22,300 mi + R_E = 22,300 mi + 4000 mi = 26,300 mi the period is T = T_M\surd(r^3/R^3) = 27.3 days\surd[{(2.630 × 10^4)^3 mi^3}/{2.39 × 10^5)^3 mi^3}] = 1.00 day. Such a satellite therefore rotates about the center of the earth with the same period as the earth rotates about its axis. In other words, if it is rotating in the equatorial plane, it is always vertically above the same point on the earth's surface.

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Question:

To measure magnetic fields, rotating coils are often used. A cir-cular coil of radius 1 cm and with 100 turns is rotated at 60 cps in a magnetic field. The emf induced in the coil has a maximum value of 12.3 volts. Calculate the Intensity of the field.

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/Users/wenhuchen/Documents/Crawler/Physics/D22-0756.htm

Solution:

The magnetic flux (\varphi) through one coil is defined as follows: \varphi_m = \int_A B^\ding{217} \bullet dA^\ding{217} = \int_A B cos \texttheta dA = B cos \texttheta \int_\texttheta dA = BA cos \texttheta B is a constant in this problem, and \texttheta is a function of time, so they were taken out of the integral. \texttheta is the angle between the plane of the coil and the direction of B^\ding{217}, and A is the area of the loop, (= \piR^2 ). Since the coil is rotating with a constant angular velocity \omega, \texttheta is related to time by the following relation; \texttheta = \omegat Therefore, the flux through N coils is, N times the flux through one coil, \varphi_m = NBA cos \omegat According to Faraday's law of induction, a change in flux through a coil with N turns will induce an EMF in the coil. The induced EMF is \epsilon = (d\varphi_m /dt) The negative sign signifies the fact that the EMF is induced in such a direction as to oppose the change in flux that created it. The EMF Induced in the coil is then \epsilon = \omega NBA sin \omegat The sine function has values ranging from +1 to -1. \epsilon will then attain its maximum value when sin \omegat = \pm 1 Then \vert\epsilon_max \vert = \vert\omega NBA \vert or, since \omega = 2\pin where n is the number of revolutions per second, \verte_max \vert = 2\pin NBA = 2\pin NB(\piR^2 ) orB = [(\epsilon_max )/{(2\pinN)/(\piR^2 )}] = [(12.3 volts)/(2\pi × 60 sec^-1 × 100 × \pi × (.01m)^2 ] = 1.04 [(volt-sec)/(m^2 )] = 1.04 (Weber/m^2 ) The unit of length of the radius, R, was changed to meters to be con-sistent with the MKS system being used.

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Question:

Describe the homeostatic mechanisms related to breathing.

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/Users/wenhuchen/Documents/Crawler/Biology/F16-0399.htm

Solution:

The control centers for breathing are located in the medulla of the brain. These centers function in maintaining respiratory rhythmicity, but it does not account for the variations in the rhythm. The question arises then, as to how the respiratory centers know what the body oxygen requirements are. The primary stimulus comes from the carbon dioxide which is released from the cells into the blood. This gas, which is generated in cellular respiration, is transported by the circulation in the form of carbonic acid (H_2CO_3), which freely dissociates to H^+ and HCO_3^- ions. The concentration of carbonic acid increases as a result of increased cellular respiration. The increased carbonic acid content of the blood stimulates the inspiratory centers of the medulla in two manners. The carbonic acid either acts directly upon the inspiratory neurons or upon nearby chemosensitive cells, from which there are neural connections to the in-spiratory neurons. The inspiratory neurons increase the rate of stimulation of the inspiratory muscles, which results in a greater respiratory rate. The result of this increased rate of breathing is the increased rate of removal of carbon dioxide from the blood upon expiration. The loss of carbon dioxide favors the reduction of car-bonic acid by the following formula, in accordance with the law of mass action. CO_2 + H_2O \rightleftarrows H_2CO_3 \rightleftarrows H^+ + HCO_3^- As the carbonic acid concentration decreases, there will be a corresponding reduction in the rate of breathing. Chemosensitive cells in the brain, called central chemo- receptors, play an integral role in the regulation of the respiratory rate. These cells are bathed by the cerebro spinal fluid, and they monitor the hydrogen ion concentra-tion of the blood. An increased carbon dioxide content ultimately yields an increased hydrogen-ion concentration through the dissociation of carbonic acid. The brain is extremely sensitive to changes in hydrogen ion (H^+) concentration, and even small fluctuations can cause serious brain malfunction. The chemosensitive cells make synaptic connections with the inspiratory neurons, and stimulate them when the H^+ concentration becomes relatively high. Chemosensitive cells are also present at the bifurcation of the common carotid artery and the aortic arch. These receptors are termed peripheral chemoreceptors. Peripheral receptors respond to changes in the concentration of H^+, CO_2 and O_2. The rate of breathing is controlled more by fluctuations in H^+ and CO_2 levels than by fluctuations in O_2 levels. Hence, the respiratory stimulation due to low O_2 tension may be overridden by other simultaneously occurring changes to which a person is more sensitive, particularly changes in CO_2 and H^+ levels in the blood. It should be noted that only the peripheral chemoreceptors respond to decreased O_2 tension. The concept that respiration is controlled at any instant by multiple factors is of great importance. Consider, for example, what happens when a person hyper-ventilates before holding his breath, and then does strenuous exercise. Hyperventilation causes CO_2 to be lost from the blood faster than it is produced by the cells of the body. The decreased CO_2 level inhibits the firing of chemoreceptors, and the respiratory rate tends to decrease. If the person then holds his breath and exercises, O_2 is lost, but CO_2 will not increase fast enough to immediately stimulate respiration. This is a very dangerous procedure. During exercise oxygen con-sumption is high. Because of our relative insensitivity to oxygen deficits, a rapidly decreasing oxygen level without a proportionately increasing CO_2 level may cause fainting, before ventilation is stimulated by the chemoreceptors. Afferent nerves from the peripheral chemoreceptors enter the medulla and synapse with the medullary centers. A low oxygen content in the blood increases the rate of firing of the receptors. Normally, however, the much greater sensitivity to CO_2 and H^+ fluctuations in the blood prevent the oxygen level from dropping to such a low level. Respiration is stimulated by other factors aside from variations in dissolved gas concentrations. Extremes of temperature, pain, initiation of exercise, and certain drugs all act to stimulate respiration. Respiration also changes with an individual's emotional, state (probably via hormonal changes), and may also change by voluntary control.

Question:

Using power series, find an approximate solution to the definitein-tegral 1\int0 sin xdx. Write a FORTRAN program segment.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G21-0524.htm

Solution:

First, we must remember that sinx = x - (x3/3!)+ (x5/5!)- (x7/7!)\bullet\bullet\bullet We may then write the integral like this: 1\int0 sin xdx\approx1\int0 [x - (x3/3!)+ (x5/5!)- (x7/7!)]dx assumingwe want to compute only 4 terms. The generalized expansion wouldbe 1\int0 [x - (x3/3!) + (x5/5!) - (x7/7!)\bullet\bullet\bullet - {(x^2x+1) / (2n + 1)!} + \bullet\bullet\bullet] dx. When integration is performed, we get the generalized primitives as [(x2/2!) - {x4/4(3!)} + {x6/6(5!)} -{x8/8(7!)} \bullet\bullet\bullet {(x^2n+1) / (2n + 2)((2n + 1)!)}]10 We can make use of theFACT(N) function to calculate the integral. Notice thatwe have not defined the accuracy parameter ERROR. You may choosethat value and insert it at the start of the program. Since the lower boundaryof this integral is zero, you need only compute the terms for x = 1. SINDX = (X\textasteriskcentered\textasteriskcentered2)/2.0 I=1 2\OIF (ABS (TERM) .LT.ERROR) GO TO 5\O N = FLOAT (I) SINDX = SINDX + TERM TERM = -1.0\textasteriskcentered(X\textasteriskcentered\textasteriskcentered(2\textasteriskcenteredN + 2)/(2\textasteriskcenteredN+2)\textasteriskcentered(FACT(2\textasteriskcenteredN+1))) CIF I IS ODD, TERM IS NEGATIVE. IF I IS CEVEN, TERM IS POSITIVE. IF (2\textasteriskcentered(I/2).EQ.I) ABS (TERM) = TERM I=1+1 GO TO 2\O 5\OWRITE (S,\textasteriskcentered) SINDX

Question:

Determine the relative abundance of each isotope in naturally occurring gallium from the following data: At. wt. Ga = 69.72. Masses if isotopes ^69Ga = 68.926, ^72Ga = 70.925.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E04-0120.htm

Solution:

The relative abundance of the various isotopes of gallium can be found by using the following equation: atomic weight of Ga = (% of ^69Ga × at. wt. of ^69Ga) + (% of ^71Ga × at wt. of ^71Ga) It is given that Ga consists only of the two isotopes ^69Ga and ^71Ga. Thus, if one lets x = fraction of ^69Ga, then 1 - x = fraction of ^71Ga. Using the above equation one can solve for x. 69.72= (x × 68.926) + (1 - x) × 70.925 69.72= 68.926x + 70.925- 70.925x -1.205 = - 1.999x x = 1.2/2.0 = 0.60. This means that the ^69Ga makes up 60 % of Ga. ^71Ga makes up 1 - .60 or 40% of Ga.

Question:

The "Meat is Neat" company wishes to enter the frozen shrimp market. They contract a researcher to investigate various methods of growing shrimp in large tanks. The researcher suspects that temperature and salinity are important factors influencing shrimp yield and conducts a factorial experiment with three levels of temperature and salinity. That is, each combination of temperature and salinity are employed and the shrimp yield for each (from identical 80 gallon tanks) is measured. The recorded yields are given in the following chart: Temperature Salinity (inppm) 700 1400 2100 60 3 5 4 70 11 10 12 80 16 21 17 Write a computer program that computes the ANOVA (analysis of variance) tables.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G19-0489.htm

Solution:

This is a two factor factorial design experiment. The logic of the program is best illustrated by showing how the above problem is solved. The interaction model is Y_ij= \mu +\alpha_i+\beta_j+\epsilon_iji= 1, 2, 3 j = 1, 2. 3 The construction of the ANOVA table requires the computation of various sums of squares. If\alpha_iis the effect due to temperature, then the row averages are: Y_1. = (3 + 5 + 4) / 3 = 4 Y_2. = (11 + 10 + 12) / 3 = 11 Y_3. = (16 + 21 + 17) / 3 = 18 The column averages are: Y.1 = (3 + 11 + 16) / 3 = 10 Y.2 = (5 + 10 + 21) / 3 = 12 Y.3 = (4 + 12 + 17) / 3 = 11 Y.. = (4 + 11 + 18) / 3 = 11 = (10 + 12 + 11) / 3 (The dots represent the index over which summation is taking place). The sum of squares about the origin is: ^3\sum_i=1 ^3\sum_j=1 y^2_ij = 3^2 + 5^2 + 4^2 + 11^2 + ... + 21^2 + 17^2 = 9 + 25 + 16 + 121 + ... + 441 + 289 = 1401. The sum of squares due to variation in temperature is: SSA= ^3\sum_i=1 ^3\sum_j=1 (Y_i. -Y..)^2 = 3\sum(Y_i. -Y..)^2 = 3[(4 - 11)^2 + (11 - 11)^2 + (18 - 11)^2] = 3[49 + 49] = 294. The sum of squares due to variation in salinity is: SSB= ^3\sum_i=1 ^3\sum_j=1 (Y._j-Y..)^2 = 3 ^3\sum_j=1 (Y._j-Y..)^2 = 3[(10 - 11)^2 + (12 - 11)^2 + (11 - 11)^2] = 3[1 + 1 + 0] = 6 The total sum of squares is: SSTO = ^3\sum_i=1 ^3\sum_j=1 (Y_ij-Y..)^2 = ^3\sum_i=1 ^3\sum_j=1Y^2_ij - 9.Y^2.. = 1401 - 9.11^2 = 1401 - 1089 = 312. Thus, the sum of squares due to the error is SSE= SSTO - SSA - SSB = 312 - 6 - 294 = 12 . The ANOVA table (Analysis of Variance) Source of Variation Sum of Squares Degrees of Freedom Mean Squares Temperature 294 3 - 1 = 2 147 Salinity 6 3 - 1 = 2 3 Error 12 (9 -1) - 4 = 4 3 The program looks as follows: CPROGRAM FOR FACTORIAL DESIGN: TWO FACTORS CSUBPROGRAMS FOLLOW AFTER MAIN PROGRAM DIMENSION A(2\O), B(2\O), C(2\O,2\O), LA(2\O), LB(2\O) DIMENSION SN(2\O,2\O), SA(2\O), AM(2\O), SB(2\O), BM(2\O) DIMENSION CLM(2\O,2\O), KH(5), K1(4), K2(4), K3(4), 1N(2\O,2\O), KF(2\O), D(1\O,2\O,5\O) ND = 2\O 99READ (2,1\O\O)KH, K1, K2, K3, K,L, ((N(I,J), J = 1,K), I = 1,L) WRITE (3,2\O\O)KH,K1,K2, K3, K,L, ((N(I,J), J = 1,K), I = 1,L) 1\O\OFORMAT (5A4, 3(4A4), 213/4\OI2) 2\O\OFORMAT (\textasteriskcentered1\textasteriskcentered, 5A4, 3(4A4), 2I3/4\OI2) READ (2,3\O\O) KF WRITE (3,3\O\O) KF 3\O\OFORMAT (2\OA4) T = 0 XX = 0 NC = 0 DO 1\O J = 1,L A(J) = 0 1\OLA(J) = 0 DO 2\O M = 1 ,K B(M) =0 LB(M) = 0 DO 3\O J = 1,L DO 3\O M = 1,K 3\OC(J,M) = 0 DO 4\O J = 1,L DO 4\O M = 1,K NUM = N(J,M) DO 4\O I = 1,NUM T = T + D(J,M,I) XX = XX + D(J,M, I) \textasteriskcentered\textasteriskcentered 2 B(M) = B(M) + D(J,M,I) A(J) = A(J) + D(J,M,I) 4\OC(J,M) = C(J,M) + D(J,M,I) DO 5\O I = 1,L DO 5\O J = 1,K LA (I) = LA (I) + N(I, J) LB(J) = LB(J) + H(I,J) NC = NC + N(I, J) 5\OSN(I,J) = N(I, J) AA = 0 BB = 0 CLL = 0 DO 6\O I = 1,L SA(I) = LA (I) AM (I) = A(I)/SA(I) AA = AA + A (I) \textasteriskcentered\textasteriskcentered 2/SA(I) DO 6\O J = 1,K CLM(I,J) = C(I,J)/SN(I,J) 6\OCLL = CLL + C(I, J) \textasteriskcentered\textasteriskcentered 2/SN(I,J) DO 7\O J = 1,K SB(J) = LB(J) BM(J) = B(J)/SB(J) 7\OBB = BB + B(J) \textasteriskcentered\textasteriskcentered 2/SB(J) CCOMPUTE SS, DF, MS, F CCC = T \textasteriskcentered\textasteriskcentered 2/NC BETA = AA - CCC BETB = BB - CCC BETC = CLL - CCC AB = BETC - BETA - BETB TOT = XX - CCC ERR = TOT - BETC DT = NC - 1 DA = L - 1 DB = K - 1 DC = L \textasteriskcentered K - 1 DAB = DA \textasteriskcentered CB DER = DT - DC AMS = BETA/DA BMS = BETB/DB ABMS = AB/DAB ERMS = ERR/DER FA = AMS/ERUS FB = BMS/ERMS FAB = ABMS/ERMS PA = PRBF(CA,DER,FA) PB = PRBF(CAB,DER,FB) WRITE (3,4\O\O) KH, TOT, DT, K1, BETA, DA, AMS, FA, 1PA,K2,BETB, BMS, FB, PB,K3, AB, DAB, ABMS, FAB, PAB, 2ERR, DER, ERMS 4\O\OFORMAT CALL PRTS (AM, L, 1, 7H AMEANS, ND) CALL PRTS (BM, K, 1, 7H BMEANS, ND) CALL PRTS (CLM, L, K, 8H ABMEANS, ND) END SUBROUTINE PRTS (X,N,M,KH,ND) CPRINT A MATRIX OR VECTOR IN 1\O-COLUMN PARTITIONS DIMENSION X(2\O,2\O) KH(2) IF (M.GT.1) GO TO 2\O DO 1\O I = 1,N,1\O J = MINO (I + 9,N) WRITE (3,5) KH, (K,K = I,J) 5FORMAT (2A4, 1\OI11) 1\OWRITE (3,15) (X(K,1), K = I,J) 15FORMAT (119X, 1\OF11.4) RETURN 2\ODO 25 I = 1,N 25WRITE (3,3\O) I, (x(I,J), J = K,L) 3\OFORMAT (11 I6, 4X, 1\OF 11.4) RETURN END

Question:

Write a BASIC program which generates and prints out an Inputted number of random digits (0 through 9 inclusive).

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G10-0243.htm

Solution:

We use the RND function. [Some texts use RND with an argument; others use no argument. Consult your BASIC manual for details of your particular implementation.] Since 0 < RND <1, 0 < 10\textasteriskcenteredRND < 10, and thus INT(10\textasteriskcenteredRND) takes values 0, 1, 2, ..., 9 1\O\O REM RANDOM DIGITS 11\O PRINT "HOW M&NY RANDOM DIGITS DO YOU WANT"; 12\O INPUT N 13\O PRINT 14\O FOR K = 1 TO N 15\O PRINT INT(1\O\textasteriskcenteredRND) 16\O NEXT K 17\O PRINT 18\O GO TO 11\O 999 END

Question:

Calculate the number of H^+ ions in the inner compartment of a single rat liver mitochondrion, assuming it is a cylinder 1.8\mu long and 1.0\mu in diameter at pH 6.0.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E22-0790.htm

Solution:

One can calculate the number of moles of H^+ ions per liter from the pH because pH = - log [H^+]. After determining the volume of the mitochondrion, one can solve for the number of H^+ ions present. pH = - log [H^+] 6.0 = - log [H^+] [H^+] = 10^-6 M. The volume of a cylinder is equal to its length times the area of its cross-section. l × \pir^2, where l is the length of the cylinder and r is the radius of the cross-section. 1\mu = 10^-4 cm. Volume = 1.8 × 10^-4 cm × \pi × (0.5 × 10^-4 cm)^2 = 1.4 × 10^-12 cm^3 = 1.4 × 10^-15 l Solving for the number of moles of H^+ ion, No. of moles = 10^-6 moles / liter × 1.4 × 10^-15 l = 1.4 × 10^-21 moles There are 6.02 × 10^23 ions per mole. No. of H^+ ions = 1.4 × 10^-21 moles × 6.02 × 10^23 (ions / mole) = 843 H^+ ions.

Question:

How can you determine whether or not a given population is in genetic equilibrium?

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/Users/wenhuchen/Documents/Crawler/Biology/F26-0699.htm

Solution:

In an equilibrium population, both gene frequencies and genotype frequencies remain constant from generation to generation. The best way to determine if a given population is in equilibrium is to compare these frequencies of two generations to see if they are indeed constant. One way to do this is to determine the gene frequencies for a given adult population representative (or a group of that population) and for their offspring. Two generations are thus represented in the sample. For more reliable results, however, more generations should be compared with respect to their gene frequencies. Suppose one wanted to know if the population of Australian aborigines were in genetic equilibrium. Data from that population concerning the distribution of MN blood groups are gathered, and are shown below: Phenotype Generation (G) M MN N Total 1 241 604 195 1040 2 183 460 154 797 MN blood groups are determined by two alleles, L^M and L^N , having intermediate or codominant inheritance. Thus M type is coded for by L^ML^M, MN type by L^ML^N, and N type by L^NL^N. We can calculate the genotype frequencies using the data given: Genotype Frequency G_1 G_2 L^ML^M 241 / 1040 = .231 183 / 797 = .230 L^ML^N 604 / 1040 = .581 460 / 797 = .577 L^NL^N 195 / 1040= .188 154 / 797 = .193 1.0 1.0 The genotype frequencies are thus found to be essentially the same for both generations. We could also have determined the allelic frequencies from the data, the number of L^M alleles in the population is twice the number of M individuals (L^ML^M) plus the number of MN individuals (L^ML^N). The number of L^N alleles is twice the number of N individuals plus the number of MN individuals. Since each individual carries two alleles, the total number of alleles in the population is twice the number of individuals, or the sum of the L^M and L^N alleles. For our population, then, we have the following: L^M = 2(M) + MN L^M = 2(M) + MN G_1 2(241) + 604 = 1086 2(195) + 604 = 994 G_2 2(183) + 460 = 826 2(154) + 604 = 768 Total Alleles(L^M+L^N) Frequency L^M Frequency L^N G_1 1086+994=2(1040)=2080 1086/2080=.522 994/2080 = .478 G_2 826+768=2( 797)=1594 826/1594=.518 768/1594 = .482 As with the genotype frequencies, the allelic frequencies remain nearly the same in both generations. Thus we can say that, with respect to the MN blood group alleles, the population is in equilibrium. Note that we can only determine equilibrium in a population in reference to a given gene or a small number of genes. Though our population may be in equilibrium for the MN alleles, other allelic systems may not be. For example, migration may be occurring from nearby populations having similar MN allelic frequencies but very different allelic frequencies for color blindness, thus affecting the equilibrium of these genes in the population.

Question:

In reference to the previous problem, classify the bugs as a)severeerrors b) errors c) warnings. What actionwould PL/I take with respect to these errors?

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G13-0360.htm

Solution:

PL/I allots several errors (through default op-tions) that other languages(FORTRAN for example) would find intolerable. Thus, in spite ofall the errors in the program above, PL/I will compile it by changing the programin such a way that it can be compiled. If a program has bugs (in the etymological, not entomo-logical sense), its output will differ from what the user desires. The changes in the programdepend on the magnitude of the error. PL/Imakesthe following distinctions: a) A statement containing a severe error cannot be executed or compliedwithout a major change. PL/I makes the change to produce the program. But execution is doubtful. b) Errors cause improper execution. After an error has been modifiedby PL/I, the resulting output is highly suspect. c) Warnings notify the programmer of certain standard actions takenby PL/I in the absence of explicit specifica-tions in the program. These are not really errors and the output of the program is not affected. Hence, no action is required. Statement 1: The missing label is filled in byPL/I. This is a severe error. Since OPTIONS ismisspelt, the com-piler deletes everything beyond PROCEDURE and then inserts a semi-colon. Thus Statement 1 reads 1DUMMY PROCEDURE; This means the procedure can be used only as part of another program andnot as a PROCEDURE MAIN. Statement 3: PUTPAGE is identified as part of an as-signment statement. PL/Iaddsan equal sign and an error signal. Statement 7: GOTO is recognized as GO TO and no cor-rection is noted. Statement 9: SQRT (40 (Y-2) -W\textasteriskcenteredX) / (X-Y) is shortened to SQRT(40. PL/Iexpectssome operator to come between 40 and Y. Since nothingis there, everything after 40 is ignored. PL/I supplies ")" and "to obtain 9SQRT(40); This is a severe error. Statement 10: The \vert is interpreted as a + and TO is interpreted as a name. The W is truncated anda ;is in-serted . Statement 11: The minus sign is correctly identified although misplaced. Statement 12: PL/I spots the missing keyword THEN and makes theappropriate insertion. Statement 16: Since the branch destination is unknown and cannot befound, the entire statement is removed from the program. This is a severeerror. Statement 17: The nonexistent label on the END state-ment is noted. The END statement cannot be linked with any PROCEDURE or DO statement. This is a severe error.

Question:

Given the necessary precautions, water can be cooled to a temperature T = -10 \textdegreeC. What mass of ice m is formed from M = 1 kg. of such water when a particle of ice is thrown into it and freezing thereby produced? Regard the specific heat of thesupercooledwater as independent of the temper-ature and equal to the specific heat of ordinary water.

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Solution:

The ice particle causes some of the water to freeze. The physics behind this process is somewhat in-volved. However, we can solve the problem if we consider what happens in the final state. The partial freezing caused by the ice particle re-leases heat into the system consisting of ice and water. This heat is the latent heat of freezing released by all freezing water and must be absorbed by the rest of the system. Therefore, the freezing can continue only as long as the water is capable of absorbing the heat released in the process. Equilibrium is attained when the water reaches 0\textdegreeC at which point water must first freeze if it is to attain a higher temperature. Since water gives off heat when it freezes and there is nothing to absorb this heat, freezing stops at this point. The latent heat of water is 80 cal/gr. Therefore the heat released in forming m grams of ice is ∆Q = (mgr) × (80 cal/gr). The same heat is used to raise the temperature of the 1 kg. of supercooledwater by 10\textdegreeC; ∆Q =cM∆T = (1 cal/grC\textdegree) × [10^3gr] × (10\textdegreeC) where c is the specific heat of water. Therefore we get (mgr) × (80 cal/gr) = 10^4 cal and the amount of ice produced is m = (10^4 / 80)gr= 125gr= 0.125 gr.

Question:

A rectangular box is 5.00 in. wide, 8.00 in. long, and 6.0 in. A rectangular box is 5.00 in. wide, 8.00 in. long, and 6.0 in. deep. Calculate the volume in both cubic centimeters and in liters.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E01-0010.htm

Solution:

The volume of a solid is found by multiplying the height times the lengthtimes the width. volume= (6.0 in) × (8.0 in) × (5.0 in) = 240 in^3. From a standard conversion table, one finds that 1 inch = 2.54 cm. One findsthe volume of cubic inches in cubic centimeters by cubing both sides ofthis equality. 1 inch = 2.54 cm (1 inch)^3= (2.54 cm)^3 1 inch^3 = 16.4 cc. Thus, one can convert the volume of the rectangle from cubic inchesto cubic centimeters by multiplying the number of cubic inches by theconversion factor, 16.4cc/1 inch^3. volumeof rectangle = 240 in^3 × 16.4 cc/1 in^3 = 3936 cc. Thereare1000 cc in 1 liter. Therefore, to convert from cubic centimeters toliters, multiply the number of cubic centimeters by 1 liter/1000 cc. volumein liters = 3936 cc × l liter/1000 cc = 3.936 liters.

Question:

A rocket has a length L = 600 m measured at rest on the earth. It moves directly away from the earth at constant velocity, and a pulse of radar is sent from earth and reflected back by devices at the nose and tail of the rocket. The reflected signal from the tail is detected on earth 200 sec after emission and the signal from the nose 17.4 × 10^-6 sec later. Calculate the distance of the rocket from the earth and the velocity of the rocket with respect to the earth.

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/Users/wenhuchen/Documents/Crawler/Physics/D32-0951.htm

Solution:

First we will make a non-relativisticcalculation. The speed of the pulse is 3 × 10^8 m/sec (equal to c, the speed of light). If the rocket is a distance R from the earth then the pulse travels 2R and (3 × 10^8 m/sec )(200 sec) = 2R , R = 3 × 10^10 m To calculate the speed v of the rocket, we note that the pulse from the front end arrived 17.4 × 10^-6 sec after that from the back. Thus this pulse was sent Into space a distance (1/2)(17.4 × 10^-6 sec)(3 × 10^8 m/sec) = 2.61 × 103m farther. This distance is equal to L + vt where t = (1/2)(17.4 × 10^6 sec) = 8.7 × 10^-6 sec Thus:L + vt = 2.61 x 10^3 m v = (2.61 × 10^3 m-L) / t = (2.61 × 10^3 m - 0.6 X 10^3 m) / (8.7 × 10^-6 sec) = 2.13 × 10^8 m/sec The factor of 1/2 arises because we are interested only in the time of the rocket to ground and not In the total time of flight from ground to ground. According to the above calculation (which we might suspect from the start to be incorrect), the ratio v/c = 0.77. thus it is very probable that relativistic effects will be Important. To the observer on the ground the Lorentz length contraction makes the rocket's length appear to be L\surd(1 - \beta^2) where \beta = v/c. Then the pulse travels a distanceL\surd(1 - \beta^2) + vt farther, which is equal to ct where t = (1/2) × 17.4 × 10^-6 sec. ThusL\surd(1 - \beta^2) + vt = ct which can be solved to obtain \beta ={(ct / L)^2 - 1} / (ct / L)^2 + 1= 0.9 The distance R will be the same as in the previous calculation.

Question:

Ignoring the motion of the earth around the sun and the motion of the sun through space, calculate (a) the angular velocity, (b) the velocity, and (c) the acceleration of a body resting on the ground at the equator.

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/Users/wenhuchen/Documents/Crawler/Physics/D04-0156.htm

Solution:

Because of the rotation of the earth the body at the equator moves in a circle whose radius is equal to the radius of the earth (see figure). r = radius of earth = 6.37 × 10^6 meters We are going to use the MKS System of units. One revolution, which is 2\pi radians, takes 1 day or [(24 hr)/(1 day)] × [(60 min)/(1 hr)] × [(60 sec)/(1 min)] [(24 hr)/(1 day)] × [(60 min)/(1 hr)] × [(60 sec)/(1 min)] = 24 × 60 ×60 secs/day = 24 × 60 ×60 secs/day (a) Since the frequency of revolution is f = 1/T, where T is the period (the time for one revolution), then 2\pif = \omega = 2\pi/T . This equals the number of radians traveled per unit time, or the ang-ular velocity \omega. \omega = 2\pi/(24 × 60 × 60) = 7.27 × 10^-5 radians per second (b) The linear velocity is, by definition v= \omegar = (7.27 × 10^-5) × (6.37 × 10^6)(red/s)\bulletm = 4.64 × 10^2 in/sec . Since 1 mph = 0.447 m/sec v= [4.64 × 10^2(m/sec)] [{1/(.447)}(sec/m)] = 1040 mph (c) The acceleration toward the center of the earth is, since the motion is circular, a= v^2/r = (4.64 × 10^2 m/sec)^2/(6.38 × 10^6 m) = 3.37 × 10^-2 m/sec^2

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Question:

A man hangs from the midpoint of a rope 1 m long, the ends of which are tied to two light rings which are free to move on a horizontal rod (see the figure). What is the maximum possible separation d of the rings when the man is hanging in equilibrium, if the relevant co-efficient of static friction is 0.35?

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/Users/wenhuchen/Documents/Crawler/Physics/D02-0048.htm

Solution:

Since the man hangs from the midpoint of the rope, by symmetry the tensions in the two portions of the rope must be equal and have magnitude T, and each portion will be inclined at the same angle \texttheta to the vertical. Thus the system of forces acting on each ring will be the same. Now consider one of the rings. Three forces are acting on it: the tensional pull on the ring due to the rope, the normal force exerted upward by the rod, and the frictional force attempting to prevent motion of the ring toward its fellow. Since the ring is light, its weight may be ignored. If the ring is too far out, slipping will occur. At the maximum distance apart, each ring is just on the point of slipping. Hence F = \mu_sN. When we resolve T\ding{217} into its horizontal and vertical components, the equations for equilibrium become \sum perpendicular force= N - T cos \texttheta = 0 \sum parallel forces= T sin 0 - F = 0 where we take the positive perpendicular direction as pointing upward and the positive parallel direction as pointing to the right. Then: N= T cos \textthetaandF = \mu_sN = T sin \texttheta. \mu_s= (\mu_sN)/(N) = (T sin\texttheta)/(T cos \texttheta) = tan \texttheta = 0.35 or\texttheta = 19.6\textdegree Finally, we solve for d: sin \texttheta= sin 19.6\textdegree =(d/2)/([(1/2)m] = d m^-1 0.33= d m^-1 d = 0.33 m which is the maximum separation permissible. Note that \texttheta and d do not depend on T and therefore the ring separation is not dependent on what it is that is hanging from the midpoint of the rope.

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Question:

For the following PL/I program, a) Explain how a condition known as 'SIZE CONDITION' arises , b) Show what the computer will print out if the program is run , c) Write out a corrected program to avoid errors due to this condition , using ON SIZE condition, and, d) Show what will be printed out by the corrected program'. EXAMPLE:PROCOPTIONS(MAIN); DCLA(5) FIXED(3), (X,Y,Z) FIXED(2); GETLIST(Z); DO I = 1 TO Z; GET LIST(X,Y); A (I) = X\textasteriskcenteredY: PUTLIST(I,A(I)); END; END EXAMPLE; The Data Card contains the following DATA: 5, 10, 10, 5, 3, 20, 49, 90, 80, 7, 8

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G13-0342.htm

Solution:

a). Notice from the program that each element of the array A has beendeclared as FIXED (3). This means that each element can be a numeralwith 3 digits to the left and digits to the right of the decimal point. The GETLIST(Z) statement reads a value 5 from the DATA CARD. This value 5 of Z becomes the upper limit of the loop head statement: 'DO I=1TO Z'. In the body of the loop, X and Y are read from the DATA CARD, and then A (I)iscalculat-ed. Thus, in the first loop, X and Y are obtainedas 10 and 10. And hence, asA(I) = X\textasteriskcenteredY, A(l) = 10\textasteriskcentered10 = 100. As Z = 5, the loop is performed 5 times. In the fourth loop, the values of X andY obtained are 90 and 80, respectively. Hence, the value of X\textasteriskcenteredY is calculatedas: X\textasteriskcenteredY = 90\textasteriskcentered80 = 7200. However, the arrayA(5) is declared as FIXED(3). Hence, each array elementcan consist of 3 digits only. And so, the computer throws away thefourth digit, viz., 7, and stores only 3 digits in the array element A(4). Thus A (4) = 200. This condition, where extra digits are thrown away is knownas a SIZE CONDITION. But this throwing away of extra digits can leadto serious errors by the user of the program if he is not aware that the computerhas dropped some digits, and that the 200 in A(4) should really havebeen 7200, for example. Thus, the computer is programmed to print out a message indicatingthat a SIZE CONDITION occurred in a particular statement numberwhenever it does. b) The print out of the computer for the given program will be as follows: 1100 215 3980 4200 556 c) The program can be corrected by adding '(SIZE) :' as a prefix to the first statementof the program, and then in-serting a statement 'ON SIZE ............'insidethe body of the program. A good place for inserting this statementis just after the DCL statement so that the computer reads it beforeit starts with the calculations, etc. Thus, the corrected program is given as: (SIZE): EXAMPLE:PROCOPTIONS(MAIN); DCLA(5) FIXED(3),(X,Y,Z) FIXED(2); ON SIZE X\textasteriskcenteredY = 999; GET LIST (Z); DO I = 1 TO Z; GET LIST(X,Y); A(I) = X\textasteriskcenteredY; PUTLIST(I,A(I)); END; END EXAMPLE; Note that in the above program, the computer has been in-structed that if a SIZE CONDITION occurs it must put X\textasteriskcenteredY = 999. Hence, in any loop, wheneverthe values of X and Y are such that X\textasteriskcenteredY becomes a number withmore than 3 digits, the com-puter makes X\textasteriskcenteredY = 999. This means the old, larger value of X\textasteriskcenteredY, is replaced by 999. And thus, the value of the cor-respondingelement of A(I) becomes 999. It is assumed that the programmer knows that whenever there is a printout of 999, it means he should not consider that value as it is the resultof a SIZE CONDITION. d) The print out of the corrected program will be as follows: 1100 215 3980 4999 556 A computer message stating that a SIZE CONDITION had occurred in a particularstatement will also be printed out. Note in the above printout that the value of A (I) cor-responding to 1 = 4 has been affected by the SIZE CONDITION.

Question:

The intensity of the deep purple color ofan acid KMnO_4, solution can be used as an indicator of theextent to which oxidation has taken place, Often there is aconsiderable time lapse before any visual evidence of reactionis noted; but once begun, the process proceeds vigorously. If, on the other hand, a small amount of the essentiallycolorless MnCl_2 salt is dissolved in this solution, theprocess is immediately vigorous, no latency period being noted.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E13-0459.htm

Solution:

The presence of the Mn^2+ in the solution is the key to the reaction. When MnCl_2 is dissolved in the solution, the reaction is vigorous since the saltimmediately dis-sociates to produce Mn^2+ ions. KMnO_4, also immediately dissociates into K^+ and MnO_4^-.with KMnO_4, however, a latency period is noticed since it takes time for Mn^2+ to beproducedfrom the reduction of the MnO_4^-+ ion. The fact that once Mn^2+ does exist, whether from MnO_4^- or MnCl_2, thereaction proceeds vigorously suggests that Mn^2+ is a catalyst. Thus, Mn^2+ increases the rate of the reaction without being consumed. Because Mn^2+ is also a reaction product, the reaction is autocatalytic.

Question:

How are nutrients and waste materials transported from the circulatory system to the body's tissues?

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/Users/wenhuchen/Documents/Crawler/Biology/F15-0373.htm

Solution:

It is in the capillaries that the most important function of circulation occurs; that is, the exchange of nutrients and waste materials between the blood and tissues. There are billions of capillaries, providing a total surface area of over 100 square meters for the exchange of material. Most functional cells in the body are never more than about 25 microns away from a capillary. Capillaries are well-suited for the exchange process. They are both numerous and branch extensively throughout the body. Unlike the arteries and veins, the capillaries have a very thin wall; it is one endothelial cell thick. This thin wall permits rapid diffusion of substances through the capillary membrane. The extensive branching increases the total \textbackslash cross-sectional area of the capillary system. This serves to slow down the flow of blood in the capillaries, allowing more time for the exchange process to occur. The very small diameter of each of the capillaries (which is not much larger than the diameter of the blood cells which pass through them) provides friction, which increases the resistance to flow. This causes a significant drop in blood pressure in the capillaries, which is important in the filtration of fluid between the capillaries and the interstitial fluid (lying in the space between cells). The most important means by which nutrients and wastes are trans-ferred between the plasma and interstitial fluid is by diffusion. Material must first diffuse into the interstitial fluid before it can enter the cells of the body's tissues. As the blood flows slowly through the capillary, large amounts of water molecules and dis-solved substances diffuse across the capillary wall. The net dif-fusion of substances is proportional to the concentration differ-ence between the two sides of the capillary membrane. For example, the concentration of oxygen in the blood is usually greater than that in the interstitial fluids. Therefore, oxygen diffuses into the in-terstitial fluid and then into the tissues. In the same way, if the concentration of nitrogenous wastes is higher in the tissues than in the blood, they will diffuse into the interstitial fluids and then into the blood. Actual diffusion across the capillary wall occurs in three ways. Water-soluble substances (sodium ions, glucose) diffuse between the plasma and interstitial fluid only through the pores in the capillary membrane. The diameter of the pores (8 - 10 \AA) allows small molecules such as water, urea, and glucose to pass through, but not larger molecules such as plasma proteins. Lipid-soluble substances diffuse directly through the cell membranes and do not go through the pores (which are filled with water). Oxygen and carbon dioxide permeate all areas of the capillary membranes. Another method by which sub-stances can be transported through the membrane ispinocytosis. Inpinocytosis, the cell membraneinvaginatesand sequesters the sur-rounding substances. Small vesicles are produced which migrate from one side of the endothelial cell to the other, where the con-tents are released. Pinocytosisaccounts for the transport of larger substances, such as plasma proteins andglycoproteins.

Question:

Why is a moss plant restricted to a height of less thanabout 15 cm?

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/Users/wenhuchen/Documents/Crawler/Biology/F07-0177.htm

Solution:

A moss plant is rarely more than 15 cm tall due to the inefficiency of the rhizoids in absorbing water and minerals and the absence of true vascular and supporting tissues. The rhizoids of the moss are small, slender filaments of cells. Because of their small size and simplicity, they an provide anchorage and absorb enough water and salts only for a small plant . Moreover, the moss plant, like other bryophytes, lacks a vascular system . With no xylem to conduct water and minerals, and no phloem to transport photosynthetic products, the moss plant is necessarily restricted to short-distance internal transport. An unusually large moss plant would soon face water and food shortage, and would eventually starve to death. In addition, since xylem and phloem also function in supporting the plant body , the moss plant lacks sufficient support to enable it to attain a large size . The reproductive process of the mosses also restricts their height. Fertilization must occur in a moist environment, since water must be present to provide a medium through which the sperm can swim. A low growing plant body, being closer to the moist soil, is ad-vantageous for the moss . A plant that is too tall would risk loss of its sperm by dispersing them into a non-fluid environment.

Question:

A motor boat can move with a maximum speed of 10 m/sec, relative to the water. A river 400 m wide flowing at 5 m/sec must be crossed in the shortest possible time to reach a point on the other bank directly opposite the starting point. In which direction must the boat be pointed and how long will it take to cross?

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/Users/wenhuchen/Documents/Crawler/Physics/D03-0085.htm

Solution:

If the boat were pointed directly at the opposite bank, then during the crossing it would drift downstream and it would not reach the other bank at a point directly opposite the starting point. It must therefore be pointed in a direction tilted in the up-stream direction as shown in the figure. As illustrated in the vector diagram PQR, the result of adding the velocity of the boat relative to the water to the velocity of the water must be a resultant velocity v\ding{217} pointing directly toward the opposite bank. We can-not draw this triangle of vectors immediately be-cause we do not know the angle \texttheta between the direction of motion and the direction straight across the stream. However, inspecting the triangle PQR and remembering that in trigonometry the since of the angle \texttheta is defined as sin \texttheta = (QR/PQ) = 5/10 = 0.5 we refer to table of since and find that the angle whose sine is 0.5 is 30\textdegree. The boat must therefore be pointed upstream at an angle of 30\textdegree from the direction perpendicular to the bank. Applying Pythagoras' theorem to the triangle PQR PQ^2 = QR^2 + PR^2 orPR^2 = PQ^2 - QR^2 that is;v^2 = 10^2 - 5^2 = 75 v = \surd75 = 8.66 m/sec The boat therefore crosses the river at a speed of 8.66 m/sec. Since the distance across the river is 400 m, the time taken is, since v = constant, t = d/v = (400 m)/(8.66 m/sec) t = 46.2 sec.

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Question:

The hard shell of crustaceans (lobsters, etc,) and insects (roaches, etc.) is a polysaccharide called chitin. On enzymatic hydrolysis of chitin, N-acetylglucosamine is obtained. This molecule resembles glucose except that at C\rule{1em}{1pt}2 a is attached instead of -OH. (a) Write an open chain formula for N-acetylglucos amine. (b) The structure of chitin is analogous to that of cellulose. Draw a formula containing two joined N-acetylglucosamine units, (c) If the molecular weight of chitin is 150,000, how many units are in the polymer?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E22-0804.htm

Solution:

A polysaccharide is a chain made up of many simple sugars. Chitin is a long unbranched chain of many molecules of the sugar, N-acetylglucosamine. (a)One is given that the only difference between glucose and N-acetylglucosamine is that agroup replaces the \rule{1em}{1pt}OH on C\rule{1em}{1pt}2. The open-chain formula of N- acetylglucosamine can therefore be written as shown in Figure A (b) From the open chain formula, one can see that the ring structure can be written as shown in Figure B for N-acetylglucosamine. For the ring structure to form, the O in the \rule{1em}{1pt} OH on C\rule{1em}{1pt}5 joins C\rule{1em}{1pt}l. From the figure of cellulose, Figure C, one can see that the structure of two N-acetylglucosamine joined to-gether can be written as shown in Figure D. Because the O in the -OH on C-l joins to a C-4 atom on the adjacent molecule and the ring structures are called glycosides, this is termed a 1,4- glycosidic linkage. (c) To determine the number of units in one strand of chitin, one can divide the total weight of the molecule by the weight of one glycoside. From the ring structure of a glycoside joined to another glycoside, one sees that there are 8c's, 5 0's, 13 H's and 1 N. The weight of one mole of this glycoside is (8 × 12) + (5 × 16) + (13 × 1) + (14) = 203 (g / mole). No. of units= (150,000 g) / (203 g/mole) (unit) = 739 units.

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Question:

A gallon of gasoline will deliver about 110,000 Btu when burned. To how many foot-pounds is this equivalent?

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/Users/wenhuchen/Documents/Crawler/Physics/D13-0460.htm

Solution:

1 Btu = 778 foot-pounds 110,000 Btu = 778[(ft-Ib)/Btu] × 110,000 Btu = 85,580,000 foot-pounds.

Question:

An electron, charged - 1.6 × 10^-19 coul, moves at some instant of time in the +x-direction with velocity v = 0.8 c. A magnetic field B = 10 W/m^2 is present in the +y-direction. What is the direction and magnitude of the magnetic force?

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/Users/wenhuchen/Documents/Crawler/Physics/D21-0707.htm

Solution:

To find the magnetic force acting on a charged particle moving through a magnetic field, one uses the formula F_m^\ding{217} = qv^\ding{217} × B^\ding{217} where F_m^\ding{217} is the resulting magnetic force, v^\ding{217} is the velocity of the particle and B^\ding{217} is the magnetic field vector. In this problem v^\ding{217} is 0.8 c = 0.8 × 3 × 10^8 m/sec in the +x-direction, B^\ding{217} is 10 W/m^2 in the +y-direction, and q is e ־ = - 1.6 × 10^-19 coul. So: F_m^\ding{217} = (- 1.6 × 10^-19 coul.) (0.8) (3 × 10^8 m/sec) \^{\i} × (10 W/m^2 ) \^{\j} = - 3.84 × 10^-10[(coul. - m - W)/(m^2 - sec)] (\^{\i} × \^{\j}) = - 3.84 × 10^-10 N K^^ where we used the fact that \^{\i} × \^{\j} = K^^,\^{\i},\^{\j}, and K^^ being the unit vectors in the +x, +y, and +z-directions, re-spectively.

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Question:

DIMENSION A(10,10,10) Assuming that elements of the array are stored in lexicographic order of their indices, find displacements of the locations of: (a) A(5,6,7)(b) A(10,9,8)(c) A(1,5,9) relative to the location of A(1,1,1).

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G12-0286.htm

Solution:

In the previous problem, it was shown that if A has dimen-sions n_1 × n_2 × n_3, the location of any element A(I,J,K) is given by: loc(I,J,K) = loc(1,1,1) + (I\rule{1em}{1pt}1)n_2n_3 + (J\rule{1em}{1pt}1)n_3 + (K\rule{1em}{1pt}1), assuming that elements are stored in lexicographic order of their indices. The displacement of the location of the element A(I,J,K) relative to the location of A(1,1,1), therefore, is given by: disp(I,J,K) = loc(I,J,K) \rule{1em}{1pt} loc(1,1,1) = (I\rule{1em}{1pt}1)n_2n_3 + (J\rule{1em}{1pt}1)n_3 + (K\rule{1em}{1pt}1)(1) We are given that n_1 = n_2 = n_3 = 10. (a) Here I = 5, J = 6, K = 7. Substitution into equation (1) yields: disp(5,6,7) = (5-1) (10\bullet10) + (6\rule{1em}{1pt}1)10 + (7\rule{1em}{1pt}1) = (4)100 + (5)10 + 6 = 456 (b) Here I = 10, J = 9, K = 8. Substituting into equation (1) gives us: disp(10,9,8) = (10\rule{1em}{1pt}1) (10 10) + (9\rule{1em}{1pt}1) 10 + (8\rule{1em}{1pt}1) = (9)100 + (8)10 + 7 = 987 (c) Here I = 1, J = 5, K = 9. Substituting into equation (1), we have: disp(1,5,9) = (1\rule{1em}{1pt}1) (10\bullet10) + (5\rule{1em}{1pt}1)10 + (9\rule{1em}{1pt}1) = (0)(100) + (4)(10) + 8 = 48

Question:

What are the chief characteristics of the subphylum Vertebrata?

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/Users/wenhuchen/Documents/Crawler/Biology/F13-0327.htm

Solution:

In addition to the three basic chordate characteristics, the vertebrates have an endoskeleton of cartilage or bone that reinforces or replaces the notochord. The notochord is the only skeletal structure present in lower chordates, but in the vertebrates there are bony or cartilaginous segmental vertebrae that surround the notochord. In the higher vertebrates the notochord is visible only during embryonic development. Later the vertebrae replace the notochord completely. A part of each vertebra consists of an arch, and all of the vertebrae together form a tunnel-like protection for the dorsal nerve cord. A brain case, skull, or cranium, composed either of cartilage or bone, develops as a protective structure around the brain of all vertebrates. The eyes in vertebrates are unique and differ both in structure and development from those of the inver-tebrates. The eyes of vertebrates develop as lateral outgrowths of the brain. Invertebrate eyes, such as those of insects, may be highly developed and quite efficient, but they develop from a folding of the skin. The formation of ears for detecting sounds is another vertebrate characteristic. The ears also function as organs of equilibrium, as is the major function of the ears in the lowest vertebrates. The circulatory system of vertebrates is distinctive in that the blood is confined to blood vessels and is pumped by a ventral, muscular heart. The higher inver-tebrates such as arthropods andmolluscstypically have hearts but they are located on the dorsal side of the body and pump blood into open spaces in the body, calledhemocoels. Vertebrates are said to have a closed circu-latory system. The invertebrate earthworm is an exception among the lower invertebrates in that it has a closed circulatory system. Arthropods andmolluscshave an open circulatory system.

Question:

How does the body control the amount of water excreted?

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/Users/wenhuchen/Documents/Crawler/Biology/F18-0455.htm

Solution:

Factors such as blood volume and glomerular capillary pressure act to regulate the amount of fluid initially absorbed by the kidney. The volume of urine excreted, however, is ultimately controlled by the permeability of the walls of the distal convoluted tubules and collecting ducts to water. This permeability can be varied, and is regulated by a hormone known as vasopressin or antidiuretic hormone (ADH). In the absence of ADH, the water permeability of the distal convoluted tubule (DCT) and collecting tubule is very low, and the final urine volume is corresponding-ly high. In the presence of ADH, water permeability is high, and the final urine volume is small. ADH has no effect on sodium absorption, but regulates the ability of water to osmotically follow ionic absorption ADH is produced by a discrete group of hypothalamic neurons whose axons terminate in the posterior pituitary, from where ADH is released into the blood. The release of ADH is regulated by osmoreceptors in the hypothalamus. In-creased plasma osmolarity causes increased secretion of ADH. Decreased osmolarity leads to decreased secretion of ADH. Blood volume also influences ADH secretion via stretch re-ceptors in the left atrium. Increased blood volume stimu-lates this baroreceptor reflex to decrease secretion of ADH. Let us look at the interaction of these regulatory mechanisms in a specific situation. If a man drinks an excess amount of water, but does not increase his sodium intake, the most logical way to maintain optimal osmotic concentrations in the body would be to excrete the excess water without altering normal salt excretion. Intake of the excess water results in an increase in extracellular and blood fluid volumes. This has a two-fold effect: First, the osmotic concentration of the blood decreases. This stimulates the osmoreceptors to cause decreased secretion of ADH. Second, the arterial baroreceptors are stimulated and send impulses to the hypo-thalamus, where ADH release is inhibited. The permeability of the collecting tubules to water is lowered, and conse-quently more water is excreted.

Question:

List the common types of registers used in the computer and statetheir purpose.

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Solution:

There are nine types of registers in the computer. 1.Program Counter:This register holds the address of the instruction tobe executed. As the name implies, it is also a counter since instructions arelisted in memory sequentially (for the most part). 2.Instruction Register:This register holds the binary code of the instructionto be executed. 3.Memory Address Register (MAR):This register holds the address ofany type of data (instruction, operand, etc.) to be stored in, or recalled from, memory. 4.Memory Buffer Register (MBR):This register holds the data written frommemory until the appropriate register is ready to accept it. The MBR alsostores information that is to be transferred from a register into main memory. 5.Accumulator:This register stores temporary data during calculations. The accumulator always holds one of the operands in arithmeticoperations. 6.General Purpose Registers:These registers usually serve as temporarystorage for data and addresses. In some computers, the user mayspecify them as accumulators or program counters. If he wishes, the programmermay also use these registers for arithmetic calculations. 7.Index Register:This register holds a relative address which, when addedto the memory address, forms an effective address, that states the actuallocation of data. Thus, if the contents of the Index Register are changed, the same instructions can be used to handle data from different locations. Generally, the contents of the index register (XR or X) are addedto the numerical ad-dress of a symbol to calculate its absolute (actual) address in main memory. This numerical address is com-prised of adisplacement and the contents of a base register. The base register holdsthe address at which the program was stored in main memory upon itsentry. The displacement is the distance (i.e. the number of bytes) which aninstruction is from the start of the program. Thus, by adding the contentsof the index register, the contents of the base register, and the dis-placement, the absolute address of an instruction can be calculated. 8.Condition Code Register:This register holds one-bit flags which representthe state of conditions inside the CPU. The states of these flags arethe basis for computer decision making in conditional instructions such as"Jump to Location 31 if Accumulator is greater than zero." 9.Stack Pointer:This register contains the address of the top of the stack. It can be incremented or decre-mented depending upon whether datais put onto, or taken off of, the stack.

Question:

The voltage across the terminals of a resistor is 6.0 volts and an ammeter connected as in the diagram reads 1.5 amp. (a) What is the resistance of the resistor? (b) What would the current be if the potential difference were raised to 8.0 volts?

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/Users/wenhuchen/Documents/Crawler/Physics/D20-0649.htm

Solution:

(a)R = (V/I) = [(6.0 volts)/(1.5 amp)] = 4.0 ohms (b)I= (V/R) = [(8.0 volts)/(4.0 ohms) = 2.0 amp . In part a of this solution we have used merely the de-finition of resistance. But in part b we have used Ohm's law, that is, the fact that R is constant.

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Question:

Explain the locomotory pattern of the earthworm.

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/Users/wenhuchen/Documents/Crawler/Biology/F12-0299.htm

Solution:

In the earthworm, two layers of muscle in the body wall and the four pairs of short bristles or chaetae of each body segment are the structures involved in locomotion. Circular muscle occupies the outer layer of the body wall, and just beneath it is the longitudinal muscle layer. Contractions of these two sets of muscles account for the changes in shape of the worm. Each segment of the body is filled with coelomic fluid that is in compressible. Thus the contraction of either layer of muscle results in a change of shape. Contractions of the longitudinal muscles make the worm shorter and thicker, while contractions of the circular muscles make it thinner and correspondingly longer (see Figure). Circular muscle contraction is most important in crawling, and always generates a pressure pulse in the coelomic fluid. Longi-tudinal muscle contraction is more important in burrowing. When the worm is crawling, longitudinal contraction may not be strong enough to generate a coelomic pressure pulse, but it does so in burrowing. Chaetae are extended during longitudinal muscle contraction, functioning to anchor the worm in its burrow and to determine the direction of locomotion. They are retracted during circular muscle contraction. Because of the combined effects of the chaetae and longitudinal muscle contraction, each segment of the body moves in steps of 2 to 3 cm, at the rate of seven to ten steps per minute. The direction of contraction can be reversed, thus enabling the worm to reverse directions. It should be remembered that without the chaetae, no forward move-ment can take place, but only a change in shape.

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Question:

Give examples of the following control constructs as they would be rendered in FORTRAN: a) IF-THEN-ELSE b) DO-WHILE c) REPEAT-UNTIL d) DO-FOR

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G08-0171.htm

Solution:

Although FORTRAN is not a structured language as such, these constructs can be rendered with the use of a few simple statements, including the GO-TO statement. a) The IF-THEN-ELSE construct can be rendered with the following statements. A verbal description is "IF X > Y, then J = J +1. Other-wise, if X \leq Y, J = J - 1". 10 IF (X.LE.Y) GO TO 40 C IF X IS NOT LESS THAN OR EQUAL TO Y THEN WE GO TO STEP 20 20 J = J + 1 30 GO TO 50 C ELSE, J = J - 1 40 J = J - 1 C END OF IF-THEN-ELSE 50 CONTINUE b) The Do-WHILE construct uses an IF statement in the first line of the construct. The last line is a GO TO statement, which creates a loop. This loop terminates only when the condition in the IF statement is satisfied. This example may be described by the following: C DO WHILE X NOT EQUAL TO ZERO 20 IF (X.EQ.O.) GO TO 30 C IF X IS EQUAL TO ZERO WE EXIT FROM THE LOOP X = X - Y Y = Y\textasteriskcentered\textasteriskcentered2 GO TO 20 C END DO-WHILE 30 CONTINUE c) The REPEAT-UNTIL construct Is similar to the DO-WHILE In that an IF statement Is used to control the loop. Because the REPEAT-UNTIL con-struct requires that the statements contained within It must be executed at least once, we place the IF statement at the bottom of the loop. C REPEAT UNTIL X BECOMES GREATER THAN 50 20 CONTINUE X = Y + 2 Y = Y \textasteriskcentered Z IF (X.LE.50.) GO TO 20 30 CONTINUE C END OF REPEAT-UNTIL d) FORTRAN has a special statement to render the DO-FOR construct. The simple DO statement has the following format: DO20 I= 1, 25, 2 statement number at end of construct index Initial value Final value Increment value If the increment value is to be 1, you may omit it. The final statement number is usually either an assignment statement or a CONTINUE statement. C DO FOR I FROM 1 TO 25 1\O DO 2\O I = 1, 25 J = I \textasteriskcentered 2 WRITE (5,\textasteriskcentered) J 2\O CONTINUE C END D0-F0R

Question:

How are eyes formed in embryos of the bird and mammals?

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/Users/wenhuchen/Documents/Crawler/Biology/F23-0601.htm

Solution:

In the area of the forebrain, there appears a lateral growth on each side of the neural tube. These saclike protrusions are termed the optic vesicles. These vesicles grow outward and come into contact with the inner surface of the overlying epidermis. When contact is made, the optic vesicle flattens out and invaginates to form a double walled optic cup. The inner, much thicker layer of the cup ultimately will become the sensory retina of the eye, and the outer layer becomes the pigment layer of the retina. When the optic cup touches the overlying epidermis it stimulates the latter to develop into a lens rudiment. In birds and mammals, the epidermis thickens and folds in to produce a pocket. This pinches off and forms a lens vesicle that lies in the opening of the optic cup and is surrounded by the iris, formed from the rim of the optic cup. The cells on the inner side of the lens vesicle become columnar and then are transformed into long fibers. The nuclei degenerate and the cytoplasm becomes hard, transparent and able to refract light.

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Question:

A ball of mass m1= 100 g traveling with a velocity v1= 50 cm/sec collides "head on" with a ball of mass m_2 = 200 g which is initially at rest. Calculate the final velocities, v ׳ 1 and v ׳ 2 in the event that the collision is elastic.

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/Users/wenhuchen/Documents/Crawler/Physics/D06-0305.htm

Solution:

In any collision there is conservation of momentum and since this is an elastic collision, kinetic energy is also conserved. First, we use momentum conservation to write p (before) = p(after) m_1v_1 + m_2v_2 = m_1v'_1 + m_2v'2. In order to prevent the equations from becoming too clumsy, we suppress the units (which are CGS throughout); then we have 100 × 50 + 0 = 100v'_1 + 200v'_2. Dividing through by 100 gives 50 = v'_1 + 2v'_2(1) From energy conservation, we have (since there is no PE involved and since the collision is elastic) KE(before) = KE(after) (1/2) m_1v ^21+ (1/2) m_2v ^2 _2 = (1/2) m_1v_1'^2_ + (1/2) m_2v_2'^2. 1/2 × 100 × (50)^2 + 0 = 1/2 × 100 v1 ׳ 2 + 1/2 × 200 v_2 ׳ 2 . Dividing through by 100/2 gives = 50 gives 2500 = v_1'^2+ 2v_2'^2 .(2) We now have two equations, (1) and (2), each of which contains both of the unknowns, v_1' and v_1' We can obtain a solution by solving Eq. 1 for v_1', v_1' = 50 - 2v_2'(3) and substituting this expression into Eq. 2 : 2500 = (50 - 2v_2')^2 + 2v_2'^2or, 2500 = 2500 - 200v_2' + 4v_2'^2 + 2v_2'^2. From this equation we find 6v_2'2= 200v_2' so that v_2' = 200/6 = 33 (1/3) cm/sec. Substituting this value into Eq. 3 we find v_1' = 50 - 2 × 33 (1/3) = - 16(2/3) cm/sec . The negative sign means that after the collision, m_1 moves in the direction opposite to its initial direction (see figure).

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Question:

What hormonal changes occur during the delivery of an infant (parturition)?

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/Users/wenhuchen/Documents/Crawler/Biology/F21-0563.htm

Solution:

Present theory holds that a shift in the balance of estrogen and progesterone is one important factor in parturition. Estrogen is known to stimulate contractions of the uterine muscles and progesterone is known to inhibit muscular contraction. It is known that just prior to parturition, estrogen secretion by the placenta rises sharply, and this increase may play a role in the contraction of the uterus. It is therefore believed that during pregnancy progesterone suppresses contraction of the uterine muscles, but the rise of estrogen late in pregnancy overcomes the effects of the progesterone and initiates the contractions necessary for birth. Oxytocin, one of the hormones released from the posterior pituitary, is an extremely potent uterine muscle stimulant and is released as a result of stimulation of the hypothalamus by receptors in the cervix. Relaxin a hormone secreted by the ovaries and placenta during pregnancy, is another hormone which may be important in parturition. Relaxin loosens the connections between the bones of the pelvis, thereby enlarging the birth canal to provide easy exit for the newborn. Prostaglandins are fatty acid derivatives secreted by animal tissues. Prostaglandins stimulate the smooth muscle of many organs, including the smooth muscle of the wall of the uterus.

Question:

A 5.0 - lb ball is fastened to the end of a flat spring. A force of 2.0 lb is sufficient to pull the ball 6.0 in. to one side. Find the force constant and the period of vibration.

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/Users/wenhuchen/Documents/Crawler/Physics/D09-0361.htm

Solution:

The restoring force on the ball as it is displaced from the equilibrium point 0 (see the figure) is, by Hooke's law, F\ding{217}= -kx\ding{217} \ding{217} \ding{217} where x is the displacement and k is the spring constant. Hence, k = \vertF\ding{217}\vert / \vertx\ding{217}\vert \ding{217} \ding{217} = 2.0 lb/0.50 ft = 4.0 lb/ft The mass of the ball is m = W/g where W is the ball's weight. Hence, m = (5.0 lb)/(32 ft/sec^2) = 0.16 slug The period of oscillation is given by T = 2\pi\surd(m/k) = 2\pi\surd[(0.16 slug) / (4.0 1b/ft) = 1.2 sec.

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Question:

One mole of H_2 is to be reacted with one mole of I_2 . Assum-ing the equilibrium constant is 45.9, what will be the final concen-trations of the chemical components in a 1 liter box at 49C\textdegreeC ?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E09-0316.htm

Solution:

The first thing to do with this type of problem is to write a balanced equation for the reaction. A reaction between H_2 and I_2 is one which produces HI , As such, you have H_2 (g) + I_2 (g) \rightarrow 2HI(g) . You are given the equilibrium constant of this reaction. To use this information, you must know the meaning of such a constant. Given the general reaction:xA+yB\rightarrowzC, the equilibrium constant, K , is defined as [C]^2 / {[B]^y [ A]^x} , where the brackets serve to indicate concentrations. For this problem, K = [HI]^2/ {[H_2 ] [I_2 ]} . You are asked to find the concentrations. Let x = the no. of moles ofH_2 that react. It follows, then, that x is also equal to the number of moles of I_2 that react and 2x is equal to the number of moles of HI that form. Substituting you obtain [HI]^2/ {[H_2] [I_2]} = 45.9 . You started with one mole of each H_2and I_2 . Therefore, at equilibrium you must have 1 \rule{1em}{1pt} x moles of each species. Since the volume of the box is one liter, this means that the concentration of each substance is equal to the number of moles of each substance. You can now substitute these values in the equilibrium equation. (2x)^2 / {(1 \rule{1em}{1pt} x)(1 \rule{1em}{1pt} x) } = 45.9 . Solving for x using the quadratic equation, you obtain x = .772. Therefore, at equilibrium you have [H_2] = 1 \rule{1em}{1pt} x = .228 mol/liter [I_2]= 1 \rule{1em}{1pt} x = .228 mol/liter [HI] = 2x = 1.554 mol/liter .

Question:

The earth acts onany bodywith a gravitational force inversely proportional to the square of the distance of the body from the center of the earth. Calculate the escape velocity from the earth, i.e., the speed with which a vertically moving body must leave the earth's surface in order to coast along without being pulled back to the earth. Also find the time it takes for a rocket projected upward with this escape velocity to attain a height above the earth's surface equal to its radius. In both cases ignore the effect of any other heavenly bodies and take the earth's radius as 6.38 × 10^6 m.

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/Users/wenhuchen/Documents/Crawler/Physics/D04-0151.htm

Solution:

Assuming that the body moves along a radial trajectory after leaving the earth, it continually ex-periences a gravitational force F = -(Gmm_e/r^2 ) where m and m_e are the mass of the body and the earth, respectively. By Newton's Second Law, F = ma, and the acceleration of m is then a = -(Gm_e/r^2 ) But, being that the trajectory is radial, a = (dv/dr)(dr/dt) = v(dv/dr) = -(Gm_e/r^2) where v is the object's velocity. vdv= -Gm_e(dr/r^2)(1) Now, we want a projectileto leave the earth (r = R_e) with a velocityv_eand to reach a destination at which it no longer feels the effect of the earth's gravitation-al force. If it reaches this point , the body can have no velocity and yet still not be accelerated towards the earth. Since the gravitational force is zero at \infty , we require that v = 0 at r = \infty. Hence ^0\int_ve (vdv) = -Gme_e^\infty\int_(R)e (dr/r^2) - v2_e/2 =Gm_e(1/r)\infty_Re v2_e = 2Gm_e/R_e(2) At the surface of the earth Gm_e/R^2 _e = g andv2_e = 2gR_e v_e= \surd2gR_e = \surd(2 × 9.81 m\bullets^2 × 6.38 × 10^6 m) v_e= 1102 × 10^3 m/s^2 The second part of the problem asks us to find out how long it takes for an object to reach a distance r = 2R_e from the center of the earth if its initial velocity isv_e. In order to do this, we must find r as a function of t. Going back to (1) vdv-Gm_e(dr/r^2) But, now, v = v at r = r, and v =v_eat r = R_e and ^v\int_ve(vdv) = -Gm_e^r\int_Re(dr/r^2) (v^2 - v^2 _e)/2 =Gm_e(1/r - 1/R_e) v^2 = v^2_e + (2Gm_e/r) - (2Gm_e/R_e)(3) However, from (2) v^2 _e = 2Gm_e/R_e Therefore, from (3) v^2 = 2Gm_e/r v = \surd(2Gm_e)/r^1/2 To find v as a function of t, note that v = (dr/dt) Therefore,\surd(2Gm_e)/r^1/2 = (dr/dt) r^1/2dr= \surd(2Gm_e)dt Since r = R_e when t = 0, and r = 2R_e when t = t, ^2Re\int_Re (r^1/2dr) = \surd2Gm_e^t\int_0dt (2/3) r^3/2^2Re\mid_Re = \surd(2Gm_e) t (2R_e)^3/2 - (R_e)^3/2 = 3/2\surd(2Gm_e) t

Question:

In figure A, a block of weight w hangs from a cord which is knotted at O to two other cords fastened to the ceiling. Find the tensions in these three cords. Let w = 50 lb, \texttheta_2 = 30\textdegree, and \texttheta_3 = 60\textdegree. The weights of the cords are negligible. (b) Forces acting on the block, on the knot, and the ceiling, (c) Forces on the knot 0 resolved into x- and y- components.

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/Users/wenhuchen/Documents/Crawler/Physics/D02-0026.htm

Solution:

In order to use the conditions of equilibrium to compute an unknown force, we must consider some body which is in equilibrium and on which the desired force acts. The hanging block is one such body and the tension in the vertical cord supporting the block is equal to the weight of the block. The inclined cords do not exert forces on the block, but they do act on the knot at 0. Hence, we consider the knot as a small body in equilibrium, whose own weight is negligible. The free body diagrams for the block and the knot are shown in figure B, where T_1\ding{217}, T_2\ding{217}, and T_3\ding{217} represent the forces exerted on the knot by the three cords and T'_1\ding{217}, T'_2\ding{217} , and T'_3\ding{217} are the reactions to these forces. Consider first the hanging block. Since it is in equilibrium, T'_1 = w = 50 lb Since T_1\ding{217} and T'_1\ding{217} form an action-reaction pair, T'_1 = T_1 HenceT_1= 50 lb. To find the forces T_2\ding{217} and T_3\ding{217}, we resolve these forces (see fig. C) into rectangular components. Then, from Newton's second law, \sumF_x= T_2 cos \texttheta_2 - T_3 cos \texttheta_3 = 0, \sumF_y= T_2 sin \texttheta_2 + T_3 sin \texttheta_3 - T_1 = 0 We have T_2 cos 30\textdegree - T_3 cos 60\textdegree = 0 T_2 sin 30\textdegree + T_3 sin 60\textdegree = 50 or0.866 T_2 - 0.500 T_3 = 0 0.500 T_2 + 0.866 T_3 = 0 Solving these equations simultaneously, we find the tensions to be T_2 = 25 lb,T_3 = 43.3 lb. Finally, we know from Newton's third law that the inclined cords exert on the ceiling the forces T'_2 and T'_3\ding{217}, equal and opposite to T_2\ding{217} and T_3\ding{217}, respectively.

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Question:

Explain the concept of coding and give the general characteristicsof the most widely used codes.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G06-0125.htm

Solution:

Coding is representation of the characters via a character code in whicheach character is assigned a unique combination of bits. The three mostpopular codes are BCDIC-Binary Coded Decimal Inter-change code, EBCDIC-Extended Binary Coded Decimal Interchangecode,and ASCII- American Standard Code for Information Interchange. The BCDIC was widely used until the advent of the IBM System/360 and similar to it systems.Being a six-bit character code, BCDIC can accumulate as much as 64 (26) characters assigning 000000 tothe first and 111111to the last character. This is sufficient for the 26 upper-caseletters, 10 digits, and 28 special characters such as period, comma, etc. If, however, distinction is to be made between the upper-case andthe lower-case letters, 64 characters are not enough. The 256 (28) "spaces" of the eight-bit codes, such as EBCDIC and ASCII, are more than enough toaccomodateall the needed characters. BCDIC and EBCDIC are very similar except that the second code contains morecharacters. Table1 ;illustrates thesethree character codes. Table 1. Character BCDIC8 EBCDIC ASCII 0 001 010 1111 0000 0011 0000 1 000 001 1111 0001 0011 0001 2 000 010 1111 0010 0011 0010 3 000 011 1111 0011 00110011 4 000 100 1111 0100 0011 0100 5 000 101 1111 0101 0011 0101 6 000 110 1111 0110 0011 0110 7 000 111 1111 0111 0011 0111 8 001 000 1111 1000 0011 1000 9 001001 1111 1001 0011 1001 A 110 001 1100 0001 0100 0001 B 110 010 1100 0010 0100 0010 C 110 011 1100 0011 0100 0011 D 110 100 1100 0100 01000100 E 110 101 1100 0101 0100 0101 F 110110 1100 0110 0100 0110 G 110 111 1100 0111 0100 0111 H 111 000 1100 1000 0100 1000 I 111 001 1100 1001 0100 1001 J 100 001 1101 0001 0100 1010 K 100 010 1101 0010 0100 1011 L 100 011 1101 0011 0100 1100 M 100100 1101 0100 0100 1101 N 100 101 1101 0101 0100 1110 O 100 110 1101 0110 0100 1111 P 100 111 1101 0111 0101 0000 Q 101 000 1101 1000 0101 0001 R 101 001 1101 1001 0101 0010 S 010010 1110 0010 0101 0011 T 010 011 1110 0011 0101 0100 U 010 100 1110 0100 01010101 V 010 101 1110 0101 0101 0110 W 010 110 1110 0110 0101 0111 X 010 111 1110 0111 0101 1000 Y 011 000 1110 1000 0101 1001 Z 011 001 1110 1001 0101 1010 blank 010 000 0100 0000 0010 0000 = 001 011 0111 1110 0011 1101 + 110 000 0100 1110 0010 1011 - 100 000 0110 0000 0010 1101 \bullet 101 100 0101 1100 0010 1010 / 010 001 0110 0001 0010 1111 ( 011 100 0100 1101 0010 1000 ) 111 100 0101 1101 0010 1001 , 011011 0110 1011 0010 1100 . 111 011 0100 1011 0010 1110 $ 101 011 0101 1011 0010 0100 , 0111 1101 0010 0111 < 111 010 0100 1100 0011 1100 > 101 111 0110 1110 0011 1110 : 0111 1101 0011 1010 ; 0101 1110 0011 1011 \textquotedblleft 111111 0111 1010 00100010

Question:

A male bacterium conjugates with a female bacterium. After conjugation , the female becomes a male. Account for this "sex-change".

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/Users/wenhuchen/Documents/Crawler/Biology/F05-0136.htm

Solution:

Conjugation occurs between bacterial cells of different mating types . Maleness in bacteria is determined by the presence of a small extra piece of DNA, the sex factor, which can replicate itself and exist autonomously (independent of the larger chromosome) in the cytoplasm. Male bacteria or donors having the sex factor, also known as the F factor, are termed F^+ if the sex factor existsextrachromosomally. F^+ bacteria can only conjugate with F^-, thefemalecounterpartsor recipients which do not posses the F factor. Genes on the F factor determine the formation of hairlike projections on the surface of the F^+ bacterium, called F or sexpili. Thepiliformcytoplamicbridges through which genetic material is transferred and aids the male bacterium in ad-hering to the female during conjugation . During conjugation of an F^+ with an F^- bacterium, the DNA that is the most likely to be transferred to the female is the F factor. Prior to transfer, the F factor undergoes replication. The female thus becomes a male by receiving one copy of the F factor, and the male retains its sex by holding on to the other copy of the sex factor. The DNA of the male chromo-some is very rarely transferred in this type of conjugation . If this were the only type of genetic exchange in conjugation, all bacteria would become males and conjuga-tion would cease. However, in F^+ bacterial cultures, a few bacteria can be isolated which have the F factor in-corporated into their chromosomes. These male bacteria that conjugate with F^- cells are calledHfr(high frequency of recombination) bacteria . They do not transfer the F factor to the female cells during conjugation , but they frequently transfer portions of their chromosomes. This process is unidirectional, and no genetic material from the F^- cell is transferred to theHfrcell.

Question:

An army recruit on a training exercise is instructed to walk due west for 5 mi, then in a northeasterly direction for 4 mi, and finally due north for 3 mi. When he completes his exercise, what is his resultant displace-ment R\ding{217}? How far will he be from where he started?

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Solution:

The recruit's path is shown in the figure, where each division on the graph represents one mile. We find R\ding{217} by first adding the components of his individ-ual displacements which we regard as vectors. We will let E\ding{217} and N\ding{217}, representing east and north, be our unit vectors, regarding western and southern displacements as being negative eastern and negative northern dis-placements, respectively. Assume north and east are given equal weights. Then N\ding{217}E\ding{217} is as shown in the diagram. Thus, the sum of the components is: E\ding{217} N\ding{217} - 5 mi 4 cos 45\textdegree mi 0 mi 0 mi 4 sin 45\textdegree mi 3 mi (4 cos 45\textdegree - 5)mi (4 sin 45\textdegree + 3)mi R\ding{217} = [ 4 (1/\surd2) - 5 ] mi E\ding{217} + [ 4 (1/\surd2) + 3 ] N\ding{217} = (2.8 - 5)mi E\ding{217} + (2.8 + 3)mi N\ding{217} = - 2.2 mi E\ding{217} + 5.8 mi N\ding{217} The recruit's final distance from the starting point will be the magnitude of R\ding{217}: R = \surd{(- 2.2 mi)^2 + (5.8 mi)^2} = 6.20 mi

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Question:

The pancreas is a mixed gland having both endocrine and exocrine functions. The exocrine portion secretes diges-tive enzymes into the duodenum via the pancreatic duct. The endocrine portion secretes two hormones (insulin and glucagon) into the blood. What are the effects of these two hormones ?

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/Users/wenhuchen/Documents/Crawler/Biology/F21-0536.htm

Solution:

Insulin is a hormone which acts directly or indirectly on most tissues of the body, with the exception of the brain. The most important action of insulin is the stimulation of the uptake of glucose by many tissues, particularly liver muscle and fat. The uptake of glucose by the cells decreases blood glucose and increases the availability of glucose for those cellular reactions in which glucose participates. Thus, glucose oxidation, fatsynthesis, and glycogen synthesis are all accentuated by an uptake of glu-cose. It is important to note that insulin does not alter glucose uptake by the brain, nor does it influence the ac-tive transport of glucose across the renal tubules and gas-trointestinal epithelium. As stated, insulin stimulates glycogen synthesis. In addition, it also increases the activity of the enzyme which catalyzes the rate-limiting step in glycogen synthesis. Insulin also increases triglyceride levels by inhibiting triglyceride breakdown, and by stimulating production of triglyceride through fatty acid and glycerophosphate synthesis. The net protein synthesis is also increased by insulin, which stimulates the active membrane transport of amino acids, particularly into muscle cells. Insulin also has effects on other liver enzymes, but the precise mechanisms by which insulin induces these changes are poorly understood. Insulin secretion is directly controlled by the glucose concentration of the blood flowing through the pancreas. This is a simple system which requires no participation of nerves or other hormones. Insulin is secreted by beta cells, which are located in the part of the pancreas known as the Islets of Langerhans. These groups of cells, which are located randomly throughout the pancreas, also consist of other secretory cells called alpha cells. It is these alpha cells which secrete glucagon. Glucagon is a hormone which has the following major effects: it increases glycogen breakdown thereby raising the plasma glucose level; it increases hepatic synthesis of glucose from pyruvate, lactate, glycerol, and amino acids, (a process called gluconeogenesis, which also raises the plasma glucose level); it increases the breakdown of adipose-tissue triglyceride, thereby raising the plasma levels of fatty acids and glycerol. The glucagon-secreting alpha cells in the pancreas, like the beta cells, respond to changes in the concentration of glucose in the blood flowing through the pancreas; no other nerves or hormones are involved. It should be noted that glucagon has the opposite effects as insulin. Glucagon elevates the plasma glucose whereas insulin stimulates its uptake and thereby reduces plasma glucose levels; glucagon elevates fatty acid concentrations whereas insulin converts fatty acids (and glycerol) into triglycerides, thereby inhibiting trigly-ceride breakdown. Thus the alpha and beta cells of the pancreas constitute a "push- pull" system for regulating the plasma glucose level.

Question:

A string 100 centimeters long has a frequency of 200 vibrations per second. What is the frequency of asimilar string under the same tension but 50 centimeters long?

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/Users/wenhuchen/Documents/Crawler/Physics/D25-0813.htm

Solution:

Since the frequency is inversely proportional to length, the frequency of the 50-centimeter string is 2 × 200, or 400vps.

Question:

Write a BASIC program to calculate the sum of cubes of integers from 1 to 100.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G10-0227.htm

Solution:

This can be accomplished by an elementary FOR ... NEXT loop. A more sophisticated approach could be based on the formula ^K\sum_i_=1 i^3 = [K^2 (K+1)^2] / [4] where ^K\sum_i_=1 i^3 = 1^3 + 2^3 + 3^3 + ... + K^3. PROGRAM # 1 1\O\O REM SUM OF CUBES 11\O LET S = \O 12\O FOR I = 1 TO 1\O\O 13\O LET S = S + I _\uparrow 3 14\O NEXT I 15\O PRINT S 16\O END PROGRAM # 2 1\O\O LET K = 1\O\O 11\O LET S = K\textasteriskcenteredK\textasteriskcentered(K+1)\textasteriskcentered(K+1)/4 12\O PRINT S 13\O END

Question:

a) the block diagram, b) the timing diagram,c) the Karnaugh Map, d) the excitation equation,e) the state table,and f) the state diagram, for an S-C type, clocked, master-slave flip-flop.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G04-0083.htm

Solution:

a) The block diagram of a clocked, master-slave SC type flip-flop is shown in figure 1. The master-slave flip-flop is a combination of two flip-flops, the master getting the direct clock pulse and the slave get-ting the inverted clock pulse. The input condition is that the Boolean product S\bulletC = 0. b) The timing diagram is shown in figure 2. Timings for four clock pulses t_1, t_2, t_3 and t_4 are shown. We assume that initially, before t_1 arrives, x and y are each = 1. Note that x and x' change state, whenever required to do so, at the leading edge of the clock pulse, but y and y' change state at the trailing edge of the clock pulse due to the inverter. c) The Karnaugh Map of the master-slave SC flip-flop is shown in figure 3. The Karnaugh Map in figure 3 is for y\rightarrow, where y\rightarrowdenotes the next state of \rightarrow \rightarrow y. Note also that as the input with both S=1 and C=1 is not allowed, and hence does not ever' occur, we insert a 'd' in the squares for S, C = 1,1. The 'd' repre-sents a "don't care" condition. d) The Karnaugh Map leads to the formulation of the exci-tation equation, i.e., the equation for y\rightarrowin terms of S, C and y. \rightarrow \therefore y\rightarrow= Sy' + C'y . \rightarrow e) The Karnaugh helps us to draw the state table, Figure 4, SCy y\rightarrow \rightarrow REMARKS 000 001 0 1 NO CHANGE 010 011 0 0 CLEARS 100 101 1 1 SETS 110 111 d d NOT ALLOWED Fig. 4 The state diagram is drawn in Figure 5: Note in the above that the input condition 11 does not occur at all.

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Question:

(a) 4NH_3 (g) + 5O_2 (g) \rightarrow 4NO (g) + 6H_2 O (l) (b) H_2S (g) + 1(1/2) O_2 (g) \rightarrow SO_2 (g) + H_2 O (l) Use the following values for ∆H\textdegree of the components Compound H\textdegree (Kcal/mole) H_2 O (l) - 68.3 SO_2 (g) - 71.0 H_2 S (g) - 5.3 NO (g) 21.6 NH_3 (g) - 11.0

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/Users/wenhuchen/Documents/Crawler/Chemistry/E14-0507.htm

Solution:

∆H\textdegree may be defined as the heat released or absorbed as a reaction proceeds. ∆H\textdegree for a particular re-action is found by subtracting the sum of the ∆H\textdegree's of the compounds reacting from the sum of the ∆H\textdegree's of the products being formed, i.e., ∆H\textdegree = (∆H\textdegree of products) - (∆H\textdegree of reactants). (1) Determination of the ∆H\textdegree's for the reactions: (a) 4NH_3(g) + 5O_2 (g) \rightarrow 4NO(g) + 6H_2O(l) . When more than one mole of a compound is reacted or formed, the ∆H\textdegree of the compound is multiplied by the number of moles present. ∆H\textdegree = (4 ×∆H\textdegree of NO + 6 ×∆H\textdegree of H_2 O) - (4 × ∆H\textdegree of NH_3 + 5 × ∆H\textdegree of O_2) The ∆H\textdegree of any element is always 0. Thus, the ∆H\textdegree of O_2 is zero. Substituting, ∆H\textdegree= [(4 × 21.6) + 6 × (- 68.3)] - [(4 × (- 11.0)) + (5 × 0)] = - 279.40 Kcal. (b) H_2 S(g) + 1(1/2) O_2 (g) \rightarrow SO_2 (g) + H_2 O(l) ∆H\textdegree= (∆H\textdegree of SO_2 + ∆H\textdegree of H_2O) - (∆H\textdegree of H_2 S + 1(1/2) ∆H\textdegree of O_2) ∆H\textdegree= [(- 71.0) + (- 68.3)] - [(- 5.3) + (1(1/2) × O)] = - 134.0 Kcal (2) Determination of ∆E: ∆E\textdegree is the change in energy of a given reaction. It is defined as the difference of the ∆H\textdegree's of the reaction and the pressure times the change in volume occurring during the reaction, i.e., ∆E\textdegree = ∆H\textdegree - P∆V, where P is the pressure and ∆V is the change in volume. One is not given the values for pressure or volume here, thus one uses the Ideal Gas Law to substitute ∆nRT for P∆V where ∆n is the change in the number of moles present, R is the gas constant (1.99 cal/mole-\textdegreeK)and T is the absolute tem-perature. The Ideal Gas Law states that P∆V = ∆nRT Here, one is told that the reactions both occur at 25\textdegreeC. This temperature can be converted to the absolute scale by adding 273 to 25\textdegreeC. T = 25 + 273 = 298\textdegreeK. ∆n is found by subtracting the number of moles of compounds reacting from the number of moles formed. Caution: Only moles of gases are taken into account. ∆n = number of moles formed - number of moles reacted. One can solve for ∆E\textdegree in these two reactions. ∆H\textdegree for this reaction was found to be - 279.40 Kcal. ∆H\textdegree must be converted to calories when R is used. Kcal are converted to cal by multiplying the number of Kcal by 1000 cal/1 Kcal. ∆n= (4 moles NO) - (4 moles NH3+ 5 moles O_2) = (4) - (4 + 5) = - 5 mole ∆E\textdegree= ∆H\textdegree - ∆nRT∆H\textdegree= - 279.40 Kcal ∆n= 5 moles R= 1.99 cal/mole-\textdegreeK T= 298\textdegreeK ∆E\textdegree = - 279.4 Kcal × [(1000 cal) / (1 kcal)] - ( - 5mole × 1.99 cal/mole \textdegreeK × 298\textdegreeK) = - 279,400 cal + 2965 cal = - 276,435 cal = - 276.4 Kcal. (b) H_2S (g) + 1(1/2) O_2 (g) \rightarrow SO_2 (g) + H_2 O (l) ∆H\textdegree = - 134.0 Kcal ∆H\textdegree should be expressed in calories ∆H\textdegree in cal = - 134.0 Kcal × 1000 cal/1 Kcal = - 134,000 cal ∆n = (no. of moles of SO_2) - (no. of moles of H_2S + no. of moles of O_2) ∆n = (1 mole of SO_2) - (1 mole of H_2S + 1 (1/2) moles of O_2) = (1) - (1 + 1(1/2)) = -1.5 moles R= 1.99 cal/mole \textdegreeK T= 298\textdegreeK ∆E\textdegree = ∆H\textdegree - ∆nRT = - 134,000 cal - (- 1.5 moles × 1.99 cal/mole \textdegreeK × 298\textdegreeK) = - 134,000 cal + 890 cal = - 133,110 cal = - 133.1 Kcal.

Question:

Write and explain ENVIRONMENT DIVISION, and section headers, along with the two paragraphs, to compile a program on a Honeywell 200, and then execute it on an IBM System/360 Model 40.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G11-0252.htm

Solution:

ENVIRONMENT DIVISION describes the environment, or equipment, on which the program will be run. Entries in this division depend on the computer available to the pro-grammer . All ENVIRONMENTAL DIVISION entries begin with margin A, and end with a period. The computer names used in the configuration section are not optional or created by the programmer; they are the official names given to the computers available. In prac-tice the ENVIRONMENT DIVISION will be constant in programs for the same installation. To allow a program to be used on different computers, the CONFIGURATION SECTION, heading allows for specification of the compilation and execution computers. To define the computer which will be used to compile the program, ENVIRONMENT DIVISION, uses the statement SOURCE-COMPUTER. In this case the SOURCE-COMPUTER. is H-200. OBJECT-COMPUTER. is where the program gets executed, and in this problem it is given to be IBM-360-40. The complete ENVIRONMENT SECTION is given in fig.1.

Question:

Starting from rest, an engine at full throttle pulls a freight train of mass 4200 slugs along a level track. After 5 min, the train reaches a speed of 5 mph. After it has picked up more freight cars, it takes 10 min to ac-quire a speed of 7 mph. What was the mass of the added freight cars? Assume that no slipping occurs and that frictional forces are the same in both cases.

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/Users/wenhuchen/Documents/Crawler/Physics/D04-0116.htm

Solution:

In the first case the train acquires a speed of 5 mph in 5 min = 1/12 hr. When one applies the formula v_1 = v_0 + a_1t_1, for constant acceleration a_1, where v_0 = 0, then a_1 = 5 mph/1/12 hr = 60 mi/hr^2. In the second case the train acquires a speed of 7 mph in 10 min = 1/6 hr. When one applies the formula v_2 = v_0 + a_2t_2, then a_2 = 7 mph/1/6 hr = 42 mi/hr^2. In both cases the engine is at full throttle and is thus applying the same net force F to the train: In the first case it is applying it to a mass of 4200 slugs and in the second case to a mass of M + 4200 slugs. Thus F = 4200 slugs × a_1 = (4200 slugs + M) × a_2. M = 4200{(a_1/a_2) - 1}slugs = 4200{(60 mi/hr^2)/(42 mi/hr^2) - 1}slugs = 4200 × (3/7) slugs = 1800 slugs. Note that although it is not normal to measure acceleration in mi/hr^2, it is a mistake to convert to more familiar units unless and until it is found to be necessary. In this case the units of acceleration cancel out and no conversion is ever necessary.

Question:

There exists the equilibrium CO_2 + 2H_2O \rightleftarrows H_3O^+ + HCO_3^-. The K_a = 4.3 × 10^-7 and pH = 7.40. Find the ratio of a concentrations of bicarbonate (HCO_3^-) to carbon dioxide (CO_2).

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/Users/wenhuchen/Documents/Crawler/Chemistry/E12-0420.htm

Solution:

The equlibrium constant expression, which indicates the ratio of the products to reactants, each raised to the power of their respective coefficients, can be written as follows: K_a = {[H_3O^+] [HCO_3^-]} / [CO_2] Note, [H_2O] is omitted, since it is considered to be constant. Since K_a = 4.3 × 10^-7, then 4.3 × 10^-7 = {[H_3O^+] [HCO_3^-]} / [CO_2] One is solving for [HCO_3^-] / [CO_2]. If one knows [H_3O^+], then one can calculate this ratio. [H_3O^+] can be found, since, as given, the pH = 7.40. pH = - log [H_3O^+] or [H_3O^+] = antilog (- pH) = antilog (- 7.40) = 4.0 × 10^-8 M. Thus,4.3 × 10^-7 = 4.0 × 10^-8 [HCO_3^-] / [CO_2] , and [HCO_3^-] / [CO_2] = [4.3 × 10^-7] [4.0 × 10^-8] = 10.75.

Question:

On either side of a pane of window glass, temperatures are 70\textdegreeF and 0\textdegreeF. How fast is heat conducted through such a pane of area 2500 cm^2 if the thickness is 2 mm?

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/Users/wenhuchen/Documents/Crawler/Physics/D16-0537.htm

Solution:

The equation of heat conduction is dQ = - KA (dT/dx)(1) wheredQ/dtis the rate at which heat is transferred across a cross-section A of a material with coefficient of thermal conductivity K.dT/dxis the temperature gradient in the material. In the steady state, the temperature at each point of the material remains constant in time. Hence, the rate of heat transfer across a cross-section is the same at all cross-sections. As a result of (1),dT/dx must be the same at all cross-sections. If T_1 is the temperature at a cross-section at x_1,and T_2 is the temperature at x_2, we obtain dT /dx= ∆T/∆x = (T_2 - T_1)(x_2 - x_1)(2) (Note that this is a direct consequence of the fact thatdT/dxis constant). Using (2) in (1) dQ/dt= -K A [(T_2 - T_1)(x_2 - x_1)] = K A [(T_1 - T_2)(x_2 - x_1)] But x_2 - x_1 is equal to L, the length of the material across which the heat conduction is taking place. dQ/dt= [K A (T_1 - T_2)]/L For the pan of glass, T_1 = 70\textdegreeF@x_1 = 0 mm T_2 = 0\textdegreeF@x_2 = 2 mm Furthermore, K = .0015 cal/cm\bullets\bullet\textdegreeC for glass, whence dQ /dt= [(.0015 cal/cm\bullets\bullet\textdegreeC)(2500 cm^2)(70\textdegreeF - 0\textdegreeF)]/(2mm - 0 mm) Since 70\textdegreeF = 5/9 \textbullet 70\textdegreeC = 350\textdegreeC/9, we obtain, dQ /dt= [(.0015 cal/cm\bullets\bullet\textdegreeC)(2500 cm^2)(350\textdegreeC/9)]/(.2 cm) dQ /dt= 729 cal/s Note that, by convention, temperature decreases as x increases. Hence, T_2 < T_1 and x_2 > x_1. As a result,dQ/dt> 0 in the direction in which dT /dx< 0.

Question:

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/Users/wenhuchen/Documents/Crawler/Chemistry/E21-0759.htm

Solution:

To predict the products, you need to identify the type of reaction that will take place and its mechanism. The first two reactions, (a) and (b), involve addition across a double bond, and the last reaction, (c) concerns the dehalogenation of an alkyl halide using the base, KOH. In other words, it is an elimination reaction. The general mechanism of addition to a double bond is as follows: Note: From the mechanism, observe that the reaction goes through a species called a carbonium ion. The stability of a carbonium ion follows the order: 3\textdegree>2\textdegree>1\textdegree. This is shown schematically as A reaction will proceed so that the product formed in greatest amounts has gone through the mechanism that gives the most stable carbonium ion possible. Thus, you have You do not have to consider carbonium ion formation here, since the final product would be the same in either case, however, for (b), the carbonium ion must be con-sidered. There are two possible reactions: Two different products are obtained. Notice that in (1), the reaction proceeds through a 3\textdegree carbonium, whereas in (2) the reaction proceeds through a 1\textdegree carbonium ion. Because 3\textdegree is more stable than 1\textdegree, (c) In considering this part, remember that potassium hydroxide (KOH) eliminates the halogen from an alkyl halide to give alkenes, compounds containing double bonds. With this in mind, two possible reactions and products are poss-ible: The first product, greatly predominates over the second. The reason stems from the fact that alkenes are more stable when the double bond is surrounded by the maximum number of alkyl groups.

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Question:

An aircraft wing requires a lift of 25.4 lb/ft^2. If the speed of flow of the air along the bottom surface of the wing is to be 500 ft/s, what must be the speed of flow over the top surface to give the required lift? The density of air is 2.54 × 10^-3 slug/ft^3.

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Solution:

Below an aircraft wing, there is little disturbance of the air flow. Because of the shape of the wing, the streamlines crowd together above it, effectively decreasing the cross-sectional area A for the air flow. The equation of continuity states that for an incompressible fluid (such as air), Av = constant. Therefore, the velocity v of the air above the wing must increase. Consider two points in the air flow at the same height, the first at a point to the left of the wing (see figure),where the flow has yet been disturbed and the second at a point above the wing. Bernoulli's equation states that P_1 + (1/2) \rhov^2_1 = P_2 + (1/2) \rhov^2_2 where the \textquotedbllefty\textquotedblright terms in the equation cancel since the two points are at the same elevation. From this equation, we see that since v_2 is greater than v_1, p_2 must be less than p_1. Since p_1 is also the pressure of the air under the wing, we see that the pressure is less above the wing than below it. This pressure differential gives rise to the lift on the wing. In Bernoulli's equation, let the subscript 1 refer to the lower surface and the subscript 2 to the upper surface of the wing. Thus the dynamic lift per unit area is P_1 = P_2 = 25.4 lb/ft^2 Solving for the speed of flow v_2 over the top of the wing in the first equation, we get v^2_2 = [2(p_1 - p_2)]/\rho+ v^2_1 = [(2)(25.4 lb/ft^2)]/[(2.54) (10^-3) slug/ft^3] + (25) (10^4)ft^2/sec^2 = [(2) (10^4) + (25) (10^4)] ft^2/sec^2 = (27) (10^4) ft^2/sec^2 v_2= 519.6 ft/sec.

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Question:

What is the resultant force on a body weighing 48 lb when its acceleration is 6 ft/sec^2?

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/Users/wenhuchen/Documents/Crawler/Physics/D04-0099.htm

Solution:

We find the resultant force by using Newton's Second Law, F ma. Here, F is the net force on a body of mass m having a net acceleration a. In order to use this law, we must first find the mass of the body. Since the weight of a body is defined as the gravitational force of attraction on it, we have w = mg where g is the acceleration due to gravity. Hence m = w/g and F = (w/g)a = [{(48 lb)(6 ft/s^2)}/{32 ft/s^2}] F = 9 lb.

Question:

Why does one experience difficulty in breathing at high altitudes ?

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/Users/wenhuchen/Documents/Crawler/Biology/F16-0404.htm

Solution:

Barometric pressure progressively decreases as altitude increases. The air is still 21% oxygen, but the partial pressure of oxygen decreases along with barometric pressure. At sea level, the barometric pres-sure is 760 mm Hg, and thus, the partial pressure of O_2 is about 160 mm Hg. Arterial blood, under these conditions, contains nearly fully saturated hemoglobin . At an alti-tude of 9000 ft. the barometric pressure is about 525 mm Hg, and thus, the partial pressure of oxygen is 110 mm Hg. Under these conditions, there is only 90% saturation of the hemoglobin in arterial blood . It must be understood that 90% saturation does not mean that some hemoglobin molecules are carrying less than four O_2 (recall that each hemoglobin molecule binds four O_2 molecules). The binding of oxygen to hemoglobin is cooperative; that is, the binding of the first oxygen molecule facilitates the binding of the second molecule, which in turn makes it easier to bind the third molecule, and so on. Therefore, through cooperative binding, a hemoglobin molecule binds four O_2 or none at all, and 90% saturation means that 10% of the hemoglobin molecules have no oxygen bound to them at all. The higher the al-titude, the lower the percentage of hemoglobin saturated with oxygen. Thus at high altitudes, hypoxia, or the deficiency of oxygen at the tissue level, becomes a problem. The effects of oxygen deprivation vary with indi-viduals. Most persons who ascend rapidly to altitudes of about 10,000 ft. experience breathlessness , nausea, fatigue, decreased mental proficiency, and ultimately become comatose if the lack of oxygen is too severe. These symptoms will disappear if an individual remains at high altitudes for several days, but full physical capacity will be diminished. The initial acute responses to high altitudes are increased rate and depth of breathing, increased cardiac output and increased heart rate. The latter increases the rate and volume of blood flow to the alveoli,enhancing enhancing CO_2 and O_2 exchange and insuring enough oxygen for every cell. As ventilation increases, thereby in-creasing the oxygen supply, the heart rate returns to normal. During this same time, erythrocyte and hemoglobin syntheses are stimulated by erythropoietin, a hormone produced by the kidney . Consequently, the total circu-lating red cell mass increases considerably . The produc-tion of more red cells, and the synthesis of more hemoglobin together compensate for the low partial pres-sure of oxygen. There are now more cells transporting oxygen, although each cell is carrying less oxygen than at atmospheric pressure. Compared to the rapid increases in respiratory rate and heart rate, erythropoiesis is a slow process. A second adap-tation of the body to an even greater altitude involves an increase in capillaries. The mechanism that stimulates the growth of new capillaries is unknown, but the result is a decrease in the distance which oxygen must diffuse to reach the tissues. Also, a compound 2,3-DPG increases in red blood cells. DPG decreases the affinity of hemoglobin for oxygen which eases the unloading of oxygen at the cells. In time, these three slow compensatory phenomena can once again allow a person to function effectively at higher altitudes.

Question:

A certain recessive gene (r) in a population has a frequency of 0.5. As a result of movement of the popu-lation to a new environment, homozygous recessive in-dividuals (rr) are now selected against, with a loss of 80% of thehomozygotes before maturity. Homozygous dominant (RR) and heterozygotes (Rr) are not affected. What is the frequency of the gene in the population after one generation? Has equilibrium in the new environment been reestablished?

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/Users/wenhuchen/Documents/Crawler/Biology/F26-0704.htm

Solution:

Since there are only two alleles involved, and we know that the sum of their frequencies must be one, we can calculate the frequency of the dominant gene (R) in the population. Letting p equal the frequency of R, and q equal the frequency of r (which is known to be 0.5): p + q = 1 p + q = 1 p = 1 - g p = 1 - g = 1 - 0.5 p = 0.5 We can now use the Hardy-Weinberg binomial expansion to determine the corresponding genotype frequencies after one generation. (p + q)^2 = p^2 + 2pq+ r^2 = (.5)^2 + 2 (.5) (.5) + (.5)^2 = .25 + .50 + .25(=1), wherep^2 = frequency RR 2pq= frequencyRr q^2 = frequencyrr The frequency q^2, however, is only the initial frequency of recessive homozygotes in the offspring. 80% of these will be lost to the population before reproductive age. Thus, the number ofrrindividuals in the population will in effect be only 20% of what it originally was. The effect that this will have on actual gene frequencies can best be seen by first converting the frequencies to actual individual numbers. For convenience sake, let us assume an original population of 100. The original number of rr individuals is (.25) (100) or 25. Loss of 80% of these would give a final effective number of 25 - (.8) (25) = 25 - 20 = 5. The new frequencies now are calculated for an effective total population of 100 - 20 or 80. The number of RR andRrindividuals, 25 and 50 respectively, remains unchanged. The new genotype frequencies are: frequency RR = 25/80= .313 frequencyRr= 50/80= .625 frequencyrr= 5/80= .062 1.0 1.0 The allelic frequencies can be calculated using a new effective allelic population of 2(80) or 160 (two alleles/individual). number R = 2 RR +Rr= 2(25) + 50= 100 number r = 2rr+Rr= 2( 5) + 50=60 160 frequency R = p = 100/160= .625 frequency r = q =60/160= .375 1.0 1.0 Notice that the actual distribution of genotypes doesnotcorrespond to the binomial expansion of the gene frequencies, (p + q)^2 (i.e., .313 \not = p, .625 \not = 2pq, .062 \not = q^2). This is because our population is not in equilibrium, and the Hardy-Weinberg principle applies only to populations in equilibrium. Selection has upset the equilibrium in the population. Until the selection factor, whatever it is, is removed, the frequency of the recessive gene will continue to decrease as it is being removed from the population with each successive genera-tion. We can see, however, that it will never be en-tirely removed, due to the presence ofheterozygotesin the population. Nonetheless, its frequency can be reduced to almost zero if the selection is strong enough, and if enough generations have gone by.

Question:

A series-wound D.C. motor has an internal resistance of 2.0 ohms. When running at full load on a 120 volt line, a current of 4.0 amp is drawn, (a) What is the emf in the armature? (b) What is the power delivered to the motor? (c) What is the rate of dissipation of energy in the motor? (d) What is the mechanical power developed?

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Solution:

(a) As the armature rotates between the poles F and F' of the magnet (see figure 1), an emf is induced in the loops C. According to Lenz's Law, this emf will always be in opposition to the emf applied to run the motor. This induced emf is called the "back emf" since its polarity is opposite to that of the line voltage (the voltage which runs the motor. See Figure 2). Kirchoff's Voltage Law states that the sum of the voltage drops around a circuit loop must be zero. Using this law along with Ohm's Law, \epsilon + IR - V_line = 0 Hence\epsilon = V_line - IR = 120 V - (2\Omega) (4A) = 112 V and opposesV_line , as shown in the diagram. (b)By definition, the power delivered to the motor is equal to the product of the line voltage and the current or P_s = IV_l = (4 amp) (120 volt) = 480 watts. (c)The power dissipated in the motor is P_d = l^2 R = (16 amp^2 ) (2 ohms) = 32 watts. (d)The mechanical power developed is the power supplied to the motor (P_s) less the power dissipated (P_d) P_m = P_s - P_d = 448 watts. The mechanical power developed may also be found from Mechanical Power = back emf × current = (112 v) × (4 amp) = 448 watts.

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Question:

If the average distance the free electrons travel between collisions in copper is 4 × 10^-8 m, how many collisions per second do the electrons make? What is the time between collisions? The average electron speed in copper is 1.21×10^6 m/s.

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/Users/wenhuchen/Documents/Crawler/Physics/D31-0916.htm

Solution:

The number of collisions per second, N, is, N = v/l= [{1.21 × 10^6 (m/s)} / {4 × 10^-8 (m/collision)] = 3.0 × 10^13 collisions/s The average time t between collisions is t = 1/N = [1/(3.0 × 10^13)]s = 3.3 × 10^-14 s.

Question:

A solution contains 1 mg per ml of myosin and 10^14 latex particles per ml. When a given volume of this solution is dried on a grid and viewed under the electron microscope, a typical field contains 122 protein molecules and 10 latex particles. Calculate the molecular weight of myosin.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E22-0802.htm

Solution:

If 1 ml of solution contains 10^14 latex particles and 10 latex particles are present in a certain volume, the following ratio can be set up to solve for this unknown volume. (10^14particles) / (1ml) = (10 particles) / (x ml) x ml = (10 particles × 1 ml) / (10^14 particles) = 10^-13 ml There is 1 mg of protein in ml of solution. Thus in 10^-13 ml, there is 10^-13 mg of protein. It is given that in 10^-13 mg of protein there are 122 molecules. It is known that there are 6.02 × 10^23 molecules per mole (Avogrado's Number), so that the molecular weight of myosin can be found by the following proportion: (10^-16 g) / (122 molecules) = (MW ) / (6.02 × 10^23 molecules / mole) MW = (10^-16 g × 6.02 × 10^23 molecules / mole) / (122 molecules) = 4.93 × 10^5 g / mole.

Question:

At what value or values of r can nodes in the wave function for a 3s electron bound to a nucleus of charge +11 be predicted? Bohr's radius = 0.0529 nm.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E18-0679.htm

Solution:

One must know two important things to answer this question: (1) The wave function expression for a one-electron atom for 3s, and (2) The fact that at a node, \psi or the wave function is zero. If \psi^2 is the probability of finding an electron in a given area and a node is an area of zero probability, then \psi must also be zero. The wave function expression relates \psi to radius r. The wave function for a 3s electron is \psi_300 = [(1)/{81\surd(\pi)] [Z/a_0]^3/2 [27 - {(18 Zr) / a_0} + {(2Z^2 r^2 )/(a^2_0)}] e^-Zr/3(a)0 For \psi300 to be equal to zero the factor [27 - {(18 Zr) /a_0} + {(2Z^2 r^2 )/(a2_0)}] must be equal to zero because the other factors in the equation can never be equal to zero. Using O = 27 - [(18 Zr)/(a_0)] + [(2Z^2 r^2)/(a2_0)] solve for r. O = 27 - [{(18 (11) (r)} / (.0529)] + [{(2(11)^2 (r)^2} / (.0529)^2] = 27 - 3.74 × 10^3 r + 8.67 × 10^4 r^2 Using the quadratic equation one finds that r is equal to 0.0092 or 0.034 nm.

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Question:

What voltage will send a current of 2 amperes through a bell circuit if the resistance of the circuit is 3 ohms?

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/Users/wenhuchen/Documents/Crawler/Physics/D20-0645.htm

Solution:

Since we wish to find the voltage, we use Ohm's Law in the form E = IR. I = 2 amperes,R = 3 ohms Therefore E = 2 amperes × 3 ohm = 6 volts.

Question:

The air pollutant, sulfur dioxide, is considered particular-ly foul. Explain why there is so much concern with this compound.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E24-0851.htm

Solution:

To solve this problem, consider the reactions that can occur between sulfur dioxide (SO_2) and other elements, and/or compounds, in the natural environment. SO_2 can enter the atmosphere from both natural and man-made sources. Besides causing eye irritation and plant damage, it can cause other health problems because of its ability to form acids in the atmosphere. When SO_2 comes in contact with water (H_2O), the following reaction occurs. SO_2 + H_2O \rightarrow H_2SO_3 sulfurous acid SO_2 can also be oxidized in air to sulfur trioxide (SO3) by the reaction 2SO_2 + O_2 \rightarrow 2SO_3. Sulfur trioxide, when exposed to water, produces sulfuric acid by the Reaction SO_3 + H_2O \rightarrow H_2SO_4 sulfuric acid These two acids, sulfuric and sulfurous, account for the corrosion of metal parts exposed to the atmosphere. They are also responsible for the decay of stone statues and buildings. This occurs because most stones consist of CaCO_3 (calcium carbonate), which, upon addition of H_2SO_4, undergoes the following reaction: CaCO_3 + H_2SO_4, \rightarrow CaSO_4, + H_20 + CO_2. CaSO_4 is soluble in water and will wash away with the rain causing the wearing away of the stone.

Question:

A chemist wants to make up a 50 ml solution of pH 1.25HCl. TheHClis labeled 11.6 M. How should she go about the preparation?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E10-0345.htm

Solution:

To answer this question, find how many moles ofHClexist in 50 ml ofpH = 1.25. Once this is determined, one can calculate the volume of concentratedHClneeded. The total volume of water to be added will be thediffer-ence between the total volume of the solution and this amount of concentratedHCl. Thus, proceed as follows: One can find the number of moles ofHClpresent from thepH. pH= - log [H^+]. As given, pH = 1.25. Thus, 1.25 = - log [H^+]. Solving, [H^+] = 5.65 × 10^-2 M. The concentrationof H^+ must be equal to the concentration ofCl^- andHCl, since HCl\rightleftarrows H^+ +Cl^-.Thus, [HCl] = 5.65 × 10^-2 M. The solution is to have a volume of 50 ml or 0.05 liters. (Note: 1000 ml= 1 liter.) M =molarity= moles/liters Thus, the number of moles is equal to the volume in liters ×molarity or (5.65 × 10^-2 M) (0.05 l) = 2.82 × 10^-3 molesHCl. It is given that themolarityof the solution is 11.6 M. One needs 2.82 × 10^-3 moles.Recalling thatmolarity= moles/liter, one finds that [(2.82 × 10^-3 moles)/(11.6 M)] = 2.44 × 10^-4 l = 0.24 ml ofconcentratedHClthat is required. If the total volume of the solution is to be50 ml, then, add (50 - .24 =) 49.76 ml of H_2O to 0.24 ml of the concentratedHCl.

Question:

Find the roots of the equation x^2 + 12 - 85 = 0.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E33-0970.htm

Solution:

The roots of this equation may be found using the quadratic formula x = [{\rule{1em}{1pt}B \pm\surd(B^2 - 4AC)} / {2A}] In this equation A = 1, B = 12, and C = -85. Hence, by the quadratic formula, x = [{\rule{1em}{1pt}12 + \surd(144 + 340)} / {2}]orx = [{\rule{1em}{1pt}12 \rule{1em}{1pt} \surd(144 + 340)} / {2}] x = (\rule{1em}{1pt}12 + 22) / 2orx = (\rule{1em}{1pt}12 - 22) / 2 Therefore x = 5 or x = \rule{1em}{1pt}17. This is equivalent to the statement that the solution set is {\rule{1em}{1pt}17,5}.

Question:

An electron is released from rest at one point in a uniform electric field and moves a distance of 10 cm in 10^-7 sec. What is the electric field strength and what is the voltage between the two points?

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/Users/wenhuchen/Documents/Crawler/Physics/D17-0560.htm

Solution:

To find the electric field strength we divide the force exerted on the test charge, the electron, by the magnitude of the electron. E = F/q We are given the distance the electron travels and the time it takes. We must try to find a relation-ship to use in solving for the unknown F, in order to determine E. Then: For a uniform electric field, the force on the electron is constant; hence, the acceleration, a = F_E/m_e, is also constant. The distance traveled is d = (1/2) at^2, so d = (1/2) at^2 = (1/2) [F_E/m_e] t^2 andF_E = (2m_ed)/t^2 The field strength is the force divided by the charge: E = [F_E/e] = (2m_ed)/et^2 = [2 × (9.1 × 10^-28 g) × 10 cm] / [(4.8 × 10^-10 statC) × (10^-7 sec)^2] = 3.8 × 10^-3 statV/cm. The voltage is V = E × d = (3.8 × 10^-3 statV/cm) × 10 cm = 0.038 statV = 0.038 × 300 = 11.4 V

Question:

Calculate the composition (molality) of an alcohol - water mixture which will not freeze above a temperature of - 10\textdegreeC (+14\textdegreeF). (MW of alcohol = 46; Freezing point constant for water (K_f) = (1.86\textdegree.)

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Solution:

For dilute solutions, the number of degrees that the freezing point is lowered by adding asoluteto a solvent is equal to themolalityof the solute times the freezing point constant (K_f) of the solvent. freezing point depression =Molality×k_f Because the solvent in this case is water, thek_ffor water is used in solving the above equation. The freezing point depression is found by subtracting the new freezing point of the water from the original one. The original freezing point of water is 0\textdegreeC. freezing pt. depression = 0 - (-10) = 10\textdegreeC. Since the freezing point depression and the freezing point constant are known, one can now solve for themolality. Molality= ( freezing point depression) / (freezing point constant) Molality= (10\textdegree) / (1.86\textdegree) = 5.4molal Molalityis defined as the number of moles of solute present per kilogram of solvent. Here, the solute is the alcohol and the solvent is water. The number of grams of the alcohol present in 1000 g of water can be found by multiplying themolalityby the molecular weight of the alcohol. (46 g / mole) × (5.4 moles / kg) of H_2O = (250 g / kg) H_2O Therefore, if 250 g of this alcohol are added to 1000 g of water, the freezing point of the water will be lowered by 10\textdegreeC.

Question:

The ciliates are the fastest moving protozoans. Explain ciliary movement in a ciliate. What other function is served by the cilia?

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/Users/wenhuchen/Documents/Crawler/Biology/F11-0274.htm

Solution:

Ciliary movement consists of an effective and a recovery stroke. As depicted in the accompanying figure, during the effective (power) stroke, the cilium is out-stretched and moves from a forward to a backward position. During the recovery stroke, the cilium, bent over to the right against the body when viewed from above and looking anteriorly, is brought back to the forward position in a counterclockwise movement. The recovery position offers less water resistance. The beating of the cilia is synchronized and waves of ciliary beat progress down the length of the body from anterior to posterior. The di-rection of the waves is slightly oblique, which causes the ciliate to swim in a spiral course and at the same time to rotate on its longitudinal axis. The beating of the cilia can be reversed in direction, and the animal can move backwards. The so called "avoidance reaction" is associated with the backward movement of the ciliate. When the organism comes in con-tact with some undesirable object, the ciliary beat is reversed. It moves backward a short distance, turns, and moves forward again. This avoidance reaction can be repeated. Detection of external stimuli is another function of the cilia and though perhaps all of the cilia can act as sensory receptors in this respect, there are certain long, stiff cilia that play no role in movement and are probably exclusively sensory. The two functions, then, of the cilia are locomotion and sensory reception.

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Question:

A skier descends a slope of 30\textdegree at a constant speed of 15 m/s. His total mass is 80 kg. How much snow melts beneath his skis in 1 min, if the latent heat of fusion of snow is 340 J/g and it is assumed that all the friction goes into melting snow?

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/Users/wenhuchen/Documents/Crawler/Physics/D13-0470.htm

Solution:

When the skier is descending the slope, the forces acting on him are his weight mg^\ding{217} vertically down-ward and the two forces exerted on him by the slope, the normal force N^\ding{217} at right angles to the slope and the frictional force F^\ding{217} opposing the motion (see figure). Since the skier does not rise from the snow and is traveling with constant speed, all forces perpendicular to the slope, and all forces parallel to the slope, must cancel out, as a result of Newton's Second Law. Hence N = mg cos \texttheta and F = mg sin \texttheta. By Newton's third law, an equal and opposite force F^\ding{217} is exerted by the skier on the snow. This equal and opposite force moves its point of application a distance v^\ding{217} in 1 s, where v^\ding{217} is the constant velocity of the skier. Hence the rate of working of the frictional force act-ing on the snow is P = F^\ding{217} \textbullet v^\ding{217} = Fv = mg sin \textthetav = 80 kg × 9.8 m/s^2 × 1/2 × 15 m/s = 5880 W. If all this power is used in melting snow, the energy available per min is Q = 5880 J/s × 60 s/min = 352,800 J/min. But if a mass m of snow is melted per min, the heat required is mL, where L is the latent heat of fusion of snow. Hence m × 340 J/g = 352,800 J/minor m = (5880 × 60 g/min) / 340 = 1038 g/min = 1.038 kg/min.

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Question:

A step-down transformer at the end of a transmission line reduces the voltage from 2400 volts to 120 volts. The power output is 9.0 kw, and the overall efficiency of the trans-former is 92 percent. The primary ("high-tension") winding has 4000 turns. How many turns has the secondary, or "low- tension," coil? What is the power input? What is the current in each of the two coils?

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Solution:

\epsilon_p and \epsilon_s are the voltages induced by the currents I_P and I_S, respectively (as shown in the figure above), \epsilon_P and \epsilon_S are related by the following formula: (\epsilon_P/\epsilon_S) = (N_P/N_S) where N_P and N_S are the number of turns in coils P and S, respectively. This is due to the configuration of the coils in the transformer. The fact that the transformer is step-down indicates \epsilon_P must be the higher voltage, and \epsilon_S the lower. We then have, (2400 v.)/(120 v.) = (4000 turns)/(N_S) Hence N_S = 200 turns. The power output P_S is the power available at the secondary terminals (S), or P_S = 9.0 kw = 9000 watts. But Efficiency = (P_S/P_P) or0.92 = (9000 watts)/(P_P) ThenP_P = 9800 watts. To find the currents, note that P_P = I_P \epsilon_P orI_P = (P_P/\epsilon_P) = (9800 watts)/(2400 volts) = 4.1 amp Similarly, P_S = I_S \epsilon_S orI_S = (P_S/\epsilon_S) = (9000 watts)/(120 volts) = 75 amp.

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Question:

What is a One-to-One Assembler?

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G09-0213.htm

Solution:

One to One Assembler: - The most direct approach to interpreting each line of an assembler language program is as one machine word during assembly. The assembler can be a one pass or two pass assembler. Often, when we hear the term one-to-one, the thought of correspondence comes to mind, in particular, one-to-one correspondence. This is also the case with the One-to-One Assembler; it makes a one-to-one correspondence. This correspondence stems from the fact that a One-to-One assembler translates each line of an assembler language program into one machine word during as-sembly. The translation is done on a sequential basis; an instruction is "mapped" into a machine word as it is en-countered, with the assembler making one or two passes over each instruction, depending on its type. The following is an example of how a One-to-One Assembler may do its trans-lation. The One-to-One Assembler is an example of one of the simplest assemblers. Actually, it forms the foundations on which most practical assemblers are built. For the assem-bler to translate assembly language into machine code the following rules must be observed: For naming locations: may be alphabetic or numeric in any desired mixture. Any other character, such as a comma, is illegal. No character of the name may be a blank. exactly once in the NAME field. be the start of the command field. similar to those of most assemblers. This is not surprising since the One-to-One Assembler forms the basis for most assemblers.

Question:

To measure the wavelength of lines in atomic emission and absorption spectra, one uses a spectroscope. Two lines in the yellow emission spectra of hot Na metal, the so-called sodium-D lines, are used to calibrate this instrument. If the wavelength of one of these lines is 5890 \AA, find the energy of the electronic transition associated with this line, h = 6.62 × 10^-27 erg-sec.

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Solution:

The energy absorbed or released in electron transition is given by the formula E =hѵ, where E = energy, h = Planck's constant and ѵ = frequency. ѵ can be found using the sodium wavelength of 5890 \AA. Wavelength (\lambda), frequency (ѵ), and speed of light (c) are related by the equation\lambdaѵ= c (c = 3.00 × 10^10 cm sec^-1). To find ѵ, substitute these values into the equation and solve. There are 10^-8 cm per \AA. Consequently, 5890 \AA = 5.89 × 10^-5 cm. Substituting, we obtain (5.89 × 10^-5 cm) ѵ = 3 × 10^10 cm/secwhich yields ѵ = 5.093 × 10^14 sec^-1. To find E, note that E =hѵ. Thus, E = h (5.093 × 10^14) = (6.62 × 10^-27 erg-sec) (5.093 × 10^14 sec^-1) = 3.37 × 10^-12 erg.

Question:

An anemometer (a gauge for measuring the pressure or velocity of the wind) is made by attaching cups to each end of a metal rod 50 cm long fixed rigidly to a central vertical column which can rotate freely. A square vertical coil of side 10 cm is attached to the column and the wind speed is measured by finding the emf induced in the coil due to rotation in the earth's magnetic field, (a) Given that the maximum wind velocity to be measured is 120 mph and the maximum induced emf cannot exceed 15 mV, how many turns must the coil have? (b) Calculate the emf induced in the metal rod during rotation at maximum speed. Assume that the horizontal and vertical components of the earth's magnetic induction are 1.5 × 10^-5 Wb \bullet m^-2 and 5.5 ×10^-5 Wb \bullet m^-2 , respectively.

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Solution:

(a) When the anemometer is moving at maximum speed, the cups are moving at 120 mph = 53.64 m \bullet s^-1 . The angular velocity of rotation of the cups is thus \omega = (v/r) [(53.64 m \bullet s^-1 )/(0.25 m)] = 214.6 rad \bullet s^-1 . The plane of the coil is vertical, and thus the vertical component of the earth's magnetic induction produces no effect during rotation, as shown in Fig. 1. Only the horizontal component is involved. The maximum emf induced during rotation of a coil in a magnetic field of induction B is \epsilon_max= n\omegaAB, where n is the number of turns of the coil and A is its area. In this case must not exceed 15 mV. Hence the number of turns on the coil is n = [(\epsilon_max )/(\omegaAB)] = (15 × 10^-3 V)/(214.6 s^-1 × 10^-2 m^2 × 1.5 × 10^-5 Wb \bullet m^-2) = 470 turns. (b) The rod which holds the cups is rotating at right angles to the vertical component of the earth's magnetic induction, and is therefore acting as a crude form of Faraday disk dynamo. (The horizontal component is in the plane of rotation and thus has no effect.) Let us consider an element, dy of the rod at a dis-tance y from the center of the rod as shown in Fig. 2. This element is moving with velocity perpendicular to the vertical component of the earth's magnetic field, he electric field Ey^\ding{217} set up in this element is, Ey^\ding{217} = V_y^\ding{217} × B_v^\ding{217} acting in a radial direction. Since V_y = \omegay, its magnitude is given by Ey = V_y B_v = y \omega B_v . The e.m.f. established between the center and either end of the rod has the magnitude \epsilon = ^a\int_ody Ey = \omegaB_v ^a\int_o dy y = (1/2) \omegaB_v a 2 = (1/2) × 214.6 s^-1 × 5.5 × 10^-5 wb \bullet m^-2 × (0.25)^2 m^2 = 3.69 × 10^-4 v.

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Question:

Write a PL/I program that reads in pairs of points on a Car-tesian coordinate system and classifies them according to the quadrant in which they lie. Make sure you take care of the possibility that one of the pairs lies on the axes of the coordinate system.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G13-0337.htm

Solution:

To solve this problem, we will use three if state-ments, each taking care of a specific task. The first, la-beled IF1, decides whether the calculation should terminate by going to the statement FINISHED, or if the coordinate point lies on the Y-axis. IF2, the second if statement, looks for points that lie on the X-axis. Finally, IF3 classifies all remaining points by quadrant location. As a refresher, the following coordinate system will point out the X- and Y-axes and the quadrant names: Assume that the program ends when the origin point is encountered, which is when X = 0. and Y = 0. Output is in tabular form. POINT_CLASSIFY: PROC OPTIONS(MAIN); PUT PAGE LIST('b'CLASSIFYING POINTS'); PUT SKIP(2) LIST ('X-COORDINATE','Y-COORDINATE', 'CLASSIFICATION'); SEARCH: GET DATA(X,Y) IF1:IF X = 0 THEN IF Y = 0 THEN GO TO FINISHED; ELSE DO; PUT LIST(X,Y,'Y-AXIS')SKIP; GO TO SEARCH; END; IF2: IF Y = 0 THEN DO; PUT LIST(X,Y,'X-AXIS')SKIP; GO TO SEARCH; END; IF3: IF X>0 THEN IF Y>0 THEN PUT LIST(X,Y,'I')SKIP; ELSE PUT LIST(X,Y,'IV')SKIP; ELSE IF Y>0 THEN PUT LIST(X,Y, 'II')SKIP; ELSE PUT LIST(X,Y, 'III')SKIP; GO TO SEARCH; FINISHED:PUT LIST('ORIGIN ENCOUNTERED: DONE')SKIP; END POINT_CLASSIFY;

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Question:

Removal of the pituitary in young animals arrests growth owing to termination of supply of growth hormone. What are the effects of growth hormone in the body? What is acromegaly?

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Solution:

The pituitary, under the influence of the hypothalamus, produces a growth-promoting hormone. One of the major effects of the growth hormone is to promote protein synthesis. It does this by increasing membrane transport of amino acids into cells, and also by stimu-lating RNA synthesis. These two events are essential for protein synthesis. Growth hormone also causes large increases in mitotic activity and cell division. Growth hormone has its most profound effect on bone. It promotes the lengthening of bones by stimulating protein synthesis in the growth centers. The cartilaginous center and bony edge of the epiphyseal plates constitute growth centers in bone. Growth hormone also lengthens bones by increasing the rate of osteoblast (young bone cells) mitosis. Should excess growth hormone be secreted by young animals, perhaps due to a tumor in the pituitary, their growth would be excessive and would result in the production of a giant. Undersecretion of growth hormone in young animals results in stunted growth. Should a tumor arise in an adult animal after the actively growing cartilaginous areas of the long bones have disappeared, further growth in length is impossible. Instead, excessive secretion of growth hormone produces bone thickening in the face, fingers, and toes, and can cause an overgrowth of other organs. Such a condition is known as acromegaly.

Question:

Explain the following statements. (1) An impulse will be initiated only when the stimulus is of a certain intensity. (2) Although a neuron exhibits all-or-none response, an animal is able to detect different degrees of intensity of a stimulus.

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/Users/wenhuchen/Documents/Crawler/Biology/F20-0528.htm

Solution:

The two statements above come from experiments done on an isolated nerve. Two electrodes are placed at points several centimeters apart on the surface of a nerve. The electrodes are connected to recording devices, which are able to detect any electrical changes that may occur at the points on the nerve the electrodes are in contact with. An extremely mild electrical stimulus is applied to the nerve initially. The recording device shows no change. Next, a slightly more intense stimulus is applied, and still there is no change. The intensity of the stimulus is increased further and this time the device records sin electrical change occurring at the point of the membrane in contact with the first electrode and a fraction f a second later, a similar electric change at the point in contact with the second electrode. The nerve is now stimulated, as evidenced by the wave of electrical changes that moves down the fiber, passing first one electrode and then the other. This shows that an impulse, will be generated in a nerve fiber only if the stimulus reaches a certain intensity. If a more intense stimulus is applied next, an impulse is again triggered and a wave of electrical changes moves down the nerve. However, the intensity and speed of this impulse are the same as those of the one recorded from the previous stimulation. Based on these experiments, we are ready to account for the two statements. (1) A potential stimulus must be above a critical intensity if it is to generate an action potential. This critical intensity is known as the threshold value, and it varies in different neurons. Physiologically, what happens is that below the threshold value, the membrane permeability to sodium ions, though increased, is still less than that of potassium ions. This means that the inside of the membrane is still too negative, and no action potential is generated. Only when the intensity of the stimulus reaches the thresh-old value does the membrane permeability of Na^+ ions in-crease so much that its inflow exceeds the potassium out-flow. This inflow of sodium ions is now sufficient to depo-larize the membrane potential to a critical level, known as the threshold potential. Consequently, an action potential is triggered. Once threshold depolarization is reached, the action potential is no longer dependent upon stimulus strength. Thus, action potentials generated by threshold stimuli are identical to those generated by above-threshold (or suprathreshold) stimuli. (2) Increasing the intensity of the stimulus above the threshold value does not alter the intensity or speed of the nervous impulse produced, that is, the nerve fires maximally or not at all, a type of reaction commonly called an all-or-none response. An animal can detect the different degrees of intensity of a stimulus in two ways. The more intense the stimulus, the more frequent are the impulses moving along the fiber. This is called the frequency code. Second, a more intense stimulus ordinarily stimulates a greater number of nerve fibers than does a weak stimulus. This is called re-cruitment or the population code. Hence, the frequency of firing of impulses and the number of axons stimulated can be interpreted by the brain, and the strength of the stimulus can be detected.

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Question:

A parallel plate capacitance is charged (linearly) to 10^\rule{1em}{1pt}3 coul in a time of 10^\rule{1em}{1pt}3 sec. Calculate the displacement current between the plates.

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Solution:

A plot of charge on one plate versus time is shown in the figure. The charge density on a surface of the parallel plate capa-citance, as a function of time, is given by, \sigma(t) = q(t) / A where q is the charge on a plate of A. The field as a function of time is E(t) = \sigma(t) / \epsilon_0 = q(t) / \epsilon_0A The electric flux is then \Phi_E = EA = q(t) / \epsilon0 A change on charge on the plate ∆q will then produce a change of electric flux given by ∆\Phi_E = ∆q / \epsilon0 Dividing both sides by ∆t, we have ∆\PhiE/ ∆t= (1 /\epsilon0)( ∆q / ∆t) This expression relates a change in flux with time to a change in charge with time, ∆q / ∆t can be viewed as a current through the capacitance. It is called a dis-- placement current, I_D to distinguish it from an actual movement of charge, From the graph we see that ∆q = 10\rule{1em}{1pt}3coul \rule{1em}{1pt} 0 coul for ∆t = 10^\rule{1em}{1pt}3 sec\rule{1em}{1pt} 0 sec Therefore I_D = ∆q / ∆t =10^\rule{1em}{1pt}3 / 10\rule{1em}{1pt}3=1 coul / sec

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Question:

A solution of 20.0 g of a non-volatile solute in 100 g of benzene at 30\textdegreeC has a vapor pressure 13.4 torr lower than the vapor pressure of pure benzene. What is the mole fraction of solute? Vapor pressure of benzene at 30\textdegreeC = 121.8 torr.

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Solution:

At constant temperature, the lowering of the vapor pressure by a non-volatile solute is proportional to the concentration of the solute in the solution (the mole fraction). This is called Raoult's Law. The mole fraction of a solute is defined as the number of moles of solute divided by the sum of the number of moles of solute and of solvent. mole fraction of solute = (moles solute) / [(moles solute) + (moles solvent)] Raoult's Law is used to solve this problem. It can be stated: P = P\textdegreeX_2, where P\textdegree is the vapor pressure of the pure solvent, P is the vapor pressure of the solution, and X_2 is the mole fraction of the solute. Using Raoult's Law for this solution: 13.4 torr = (121.8 torr)(X_2) 0.111 = X_2 The mole fraction of the solute is 0.111.

Question:

How much work is done when a man weighing 75 kg (165 lb) climbs the Washington monument, 555 ft high? How many kilocalories must be supplied to do this muscular work, assuming that 25 % of the energy produced by the oxidation of food in the body can be converted into muscular mechanical work?

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Solution:

The energy (work) needed is equal to the potential energy the man attains once he is 555 ft. above the ground. The formula for the potential energy is P.E. = \omega =mgh, where P.E. = \omega, which is the work done, m = mass, g is the gravitational acceleration constant = 9.8 m/s^2, and h is height. Substituting the given values \omega = (75 kg) (9.8 m/s^2) (555 ft) (12 in/ft) [(1 m) / (39.37 in)] = 1.24 × 10^5 (kg m^2 / s^2) = 1.24 × 10^5 J The energy needed is four times greater than the work done. E = 4(1.24 × 10^5 J) / (4.18 J/cal) (10^3 cal/Kcal) = 119 Kcal.

Question:

Develop a BASIC program to compute the least-squares regression line Y = A + BX corresponding to data points (X_i , Y_i) fori= 1,2,...,N. Apply the program to the five data points (X_i , Y_i) given by (1, 1), (2, 2.5), (1.8, 2), (3.1, 2.9), and (0.9,1.1).

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Solution:

We quote without proof the basic results needed from the statistical theory of regression analysis. The coefficients of the least squares line are given by A = [\sumY \sumX^2 - \sumXY \sumX] / [N\sumX^2 - (\sumX)^2] , B = [N \sumXY - \sumX \sumY] / [N\sumX^2 - (\sumX)^2] It is understood that all sums (\sum) apply fori= 1,2,...,n . The program should allow for the input of N, to be followed by N pairs of data, each pair consisting of a value of X and the associated value of Y. Then, compute the least squares approximation to the data. Since we cannot use the sigma notation for variable names in BASIC, we will use the symbol substitution as follows: S1 for \sumX, S2 for \sumY, S3 for \sumXY, S4 for \sumX^2, S5 for (\sumX)^2, S6 for N\sumX^2 . 1\O\O REM LEAST SQUARES ANALYSIS 11\O LET S1 = \O 111 LET S2 = \O 112 LET S3 = \O 113 LET S4 = \O 12\O READ N 131 FOR I=1 TO N 14\O READ X,Y 15\O LET S1 = S1 + X 151 LET S2 = S2 + Y 152 LET S3 = S3 + X \textasteriskcentered Y 153 LET S4 = S4 + X\uparrow2 154 NEXT I 16\O LET S5 = S1\uparrow2 161 LET S6 = N \textasteriskcentered S4 17\O LET A = (S2\textasteriskcenteredS4-S3\textasteriskcenteredS1) / (S6-S5) 171 LET B = (N\textasteriskcenteredS3-S1\textasteriskcenteredS2) / (S6-S5) 184 PRINT "FOR"; N; "SETS OF DATA" 181 PRINT "THE LEAST SQUARES LINEAR" 182 PRINT "EQUATION IS" 183 PRINT "Y ="; A; "+"; B; "X" 8\O\O DATA 5 8\O1 DATA 1,1 8\O2 DATA 2,2.5 8\O3 DATA 1.8,2 8\O4 DATA 3.1,2.9 8\O5 DATA \O.9,1.1 999 END

Question:

Analysis of the exhaust composition of the supersonic transport for one hour of flight yielded the following information: Compound Molecular Weight (g/mole) Mass (g) H_2O 18 8.0 × 10^7 CO_2 44 6.6 × 10^7 CO 28 3.6 × 10^6 NO 30 3.6 × 10^6 Determine the mole fraction of CO_2 in this mixture.

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Solution:

To determine the mole fraction of CO_2 in the mixture, we divide the number of moles of CO_2 by the total number of moles present. The number of moles of each component is found by dividing the mass of that component by its molecular weight. Thus, for each component: moles H_2O = 8.0 × 10^7 g/18 g/mole = 4.4 × 10^6 moles moles CO_2 = 6.6 × 10^7 g/44 g/mole = 1.5 × 10^6 moles moles CO = 3.6 × 10^6 g/28 g/mole = 1.3 × 10^5 moles moles NO = 3.6 × 10^6 g/30 g/mole = 1.2 × 10^5 moles The total number of moles is 4.4 × 10^6 + 1.5 × 10^6 + 1.3 × 10^5 + 1.2 × 105 = 6.2 × 106moles. The mole fraction of CO_2 is then Moles fraction CO_2= (moles CO_2)/(total moles) = (1.5 × 10^6 moles)/(6.2 × 10^6 moles) = 0.24.

Question:

Solve x^2 - 7x - 10 = 0.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E33-0974.htm

Solution:

x^2 - 7x + 10 = 0 is a quadratic equation of the form ax^2 + bx + c = 0 with a = 1, b = \rule{1em}{1pt}7, c = 10. The roots of the equation may be found using the quadratic formula: x = [{\rule{1em}{1pt}b \pm \surd(b^2 - 4ac)} / {2a}]. Substituting values for a,b, and c x = {\rule{1em}{1pt}(\rule{1em}{1pt}7) \pm \surd[(\rule{1em}{1pt}7)^2 - 4(1) (10)]} / {2(1)}. x = {7 \pm \surd[(49 \rule{1em}{1pt} 4(1) (10)]} / {2} x = [{7 \pm \surd9} / {2}] = (7 \pm 3) / 2 x = [{7 + 3} / {2}] = 5 ; x = (7 - 3) / 2 = 2. Check: for x = 5, (5)^2 - 7(5) + 10= 0 25 - 35 + 10= 0 0= 0 for x = 2, (2)^2 - 7(2) + 10= 0 4 - 14 + 10= 0 0= 0 More simply, the problem could have been solved by factoring: x^2 - 7x + 10 = 0 (x - 5) (x - 2) = 0. Set each factor equal to zero to find all values of x which make the product = 0. x - 5 = 0\vertx - 2 = 0 x = 5\vertx = 2

Question:

A man is exposed to a virus. The virus breaks through the body's first and second lines of defense. Does the body have a third line of defense? Outline the sequence of events leading to the destruction of the virus.

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/Users/wenhuchen/Documents/Crawler/Biology/F14-0360.htm

Solution:

The first line of defense against invading pathogens, such as viruses, is the skin and mucous mem-branes. If this line is broken, the second defense mechanism,phagocytosis, takes over. If the virus breaks this line of defense, it enters the circulatory system. Its presence here stimulates specific white blood cells, the lymphocytes, to produce antibodies. These anti-bodies combine specifically with the virus to prevent it from further spreading. Any substance, such as a virus, which stimulates antibody synthesis is called an antigen. Antibodies are not produced because the body realizes that the virus will produce a disease, but because the virus is a foreign sub-stance. An individual is "immune" to a virus or any antigen as long as the specific antibody for that antigen is present in the circulatory system. The study of antigen-antibody interactions is called immunology. The basic sequence of events leading to the destruc-tion of the virus is similar to most antigen-antibody in-teractions. The virus makes contact with a lymphocyte which has a recognition site specific for the virus. The lymphocyte is then stimulated to reproduce rapidly, causing subsequent increased production of antibodies. The antibodies are released and form insoluble complexes with the virus. These insoluble aggregates are then engulfed and destroyed by macrophages.

Question:

During the takeoff roll, a Boeing 747 jumbo jet is accelerating at 4 m/sec^2. If it requires 40 sec to reach takeoff speed, determine the takeoff speed and how far the jet travels on the ground.

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/Users/wenhuchen/Documents/Crawler/Physics/D03-0072.htm

Solution:

The initial speed, v_0 = 0, the acceleration a = 4 m/sec^2, and the time interval of the takeoff, t = 40 sec are given. The unknown observables are the final speed, v, and the distance the plane traveled, d. From the laws of kinematics for constant acceleration v_f = v_0 + at, v_0 = 0 andv_fis the plane's final velocity. Therefore,v_f= at = (4 m/sec^2) (40 sec) = 160 m/sec The plane's takeoff velocity is 160 m/sec in the same direction as the acceleration. The distance s an object with constant acceleration travels in time t is: s= v_0t + (1/2)at^2,v_0 = 0 Hence, s = {(1/2)(4 m/sec^2)} (40 sec)^2 = 3,200 m The plane travels a distance of 3.2 km during the takeoff.

Question:

A is any 20 × 20 array. Write a FUNCTION subprogram to compute PD (A, I, J) = [ A(I-1,J) + A(I+1,J) + A(I,J-1) + A(I,J+1)] / 4 Then use it to compute B_ij = (1-\alpha)B_ij+ \alpha [{Bi-1, j+Bi+1, j+ Bi, j-1+ Bi, j+1} / 4].

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G20-0492.htm

Solution:

There are no special tricks involved here. The function looks as follows: FUNCTION PD (A, I, J) DIMENSION A (20, 20) PD = (A (I-1, J) + A (I+1, J) + A (I, J-1) + A (I, J+1)) / 4.0] RETURN END To computeB_ijthe following statement will appear in the main program. Note, that ALPHA has to be defined before this statement occurs. B(I, J) = [(1.-ALPHA)\textasteriskcenteredB(I, J) + ALPHA\textasteriskcenteredPD(B, I, J)].

Question:

Design a BASIC program to calculate the mean, variance, and standard deviation for a number of observations.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G19-0478.htm

Solution:

The arithmetic mean is known as the first moment about the origin, a measure of central tendency. The standard deviation is the second moment about the origin, and it is a measure of dispersion. The variance is simply the square of the standard deviation. The equations are given below, with M as the mean, V as the variance, S as the standard deviation, and N as the number of observations, or sample size: M =^N\sum_i_=1 x_i / N V = (^N\sum_i_=1 x^2_i / N) - M^2 S = \surd[(^N\sum_i_=1 x^2_i / N) - M^2] The variance is actually the average of squared deviations from the mean, i.e., V = [^N\sum_i(x -x)^2] / N . This may be rewritten as V = {[^N\sum_ix^2] / N} - 2{[^N\sum_ixx] / n} + {[nx^2]/n} = {[^N\sum_ix^2] / N} - 2x^2 +x^2. = {[^N\sum_ix^2] / N} -x^2. This form is easier for computations. Remember that these formulae are to be applied only for observed, ungrouped data. Other formulae are needed for other types of data. 1\O REM N WILL BE COMPUTED BY KEEPING 11 REM A COUNT OF THE NUNBER OF TIMES WE GO THROUGH 12 REM THE LOOP BELOW (STEPS 50 TO 100). T REPRESENTS THE 13 REM SUM \sumx_i , AND D REPRESENTS THE SUM OF THE SQUARES 14 REM \sumx^2_i . N,T AND D ARE INITIALIZED TO ZERO. 15 REM x ONLY TAKES POSITIVE VALUES. HENCE, ON THE LAST DATA CARD 16 REM A NEGATIVE VALUE OF x IS USED TO INSTRUCT THE COMPUTER 17 REM TO EXIT FROM THE 18 REM LOOP AND GO TO STEP 12\O. 2\O LET N = \O 3\O LET T = \O 4\O LET D = \O 5\O READ X 6\O IF X < \O THEN 12\O 7\O LET T = T + X 8\O LET D = D + X\uparrow2 9\O LET N = N + 1 1\O\O GO TO 5\O 11\O REM NOW COMPUTE MEAN, VARIANCE, AND STD.DEV. 12\O LET M = T/N 13\O LET V = D/N - M\uparrow2 14\O LET S = SQR(V) 15\O REM PRINT OUT RESULTS 16\O PRINT "N ="; N 17\O PRINT "MEAN ="; M 18\O PRINT "VARIANCE ="; V 19\O PRINT "STANDARD DEVIATION ="; S 2\O\O DATA 75, 22, 14, 83, 16, 12, 17, -1 21\O END [Note: for small samples, N may have to be amended to N - 1.]

Question:

Consider a point mass M at rest in a non-inertial frame, so that in this frame a = 0. The non-inertial frame rotates uniformly about an axis fixed with respect to an inertial frame. What is the acceleration of M with respect to an inertial reference frame? What is the fictitious force acting on M?

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Solution:

Viewed from an inertial reference frame, M seems to be travelling in a circular path with constant speed. This is so because the non inertial frame, in which M is at rest, is rotating uniformly with respect to the inertial frame. Hence, to restate the first part of the problem, we are looking for the acceleration of a particle undergoing circular motion with constant speed. Now, acceleration is defined as a^\ding{217} = (d^2r^\ding{217})/(dt^2)(1) where r^\ding{217} is the vector from the origin to the particle. We may write r^\ding{217} in terms of its x and y components as r^\ding{217} = xȊ + y\^{\j}(2) where Ȋ and \^{\j} are unit vectors (vectors of magnitude 1) in the positive x and y directions, respectively. We may also write x = r cos \varphi y = r sin \varphi(3) where r is the radius of the circle in which M travels. Substituting (3) in (2) we obtain: r^\ding{217} = r (Ȋ cos \varphi+ \^{\j} sin \varphi) Noting that r, Ȋ, and \^{\j} are constant, and differentiating twice with respect to time, we obtains: (dr^\ding{217}/dt) = r(- Ȋ sin \varphi + \^{\j} cos \varphi)(d\varphi/dt) (d^2r^\ding{217}/dt^2) = r(- Ȋ cos \varphi + \^{\j} sin \varphi)(d\varphi/dt)^2 (d^2r^\ding{217}/dt^2) = - r^\ding{217} (d\varphi/dt)^2(4) We now define (d\varphi/dt) to be equal to \omega, the angular velocity of the particle. (Physically, this is the number of radians the particle traces out per second.) Then, from (4) and (1) a^\ding{217} = (d^2r^\ding{217}/dt^2) = -\omega^2 r^\ding{217}(5) This acceleration, which always points in the -r^\ding{217} direction (towards the center of the circular path), is called the centripetal acceleration. If we now choose to analyze the forces acting on M from the point of view of an observer sitting in the non-inertial frame, we must use a modified form of Newton's Second Law, because the usual form (F^\ding{217} = ma^\ding{217}) only holds in inertial reference frames. The new form of the Second Law is F^\ding{217} - F'^\ding{217} = Ma"^\ding{217}(6) where a"^\ding{217} is the acceleration of the system being examined as recorded in the non-inertial frame, F^\ding{217} is the sum of all "real" forces acting on the system (i.e., tensions, gravity), - F'^\ding{217} is the sum of all "fictitious" forces acting on the system (Carioles forces, centrifugal forces, etc.), and M is the mass of the system. In our example, since M is at rest in the non- inertial system, a"^\ding{217} = 0. Substitut-ing this in equation (6), we find: F^\ding{217} = F'^\ding{217}(7) Now, of course, we must obtain the same value for F^\ding{217} no matter which frame we examine the mass M from, whether it be the rotating frame or the non-rotating frame, because F^\ding{217}represents all real forces acting on M. Since we know nothing about - F'^\ding{217}, let us find F^\ding{217} in the inertial frame, and substitute this F^\ding{217} into (7). In this way, we will be using our knowledge of dynamics in non-rotating frames to foster our knowledge of dynamics in rotating frames, and we will obtain - F'^\ding{217}. In the inertial frame, we may use the standard form of the Second Law, and obtain: F^\ding{217} = Ma^\ding{217}(8) From the first part of this solution we known that a^\ding{217} = - \omega^2r^\ding{217}. Substituting in (8), we find F^\ding{217} = - M\omega^2r^\ding{217}(9) Now, using (7), we obtain F^\ding{217} = F'^\ding{217} = - M\omega^2r^\ding{217} But, - F'^\ding{217} = M\omega^2r^\ding{217}, and this is just the fictitious force acting on M. Note that it points away from the axis of rotation, and, hence, is called a centrifugal force.

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Question:

In one of Edgar Allan Poe's stories, the author describes his terrifying experience of focusing a telescope on a distant hill and observing a "dragon" crawling up it. The punch line comes when he realizes that the dragon is an ant crawling up the windowpane through which he is observing the hill. Explain why this is impossible.

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Solution:

The distant hill is, for all practical purposes, at infinity. The objective lens of the telescope forms an image of it at its second focal point, which is also the first focal point of the eye lens (see figure). The final image is being viewed at infinity. Unless the ant is at infinity or at the second focal point of the objective lens, i.e., unless it occupies the same position as the object or the intermediate image, its final image cannot be at in-finity also, and thus it will not be viewed at the same position as the image of the hill. In fact, assuming that the telescope has focal lengths of 100 cm and 10 cm for its objective and eye lenses, respectively, and that the ant is 200 cm from the objective, we can work out where the final image of the ant lies. Treating the objective, we find that 1/s-_1 + 1/si= 1/f1 where s_1 and s_i are the object and image distances, respectively (1/200cm) + (1/ s_i) = (1/100cm) . 1/s_i= (2/200cm) - (1/200cm) = (1/200cm) si= 200cm . The objective lens and eye lens are 110 cm apart (see figure) and the image formed by the objective will act as a virtual object for the eye lens. The object distance is s2= (110 - 200) cm = - 90cm. -(1/90cm) + 1/ s1/2= 1/10cm 1/s_1/2 = (9/90cm) + (1/90cm) = (10/90cm) ors1/2= 9 cm The final image is a real one on the observer's side of the eye lens and therefore Poe, when he looked through the telescope, could not see an image of the ant.

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Question:

A coil has a self-inductance of 1.26 millihenrys. If the current in the coil increases uniformly from zero to 1 amp in 0.1 sec, find the magnitude and direction of the self- induced emf.

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Solution:

The self-induced emf on the inductance is given by \epsilon = -L (di/dt) \epsilon = (-1.26 × 10^-3 henry) (lamp/.1 sec) . Since 1 henry= [(1 volt \bullet sec)/(amp)] \epsilon = [-1.26 × 10^-3 {(volt \bullet sec)/amp}] (10 amp/sec) = -12.6 millivolts. Since the current is increasing, the direction of this emf is opposite to that of the current.

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Question:

The coil of a galvanometer has 150 turns of mean area 1 cm^2 and the restoring couple of the suspension is 10^-6 N \bullet m per radian. The magnitude of the radial magnetic induction in which the coil swings is 0.2 Wb \bullet m^-2. What deflection will be produced when a current of 10 \muA passes through the coil?

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Solution:

Our first task is to relate the deflection of the galvanometer coil (measured by the angle \texttheta in figure (a)) to the coil characteristics. Before we do this, we briefly review the method of operation of a galvanometer. The coil of the galvanometer is wrapped around a cylindrical form. The shaft of this form (figure (a)) is mechanically connected to two restoring coil springs, present at supports A (figure (b)). The entire coil apparatus is placed between the poles of a permanent magnet. When no current passes through the coil, it is in equilibrium between the poles, as shown by the dotted line in figure (a). Now, any current carrying coil possesses a magnetic moment \mu^\ding{217}. If the current passes through the coil from left to right in figure (a), \mu^\ding{217} will be in the direction of the pointer. How-ever, a magnetic moment in a field of magnetic induction B^\ding{217} will experience a torque T^\ding{217} given by T^\ding{217} = \mu^\ding{217} × B^\ding{217}(1) Since the field lines of B^\ding{217} are radial, the coil will rotate about the form axis. Since the shaft of the form is attached to restoring springs, the motion of the coil will not be unimpeded, and it will tend to return to its equilibrium position. We now show that \texttheta is related to the current I flowing in the coil. From (1), the magnitude of the torque experienced by the coil due to B^\ding{217} is T^\ding{217} = \muB sin \varphi(2) where \varphi is the angle between \mu^\ding{217} and B^\ding{217}. When the coil has attained its maximum deflection, it will rotate no more. Hence, the torque in (2) will then equal the restoring torque due to the springs. This latter torque is given by T' = - k\texttheta(3) where k is the torsional constant of the spring. At \texttheta_max' the net torque is zero, and max T + T' = 0 ork \texttheta_max = \mu B sin 90\textdegree = \muB Now, for a planar coil \mu = nIA where n is the number of turns the coil has, and A is its cross-sectional area. Hence k \texttheta_max = nlABor\texttheta_max = [(nIAB)/(K)] We see that the maximum coil deflection is related to the current it carries. We may therefore use the galvanometer as the basis of a current-measuring device. In our case, \texttheta_max = [(150 × 10^-4m^2 × 10^-5 A × 0.2 Wb \bullet m^-2 )/ (10^-6 N \bullet m)] = 0.03 rad.

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Question:

A rubber ball bounces off a brick wall. It's incident velocity makes a 45\textdegree angle with the normal to the wall at the point of contact. If the collision is inelastic and the ball loses 20% of its kinetic energy during collision what will be its final momentum? Assume there is no gravity or sliding friction between the ball and the wall.

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Solution:

As can be seen in figure A, before the ball strikes the wall it has components of linear momentum, in the - x and -y-directions, of equal magnitude p. In this problem we assume that the ball touches the wall for an infinitesimal time, so that there is no sliding fric-tion between the ball and the wall resulting from the y- component of the ball's momentum. Thus, only the x-component of the balls' momentum is affected by the energy loss. When the ball leaves the wall, its x-component of momentum must be reduced by an amount that will cause the kinetic energy to be reduced by 20%. Kinetic energy and momentum are related by: E_k = 1/2 mv^2 = (mv)^2/2m = (P2_R)/(2m) where p_R is the magnitude of the resultant momentum, kinetic energy after collision = 80% kinetic energy before collision (p2xf+ p^2)/2m = 0.8 [(p^2 + p^2)/2m] + [(0.8 p^2)/m] 0.5 p_xf = 0.3 p^2 p2xf= 3/5 p^2 p_xf = 0.78 p where p_xf is the ball's momentum in the x-direction after collision and p2_xf + p^2 is the square of the magnitude of the final resultant momentum. We now find the balls' final resultant momentum, magnitude of final momentum = \surd(p2_xf + p^2) = \surd[(3/5) p^2 + p^2]^ = \surd[(8/5) p^2] = 1.27 p To find the final momentum's orientation with respect to the normal, we use the laws of geometry (see diagram b). sin \texttheta = p/(1.27 p) = 0.7873 \texttheta \cong 52\textdegree.

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Question:

Give examples of the following editing procedures in APL. i ) Perform editing on a single instruction of a function. ii) Change the name of the header of a given function. iii) Open a function, exhibit a specific line, then direct program control to this line. iv) Erase a function that is contained in your active workspace.

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Solution:

Assume the following program is in the active workspace: \nablaVARIANCE [1]'ENTER SAMPLE ELEMENTS' [2]S \leftarrow X1 X2 .... XN [3]AVG \leftarrow (+/S) \div \Rhos [4]RANGE \leftarrow (\Gamma/S) - L/S [5]NS \leftarrow AVG × (+/S) [6]S [7]S2 \leftarrow S\textasteriskcentered2 [8]SSQ \leftarrow (+/S2) [9]VAR \leftarrow SS2 - NS [10]'THE VARIANCE IS:';VAR [11]\nabla This program finds the variance of a set of numbers. The procedure is as follows: 1) Let X1, X2,......, XN be a series of observations: X =^N\sum_i= 1X_i / N VAR = [\sum(X_i - x)^2]/ [N] = \sumX_i^2 - [\sumX_i]^2 / N. The second expression is more easily evaluated and is also more amenable to treatment by APL than the first. i ) Suppose it is desired to change NS to MSQ (Mean Squares) a more descriptive acronym. Key in \nabla VARIANCE [5 [] 10]. Here [5] is the line number and 10 is the print position, where editing is to begin. Actually NS may be in any print posi-tion (<10) . Backspace until the type ball is below N and enter two slashes followed by a 3. The two slashes erase NS and the 3 allows 3 characters for MSQ. Now, depress the RETURN key. At this point, the fifth line looks as follows: [5]\leftarrow AVG × (+/S) Key in MSQ and then ask for a display of [5]: [5 [] ], which should be [5]MSQ \leftarrow AVG × (+/S) A similar procedure is followed for [9]. ii) Next, suppose it is required to find the standard deviation of the set of numbers. The header can be changed from \nabla VARIANCE to \nablaSDEV as follows: \nabla VARIANCE [0 [] 10] [0]VARIANCE Backspace until the ball is under V and print 8 slashes, followed by 4. Depress the RETURN key to display a blank [0]-line with 4 vacant spaces for a new name. [0] Type SDEV and display the line [0] for a check. It should look like the following [0]SDEV Type DEL to terminate the procedure. To add a line that finds the standard deviation, key in \nablaSDEV[9.5] SDEV \leftarrow VAR\textasteriskcentered.5\nabla iii) Line [10] should be amended to 'THE STANDARD DEVIATION IS:';SDEV Key in \nablaSDEV[10 [] ] to get a display of the tenth line: [10]'THE VARIANCE IS:';VAR Key in [10]'THE SDEV IS:';SDEV [11]\nabla iv) Suppose SDEV is no longer required. The command ) ERASE SDEV deletes it from the active workspace.

Question:

Discuss the absorption of glucose and amino acids. How does it differ from the mechanism of absorption of fatty acids?

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/Users/wenhuchen/Documents/Crawler/Biology/F17-0431.htm

Solution:

Almost all the nutrient products obtained from digestion are absorbed through the walls of the small intestine. The total volume occupied by the small intes-tine is relatively small. However, the internal surface area of the small intestine is immense, allowing for very efficient absorption of digested nutrients. This large surface area is a result of the folds and ridges of the small intestine, compounded with the villi and microvilli that cover its internal surface. The villi are finger-like processes that extend into the lumen of the intestine. The microvilli are hair-like projections of the membrane of cells lining the villi. It is here that specific ab-sorption takes place. When glucose or free amino acids approach the cell membrane, they are bound by special groups of enzymes that are within the membrane. These enzymes quickly transport the molecules through the membrane into the cytoplasm. The nutrients are then transported across the intestinal epithelium into the blood. This occurs in part by simple diffusion, in part by facilitated diffusion, and in part by active transport. The presence of sodium in the lumen is required for monosaccharide transport. The various hexoses are absorbed by active transport. Amino acids are also absorbed by means of active transport. These nutrients are then circulated to the cells of the body, or transported to the liver for storage. Glycerol and fatty acids are not selectively absorbed. After fats are digested in the intestinal lumen, the majority of their products, primarily fatty acids and monoglycerides, simply diffuse across the cell membrane because of their high solubility in membrane lipids. These are then brought into the endoplasmic reticulum (E.R.) of the epithelial cells, where they are resynthesized into triglycerides. These lipids, in turn, combine with speci-fic proteins, forming lipoproteins. The lipoproteins aggregate into droplets called chylomicra within the E.R. The E.R. then fuses with the cell surface and empties the chylomicra into the extracellular space. Here, they are absorbed by the lacteals of the lymphatic system and even-tually transferred into the bloodstream. After a high- fat meal, one's blood may become "milky-looking", due to the high concentration of chylomicra. Note that no energy is required for fat absorption. Another important aspect of absorption is the uptake of partially digested material. When disaccharides and dipeptides are within the vicinity of the cell membrane, the membrane extends itself, forming a pocket around the partially digested molecules. This pocket invaginates into the cell in a process known as pinocytosis. The pinocytotic vesicle fuses with another vesicle called a lysosome. The lysosome contains the cells own digestive enzymes. As these enzymes digest the vesicle's contents, the degraded nutrients diffuse through the membrane into the cytoplasm and are used by the cell.

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Question:

A clock is controlled by a pendulum which correctly beats seconds at 20\textdegreeC. The pendulum is a light iron rod, of coefficient of linear expansion 16X10^-6 C deg^-1 , with a concentrated mass at one end. How many seconds does it lose in a week if the temper-ature is kept at 30\textdegreeC?

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Solution:

If the length of the pendulum is l at 20\textdegreeC, at 30\textdegreeC its length will be l (1 + \alpha × 10 deg), where \alpha is the coefficient of linear expansion of iron. The period of the pendulum is given by T = 2\pi\surd(l/g). Since it is easier to differentiate the log of T with respect to l than to differentiate T directly with respect to l, we obtain logT = log 2\pi (l/g)^1/2 = log (4\pil/g)^1/2 logT = (1/2) log (4\pil/g) (d/dl) [logT] = (d/dl) [(1/2) log (4\pil/g)] (1/T) (dT/dl) = (1/2) (g/4\pil) \textbullet (4\pi/g) (dT/T) = (dl/2l) This is an exact relation betweendTanddl. If \DeltaT and\Deltalrepresent small changes in T and in l re-spectively, then, (\DeltaT/T) \approx (1/2) (\Deltal/l) = (1/2) \alpha × 10 deg. \DeltaT = 2 s × (1/2) × 16 × 10^-6 deg^-1 × 10 deg = 1.6 × 10^-4 s. Note that T = 2s, since each "tick" of the pendulum encompasses 1/2 of its periodic motion. The number of seconds lost in a week is thus (\DeltaT) (#secs, in 1 wk.) = (1.6 × 10^-4) × (302400 s) = 48.4 s.

Question:

What is behavior? Are behavior patterns solely inherited or canthey be modified?

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/Users/wenhuchen/Documents/Crawler/Biology/F31-0789.htm

Solution:

The term behavior refers to the patterns by which organisms surviveand reproduce. The more an organism must actively search the environmentto maintain life, the more advanced are its behavior patterns. Thus a deer shows more complex behavior patterns than does aplanaria. However, even sedentary organisms such as sea coral display forms of behaviornecessary for survival. Behavior is not limited to the animal kingdom; plants display simple behavior, such as the capture of prey by thevenusflytrap. Behavior has a genetic or hereditary basis that is controlled by an organism'sDNA. The anatomical aspects of both the nervous and endocrinesystems are chiefly determined by the genetic composition. These two systems are responsible for most behavioralpheneomena. In general, an organisms 's behavior is principally an ex-pression of the capabilitiesof its nervous system, modi-fied to various extents by its endocrinesystem. Behavioral development is often influenced by experience. This modificationof behavioral patterns through an organism's particular experiencesis called learning. For example, some birds transmit their species-specificsong to their offspring solely by heredity, while others, suchas the chaffinch, must experience the singing of the song by other chaffinchesin order for them to acquire the song. Bothinheritanceandlearning are therefore fun-damental in the determinationof behavioral patterns. Inheritance determines the limits withinwhich a pattern can be modified. The nervous system of a primitive animalcan limit the modification of existing behavior patterns, allowing onlysimple rigid behavioral responses. In highly developed animals such asman, there are fewer purely inherited limits, permitting learning to play asignificant role in determining behavior.

Question:

What is the lowest frequency of the standing sound wave that can be set up between walls that are separated by 25 ft?

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/Users/wenhuchen/Documents/Crawler/Physics/D26-0834.htm

Solution:

A sinusoidal wave that maintains its overall shape between two termination points is called a standing wave. Its amplitude does change with time. For such a wave, the end points have zero amplitude at all times. Zero amplitude points are called nodes and occur every half wavelength. The wave of lowest frequency has the longest wave-length. For a given distance L, the standing wave of longest wavelength has \lambda = 2L with the only nodes occurring at the termination points. Therefore, \lambda = 2L = 2 × 25 ft = 50 ft The frequency of the wave is v = \upsilon/\lambda = (1100 ft/sec) / (50 ft) = 22 Hz which is close to the lowest frequency that can be heard by a human ear. Therefore, a room somewhat larger than 25 ft is necessary in order to set up standing waves of the lowest audible frequency (for example, organ notes of 16 Hz).

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Question:

A mixture consisting only of KOH and Ca(OH)_2 is neutralized by acid. If it takes exactly 0.100 equivalents to neutral-ize 4.221 g of the mixture, what must have been its initial composition by weight?

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Solution:

One equivalent of an acid is the mass of acid required to furnish one mole of H_3O^+ ; one equivalent of a base is the mass of base required to furnish one mole of OH^- or to accept one mole of H_3O^+. Here one must find the mixture of KOH and Ca(OH)_2 that contains 0.100 equivalent of base. There is one equivalent for each OH^- in a molecule of base. Thus, there is 1 equivalent of base for each KOH. There are two for each Ca(0H)_2. Since one is given that there are 0.100 equivalent in 4.221 g of the mixture, one should solve for the equivalent per gram for KOH and Ca(OH)_2. for KOH: MW = 56.1 equiv = 1 (equiv/g) = (1 / 56.1) = 1.78 × 10^-2 equiv/g For Ca(OH)_2 MW = 74.06 equiv = 2 (equiv/g) = (2 / 74.06) = 2.70 × 10^-2 equiv/g. There must be 0.100 equiv present. Let x = number of grams of KOH 4.221 - x = number of grams of Ca(OH)_2 0.100equiv. = [1.78 × 10^-2 (equiv/g)] (x) + [2.70 × 10^-2 (equiv/g)] × (4.221 - x) 0.100equiv. = 0.0178x (equiv/g) + .114 equiv - .0270x (equiv/g) 0.100equiv = - .0092x (equiv/g) + .114 equiv -.014equiv. = - .0092x (equiv/g) [(- .014 equiv)/(- .0092 equiv/g)] = x 1.52 g = x 2.701 g = 4.221 - x The original mixture contains 1.52 g-KOH and 2,701 g Ca(OH)_2 .

Question:

There are genes on theE.colichromosome which determine amino acid synthesis. For one particular strain of E .coli, how can one determine the order of genes for the synthesis of threonine (thr^+), methionine (met^+), histidine (his^+) and arginine (arg^+)? Use your knowledge of con- jugation.

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/Users/wenhuchen/Documents/Crawler/Biology/F05-0138.htm

Solution:

To become an Hfr bacterial cell, the F factor must become integrated into the chromosome. When an Hfr cell and an F^- cell begin conjugation, the F factor portion of the circular Hfr chromosome initiates synthesis of a linear chromosome. This linear chromosome carries a frag- ment of F factor on both its ends, with one end acting as the origin for the transfer of the chromosome. Since the other end of the chromosome contains the remaining portion of the F factor, and since the whole chromosome rarely gets transferred, the F-cell usually does not receive a complete F factor to become F^+. In a particular Hfr strain, the F factor inserts at the same place on the chromosome, so that the origin of transfer is at the same site in all bacteria of this strain. During conjugation, the origin is the first part to travel through the conjugation tube and enter the F^- cell. Usually the conjugating cells separate before the complete transfer of the chromosome. The part that enters the F^- cell may or may not be incorporated into its chromosome. If it does become incorporated, recombination occurs and the F^- cell acquires new traits that are determined by the newly transferred Hfr genes. To determine the order of the genes on the chromosome of a particular Hfr strain, one can use the interrupted mating technique. We must first acquire an F^- strain that is auxotrophic (demonstrating a nutritional requirement) for the four amino acids but is resistant to streptomycin, an antibiotic which killsE.coli. The F^- genotype in con- sideration would be as follows: thr^-, met^-, his^-, arg^-, str^+. The Hfr strain must be prototrophic (can synthesize the amino acids in question) and be sensitive to strepto-mycin. It is also necessary that in the Hfr strain selected, the gene for streptomycin sensitivity be located far from the origin of transfer of the linear, chromosome to avoid its transfer. The Hfr genotype would be thr^+ , met^+ , his^+, arg^+, str^-. The reason for selecting streptomycin sensi-tive and resistant strains will be made evident later. The two cultures of F^- and Hfr cells are then mixed and incubated. At specific time intervals (e.g. every ten minutes), samples are removed from the conjugating mixture and agitated in a blender. This separates the conjugating bacteria to prevent genetic transfer. Since the linear Hfr chromosome always enters the F^- cell in a regular sequence, we can order the genes according to the length of time it took for the Hfr cells to transfer the genes to the F^- cells. For example, in the first ten minutes, only the thr^+ gene might be transferred. In the first fifty minutes, both the thr^+ and his^+ genes get transferred, the arg^+ gene after 80 minutes. The genes can then be located se-quentially on the chromosome by noting their time of trans-fer. For this particular Hfr strain, the chromosome map is as follows: In order to find out which gene is transferred after a certain period of time, the bacterial mixture is plated on four special media: The sample produced after ten minutes will only grow on plate I since the only gene transferred is the thr^+ gene. It cannot grow on the others since it is still auxotrophic for the other three amino acids. The sample after the fifty minutes time interval would grow on both plates I and II. We thus know that the his^+ gene follows the thr^+ gene. The reason why all the plates contain streptomycin is that any Hfr bacteria remaining in the sample placed on the special media plates will be killed. Although F^- cells are strepto-mycin resistant, the only F^- cells that will grow on the special plates are those that have conjugated with Hfr cells and have received and incorporated the genes for determin-ing amino acid synthesis. The streptomycin thus acts as a control to allow only newly-made recombinants to live. An-other relevant point is that in an actual experiment, more media plates with different combinations of amino acid supplements would be required. For example, Plate I, which selected for thr^+, should also check for his^+, or for arg^+, or for met^+ , since we would not know which gene is really first.

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Question:

What is recursion? WhydoesPL/I have a recursive procedurewhile FORTRAN does not?

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Solution:

In pure mathematics, the Fibonacci sequence is a recursive procedure. If you recall, Fibonacci numbers are generated by using prcedingnumbers to continue the sequence. For example, F_0 = 0 F_1 = 1 F_n = F_n - _1 + F_n - _2 for n = 2,3,... isthe method for obtaining Fibonacci numbers. In computer science, recursion takes on a slightly dif-ferent meaning. We may want, in the course of a program, a certain procedure to invokeitself by means of a CALL state-ment. By repeating the calling procedure, you are actually reducing the size of the program. For example, in this PL/I code segment, we want to store values in the array VALUE.We are assuming that we are filling the array from top to bottom bydecrementing the variable KOUNT. When KOUNT is 0, we want to terminate. STORAGE:PROC RECURSIVE IF KOUNT> = 1. THEN DO; VALUE(KOUNT) = INDEX; KOUNT=KOUNT - 1; CALL STORAGE; END; END STORAGE; In this procedure, each stage of processing contains all subsequent stages, i.e., the first stage is incomplete until the final stage has been completed. Using recursion, one can streamline the code, cutting down on repeatedlines of information. At the machine level, recursive procedures mustsave their addresses for subsequent invocations. This is generally accomplishedby a stacking procedure, in which addresses are pushed ontothe stack after an invocation and popped up each time the procedure isin-voked again. There is a danger when using recursive procedures, the danger beingrelated to the saving of machine addresses. Each time a procedure isinvoked recursively, the memory space used to save the addresses containsall of the previous information from the other invocations. If a recursivepro-cedure is not terminated correctly, the memory block will grow indefinitely, consuming valuable space in the machine.

Question:

Besides their function in locomotion and support, bones also serve several other important functions. What are they?

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Solution:

Bones are an important reservoir for certain minerals. The mineral content of bones is constantly being renewed. Roughly all the mineral content of bone is removed and replaced every nine months. Calcium and phosphorus are especially abundant in the bones and these are minerals which must be maintained in the blood at a constant level. When the diet is low in these minerals, they can be withdrawn from the bones to maintain the proper concentration in the blood. Stress seems to be necessary for the maintenance of calcium and phosphate in the bones, for in the absence of stress these minerals pass from the bones into the blood faster than they are taken in. This elevates the blood concentration of these minerals to a very high level, which may ultimately lead to the development of kidney stones. Before special stress exercise programs were developed, astronauts in space often became victims of this type of kidney trouble. During pregnancy, when the demand for minerals to form bones of a growing fetus is great, a woman's own bones may become depleted unless her diet contains more of these minerals than is normally needed. During starvation, the blood can draw on the storehouse of minerals in the bones and maintain life much longer than would be possible without this means of storage. Bones are also important in that they give rise to the fundamental elements of the circulatory system. Bone marrow is the site of production of lymphocyte precursor cells, which play an integral role in the body's immune response system. Red blood cells, or erythrocytes, also originate in the bone marrow. As the erythrocytes mature, they accumulate hemoglobin, the oxygen carrier of blood. Mature erythrocytes, however, are incomplete cells lacking nuclei and the metabolic machinery to synthesize new protein. They are released into the blood-stream, where they circulate for approximately 120 days before being destroyed by the phagocytes. Thus, the bone marrow must perform the constant task of maintaining the level of erythrocytes for the packaging of hemoglobin.

Question:

You are given .01 moles ofAgCl. Assuming theKeqfor Ag(NH_3 )_2 is 1.74 × 10^7 , how much 1M ammonia solution is required to dissolve thisAgCl? Assume, also,K_spofAgClis 2.8 × 10^-10 .

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Solution:

The first thing to do is to write the overall reaction that occurs when ammonia is mixed withAgCl. From this, write an equilibrium expression and equate this expression with the equilibrium constant. Next, calculate the concentrations of species in solution. After this point, there is a volume determination. Following this procedure: whenAgClis dissolved in a solution, it dissociates into ions, since it is a salt. The reaction isAgCl(s) \rightleftarrows Ag+ +Cl- . TheK_sp= 2.8 × 10^10 . The silver ion, Ag+, produced will react with ammonia (NH_3 ) to produce the silver ammonia complex, Ag(NH_3 )_2+ via the re-action Ag+ + 2NH_3 \rightleftarrows Ag(NH_3 )_2^+ with K = 1.74 × 10^7 . The overall reaction is the sum of these two. The K for this will be the pro-duct of the K's for the two reactions just written. Thus, overall re-action is: AgCl_(s) + 2NH_3 \rightleftarrows Ag(NH_3 )_2^+ +Cl- with K = (2.8 × 10^-10 ) ( 1.74 × 10^7 ) = 4.9 × 10-^3 . The equilibrium expression indicates the ratio of products to reactants, each raised to the power of their coefficients in the overall reaction. Thus, K = 4.9 × 10-^3 = {[Ag(NH_3 )_2^+ ] [Cl-]} / [NH_3 ]^2 . One must now find these concentrations. Let x = [Ag(NH_3 )_2^+ ], thus x = [Cl- ], since the overall reaction indicates they are formed inequimolar amounts. Since 2 moles of NH_3 is consumed per mole of complex, [NH_3 ] = 1\Elzbar 2x, where 1 is the originalmolarityof the am-monia solution. Substituting and solving for x , 4.9 × 10-^3 = (x \textbullet x) / (1\Elzbar2x) ; thus, x = .065M. From the initial .01 moles ofAgCl, .01molesofCl- must be produced. This concentration ofCl- is .065M. Since M =molarity = moles/volume, the volume required to dissolve theAgClis (.01moles) / (.065 moles/liter ) = .154 liters = 154 ml.

Question:

A simple pendulum consists of a small object (a so- called bob) hanging from a relatively long cord whose weight is negligible with respect to the bob. The to- and-fro motion of this bob in a vertical plane is called pendulum motion. If the cord is 3 ft long and the suspended bob is drawn back so as to allow the cord to make an angle of 10\textdegree with the vertical before being released, calculate the speed of the bob as it passes through its lowest position.

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Solution:

This problem can be solved by force analysis, but it lends itself most readily to a solution by the energy method. By the principle of conservation of energy, the energy of the bob at the top of its swing must equal its energy at the bottom of its swing. At the top of its swing, the bob is momentarily at rest and it has only potential energy. Taking point A as the reference level for potential energy (see figure), and letting h be the height of the bob at the top of its swing, we may write E_top = mgh At the bottom of its swing (that is, point A), the bob has only kinetic energy. Hence E_bottom = (1/2) mv^2 where v is the bob's speed at A. But E_top = E_bottom ormgh = (1/2) mv^2 whencev = \surd(2gh) To determine h, use the figure and note that it is l - lcos \texttheta = 3 - 3 cos 10\textdegree = 3 - 3 (.985) = 3 - 2.96 = .04 ft. whencev = \surd[2(32 ft/s^2)(.04 ft)] v = \surd[2.56 ft^2/s^2] v = 1.6 ft/s

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Question:

A person suffers a laceration on his arm. How does one know whether an artery or a vein has been severed?

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Solution:

One can distinguish whether an artery or a vein has been cut by observing the flow of blood as it leaves the wound. A severed artery will spurt blood intermittently, whereas in a severed vein, the blood will ooze out smoothly. The difference is due to the existence of pressure pulses in the arteries, which are absent in the veins. The heart forces blood into the arteries only during ventricular systole. Therefore, blood in the arteries tends to flow rapidly during systole and more slowly during diastole. During systole, the force of the contracting left ventricle pushes the blood forward into the aorta. The extra volume of blood distends the walls of the arteries. These distended walls soon contract (during diastole) and squeeze the blood along the artery. The walls do not actively con-tract, but recoil passively as a stretched elastic band does upon release. The arterial pressure continues to propel the blood for-ward. During the next systole, another surge of blood causes the arterial walls to distend. This alternating distention and con-traction along the arterial wall is called the pressure pulse. At the height of the pulse, the pressure (systolic) is usually 120 mm Hg, while at its lowest point, the pressure (diastolic) is usually 80 mm Hg. The difference between the two pressures, 40mm Hg, is called the pulse pressure. These however, are only average values and depend on the interaction of many physical factors. The pressure pulse wave of distention and contraction is trans- mitted along the arterial system. The velocity of transmission along the aorta is about 4 meters per second, and increases to about 8 meters per second along the large arterial branches and up to 35 meters per second in the smallest arteries. The velocity of trans-mission of the pulse is much greater than the velocity of the blood, which is about .3 meter/sec. in the arteries. As the pulse wave moves along the artery, the blood is sped up somewhat by the force of the wave. But the speed of the pulse can be 100 times as fast as the speed of the blood. The pressure pulse becomes less intense as it passes through the smaller arteries and arterioles until it almost disappears in the capillaries. This damping effect is mainly caused by: (1) vascular distensibility, and (2) vascular resistance. The distensibility (expansion or stretching of the vascular walls) of the smaller vessels is greater, so that the small extra volume of blood within the distended section of vessel produces less of a pressure rise. The resistance in the smaller arteries and arterioles impedes the flow of blood and consequently the transmission of pressure. By the time the blood reaches the veins, there is no pressure pulse. The velocity of blood flow is also less in the veins because they have a greater cross-sectional area than do the arteries. Therefore, the blood flow from a severed artery is both spasmodic, due to the pressure pulse, and faster than that from a severed vein.

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Question:

Describe the life cycle of a pine tree.

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Solution:

The pine tree, being a conifer, produces both male and female cones on the same plant. Each cone bears two sporangia on its surface. Within each ovule of the female cone, the nucellus, or megasporangia, contains a single megaspore mother cell. This cell di-vides by meiosis to form four haploid megaspores. Three of the megaspores disintegrate, leaving one functional megaspore which divides mitotically to form a multicellular haploid megagametophyte. The megagametophyte will form two to five archegonia (female sex organs) each containing a single large egg. A similar process of gamete production occurs in the staminate (male) cone. Within each microsporangium are many microspore mother cells, each of which undergoes meiosis to form four haploid microspores. While still within the microsporangium, the microspores divide mitotically to form a four-celled microgametophyte or pollen grain. Upon maturity, the microsporangia burst open and the pollen grains are released and carried by the wind. When a pollen grain reaches an ovulate cone it may shift down between the scales and land on the region of the ovule near the micropyle, which is sticky due to a secretion from the nucellus. As the sticky secretion dries, the pollen grain is pulled through the micropyle. The integument swells and closes around the micropyle. Once inside the micropyle, the pollen grain comes in contact with the end of the nucellus. At this point, one cell of the four-celled pollen grain elongates into a pollen tube, which grows through the nucellus toward the megagametophyte. This cell is known as the tube nucleus. A second cell, known as the generative cell, enters the pollen tube and undergoes a mitotic division. Only one of the daughter nuclei is functional, and undergoes division to form two sperm nuclei. When the end of the pollen tube reaches the neck of an archegonia, it bursts open and discharges its sperm nuclei near the egg. One sperm fuses with the egg to form the diploid nucleus, and the other disintegrates. After fertilization, the zygote, the surrounding megagametophyte tissue, the nucellus, and the integument develop into the seed. The haploid endosperm is derived from megagametophytic tissue. It will nourish the embryo during its early growth and development, in which several leaf like cotyledons, an epicotyl and a hypocotyl form. The embryo then remains dormant until the seed is shed and germinates. Upon germination, it will develop into a mature sporophyte, which is the pine tree.

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Question:

What is the kinetic energy of a 3.0-kg ball whose diameter is 15 cm if it rolls across a level surface with a speed of 2.0 m/sec? (Assume that I for the ball is equal to 2/5 mR^2 , where R is the radius of the ball and m its mass).

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Solution:

Consider a point P of the ball, as shown in the figure. Since the ball is rolling and translating, P has rotational kinetic energy and translational kinetic energy. This is true for every point of the ball. Hence, we may represent the total kinetic energy of the ball as E_k = (1/2) mv^2 + (1/2)I\omega^2 where m is the mass of the ball, v is its velocity, \omega its angular velocity, and I is its moment of inertia about its axis of rotation, O. By definition \omega = v/R = (2.0 m/sec) / (0.075 m) = 27 rad/sec I= (2/5) mR^2 = (2/5) (3.0 kg) (0.075 m)^2 = 6.8 × 10^-3 kg-m^2 E_k = (1/2)(3.0 kg)(2.0 m/sec)^2 + (1/2)(6.8 × 10^-3 kg-m^2) (27 rad/sec)^2 = 2.5 joules

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Question:

Write a basic assembler language program to compute a wo-man's weekly work hours in fixed point binary, and convert the result back to decimal.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G09-0202.htm

Solution:

L4,= F '6'PLACE NO. OF DAYS WORKED IN L4,= F '6'PLACE NO. OF DAYS WORKED IN REGISTER4. SR5, 5CLEAR REGISTER 5 - IT WILL STORE TOTAL HOURS WORKED NXTHRSREADCARD CRDIN, CRDEOFREAD HOURS WORKED PACK DBLE, CRDHRSEACH DAY CONVERT EBCDIC DATA TO PACKED DECIMAL IN DOUBLE-WORD AREA CVB7, DBLCONVERT PACKED DECIMAL TO BINARY IN REGISTER 7 AR5, 7STORE SUM OF HOURS IN RE-GISTER 5 BCT4, NXTHRSDECREASE REGISTER 4 BY 1 AND GO TO INSTRUCTION LABELLED NXTHRS.- CVD 5, DBLECONVERT BINARY HOURS BACK TO DECIMAL. ZAPWKHRS, DBLESTORE HOURS IN AREA CALLED WKHRS CRDE OF \textbullet \textbullet \textbullet \textasteriskcenteredDATA DEFINITION FOR INPUT AREAS CARDINDS\varphiCL80 \varphi \varphi CRDHRSDSCL3 DSCL77 \textasteriskcenteredDATADEFINITION FOR WORK AREAS DBLEDSD WKHRSDSDC 3 You should note that the CVB (convert to binary) and the CVD (convert to decimal) instructions must have a double word (indicated by D in data definition) as the se-cond operand. For the CVB instruction the data in the doubleword is converted to binary and then placed in the register specified as operand 1. The reverse is done for the CVD instruction where data in the register is converted to packed decimal form and placed in the doubleword. The BCT (branch and count) instruction will decrement the con-tents of Register 4 by one each time it is encountered. It will then branch to the label specified as operand 2. When Register 4 finally contains a zero, control will go to the next sequential instruction; the CVD instruction. At this point we have found the total sum of the hours worked for that week.

Question:

List the antagonistic activities of the sympathetic and parasympathetic systems.

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/Users/wenhuchen/Documents/Crawler/Biology/F20-0514.htm

Solution:

Below is tabulated the antagonistic activities of the sympathetic and parasympathetic systems. Organs innervated Sympathetic action Parasympathetic action Heart Strengthens and speeds up heart beat Weakens and slows down heart beat Arteries constricts lumen and raises blood pressure No innervations Digestive track slows peristalsis speeds peristalsis Digestive glands decreases secretion increases secretion Urinary bladder relaxes bladder constricts bladder Bronchial muscles dilates passages, facilitating breathing constricts passages Muscles of iris dilates pupil constricts pupil Muscles attached to hair causes erection of hair No innervations Sweet glands increases secretion No innervations In general, the sympathetic system produces the effects which prepare an animal for emergency situations, such as quickening the heart and breathing rates and dilating the pupil. These alert responses are together termed the fight-or-flight reactions. The parasympathetic systems reverse the fight-or-flight responses to restore an animal to a calm state.

Question:

Describe the parts of a typical flower. What are their functions?

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Solution:

The flower of an angiosperm is a modified stem on which concentric rings of modified leaves are attached. A typical flower consists of four such rings of structures borne on the receptacle, the expanded end of the flower stalk. The stalk of the flower is also known as the pedicel. The outermost ring, usually green and most like ordinary leaves, is composed of sepals. These enclose and protect the flower bud until it is ready to open. Collectively, the sepals are known as the calyx. Internal to the calyx are the petals, which are often brilliantly colored in order to attract those insects or birds which promote pollination. Flowers which are pollinated by the wind need not be attractive to birds and insects, and so their petals tend to be less showy, and may even be absent. Inside the corolla lie the stamens, the male reproductive organs of the flower. Each stamen consists of a slender stalk called the filament, which supports an expanded anther at the tip. The immature anther is composed primarily of pollen sacs. After reaching maturity, the anther contains numerous pollen grains, which produce haploid male gametes upon germination. In the center of the flower is a pistil (carpel) or pistils, the female reproductive organ. Each pistil is composed of a swollen portion, the ovary, at its base; a long, slender stalk which rises from the ovary, the style; and on top of this, an enlarged flattened crown called the stigma. The ovary contains one or more ovules, which are the future seeds of the angiosperm. The stigma's function is to secrete a moist, sticky substance to which the pollen grains will adhere. The style provides a lubricated pathway through which the pollen tube of the germinated pollen grain can grow on its way towards the egg cell.

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Question:

The four-legged land vertebrates (theamphibia, reptiles, birds and mammals) are placed together in thesuperclass Tetrapoda. Which animal was the firsttetrapodand what did it give rise to?

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Solution:

The first successful land vertebrates were the labyrinthodonts, which became extinct in the first part of the Mesozoic Era. They were clumsy, salamander like, ancient amphibians with short necks and heavy muscular tails. These amphibians closely resembled the ancestral lobe- finned fish, but the labyrinthodonts had evolved pentadactyl limbs strong enough to support the weight of the body on land. The labyrinthodonts ranged in size from small, salamander-size animals up to ones as large as crocodiles. They gave rise to other primitive amphibians, to modern frogs and salamanders, and to the earliest reptiles, thecotylosaurs.

Question:

You pass a 1.0 amp current through an electrolytic cell for 1.0 hr. There are 96,500 coul in a Faraday (F). Calculate the number of grams of each of the following that would be deposited at the cathode: (1) Ag from an Ag+ solution, (2) Cu from a Cu^+2 solution and (3) A1 from an Al^3+ solution.

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Solution:

To answer this problem, you can use Faraday's Laws of Electro-lysis. Electrolysis is the phenomenon that occurs when electricity is passed through a solution, such that ions are generated and move toward an anode or cathode. The laws are as follows: Masses of substances in-volved are proportional to the quantity of electricity that flows through the electrolytic cell. Masses of different substances produced during the process are proportional to their equivalent weight. The electrical equivalent is defined as a Faraday (F). It is capable of reducing one equivalent of positive charge, i.e., Avogadro's number of individual unit electric charges. With this information, you can calculate how many Faraday's of electricity were passed through the solution. This tells you the equivalents of the substance that are reduced (recall, one Faraday reduces one equivalent of positive charge). From this number, the weight deposited can be determined. For all 3 parts, the Faradays generated = (1.00 amp)[1coul / sec.) / (amp)](1hr) times (F / 96,500 coul) = .0373 F. (1) Ag from an Ag^+ solution. 1 electron is transferred in Ag^+ + e^- \rightarrow Ag(s). Since one electron is transferred per Ag atom, 1 mole of Ag atoms requires one mole of electrons or 1F. 1 mole of Ag atoms weighs 107.87g (atomic weight - see Periodic Table). You have, though, only .0373 F. Thus, (.0373 mole) (107.87g / mole) = 4.02g of Ag is deposited. (2) Here, 2 electrons are transferred in Cu^2+ + 2e^- \rightarrow Cu(s). Thus, 1F is required for 1/2 of a mole of Cu(s) to be deposited. 1 mole weighs 63.55, 1/2 of a mole = 63.55/2 = 31.775g. Thus, with .0373 F, you can deposit .0373(31.775) = 1.19g. (3) Here, 3 electrons are transferred. Thus, 1F can only deposit 1/3 of a mole. weight deposited = (mw/3) × (.0373) = (26.98/3)(.0373) = .335g.

Question:

A recursive function is the one which uses itself in the body of its definition. Use APL to write anonrecursiveand a recursive function for computing the factorial of a number.

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Solution:

Using the most commonly used mathematical convention, the factorial function is written as n!, where n is a non-negative integer, n! is defined as follows: \mid1,if n = 0 n! =\midn × (n-1) × (n-2) × ..... × 2 × 1,if n \not = 0 Anonrecursivefunction for computing the factorial is as follows: \nablaZ \leftarrow FAC J\mid [1]Z \leftarrow I \leftarrow 1\mid [2]\rightarrow (I \geq J) / 6\mid [3]I \leftarrow I + 1\mid(1) [4]Z \leftarrow Z × I\mid [5]\rightarrow 2\mid [6]Z\mid \nabla\mid An explicit result function was used above to compute factorials nonrecursively . [1] initializes both Z and I to the value 1. If the value of I is greater than or equal to the value of J, then Z is already the value of J! and the program branches to statement [6]. In statement [6], the value of Z which is computed is printed out. The program then halts. But, if the value of I is smaller than J, the condition-al branching does not take place. The program then sequences to statement [3]. In statement [3], the value of I is incre-mented by 1. Statement [4] replaces the value of Z with Z × I. Then statement [5] sends the program back to state-ment [2] to repeat the comparison of I and J. This process is continued until the value of I is greater than or equal to the value of J; J! is then Z. A jump is made to state-ment [6] , the value of Z is printed out, and the program ends. A recursive function for computing the factorial is as follows: \nablaZ \leftarrow REFAC J\mid [1]\rightarrow (J = 0) / GO\mid [2]Z \leftarrow J × REFAC J-1\mid(2) [3]\rightarrow 0\mid [4]GO: Z \leftarrow 1\mid \nabla\mid Again, an explicit result function was used. In [1], if J=0 the program goes to the statementlabelledGO. Here, the value of Z is made equal to 1. The program ends. Thus, the program obtains the value of 0! equal to 1. Note that the program does not ask for the value of Z to be printed out. The value of Z is just stored in memory for future use when required by the programmer. If J \not = O, then step [2] indicates how J! is computed recursively. Notice how the REFAC function calls itself in step [2].

Question:

Aluminum and oxygen react to form AI_2O_3. This oxide has a density = 3.97 g/ml and by chemical analysis is 47.1 weight percent oxygen. The atomic mass of oxygen is 15.9999, what is the atomic mass of aluminum?

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Solution:

You can answer this question by setting up a proportion between the relative number of atoms in the oxide and the percentages by weight of the atoms in the compound. From AI_2O_3, you see there must be 2 atoms of A1 for every 3 atoms of O. You are told that the atomic mass of oxygen is 15.9999. There exist only 2 elements in AI_2O_3. Thus, the weight percent of the aluminum is 100 - 47.1 = 52.9. You have, 2 A1 / 3 O =(weight-percent Al) / (weight-percent O) 2Al / {3(15.9999)} = 52.2 / 47.1 Solving for Al, which is the atomic mass, you obtain Al = 3/2 (15.9999) (52.9 / 47.1) = 26.9 g/mole.

Question:

(a) Calculate the acceleration experienced by the two weights, shown in the figure, if the coefficient of friction between the 32 lb. weight and the plane is 0.2. (b) Calculate also the tension in the cable, whose weight we assume to be negligible.

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Solution:

Consider the two weights as a system. This implies that the cable does not stretch and no internal forces have to be considered, since they consist of action-reaction forces and therefore cancel. The external forces acting on the system are the frictional force on the 32 lb block and the gravitational force on both weights. Since the frictional forceF_f is proportional to the normal force, we first find N. The 32 lb block has no movement perpendicular to the plane. Setting the sum of the forces in this direction equal to zero, we get from the diagram, N - mg cos 37o = 0 Therefore N = 32 cos 37o and F_f = \muN = (0.2)(32 cos 37o) with direction down the plane, since it opposes the motion. All the cable does is change the direction of forces which are applied to each weight and which are in line with the cable. In the same direction as the 64 lb. force on m_1, (the cable makes it so), the 32 lb, weight experiences the frictional force and the component of the gravitational force parallel to the cable or 32 sin 37\textdegree. Applying Newton's second law to the system, we have \sumF = (m_1 + m_2)a since both weights experience the same acceleration. Then 64 - 32 sin 37o - (0.2)(32 cos 37o) = {(64/32) + (32/32)}a 64 - 19.2 - 5.1 = 3a 39.7 = 3a a = 13.2 ft/sec^2(a) The tension is obtained by isolating the 64 lb weight and noting that the only two forces acting on it are T and the force of gravity. Calling the downward direction positive, we obtain from Newton's second law, 64 - T = m_1a = (64/32)(13.2) = 26.4 T= 64 - 26.4 + 37.6 lbs.(b)

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Question:

What volume of ammonia at STP can be obtained when steam is passed over 4000 g of calciumcyanamide? The balanced reaction is CaCN_2 + 3H_2 O \rightarrow 2NH_3 + CaCO_3 (Molecular weight of CaCN_2 = 80, MW of NH_3 = 17.)

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Solution:

From thestoichiometryof the equation 1 mole of CaCN_2 produces 2 moles of NH_3. Thus, if one calculates the number of moles of CaCN_2, then one knows the number of moles of NH_3. To calculate the number of moles of CaCN_2 divide its weight by its molecular weight, thus, the number of moles of CaCN_2 equals [(4000 g CaCN_2 ) / (80 g/mole)] = 50 moles. Therefore, the number of moles of NH_3 produced equals 100 moles. At STP, one mole of any gas occupies 22.4 liters. Hence, the total volume of NH_3 produced at STP is V_(NH)3 = (22.4 liter/mole) (100 mole NH_3) = 2240 liter.

Question:

Would you expect the transfer RNA molecules and the messenger RNA molecules to be the same in the cells of a horse as in similar human cells? Give reasons for your answer in each case.

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Solution:

We have to realize that all tRNA and mRNA molecules are composed of the same chemical constituents, regardless of what species they come from. That is, they all contain the bases adenine, thymine, uracil and cytosine, (tRNA may have other bases in addition) ribose, and phosphate. Observations from biochemical experiments aimed at elucidating the sequence and chemical composition of tRNA from different species have demonstrated that between spe-cies, the different types of tRNA molecules have basically the same nucleotide sequence and the same three-dimensional configuration (cloverleaf-shaped). The reason for this lies in the fact that the function of tRNA is to transfer the amino acids to their correct positions specified by the mRNA. Since all organisms make use of the same 20 amino acids to make their proteins, the tRNA used to transfer these amino acids are basically the same for different species. The different mRNA molecules made by the cells of the same animal differ considerably in length. The great variety in lengths can be explained when we recall that mRNA carries the information for synthesizing different proteins. Different proteins are of different lengths, and therefore, there is a corresponding difference in lengths of the mRNA. When we compare the mRNA from different species, such as man and horse, we have to bear in mind that there exist both equivalent and contrasting systems in the two species. Equivalent systems such as the Krebs cycle and electron transport system, which are almost universal among higher organisms utilize similar enzymes. The digestive system is an example of a contrasting system. The horse is a herbivore, whereas man is an omnivore. Because of the different modes of nutrition, some very different enzymes are involved. We know enzymes, which are protein molecules, are the products of translation of mRNA. When translated, different mRNA molecules will give rise to different enzymes. Because horses and men rely on different enzymes, at least for digestion, we would infer that the mRNA from the two animals are different. However, we must not forget that both animals also use similar enzymes, as in Krebs cycle and electron transport system. Therefore they would also possess some similar if not identical mRNA. In summary, we have determined that tRNA from cells of horse and man are basically the same, whereas the mRNA from the cells of these two animals would show much more difference.

Question:

The first two lines in the Lyman series for hydrogen are 1215.56 \AA and 1025.83 \AA. These lines lie in the ultraviolet region of the spectrum. For each of these lines calculate the following: (a) the corresponding energy in ergs; (b) the corresponding energy in Kcal/mole; (c) the frequency in sec^-1.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E18-0699.htm

Solution:

The probleminvolesthe application of two basicformulas.Thefirst is the relationship between wave-length (\lambda) and frequency (ѵ). The product of the two is the speed of light c,\lambdaѵ= c (or ѵ = c/\lambda). The second is the relationship between frequency and energy (E), E =hѵ, where h is Planck's constant. First calculate the frequencies for the two lines. Distinguish \lambda, ѵ, and E for these lines with the subscripts "1\textquotedblright and ''2". Thus, \lambda_1 = 1215.56 \AAand\lambda_2 = 1025.83 \AA. To express these in the more convenient units of centimeters, use the relationship 1 \AA = 10^-8 cm. Then, \lambda_1 = 1215.56 × 10^-8 cmand\lambda_2 = 1025.83 × 10^-8 cm. Using the first of the two formulas, one obtains: ѵ_1 = (c/\lambda_1) = [(2.9979 × 10^10 cm/sec) / (1215.56 × 10^-8 cm)] = 2.4663 × 10^15 sec^-1 ѵ_2 = (c/\lambda_2) = [(2.9979 × 10^10 cm/sec) / (1025.83 × 10^-8 cm)] = 2.9224 × 10^15 sec^-1. The energies for these two lines are calculated using the second of the two relationships. Planck's constant is h = 6.626 × 10^-27 erg-sec. The units of frequency are sec^-1, the units of energy (found by using E =hѵ), will then be erg-sec × sec^-1 = erg. To convert from ergs to Kcal/mole, first convert to joules (1 joule = 10^7 ergs) , then from joules to Kcal (1 Kcal = 4.184 × 10^3 joules), and finally multiply by Avogadro's number (6.022 × 10^23/mole) in order to put this on a per mole basis. Hence, the conversion factor from ergs to Kcal/mole is 1 erg = 1 erg × 1 joule/10^7 ergs × 1 Kcal/4.184 × 10^3 joules × 6 022 × 10^23/mole = 1.439 × 10^13 Kcal/mole, Applying the second equation one obtains: E_1 = hѵ_1 = 6.626 × 10^-27 erg-sec × 2.4663 × 10^15 sec^-1 = 1.6342 × 10^-11 erg = 1.6342 × 10^-11 erg × 1.439 × 10^13 Kcal/mole-erg = 235.2 Kcal/mole. E_2 = hѵ_2 = 6.626 × 10^-27 erg-sec × 2.9224 × 10^15 sec^-1 = 1.9364 × 10^-11 erg = 1.9364 × 10^-11 erg × 1.439 × 10^13 Kcal/mole-erg = 278.6 Kcal/mole.

Question:

Verify that the following data confirm the law of equivalent proportions: Nitrogen and oxygen react with hydrogen to form ammonia and water, respectively. 4.66 g of nitrogen is required for every gram of hydrogen in ammonia, and 8 g of oxygen for every gram of hydrogen in water. Nitrogen plus oxygen yields NO. Here, 14 g of nitrogen is required for every 16 g of oxygen.

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Solution:

To verify, you must show that when two elements (nitrogen and oxygen) combine with a third element (hydrogen), they will do so in a simple multiple of the proportions in which they combine with each other, Thus, thenitrpgento oxygen ratio in NO must be a ratio of small integers with the nitrogen and oxygen ratio in H_3 N (ammonia)and H_2 O (water). For ammonia and water,(N/O) = [(4.66 g) / (8.00)] = .582 For NO,(N/O) = [(14 g) / (16 g)] = .875. If .582 and .875 are a ratio of small integers to each other, you verify the law of equivalent proportions. Therefore, (.582/.875) = .665 \approx (2/3), concluding that they are a ratio.

Question:

The string of a conical pendulum is 10 ft long and the bob has a mass of 1/2 slug. The pendulum is rotating at 1⁄2 rev\bullets^-1. Find the angle the string makes with the vertical, and also the tension in the string.

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Solution:

Let r be the radius of the horizontal circle traversed by the bob of mass m, l be the length of the string, and T^\ding{217} be the tension which the string exerts on the mass. The forces acting on the bob are the weight mg^\ding{217} downward and the tension T^\ding{217} at an angle \texttheta to the vertical. Resolve T^\ding{217} into horizontal and vertical com-ponents. Applying Newton's Second Law to the horizontal direction of motion F_net = ma where a is the horizontal acceleration of m, and F_net is the net horizontal force on m. Since m is in uniform circular motion, T sin \texttheta provides the centripetal force necessary to keep the bob in the circle. Thus T sin \texttheta = mv^2/r, where v is the velocity of the bob. But v is the distance traveled in 1 s. That is, v = n × 2\pir = 2\pirn, where n is the angular speed in rev\bullets^-1. Also, from the figure, sin \texttheta = r/l. \thereforeT= ( T= ( 4\pi^2 rmn^2)/(r/l) = 4\pi^2mln^2 = 4\pi^2 × 1/2 slug × 10 ft × (1/2 s^-1)^2 = 49 lb. The bob stays in the same horizontal plane, so that the vertical forces must balance. Thus, from Newton's Second Law, T cos \texttheta = mg. \therefore cos \texttheta= mg/(4\pi^2mln^2) \therefore cos \texttheta= mg/(4\pi^2mln^2) = 32/[4\pi^2 × 10 ft × (1/2 s^-1)^2] = 32/[4\pi^2 × 10 ft × (1/2 s^-1)^2] = 0327; = 0327; \therefore\texttheta= 71\textdegree.

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Question:

The molecular weight of nicotine, a colorless oil, is 162.1 and it contains 74.0 % carbon, 8.7 % hydrogen, and 17.3 % nitrogen. Using three significant figures for the atomic weights, calculate the molecular formula of nicotine.

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Solution:

A molecular formula gives the actual composition in number of atoms per molecule. It also summarizes the weight composition of the substance. Hence, to calculate such a formula, the molecular weight, the weight composition of the substance, and the atomic weights of the constituent elements must be known. The weight composition of the substances in nicotine, assuming there are 100 g of it, are C = 74.0 g, H = 8.7 g, N = (110 - 74.0 - 8.7) = 17.3 g. The atomic weights of these substances are C = 12.0 g / mole, H = 1.01 g / mole, N = 14.0 g / mole. The number of moles of each constituent is, therefore, moles C= [(74.0 g) / (12.0 g / mole)] = 6.17 moles moles H= [(8.7 g) / (1.01 g / mole)]= 8.61 moles moles N= [(17.3 g) / (14.0 g / mole)] = 1.24 moles Thus, the mole ratios of C : H : N are 6.17 : 8.61 : 1.24. The empirical formula is C_6.17 H_8.61 N1.24or C_[(6.17)/(1.24)] H_[(8.61)/(1.24)] N_[(1.24)/(1.24)] = C_5H_7N_1. To determine the molecular formula, which is related to the formula weight of the empirical formula by a whole number, divide the molecular weight (MW Nicotine = 162.1) by the formula weight and multiply this whole number times the ratio of atoms in the empirical formula. Thus, the formula weight of C_5H_7N_1= 81.05 the molecular weight of nicotine= 162.1 the whole number =(1.62 / 81.05)= 2. The molecular formula is (C_5H_7N_1)_2 = C_10H_14N_2.

Question:

What is meant by an antagonistic muscle? Give examples.

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Solution:

Muscle can exert a pull but not a push. For this reason, muscles are typically arranged in antagonistic pairs: one pulls a bone in one direction and the other pulls it in the opposite direction. The biceps, for example, bends or flexes the arm and is termed a flexor. Its antagonist, the triceps, straightens or extends the arm and is termed an extensor (see accompanying figure.) Such pairs of opposing extensors and flexors are found at the wrist, ankle and knee, as well as at other joints. When either the flexor or the extensor contracts, its antago-nistic muscle must relax to permit the bone to move. The proper coordination of nerve impulses is necessary for antagonistic pairs to function properly. Other antagonistic pairs of muscles are adductors and abductors, which move parts of the body toward or away from the central axis of the body, respectively; levators and depressors raise or lower parts of the body; and while pronators rotate parts of the body downward and backward, supinators rotate them upward and forward.

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Question:

A car engine working at the rate of 33 horsepower is moving the car at a rate of 45 mi/hr. What force is the engine exerting?

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Solution:

Force can be calculated from the expression for power: P = Fv. Converting horsepower to ft-lb/sec we have P = 33 hp × 550[(ft-lb/sec)/hp] = 18150 ft-lb/sec Since 60 mph = 88ft/sec, 45 mph = 66 ft/sec ThenP= Fv 18150 ft-lb/s= F 66 ft/s F= 275 lbs.

Question:

Develop a program in FORTRAN for the sort.

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Solution:

Say we have two arrays, A and B, and we wish to merge them into one master array C. Assume that A and B are already sorted in ascending order. For example, A (I)B (J)IJ 2611 31122 41333 51444 121955 To merge the two, you must first compare A (l) with B (l). The Smaller of the two will go in position C (1). If they are equal, a random device can decide which element goes first. Next, compare A (2) with B (1) to find the smaller element. This process continues until all of the elements have been compared. The followingpseudocodeillustrates the algorithm For simplicity, let us assume that A and B are 6 elements long; C will contain 12 elements: 1 = 6, J = 6, N = 1 II = 1; JJ = J do while N \leq (length of array C.) if (elements of A have been sorted into C) C (N) = B (JJ) JJ = JJ + 1 if (elements of B have been sorted into C) C (N) = A (II) II = II + 1 else C (N) = B (JJ) JJ = JJ + 1 end if N = N + 1 end do while end program The FORTRAN version of this algorithm is presented below Appropriate comments have been placed throughout the program. As it is written, the program assumes the data has already been stored in the arrays A and B. DIMENSION A (6), B (6), C (12) INTEGER A, B, C DATA II, JJ, N / 6, 6, 1 / CDO WHILE N LESS THAN OR EQUAL TO 12 20IF (N.GT.12) GO TO 98 CIF FINISHED WITH ARRAY A, THEN IF (II.LB.6) GO TO 25 C (N)=B (JJ) JJ=JJ + 1 GO TO 40 25CONTINUE CIF FINISHED WITH ARRAY B, THEN IF (JJ.LE.6) GO TO 30 C (N) = A (II) II = II + l GO TO 40 30CONTINUE CIF A (II) LESS THAN B (JJ), THEN IF (A (II).GE.B (JJ)) GO TO 35 C (N) = A (II) II = II + 1 GO TO 40 CELSE 35CONTINUE C (N) = B (JJ) JJ = JJ + 1 CENDIF 40N = N + 1 GO TO 20 CEND DO-WHILE 98WRITE (5, 100) (C (N), N = 1, 12) 100FORMAT (12 (2x, 13))

Question:

Discuss the mechanism by which the guard cells regulate the opening and closing of a stoma.

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Solution:

The opening and closing of a stoma is regulated by changes in the turgor pressure within the two guard cells that surround the stomatal opening. Each bean-shaped guard cell has a thicker wall on the side toward the stoma called the inner wall, and a thinner wall on the side away from the stomatal opening called the outer wall. Increased turgor pressure causes the cells to bulge. Since the outer walls are thinner than the inner walls, the former stretch more than the latter, causing the cells to bow in and the stoma to open. When the turgor pressure in the guard cells decreases, the inner walls regain their original shape and the cells unbend, closing the stoma as a result. Thus the opening and closing of a stoma depends on a mechanism which varies the turgor pressure within the guard cells. This variation of turgor involves in part the production of glucose and other osmotically active substances in the guard cells themselves. It is believed that light initiates a sequence of enzymatic reactions that lead to the conversion of starch stored in the guard cells into osmotically active glucose molecules, thus increasing the turgor pressure of the guard cells. As light increases the rate of photosynthesis, carbon dioxide is utilized and the decrease in carbon dioxide concent-ration, increases the pH of the guard cells (remember that guard cells contain chloroplasts and so they are able to undergo photosynthesis). This increased basicity stimulates the enzyme phosphorylase to convert starch to glucose-l-phosphate. The glucose-l- phosphate is sub-sequently converted to glucose, increasing the concentration of glucose in the guard cells and causing the influx of water into the cells by osmosis. (Another osmotically active solute may be K+ -potassium ions- which are pumped into the guard cell from the epidermal cell as CO_2 levels decrease.) This phenomenon raises the turgor pressure in the guard cells and causes the stoma to open (see Fig.) In the dark the process is reversed. Thus under normal conditions, the stomata in many species open regularly in the morning and close in the evening.

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Question:

Given the function f(x,y,z) below, write f(x,y,z) as a sum of minterms. f (x,y,z) =xz+yz+ xyz

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G03-0054.htm

Solution:

If a switching function is given in Sum of Products form, it may be expanded to the canonical Sum of Products through repeated use of the Theorem below. THEOREM 1:ab+ab= a Therefore given the function f(x,y,z) =xz+yz+ xyz , f(x,y,z) = xyz+ xyz+ yz + xyz can be obtained, by using theorem 1, and sayingxz= xyz+ xyz. Using Theorem 1 a second time and making the equalityyz= xyz+xyz, the function f(x,y,z) becomes: f(x,y,z) = xyz+ xyz+ xyz+xyz+ xyz At this point a second theorem may be used to reduce the number of terms present; THEOREM 2: a + a = a In the final form of the function it can easily be seen that the first and the third terms are in the same form, xyz. Using Theorem 2 these terms can be reduced as follows; xyz+ xyz= xyz then the final form of the function is; f(x,y,z) = xyz+ xyz+xyz+ xyz. After the desired reduced form is obtained where each term has the same number of variables, a specialmintermcode is used to CODE the function.Mintermcode is basically giving a value of 1 for each true variable, and a value of 0 for each false variable. Taking this definition and applying it to the function gives; xyz+ xyz+xyz+ xyz 110100010111(mintermcode) In decimal arithmetic these terms represent the numbers 6,4,2, and 7. Therefore the desired Sum ofMintermsform is written as; f(x,y,z) = m_6 + m_4 + m_2 + m_7 or f(x,y,z) = \summ (2,4,6,7)

Question:

The elements of a victor V are entered as follows on an APL computer. V \leftarrow 3 4 2 -1. Show what will be displayed as a result of each of the Following statements: a) ×/V,b) \div/V,c) \textasteriskcentered/V d) -/V,e) +/V.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G14-0364.htm

Solution:

a) The value of x/v is obtained as follows: 2 × (-1) = -2 4 × (-2) = -8 3 × (-8) = -24. Therefore -24 is displayed. b) The value of \div/V is obtained as follows: 2 \div (-1) = 2 /-1 = -(2/1) = -2 4 \div (-2) = 4 / -2 = -(4/2) = -2 3 \div (-2) = 3 / -2 = -(3/2) = -1.5. Therefore -1.5 is displayed c) The value of \textasteriskcentered/V is obtained as follows: 2 \textasteriskcentered (-1) = 2^-1 = 1/2^1 = 1/2 = 0.5 4 \textasteriskcentered 0.5 - 4^0.5 = \surd4 = 2. 3 \textasteriskcentered 2 = 3^2 = 9. Therefore 9 is displayed. d) The value of -/V is obtained as follows: 2 - (-1) =2 + 1 = 3 4 - 3 = 1 3 - 1=2. Therefore 2 is displayed. e) The value of +/V is obtained as follows: 2 + (-1) = 2 - 1=1 4 + 1 = 5 3 + 5 = 8. Therefore 8 is displayed.

Question:

At what point should a uniform board 100 cm long be supported so that it balances a 10 gram mass placed at one end, a 60 gram mass on the other end, and a 40 gram mass 30 cm from the 10 gram mass (see figure), (b) What is the magnitude of the supporting force F\ding{217} ?

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/Users/wenhuchen/Documents/Crawler/Physics/D02-0031.htm

Solution:

If the board is to balance, the sum of the moment about any point along the board must equal zero. A torque \cyrchar\cyrt with respect to the point is defined as \cyrchar\cyrt\ding{217} = r\ding{217} × F\ding{217} where the direction of the torque is given by the right hand rule for the vector product of two vectors. For a two dimensional problem, such as this one, we note whether each torque produced is clockwise or count- erclockwise ; with clockwise torques taken as negative. This is done by noting the direction in which the fingers of the right hand must curl in order to swing r\ding{217} into F\ding{217} through the smaller angle \texttheta between them. The magnitude of the torque is given by \cyrchar\cyrt = rF sin \texttheta . For this problem, r\ding{217} and F\ding{217} are perpendicular so that the magnitudes of the torques are just rF. Since the force F\ding{217} produced at the support is unknown, we take moments about this point. In this case, F\ding{217} does not contribute to the net torque since its displacement vector is zero. We have (10)(g)(100 - A) + (40)(g)(100 - A - 30) - (60)(g)(A) = 0 where g is the acceleration due to gravity. We solve for A, the distance of point of support from the 60gm mass: 1000 - 10A + 4000 - 40A - 1200 - 60A = 0 110A = 3800 A = 34.5 cm (b) The force F\ding{217} can be found by applying the first condition of equilibrium in the vertical direction. \sumF_y = 0 = F - (10) (g) - (40) (g) - (60) (g) F = (110) (g) = (110 gm)(980 cm/sec^2) = 1.078 × 10^5 dynes = 1.078 Newtons

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Question:

The vapor pressure of benzene at 75\textdegreeC is 640 torr. A solution of 3.68 g of a solute in 53.0 g benzene has a vapor pressure of 615 torr. Calculate the molecular weight of the solute. (MW of benzene = 78.0.)

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/Users/wenhuchen/Documents/Crawler/Chemistry/E07-0259.htm

Solution:

At constant temperature, the lowering of the vapor pressure by a non-volatile solute is proportional to the concentration of the solute in the solution. Hence, P\textdegree - P = P\textdegreeX_2, where P\textdegree is the original pressure, P is the final pressure and X_2 is the mole fraction of the solute. Here, one solves for the mole fraction of the solute, and from that, one can determine the molecular weight. P\textdegree - P = P\textdegreeX_2X_2 = mole fraction P\textdegree = 640 torr P= 615 torr 640 torr - 615 torr = (640 torr) \bullet X_2 25 torr = (640 torr)\textbullet X_2 25 torr= X_2 mole fraction = .039 Mole fraction is defined as the number of moles of each component divided by the sum of the number of moles in the solution. mole fraction of solute = = (no. of moles of solute) / [(no. of moles of solute) + (no. of moles of benzene)] To find the number of moles of solute, one must first know the number of moles of benzene present. This is found by dividing the number of grams of benzene by the molecular weight of benzene. (MW = 78.) no. of moles of benzene = (no. of grams) / MW no. of moles = (53.0 g) / (78.0 g / mole) = .679 moles One can now solve for the number of moles of solute. Let X = moles of solute mole fraction of solute = (moles solute) / (moles benzene + moles solute) 0.39 = X / (.679 + X) .039(.679 + X) = X 0.265 + .039X = X .265 = .961X .275 = X Thus there are .275 moles of solute present. One is told that there are 3.68 g of this solute, therefore, there are .275 moles in 3.68 g. The molecular weight is found by dividing 3.68 g by 0.275 moles. Molecular weight = (3.68 g) / (0.275) mole = 133.8 (g / mole).

Question:

The teeth are usually considered as a part of the skull although they are formed from invaginations of the outer ectoderm in the embryo and hence should be part of the integumentary system. What are the different kinds of mammalian teeth? What is the structure of a typical human tooth?

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/Users/wenhuchen/Documents/Crawler/Biology/F19-0480.htm

Solution:

The teeth of mammals are differentiated into four types: incisors, canines, premolars, and molars. The incisors are usually simple teeth with a chisel-shaped cutting edge used for biting. In man, there are four incisors in the upper jaw and four in the lower. The canines are usually simple conical teeth which are often greatly enlarged, as in the tusks of the walrus, and are used for tearing food. Man has four canines, one on the side of each jaw. The premolars and molars are often spoken of together as cheek teeth. These teeth have flattened, ridged surfaces, and function in grinding, pounding, and crushing food. Behind the canines in man, there are two premolars and three molars on each side of each jaw. A child's first set of teeth does not include all those mentioned above; the first (or milk) teeth are lost as the child gets older, being replaced with permanent teeth that have been growing in the gums. The final permanent set of teeth totals 32 in man. A typical tooth consists of a crown, neck, and root. The part projecting above the gum is the crown; that surrounded by the gum is the neck; below the neck is the root, embedded in the jawbone. The body of a tooth is composed of dentin, which resembles bone in its structure, chemical components, hardness, and development. The crown is covered with a layer of enamel the hardest substance in the body, and the root is fastened to the jawbone by a layer of cement. The central region of each tooth is the pulp cavity, which contains connective tissue with nerves and blood vessels.

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Question:

Differentiate between hermaphroditism and parthenogenesis.

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/Users/wenhuchen/Documents/Crawler/Biology/F22-0570.htm

Solution:

Parthenogenesis is defined as the development of an egg without fertilization. A classic example of parthenogenesis is found among the Hymenoptera (ants, bees, and wasps). In the honeybees, males (or drones) develop from unfertilized eggs. The queen bee, though having been inseminated, has the ability to lay some eggs that have not been fertilized. The sperm she receives from a male bee (males are all haploid and produce gametes by mitosis) are stored in a pouch connected to her genital tract but closed off by a muscular valve. As the queen lays eggs, she may either open this valve, permitting the sperm to fertilize the eggs, or keep the valve closed, so that the eggs are laid without being fertilized. Fertiliza-tion usually occurs in the fall, and the fertilized eggs are quiescent during the winter. The fertilized eggs become females (queens and workers) and the unfertilized eggs become males (drones). Eggs from species that normally do not exhibit parthenogenetic development may be stimulated artificially to develop without fertilization. This can be done by changing the temperature, pH or salinity of the surrounding media, or by chemical or mechanical stimulation of the egg. Eggs of the sea urchin (an echinoderm) will develop if they are treated with strong salt water, pricked with a needle, or exposed to organic acids. Frog eggs can also be stimu-lated. In addition, it has been possible to stimulate rabbit eggs to successfully develop parthenogenetically provided the cleaving egg that results is placed in the uterus of a hormonally primed female. However only weak and sterile female rabbits have been produced parthenogenetically. Hermaphroditism refers to the presence within a given individual of both male and female sex organs. Many of the lower animals are hermaphroditic. Some, such as the parasitic flukes, are self-fertilizing, but more often two hermaphroditic animals copulate and each inseminates the other. This latter method is used by earthworms. In other species, self-fertilization is prevented by the development of the testes and ovaries at different times.

Question:

When silver is irradiated with ultraviolet light of wavelength 1000 \AA, a potential of 7.7 volts is required to retard completely the photoelectrons. What is the work function of silver?

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/Users/wenhuchen/Documents/Crawler/Physics/D33-0978.htm

Solution:

First, the energy of a 1000-\AA photon is E = h\Elzpscrv = (hc/\lambda) = [{(6.6×10^-27 erg-sec) × (3×10^10 cm/sec)} / {(1.0×10^-5 cm) × (1.6 ×10^-12 erg/eV)}] = 12.4 eV Note that if 7.7 eV stops all photoelectrons, it will stop the surface photoelectrons also. Hence we may use the photoelectric effect equation, \textphi = h\Elzpscrv - (1/2) mv^2_max(1) The electron looses kinetic energy and gains potential energy as it moves through the retarding potential (similar to the conversion of kinetic energy to gravitational po-tential energy when an object is thrown up from the ground). At plate A, the electron has only kinetic energy. At point B, the electron has both kinetic energy and potential energy. At plate C, the electron has only potential energy, because we are told that a potential of 7.7 volts will completely retard the motion of the photoelectrons. Hence, by the principle of conservation of energy, we relate the energy of the photoelectrons at A and C, or (1/2) mv^2_max = P.E.@ C. But, by the definition of potential energy, (W), we have V_c - V_a = (W_ac/q) where V_c - V_a is the potential difference of the plates, and q is the charge transported between A and C (a photo-electron, in our case). Hence (1/2) mv^2_max = q (V_c - V_a) Substituting this in (1), we obtain \textphi = h\Elzpscrv - e (V_c - V_a)(2) Putting the given data in (2), \textphi = 12.4 eV - [(-1) (0 - 7.7)] eV where we have used the fact that e = - 1.6 × 10^-19 C. Because 1.6 × 10^-19 J = 1 eV, \textphi = 12.4 eV - 7.7 eV = 4.7 eV (Note that we have used the fact that the energy of an electron that is completely stopped by a potential of 7.7 volts is just 7.7 electron volts.)

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Question:

Describe the various types of evidence from living organisms whichsupport the theory of evolution.

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/Users/wenhuchen/Documents/Crawler/Biology/F28-0742.htm

Solution:

There are five lines of evidence from living organisms that support thetheory of evolution. First, there is the evidence from taxonomy. The charac-teristicsof living things differ in so orderly a pattern that they can be fittedinto a hierarchical scheme of categories. Our present, well- establishedclassification scheme of living organisms, developed by CarolusLinnaeus in the 1750's, groups organisms into the Kingdom, Phylum or Division, Class, Order, Family, Genus, and Species.The relationshipsbetween organisms evident in this scheme indicates evolutionarydevelopment. If the kinds of plants and animals were not relatedby evolution-ary descent, their characteristics would most probably bedistributed in a confused, random fashion, and a well- organized classificationscheme would be impossible. Secondly, there is the evidencefrom morphology. Comparisons of the structures of groups of organismsshow that their organ systems have a fundamentally similar patternthat is varied to some extent among the members of a given phylum. This is readily exemplified by the structures of the skeletal, circulatory, and excretory systems of the verteb-rates. The observation of homologousorgans - organs that are basically similar in their structures, siteof occur-rence in the body, and embryonic development, but are adaptedfor quite different functions, provides a strong argument for a commonancestral origin. In addition, the presence of vestigial organs, whichare useless or degenerate structures found in the body, points to theexistence of some ancestral forms in which these organs were once functional. Thirdly, there is the evidence from comparative biochemistry. For example, the degree of similarity between the plasma proteins of variousanimal groups, tested by an antigen-antibody technique, indicates anevolutionary relationship between these groups. Fourthly, embryologicalstructures and development further support the occurrence ofevolution. Different animal groups have been shown to have a similar embryologicalform. It is now clear that at certain stages of development, theembryos of the higher animals resemble the embryos of lower forms. The similarity in the early developmental stage of all vertebrate embryos indicatesthat the various vertebrate groups must have evolved from a commonancestral form. Finally, there is the evidence from genetics. Breeding experiments and results demonstrate that species are not unchangeablebiologic entities which were created separately, but groups oforganisms that have arisen from other species and that can give rise to stillothers.

Question:

A hunter enters a lion's den and stands on the end of a concealed uniform trapdoor of weight 50 lb freely pivoted at a distance x from the other end. Given that the hunter's weight is 150 lb, what fraction of the total length must x be in order that he and the end of the trap-door shall start dropping into the depths with acceleration g when the trapdoor is released?

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/Users/wenhuchen/Documents/Crawler/Physics/D04-0130.htm

Solution:

The forces acting on the trapdoor of length a are its weight Mg acting downward at the center, since it is uniform, the hunter's weight mg acting downward at the end, and the normal force N exerted by the pivot upward (see the figure ). When we take moments about the pivot, counterclockwise moments being taken as positive, the moments causing rotational acceleration are -Mg(a/2 - x) - mg(a - x), and thus using the rigid body analog of Newton's second law, if T is the resultant torque acting on the rigid bar, I_T, the moment of inertia of the man-bar system about the pivot, and \alpha its angu-lar acceleration, then T = I_T\alpha a or -Mg(a/2 - x) - mg(a - x) = I_T\alpha The moment of inertia of the trapdoor about a horizontal line parallel to the pivot and passing through the center of gravity is (1/12)Ma^2 . By the parallel-axis theorem, if I_cmis the moment of inertia of the bar (or any rigid body) about its center of gravity and I_b is its moment of iner-tia about any axis, where h is the distance separating the two axes, then I_b = I_cm + Mh^2. Thus, I-_b about the pivot is (1/12) Ma^2 + M[(a/2) - x]^2. The moment of inertia of the hunter about the pivot is I_m = m(a-x)^2 . Hence T = (I_b + I_m)\alpha for I_T = I_b + I_m -Mg(a/2 - x) -mg(a - x) = \alpha[(1/12)Ma^2 + M(a/2 - x)^2 + m(a - x)^2].(1) If the hunter and the end of the trapdoor are to have a linear acceleration g downward, then -g = (a - x)\alpha. Where (a - x) is the radius of the circular arc path the man fol-lows about the pivot (see the figure). Therefore, upon division of both sides of equation (1) by \alpha and using the above relation between g and \alpha, M(a/2 - x)(a - x) + m(a - x)^2 = (1/12)Ma^2 + M(a/2 - x)^2 + m(a - x)^2. M(a/2 - x)(a - x) = (1/12)Ma^2 + M(a/2 - x)^2 a^2/2 - ax - ax/2 + x^2 = (1/12)a^2 + a^2/4 - ax + x^2 (a^2/2 - a^2/4) - (3/2)ax + ax = (1/12)a^2 a^2/4 - ax/2 = (1/12)a^2 \therefore a/2(a/2 - x) = (1/12)a^2 \therefore 3a^2 - 6ax = a^2 \therefore x = (1/3)a The pivot must be located one-third of the length of the trapdoor from the end.

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Question:

What is the Lorentz contraction of an automobile traveling at 60 mph ? (60 mph is equivalent to 2682 cm/sec.)

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/Users/wenhuchen/Documents/Crawler/Physics/D32-0937.htm

Solution:

Suppose we are given two frames of reference moving relative to one another with a velocity v (see figure). If we are dealing with classical physics and want to relate the coordinates of an event occurring in the S - frame (x, y, z , t) to the coordinates of an event occurring in the S'-frame (x', y', z', t'), we use the Galilean trans-formation , or x' = x - vt y' = y (If v is in the x-direction only.) z' = z t' = t. In relativistic physics, this transformation is invalid, and must be replaced by the Lorentz transformation , or x' = (x - vt) / \surd{1 - (v^2 / c^2)} y' = y z' = z t' = {t - (vx / c^2)} / \surd{1 - (v2/ c^2)} . Now, we may relate distances measured in S' to distances as measured in S. Let us imagine the measurement of a dis-tance parallel to the x'-axis in the S' frame. In order to measure the length of a rod in S, we must locate both ends of the rod (x_1, x_2) at the same time (t_1 = t_2 ) in S. Hence, the length in S' is x'_2 -x'1= [(x_2 - x_1) - {v (t_2 - t_1)}] / [\surd{1 - (v^2 / c^2)}] . But t1= t2, therefore x'_2 -x'1= (x_2 - x_1) / \surd{1 - (v^2 / c^2)} . Hence (x_2 - x_1) = x'2-x'1\surd{1 - (v^2 / c^2) }.(1) Since\surd{1 - (v^2 / c^2)} < 1 we then have x_2 - x1< x'2-x'_1 . The observerin S measures a smaller rod length (contracted) than the observer in the rod's rest frame, S'. Now, we calculate the length of the car in S, ( x_2 - x_1)If V_r is 2682 cm/sec. Vr/ c = 2682 / (3 × 10^10) = 8.94 × 10^-8 {V_r/ c}^2 = 8.0 × 10^-15 . When x is very much less than 1, \surd(1 - x) = [1 - {(1/2)x} approximately . Therefore , \surd{1 - (v_r / c)^2 } \approx [1 - (4.0 × 10^-15)]. Substituting in (1) x_2 - x_1\approx (x'2-x'_1) (1 - 4.0 × 10^-15) . This means that the change in length of a meter rule is only 4.0 × 10-15 meters, or 4.0 × 10^-13 cm. Since the diameter of an atom is about 10^-8 cm, the diameter of a nucleus is about 10^-12 cm and the size of the electron is about 10^-13 cm, this contraction is clearly negligible. Again we see that the difference between relativistic and classical physics is not important for the velocities we are normally concerned with.

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Question:

Exactly one mole of gaseous methane is oxidized at fixed volume and at 25\textdegreeC according to the reaction CH_4(g) + 2O_2 (g) \rightarrow CO_2(g) + 2H_2O(l). If 212 Kcal is liberated, what is the change in enthalpy, ∆H?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E14-0513.htm

Solution:

We must first find the change in internal energy, ∆E, before applying the equation, ∆H = ∆E + ∆nRT, to find the enthalpy change (∆n = number of moles of gaseous products - number of moles of gaseous reactants, R = 1.987 cal/deg-mole, T = absolute temperature). Since heat was evolved (exothermic reaction) , ∆E must be negative. Furthermore, since no useful work was done (∆V \not = 0 if work is done), ∆E is equal in magnitude to the quantity of heat evolved. Thus ∆E = - 212 Kcal. Since there is one mole of gaseous product (1 CO_2 (g)) and three moles of gaseous reactants (2O_2(g) + 1 CH_4 (g)), ∆n - 1 - 3 = - 2 mole. The absolute temperature is T = 25 + 273 = 298\textdegreeK. Hence ∆H= ∆E + ∆nRT = - 212 Kcal + (- 2 mole)(1.987 cal/deg-mole)(298\textdegreeK) = - 212 Kcal - 1180 cal = - 212 Kcal - 1.180 Kcal \cong - 213 Kcal.

Question:

One cubic millimeter of oil is spread on the surface of water so that the oil film has an area of 1 square meter. What is the thickness of the oil film in angstrom units?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E01-0003.htm

Solution:

Since one is asked to give the final thickness of the film in angstroms, it is useful to convert the other dimensions given to angstroms first. 1\AA = 10\Elzbar10m = 10\Elzbar7mm. Therefore, 1 mm = 1\AA/10\Elzbar7= 10^7\AA. Cubing both sides of this equation gives the number of cubic angstroms in 1 cubic millimeter. (1 mm)^3 = (10^7\AA)^3 1 mm^3 =10^21 \AA^3= volume of oil. The final area of the film is given as 1 m^2. One knows that 1 m = 10^10 \AA; therefore, 1 m^2 = (10^10 \AA)^2 =10^20\AA^2. The volume is equal to the area of the film multiplied by the thickness. Thus, one can find the thickness of the film by dividing the volume by the area. thickness = (10^21 \AA^3 ) / (10^20 \AA^2 ) = 10 \AA.

Question:

Write FORTRAN subprograms to simulate the action of the following forcing inputs, E(t): Case 1:E_1 (t) = h where h is constant. E_2(t)is a pulse of height h and width (t_2 - t_1) . Case 3:E_3(t) = A sin(\omegat + \varphi) where A,\omega, \varphi are constants. \omega \omega E_3(t) is a sinusoid of amplitude A, period 2\pi/\omegaand phase \omega angle \varphi. and E_4(t) = E,(t-PW), if t \geq t---_1 + PW . E_4(t) is a pulse train with height h, width W and period PW.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G24-0575.htm

Solution:

Use FUNCTION subprograms with t as the argument. The parameters can be read in, Case 1:FUNCTION CONST(T) READ, H CONST = H RETURN END Case 2:FUNCTION PULSE (T) READ, H, T1, T2 PULSE = 0 IF(T.GE.T1. AND. T.LT,T2) PULSE = H RETURN END Case 3:FUNCTION SINE (T) READ A, OMEGA, PHI SINE = A \textasteriskcentered SIN(OMEGA\textasteriskcenteredT + PHI) RETURN END Case 4:The formula for E_4(t) is a recursion formula that defines E_4(t) for t < t_1 + PW and equates E_4(t) with E_4(t-PW) for t \geq t_1 + PW, thus requiring repeated subtraction of PW until t < t_1 + PW. Then t will be mapped into t\textasteriskcentered wheret_1 \leq t\textasteriskcentered < T_1 + PW. This mapping is a translation of t along the time axis to the point t\textasteriskcentered; and the distance of the translation is t - t\textasteriskcentered = nPW(1) where n is the number of subtractions required. (See Fig. 2). But repeated subtraction is the operation of division. Therefore, the same translation mapping can be achieved by dividing the distance t - t_1 by PW: (t - t_1) / PW = n + (r +PW)(2) where r is the remainder. Multiplying both sides of equation (2) by PW: t - t_1 = nPW + r . Substituting, using equation (1) t - t_1 = t - t\textasteriskcentered + r - t_1 = - t\textasteriskcentered + r r=t\textasteriskcentered - t_1.(3) Equation (3) indicates that the remainder gives t\textasteriskcentered's distance from t_1. Therefore, adding r to t_1 gives the absolute location of t\textasteriskcentered. FUNCTION TRAIN(T) READ H, T1, W, P TIME = T IF (T.GE. (T1 + P\textasteriskcenteredW)) GO TO 5 GO TO 6 TIME = AM0D ((T-T1)/(P\textasteriskcenteredW))\textasteriskcenteredP\textasteriskcenteredW + T1 IF( (TIME.LT.T1) .0R. (TIME.GE. (T1-W))) TRAIN = 0 . IF((TIME. GE.T1). AND. (TIME.LT. (T1+W))) TRAIN = H. RETURN END

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Question:

A sonar device emits waves of frequency 40,000 cycles\bullet s^-1. The velocities of the wave in air and water are 1100 ft \bullet s^-1 and 4200 ft \bullet s^-1, respectively. What are the frequency of the wave in air and the wavelengths in air and water? Suppose that the device is fixed to the bottom of a ship. It emits a signal and the echo from the ocean bed returns 0.8 s later. What is the depth of the ocean at that point?

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Solution:

The frequency of the waves emitted is the same in air or water. The surrounding medium has no influence on the vibration mechanism. Since \lambda = c/f where c is thewavespeed, and f and \lambda are the fre-quency and wavelength of the sound, respectively, we have (a) in air \lambda = (1100 ft \bullet s^-1)/(4 × 10^4 s^-1) = 2.75 × 10^-2 ft and (b) in water \lambda' = (4200 ft \bullet s^-1)/( 4 × 10^4 s^-1) = 10.50 × 10^-2 ft. Since the velocity of sound in water is 4200 ft \bullet s^-1 and the echo returns in 0.8 s after traversing 2d, where d is the ocean depth at that point, we have s = ct or 2d = 4200 ft \bullets^-1 × 0.8 s. \therefored = 1680 ft.

Question:

Astronomers have found that many stellar objects contain hydrogen atoms that emit photons. Find the wavelength of this radiation and express the value in centimeters (the conventional unit used by astronomers to measure radio photon spectra) if the approximate value for the frequency of theemmittedphotons is 1.42 × 10^9 Hz.

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Solution:

The wavelength \lambda of the photon and the frequency f are related in the following manner: \lambda = (c/f) where c is the speed of light. Hence, \lambda = [(3 × 10^8 m/s)/(1.42 × 10^9Hz) = 2.1 × 10^-1 m Now 1 m is equivalent to 10^2 centimeters (cm); therefore, \lambda =2.1 × 10^-1n10^2 cm/n= 21 cm

Question:

The sedimentation and diffusion coefficients for hemoglobin corrected to 20\textdegree in water are 4.41 × 10^-13 sec and (6.3 × 10^-11 m^2/s), respectively. Ifѵ= (.749 cm^3/g) and \Rho_(H)2 O = (0.998 g/cm^3) at this temperature, calculate the molecular weight of the protein. If there is 1 g-atom of iron per 17,000g of protein, how many atoms of iron are there per hemoglobin molecule?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E22-0821.htm

Solution:

One can calculate molecular weight by using the Svedberg equation. This equation is written M = [RTs] / [D (1 -ѵ\rho)] where M is the molecular weight, R is the gas constant in ergs (8.314 × 10^7 ergs/mole \textdegreeK) , T is the absolute tem-perature, s is the sedimentation coefficient, D is the diffusion coefficient, \rho is the density of the solvent and ѵ is the partial specific volume of the protein. Solving for Ms M = [(8.314 × 10^7 ergs/mole \textdegreeK) (293\textdegreeK) (4.41 × 10^-13 sec)] / [(6.3 × 10^-11 m^2/s) {1-(.0.749 cm^3/g) (,998 g/cm^3)}] = 6.75 × 10^8 ergs (s^2 / m^2 mole) = 6.75 × 10^8 [g cm^2 / sec^2] [sec2/ m^2 mole][1 m^2 / 10^4 cm^2] = 6.75 × 10^4 (g/mole) 1ergs = (g cm^2 / s^2);1 m^2 = 10^9 cm^2 To determine the number of Fe atoms per molecule of hemoglobin, calculate the number of moles of hemoglobin in 17000 g. It is given that there is 1 mole of Fe in g of Hgb. The number of Fe atoms per molecule is then equal to the number of moles of Fe divided by the number of moles of hemoglobin. No. of moles of Hgb= (17,00 0 g) / (6.75 × 10^4 g/mole) = 2.52 × 10^-1 moles No. of Fe atoms per Hgb molecule = - [(1mole) / (2.52 × 10^-1) = 4.0 atoms.

Question:

How is the Bohr effect of physiological importance?

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Solution:

Both free protons (H^+) and carbon dioxide (CO_2) are known to promote the release of oxygen from hemo-globin. This is known as the Bohr effect. Myoglobin shows no change in oxygen binding over a range of pHs or carbon dioxide concentrations. But in hemoglobin, acidity enhances the release of oxygen. Lowering the pH shifts the oxygen dissociation curve to the right (see Figure). A shift to the right enhances oxyhemoglobin dissociation. For example, at any partial pressure of oxygen, the curve at the lower pH is lower, signifying that less oxygen is bound to the hemoglobin molecule (less % saturation) as a result of a decreased O_2 affinity. An increasein the CO_2 concentration also lowers oxygen affinity. This could be due to, in part, a pH effect. CO_2 can react with the water in the erythrocyte to form carbonic acid: CO_2 + H_2O \rightarrow H_2CO_3 Carbonic acid then dissociates to form H^+ ions which lower the pH: H_2CO_3 \rightarrow H^+ +HCO^-_3 However, increasing the CO_2 concentration, while keeping pH constant, also lowers the oxygen affinity. Both H^+ and CO_2 bind to the hemoglobin molecule at special regulatory sites spatially distinct from the heme site. The binding alters the structure of the molecule by changing the arrangement of its polypeptide chains. This altered conformation lowers hemoglobin's affinity for oxygen. Since H^+ and CO_2 affect hemoglobin's binding of oxygen by allosteric interaction between the hemoglobin subunits, they have no similar effect on the myoglobin molecule, since it has only one polypeptide chain. Now that we understand the Bohr effect, what is its physiological significance? Rapid metabolism occurs in highly active tissue such as contracting muscle. Here much CO_2 is produced from cellular respiration. Acid production is also high, resulting from the increased CO_2 concentration and the reduction of pyruvate to lactic acid when the amount of oxygen is limiting. Thus, the presence of high levels of CO_2 and H^+ in actively metabolizing tissue enhances the release of oxygen from oxyhemoglobin. More oxygen is thus provided to the tissues where it is needed most. The opposite process occurs in the alveoli of the lungs. The increased concentration of oxygen drives off H^+ and CO_2, so that more oxygen may bind.

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Question:

There exists a carbohydrate that is slightly soluble in water, does not have a sweet taste, but does dissolve slowly when heated with diluteHCl. To what class of carbohydrates does it probably belong?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E23-0835.htm

Solution:

Carbohydrates may be defined as simple sugars and the substances that hydrolyze to yield simple sugars. Carbohydrates are placed in three classes: (1)Mono-saccharides- those that do not undergo hydrolysis; (2) Disaccharides - those that may be hydrolyzed to two monosaccharide molecules; and (3) Polysaccharides - those which form many monosaccharide molecules after hydrolysis. It is given that the carbohydrate dissolves in diluteHCl. Thus, one can immediately eliminatemono-saccharidesas a possible case. The fact that it dissolves whenHClis present suggests that a hydrolysis reaction has taken place. Only disaccharides and polysaccharides have the ability to undergo such a reaction. Thus, the carbohydrate belongs to either the class of disaccharides or polysaccharides. To determine which one it is, use the information that it is not sweet or very soluble in water. The more carbons in a molecule the more unlikely it will be soluble in water. Carbon -carbon bonds arenonpolar, while water is a polar molecule. Thus, a polysaccharide, which possesses many more non-polar bonds than a disaccharide, should have a tendency to be less soluble in water. Note also that among the disaccharides exist the very sweet sugars of sucrose (table sugar), maltose (malt sugar), and lactose (milk sugar). The classic examples of polysaccharides are, however, starch and cellulose, which are not sweet at all. Thus, one can con-clude that this carbohydrate probably belongs to the class of polysaccharides.

Question:

A glass bulb with volumetric expansion coefficient\beta_Bis weighed in water at temperatures T and T_1 . The weights of the displaced water are W and W_1, respectively. Find the volumetric expansion co-efficient\beta_Wof the water in the temperature interval from T to T_1.

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Solution:

The volumetric expansion coefficient \beta relates the change \DeltaV in the volume of a substance to a small change \DeltaT in the tempera-ture of that substance: (\DeltaV/V) = \beta \DeltaT where V is the initial volume. The volume of the displaced water equals the volume of the bulb since the bulb is completely immersed in the water while being weighed. The change in the volume of the bulb is due to the change in the volume of the glass since the gas inside the bulb cannot appreciably enlarge the glass. If the specific weights of water at T and T_1 are\rho_\omegaand \rho_(\omega)1, re-spectively, then V_B = (W/\rho_\omega) ,(1) V_(B)1 = V_B + \DeltaV_B = (W_1/\rho_(\omega)1 )(2) are respectively the volumes of the bulb at T and T_1 . The specific weight of water will decrease as a result of the thermal expansion of its volume since its weight remains constant. If the weight of water is W_W and its volume at T is V_W , we can write W_W =V_W\rho_\omega= (V_W + \DeltaV_W)\rho_(\omega)1 where \DeltaV_W is the volumetric expansion of water. Hence \rho_(\omega)1 = [V_W/(V_W + \DeltaV_W)]\rho_\omega= [(1)/{1 +(\DeltaVW/ V_W)}]\rho_\omega(3) The volumetric expansions \DeltaV_B and \DeltaV_W are given as \DeltaV_B =\beta_BV_B\DeltaT=\beta_B(W/\rho_\omega)(T_1 - T)(4) \DeltaV_W =\beta_WV_W\DeltaT=\beta_WV_W(T_1 - T)(5) From (1) and (2), we get \DeltaV_B = (W_1/\rho_(\omega)1) - (W/\rho_\omega) or, using (3), \DeltaV_B = (W_1/\rho_\omega){1 + (\DeltaW_W/V_W)} - (W/\rho_\omega) = (1/\rho_\omega)(W_1 - W) + (W_1/\rho_\omega)(\DeltaV_W/V_W) Substituting the expressions (4) and (5) for \DeltaV_B and \DeltaV_W in the above equation, we get \beta_B(W/\rho_\omega)(T_1 - T) = (W_1/\rho_\omega)\beta_W(T_1 - T) + {(W_1 - W)/\rho_\omega} or\beta_B= (W_1/W)\beta_W+ [(W_1 - W) / {W(T_1 - T)}] The volumetric expansion coefficient for water is \beta_W= (W/W_1)\beta_B+ [(W - W_1)/{W_1(T_1 - T)}] = [\beta_B+ {(W - W_1)/W_1} {\beta_B+ (1/\DeltaT)}] The above relation will hold for small \DeltaT = (T_1 - T). This corresponds to a small volumetric change for the bulb in the sense that (\DeltaV_B/V_B)<< 1.

Question:

Using the data from the accompanying figure, calculate the pressure exerted by .250 moles of CO_2 in .275 liters a 100\textdegreeC. Compare this value with that expected for an ideal gas. van der Waals constants Gasa, liter^2 atm/mol^2b,liter/mol Helium0.03410.0237 Argon1.350.0322 Nitrogen1.390.0391 Carbon dioxide3.590.0427 Acetylene4.390.0514 Carbon tetrachloride20.390.1383

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Solution:

To solve this problem you must understand the concept of an "ideal gas" and the formulas associated with it. Boyle's law states that PV = constant, if the temperature is fixed. In other words, if T is constant, then a fixed mass of gas occupies a volume inversely pro-portional to the pressure exerted on it. if a gas obeys this law, it is termed an "ideal gas." When a gas is ideal, PV = nRT, where P = pressure, V = volume, n = moles, R = universal gas constant, and T = temperature in Kelvin. When a gas is not ideal, it doesn't obey Boyle's Law; and instead of the Ideal Gas law you use Van der Waal's equation, [P + (n^2 a/V^2 ) (V - nb) = nRT,where, for CO_2, a = 3.59 liter^2 - atm/mole^2 and b = .0427 liter/mole. These values are called Van der Waal's constants. Substitute the values into these equations and solve for P. As such, [P + {(.250)^2 (3.59)/(.275)^2 }] [.275 - (.250) (.0427)] = (.250) (.08206) (373). Solving for P, you obtain P = 26.0 atm. If you had considered the gas as ideal, then PV = nRT, and P = 27.8 atm.

Question:

Ethylene glycol, C_2H_4(OH)_2, which is widely used as anti-freeze, may be converted into an explosive in a manner similar to the manufacture of nitroglycerine, (a) Write an equation for the preparation of C_2H_4(NO_3)_2. (b) Write an equation for the detonation of the latter compound into CO_2, N_2, and H_2O.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E27-0901.htm

Solution:

The manufacture of most explosives involves the use of concentrated sulfuric acid. The action of concentrated nitric acid on such compounds as glycerin seems to be markedly hastened by concentrated sulfuric acid because water is re-moved from the reaction. Most of the compounds formed by the action of nitric acid on organic substances are explosive. Thus, for part (a), An explosion occurs as a result of a rapid chemical reaction attended by the formation of a large volume of gas. The equation for the detonation of nitroglycerine is 4C_3H_5 (NO_3)_3 \ding{217} 12CO_2 + 10H_2O + 6N_2 + O_2 For the detonation of C_2H_4 (NO_3)_2, the products CO_2, N_2, and H_2O are known, thus only balancing need be done. after balancing: C_2H_4 (NO_3)_2------------------\rightarrow 2CO_2 + N_2 + 2H_2O.

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Question:

What is the de Broglie wavelength of an 88 kg man skiing down Mt.Tremblantin Quebec at 5.0 × 10^5 cm/sec?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E30-0920.htm

Solution:

The de Broglie wavelength, \lambda, of a body of mass moving at speed v is \lambda = h/mv, where h is Planck's constant, h = 6.6 × 10-^7 erg-sec = 6.6 × 10-^7 g-cm^2/sec. Hence, \lambda = (h/mv) = (6.6 × 10-^7 g-cm^2/sec) / (88 kg × 5.0 × 10^5 cm/sec) = (6.6 × 10-^7 g-cm^2/sec) / (88000 g × 5.0 × 10^5 cm/sec) = 1.5 × 10-^37 cm. The de Broglie wavelength of theskiier, 1.5 × 10-^37 cm, is in the microwave region of the spectrum.

Question:

An automobile of mass 50 slugs is traveling at 30 ft/sec. The driver applies the brakes in such a way that the velo-city decreases to zero according to the relation v = v_0 - kt^2, where v_0 = 30 ft/sec, k = 0.30 ft/sec^3, and t is the time in seconds after the brakes are applied. Find the resultant force decelerating the automobile, 5 sec after the brakes are applied.

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/Users/wenhuchen/Documents/Crawler/Physics/D04-0112.htm

Solution:

With constant force, (and, hence, constant acceleration), the velocity can be expressed as v = v_0 + at. In this problem, velocity is proportional to the time squared. We can find the force at any instant by finding the acceleration at that instant and, then using Newton's Second Law, F = ma, to relate acceleration to force. Hence a =dv/dt= d/dt(v_0 - kt^2) = -2kt. Hence when t = 5 sec, a = -2 × 0.3(ft/sec^3) × 5 sec = -3(ft/sec^2). Therefore, at this instant \sumF = ma = 50 slugs × [-3 (ft/sec^2)] = -150 lb. The negative sign indicates the force opposes the motion of the car and therefore acts to decelerate it.

Question:

Develop a BASIC program that produces a horizontal bar graph for some given distribution. Use the DIM statement to dimension any arrays that might occur.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G19-0479.htm

Solution:

This program will produce a bar graph (a histogram) of a distribution. In this case, the distribution will represent student's grades for a math course. The frequency of a particular grade will be given in the graph as a series ofx's. The array A(J) will serve as the counter for each class mark. The program terminates upon the reading of a negative data point. 1\O REM GRADE DISTRIBUTION GRAPH 2\O DIM A (4) 25 REM INITIALIZE ALL A(L) TO ZERO 3\O FOR L = 1 TO 4 4\O LET A(L) = \O 5\O NEXT L 6\O READ X 7\O IF X = -1 THEN 12\O 75 REM STANDARDIZE X VALUES 8\O LET J = (X/10) - 5 9\O LET A(J) = A(J) + 1 1\O\O GO TO 6\O 11\O FOR G = 1 TO 4 12\O PRINT 5\O + 1\O \textasteriskcentered G ; "TO"; 5\O + 1\O \textasteriskcentered G + 9; 13\O FOR W = 1 TO A(G) 14\O PRINT "X"; 15\O NEXT W 16\O PRINT 17\O NEXT G 18\O DATA 77, 85, 96, 75, 81, 68, 73, 78 19\O DATA 91, 72, 82, 74, 61, -1 2\O\O END Sample Output: RUN 6\O TO 69 XX 7\O TO 79 XXXXXX 8\O TO 89 XXX 9\O TO 99 XX READY A word about format: you may wish to include letter correlates for the number grades, such as A = 90 - 100, etc. This will make clear the ranges for the various class marks. By adding a PRINT statement with this information, you can make clear to the reader what the histogram is trying to show.

Question:

Consider the simple circuit in the figure grounded at point b. Compute the potentials of points a and c.

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/Users/wenhuchen/Documents/Crawler/Physics/D20-0668.htm

Solution:

In this circuit point B is grounded. When dealing with circuits, the grounded point is considered as zero potential and we express potentials relative to this refer-ence level. The direction of current flow is counterclock-wise so we will traverse the circuit in this direction. Kirchoff's loop theorem states that the sum of the changes in potential around any closed loop must be equal to zero. Note that this is just a statement of conservation of energy. Applying this theorem gives the following equation: \epsilon - ir - i (3 ohms) - i (1 ohms) = 0 Observe that although there is 10 volts across the battery by itself, when it is connected to the circuit its internal resistance, r, causes a potential drop equal to -ir. Substituting 1\Omega for r and solving for i, yields 2 am-peres as the value for the current. Since the potential at point b is zero, we start there and proceed to points a or c to find V_a or V_c respectively. We may go in either direction so long as we take into account the proper sign for the current (negative if counterclockwise, positive if clockwise). Proceeding counterclockwise from b to c yields: Vc = -i (1\Omega) = -2 volts Proceeding clockwise from b to a and remembering that in this direction the current is -2 amp, we see that V_a = - (-2)(3) = +6 volts. That is, point a is 6 volts above ground and point c is 2 volts below ground. The potential difference V_ac can now be found by subtraction ground. The potential difference V_ac can now be found by subtraction V_ac = V_a - V_c = 6 - (-2) = +8 volts

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Question:

The density of concentrated sulfuric acid is 1.85 g/ml. What volume of the acid would weigh 74.0 g?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E01-0016.htm

Solution:

Density is defined as weight per unit volume. density = (weight / volume) = g/ml. Therefore: volume =(weight / density) . Solving for the volume: volume = (74.0 g / 1.85 g/ml) = 40.0 ml.

Question:

Describe the events of meiosis and account for its importance .

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Solution:

Meiosis is the process of gamete formation in which two cell divisions occur in succession to give rise to four haploid daughter cells with each bearing only one copy of an individual chromosome, unlike the diploid parent cell which carried two copies. The sequence of events in meiosis begins as in mitosis. Chromosomes that have already duplicated in interphase begin to shorten in the first meiotic prophase to form con-densed, easily visible structures; each consists of two identical chromatids joined together at the centromere. It is during this condensation of the chromosomes that the first unique event of meiosis occurs: the synapsis (pairing) of homologous chromosomes, each made up of two chromatids. Each joined pair of homologous chromosomes is known as a tetrad. As the chromatids become more and more condensed, it is possible to see points at which they connect to one another, forming cross-bridges called chiasmata. These may be considered to be the visible manifestations of crossing over between a pair of chroma-tids. Only two of the four chromatids (non-sister chromatids) present in the tetrad can be involved in the formation of a chiasma. The end of the first meiotic prophase is marked by the disappearance of the nuclear membrane and the nucleolus. At the first meiotic metaphase, the homologous pairs (tetrads), having reached maximum condensation, line up along the equator of the meiotic spindle. This is followed by the first meiotic anaphase, during which the chromosomes separate from their homologs without separation of the sister chromatids, which remain attached at the centromere. Gradually, the duplicated chromosomes, now known as dyads, move to the two poles of the meiotic spindle. At telophase, when the chromosomes have reached the two poles, cytoplasmic division occurs. At this point, the chromosomes may become less condensed and enter into a brief, second meiotic interphase, during which the nuclear membrane forms again. The chromosomes grow diffuse but do not replicate. A second prophase follows rapidly, and the chromosomes again condense. There is, however, no synapsis, because the homologous chromosomes had separated at the first meiotic anaphase. The nuclear membrane and the nucleolus now disappear as during the first meiotic division. Those species which do not exhibit a second interphase go directly from the first telophase into a brief second prophase, during which spindle formation occurs. During metaphase, the two members of each dyad line up at the equator. At the second meiotic anaphase, the sister chromatids separate at the centromere, one going to each pole. When they have reached the poles, the nuclear membrane and nucleolus reappear. Cytoplasmic cleavage follows, giving rise to distinct daughter cells. After the second meiotic division, the number of chromosomes in each cell is half the original number, and reduction is complete. The process of meiosis began with a single cell (2n) which doubled its chromosomes, resulting in four copies of each type of chromosome (4n). Two meiotic cell divisions then produced four cells (1n) from this one cell. Because there was no chromosome doubling between the two divisions, there is necessarily a halving of the original chromosome number. In any sexual organism, there are two genetic forms, the haploid (1n) and the diploid (2n), which alternate from generation to generation. The diploid generation gives rise to the haploid generation by meiosis, and the haploid generation gives rise to the diploid generation by fertilization, which is the fusion of two haploid cells. Meiosis is thus important in ensuring that, after fertilization, the original diploid state is retained. Another extremely important factor in meiosis is the possibility of getting new combinations of genes as a result not only of crossing over between synapsing chromosomes in the first meiotic division, but also of the random segregation of the homologous chromosomes during the first anaphase. These factors contribute to the genetic variation which forms the basis for natural selection and thus evolution. If the new combination of genes confer some advantage to the organism possessing it, that organism will have a better chance of survival.

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Question:

On the basis of valence, predict the formulas of the compounds formed between the following pairs of elements: (a)Snand F. (b) P and H. (c) Si and O.

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Solution:

Valence may be defined as a number which represents the combining capacity of an atom or radical, based on hydrogen as a standard. For molecules containing two kinds of atoms, the product of the number of times one kind of atom appears in a molecule and the valence of that kind of atom, must be equal to the product of the number of times the other kind of atom appears multiplied by the valence of this second kind of atom. (a) The valence ofSnis 4 and that of F is 1. Hence, the compound is SnF_4 (1 atomSn× 4 = 4 atoms F × 1). (b) The valence of P is 3 and that of H is 1. Hence, the compound is PH_3 (1 atom P × 3 = 3 atoms H × 1). (c) The valence of Si is 4 and that of O is 2. Hence, the compound is SiO_2 (1 atom Si × 4 = 2 atoms O × 2).

Question:

You have the reaction A_2 + B_2 \rightleftarrows 2AB with AB initially at 5 liters, 27\textdegreeC, and 25 atm. K = 50. Find (a) the initial concentration of AB, (b) the A_2 and B_2 concentrations at equilibrium and (c) the partial pressures of A_2 and B_2 and AB at equilibrium.

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Solution:

Use the equation of state, which indicates PV = NRT, where P = pressure, V = volume, N = moles, R= universal gas constant (.0821 liter- atm/mole-K), and T = temperature in degreeskelvin(Celsius plus 273\textdegree). You can use this equation to find N, the moles of AB initially present. (a) PV = NRT or N = PV/RT = {(25) (5)} / {(.0821) (300)} = 5.08 moles. Thus, the initial concentration = moles/liters = 5.08/5 = 1.02. (b) K = 50 (given).By definition, K = ( [AB]^2 / { [A_2 ] [B_2 ]}). At equilibrium, let x = moles/liter of AB that are dissociated. If its initial con-centration is 1.02M, then the concentration is (1.02-x)M at equilibrium. From the reaction, 1 mole each of A_2 and B_2 is produced by every 2 moles of AB which react. Thus, if x moles/liter of AB dissociate, (1/2) x moles/liter of both A_2 and B_2 are produced.Recalling,K= ([AB]^2 / {[A_2] [B_2]}) = 50, you now substitute to obtain (1.02-x)^2 / (.5x)^2 = 50 . Solving for x, x = .225 mole/liter. Therefore, at equilibrium, [A_2] = [B_2] = x/2 = .1125 mole/literand [AB] = 1.02 - .225 = .795 mole/liter. (c) The partial pressures of each will be P = (N/V) RT from the equation of state, where (N/V) is the concentration at equilibrium. Thus, P_(A)2 = .1125(.0821)(300) = 2.77atm, P_(A)2 = P_(B)2 , since these concentrations are the same at equilibrium, P_AB = .795(.0821) (300) = 19.5 atm.

Question:

Explain how an estimate of the age of a rock is made on the basisof the radioactive elements present.

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Solution:

Certain radioactive elements are spon-taneously transformed into otherelements at rates which are slow and essentially unaffected by externalfactors, such as the temperatures and pressures to which the elementsare subjected. The transformation, or decay, of each individual elementtakes place at a rate which can be measured. For example, half ofa given sample of the element uranium will be converted into lead in 4.5 billionyears. Thus, 4.5 billion years is the half-life of uranium. By measuringthe proportion of uranium and lead in a given rock, we can estimatewith a high degree of accuracy the absolute age of the rock. For instance, assume we determine that the ratio of uranium to lead in a given sampleis 1:1. We know, that the rock has existed long enough for half of theoriginal amount of uranium to be converted into lead, or a total of 4.5 billionyears, which is equal to one half-life of uranium. If the ratio is 1:3, three-fourths of the original amount of uranium is now lead, and one-fourth remains. Therefore, two half- lives must have passed, and half of the uraniumremaining after one half-life has turned into lead [(1/2) × (1/4)X = (1/4) X, where x equals the original amount of uranium in thesample ]. Two half-lives are equivalent to a time period of 9 (= 2 × 4.5) billion years, and the age of the rock, then, is 9 billion years. Other methods are used to determine the age of rocks which are notold enough to use the uranium - leadmetodof dating. One of these is thepotassium 40 - argon (K40 -Ar) method. Potassium - 40 decays into argonand has a half-life of 1.3 billion years. Because these two elements arecommonly found in volcanic rock, this method is particularly useful in areaswhere volcanic activity is known to have occurred in the past. Other absolutedating techniques include fission - track andpaleomagnetic reversals. These, along with others being developed, promise to provide a firmbasis in understanding the chronology of the events in the evolutionary record.

Question:

A cylinder contains oxygen at a pressure of 10 atm and a temperature of 300\textdegreeK. The volume of the cylinder is 10 liters. What is the mass of the oxygen?

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Solution:

In this problem, we are dealing with 4 variables M = mass, V = volume, P = pressure, and T = temperature. Three of them are given; we must calculate the fourth. These variables are governed by the ideal gas law. An ideal gas is one in which the molecules have no attraction for one another and the molecules themselves occupy no space (a situation contrary to fact). Real gases do not completely satisfy either of these conditions, but under conditions close to STP (0\textdegreeC and 1 atm), the real gases come very close to meeting these conditions. Therefore, the difference between the real and ideal values is small enough such that the ideal gas law equation can be used, it is: PV = n RT where n is the number of moles. Moles is defined as grams/molecular weight. R is a gas constant equal to .082 (liter) - (atm)/(\textdegreeK) - (mole). We are asked to find the mass of oxygen, yet there is no value for mass in the ideal gas equation. There-fore, we must find an equation that involves mass and moles. That equation is moles = grams/M.W. The molecular weight of O_2 = 32 g/ mole. Therefore, substitute the mole equation into the ideal gas equation PV = nRT = (M/M.W.) RT The only unknown value is mass and the solution is arrived at by substituting in the appropriate values (10 atm) (10 liter) = (m/32) (.082) (300\textdegreeK) m = [(32 × 10 × 10)/(.082 × 300)] = 130.08 grams.

Question:

For the following oxidation-reduction reaction, (a) write out the two half-reactions and balance the equation, (b) calculate ∆E\textdegree, and (c) determine whether the reaction will proceed spontaneously as written; Fe^2+ + MnO^-_4 + H^+ \rightarrow Mn^2+ + Fe^3+ + H_2O. (1) Fe^3+ + e^- \leftrightarrowsFe^2+, E\textdegree = 0.77 eV (2) MnO^-_4 + 8H^+ + 6e^- \leftrightarrows Mn^2+ + 4H_2O, E\textdegree = 1.51 eV.

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Solution:

(a) The two half-reactions of an oxidation- reduction reaction are the equations for the oxidation process (loss of electrons) and the reduction process (gain of electrons).In the overall reaction, you begin with Fe^2+ and end up with Fe^3+. It had to lose an electron to accomplish this. Thus you have oxidation: Fe^2+ \rightarrow Fe^3+ + e^-. Notice: this is the reverse of the reaction given with E\textdegree = .77. As such, the oxidation reaction in this problem has E\textdegree = -.77 eV. The reduction must be MnO^-_4 + 8H^+ + 5e^- \rightarrow Mn^2+ + 4H_2O, since in the overall reaction, you see MnO^-_4 + H^+ go to Mn^2+, which suggests a gain of electrons. This is the same reaction as the one given in the problem, E\textdegree = 1.51 eV. To balance the overall reaction, add the oxidation reaction to the reduction reaction, such that all electron charges disappear. If you multiply the oxidation reaction by 5, you obtain: 5Fe^2+ \rightarrow 5Fe^3+ + 5e^- MnO^-_4+ 8H^++ 5e^- \rightarrow Mn^2+ + 4H_2O 5Fe^2+ + MnO^-_4 + 8H^+ \rightarrow 5Fe^3+ + Mn^2+ + 4H_2O Notices Since both equations contained 5e^- on different sides, they cancelled out. This explains why the oxidation reaction is multiplied by five. Thus you have written the balanced equation. (b) The ∆E\textdegree for the overall reaction is the sum of the E\textdegree for the half- reactions, i.e., ∆E\textdegree = E_red + E_oxid. You know that Eredand E_oxid ; ∆E\textdegree = 1.51 - (.77) = 0.74 eV. (c) A reaction will only proceed spontaneously when ∆E\textdegree = a positive value. You calculated a positive ∆E\textdegree, which means the reaction proceeds spontaneously.

Question:

Starting from Planck's radiation law, show that one can obtain Wien's displacement law, \lambda_max T = constant. It is known that this constant is equal to 2.891 × 10^6 nm \textbullet K deg. Obtain the value of Planck's constant, given that Boltzmann's constant is 1.380 × 10^-23 J \textbullet K deg^-1.

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Solution:

As shown in the figure, the \lambda_max appearing in Wien's displacement law is the value of \lambda at which \psi (\lambda) is a maximum. In order to show that Wien's Law follows from Planck's radiation law, we calculate the values of \lambda for which \psi (\lambda) is a maximum. This will yield the value of \lambda _max as a function T. Planck's radiation law connects the monochromatic energy density in an isothermal blackbody enclosure with the wavelength and absolute temperature T \psi = [(8\pich\lambda^-5)/{exp(ch/\lambdakT) - 1}] . where h is Planck's constant, and k is Boltzmann's constant. Thus (d\psi/d\lambda) = - [(40\pich\lambda^-6)/{exp(ch/\lambdakT) - 1}] - [8\pich\lambda^-5 × { - (ch/kT\lambda^2) exp (ch/\lambdakT)}/{exp (ch/\lambdakT) - 1}^2] = [(8\pich\lambda^-6)/{ exp (ch/\lambdakT) - 1}] [ - 5 + {(ch/\lambdakT) exp (ch/\lambdakT)} / { exp (ch/\lambdakT) - 1}] . For maxima or minima of \psi, d\psi/d\lambda must be zero. The term outside the bracket becomes zero for either \lambda = 0 or \lambda = \infty. Both of these are minima, as can be verified by differentiating again. When the expression inside the bracket becomes zero, we have a maximum, and for this [{(ch/\lambda_max kT) exp (ch/_max kT)} / {exp (ch/\lambda_max kT) - 1}] = 5 (ch/\lambda_max kT) exp (ch/\lambda_max kT) = 5 [exp (ch/\lambda_max kT) -1] exp (ch/\lambda_max kT) (5 - ch/\lambda_max kT) = 5 orexp (- ch/\lambda_max kT) = 1 - ch/5\lambda_max kT(1) This is a transcendental equation, which must be solved graphically. Letting x = -ch/\lambda_max kT equation (1) becomes exp (x) = 1 + (x/5)(2) In order to find x, we draw a graph of y = exp (x), and another graph y = 1 + x/5. The intersection points of these 2 graphs are the values of x which satisfy (2). From figure (2), we obtain x = - 4.965 as the solution set of (2). Thus for a maximum of the radiation curve, exp(- ch/\lambda_max kT) = 1 - ch/5\lambda_max kT and this implies that ch/\lambda_max kT = 4.965. \lambda_max T = (ch/4.965 k) = const, which is Wien's displacement law. Experimentally, \lambda_max T = 2.891 × 10^-3 m \textbulletK deg. Solving for h h = [{(4.965 k) (\lambda_max T)} / c] h = [(4.965 k × 2.891 × 10^-3 m \textbullet k deg) / c] = [(4.965 × 1.380 × 10^-23 J \textbullet k deg^-1 × 2.891 × 10^-3 m\textbulletk deg)/(2.998 × 10^8 m \textbullet s^-1)] = 6.607 × 10^-34 J \textbullet s.

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Question:

A swimmer whose mass is 60 kg dives from a 3-m high platform. What is the acceleration of the earth as the swimmer falls toward the water? The earth's mass is approximately 6 × 10^24 kg.

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Solution:

The diver's mass,m_d= 60 kg, the accelera-tion of the diver, a_d = 9.8 m/s^2, and the mass of the earth, m_e = 6 × 10^24 kg, are the known observables. The earth's acceleration can be determined using the fact the force on the earth on the diver is equal in magnitude and opposite in direction to the force of the diver on the earth. Letting the subscripts d and e refer to the diver and earth, respectively, we obtain m_da\ding{217}_d = -m_ea\ding{217}_e Considering just the magnitude of the two vectors, we find (60 kg)(9.8 m/s^2) = (6 × 10^24kg)a_e a_e= {(60 kg)(9.8 m/s^2)} / (6 × 10^24kg) = 9.8 × 10^-23m/s^2 Since the diver is accelerated downward, the earth is accelerated upward, toward the falling diver.

Question:

A chemist wants to dilute 50 ml of 3.50 M H_2SO_4 to 2.00 M H_2SO_4. To what volume must it be diluted?

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Solution:

Molarityis defined as the number of moles of solute per liter of solution. In other words, molarity= [(no, of moles of solute)/(liters of solution)] In this problem you have a 50 ml solution of 3.50 M. In a liter there are 1000 ml. Thus, the number of liters in this solution is .05. Substituting, you have 3.50 M = [(no. of moles of H_2SO_4)/(.05 l)] Solving for the number of moles, you obtain .175 moles of H_2SO_4 . When you dilute this mixture, you will still have .175 moles of the solute, H_2SO_4 . You are only increasing the volume of the solvent, water. Therefore, in the diluted 2.0 M solution you wish to have 2.00 = [(.175)/(liters of solvent)] . Solving for liters of solvent, you obtain .0875 liters or 87.5 ml. Thus, to dilute to amolarityof 2, the total volume must be 87.5 ml.

Question:

Find the pH of a 0.25 M solution of Na_2CO_3, a strong electrolyte.

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Solution:

The crux of this problem is to realize that the solution contains the diproticbase CO_3^-2. Since Na_2CO_3 is a strong electrolyte, it dissociates completely.Diproticmeans that 2 hydrogen ions will dissociate. Thus, there are 2 acid equilibrium constants. H_2CO_3 is the fullyprotonatedacid. Fordiproticacids, the second base ion-ization constant, K_(b)2 , is related to the first acid ion-ization constant, K_(a)1 , and is related to K_(b)1 is related to K_(a)2. The first H^+ ion neutralized behaves as a stronger acid then the second H^+ ion, because in each successive stage, ionization is less complete. The ionization and equilibrium constant equations are: CO_3^-2 + H_2O \rightleftarrows HOC_3^- + OH^- K_(b)1 = {[HOC_3^-][OH^-]}/[CO_3^-2]and HOC_3^- + H_2O \rightleftarrows H_2OC_3 + OH^- K_(b)2 = {[H_2OC_3][OH^-]}/[HOC_3^-] Since it is given that the [CO_3^-2] = 0.25 M, attention shall be limited to the first ionization. Each x moles of [CO_3^-2] that reacts with H_2O forms x moles each of HCO_3^- and OH^-. CO_3^-2 + H_2O \rightleftarrowsHCO_3^- + OH^- Before:0.25OO After:0.25-xxx If the concentration of CO_3^-2 is initially 0.25 M, and x moles of it react, then, at equilibrium, there are 0.25 - x. Before substituting these values into the K_(b)1 equation first find K_(b)1. This is accomplished by referring to a table of K_a ionization constants. K_(a)2 has a value of 5.62 × 10^-11. One can then use this information in the following equation: K_(a)2 K_(b)1 = K_W which relates the acid and base ionization constants to the constant for theautodissociationof water, K_W .Therefore, K_(b)1 = (K_W/ K_(a)1) = [(1.0 × 10^-14)/(5.62 × 10^-11)] = 1.8 × 10^-4 One can find [OH^-] by writing the equilibrium constant expression and substituting the values for the concentrations. K_(b)1 = {[HOC_3^-][OH^-]}/[CO_3^-2] [{(x) (x)}/(0.25)] = 1.8 × 10^-4 x = [OH^-] = 6.7 × 10^-3 . The x from(0.25 - x)was eliminated since so few moles react, as CO_3^-2 is a weak base. In order to calculate pH, first obtainpOHusing pOH= - log [OH^-] = - log [6.7 × 10^-3] = 3 - log 6.7 = 2.17. pH =pK_W-pOH= 14 - 2.17 = 11.83. It makes sense that K_a K_b = K_W since K_a is a measure of [H^+] and K_b is a measure of [OH^-], and one knows that [H^+][OH^-] = K_W = 10^-14 .

Question:

An experiment is done using Thomson's apparatus for positive-ray analysis. A set of positive-ray parabolas are examined and it is found that for the same horizont-al displacement y, the corresponding vertical displace-ments for the three parabolas observed are 3.24 mm, 3.00 mm, and 2.81 mm. The parabola with the largest displacement is known to correspond to C^12 and all ions are singly charged. What are the other ions present?

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Solution:

In Thompson's apparatus [figures (a) and (b)] a gas of the element

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Question:

Determine the value of (0.0081)^\rule{1em}{1pt}3/4.

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Solution:

(0.0081) = .3 × .3 × .3 × .3 = (.3)^4 , therefore (0.0081)^\rule{1em}{1pt}3/4 = (.3^4)\rule{1em}{1pt}3/4 Recalling the property of exponents, (a^x)^y = ax \textbullet y we have, (.3^4)^ \rule{1em}{1pt}3/4 = .3(4) (\rule{1em}{1pt}3/4)= .3^\rule{1em}{1pt}3. Since a^\rule{1em}{1pt}x = [1/(a^x)], .3^\rule{1em}{1pt}3 = [1/(.3^3)] = [1/(0.027)] = [(1/27) / 1000] Division by a fraction is equivalent to multiplication by its reciprocal, thus, [(1/27) / 1000)] =[(1000) / (27)]. Hence,(0.0081)^\rule{1em}{1pt}3/4 = [(1000) / (27)].

Question:

The rate of a certain biochemical reaction at body temperature in the absence of enzyme has been measured in the laboratory. The rate of the same reaction, when enzyme-catalyzed in the human body, is 10^6 times faster. Explain this phenomenon.

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Solution:

For any reaction to proceed, a certain amount of energy is needed. This energy is called the energy of activation. When a catalyst, such as an enzyme, is added to the system, the energy of activation is lowered. Because the energy requirement for the reaction is less, the reaction proceeds more quickly.

Question:

How many calories of heat must be added to 3.0 liters of water to raise the temperature of the water from 20\textdegreeC to 80\textdegreeC?

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Solution:

A calorie is defined as the amount of heat necessary to raise the temperature of one gram of water one degree Centigrade. One can find the mass of 3.0 liters of water by using the density. The density of water is 1 g/ml. 1 liter = 1000 ml. Therefore, 3.0 liters = 3000 ml. Mass = density × volume Mass = 1 g/ml × 3000 ml = 3000 g The temperature change is found by subtracting 20 from 80, which is 60\textdegreeC. One can find the calories necessary to raise 3000 g of water 60\textdegreeC by multiplying 1 cal/g\textdegreeC by 3000 g by 60\textdegreeC. No. of calories= 1 cal/g\textdegreeC × 3000 g × 60\textdegreeC = 180000 cal. = 180 Kcal.

Question:

In what ways do the bryophytes resemble thetracheophytes, and in what ways are they different?

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Solution:

Botanists place the bryophytes and thetracheophytes(the vascular plants ) in the same sub-kingdomEmbryophyta. This is due to the fact that these two divisions of plants have evolved a life cycle in which the zygote is borne within the female sex organ of the gametophyte. Here it acquires protection , water and nutrients while it develops into amulticellular sporophyte embryo. Besides this common characteristic, the bryophytes and tracheophytesresemble each other in other ways. They have structures which are analogous and serve similar purposes. For example, the tracheophyteshave roots and the bryophytes have rhizoids, both of which serve for anchorage and water absorption; the former have stomatae and the latter have pores for gaseous diffusion; the tracheophytes have well developed leaves to capture solar energy and manufacture food and the bryophytes have smaller, simpler leaf-like structures for the same purpose; both have an epidermis, which, in the tracheophytes and certain mosses is impregnated withcutinto prevent exce-sive water loss; and finally, both groups live on the land. Despite these similarities, thetracheophytesand bryophytes are believed to have diverged in their evo-lutionary development long ago, and so also display significant differences between them. Thetracheophytes are taller, larger, and more complicated than the bryophytes. They are organized into complex plant bodies in which many different kinds of tissues and structures can be recognized. They have developed a root system with an extensive surface area to extract large quantities of water and minerals from the soil and have evolved an efficient vascular system to conduct water and nutrients to different parts of the plant. The bryophytes , on the other hand, have only simple rhizoids and no true vascular tissues, and hence have to compensate for these dis-advantages with a small body size and height. In addition, the highertracheophytes (gymospermsand angiosperms) produce non-flagellated sperm, and are not dependent on moist conditions in order for fertilization to occur. Thus, while the bryophytes and lower vascular plants rigidly require a moist environment for sperm transport, the highertracheophyteshave evolved a greater variety of means by which the male sex cell can be carried to the female over long distances, such as by wind, water, insects, and other animals . Thetracheophytesalso differ from the bryophytes in their reproductive cycles. The bryophytes have a pattern of alternation of generations in which the game-tophyte is dominant; thesporophyteis greatly de-pendent upon it. On the contrary, thesporophytedom-inates in the higher plants and, except for the earliest stages in the formation of an embryo , it is independent of the gametophyte. Correspondingly, the tracheo- ga -metophyteis highly reduced in size and complexity, and, in the conifers and flowering plants, it is completely dependent upon the sporophyte . The highertracheophyteshave also managed to enhance embryonic protection by the evolution of seeds. Seeds protect the embryonic sporophytefrom the rigors of a life on land, nourish it, and allow it to travel over great distances for dispersal of the species. The similarities and differences between thetracheophytesand bryophytes discussed above lead us to conclude that although both are land plants, the former are more advanced and better adapted to a terrestrial life than the latter.

Question:

A 3-liter bulb containing oxygen at 195 torr is connected to a 2-liter bulb containing nitrogen at 530 torr by means of a closed valve. The valve is opened and the two gases are allowed to equilibrate at constant temperature, T. Calculate the equilibrium pressure.

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Solution:

we will solve for the final or equilibrium pressure, P_f, by using the ideal gas law, P_f = (nRT/V_f ) where n is the number of moles of O_2 (n_O(2)) Plus the number of moles of N_2 (n_N(2)), that is, the total number of moles of gas, R is the gas constant, T the absolute temperature, and V_f the final volume (3 liters + 2 liters = 5 liters). We cannot solve this equation directly, since we do not know T. The quantities n_O(2) and n_N(2) are calculated using the initial conditions and the ideal gas law, na = [(P_a V_a )/(RT)] where P_a is the original partial pressure of gas a V_a is the volume initially occupied by gas a. Thus, n_O(2) = [(P_O(2) V_O(2) )/(RT)] = [(195 torr × 3 liters)/(RT)],and n_N(2) = [(P_N(2) V_N(2) )/(RT)] = [(530 torr × 2 liters)/(RT)] Multiplying both equation by RT and then adding, n_O(2) RT + n_N(2)RT = [(195 torr × 3 liters)/(RT)] RT + [(530 torr × 2 liters)/(RT)] RT [n_O(2) + n_N(2) ] RT = 195 torr × 3 liters + 530 torr × 2 liters. Dividing both sides by V_f , we obtain [n_O(2) + n_N(2) ] (RT/V_f ) = [(195 torr × 3 liters + 530 torr × 2 liters)/(V_f )]. But n_O(2) + n_N(2) = n, and P_f = nRt/V_f . Hence, P_f = (nRT/V_f ) = [n_O(2) + n_N(2) ] (RT/V_f ) = [(195 torr × 3 litrers + 530 torr × 2 liters)/(V_f )] = [(195 torr × 3 litrers + 530 torr × 2 liters)/(5 liters)] = 329 torr.

Question:

In a hydraulic press the small cylinder has a diameter of 1.0 in., while the large piston has a diameter of 8.0 in. If a force of 120 lb is applied to the small piston, what is the force on the large piston, neglecting friction?

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Solution:

Pascal's Law states that pressure applied to an enclosed fluid is transmitted throughout the fluid in all directions without loss. In the hydraulic press shown, this means that the pressure applied to the smaller piston is transmitted unchanged to the larger piston. Since it has a larger area, it experiences a greater force since F = PA. Hence, we have P_2 = P_1F_2/A_2 = F_1/A_1 F_2 = (A_2/A_1) F_1 = [\pi(4.0 in.)^2]/[ \pi(0.50 in.)^2] 120 lb = 7.7 × 10^3 lb

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Question:

The ionization constant for acetic acid is 1.8 × 10^-5 . a) Calculate the concentration of H^+ ions in a 0.10 molar solution of acetic acid. b) Calculate the concentration of H^+ ions in a 0.10 molar solution of acetic acid in which the concentration of acetate ions has been in-creased to 1.0 molar by addition of sodium acetate.

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Solution:

The ionization constant (Ka) is defined as the concentration of H^+ ions times the concentration of the conjugate base ions of a given acid divided by the concentration of unionized acid. For an acid, HA., Ka = {[H^+] [A^-]}/[HA] , where Ka is the ionization constant, [H^+] is the concentration of H^+ ions, [A^- ] is the concentration of the conjugate base ions and [HA ] is the concentration of unionized acid. The Ka for acetic acid is stated as Ka = {[H^+] [acetate ion]}/[acetic acid] = 1.8 × 10^-5 . The chemical formula for acetic acid is HC_2H_3O_2 . When it is ionized, one H^+ is formed and one C_2H_3O^- (acetate) is formed, thus the concentration of H^+ equals the concentration of C_2H_3O^- . [H^+] = [C_2H_3O^-] . The concentration of unionized acid is decreased when ionization occurs. The new concentration is equal to the concentration of H^+ subtracted from the concentration of unionized acid. [HC_2H_3O] = 0.10 - [H^+] . Since [H^+] is small relative to 0.10, one may assume that 0.10 -[H^+] is approximately equal to 0.10. 0.10 - [H^+] \cong 0.10 . Using this assumption, and the fact that [H^+] = [C_2H_3O^-] can berewriteen as Ka = {[H^+] [H^+]}/(0.10) = 1.8 × 10^-5 . Solving for the concentration of H^+ : [H^+]^2 = (1.0 × 10^-1)(1.8 × 10^-5) = 1.8 × 10^-6 [H^+] = \surd(1.8 × 10^-6) = 1.3 × 10^-3 . The concentration of H^+ is thus 1.3 × 10^-3 M. b) When the acetate concentration is increased, the concentration of H^+ is lowered to maintain the same Ka . The Ka for acetic acid is stated as Ka = {[H^+][C_2H_3O^-]}/[HC_2H_3O] = 1.8 × 10^-5 As previously shown for acetic acidequilibriain a solution of 0.10 molar acid, the concentration of acid after ionization is [HC_2H_3O] = 0.10 - [H^+] . Because [H^+] is very small compared to 0.10, 0.10 - [H^+] \cong 0.10 and [HC_2H_3O] = 0.10 . In this problem, we are told that the concentration of acetate is held constant at 1.0 molar by addition of sodium acetate. Because one now knows the concentrations of the acetate and the acid, the concentration of H^+ can be found. {[H^+][C_2H_3O^-]}/[HC_2H_3O] = 1.8 × 10^-5 {[H^+] [1.0]}/[0.10] = 1.8 × 10^-5 [H^+] = 1.8 × 10^-6

Question:

Evaluate I for a sphere of uniform density \rho, for an axis through the center of the sphere.

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Solution:

We may assemble the sphere from spherical shells of thickness dr and having mass density \sigma = \rho dr per unit area as shown in the figure. The moment of inertia, dl, of the thin spherical shell of mass dm shown in the figure, about a radius, is dI = (2/3)r^2 dM =(2/3)r^2(4\pir^2 \rho dr) =(8/3)\pir^4 \rho dr The total moment of inertia of the solid sphere therefore becomes I = \int dI = _0\int^R (8/3) \pir^4\rho dr =(8/15) \pi \sigma R^5 = (2/5)[(4/3) \pi \sigmaR^3 ]R^2 = (2/5)MR^2, where M is the mass of the sphere.

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Question:

Paper electrophoresis at pH 6.0 was carried out on a mixture of glycine, alanine, glutamic acid, lysine, arginine and serine, (a) Which compound moved toward the anode? (b) Which moved toward the cathode? (c) Which re-mained at the origin?

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Solution:

A method of separating amino acids is paper electrophoresis. A drop of the solution containing the amino acids is placed on a filter paper sheet, which is then moistened with a buffer of a given pH. The ends of the sheet dip into electrode vessels, and a high-voltage electrical field is applied while cooling. Because of their different pK^1 values, the amino acids migrate in different directions and at different rates, depending on the pH of the system and the emf (.electromotive force) applied. To determine whether an amino acid will move towards the anode (positive end) or the cathode (negative end) , one must determine the charge on the particular amino acid at. pH = 6. This is done by calculating the isoelectric pH_I (the pH_I, where the amino acid is electrical-ly 1y neutral) of each amino acid. pH_I = (1/2) (pK'_1 + pK'_2) . If pH_I is less than pH = 6, then the amino acid will be negatively charged at pH = 6 and will move towards the cathode. When the pH_I of an amino acid is near 6, the amino acid will not move. glycine; pH_I = (1/2) (2.34 + 9.6) = 5.97. This is very near pH = 6,and glycine will remain in the center. alanine:pH_I = (1/2) (2.34 + 9.69) = 6.02. This is also very close to 6 and alanine will not move from the center. glutamic acid: As glutamic acid is titrated with base, the following species come about: Species B is the neutral species, here. Therefore, pH_I = (1/2) (pK^1+ pK^1_R) = (1/2) (2.19 + 4.25) = 3.22. 3.22 is much less than 6 and therefore glutamic acid will be negatively charged at pH = 6 and will move towards the anode. lysine: Using a method similar to the one used above for glutamic acid, one can determine that pH_I = (1/2)(pK^1_2 + PK^1_R)pH_I = (1 / 2)(8.95 + 10.53) = 9.74. Lysine will be posi-tively charged at pH = 6 and will be attracted towards the cathode. Arginine: pH_I = (1/2)(pK'_2 + pK'_R) = (1/2) (9.04 + 12.48) = 10.76.Arginine will be positively charged and will move towards the cathode move towards the cathode serine: pH_I = (1/2) (pK^1_1 + pK^1_2) = (1/2) (2.21 + 9.15) = 5.68. 5.68 is close to 6.0 and serine will remain close to the center of the strip.

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Question:

Find the pressure due to a column of mercury 74.0 cm height.

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Solution:

The total force F acting at the bottom of the column of mercury is due to the weight of the mercury. Or, by Newton's Second Law F = W = mg. SinceDensity (p) = mass(m)/volume(v) and V = Ah where A is the cross sectional area of the column and h is its height. We then have F = \rhoVg = \rhoAhg The pressure P at the bottom of the column is defined as P = F/A = (pAhg)/A = hpg = (0.740 m)(1.36 × 10^4 kg/m^3)(9.80m/sec^2) = 9.86 × 10^4 nt/m^2. In the equation derived above, the pressure is that due to the liquid alone. If there is a pressure on the surface of the liquid, this pressure must be added to that due to the liquid to find the pressure at a given level. The pressure at any level in the liquid is then P = P_s + hpg where P_s is the pressure at the surface of the liquid.

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Question:

The two protons in a helium nucleus are separated by about 2 × 10-15m. Calculate the electrostatic potential energy and thus estimate the fractional change in the mass of helium.

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Solution:

The electrostatic potential energy of a system of two charges separated in space by a distance R is V = K[(q_1 q_2) / R], where K is a constant having the value 9 × 10^9 {[N\textbulletm^2]/ c^2}. For two protons separated by 2 × 10^-15 m, the potential energy is V ={[9 × 10^9 {(N\textbulletm^2) / c^2}] (1.6 × 10^-19 c) 2 } / (2 × 10^-15 m) V = 1.15 × 10^-13 Joules By E = mc^2, this energy is equivalent to a mass \Deltam = (1.15 × 10^-13 J) / {9 × 10^16(m^2/s^2)} \Deltam = 1.28 × 10^-30 kg ∆m represents a mass increase of the helium atom due to electrostatic potential energy electrostatic potential energy . In calculating the mass of a helium atom, we may neglect the mass of the electrons, because they have a small mass when compared with the nuclear components of the atom (protons and neutrons). Hence, the mass of the helium atom is approximately equal to the mass of the helium nucleus. The helium nucleus consists of two protons and two neutrons of total mass m = (2 × 1.672 × 10^-27 + 2 × 1.675 × 10^-27)kg = 6.69 × 10^-27 kg The fractional change in mass of the helium nucleus is \Deltam / m = (1.28 × 10^-30) / (6.69 ×10^-27 ) = 1.9 × 10^-4 .

Question:

Using a clocked JK flip-flop construct a; a) D flip-flop.b) T flip-flop. c) Clocked T flip-flop.d) SR flip-flop.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G04-0087.htm

Solution:

The construction shown in fig. 1 gives a D flip-flop realization. Note that both S and R inputs are zero, and the circuit obeys the truth table given below; b) By applying 0 to both inputs S and R and 1 to inputs J and K, a T flip-flop may be constructed as shown in fig. 2. Note that the C (clock) input becomes the T input and the circuit constructed obeys the truth table of a T flip-flop. c) Shown in figure 3 is a realization of the clocked T flip-flop. The control signal T_c allows the clock pulses to be selectively applied to the input terminal. Each clock pulse that arrives at T causes the flip-flop to a state change. d) Applying 0 to J, K, and Clock inputs gives an SR flip-flop realization, shown in fig. 4.

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Question:

An automobile driver, A, traveling relative to the earth at 65 mi/hr on a straight, level road, is ahead of motorcycle officer B, traveling in the same direction at 80 mi/hr. What is the velocity of B relative to A? Find the same quantity if B is ahead of A.

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/Users/wenhuchen/Documents/Crawler/Physics/D01-0014.htm

Solution:

The velocity of B relative to A is equal to the velocity of B relative to the earth minus the velocity of A relative to the earth, or V_BA = V_BE - V_AE = 80 mi/hr - 65 mi/hr = 15 mi/hr If B is ahead of A, the velocity of B relative to A is still the velocity of B relative to the earth minus the velocity of A relative to the earth or 15mi/hr. In the first case, B is overtaking A, and, in the second, B is pulling ahead of A.

Question:

An eastbound car travels a distance of 40 m in 5 s. Determine the speed of the car.

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/Users/wenhuchen/Documents/Crawler/Physics/D03-0057.htm

Solution:

The observables of distance, d = 40 m, and time interval, t = 5 s, are given. We know that, since the velocity v of the car is constant, v = d/t = (40 m)/(5 s) = 8(m/s) Here, d is the distance travelled in time t. The speed of the car is 8 m/s.

Question:

In an experiment involving the gravitational red shift, two identical nuclei are placed at different heights in a tower. The nuclei each emit gamma rays. The difference in height of the two nuclei is 2.2 × 10^3 cm. What is the fractional difference in the frequency of the \Upsilon-rays?

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/Users/wenhuchen/Documents/Crawler/Physics/D33-0995.htm

Solution:

Consider a photon emitted by nucleus A located at the lower gravitational potential. Its energy upon emission is E_1 = hf_0 where h is Plank's constant and f_0 is the frequency associated with the photon. As the photon rises, it gains potential energy. The law of conservation of energy demands that the total energy of the photon remain con-stant in time. The frequency of the photon must then decrease to offset the increase in potential energy to maintain the total energy constant. After travelling a height H, its energy is E_2 = hf + mgH m is the mass equivalent of its energy which is given by Einstein's mass- energy relationship m = (hf/c^2) To satisfy the conservation of energy, we must have E_1 = E_2 hf_0 = hf + (hfgH)/c^2 or(f_0/f) = 1 + (gH/c^2) = 1 + [(980 cm/sec^2 × 2.2 × 10^3 cm)/{(3 × 10^10 cm/sec)^2}] = 1 + 2.3 × 10^-15 The frequencies differ by 2.3 parts in 10^15. This minute change in frequency is termed the gravitational red shift (for the frequency decreases and shifts towards the red part of the spectrum as the light leaves the Earth).

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Question:

Calculate the most probable speed ѵ_p the arithmetic mean speedѵ, and the root mean square speed ѵ_rms for hydrogen molecules at 0\textdegreeC.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E03-0100.htm

Solution:

Before beginning this problem, we must first understand the basic differences in the three types of speeds. The arithmetic mean speedvis obtained by summing all the speeds and dividing by the total number of molecules N. v= 1/N^N\sumi = 1v_i The symbol \sum v_i indicates the sum of the speeds v_1, v_2, v_3, ... v_N of all the N individual molecules. The equation for mean speed is v= [(8kT)/(\pi m)]^1/2 =[(8RT)/(\pi M)]^1/2 where k is Boltzmann's constant, R is the gas constant, T is the absolute temperature, m is the weight of 1 molecule, and M is the molecular weight. The most probable speed is obtained from the plot of f(v) versus v, where f(v) is the fraction of molecules with speed v. The equation of this curve f(v) = [m /(2\pikT)]^3/2 e-(mv)2 / 2kT4\piv^2 is differentiated with respect to v and set equal to zero [{df(v)}/{dv} = 0]. The shape of the curve of f (v) versus v is indicated in the accompanying figure. {df(v)} / {dv} = [m/2\pikT]^3/2 e-(mv)2 / 2kT[8\piv + 4\piv^2 {(-mv)/(kT)}] This derivative vanishes at v = v_p , called the most probable speed, where v_p = [(2kT)/(m)]^1/2 = [(2RT)/(M)]^1/2 The root-mean-square speed, v_rms, is defined by v_rms = [1/N ^N\sumi =1v2_i]^1/2 The symbol \sum v2_i indicates the sum of the squares ofthe the velocities,v_1, v_2, v_3, ..., v_n of all the N individual molecules. Since the velocity distribution is continuous, the root-mean-square velocity is obtained by multiplying each velocity squared by the probability of that velocity, integrating over all velocities, and taking the square root: v_rms = [^\infty\int_o v^2f(v) dv]^1/2 Substituting for f (v) and simplifying one obtains v_rms = [(3kT)/m]^1/2 = [(3RT)/M]^1/2 At any given temperature, these velocities are in-versely proportional to the square root of the molecular weight. Lighter molecules move more rapidly so that their average kinetic energies are exactly equal to those for the heavier molecules. To solve this problem for hydrogen at 0\textdegreeC one must 1) find the molecular weight of hydrogen in kg, 2) convert 0\textdegreeC into the absolute temperature scale, 3) substitute these values into the equations for v_p,v, and v_rms. The molecular weight of hydrogen is 2.016 g/mole, which is 2.016 × 10^-3 kg/mole. The temperature in degrees Kelvin is 273\textdegreeK. Other pertinent information R = 8.314 J/\textdegreeK -mole, \pi = 3.142. Thus, v_p = [(2RT)/M]^1/2 = [(2) (8.314 J/0K-mol) (273\textdegreeK) / (2.016 × 10^-3 Kg/mol)]^1/2 = 1.50 × 10^3 m/s v= [(8RT)/(\piM)]^1/2 = [(8)(8.314 J/0K-mol) (273\textdegreeK) / (3.142) (2.016 × l0^-3 Kg/mol)]^1/2 = 1.69 × 10^3 m/s. v_rms = [(3RT)/M]^1/2 = [(3) (8.314 J/0K-mol) (273\textdegreeK) / (2.016 × 10^-3 Kg/mol)]^1/2 = 1.84 x 10^3 m/s. The relative ratios of v_p :v: v_rms are 1 ; 1.13 : 1.23.

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Question:

Write a program that into an array. Maximum length of the string is 60 characters.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G17-0428.htm

Solution:

Before defining the program, the following points are to be noted: a) #include :It is a standard file used for reference during character I/0 b) #define statement:It is a statement used for symbol and constant declarations. c) Print a string usingTo print a string usingprintf( ), we just have to printf( ):provide the name of the array (toprintf( )) containing the string. #include /\textasteriskcenteredconstants for character I/O\textasteriskcentered/ #define MAXLINE 61 /\textasteriskcenteredmaximum size of the string\textasteriskcentered/ #define EOL '\textbackslashn' /\textasteriskcenteredassume carriage return is the EOL\textasteriskcentered/ #define NULL '\textbackslash0' /\textasteriskcenteredappend NULL to the end of the string\textasteriskcentered/ main ( ) { int index; /\textasteriskcenteredindex to count characters \textasteriskcentered/ int letter; /\textasteriskcenteredletter entered by the user\textasteriskcentered/ char string [MAXLINE]; /\textasteriskcenteredstring to be read in and stored in an array\textasteriskcentered/ /\textasteriskcenteredFollowing operations are to be performed\textasteriskcentered/ /\textasteriskcentered1. Read a character into string [ ] until\textasteriskcentered/ /\textasteriskcentered an EOL is encountered.\textasteriskcentered/ /\textasteriskcentered2. Append a NULL character to the string.\textasteriskcentered/ /\textasteriskcentered3. Useprintf( ) to write the string.\textasteriskcentered/ for (index = 0; (letter = get char ( )) ! = EOL; ++index) string [index] = letter; /\textasteriskcenteredput letter in the string\textasteriskcentered/ string [index] = NULL /\textasteriskcenteredappend a NULL character to the string\textasteriskcentered/ printf ("The string entered was %s\textbackslashn", string); /\textasteriskcenteredwrite the string\textasteriskcentered/ printf ("The number of characters is %d\textbackslashn", index); /\textasteriskcenteredwrite the no. of characters in the string\textasteriskcentered/ }

Question:

A cylinder contains an ideal gas at a pressure of 2 atm, the volume being 5 liters at a temperature of 250\textdegreeK. The gas is heated at constant volume to a pressure of 4 atm, and then at constant pressure to a temperature of 650\textdegreeK. Calculate the total heat input during these processes. For the gas, C_v is 21.0 J \bullet mole^-1\bullet K deg^-1. The gas is then cooled at constant volume to its original pressure and then at constant pressure to its original volume. Find the total heat output during these processes and the total work done by the gas in the whole cyclic process.

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/Users/wenhuchen/Documents/Crawler/Physics/D15-0532.htm

Solution:

The number of moles, n, originally present at point (1) in the P-V diagram can be calculated from the gas equation n= pV/RT = (2 atm × 5 liters)/(0.0821 liter\textbulletatm\bulletmole^-1\bulletK deg^-1 × 250\textdegreeK) = 0.487 mole. The specific heat C_p at constant pressure is C_p = C_v + R = (21.0 + 8.317)J \bullet mole^-1 \bullet K deg^-1 = 29.317 J \bullet mole^-1 \bullet K deg^-1. In going from (1) to (2), V is constant. There-fore in the first change, P/T remains constant: (from the universal gas law) P_1/T_1 = P_2/T_2 orT_2 = T_1(P_2/P_1) SinceP_2/P_1 = 4 atm/2 atm = 2, we have T_2 = 2T_1 = 2 × 250\textdegreeK = 500\textdegreeK The heat input along the change of state from (1) to (2) is H_12 = nC_v(T_2 - T_1) = 0.487 mole × 21.0 J\bulletmole^-1\bullet K deg^-1 × (500 - 250) \textdegreeK = 2558 J In the second change, from (2) to (3), P and there-fore V/T are constant V_2/T_2 = V_3/T_3 orV_3 = (T_3/T_2)V_2 = (T_3/T_2)V_1 = (650\textdegreeK/500\textdegreeK) 5 lit = 6.5 lit. Heat input during this change is H_23 = nC_p(T_3 - T_2) = 0.487 mole × 29.317 J\bulletmole^-1\bulletK deg^-1 × (650 - 500)\textdegreeK = 2143 J. The total heat input during these two processes is thus H = H_12 + H_23 = 4701 J. During the change from (3) to (4), the gas cooled at constant volume. Hence P_3/T_3 = P_4/T_4 orT_4 = (P_4/P_3)T_3 SinceP_4/P_3 = 1/2 we get T_4 = (1/2)T_3 = (1/2) × 650\textdegreeK = 325\textdegreeK. The heat rejected by the gas during this process is H_34 = nC_v(T_3 - T_4) = 0.487 mole × 21.0 J\bulletmole^-1\bulletK deg^-1 × (650 - 325)\textdegreeK = 3325 J In the second cooling process; from (4) to (1), P is kept constant; V_4/T_4 = V_1/T_1 orT_1 = (V_1/V_4)T_4 SinceV_1/V_4 = V_1/V_3 = 5 lit/6.5 lit, we get T_1 = (5/6.5) × 325\textdegreeK = 250\textdegreeK as expected. The heat output during this change is H_41 = nC_p(T_4 - T_1) = 0.487 mole × 29.317 J\bulletmole^-1 K deg^-1 × (325 - 250)\textdegreeK = 1072 J. The total heat output during the cooling processes is thus H' = H_34 + H_41 = 4397 J. The difference between heat input and heat output is 304 J. This must appear as work done by the gas, since the internal energy of the gas must be the same at the beginning and at the end of a cyclic process. The mechanical work done during the cycle is given by W = ^2\int_1 P dV + ^3\int_2 P dV + ^4\int_3 P dV + ^1\int_4 P dV, which is the area enclosed by the rectangular figure in the P-V diagram. This is a rectangle of height 2 atm and length 1.5 liters. The area under the curve is thus W = 2 × 1.013 × 10^6 dynes\bulletcm^-2 × 1.5 × 10^3 cm^3 = 3.04 × 10^9 ergs = 304 J, which agrees with the net heat input.

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Question:

(a) Set up the truth table for each of the following Boolean expressions: (i)\textasciitilde A (ii)A + B(inclusive or) (iii)A \textbullet B(and) (b) Show the equivalent switching circuits for (i) - (iii). (c) Show the gate representations of (i) - (iii).

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G03-0049.htm

Solution:

(a) A truth table is set up by specifying all the possible combinations of values of the Boolean variables, then evaluating each term according to the rule of relation of the operation. There are 2 possible values for each variable: 0 or 1. Then, if there are n variables, there will be 2^n possible combinations of the variables, and therefore, the truth table will have 2^n rows of entries. (i) Since there is only one variable, the truth table has only 2^1 = 2 rows of entries. A \backsimA 0 1 1 0 The \backsim operation (called complement or invertion ) changes 0 to 1 or 1 to 0. (ii) Since there are 2 variables, the truth table has 2^2 = 4 rows of entries: A B A + B 0 0 1 1 0 1 0 1 0 1 1 1 The + operation (called disjunction) yields a value of 0 if and only if both variables have a value of 0. Otherwise, it yields a value of 1, (iii) Since there are 2 variables, the truth table has 2^2 = 4 rows of entries: A B A \bullet B 0 0 1 1 0 1 0 1 0 0 0 1 The \textbullet operation (called conjunction) yields a value of 1 if and only if both variables have a value of 1. Otherwise, it yields a value of 0. (b) There is a one-to-one correspondence between Boolean expressions and switching circuits. An open switch or nonconductor can represent the 0-condition and a closed switch or conductor can represent the 1- condition. Boolean algebra can be used to determine what circuits are necessary to perform a specific logic function. The action of a two-position switch with an open position and a closed position can be represented by a Boolean variable. (i) If A represents a switch in the closed position, then \textasciitilde A would represent the same switch in the open position, as shown in figure 1. (ii) The + operation yields a value of 0 if and only if both vari-ables have a value of 0. Therefore, the switching circuit needed to perform the + operation must be open (nonconducting) if and only if both switches are open. The parallel connection is the switching circuit realization of the + operation; as shown in figure 2. (iii) The \bullet operation yields a value of 1 if and only if both variables have a value of 1. The switching circuit needed to perform the \bullet operation, therefore, must be closed (conducting) if and only if both switches are closed. The series connection is the switching circuit realization of the \bullet operation, as shown in figure 3. (c) Gates are switches that are sensitive to high and low voltages. If 0 represents a low voltage and 1 represents a high voltage, the output of a gate is either 0 or 1, depending on the type of gate. (i) The \textasciitilde operation changes 0 to 1 and 1 to 0. The gate needed to perform the \textasciitilde operation, therefore, must output a 1 for a 0 input and output a 0 for a 1 input. This is the operation of a NOT gate shown in figure 4. (ii) An OR-gate outputs a 0 if and only if all inputs are 0. Other-wise, it outputs a 1. This corresponds to the + operation and is shown in figure 5 (iii) An AND-gate outputs a 1 if and only if all inputs are 1. Otherwise, it outputs a 0. This corresponds to the \textbullet operation; shown in figure 6.

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Question:

What are the major types of food and how much energy doeseach type yield?

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/Users/wenhuchen/Documents/Crawler/Biology/F17-0413.htm

Solution:

The food consumed by a human is a combination of carbohydrates, fats, proteins, water, mineral salts, and vitamins. All of these are essential tolife but only the carbohydrates, fats and proteins can provide energy. The vitamins and minerals act as coenzymes and chemical cofactors in manymetabolic reactions. The standard chemical unit of energy is the calorie. One calorie is theamount of energy required to raise the temperature of one gram of waterby one degree Celsius. In biology, kilocalorie (kcal.) is used and is equi-valentto 1000 calories. This is enough energy to raise the temperatureof a liter (about one quart) of water by one degree Celsius. Most proteins and all carbohydrates and fats can be degraded into carbon dioxideand water. The energy released from this breakdown is called the heatof combustion. The heat of combustion of the three major types of compoundsare as follows: fat - 9.5 kcal/gram, protein - 4.5 kcal/gram, carbohydrate- 4.0 kcal/gram. Thus, fats contain more energy per gram thancarbohydrates and proteins. Physiologically, more than one half of the energy of combustion is lostas heat while the rest (about 38%) is converted into cellular energy in theform of ATP. The actual yield from one mole of glucose is 36 ATP plus heat, and the yield from a mole of fat is 463 ATP plus heat. These are equivalentto .21 mole ATP/(gram glucose) and .55 moles ATP/(gram fat) respectively. In terms of the energy actually produced in the cell, fats producetwice that of carbohydrates. Proteins yield about as much energy ascarbohydrates on a per gram basis. Proteins, however, are rarely fully metabolized. Their constituent amino acids are used for cellular protein pro-duction. It takes less energy for the cell to utilize existing amino acids thanto degrade them and then syn-thesize new ones.

Question:

Two moles of gaseous NH_3 are introduced into a 1.0-liter vessel and allowed to undergo partial decomposition at high temperature ac-cording to the reaction 2NH_3 (g) \rightleftarrows N_2 (g) + 3H_2 (g) . At equilibrium, 1.0 mole of NH_3 (g) remains. What is the value of the equilibrium constant?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E09-0312.htm

Solution:

This problem involves substitution into the expression for the equilibrium constant for this reaction, K = {[N_2 ] [H_2 ]^3} / [NH_3]^2 Since 1.0 mole of NH_3 remains, 2.0\Elzbar 1.0 = 1.0 mole of NH_3 was consumed. Also, since one mole of N_2 and three moles of H_2 are formed per 2 moles NH_3 consumed, at equilibrium, there are 3/2molesof H_2 and 1/2 mole of N_2 . The equilibrium concentra-tions are therefore [NH_3 ] = 1.0 mole/1.0 liter = 1.0m, [N_2 ] =(1/2 mole) / (1.0 liter) = 0.5M, and [H_2] = (3/2moles) / (1.0 liter) =1.5M. Substituting these into the expression for K gives K = {[N_2] [H_2]^3} / [NH_3]^2 = {(0.5) (1.5)^3} / (1)^2 = 1.6875 .

Question:

Find the difference in characteristicvibrationalfrequency of a diatomic molecule of mass 20amuif (a) the mass is equally distributed between the two ends, and (b) one atom of 1amu is at one end and an atom of 19amuis at the other.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E18-0701.htm

Solution:

Thevibrationalfrequency is frequency of the oscillation between the two atoms of the molecule. It is expressed quantitatively by ѵ = (1/2) \pi \surd(K/M_eff) where ѵ is thevibrationalfrequency, K a force constant, andM_effthe effective mass of the molecule. M_eff = [(M_A M_B)/(M_A + M_B)] where M_A is the mass of one atom and M_B is the other. (a) If the mass is equally distributed between both atoms, each weighs 10 amu. M_eff = [(10 × 10)/(10 + 10)] = 5 Assume that K = 1 ѵ = (1/2\pi)\surd(1/5) = 0.0711 ѵ = (1/2\pi)\surd(1/5) = 0.0711 (b)M_eff= [(1 × 19) / (20) = (19/20) ѵ = (1/2\pi)\surd(19/20) = 0.1632 Therefore, the atoms in the molecule described in b are vibrating more quickly than those in molecule a.

Question:

Write apseudocodedprogram to set up a simple system of computer- aided instruction (CAI). It should keep track of the student's grades for each lesson.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G12-0288.htm

Solution:

Computer teaching methods are used in many schools and businesses. It allows students to study when they want, and it eliminates some of the distractions of the classroom. The cathode ray tube (CRT) is a very useful device for these systems. This device, which looks like a TV screen, lets a student see segments of a lesson, a few at a time. After a sufficient amount of material has been presented, a short quiz of one question appears. The student types in his choice of answer, and the computer evaluates it. If the answer is correct, the lesson continues. If incorrect, a message pointing out the mistake comes on the tube. The structure of a lesson is a series of pages. Each page has a question associated with a set of possible answers. The answer selected determines what page will appear next. The organization of a simple CAI system is outlined below. Pages are numbered 1 to M. The page following a correct answer J is given in the array element PAGE (I,J). We store O in PAGE (I,J) if J corresponds to the end of the session. We could also say that PAGE (I,J) equals \rule{1em}{1pt}1 if J is the wrong answer. The array SCORE keeps track of the student's performance throughout the course. Integer I,J,K,L,M,N, STUD \rule{1em}{1pt} SCORE, SCORE(M,N), PAGE (M,N) STUD\rule{1em}{1pt}SCORE \leftarrow 0 I \leftarrow 1 do while1\not = 0 output K 4\leftarrow \rule{1em}{1pt}1 PAGE I do while K = \rule{1em}{1pt}1 input response J K \leftarrow PAGE (I,J) If K = \rule{1em}{1pt}1, output 'TRY AGAIN, PLEASE' end do while STUD\rule{1em}{1pt}SCORE \leftarrow STUD\rule{1em}{1pt}SCORE + SCORE (I,J) I \leftarrow K end do while end program

Question:

The graph shows a displacement-time curve for a motion along a straight line. What are the average velocities from A to B and from A to C?

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/Users/wenhuchen/Documents/Crawler/Physics/D03-0065.htm

Solution:

The average velocity of an object in motion is the distance d it travels divided by the time t it takes in transit. v_av = d/t Looking at the figure: from A to Bv_av= d_AB/t_AB = (3 m)/(2 s) = 1.5 m/sec from A to Cv_av= d_AC/t_AC = (5 m)/(6 s) = 0.83 m/sec.

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Question:

The variable capacitor in the figure is connected to a battery of e.m.f. V and internal resistance r. The current in the circuit is kept constant by changing the capacitance. (Assume that we are able to increase the capacitance in-definitely in order to accomplish this). a)Calculate the power supplied by the battery. b)Compare the rate at which energy is supplied by the battery with the rate of change of the energy stored in the capacitor.

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/Users/wenhuchen/Documents/Crawler/Physics/D20-0658.htm

Solution:

a)The current in the circuit is related to the voltage across the capacitor as C (dV_c /dt) = I As the capacitor charges up, V_C rises toward its limiting value and its rate of change goes to zero. Therefore, in order to keep I constant, capacitance, C, should compensate for this drop in (dV_c /dt) (observe that, as time t \rightarrow \infty, (dV_c /dt) \rightarrow 0 and we should have C_\infty \rightarrow \infty ). Ignoring the practical difficulties of maintaining a constant I, asymp-totically we have a continuous charge flow in the circuit [(dq)/(dt)] = I, andV_C = V. The battery supplies a constant power since its e.m.f. is constant, P_B = VI . b)The capacitor is being continuously charged. Once the voltage across the capacitor stabilizes at t \rightarrow \infty, we have V = Ir + V_C, \infty orV_C, \infty= V - Ir. The energy stored in the capacitor becomes W = (1/2) q_\infty V_C, \infty = (1/2) q_\infty (V - Ir). The rate of change of W with time is P_C = [(dE)/(dt)] = (1/2) [(dq_\infty )/(dt)] VC , \infty= (1/2) I (V - Ir) The rate of dissipation of energy in the resistor r is P_r = I^2 r, Therefore the power supplied directly to the capacitor is P = P_B - P_r = IV - I^2 r = I(V - Ir) , which is twice as large as that used in charging the capacitor: P = 2P_C . The battery supplies twice as much energy to the capacitor as is stored in the capacitor. The difference is the work done by the capacitor against the external agent that is changing the capacitance.

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Question:

Calculate the sedimentation coefficient of tobacco mosaic virus from the fact that the boundary moves with a velocity of 0.454 (cm/hr) in an ultracentrifuge at a speed of 10,000 rpm at a distance of 6.5 cm from the axis of the centrifuge rotor.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E22-0818.htm

Solution:

When a sedimenting boundary of protein moves at a constant rate, the centrifugal force just counter-balances the frictional resistance of the solvent. The rate of sedimentation is expressed as the sedimentation coefficient, s s = [dx/dt] / [\omega^2 x], where x is the distance from the center of rotation, \omega is the angular velocity in radians per second, and t is the time in seconds. One is given (dx/dt), \omega and x in the problem, but they are not in units that are used in determining s. Converting to proper units: dx/dt; This should be in cm/sec but is given in cm/hr. 0.454 cm/hr = 0.454 cm/hr × 1 hr/3600 sec = 1.26 × 10^-4 cm/sec. \omega; \omega is given as 10,000 rpm (revolutions per minute) but must be expressed in radians per second for use. There are 2\pi radians in one revolution. (10,000 rmp × 2\pi) / (60 sec) = 1.05 × 103(radians/sec) Solving for s: s= [dx/dt] / [\omega^2 x] = [(1.26 × 10^-4 cm/sec)] / [(1.05 × 10^3/sec)^2 6.5cm] = 1.76 × 10^-11 sec.

Question:

Use the uncertainty principle to show that the rest mass of a photon is zero.

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/Users/wenhuchen/Documents/Crawler/Physics/D33-0993.htm

Solution:

Let us assume the photon rest massm_\Upsilonis not zero. The Heisenberg uncertainty principle isp_x\Deltax\geq (h/4\pi) wherep_xis the x component of the photon momentum. Hence, m_\Upsilonv\Deltax\geq (h/4\pi). But, for a photon, v = c and m\Upsilon\geq [h/{4\pi(\Deltax)c}]. Coulomb's law for the force between two charged particles is F = (kq_1 q_2/d^2) and this force is exerted over arbitrarily large values of d. Consider 2 charges separated by a distance equal to the diameter of the Milky Way galaxy (1 × 10^21 m). According to the exchange picture, the electric force between the charged particles is achieved by the exchange of virtual (unobservable) photons between two electrically charged particles. Any measurements made to detect the virtual photon's presence would require measurement precisions ex-ceeding those permitted by the uncertainty principle. The uncertainty in the position of these photons is then the distance over which they are exchanged, ( in this instance 1 × 10^21 m ) for we know that the photons must be somewhere between the 2 charges. Thus m = [h/{4\pi(\Deltax)c}] = [(6.6 × 10^-34 J s)/{(4) (3.14) (1 × 10^21 m)(3 × 10^8 m/s)}] = 1.8 × 10^-64 kg. This mass is very much smaller than the mass of any other particle and for all practical purposes is zero. The dia-meter of the Milky Way does not have to be used, however, because according to Coulomb's law, force still exists for even larger distances d. The force only vanishes as d tends toward infinity, but if the distance becomes very large, so does the uncertainty in the position of the photon; hence the photon mass must be precisely zero.

Question:

The rate of steady precession \varphi ̇of the cone shown about the vertical is observed to be 20 rpm. Knowing that r = 100 mm and h = 200 mm, determine the rate of spin \Psi̇ of the cone about its axis of symmetry if \beta = 135\textdegree.

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/Users/wenhuchen/Documents/Crawler/Physics/D37-1074.htm

Solution:

Choose a coordinate system \^{\i}, \^{\j}0, k ̂ which is fixed in the cone with k ̂ along the axis of symmetry. K̂ is a unit vector fixed in space along the vertical. The cone has radius r, height h, and its center of mass is located a dis-tance a^\ding{217} = ak˄ from 0 on the symmetry axis. The cone precesses about the vertical with a constant angular velocity p = \varphi ̇. The spin about the symmetry axis is s = \Psi̇. From figure 1, K˄ = sin \beta\^{\i} + cos \betak˄. The angular momen-tum vector lies in the x, z plane. \omega^\ding{217} = \omega_1\^{\i} + sk˄. The center of mass, C, rotates about \omega^\ding{217} with a velocity v^\ding{217}_C = \omega^\ding{217} × a^\ding{217} = (\omega_1\^{\i} + sk˄) × ak˄ = \rule{1em}{1pt}a\omega_1\^{\j}. In the fixed system, C rotates around the vertical with velocity v^\ding{217}_c = pk˄ × a^\ding{217} = (P sin \beta\^{\i} + p cos \betak˄) × ak˄ = \rule{1em}{1pt}ap sin \beta\^{\j}. Since both velocities are for the same point, a comparison gives \omega_1 = p sin\beta. The only force acting on the top is the weight W^\ding{217} =mgk˄. It produces a torque, \tau^\ding{217}, about 0 of \tau^\ding{217} = a^\ding{217} × W^\ding{217} = ak˄ × mgk˄ = ak˄ × mg(sin \beta\^{\i} + cos \betak˄) = amg sin \beta\^{\j}.(1) The angular momentum is given by L^\ding{217} = I_x\omega_1\^{\i} + I_zsk˄. L^\ding{217} also lies in the x, z plane and rotates with it about the vertical. From Newton's Second Law \tau^\ding{217} = (dL^\ding{217} / dt)the rate of change of a fixed vector in a rotating coordinate system is dA^\ding{217} / dt = \Omega^\ding{217} × A^\ding{217}. In this case \tau^\ding{217} = pk˄ × L^\ding{217} = P(sin \beta\^{\i} + cos \betak˄) × (I_x\omega_1\^{\i} + I_Zsk˄) = (p^2I_xcos \beta sin \beta \rule{1em}{1pt} pI_Zs sin \beta)\^{\j}. Comparing this with equation (1) and substituting in \omega_1 = p sin \beta gives amg sin \beta = p^2I_x cos \beta sin \beta \rule{1em}{1pt} psI_z sin\beta. Solving for the spin gives s = [(pI_x cos \beta) / (I_z)] \rule{1em}{1pt}(amg / pI_z).(2) Next, determine the moments of inertia. For the right circular cone shown in Figure 2 the center of mass is at z^\ding{217} = (h / 4) and the moments of inertia are I_z = (3 / 10)mR^2andI_x' = (m / 20)(3R^2 + 2h^2). For this problem we need I_x, the moment of inertia about the x-axis through the vertex. To obtain this use the parallel axis Theorem, I_0 = I_c + mr^2 where r is the perpen-dicular distance between the parallel lines through 0 and C. I_C = I_x' \rule{1em}{1pt}mr21= I_x' \rule{1em}{1pt}m(h^2/16) Ix= I_C + mr2_2= I_C + (9/16)mh^2 = Ix'\rule{1em}{1pt}(1/16)mh^2+ (9/16)mh^2 =I_x' +(1/2)mh^2 = (3/20)mR^2 + (1/10)mh^2 + (1/2) mh^2 = (3/20)m(R^2 + 4h^2). So, I_x / I_z = [{3/20m(R^2 + 4h^2)} / (3/10 mR^2)] = 1/2[1 + 4(h / R)^2]. For this problem p = 20 rpm = 0.33 (rad / sec), R = 10 cm, h = 20 cm, \beta = 135\textdegree and a = h \rule{1em}{1pt}(1 / 4)h = (3 / 4)h = 15 cm. Substitu-tion of these values into equation (2) gives s = (\pi/2)[1 + 4(20/10)^2]cos 135\textdegree\rule{1em}{1pt} [(15cm 980cm/sec^2) / {\pi rad/sec(3/10)(10 cm^2)}] = \rule{1em}{1pt}18.88 rad / sec \rule{1em}{1pt} 155.97 rad / sec = \rule{1em}{1pt}174.85 rad / sec = \rule{1em}{1pt}1670 rpm. This is the angular velocity of the cone about its symmetry axis. The negative sign means s^\ding{217} points in the negative k˄ direction.

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Question:

The properties of diamond and graphite differ vastly. Explain these differences on the basis of their fundamentally different space lattices.

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Solution:

The arrangement (location) of all atoms in a unit of a crystal is termed a space lattice, i.e., those patterns of atoms describing the atomic or molecular arrangements. This problem calls for a comparison of such lattices in a diamond and graphite. The accompany-ing figures show the space lattices in question. The diamond lattice surrounds each carbon atom by 4 others in a tetrahedral configuration. There exist covalent bonds between the carbon atoms, which means they share electrons. These covalent forces are responsible for the hard nature and high melting point of the substance. On the other hand, graphite has covalent bonding in two-dimensional, planar, hexagonal rings. As the figure suggests, the sheets of these atoms are held together by weak forces of attraction. This means, that the layers can slide easily over each other. This, then, accounts for the lubricating proper-ties and flakiness associated with this material.

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Question:

Insect flight differs from bird flight in that insects have no muscles attached to the wings. Describe the mechanism of insect flight.

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Solution:

Insect wings are attached directly to the body wall over a fulcrum. The fulcrum system of the wings and body wall function like a seesaw mechanism. As contractions of the muscles occur changing the shape of the thorax, the wings respond by up and down motions (Figure 1) The dorsal plate of a segment in the insect body is termed the tergum, and the ventral plate is the sternum. Contraction of vertical muscles of a segment pulls the tergum to the sternum, causing the wings, on the oppo-site side of the fulcrum, to be raised. The contraction of the longitudinal muscles of a segment causes the tergum to bulge upward, and the wing moves downward. The movements of the thoracic body wall segments are barely perceptible, but the lengths of the levers on the two sides of the fulcrum are very different, and the movement of the wings is great in comparison (see Figure 2). The vertical muscles are referred to in some textbooks as the tergo sternal muscles. Their name indicates their function, which is to move the tergum and sternum towards and away from each other. In most insects, two pairs of wings are present, both usually function in flying. The anterior pair of wings functions in propulsion, while the posterior wings are balancing organs. They are termed halteres and They are termed halteres and function as gyroscopes for control of flight stability. This wing arrangement is seen in the mechanism of flight for mosquitoes and flies. In the grasshoppers and beetles, only the posterior wings are flying organs. The anterior wings are protective devices for the flying pair. Insect flight muscles are very powerful. The fibrils are relatively Insect flight muscles are very powerful. The fibrils are relatively large and the mitochondria are huge - about half the size of a human red blood cell.

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Question:

What process in the plant is responsible for the phenomenon ofguttation?

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Solution:

Droplets on a leaf commonly seen early in the morning are usually assumed to be dew, which is water that has condensed from the air. Actually, these droplets may have resulted fromguttation, and come from the leaf itself through special openings on the leaf margins (and not through the stomata). Like trans-piration,guttationis the loss of water through plant leaves, yetguttationoccurs only in those plants with short stems. The phenomenon ofguttationcan be explained by the effect of root pressure. Root pressure is a term given to the forces opera-ting at the junction of root and stem in pushing a column of water up the xylem. The existence of root pressure has been experimentally demonstrated by removing the upper part of the stem of a well-watered plant, and attaching an air-tight piece of glass tubing to the remaining stump. Water actually is seen to rise in the tubing from the stump up to a height of one meter or more. If roots are killed or deprived of oxygen, root pressure falls to zero, indicating that the pushing force must be generated by the root. Root pressure has been found to equal 6 to 10 or more atmospheres of pressure even in plants that raise water to a height of less than 1 meter. Guttation usually occurs in plants when little or no transpiration is taking place. At night, for instance, when the stomata are closed, transpiration is prevented or kept to a minimum. When the air is very moist or when the air currents are still, transpiration occurs only slowly. In these cases,guttationreplaces trans-piration, and is the chief force pushing water up the stem to the leaves. A small part of the water is actually utilized by themesophyllin photosynthesis, and a large part of it passes through special pores of the leaf and appears as droplets on its surface.

Question:

In a car which is accelerating, a plumb line hanging from the roof maintains a constant angle of 30 with the vertical. What is the acceleration value?

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Solution:

Since the plumb line maintains a constant angle, the acceleration of the car must be constant. There are only two forces acting on the bob of the plumb line, the weight W^\ding{217} = mg acting downward, and the tension T^\ding{217} in the string. Splitting T^\ding{217} into its vertical and horizontal components (see figure) one obtains for the vertical direction T cos 30o = mg,(1) This results because the bob does not move in the vertical direction, hence the vertical forces must balance. By Newton's second law, T sin 30o = ma,(2) since the horizontal force must produce acceleration a to match the motion of the car. Dividing equation (2) by equation (1) T sin 30o/T cos 30o = tan 30o = a/g Thusa = g tan 30o = g(1/\surd3) = (32 ft/sec^2)/(1.732) = 18.47 ft/sec^2 .

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Question:

In a given chemical compound, there is a bond between "A" and "B". Theelectronegativitiesofthe involved elements are , respectively, 1.2 and 2.1 What is the probable nature of the bond between "A" and "B"?

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Solution:

ELectronegativityis a measure of the tendency of an atom to attract shared electrons in a molecule. The greater the difference in electronegativity between two elements, the greater the ionic character of the bond. When the difference inelectronegativitiesis greater than 2.0, the bond is considered ionic. Theelectronegativitydifference in this problem is 2.1 - 1.2 or 0.9. Thus, while the bond is not ionic, the difference suggests a degree of polarity in this covalent bond.

Question:

Write a program that calculates the number of permutations of N things taken R at a time.

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Solution:

The factorial of N is written as N! and is equal to N(N-1) (N-2) ... (3) (2) (1). If N things are taken R at a time there are N choices for the first item, N-1 choices for the second, N-2 choices for the third, N-3 choices for the fourth and, continu-ing, N-R+1 choices for theRthitem. By the Fundamental principle of Counting, the final number of permutations is N(N-1)(N-2)(N-3) ... (N-R+1). This number is denoted by ^NP_R . A more compact representation is N! / (N-R)! since this equals N(N-1) ... (N-R+1). The program appears as follows: 1\OREAD N,R 2\OIF N < \O THEN 1\O\O 3\OLET P = 1 4\OFOR X = N TO N-R+1STEP -1 5\OLET P = P \textasteriskcentered X 6\ONEXT X 7\OPRINT N; "THINGS"; R; "AT A TIME"; P; "PERMUTATIONS" 8\OGO TO 1\O 9\ODATA 8, 3, 4, 4, \O, \O 1\O\OEND

Question:

Give the formation of a Procedure in COBOL and explain in brief the terms statements, sentences, paragraph and section as applied to COBOL.

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Solution:

Procedure Formation: Procedures are formed by combining one or more sentences into a paragraph and one or more paragraphs into a section. Each paragraph or section must be preceded by a procedure name. These may be either numeric, alphabetic, or alphanumeric. If numeric, leading zeros are significant (i.e., 18 is not the same as 018). Allprecedurenames must start at position A (column 8) on the COBOL programming form, be a maximum of 30 characters in length and be followed by a period. e.g.,010200 PROCEDURE DIVISION. 030000 CHECK-STOCK-NUMBER. 060100 FINISH-MASTER. The characters underlined by the break lines are procedure names. Statements and Sentences: Statementsformthe basic functional component of COBOL procedures. Just as clauses make up sentences in normal English-language construction, so statements make up COBOL sentences. A sentence may con-tain one or more statements. The rules governing formation of sentences are: a) Each sentence may be made up of one or more statements. b) Each sentence must be terminated by a period. c) Separators are optional and used to enhance readability. The allowable separators are: b(space, blank) b b ; (semicolon) , (comma) d) Two contiguous separators are not permissible. e) Separators may be used in the following places: -Between statements e.g., ADD A TO B GIVING C; PERFORM 180 THRU 198 -Between a condition and statement-1 in a condition sentence. e.g., IF condition; statement-1 -Between statement-1 and ELSE in a conditional statement. e.g., statement-1; ELSE statement 2. Paragraphs: One or more sentences may be combined to form a paragraph. Essentially, a paragraph expresses a single procedure to be carried through in the main program. Each paragraph must be preceded by a procedure name. When-ever reference is made, it is to an entire paragraph and not to individual sentences contained therein. If reference is made to a single sentence, that sentence must be defined as a complete paragraph and must be preceded by a procedure name. Sections: One or more paragraphs can be grouped into a section. The section is the largest unit in COBOL to which a procedure-name may be assigned. This is done by writing a procedure-name, followed by a key word SECTION, followed by a period; the remainder of the line must be left blank.

Question:

A solution is prepared by dissolving 464 gNaOHin water and then diluting to one liter. The density of the re-sulting solution is 1.37 g/ml. Express the concentration ofNaOHas (a) percentage by weight, (b)molarity, (c) normality, (d) molality, (e) mole fraction.

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Solution:

(a) The percentage by weight ofNaOHin this solution is found by dividing the weight ofNaOHpresent, 464 g, by the weight of the solution and multiplying by 100. The weight of the solution is found by using the density of the solution. The density, 1.37 g/ml, tells the weight of 1 ml of solution, namely, 1.37 g. In one liter, there are 1000 ml, thus the weight of the solution is 1000 times the weight of one ml. weight of 1 liter = 1000 ml × 1.37 g/ml = 1370 g. The percentage by weight of theNaOHin the solution can now be found. percentage ofNaOH= [(weight ofNaOH)/(weight of solution)] × 100 = (464 g/1370 g) × 100 = 33.9 %. (b) Themolarityis defined as the number of moles in one liter of solution. Themolarityin this case can be found by determining the number of moles in 464 g ofNaOH, which is the amount ofNaOHin one liter. The number of moles can be found by dividing 464 g by the molecular weight ofNaOH. (MW ofNaOH= 40.) no. of moles =(464 g/MW) = [(464 g)/(40 g/mole)] = 11.6 moles. Themolarityof this solution is thus 11.6 M. (c) The normality of a basic solution is the number of moles of ionizableOH^- ions in one liter of solution. There is oneionizableOH^- ion in eachNaOHas shown by the equation: NaOH\rightleftarrows Na^+ + OH^- Therefore, there are the same number of OH^- ions in the solution as NaOHmolecules dissolved, Thus, themolarityequals the normality. normality =molarity× 1ionizableOH^- normality = 11.6 N (d) Themolalityis defined as the number of moles present in 1 kg of solvent. One has already found that the solution weighs 1370 g or 1.37 kg and that there are 11.6 moles ofNaOHpresent. Therefore themolality can now be found. molality= [(no. of moles present)/(no. of kg present)] = [(11.6moles)/ = [(11.6moles)/ {1.37 kg -.339(1.370 kg)}] = 12.8 m {1.37 kg -.339(1.370 kg)}] = 12.8 m (e) The mole fraction is equal to the number of moles of each component divided by the total number of moles in the system. The components in this system are H_2O andNaOH. One already has found that there are 11.6 moles ofNaOHpresent, but not the number of moles of H_2O. This can be found by determining the weight of the water and dividing it by its molecular weight. This solution weighs 1370 g, theNaOH weighs 464 g, thus the weight of the water is equal to the difference of these two figures, weight of H_2O = weight of solution - weight ofNaOH = 1370 g - 464 g = 906 g. One can now find the number of moles of H_2O present. (MW of H_2O = 18.) no. of moles = [(906 g)/(18 g/mole)] = 50.3 moles The total number of moles in the system is the sum of the number of moles of H_2O and of theNaOH. no. of moles in system = moles H_2O + molesNaOH = 50.3 + 11.6 = 61.9 moles One can now find the mole fractions. mole fraction = [(no. of moles of each component)/(total no. of moles in system)] mole fraction of H_2O = (50.3 / 61.9) = .81 mole fraction ofNaOH= (11.6 / 61.9) = .19.

Question:

The simple harmonic motion of the pendulum in figure (a) is slowed as a result of air friction. If the frictional force is proportional to the velocity of the bob, (a) find the displacement of the pendulum as a function of time, (b) calculate the rate of energy dissipation by the damped harmonic motion of the pendulum. Assume that the oscillation fre-quency of the pendulum is much greater than the rate of damping (the weak damping limit).

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Solution:

(a) The restoring force on the pendulum for a small angular displacement \texttheta is F = - mg sin \texttheta \approx -mg \texttheta where m and g are the mass of the pendulum and the gravitational acceleration, respectively. The frictional force is f = - bv where b is a proportionality constant, and v is the velocity. The minus sign in the expressions for F and f means that they oppose an increase in \texttheta and v, respectively. The displacement along the arc is x = l\texttheta The equation of motion for displacement x is m[(d^2x)/(dt^2)]= F + f = - mg \texttheta - bv = - mg (x/l) - b (dx/dt) which gives the following differential equation for x: [(d^2x)/(dt^2)] + (b/m)(dx/dt) + (g/l)x = 0(1) In the weak damping limit, b is small and the amplitude of the oscillation is damped as shown in Fig. b. Therefore, we try the solution x(t) = A e^-t/\tau sin \omegat in (1), where e^-t/\tau is the equation of the amplitude envel-ope. We assumed that at t = 0, x = 0. We have [(d^2x)/(dt^2)] = A e^-t/\tau [{(1/\tau^2) - \omega^2} sin \omegat - (2\omega/\tau) cos \omegat] (b/m)(dx/dt) = A e^-t/\tau [(b\omega/m) cos \omegat -(b/m\tau) sin \omegat] (g/l) × = A e^-t/\tau (g/l) sin \omegat. Substituting in (1), we obtain A e^-t/\tau [{(1/\tau^2) - \omega^2 - (b/m\tau) + g/l} sin \omegat + {(b\omega/m) - (2\omega/\tau)} cos \omegat] = 0 The coefficients of sin \omegat and cos \omegat must equal zero separately; (b\omega/m) - (2\omega/\tau) = 0 (1/\tau^2) - \omega^2 - (b/m\tau) + (g/l) = 0 or(1/\tau) = (b/2m)(2) \omega^2 = (g/l) + (1/\tau^2) - (b/m\tau) = (g/l) + (b^2/4m^2) - (b^2/2m^2) = (g/l) - (b^2/4m^2). Hence the angular velocity of the motion is \omega = \surd{\omega^2 _0 - (b^2/4m^2)} = \omega_0 \surd{1 - (1/2m\omega_0)}^2,(3) where \omega^2 _0 = g/l. For weak damping, (b/2m\omega_0) = (1/\omega_0\tau) < < 1, and (3) becomes \omega \approx \omega_0 {1 - (b/m\omega_0)} = \omega_0 - (b/m).(4) (b) The kinetic energy \cyrchar\cyrk is \cyrchar\cyrk= 1/2 mv^2 = 1/2 m(dx/dt)^2 = 1/2 m[A e^-t/\tau {\omega cos \omegat - (1/\tau) sin \omegat}] = 1/2 m A^2 e^-t/\tau [\omega^2 cos^2 \omegat - (1/\tau^2) sin^2 \omegat - (2\omega/\tau) sin \omegat cos \omegat](5) Now, let us consider the time average of \cyrchar\cyrk. For this we substitute the following trigonometric identities in (5). 2 sin \omegat cos \omegat= sin 2\omegat, cos^2 \omegat= (1 + cos 2 \omegat)/2 sin^2 \omegat= (1 - cos 2 \omegat)/2 \cyrchar\cyrk = 1/2 m A^2 e^-2t/\tau[1/2 \omega^2 + (1/2\tau^2) + 1/2 \omega^2 cos 2\omegat - (1/2\tau^2) sin 2\omegat - (\omega/\tau) sin 2\omegat] The time average of a pure sine wave is zero, i.e. < cos 2\omegat > = < sin 2\omegat > = 0 Therefore, only constant terms in the expression for \cyrchar\cyrk survive when we take its time average < \cyrchar\cyrk > = 1/4 mA^2 e^-2t/\tau[\omega^2 + (1/\tau^2)]. We see that the average kinetic energy decays exponentially. The potential energy U is U= mgh = mgl (1 - cos \texttheta) \approx 1/2 mgl \texttheta2 \approx 1/2 mgl (x/l)^2 = 1/2 (mg/l) x^2 \approx 1/2 (mg/l) A^2 e^-2t/\tau sin^2 \omegat \approx 1/4 (mg/l) A^2 e^-2t/\tau (1 - cos 2\omegat) The time average of U is < U >\approx 1/4 (mg/l) A^2 e^-2t/\tau Therefore, the time average of total energy is < E >= < U > + < \cyrchar\cyrk > = 1/4 A^2 e^-2t/\tau[(mg/l) + m\omega^2 + (m/\tau^2)] = 1/4 A^2 e^-2t/\tau[(mg/l) + (mg/l) - (b^2/4m) + (b^2/4m)] = 1/2 \omega^2 _0 A^2 e^-2t/\tau. The rate of change of energy with time has a negative sign since the energy decreases. Therefore, the rate of dissipation, P, is positive since it is the rate of decrease (negative rate of increase) of the energy < P >= -(d/dt) < E > = {(\omega2_0 A^2)/\tau} e^-2t/\tau = (2 < E >/\tau).

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Question:

The flywheel of a cutting machine has a mass of 62.5 slugs and a radius of gyration of 3 ft. At the beginning of the cutting stroke it has a rotational speed of 120 rev\bulletmin^-1, and at the end a speed of 90 rev\bulletmin^-1. If it cuts in a 6-in. stroke, what is the average cutting force exerted?

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Solution:

The energy lost during the stroke is the difference between the rotational kinetic energies of the flywheel at the beginning and at the end of the operation. If I = Mk^2 is the moment of inertia of the flywheel where k is its radius of gyration, and \omega_0 and \omega the initial and final rotational speeds, then the energy lost is (1/2)I(\omega^2_0 - \omega^2) = (1/2) Mk2(\omega^2_0 - \omega^2). This energy is lost in producing the cutting stroke. If F^\ding{217} is the average cutting force exerted over the distance d^\ding{217}, by the work energy theorem, the work done by F^\ding{217} equals the change in kinetic energy of the flywheel. Hence F^\ding{217} \textbullet d^\ding{217} =Fd= (1/2) Mk2(\omega^2_0 - \omega^2) orF = [(1/2) Mk2(\omega^2_0 - \omega^2)]/d = [(1/2)(62.5sl)(9 ft^2)(14400 - 8100) rev^2\bulletmin^-2]/[(1/2)ft] In order to obtain F in conventional units, note that 1 rev\bulletmin^-1 = [(2\pi)/(60)] red\bullets^-1 Hence F = [(1/2)(62.5sl)(9 ft^2)(6300)(4\pi^2/3600 rad^2\bullets^-2 )]/[(1/2)ft] F \approx 38861.6 lb.

Question:

The yellow light of a sodium lamp has an average wavelength of 5890 \AA. Calculate the energy in (a) electron volts and (b) kilocalories per mole.

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Solution:

Experiments involving light suggest that it is either made up of waves or particles. This problem deals with the particle nature of light. The following equations are used. E =hѵandc =\lambdaѵ, where E is the energy of the particle of light, h Planck's constant, 6.626 × 10^-34 Js, ѵ the frequency of the light, c the speed of light and \lambda is the wavelength of the light. One can derive a suitable expression to solve this problem from these equations. This is done by substituting c/\lambda for ѵ. Thus, E = (hc/\lambda). \lambda = 5890 \AA h = 6.626 × 10^-34 Js c = 3.0 × 10^8 m/s Before placing these values into the equation, one must convert the wavelength of light from angstroms, \AA, to meters, m. 1 \AA = 1 × 10^-10 m; thus, the wavelength is 5890 × 10^-10 m. Hence, E = [(6.626 × 10^-34 Js) (3.0 × 10^8 m/s)] / [(5890 × 10^-10 m)] = 3.4 × 10^-19 J. Once the energy is known, one must convert joules to electron volts and then to Kcal/mole. The conversion factors are: 1eV= 1.6 × 10^-19 J 1eV= 23.06 Kcal/mole. Thus, (a)E = [(3.4 × 10^-19 J) / (1.6 × 10^-19 J/eV)] = 2.1eVand (b)E = (2.1eV) (23.06 Kcal/mole-eV) = 48.4 Kcal/mole.

Question:

A single DNA strand begins with the sequence ATGGACGTATTC. Write the complementary DNA strand sequence. (A is adenine, T thymine, G guanine and C cytosine.)

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Solution:

DNA, deoxyribonucleic acid, may be described as the genetic blueprint of cells and is responsible for heredity. DNA strands consist of a sugar (deoxyribose) - phosphate "backbone\textquotedblright and four important bases.The bases are divided into two main classes: purines and pyrimidines. There exists two bases under each class. The purines, are adenine and guanine? the pyrimidines are cytosine and thymine. In a particular DNA strand, these bases alter-nate with each other in a specific way - the sequence. DNA structures consist of 2 strands, each the complement of the other. Each complement refers to the fact that the bases on one strand pair up with the bases on the other strand in a particular fashion through hydrogen bonding. This bonding holds the two strands together. Adenine (A) will pair with thymine (T) . Guanine (G) will pair with cytosine (C) . Only these pairings produce effective hydrogen bonding. Thus, to determine the complementary DNA strand sequence of ATGGACGTATTC, determine the bases that corre-spond to the bases present for effective hydrogen bond-ing to occur. In the complementary strand thymine will occupy the same spaces as adenine, adenine will occupy same positions as thymine, quanine the same positions as cytosine and cytosine the same as guanine. Thus the complementary strand is TACCTGCATAAG. This is seen in the accompanying figure.

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Question:

Alpha particles are subatomic objects that have a mass of about 6.7× 10^\rule{1em}{1pt}27kg and a charge of 3.2 × 10^\rule{1em}{1pt}19 C. Calcu-late the force on an alpha particle when it is in an electric field \xi whose strength is 1 x 10^3 N / C.

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/Users/wenhuchen/Documents/Crawler/Physics/D18-0580.htm

Solution:

The charge, q = 3.2 x 10^\rule{1em}{1pt}19 C, and the electric field, \xi = 1 x 10^3 N/C, are given. By definition of the electric field, \xi = F / q where F is the force on the charge. Therefore, F =q\xi= (3.2 x 10^\rule{1em}{1pt}19C) (1 x 10^3 N /C) = 3.2 x 10^\rule{1em}{1pt}16 N. The weight of the alpha particle is W = mg = (6.7 x 10^\rule{1em}{1pt}27 kg) [(9.8)(m / sec^2)] \approx 7× 10\rule{1em}{1pt}26N. We see that the force due to the electric field is very much larger than the gravitational force on the particle(10^10 orders of magnitude greater!).

Question:

A certain weight of alcohol has a volume of 100 cm^3 at 0\textdegreeC. What is its volume at 50\textdegreeC?

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Solution:

The coefficient of volume expansion of alcohol is 0.00112/\textdegreeC. Thus, the increase of 100 cubic centimeter for a 50\textdegreeC rise is (0.00112/\textdegreeC) × 100 cm^3 x 50\textdegreeC = 5.60 cm^3 The new volume is therefore 105.60 cubic centimeters.

Question:

The parathyroid glands in man are small, pealike organs. They are ususally four in number and are located on the surface of the thyroid. What is the function of these glands?

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/Users/wenhuchen/Documents/Crawler/Biology/F21-0544.htm

Solution:

The parathyroids were long thought to be part of the thyroid or to be functionally associated with it. Now, however, we know that their close proximity to the thyroid is misleading; both developmentally and functionally, they are totally distinct from the thyroid. The parathyroid hormone, called parathormone, regu-lates the calcium-phosphate balance between the blood and other tissues. Production of this hormone is directly controlled by the calcium concentration of the extracellular fluid bathing the cells of these glands. Parathormone exerts at least the following four effects: (1) it increases gastrointestinal absorption of calcium by stimulating the active-transport system which moves calcium from the gut lumen into the blood; (2) it increases the movement of calcium and phosphate from bone into extracellular fluid. This is accomplished by stimulating osteoclasts to break down bone structure, thus liberating calcium phosphate into the blood. In this way, the store of calcium contained in bone is tapped; (3) it increases reabsorption of calcium by the renal tubules, thereby decreasing urinary calcium excretion; (4) it reduces the reabsorption of phosphate by the renal tubules. The first three effects result in a higher extra-cellular calcium concentration. The adaptive value of the fourth is to prevent the formation of kidney stones. Should the parathyroids be removed accidentally during surgery on the thyroid, there would be a rise in the phosphate concentration in the blood. There would also be a drop in the calcium concentration as more calcium is excreted by the kidneys and intestines, and more is incorporated into bone. This can produce serious dis-turbances, particularly in the muscles and nerve, which utilize calcium ions for normal functioning. Overactivity of the parathyroids, which can result from a tumor on the glands, produces a weakening of the bones. This is a condition which makes them much more vulnerable to frac-turing, because of excessive withdrawal of calcium from the bones.

Question:

It is known that the anti conformation of n-butane is .8Kcal/mole more stable than the Gauche conformation. Explain why.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E20-0729.htm

Solution:

Different arrangements of atoms that can be converted into one another by rotation about single bonds are called conformations. In the anti conformation of n-butane the methyl groups (CH_3) are as far apart as possible. The gauche conformations of n-butane are with the methyl groups only 60\textdegree apart. This can be seen more clearly by using Newman projections for the conformations of n-butane. See figure A. Notice that in the gauche conformations, the methyl groups are closer together than in the anticonformation. When these groups are close together, as in the gauche conformations, steric hindrance occurs between them. This means that there is not really enough room for these two groups to be so close together. This is due to the fact that around any nuclei there is a certain volume which is taken up by the atoms electrons. This volume is determined by the van der Waal's radii of a particular atom. When atoms are too close to permit this volume, a repulsive force called van der Waals' repulsion occurs. This force acts to create a greater distance between the nucleii. This, then, raises the energy (level and lowers the stability) of the gauche conformation. In the anti conformation, the methyls are much further apart, so that you don't have van der Waal's repulsion, thus the molecule is much more stable.

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Question:

Carbon dioxide is reduced by hydrogen according to the equation CO_2(g) + H_2(g) \rightleftarrows CO(g) + H_2O(g) . One mole of CO_2 is reacted with one mole of H_2 at 825K and it is found that, at equilibrium, 0.27atmof CO is present. Determine the equilibrium constant, K, at 825K .

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/Users/wenhuchen/Documents/Crawler/Chemistry/E09-0311.htm

Solution:

This problem makes use\deltafthe fact that since there are equal numbers of moles of gaseous products and reactants, the partial pressure of each component is equal to the number of moles present at equilibrium. For each mole of CO_2 and each mole of H_2 reacted, one mole of CO and one mole of H_2 O are produced. Thus, if at equilibrium there is 0.27atm= 0.27 mole of CO, then there is 0.27 mole of H_20 , 1\Elzbar 0.27 = 0.73 mole of C0_2 , and 1\Elzbar 0.27 = 0.73 mole of H_2 . If we let v denote the volume (in liters) in which the re-action takes place, then: [CO] = [H_2 O] = 0.27 mole/v , [CO_2 ] = [H_2 ] = 0.73 mole/v . Substituting these values into the expression for the equilibrium constant gives K = {[CO] [H_2 O]} / {[CO_2 ] [H_2 ]} = {(0.27/v) (0.27/v)}/ {(0.73/v)(0.73/v)} = (0.27)^2/ (0.73)^2= 0.137.

Question:

The density of liquid NH_3 = 0.68 g/ml. Liquid NH_3, dissociates according to the reaction 2NH_3 (l) \rightleftarrows NH_4^+ + NH_2^-, for which K = 1.0 × 10-3 3at - 33.4\textdegreeC. Determine the con-centrations of NH_4^+ and NH_2^- at equilibrium.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E10-0357.htm

Solution:

To solve this problem, write the equilibrium constant expression for the dissociation of liquid NH_3. This equates the equilibrium constant, K, to the ratio of the concentrations of products to reactants, each raised to the power of its coefficient in the chemical reaction. K = 1.0 × 10^-3 3 = {[NH_4^+][NH_2^-]/{[NH_3]^2}] One is asked to find [NH_4^+] and [NH_2^-]. To do this, one needs the initial concentration of the liquid ammonia, which can be found from the density, (density = mass/volume). It is given that density = 0.68 g/ml. The molecular weight of ammonia (NH_3) is 17 g/mole. For the correct units, con-vert 0.68 g/ml to 680 g/l. Thus concentration (moles/liter) = (680 g/l) / (17 g/mole) = 40 molar NH_3. Let x = [NH_4^+]. From the chemical reaction, it can be seen that NH_4^+ and NH_2^- are formed inequimolaramounts, which means [NH_4^+] = [NH_2^-], so that [NH_2^-] = x. If the original [NH_3] = 40 M, and x moles/liter form [NH_4^+] and [NH_2^-] each, then, at equilibrium, [NH_3] = 40 - x. Substituting these values into the equilibrium constant expression, one obtains 1.0 × 10^-3 3 = [(x \textbullet x)/(40 - x)^2] . Solving for x, using the quadratic formula: x = 1.26 × 10^-15 = [NH_4^+] = [NH_2^-] at equilibrium.

Question:

Let the mass of the body in the diagram be 25 gm, the force constant k be 400 dynes/ cm, and let the motion be started at t = 0 by displacing the body 10 cm to the right of its equilibrium position and imparting to it a velocity toward the right of 40 cm/sec. Compute (a) the period T, (b) the frequency f, (c) the angular frequency \omega, (d) the total energy E, (e) the amplitude A, (f) the phase angle \texttheta_0, (g) the maximum velocity v_max' (h) the maximum accelerationa_max' (j) the coordinate, velocity, and acceleration at a time \pi/8 sec after the start of the motion.

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Solution:

Note that the period is independent of amplitude or by implication, the initial velocity. (a) T = 2\pi\surd(m/k) = 2\pi\surd[(25 gm)/(400 dynes/cm)] = \pi/2 sec = 1.57 sec. (b) f= 1/T = (2/\pi) [(vib)/(sec)] = 0.638 vib/sec. (c) \omega = 2\pif = 4 rad/sec (d) The total energy, (that is, the sum of potential and kinetic energies), stays constant as a spring oscillates . To find the spring's energy, we will compute its energy at time t = 0. The object has both kinetic and potential energy. Its kinetic energy is 1/2mv2_0, with m, the mass of the body, and v_0, its initial velocity. Its potential energy is 1/2 kx2_0, where x_0 is the object's distance from equi-librium, and k is the spring constant. Then E_total = 1/2 mv2_0 + kx2_0 = 1/2 (25 gm)(40 cm/sec)^2 + 1/2 (400dynes/cm)(10 cm)^2 = 20,000ergs + 20,000 ergs=40,000 ergs (e) The amplitude of oscillation is the distance between the equilibrium position x_eq and the point of maximum extension. To find the maximum extension we find the position at which the system's mechanical energy of 40,000 ergs is completely potential. This occurs when the kinetic energy is zero. Then 1/2 kA^2 = 40,000 ergs 1/2 (400 dynes/cm)A^2 = 40,000 ergs A^2 = 200 cm^2,A = 10\surd2 cm (f) If x_0 is the displacement and A the amplitude, the initial angular displacement or phase angle \texttheta_0 is defined by: sin \texttheta_0 = x_0/A = 1/\surd2,\texttheta_0 = \pi/4 rad. (g) To obtain v_max we find the point at which the energy is all kinetic. This occurs at the equilibrium point, x = 0. Then E = KE + PE = KE + 0 = 1/2 mv2_max 1/2 mv^2_max = 40,000 ergs 1/2 (25 gm)(v^2_max) = 40,000 ergs v^2max= 3,200 cm^2/sec2 vmax= 40\surd2 cm/sec (h) The maximum acceleration occurs at the ends of the path where the force is a maximum. This force is given by Hooke's Law, F = -kx, where k is the spring constant and x is the displacement of the spring from equilibrium. At maximum extension, the displacement from equilibrium is just the amplitude. Hence, F_max = - kx_max = - kA = -(400 dynes/cm)(10\surd2 cm) = - 4000\surd2 dynes Forces produce accelerations according to the law F = ma. Therefore F_max = - 4000\surd2 dynes = ma_max = (25 gm)(a) a_max = - 160\surd2 cm/sec^2 At maximum extension, the acceleration is greatest and in the direction of the equilibrium point of the spring and hence in a direction which we define as negative. The motion of an oscillating spring is highly symmetric and at the point of maximum compression, the acceleration will again reach this maximum but will this time be in the positive direction. (j) The equation for the object's position is x = A sin (\omegat + \texttheta_0) where A is the amplitude, \cyrchar\cyromegais the angular velocity, t is the time variable, and \texttheta_0 is the initial angular displacement. The velocity, v, and acceleration, a, are found from v = dx/dt;a = dv/dt = d^2x / dt2 We then have x = 10\surd2 sin [4t +\pi/4], v = 40\surd2 cos [4t +\pi/4], a = -160\surd2 sin [4t +\pi/4]. When t = (\pi/8) sec, the phase angle is [4t +\pi/4] = 3\pi/4 rad, x = 10\surd2 sin (3\pi/4) = 10 cm, v = 40\surd2 cos (3\pi/4) = -40 cm/sec, a = -160\surd2 sin (3\pi/4) = -160 cm/sec^2.

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Question:

a_11x_1 + a_12x2+ a_13x3+ a_14x_4 = y_1 a_21x_1 + a_22x_2 + a_23x_3 + a_24x_4 = y_2 a_31x_1 + a_32x_2 + a_33x_3 + a_34x_4 = y_3 a_41x_1 + a_42x_2 + a_43x_3 + a_44x_4 = y_4 Hint: feel free to make use of built-in matrix instructions.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G20-0504.htm

Solution:

The above system of equations can be written in the matrix form as A\bulletX = Y, and the solution is X = (A-1)\bulletY, where A is a 4×4 matrix of the coefficients of X's, and both X and Y are 4×1 vectors. The symbol A^-1 stands for the inverse matrix of A. Here is an example of the validity of the above procedure: The system of equation x_1 - x_2 = 5 2x_1 - 3x_2 = 13 is solved to obtain the values of x_1 and x_2: x_1 = 5 + x_2; 2(5 + x_2) -3x_2 = 13; x_2 = (13 - 10) / -1 = -3 \thereforex_1 = 5 -3 = 2. Now, using the matrix procedure: A = \mid 1-1 \mid ;A^-1 =\mid 3-1 \midY = \mid5\mid \mid 2-3 \mid\mid 2-1 \mid\mid 13 \mid X = (A^-1)\bullet Y = \mid 3-1 \mid \bullet\mid5\mid = \mid 3×5 + (-1)13 \mid = \mid15-13 \mid \mid 2-1 \mid\mid13\mid\mid 2×5 + (-1)13 \mid\mid10-13 \mid X = \mid 2\mid \Rightarrow 7, X_1 = 2, X_2 = -3. \mid-3 \mid Both methods yield the same results. The program looks as follows: 10DIM X (4, 1), Y (4, 1), A (4, 4), B (4, 4) 20MAT READ Y 30MAT READ A 40MAT B = INV (A) 50MAT X = B\textasteriskcenteredY 60MAT PRINT X; 70END

Question:

Design an 8 × 1 multiplexer.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G04-0074.htm

Solution:

A multiplexer is a device that receives binary information from 2^n (in this case eight) input lines and transmits the information, selected from one input line, on a single output line. N (in this case 3) control lines, called select lines, determine which of the eight input lines will be selected. The block diagram of an 8 × 1 multiplexer is shown in fig. 1. Select lines S_0S_1S_2 Output X 000 I_0 001 I_1 010 I_2 011 I_3 100 I_4 101 I_5 110 I_6 111 I_7 Fig. 2 The truth table for the multiplexer is shown in figure 2. From the truth table it is seen that X = I_0S_0S_1S_2 + I_1S_0S_1S_2 + I_2S_0S_1S_2 + I_3S_0S_1S_2 + I_4S_0S_1S_2 + I_5S_0S_1S_2 + I_6S_0S_1S_2 + I_7S_0S_1S_2 The logic diagram of the 8 × 1 multiplexer can be derived from this equation. Figure 3 shows the logic diagram.

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Question:

A mass of 100 kilograms is pushed along the floor at a constant rate of 2 m/sec. If the coefficient of sliding friction is 0.25, at what rate is work being done in watts, in horsepower?

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/Users/wenhuchen/Documents/Crawler/Physics/D05-0284.htm

Solution:

The weight of the mass is 100 kg × 9.8 m/sec^2 = 980nt= N The force of friction = F = \mu × N = 0.25 × 980nt= 245nt Power = Fv = 245nt× 2 m/sec = 490 watts = (490 watts)/(746 watts/hp) = 0.66 hp

Question:

Give qualitative explanation for the observed splitting of the ground state energy of the hydrogen atom in the presence of a magnetic field.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E18-0683.htm

Solution:

This phenomenon is called Zeeman splitting, after the Dutch physicist Pieter Zeeman who first discovered the effect in 1896. Consider an electron moving in the ground state orbital of a hydrogen atom and describing a circular orbit (labelled "O" in the diagram). This motion will give rise to an angular momentum vector L^\rightarrow directed along the axis of rotation. In the presence of a magnetic field H^\rightarrow, the angular momentum vector will precess about the magnetic field axis. Now, since the angular momentum of an electron must be quantized, the projection of L^\rightarrow on to H^\rightarrow (that is, the com-ponent of the vector L^\rightarrow along axis H^\rightarrow) must also be quantized. From quantum theory, we know that this projection must take on values which are integral multiples (positive, negative, or zero) of h/2\pi, where h is Planck's constant. These in-tegers are the possible values of the magnetic quantum number, m. For the hydrogen atom, the magnetic quantum number can take on the values m = - 1, 0, + 1. Thus, we observe three lines for hydrogen, in a magnetic field. The high energy line corresponds to the case where the projection of L^\rightarrow onto H^\rightarrow has the same di-rection as H^\rightarrow (m = 1), the intermediate line occurs when L^\rightarrow and H^\rightarrow are perpendicular and the projection of L^\rightarrow onto H^\rightarrow is thus zero (m = 0) , and the low energy line corresponds to the case where the projection of L^\rightarrow onto H^\rightarrow has the opposite direction of H^\rightarrow (m = - 1).

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Question:

The citrate ion forms a tight bond with calcium ions so that the calcium is unable to react with other substances In view of this, why is sodium citrate added to whole blood immediately after it is drawn from a blood donor?

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/Users/wenhuchen/Documents/Crawler/Biology/F14-0349.htm

Solution:

Sodium citrate dissociates into sodium and citrate ions, the latter of which binds with calcium ions to remove them from solution. Calcium ions must be present forthromboplastin, a substance released by damaged tissues, to be converted into the enzymeprothrombinase. Calcium must also be present forprothrombinaseto convertprothrombininto thrombin, and for thrombin to catalyze the formation of fibrin. Fibrin ultimately forms the fibers which trap the cellular constituents of the blood, thus forming a clot. Without calcium ions, this process cannot occur. Blood drawn from a donor is treated with sodium citrate to prevent coagulation. The blood can then be stored for weeks in a blood bank by keeping the blood at a temperature of about 4\textdegreeC. After injection into a recipient, the citrate ion is removed from the blood within minutes by the liver. The liver polymerizes the citrate into glucose which is then metabolized for energy. If the liver is damaged or if too large a quantity of citrated blood is injected, the level of citrate will depress the level of calcium ions in the blood. If the level drops significantly, a disease called tetany can result.Tetanyis characterized by muscle spasms, which can be severe enough to cause death by asphyxia (oxygen exhaustion).

Question:

Adrenalin is a hormone which stimulates the sympathetic system . What effects would you expect adrenaline tohave on (a) the digestive tract, (b) the iris of the eye, and (c) the heart beat?

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Solution:

The sympathetic nervous system is comprised of nerves coming out of the spinal cord at the thoracic and lumbar regions. After each neuron (thepreganglionicneuron) leaves the spinal cord, it synapses with another neuron (the postganglionic neuron). It is the post-gan-glionic neuron that innervates the effectors supplied by the sympathetic system. (a) The action of the sympathetic nerves on the digestive tract is to slow peristalsis and reduce secretion to decrease the rate of digestion. Adrenaline, which stimulates the sympathetic system will hence cause peristalsis and digestion to slow down. (b) The iris of the eye consists of two sets of smooth muscle - circular and radial. Sympathetic stimulation re-sults in dilation by causing the radial muscles to contract. Adrenaline, which excites the sympathetic system , produces exactly this effect. (c) Sympathetic stimulation to the heart is twofold: it stimulates the pacemaker to increase heart rate and it stimulates the myocardium (muscularlayer of heart wall) to beat more forcefully. Thus adrenaline has the effect of ac-celerating and strengthening the heartbeat.

Question:

An airplane, whose ground speed in still air is 200 mi/hr, is flying with its nose pointed due north. If there is a cross wind of 50 mi/hr in an easterly direction, what is the ground speed of the airplane?

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/Users/wenhuchen/Documents/Crawler/Physics/D01-0019.htm

Solution:

The cross wind causes the plane to travel 50 mi/hr to the east in addition to its speed of 200 mi/hr to the north. To find its speed with respect to ground, use vector addition. Vectors are quantities that have both magnitude and direction; and velocity fits this specification. Using the Pythagorean theorem, we can find the magnitude of the resultant velocity v. This magnitude is the plane's speed. Speed does not have direction (note that speed is not a vector). v = \surd(v^2_airplane + v^2_wind) = \surd[(200)^2 + (50)^2] = \surd42,500 = 206 mi/hr.

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Question:

The mass of a rectangular bar of iron is 320 grams. It is 2 × 2 × 10 centimeters^3. What is its specific gravity?

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Solution:

Specific gravity for solids is the ratio of the density of the solid to the density of water (approximately 1 gram per cubic centimeter). The specific gravity of iron is then S =\rho_i/\rho_\omega But\rho_i= (320 gm)/(2 × 2 × 10 cm^3) = 8 gm/cm^3 Since \rho_\omega = 1 gm/cm^3 S= (8 gm/cm^3)/(1 gm/cm^3) S= (8 gm/cm^3)/(1 gm/cm^3) = 8. = 8.

Question:

When air is let out of an automobile tire, the air feels cold to the touch. Explain why you feel this cooling sensation.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E29-0917.htm

Solution:

To solve this problem, consider what is happening when the air is being expelled. For the air to be released, it must displace the atmosphere that immediately surrounds the tire. To displace the atmosphere, it must do work on the atmosphere. In other words, the air must expend energy. There exists a correlation between the energy of a substance and its temperature. As the air is being released, it is doing work. Since it consumes energy, its temperature must fall. As such, you feel a cooling sensation.

Question:

Calculate the normality of a solution 40 ml of which is required to neutralize 0.56 g of KOH.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E08-0296.htm

Solution:

KOH ionizes as shown in this equation. KOH \leftrightarrows K^+ + OH^- This means that for each KOH molecule ionized, one OH^- ion is formed. Thus to neutralize the KOH one must have 1 H^+ ion for each OH^- ion. Thus, one must first determine the number of moles of KOH present in 0.56 g. This is done by dividing 0.56 g by the molecular weight of KOH (MW of KOH = 56) number of moles of KOH = number of moles of OH^- number of moles of KOH = (0.56 g/MW) = [(0.56 g)/(56 g/mole)] = .01 mole Therefore, in the 40 ml of the solution there must be .01 moles of H^+ ions. The normality of an acid is defined as the number of equivalents of H^+ in 1 liter of solution. An equivalent may be defined as the weight of acid or base that produces 1 mole of H^+ or OH^- ions. In this problem, equivalents = moles. There are 1000 ml in 1 liter, thus in 40 ml there are 40 ml × (1 liter/1000 ml) number of liters = 40 ml × (1 liter/1000 ml) = .04 liters The normality can now be found. normality = [(number of equivalents)/( number of liters)] = [(.01 equivalents) / (.04 liters)] =.25 N.

Question:

Phylum Hemichordate is believed to be the most closely related to our own phylum, Chordate. What is the evidence for this? How are the echinoderms linked to chordates?

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Solution:

The hemichordates are marine animals, often called acorn worms. They live in U-shaped burrows in sand or mud. The body is bilaterally symmetrical and consists of a conical proboscis, a short collar, and a long trunk. The hemichordates' embryonic development is similar to that of the chordates. The echinoderms, hemichordates, and chordates are deuterostomes. The characteristics shared by these phyla are the formation of the anus from the embryonic blastopore. They also undergo radial and in-determinate cleavage, with the origin of mesoderm being from pouches of endoderm, and the formation of the coelom as the cavities in the mesodermal pouches. The other in-vertebrate phyla (mollusks, annelids and arthropods) are protosomes. In protosomes, the mouth forms from the blasto-pore, cleavage is determinate and spiral, mesoderm arises from ingrowth of a single cell near the blastopore, and the coelom developes from a split in an initially solid mass of mesoderm. The Deuterostomia and the Protosomia represent two important diverging lines of development from a primi-tive bilaterally symmetrical ancestor. Adult hemichordates resemble chordates in two other ways. The hemichordates have pharyngeal gill slits. Water drawn into the mouth is forced out over these slits in the wall of the pharynx. Oxygen is removed from the water and carbon dioxide is released into it from the blood. Capillary beds are located in the septa between the gill slits. Only chordates and hemichordates have pharyngeal clefts. Hemichordates also have a thick, hollow, dorsal nerve cord in the collar. The anterior region contains a ventral nerve cord as well as a dorsal nerve cord. All chordates have a dorsal hollow nerve cord. The echinoderms are related to the hemichordates. Hemichordates have a hydrostatic skeleton similar to the water vascular system of echinoderms. In addition, there are similarities between their larval stages. The dipleurula larvae of echinoderms and hemichordates are quite alike, and differ from the trochophore larvae of molluscs and annelids. Chordates have no ciliated larval stages similar to both of these. However, this ciliated larvae link between echino-derms and hemichordates is additional indirect evidence of the link between echinoderms and chordates.

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Question:

A dynamo driven by a steam engine which uses 10^3 kg of coal per day produces a current I = 200 A at anemfV of 240 V. What is the efficiency of the system if the calorific value of coal is 6.6 × 10^3 cal \bullet g^-1 ?

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Solution:

The energy supplied by the coal per second is equal to the product of the calorific value of coal and the mass of coal used, divided by the time it takes to burn the coal. Hence, E_0 = [(6.6 × 10^3 cal \bullet g^-1 × 10^6 g)/(24 × 60 × 60 s)] = [(4.2 × 6.6 × 10^9 )/ (24 × 60 × 60)] J \bullet s^-1 = 3.2 × 10^5 W. The electric power supplied by the dynamo is P = IV = 200 A × 240 V = 4.8 × 10^4 W The efficiency of the system is thus (P/E_0 ) × 100% = [(4.8 × 10^4 )/(3.2 ×10^5 )] % = 15% .

Question:

Design a Program usingBASIC to simulate the rolls of a die.

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Solution:

For this problem, the random number generator RND can be used. This function generates random numbers between 0 and 1. We want to generate integers from 1 to 6 inclusive, and find out how many 6's appeared after 100 tosses. A counter S is set up to sum the number of times a 6 is encountered. 1\O REM DIE THROWING SIMULATION 2\O LET S = \O 3\O FOR X = 1 TO 1\O\O 4\O LET D = INT(1 + 6\textasteriskcenteredRND) 5\O IF D = 6 THEN 8\O 6\O GO TO 1\O\O 7\O REM S COUNTS SIXES 8\O LET S = S + 1 9\O NEXT X 1\O\O PRINT 11\O PRINT S; "SIXES OUT OF 1\O\O ROLLS" 12\O END

Question:

Write a BASIC program which keeps track of the numbers of items in stock for four different classes of items. To be specific, let the four classes of items be denoted by 1\O\O1, 1\O\O2, 1\O\O3 and 1\O\O4. Consider the case where the following 7 (seven) items are in stock: 1\O\O1, 1\O\O2, 1\O\O2, 1\O\O3, 1\O\O4, 1\O\O3, 1\O\O1.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G18-0450.htm

Solution:

It is convenient to introduce separate counters K1, K2, K3, and K4 for the four classes. It is also convenient to utilize the ON-GO TO statement. Thus if we say 5\O ON X GO TO 9\O, 7\O, 11\O control will be transferred to statements 9\O, 7\O, or 11\O if the truncated value of X is 1, 2, or 3 respectively 1\OREM INVENTORY PROGRAM FOR 4 CLASSES 2\OREM K1, K2, K3, K4 ARE THE COUNTERS 3\ODATA 1\O\O1, 1\O\O2, 1\O\O2, 1\O\O3, 1\O\O4, 1\O\O3, 1\O\O1 4\OLET K1 = K2 = K3 = K4 = \O 5\OFOR I = 1 TO 7 6\OREAD X 7\OON X - 1\O\O\O GO TO 8\O, 1\O\O, 12\O, 14\O 8\OLET K1 = K1 + 1 9\OGO TO 16\O 1\O\OLET K2 = K2 + 1 11\OGO TO 16\O 12\OLET K3 = K3 + 1 13\OGO TO 16\O 14\OLET K4 = K4 + 1 15\OGO TO 16\O 16\ONEXT I 17\OPRINT "ITEM 1\O\O1;\textquotedblright; Kl;"IN STOCK" 18\OPRINT "ITEM 1\O\O2;"; K2; "IN STOCK" 19\OPRINT "ITEM 1\O\O3;"; K3; "IN STOCK" 2\O\OPRINT "ITEM 1\O\O4;"; K4; "IN STOCK" 21\OEND Some computers do not accept multiple assignment state-ments. In that case, in statement 40 each variable should be initialized to zero separately.

Question:

a) F =V + A + L b) F =A+B+C+D c) F =WXYZ d) F =ABC+ D

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G03-0041.htm

Solution:

DeMorgan's theorem states that a logical expression can be complemented by complementing all its variables and exchanging AND (\bullet) operations with OR (+) operations. For example the complement of isF = AB F=A+B. Another expression for F is found by complementingF. Thus a) Complementing variables V, A, and L and changing + to \textbullet , we get Complement the entire expression to find F, Thus,V + A + L=VAL b) Again, complementing A,B,C, and D and changing + to \textbullet we get F= ABCD c) This time exchanging \textbullet for +, we have d) This time split the function into two parts F = X + D where X =ABC. Apply DeMorgan's theorem to X. thus,X =A+B+C F =A+B+C+ D

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Question:

(a)If the inputs to half-adder are x and y, show that the sum output could be given by the Boolean expression: (x + y) \textbullet (\simx + \simy)(1) and the carry output could be given by: \sim (\simx + \simy)(2) (b)Set up a new circuit for a half-adder using the alternate expressions for sum and carry, given by (1) and (2) and OR, AND and NOT gates. (c)Show that (x + y) \textbullet \sim(x \bullet y)(3) is equivalent to (1) and set up a new half-adder circuit using x \bullet y as the expression for the carry output and (3) as the expression for the sum output.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G05-0097.htm

Solution:

(a) To show that the sum output is given by expression (1) and the carry output is given by expres-sion (2), we could set up their truth tables and note that the results agree with the addition table. But since this approach is used to show that the sum output could be given by; (x \textbullet \simy) + (x\sim \textbullet y)(4) and that the carry output could be given by; x \bullet y(5) we will use the laws of Boolean algebra to show that (1) is equivalent to (4) and (2) is equivalent to (5) : (x + y) \bullet (\simx + \simy) = [x \bullet (\simx + \simy)] + [y \bullet (\simx + \simy)] Distributive law of \textbullet over + = [ (x \textbullet \simx) + (x \textbullet \simy)] + [(y \textbullet \simx) + (y \textbullet \simy)] Distributive law of \textbullet over + = [0 + (x \textbullet \simy)] + [(y \textbullet \simx) + 0] Law of the excluded middle = (x \textbullet \simy) + (y \textbullet \simx) Identity of + = (x \textbullet \simy) + (\simx \textbullet y) Commutativity of \textbullet To show that (2) is equivalent to (5) : \sim(\simx + \simy) = \sim(\simx) \textbullet \sim(\simy) de Morgan's Law = x \bullet yInvolution Law (b)Setting up the terms in parentheses of equation 1 first, we create the circuit of Figure 1. Since both x + y and \simx + \simy are operands of the\textbackslash operation they will be inputs to an AND gate whose output is the sum digit. Using \simx + \simy as input to a NOT gate gives an output that is the carry digit. The complete implementation of equation (4) is shown in Figure 2. (c)De Morgan's law allows us to write: (x + y) \textbullet \sim(x \bullet y) = (x + y) \textbullet (\simx + \simy) The left and right terms in parentheses of (3) are simply outputs of OR and AND gates, respectively, where x and y are the inputs. But x \bullet y is also the expression for the carry digit. So the complete half-adder circuit using (3) is the simplest of those we've constructed.(See Figure 3).

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Question:

A train whose speed is 60 mi/hr rounds a curve whose radius of curvature is 484 ft. What is its acceleration?

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/Users/wenhuchen/Documents/Crawler/Physics/D04-0164.htm

Solution:

For uniform circular motion, we have an acceleration directed towards the center of curvature of magnitude a = v^2/R where v is the speed of the object, and R is the radius of the circle. To keep the units consistent, we have to con-vert the speed from mi/hr to ft/sec, since R is in feet. v = [60(mi/hr)][(5280 ft/mi)/(3600 sec/hr)] = 88 ft/sec. Substituting the appropriate values, we find the accelera-tion to be a = v^2/R = (88 ft/sec)^2 / (484 ft) = 16 ft/sec^2 .

Question:

A 10-g lead bullet is traveling with a velocity of 10^4 cm/sec and strikes a heavy wood block. If, in coming to rest in the block, half of the initial kinetic energy of the bullet is transformed into thermal energy in the block and half into thermal energy in the bul-let, calculate the rise of temperature of the bullet. (The block remains stationary during the collision.)

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/Users/wenhuchen/Documents/Crawler/Physics/D13-0468.htm

Solution:

This problem involves the transfer of energy of one form (kinetic energy) into energy of another font (heat). The total energy of the bullet- block system before the bullet strikes the block is just the kinetic energy of the bullet. Or E_T = (1/2)mv^2 = (1/2) × (10 g) × (10^4 cm/sec)^2 = 5 × 10^8 ergs After the bullet strikes the block, it loses its kinetic energy. The law of conservation of energy demands that this energy appear as ther-mal energy in the block and bullet. Or E_T = KE = Q_block + Q_bullet = 2Q_bullet for we are given that Q_block = Q_bullet . Hence, Q_bullet = E_T/2 . But Q_bullet = cm∆T where c is the specific heat of the lead bul-let (the amount of heat required to raise the temperature of one gram of the substance one degree centigrade), m is the mass of the bullet, and ∆T is the rise in temperature of the bullet due to the thermal energy. The specific heat for lead is 0.0310 cal/g.c. We then have ∆T = (Q_bullet)/mc = {(1/2)ET}/mc = [{(1/2) × 5 × 10^8 ergs} / {(4.186 x 10^7 ergs/cal)(0.0310 cal/g- 0c) × 10 g}] = 19.30C

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Question:

Even though there is no such thing as a "typical cell" - for there are too many diverse kinds of cells - biologists have determined that there are two basic cell types. What are these two types of cells?

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/Users/wenhuchen/Documents/Crawler/Biology/F02-0031.htm

Solution:

Cells are classified as either procaryotic or eucaryotic. Procaryotes are strikingly different from eucaryotes in their ultrastructural characteris-tics. A key difference between the two cell types is that procaryotic cells lack the nuclear membrane characteristic of eucaryotic cells. Procaryotic cells have a nuclear region, which consists of nucleic acids. Eucaryotic cells have a nucleus, bounded by a double-layered membrane. The eucaryotic nucleus consists of DNA which is bound to proteins and organized into chromosomes. Bacteria and blue-green algae are procaryotic unicellular organisms. Other organisms, for example, protozoa, algae, fungi, higher plants and animals are eucaryotic. Within eucaryotic cells are found discrete regions that are usually delimited from the rest of the cell by membranes. These are called membrane-bounded subcellular organelles. They perform specific cellular functions, for example, respiration and photo-synthesis. The enzymes for these processes are located within membrane-bounded mitochondria and chloroplasts, respectively. In procaryotic cells, there are no such membrane-bounded organelles. Respiratory and photo-synthetic enzymes are not segregated into discrete organelles although they have an orderly arrangement. Procaryotic cells lack endoplasmic reticulum, Golgi apparatus, lysosomes, and vacuoles. In short, pro-caryotic cells lack the internal membranous structure characteristic of eucaryotic cells. There are other differences between procaryotic cells and eucaryotic cells. The ribosomes of bacteria and blue-green algae are smaller than the ribosomes of eucaryotes. The flagella of bacteria are structural-ly different from eucaryotic flagella. The cell wall of bacteria and blue-green algae usually contains muramic acid, a substance that plant cell walls and the cell walls of fungi do not contain.

Question:

A parallel plate capacitor of 2 meter^2 in area and with charge q = 3.54 × 10^\rule{1em}{1pt}5 coulomb is insulated while a sheet of dielectric 5 mm thick, of dielectric coefficient 5, is in-serted between the plates. Compute (a) the electric inten-sity in the dielectric, (b) the potential difference across the capacitor, (c) its capacitance.

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/Users/wenhuchen/Documents/Crawler/Physics/D18-0617.htm

Solution:

(a) The insertion of a dielectric between the capacitor plates alters the electric intensity because of the reversed field set up by the induced charges on the dielectric. The electric intensity is E = (\sigma / K\epsilon_0) = ( 1 / K\epsilon_0)(q / A) = [1 / (5 × 8.85 × 10^\rule{1em}{1pt}12 )] [(3.54 × 10^\rule{1em}{1pt}5) / (2)] 4 × 10^5 (volts / meter). (b) The potential difference across the capacitor is re-duced to V_ab = Ed = 4 × 10^5 × 5 × 10^\rule{1em}{1pt}3 = 2000 volts. (c) The capacitance is increased to C = q / V_ab = (3.54 × 10^\rule{1em}{1pt}5) / (2000) = 1.77 × 10^\rule{1em}{1pt}8 farad = 17,700 \mu\muf. The capacitance may also be computed from C = \epsilon(A / d) = k \epsilon_0 (A / d) = 5 × 8.85 × 10^\rule{1em}{1pt}12 [2 / (5 × 10^\rule{1em}{1pt}3)] = 17,700 \mu\muf.

Question:

If the carbon dioxide content of the air doubled, what would happen to the temperature of the earth. Explain your answer.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E24-0852.htm

Solution:

This problem requires a knowledge of the greenhouse effect. The light energy from the sun tends to be of short wavelengths that are not absorbed by the atmosphere. Upon striking the earth's surface much of this energy is converted to thermal or infrared energy. Infrared energy can be absorbed by carbon dioxide and thus is not permitted to escape from the earth. As the concentration of CO_2 (carbon dioxide) is increased, the efficiency of this light energy-trapping is increased, which causes a rise in temperature on the earth. This phenomenon is termed the greenhouse effect. If the CO_2 concentration were doubled a rise in the earth's temperature would be expected. It has been estimated that the temperature increase would be 2-4\textdegreeC.

Question:

It is generally believed that higher plants evolved from the green algae. What is the reason for this belief?

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/Users/wenhuchen/Documents/Crawler/Biology/F06-0175.htm

Solution:

The green algae share a number of charac-teristics with the higher plants. The other algal phyla have unique characteristics which are not shared with higher plants. There are two types of chlorophyll present in most algae and plants. All photosynthetic organisms, except the photosynthetic bacteria, have chlorophyll a. The green algae and the higher plants have chlorophyll b .in addition to chlorophyll a. Neither the brown algae nor the red algae have chlorophyll b. Instead, the brown algae have chlorophyll c and the red algae possess chlorophyll d. The principal reserve material of the Chlorophyta (green algae) and of higher plants is usually starch. The dinoflagellates utilize starch and oils. The diatoms and the golden algae store food as oils and as the poly saccharide leucosin, which is a white substance. Brown algae utilize a variety of unusual carbohydrates as food reserves, one of which is laminarin. Red algae do not utilize starch, but a polysaccharide similar to it called the Floridean starch. The Euglenoids utilize the carbo-hydrate polymer paramylan as their reserve material. The cell walls of Chlorophyta and the higher plants are composed chiefly of cellulose, with very little other materials. The Euglenoids lack a cellulose cell wall, and have instead a proteinaceous pellicle, which is internal to the plasma membrane. The cell walls of the Chrysophyta (diatoms, yellow-green, and golden algae) usually consist of pectic materials impregnated with silica. These siliceous cell walls consist of two halves, which fit together like the two parts of a pill box. Some unicellular dinoflagellates may lack cell walls, but most have thick cellulose cell walls that fit together as interlocking plates to form a shell. The cell walls of red algae have large quantities of mucilaginous material. The higher plants are biochemically similar to the green algae and biochemically different from other algae. This leads biologists to propose that the ancestor of higher plants was similar to the present green algae. Also, among the Chlorophyta, there are many multicellular non-motile forms. Higher plants are always multicellular and non-motile. The Phaeophyta and Rhodophyta (brown and red algae) are also mostly multicellular and non-motile, but this is not true of the Euglenophyta, Chrysophyta, and dinoflagellates.

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Question:

Hydrogen chloride (HCl) is used industrially in the manufacture of fabric dyes. The acidic properties of HCl arise because the H-Cl bond is polar. The observed dipole moment, u_obs, of HCl is 1.03 D and the H-Cl bond distance is 1.28 \AA. Estimate the percent ionic character of the HCl bond. Electron charge, e = 4.8 × 10^-18 esu.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E26-0886.htm

Solution:

A dipole moment is an indication of bond polarity. The magnitude of the dipole moment is the product of the electronic charge and the distance between the charge centers. To determine the ionic character of HCL, the magnitude of the dipole moment is needed. Percent ionic character = (U_obs) / (U_ionic) × 100, where Uionic= the magnitude of dipole moment for total ionic character. If HCl were ionic, one electron would be trans-ferred from H to Cl to give the ionic structure H^+Cl^-. Thus, the electron charge would be the charge on one electron, 4.8 × 10^-18 esu. The distance between the two atoms is given as 1.28 \AA or 1.28 × 10^-8 cm. (There are 10^-8 cm per ^---\AA.) Recalling the definition of dipole moment, u_ionic = 4.80 ×10^-10 esu × 1.28 ×10^-8 cm = 6.14 ×10^-18 esu-cm. Dipole moment is usually expressed, however, in Debyes. There are 10^18 Debyes per esu-cm. Thus, 6.14 ×10^-18 esu-cm = 6.14 Debyes = u_ionic. Therefore, percent ionic character = (1.03) / (6.14) × 100 % = 16.8 %.

Question:

Write a program which handles a credit card operation. For simplicity, assume that there are only six customers. The set ofsix data cards is 511'MARIAN GREEN'0.00 128'WILLIAM BROWN'-5.89 094'GARY WHITE'48.50 407'JUDY GREY'234.00 351'ALLEN BLACK'0.00 013'JOHN GREEN'127.14 wherethe first (three digit) integer is an account number, the seconditem is the customer's name, and the third num-ber is theunpaid balance of previous purchases. The minus sign meansBROWN overpaid and has positive balance. The second set of data cards is 013127.1483.25Yourprogram must Your program must 407100.000.0be able to handle 5110.0141.50up to 6 sets of data 1280.047.10here where the first integer is again an account number, the se-condnumber represents the payment since the last bill, andthe third number represents new purchases made with thecredit card. Write a PL/I program which reads in this data and printsout a five column table showing, for each customer, theaccount number, name, previous unpaid balance, the newunpaid balance, and the word DELINQUENT in a case wherethe payment made is less than one half the previous balance. Print out the above table in numerical order by accountnumber with the lowest number first:

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G13-0351.htm

Solution:

This is a program which requires the use of ar-rays. The program illustrateshow to use DECLARE, GET LIST, manipulate data and PUT LIST statements.The problem also involves sorting. There are various waysof sorting, list; one technique is dealt with in the program. The programis as follows. Let the arrays beA- ACCOUNT NUMBER B - NAME OP CUSTOMER C - PREVIOUS UNPAID BALANCE G - NEW UNPAID BALANCE H - FOR REMARK Arrays A, C, G consist of numeric values and arrays B, H have character stringsas its element. D - SPECIFIES ACCOUNT NUMBER E - PAYMENT MADE FOR PREVIOUS BILL F - IS NEW PURCHASES MADE. These three D, E and F are not arrays, but simple variables as designated. SI, S2, S3, S4, S5 are simple variables used for sorting the creditcard operation in ascending order. The new unpaid balanceG[I] = Previous unpaid balance C[l] + Purchases F - Payment E. To check for condition of 'DELINQUENT' payment made E < (1/2)Previous unpaid balance{ -[C (I)/2]} Previous unpaid balance{ -[C (I)/2]} In sorting, the procedure applied here is to compare two adjacent elements, A[J] and A[J + 1]. IfA[J]>A[J+1] i.e. the number in A[J+1] is less than number in A[J], henceswitch these two numbers. For exampleA[1] = 5, A[2] = 3. SinceA[1] > A[2] therefore3 must be placed in A[1] and 5 in A[2] to have ascending sequenceorder. CREDIT_CARD:PROCEDUREOPTIONS(MAIN); DCL(((A,C,G)(6)) FIXED(6,2)),(B,H)(6)CHAR(15) VAR; DCL(D,E,F,S1,S3,S4) FIXED DEC(6,2),(S2,S5) CHAR(15) VAR; /\textasteriskcentered FORMS THE HEADING \textasteriskcentered/ PUT LIST ('ACCOUNTNUMBER' ,'NAME', 'PRE UNPAID BAL ' , 'NEW UNPAID BAL','REMARK'); PUT LIST ('- -----', '- ---' , '- --------' , '- --------' , '- -----'); GET LIST( (A (M) ,B(M) ,C (M) DO M = 1 TO 6)); /\textasteriskcentered TAKES IN THE FIRST SET OP CARDS \textasteriskcentered/ /\textasteriskcentered INITIALIZE ELEMENTS OF ARRAY H TO DELINQUENT EXCEPT FORC[I] \leq0 \textasteriskcentered/ DO K = 1 TO 6; IF ([K] > 0) THEN H [K] = 'DELINQUENT'; END; /\textasteriskcentered CALCULATES NEW UNPAID BAL, CHECKS FOR DELINQUENT \textasteriskcentered/ DO WHILE [1 = 1] ; ONENDFILE(SYSIN) GOTO LABEL; GETLIST[D,E,F]; DO I = 1 TO 6; IF D =A[I]THEN DO; G[I] = C(I) + F - E; IF EA(J + 1)THEN DO; S1 =A(J); A(J) = A[J+1] ; A[J + 1] = S1; S2 =B(J); B(J) = B[J + 1]; B[J + 1] = S2; S3 =C[J]; C[J] = C[J + 1]; C[J + 1] = S3; S4 =G(J); G[J] = G(J + 1); G(J + 1) = S4; S5 =H(J); H(J) = H(J + 1); H(J + 1) = S5; J = 0; END; END; DO N = 1 TO 6; PUT LIST(A(N),B(N),C(N),G(N),H(N)) SKIP (2); END; END CREDIT_CARD;

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Question:

Define the term vitamin. Describe the function of each essential vitamin needed by the human body.

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Solution:

Vitamins are relatively simple organic com-pounds required in minute quantities and not used as an energy source. Vitamins are required in very small quantities because they serve as coenzymes or as parts of coenzymes. A coenzyme is a substance required by an enzyme for catalytic activity. Since coenzymes, like enzymes, are not altered in a reaction and may be used repeatedly, and for more than one enzyme, only small amounts are needed. There are two principal groups of vitamins: the fat- soluble vitamins (A, D, E, K) and the water-soluble vitamins (B complex and C). Vitamin A is necessary for the growth and maintenance of the epithelial cells of the skin, various tracts in the body and the eye. It is also a precursor to the visual pigment,retinaldehyde, in the eye. A deficiency of vitamin A results in lowered resistance of skin to infection and in night blindness. Vitamin D can be synthesized by the skin in the presence of ultraviolet light. Vitamin D is required for normal growth of the skeleton and teeth because it controls the levels of calcium and phosphorus in the body. Malforma-tion of the bones, known as rickets, is associated with a lack of vitamin D. Some vitamins needed by man vitamin Some deficiency symptoms Important sources FAT-SOLUBLE Vitamin A (retinol) Dry. brittle epithelia of skin, respiratory system, andurogenitaltract; night blindness and malformed rods Green and yellow vegetables and fruit, dairy products, egg yolk, fish-liver oil Vitamin D (calciferol) Rickets orosteomalacia( very low blood calcium level, soft bones, distorted skeleton, poor muscular development) Egg yolk, milk, fish oils Vitamin E (tocopherol) Male sterility in rats (and perhaps other animals); muscular dystrophy in some animals; abnormal red blood cells in infants; death of rat and chicken embryos Widely distributed in both plant and animal food, e.g. meat, egg yolk, green vegetables, seed oils Vitamin K Slow blood clotting and hemorrhage Green vegetables WATER - SOLUBLE Thiamine (B1) Beriberi (muscle atrophy, paralysis, mental confusion, congestive heart failure) Whole-grain cereals, yeast, nuts, liver, pork Riboflavin (B2) Vascularization of the cornea, conjunctivitis, and disturbances of vision; sores on the lips and tongue; disorders of liver and nerves in experimental animals Milk, cheese, eggs, yeast, liver, wheat germ, leafy vegetables Pyridoxine (B6) Convulsions, dermatitis, impairment of antibody synthesis Whole grains, fresh meat, eggs, liver, fresh vegetables Pantothenic acid Impairment of adrenal cortical function, numbness and pain in toes and feet, impairment of antibody synthesis Present in almost all foods, especially fresh vegetables and meat, whole grains, eggs Biotin Clinical symptoms in man are extremely rare, but can be produced by great excess of raw egg white in diet; symptoms are dermatitis, conjunctivitis Present in many foods, including liver, yeast, fresh vegetables Nicotinamide Pellagra (dermatitis, diarrhea, irritability, abdominal pain, numbness, mental disturbance) Meat, yeast, whole wheat Folic acid Anemia, impairment of antibody synthesis, stunted growth in young animals Leafy vegetables, liver Cobalamin (B12) Pernicious anemia Liver and other meats Ascorbic acid (C) Scurvy (bleeding gums, loose teeth, anemia, painful and swollen joints, delayed healing of wounds, emaciation) Citrus fruits, tomatoes Vitamin E is important in some animals in maintaining good muscle condition and normal liver function. In addi-tion, vitamin E has been shown to be necessary in prevent-ing sterility in many test animals including rats, chickens and ducks. A deficiency of vitamin E produces progressive deterioration and paralysis of the muscles, abnormal red blood cells in infants, and male sterility in some animals. A deficiency of this vitamin rarely occurs in man. Vitamin K plays an important role in the normal co-agulation of blood by promoting synthesis in the liver ofprothrombinandproconvertin. These are two factors of the blood-clotting system. Vitamin K-deficient patients are highly susceptible to hemorrhages. Bacteria living in the large intestine of man are capable of producing vitamin K. Vitamin C, a water-soluble vitamin, is required in cellular metabolism, specifically in cellular reductions. It is also important in the formation ofhydroxyproline, one of the constituents of collagen. The disease scurvy, resulting from a lack of this vitamin, is characterized by bleeding gums, bruised skin, painful swollen joints and general weakness. Vitamin B complex is a group of water-soluble com-pounds, unrelated chemically but which tend to be associ-ated together. Several of them act as parts of coenzymes functioning in cellular respiration. For example, thiamine (vitamin B1) is a principal part of the coenzyme that catalyzes the oxidation ofpyruvicacid. Riboflavin (vitamin B2) is one of the hydrogen-carriers in the oxida-tivecytochromesystem. Pyridoxine (vitamin B6) is a com-ponent of a coenzyme involved intransamination-reactions transferring amino groups from one compound to another. A list of vitamins, their sources and deficiency symptoms appear in the table.

Question:

A nurse mistakenly administers a dilute Ba(NO_3)_2 (a soluble, strong electrolyte) solution to a patient for radiographic investigation. What treatment would you provide to prevent the absorption of soluble barium and subsequent barium poisoning?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E26-0877.htm

Solution:

An electrolyte is a compound that conducts electricity when dissolved in water. Electrolytes are able to do this by dissociating into ions in water. Because Ba(NO_3)_2 is a strong electrolyte, it will com-pletely dissociate into ions according to the reaction Ba(NO_3)_2 (S) \rightarrow Ba^2+ (aq) + 2NO^-_3 (aq) To prevent barium poisoning, the Ba^2+ ions, must be removed from the solution as an insoluble compound so that they cannot be absorbed. One way to precipitate out the barium ion (Ba^2+) is by adding a dilute solution of Na_2SO_4. The reason is that Na_2SO_4, reacts with Ba2+and NO^-_3 ions to produce BaSO_4, and NaNO_3. The barium sulfate is extremely insoluble as indicated by a solubility product constant of 1.6 × 10^-9 mole^2 liter^-2. (A solubility product constant is indicative of a salt's solubility in solution. If this value is low, the salt is insoluble or barely soluble.) Therefore, by adding the Na_2SO_4, the Ba^2+ pre-cipitates out and is not absorbed.

Question:

Write a program that a) Has a routinefibonacci( ). This routine calculates Fibonacci numbers using recursive algorithm. b) Assume in the main program that 10th Fibonacci number is to be calculated. c) Print the 10th Fibonacci number.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G17-0432.htm

Solution:

Fibonacci number is the sum of the previous two Fibonacci numbers,eg. Fib(0) = 0 Fib(1) = 1 Fib(n + 1) = Fib(n) + Fib(n - 1) The first two Fibonacci numbers are 1 just as factorial of 1 is 1. The algorithm followed is 1) Check if value received byfibonacci( ) is equal to 1 or2. 2) If value not equal to 1 or 2 the routine calls itself with (value - 1). 3) This continues until the condition value equals 1 or 2. This is the end condition. main ( ) {intFib _ num; Fib _ num =fibonacci(10); printf("The 10th Fibonacci number is = %d\textbackslashn", Fib _ num); } fibonacci(n) ,intn; { intresult; /\textasteriskcenteredfor result \textasteriskcentered/ if(n == 1 \vert\vert n == 2) result =1; else result =fibonacci(n - 1) +fibonacci(n - 2); return(result); }

Question:

What is the mass of 1 liter of carbon monoxide (CO) at standard temperature and pressure (STP).

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Solution:

At STP, 1 mole of any gas occupies a volume of 22.4 l. The weight of any substance divided by its molecular weight (MW) is equal to the number of moles of that substance. Thus, the weight of 1 mole of any substance weighs its molecular weight (in grams). The molecular weight of CO = 28 g/mole. Thus, 22.4 liters of CO at STP weigh 28 g. The mass of one liter of CO can then be found from the following proportion [(28 g) / ( 22.4 liters )] = [(X grams of CO) / (1 liter)] Solving for X,X = 1.25 g.

Question:

When a cathode\rule{1em}{1pt}ray tube has its deflector plates con-nected across a 100\rule{1em}{1pt}V battery, the spot on the fluo-rescent screen is deflected 12 cm. If the plates are now connected across a resistance of 20 \Omega in parallel with an ac voltmeter and in series with a 50\rule{1em}{1pt}cycle \bullet s^\rule{1em}{1pt}1 ac generator, the length of trace on the screen is 17 cm and the voltmeter reads 50 V. How can these apparently contradictory figures be explained? Suppose that the resistance is replaced by a coil of resistance 5 \Omega and reactance 0.1 H, and that the current drawn from the generator is adjusted to the same value as before. What length of trace will be obtained and what is the rate of heat production in the coil?

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/Users/wenhuchen/Documents/Crawler/Physics/D23-0779.htm

Solution:

When the 100\rule{1em}{1pt}V battery is applied to the deflector plates, a deflection of 12 cm (= d_1 in Fig. A) is obtained. The displacement per volt is thus 1.2 mm. (See Fig. A). When the ac voltage across the resistor is applied to the cathode ray tube, a trace of length 17 cm is obtained and the voltmeter reads only 50 V. At first sight this might appear to give a different displace-ment per volt, but this is not so. The reading on the voltmeter is the root\rule{1em}{1pt}mean\rule{1em}{1pt}square (rms) value of the alternating voltage. The peak value is given by the rms value multiplied by \surd2 or 50\surd2 = 70.7 V. The trace on the screen is responsive to the instantane-ous voltage applied to the plates. It therefore marks a movement of the spot from the position where it is subjected to the peak value to the position where it is subjected to minus the peak voltage. (See Figure B). In other words, the length of the trace corres-ponds to a deflection due to a change of 70.7 \rule{1em}{1pt} (\rule{1em}{1pt} 70.7) = 141.4 V. The displacement per volt is 17 cm/141.4 = 1.2 mm, in agreement with the dc measurement. The rms current through the resistor is I = (V/R) = (50 V/20 \Omega) = 2.5 A. This is the current an ammeter would read. When the resistor is replaced by a coil, we are given that the (rms) current remains the same (see Fig. C). The rms voltage across the coil is V = IZ, where Z is the magnitude of the impedance of the series combination of the resistance and the inductance. If X_L = 2\pifL is the reactance of the inductance (where f is the frequency of voltage variation) then Z = \surd(R^2 + X^2_L) = \surd{R^2 + (2\pifL)^2} andV = IZ = 2.5 A × \surd{5^2 \Omega^2 + (2\pi × 50 × 0.1)^2 \Omega^2} = 2.5 × 31.8 V = 79.5 V. The peak value of voltage is the rms voltage times \surd2 or V' = \surd2 × 79.5 V = 112.4 V. The peak to peak value is twice this or 224.8 V. From the discussion above, the displacement per volt is 1.2 mm = 0.12 cm \textbullet V^\rule{1em}{1pt}1 and the trace on the screen is 224.8 V × 0.12 cm \textbullet V^\rule{1em}{1pt}1 = 27 cm in length. The rate of heat production in the coil is due solely to the resistive part of the impedance. Thus W = I^2R = 2.5^2 A^2 × 5 \Omega = 31.25 W.

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Question:

Name each of the following alkanes. Indicated which, if any, are isomers.

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Solution:

Isomers are related compounds that have the same molecular formula but different structural formulas. Isomerism is not possible among the alkanes until there are enough carbon atoms to permit more than one arrangement of the carbon chain. To name the above compounds, one uses a set rules to provide each compound with a clear name. These rules for nomenclature are the IUPAC rules (International Union of Pure and Applied Chemistry) , and are referred to as systematic nomenclature. One of the rules of the IUPAC system is to choose the largest chain of carbon atoms in the molecule and call it parent compound. Thus, (a) has a 6 carbon parent chain with methyl groups bonded to the second and fourth carbon atoms of the parent chain. There exists, also, an ethyl group bonded to the third carbon atom. As such, the name of this organic molecule is 2,4-dimethy1-3- ethylhexane. (b) has 5 carbon parent molecule and 2 methyl groups on carbon number 3. Therefore, the name is 3,3-dimethylpentane. (c) has a 5 carbon parent molecule with 2 methyl groups bonded to the third carbon of the parent chain. Therefore, the name of this structure becomes 3,3 dimethylpentane. (d) has a 5 carbon parent molecule, and 2 methyl groups attached to carbon numbers 2 and 3. Therefore, the name is 2,3-dimethylpentane. (e) has a 5 carbon parent molecule, and 2 methyl groups attached to carbon number 3. Therefore, the name is 3,3-dimethylpentane. (f) has a 6 carbon parent molecule, and 2 methyl groups attached to carbon numbers 3 and 4. Therefore, the name is 3,4-dimethylhexane. To find which compounds are isomers, one counts the number of carbon and hydrogen atoms contained in the molecule. If the total number of both carbon and hydrogen atoms in one molecule is the same as in another molecule, but they have different structural formulas, they are isomers. If the calculation of total carbon and hydrogen molecules is made, b, c, d, and e become isomers, with b and e being same compound.

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Question:

In the proceeding problem what would be correct forms for the circuits in parts a) and b)?

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Solution:

a) For this circuit we could feed the outputs of the two AND-gates into a second level AND- or OR-gate as shown in Fig. 1. b) For this circuit the output Y of the OR-gate can be fed back into one of its inputs by passing the output Y through a better amplifier. The advantages of using the buffer (with amplification >1 or <1) are listed below: i) It matches the output impedance of the OR-gate, to the input impedance of the OR-gate. ii) It balances any phase differences in the signals between the output side and the input side of the gate. iii) It also facilitates amplification or attenuation of the signal fed back from the output to the input so that the amount of signal fed back is the right amount. The possible circuit as shown in Fig. 2.

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Question:

Would you expect the mobility of an electrolytic ion to increase or decrease for each of the following? (a) increasing the size of the ion;(b) increasing charge on the ion; (c) increasing temperature and (d) placing it in a solvent of lower viscosity.

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Solution:

To see how the mobility of the electrolytic ion changes when you alter these parameters, you must know what happens to these ions when an electric field is applied. Theionsare free to move randomly around in solution before an electric field is applied. However, when the field is present, the positive ions experience a force in one dir-ection, while the negative ions experience a force in the opposite dir-ection. With this in mind, you proceed as follows: a) when you increase the size of the ion, you would anticipate a decrease in mobility. Remember, the ion must move when the electric current is applied; additional mass and volume, therefore, inhibits this movement, since there is greater resistance from the solvent mole-cules . b) If you increase the charge on the ion, you expect mobility to in-crease due to the fact that the force which results in movement is dir-ectly proportional to the attraction of the ions to the poles. Thus, by increasing the charge, you increase this force, which, in turn, in-creases movement. c) When you increase the temperature, you increase the mobility of the ions. By increasing the temperature, the average kinetic energy of the ion increases. Kinetic energy is a measure of movement. d) A solution of lower viscosity is expected to have an increase in mobility. This stems from the fact that viscosity measures internal resistance to flow. Thus, if the viscosity decreases, there is less resistance to flow, which means that the ions can move more freely.

Question:

The atomic weight of element X is 58.7. 7.34 g of X displaces 0.25 g of hydrogen from hydrochloric acid (HCl). What is the valence of element X?

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Solution:

Chlorine has a valence of -1, thus to determine the valence of X, one must calculate the number of moles ofCl^- that will bind to each mole of X. The valence will be equal to this number. There are 2 moles ofCl- present for every mole of H_2 formed. To find the number of moles of H2 formed, one must divide 0.25 g by the molecular weight of H_2(MW of H2= 2). moles of H_2 = 0.25 / (2 g / mole) = .125 moles of H_2. no. of moles ofCl^- = 2 × .125 = .250 moles. One should now determine the number of moles of X present. This is done by dividing the number of grams by the molecular weight. (MW of X = 58.7). no. of moles = 7.34 g / (58.7 g / mole) = .125 moles. From this, one sees that .125 moles of X combines with .250 moles ofCl^-. The number of moles ofCl^- that bind to each mole of X is equal to the number of moles ofCl^- present divided by the number of moles of X. no. ofCl^- that combine with each x = (.250molesofCl^-) /(.125molesof X) = 2 moles /Cl^- / mole X The formula for the resulting compound is XCL_2. BecauseCl^- has a valence of -1 and 2Cl^- combine with each X, X must have a valence of +2 for a neutral molecule to be formed.

Question:

A voltaic cell is made up of a silver electrode in a 1M silver nitrate solution and an aluminum electrode in a 1M aluminum nitrate solution. The half reactions are 1) Ag^+ + e^- \rightarrow Ag (s) E\textdegree = .80 volt, and (2) Al^3+ +3e^-\rightarrow Al(s) E\textdegree = -1.66 volt, calculate the standard free energy change, ∆G\textdegree, in Kcal / mole. 1 cal = 4.184 joule 1 F = 96,500 coul, 1 Kcal = 1000 cal.

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Solution:

The standard free energy change, ∆G\textdegree , can be related to standard cell potential change, ∆E\textdegree , via ∆G\textdegree = n F∆E\textdegree, where n = number of electrons transferred per mole. To find ∆G\textdegree, calculate (1) ∆E\textdegree, (2) n. (1) You are told a voltaic cell exists. This means that electricity is generated. You are given the E\textdegree values for the half-reactions. For electricity to be generated, how-ever, the ∆E\textdegree for the overall reaction must be positive. Both half-reactions represent reduction processes (gain of electrons) . But ∆E\textdegree = E\textdegree_oxid + E\textdegree_red. Thus, to find ∆E\textdegree, you must reverse one of these reactions to oxidation (loss of electrons) so that ∆E\textdegree for the overall reaction, is positive. This condition can only be met if (2) is reversed to Al(s) \rightarrow Al^3+ + 3e^-, E\textdegree = -(-1.66) = 1.66 and (1) remains as Ag^+ + e^- \rightarrow Ag(s). Before adding these two to obtain the overall reaction, multiply (1) by 3, so that all electrons in the overall equation can be cancelled. Thus Al(s) \rightarrow Al^3+ + 3e^- 3Ag^+ + 3e^- \rightarrow 3Ag(s) Al(s) + 3Ag^+ \rightarrow Al^3+ + Ag(s) with ∆E\textdegree = E\textdegree_oxid + E\textdegree_red= 1.66 + .80 = 2.46 volts. (2) The number of electrons being transferred = 3, so that n = 3. ∆G\textdegree = n F∆E\textdegree. Substitute in values to obtain ∆G\textdegree = (3)(96500) (2.46)(cal / 4.184 joules) × (Kcal / 1000 cal) = + 170( Kcal / mole).

Question:

What is the acceleration of gravity at a place where the period of a simple pendulum 100 cm long is exactly 2 sec?

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Solution:

We assume that the pendulum bob are the tension of the string and the weight of the bob (it's mass is m). We assume that the displacement of the bob from the equilibrium position x = 0 (see figure) is small enough that the bob undergoes only a horizontal motion (i.e., the arc of motion of the bob Is approximated by a straight line). The y direction of motion is then zero. Therefore there is no net force acting in the y direction. Hence T cos \texttheta= mg or T = mg/cos \texttheta(1) The net restoring force acting in the x direction is F= T sin \texttheta = mg[(sin \texttheta)/(cos \texttheta)] = mg tan \texttheta Here we made use of equation (1). Since the displacement of the bob is very small, then the following approximations can be made. sin \texttheta\approx \texttheta and cos \texttheta \approx 1 F\cong mg \texttheta The restoring force is then directly proportional to the angular displacement \texttheta. Since the horizontal displacement x is approximately equal to the arc length of a circle of radius 1, then x \cong 1\texttheta where \texttheta is in radians. Therefore F \cong (mg/l)x The restoring force on the particle is then directly proportional to the displacement from the equilibrium position. This is the definition of harmonic motion. By Newton's second law, F = ma = m[(d^2x)/(dt^2)] = (mg/l)x [(d^2x)/(dt^2)] = (g/l)x The solution of this differential equation, is x = A sin[\surd(g/l)] t + B cos \surd(g/l) = C sin[{\surd(g/1)} t + \Phi](1) This can be verified by substitution into the differential equation. The constants A and B (or alternatively c and \Phi) can be found from the initial conditions of the problem (i.e., where the bob was located and its velocity, att = 0). Equation (1) is of the form x = c sin (\omegat + \Phi) Therefore\omega = 2\pif = \surd(g/l) where f is the frequency of motion. However f = 1/T where T is the time taken by the bob to undergo one oscillation. Then 2\pi/T= (g/l) T= 2\pi\surd(l/g) 2^2= [(2\pi)^2 (100)]/g thereforeg= (4\bullet\pi^2\bullet100 cm)/4s^2 = 1000 cm/sec^2.

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Question:

How does respiration occur in insects?

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Solution:

The unique respiratory system of insects con-sists of a network of tubes called trachea (see figure 1). These trachea open to the outside by means of holes in the insects body. These holes are called spiracles and are conspicuous on the sides of the thorax and abdomen. Each spiracle is guarded by a valve which can be opened or closed to regulate air flow. The tracheal tubes extend to all the internal organs. The tubes terminate in microscopic fluid-filled tracheoles; oxygen and carbon dioxide diffuse through these into the adjacent cells. The body wall of the insect pulsates, drawing air into the trachea when the body expands, and forcing air out when the body contracts. Grasshoppers draw air into the body through the first four pairs of spiracles when the abdomen expands and expel it through the last six pairs when the body contracts. Therefore in contrast to a fish or a crab, which respire through gills, the tracheal system conducts air deep within the insect's body. The air is brought near enough to each cell so that gases can diffuse across the wall of the tracheal tube. For this reason, insects need not maintain a rapid blood flow, as vertebrates must, to supply their cells with oxygen.

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Question:

A ball of mass 100 gm is thrown against a brick wall. When it strikes the wall it is moving horizontally to the left at 3000 cm/sec, and it rebounds horizontally to the right at 2000 cm/sec. Find the im-pulse of the force exerted on the ball by the wall.

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Solution:

The impulse of a force on an object is defined as the change in momentum of the object during the time that the force acts. The initial momentum of the ball is equal to the product of its mass and initial velocity, or 100 gm × -3000 cm/sec = -30 × 10^4 gm.cm/sec. The final momentum is equal to the product of the ball's mass and final velocity, or +20 × 10^4 gm\bulletcm/sec Note that the final and initial momenta are in opposite directions. We have defined the final direction of travel of the ball to be the positive direction. The change in momentum is then mvf- mv_i = 20 × 10^4 gm.cm/sec - (-30 × 10^4 gm\bulletcm/sec) = 50 × 10^4 gm\bulletcm/sec. Hence the impulse of the force exerted on the ball was 50 × 10^4 dyne^.sec. Since the impulse is positive, the force is toward the right. Note that the force exerted on the ball cannot be found without further information regarding the collision. The general nature of the force- time graph is shown by one of the curves in the figure. The force is zero before impact, rises to a maximum, and decreases to zero when the ball leaves the wall. If the ball is relatively rigid, like a baseball, the time of collision is small and the maximum force is large, as in curve (a). If the ball is more yielding, like a tennis ball, the col-lision time is larger and the maximum force is less, as in curve (b). In any event, the area under the force-time graph must equal 50 × 10^4 dyne\bulletsec.

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Question:

Ordinarily, the bodies ofmulticellularorganisms are organizedon the basis of tissues, organs, and systems. Distinguish between these terms.

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Solution:

A tissue may be defined as a group or layer of similarly specialized cellswhich together perform a certain specific function or functions. Each kindof tissue is composed of cells which have a characteristic size, shape andarrangement, and which are bound together by an intracellular substanceallowing communication between adjacent cells. Some examplesof tissues are epithelial tissue, which separates an organism fromits external environments; muscle tissue, which contracts and relaxes; and nervous tissue, which is specialized to conduct information. An organ is composed of various combinations of tis-sues, grouped togetherinto a structural and functional unit. An organ can do a specific job: the heart pumps bloodthrough outthe circulatory system; or an organ canhave several different functions: the human liver produces bile salts, breaksdown red blood cells and stores glucose as glycogen. A group of organs interacting and cooperating as a functional complexin the life of an organism is termed an organ system. In the humanand other vertebrates, the organ systems are as follows: (1)thecirculatory system, which is the internal trans-port system of animals; (2)therespiratory system, which provides a means for gas exchange betweenthe blood and the environment; (3)thedigestive system, which functions inprocurringand processing nutrients; (4)theexcretory system, which eliminates the waste products of metabolism; (5)theintegumentarysystem, which covers and protects the entire body; (6)theskeletal system, which supports the body and pro-vides for body shapeand locomotion; (7)themuscular system, which functions in movement andlocomotions; (8)thenervous system, which is a control system essen-tial in coordinating andintegrating the activities of the other systems with themselves and withthe external environment; (9)theendocrine system, which serves as an additional coordinator of the bodyfunctions; and (10)thereproductive system, which functions in the production of new individuals for the continuation of the species.

Question:

An instrument that has been used to measure atomic masses with great accuracy is the 180\textdegree mass spectrometer shown in the figure. Ionized atoms from a source pass through a potential difference and enter a region in which there is a uniform magnetic field B. The magnetic force acts at right angles to the direction of the velocity vector and causes the ions to move along a circular path. After completing one-half revolution (180\textdegree), the ions strike a detector, usually a photographic film. (a) Show that the charge/mass ratio, q/m, of these ions is given by q / m = v / BR where v is the speed of the ions, B the magnetic field strength, and R the orbit radius. (b) In an experiment the orbit radius is observed to be 0.2 m for singly ionized atoms of speed 2.1 x 10^5 m/s in a magnetic field of 0.13 T. What is the charge/mass ratio of these ions? (c) What is the mass of these atoms?

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Solution:

(a) For circular motion the centripetal acceleration is a = v^2 / R. Newton's second law of motion requires that the force exerted on the ion be F = ma = m(v^2 / R^2). The magnitudes of the electric potential and magnetic fields determine the value of q/m for protons. The charge of the proton, q = 1.6 x 10^\rule{1em}{1pt}19 C, is known; therefore, the mass may be computed as follows: An ion of charge q is accelerated through a potential difference V. It attains a velocity given by (1 / 2)mv^2 = Vq since the potential energy of the ion at the source is Vq with respect to the inlet of the magnetic deflection area. The deflecting force on the charge due to the magne-tic field is F = qvB = m(v^2 / R^2) (by Newton's second law) giving V = qBR / m. Therefore the ratio q / m for the spectrometer is , q / m = v / BR. The value of q/m for protons from this type of experiment is approximately 9.57 × 10^7 c / kg. We find that for the proton q / m = 9.57 × 107C/kg but q = 1.67 × 10^\rule{1em}{1pt}19c m =(1.6× 10^\rule{1em}{1pt}19c) / (9.57 × 107c/ kg) = 1.67 ×10^\rule{1em}{1pt}27kg. (b) The given quantities are v = 2.1 x 10^5 m/s, B = 0.13 T, and R= 0.2 m. The charge/mass ratio for these ions is found to be q / m = (2.1 × 10^5m/ s) / [(1.3 × 10^\rule{1em}{1pt}1 T)(2 × 10^\rule{1em}{1pt}1m)] = 8.1 × 10^6 c / kg. (c) The charge q of the ions is 1.6× 10^\rule{1em}{1pt}19 c, therefore, the mass is given by m = q / (q / m) = (1.6 × 10\rule{1em}{1pt}19c)/ (8.1 × 106c/ kg) = 2.0 × 10\rule{1em}{1pt}26kg.

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Question:

A chemist possesses 1 cm^3 of O_2 gas and 1 cm^3 of N_2 gas both at Standard Temperature and Pressure (STP). Compare these gases with respect to (a) number of molecules, and (b) average speed of molecules.

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Solution:

This problem requires the use of the kinetic theory of gases and Avogadro's principle. Avogadro proposed the principle that equal volumes of gases at the same temperature and pressure contain equal numbers of molecules. Both gases have equal volumes, 1 cm^3, and are at the same temperature and pressure (STP). According to Avogadro's Principle, both have the same number of molecules. The average speed of these molecules can be found from kinetic theory of gases. The rate of diffusion of a gas is directly proportional to the average speed of the molecules. According to Graham's law, there is an inverse proportionality between diffusion rate and the square root of the molecular weight (mass). In other words, rate of diffusion of N_2 / rate of diffusion of O_2 = V_(N)2 / V_(O)2 = \surd[M_(O)2 / M_(N)2] , where V = velocity and M =masM_(N)2 = 28, M(O)2= 32. In this case, mass = molecular weight. Thus, V_(N)2 / V_(O)2 = \surd[32/28] = 1.07. Thus, if the average speed of O_2 is one and the average speed of N_2 is 1.07.

Question:

Draw the structure of 4-ethyl-3, 4-dimethyl-2-hexene.

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Solution:

To draw the structure of more complex compounds, such as this one, certain steps must be followed. (1) Identify the parent compound that associated with the longest carbon chain that contains the functional group. In 4-ethyl-3, 4-dimethyl-2 hexene, the parent compound is hexene. (2) Draw the parent carbon skeleton, in this case, a six carbon chain. Do not put any hydrogen atoms in yet. (3) Number the carbon atoms starting at either end. This is important; otherwise it may get confusing when one adds the functionality. (4) Add the suffix functionality, in this case -2-ene. "Ene" tells one that a double bond is present, while "2" indicates that it is at the second carbon. (5) Add the prefix functionality, starting at the beginning of the name and continuing until the parent name is reached. Here, the prefixes are 4-ethyl-3, 4-dimethyl-. Now, the hydrogen atoms can be added to give a complete structure.

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Question:

Of the many compounds present in cigarette smoke, such as the carcinogen 3,4-benzo [a]pyrene, some of the more abundant are listed in the following table along with their mole fractions: ComponentMole Fraction H_20.016 O_20.12 CO0.036 CO_20.079 What is the mass of carbon monoxide in a 35.0 ml puff of smoke at standard temperature and pressure (STP)?

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Solution:

Assuming ideal gas behavior, we will calculate the number of moles of ideal gas in a 35.0 ml volume. Using the mole fraction of CO in smoke, we will then obtain the number of moles of CO in a 35.0 ml volume and finally convert this to a mass. The molar volume of an ideal gas is 22.4 liter/mole when STP conditions exist, i.e. when temp. = 0\textdegreeC and pressure = 1 atm. Hence, the number of moles of ideal gas in a 35.0 ml volume is obtained by dividing this volume by the molar volume of gas, or 35.0 ml/22.4 liter/mole = 0.035 liter/22.4 liter/mole = 1.56 × 10^-3 mole. The mole fraction is defined by the equation mole fraction CO = moles CO / total number of moles solving for the number of moles of CO, moles CO = mole fraction CO × total number of moles = 0.036 × 1.56 × 10^-3 mole = 5.6 × 10^-5 mole. This is converted to a massoymultiplying by the molecular weight of CO (28 g/mole). Hence, mass of CO = moles of CO × molecular weight of CO = 5.6 × 10^-5 mole × 28 g/mole = 1.6 × 10^-3 g.

Question:

Write an APL program to find the wavelengths for the first four spectral lines of hydrogen in thePaschenseries.

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Solution:

ThePaschenseries deals with the infrared region of the hydrogen spectra. The equation used to find the lines is given in terms of the wavelength's reciprocal. NU = 109,677.58 [(1/9) - (1/B^2)]cm^-1, for B = 4,5,6,7. APL is useful in that we can design a program, and then enter the values for B in sequence. Remember also that in APL, operations are performed right-to-left. The program logic is quite simple: \nablaPASCHEN [1]NU \leftarrow 109,677.58 (\div9 - \divB\textasteriskcentered2) [2]LAMBDA \leftarrow \divNU [3]LAMBDA \nabla Now, when returned to the execution mode, you can enter the integers for B. Sample output looks like this: B \leftarrow 4.0 PASCHEN 0.0001875 B \leftarrow 5.0 PASCHEN 0.0001282 B \leftarrow 6.0 PASCHEN 0.0001094 B \leftarrow 7.0 PASCHEN 0.0001005 Note that this is the crudest way of entering the values for B. This process can be made easier by introducing vec-tors. If you were to enter: B \leftarrow 4.05.06.07.0 PASCHEN the resulting output would be 0.00018750.00012820.00010940.0001005 Note, that only spaces are to be inserted between successive values of B, no commas.

Question:

What is the maximum efficiency of a steam engine if the temperature of the input steam is 175\textdegree C and the temperature of the exhaust is 75\textdegree C?

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Solution:

Carnot's Theorem states that the efficiency of all reversible engines operating between the same 2 temperatures is the same. Furthermore, no irreversible engine (including our steam engine) can have an efficiency greater than this. The efficiency of a reversible engine is e = 1 - (T_2/T_1) where T_2 and T_1 are the Kelvin temperatures of the low and high temperature sinks, respectively, of the engine. Hence, for our steam engine e_max= 1 - (T_2/T_1) = 1 - [(75 + 273)\textdegreeK/(175 + 273)\textdegreeK] We have used the fact that T (\textdegreeK) = T (\textdegreeC) + 273\textdegree Thene_max= 1 - (348/448) = 100/448 = .223 In terms of percentage e_max= 22.3%.

Question:

A piece of iron weighing 20.0 g at a temperature of 95.0\textdegreeC was placed in 100.0 g of water at 25.0\textdegreeC. Assuming that no heat is lost to the surroundings, what is the resulting temperature of the iron and water? Specific heats: iron = .108 cal/g\rule{1em}{1pt}C\textdegree; water = 1.0 cal/g\rule{1em}{1pt}C\textdegree.

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Solution:

The heat lost by the iron must be equal to the heat gained by the water. One solves for the heat lost by the iron by multiplying the number of grams of Fe by the number of degrees the temperature dropped by the specific heat of iron. The specific heat of a substance is defined as the amount of heat energy required to raise the tem-perature of 1 g of a substance by 1\textdegreeC. The specific heat for iron is .108 cal/g\textdegree C. Let t = the final temperature of the system. amount of heat lost by the iron = .108 cal/g\rule{1em}{1pt}\textdegreeC × 20.0 g × (95.0\textdegreeC \rule{1em}{1pt} t) The amount of heat gained by the water is the specific heat of water multiplied by the weight of the water multiplied by the rise in the temperature. The specific heat of water is 1.0 cal/g\rule{1em}{1pt}C\textdegree. Let t = final temperature of the system. amount of heat gained by water = 1.0 cal/g\rule{1em}{1pt}\textdegreeC × 100 g × (t - 25\textdegreeC) Solving for t: amount of heat lost by the iron = amount of heat gained by the water. Therefore, (.108 cal/g\textdegree C) (20.0 g) (95\textdegreeC\rule{1em}{1pt}t) = (1 cal/g\textdegree C) (100 g) (t\rule{1em}{1pt}25\textdegreeC) 205.2 cal \rule{1em}{1pt} (2.16 cal/\textdegreeC)t= (100 cal/\textdegreeC)t - 2500 cal 2705.2 cal= (102.16 cal/\textdegreeC)t (2705.2 cal) / (102.16 cal/\textdegreeC)= t 26.48\textdegreeC= t.

Question:

A 20 gallon automobile gasoline tank is filled exactly to the top at 0\textdegreeF just before the automobile is parked in a garage where the temperature is maintained at 70\textdegreeF. How much gasoline is lost due to expansion as the car warms up? Assume the coefficient of volume expansion of gasoline to be .0012/\textdegreeC.

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/Users/wenhuchen/Documents/Crawler/Physics/D12-0452.htm

Solution:

Here, we assume that the tank doesn't expand or contract. By definition of the coefficient of volume expansion \beta = \DeltaV / (V_0 \DeltaT) where V_0 is the original volume occupied by the liquid, and \DeltaV is the change in volume of the liquid due to a temperature change \DeltaT. The gasoline lost is then \DeltaV = \beta V_0 \DeltaT \DeltaV = (.0012 \textdegreeC^-1) (20 gal) (70\textdegreeF) But 70\textdegreeF = (5/9) \textbullet (70\textdegreeC) =(350/9)\textdegreeC and \DeltaV = (.0012)(20 gal)(350/9) \DeltaV = .94 gal.

Question:

List the structural characteristics of the blue-green algae. Where may they be found?

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/Users/wenhuchen/Documents/Crawler/Biology/F06-0164.htm

Solution:

The blue-green algae, (Cyanophyta or cyanobac-teria) are prokaryotic unicellular or filamentous organ-isms. The filamentous forms are strings of individual cells held together by fused walls. (See Figure 1.) In most species, there are no protoplasmic connections, such as plasmodesmata, between the adjacent cells. The uni-cellular forms exist as either single rods or spheres, and each individual cell is capable of carrying out all necessary life processes. The Cyanophyta are the most primitive chlorophyll- containing autotrophic organisms now living. The Cyanophyta are quite different from other algae, being structurally similar to bacteria. These similarities have formed the basis for grouping them with the bacteria in the kingdom Monera. All blue-green algae possess photosynthetic pigments, which are located in folds or convolutions of the cell membrane extending into the interior of the cell. These photosynthetic pigments are chlorophyll a, carotenoids, phycocyanin (blue pigment) and sometimes phycoerythrin (red pigment). Like the bacteria, the blue-green algae lack mitochondria, Golgi apparatus, a nuclear membrane, endoplasmic reticulum and the large cell vacuole charac-teristic of higher plants. Ribosomes are present along with many proteinaceous granules and granules of a stored carbohydrate material known as cyanophycean starch, which is very similar to glycogen. The nuclear region consists of a single circular chromosome composed of double--stranded DNA. The cell walls of blue-green algae contain some muramic acid and cellulose. Outside this cell wall is a sticky gelatinous sheath composed of pectic materials. It covers the entire cell. The cytoplasm of blue-green algae is unusually viscous, being composed of a very dense colloidal material. Cell division in blue-green algae is accomplished by binary fission. Reproduction also occurs frequently by fragmentation of filaments. Sexual reproduction has never been observed in the Cyanophyta but is thought to occur infrequently. No blue-green algae possess flagella, yet many species are capable of movement. The filamentous forms exhibit a peculiar slow gliding or oscillatory motion. The blue-green algae are found in many varied habitats. Most are found in fresh water pools and ponds. A few species are found in hot springs, at temperatures up to 85\textdegreeC. Some species are marine while others are common in soils, the banks of trees and the sides of damp rocks. Some species of blue-green algae exhibit a symbiosis with fungi, in which both act together to form a lichen.

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Question:

Calculate the concentration of calcium ion and fluoride ion in a concentrated solution of calcium fluoride, CaF_2 (K_sp= 4.00 × 10-^11 ).

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/Users/wenhuchen/Documents/Crawler/Chemistry/E11-0398.htm

Solution:

Problems of this type are solved by making use of the fact that the K_sp, which is known, is equal to the product of the concentrations of Ca^++ and F- . If we know the concentration of one ion then, from the proportions in which the ions occur in the molecule, we can determine the concentrations of all the ions formed by dis-sociation of the molecule. When calcium fluoride dissociates, CaF_2 (s) \rightleftarrows Ca^2+ + 2F-, one ion of Ca^2+ and two ions of F- form per molecule of CaF_2 . Hence, if we let x denote the concentration of calcium ions ([Ca^2+] = x), the concentration of fluoride ions is twice as great, ([F-] = 2x). The expression for the solubility product constant is K_sp= [Ca^2+] [F-]^2 = 4.00 × 10-^11 . Substituting [Ca^2+] = x and [F-] = 2x gives (x)(2x)^2 = 4.00 × 10-^11 , (x)(4x^2) = 4.00 × 10-^11 , 4x^3 = 4.00 × 10-^11 , or, x = [(4.00 × 10-^11 ) / 4]^1/3 = 3.42 × 10-^4 Hence, [Ca^2+] = x = 3.42 × 10-^4and[F-] = 2x = 2 × 3.42 × 10-^4 = 6.84 ×10-^4 .

Question:

An electron is observed moving at 50 per cent of the speed of light, v \textemdash 1.5 × 10^8 m/s. (a) What is the relativistic mass of the electron? (b) What is the kinetic energy of the electron?

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/Users/wenhuchen/Documents/Crawler/Physics/D32-0946.htm

Solution:

(a) The known observables are v = 1.5 × 10^8 m/s, and m_0 = 9.1 × 10 ^-31 kg. The relativistic expression for mass as a function of velocity is m = m0/\surd{1 - (v/c)^2} = (9.1 × 10^-31 kg) / \surd{1 - (1.5 × 108m/s^2) / 3 × 108m/s} = (9.1 × 10^-31 kg) / \surd{1 - (0.5)^2} (9.1 × 10 ^-31 kg) / \surd0.75 = (9.1 × 10 ^-31 kg) / (8.7 × 10^-1 ) = 1.05 × 10^-30 kg (b) The relativistic energy , E , of the particle is the sum of its kinetics energy , T , and its rest mass energy E = m_0c^2 + T T = E - m_0c^2 T = (m - m_0)c2 T = (10.5 × 10-31kg - 9.1 × 10^-31 kg) (3 × 10^8 m/s)2 = (1.4 × 10-31kg )( 9 × 10^16 m^2s^2) = 1.26 × 10^-14 J.

Question:

Using the information in the following table, determine ∆H\textdegree for the reactions: a) 3Fe_2O_3 (s) + CO(g) \rightarrow 2Fe_2O_4 (s) + CO_2 (g) b) Fe_3O_4 (s) + CO(g) \rightarrow 3FeO(s) + CO_2 (g) c) FeO(s) + CO(g) \rightarrow Fe(s) + CO_2 (g) Heats of Formation Compound ∆H\textdegree (Kcal/mole) CO (g) \rule{1em}{1pt} 26.4 CO_2 (g) \rule{1em}{1pt} 94.1 Fe_2O_3(s) \rule{1em}{1pt} 197 Fe_3O_4 (s) \rule{1em}{1pt} 267 FeO (s) \rule{1em}{1pt} 63.7

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/Users/wenhuchen/Documents/Crawler/Chemistry/E14-0490.htm

Solution:

The energy change involved in the formation of one mole of a compound from its elements in their normal state is called the heat of formation. Thus, ∆H\textdegree for a re-action can be found by subtracting the sum of the ∆H\textdegree's for the reactants from the sum of the ∆H\textdegree's for the products. ∆H\textdegree = ∆H\textdegree of the products - ∆H\textdegree of the reactants a) 3 Fe_2O_3 (s) + CO(g) \rightarrow 2 Fe_3O_4 (s) + CO_2 (g) When more than one mole of a compound is either re-acted or formed in a reaction, the ∆H\textdegree for that compound is multiplied by the number of moles present in solving for the ∆H\textdegree of the reaction. For this reaction: ∆H\textdegree = (2 × ∆H\textdegree of Fe_3O_4 + ∆H\textdegree of CO_2) - (3 × ∆H\textdegree of Fe_2O_3+ ∆H\textdegree of CO) ∆H\textdegree = [2 × (\rule{1em}{1pt}267) + (\rule{1em}{1pt} 94.1)] - [3 × (\rule{1em}{1pt}197) + \rule{1em}{1pt}26.4)] = \rule{1em}{1pt} 10.7 Kcal. b) Fe_3O_4 (s) + CO(g) \rightarrow 3FeO(s) + CO_2(g) ∆H\textdegree = (3 × ∆H\textdegree of FeO + ∆H\textdegree of CO_2) - (∆H\textdegree of Fe_3O_4 + ∆H\textdegree of CO) ∆H\textdegree = (3 × (\rule{1em}{1pt} 63.7) + (\rule{1em}{1pt} 94.1)) - ((\rule{1em}{1pt} 267) + (\rule{1em}{1pt} 26.4)) =8.2 Kcal. c) FeO(s) + CO(g) \rightarrow Fe(s) + CO_2(g) ∆H\textdegree = (∆H\textdegree of Fe + ∆H\textdegree of CO_2) - (∆H\textdegree of FeO + ∆H\textdegree of CO) The ∆H\textdegree of any element is 0. Thus, the ∆H\textdegree of Fe is 0. ∆H\textdegree = (0 + (\rule{1em}{1pt} 94.1)) \rule{1em}{1pt} ((\rule{1em}{1pt} 63.7) + (\rule{1em}{1pt} 26.4)) = \rule{1em}{1pt} 4.0 Kcal.

Question:

What is the simplest formula of a compound that is composed of 72.4 % iron and 27.6 % oxygen by weight?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E04-0138.htm

Solution:

For purposes of calculation, let us assume that there is 100 g of this compound present. This means that there are 72.4 g of Fe and 27.6 g O. The simplest formula for this compound isFe_nO_m, where n is the number of moles of Fe present and m is the number of moles of O. One finds the number of moles by dividing the number of grams by the molecular weight. number of moles =(number of grams) / MW For Fe: n = number of moles present. MW = 55.8. n = (72.4 g) / (55.8 g/mole) = 1.30 moles For O: m = number of moles present. MW = 16.0. m = (27.6 g) / (16.0 g/mole) = 1.73 moles. One solves for the simplest formula by finding the ratio of Fe : O. Fe / O =1.30 / 1.73 = 0.75 = 3/4 Therefore, n = 3 and m = 4. The simplest formula is Fe_3O4.

Question:

What controls the secretion of the digestive enzymes? Activities of the gastrointestinal hormones Secretin CCK Gastrin Secreted by: Duodenum Duodenum Antrum of the stomach Primary stimulus for hormone release Acid in duodenum Amino acids and fatty acids in duodenum Peptides in stomach Parasympathetic nerves to stomach Effect on: Gastric motility Inhibits Inhibits Stimulates GastricHClsecretion Pancreatic secretion Inhibits Inhibits STIMULATES\textasteriskcentered Bicarbonate STIMULATES Stimulates Stimulates Enzymes Stimulates STIMULATES Stimulates Bile secretion STIMULATES Stimulates Stimulates Gallbladder contraction Stimulates STIMULATES Stimulates Intestinal juice secretion stimulates stimulates STIMULATES denotes that this hormone is quantitatively more important than the other two.

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/Users/wenhuchen/Documents/Crawler/Biology/F17-0433.htm

Solution:

Digestive enzyme secretion is regulated by two kinds of factors: neuralandhumoral(hormonal). Neural control is usually based upon the sensingof a physical mass of food. The stretching of the digestive tract (distension), is an example of thistyupeof stimulation. It is interest-ing to notethat most of the salivary and sane of the gas-tric secretions sure stimulatedby smelling, tasting and even thinking of food. Hormonal control, on the other hand, is much more specific. The presence of specific kindsof mole-cules in the ingested food stimulates receptor cells to producetheir specific hormone. The hormone circulates in the blood until itreaches thesecretoryorgans which it con-trols . Allsecretoryorgans are under the influence of several factors. The presenceof food in the mouth initiates gastric secretions by the stomach. When the food reaches the stomach, the distension of the stomach walls stimulatesan increase in stomach motion and the partial production of gastrin. The hormonegastrinhas several effects, the major of which is the stimulationof gastric juice secretion. It achieves this by stimulating gastric HClsecretion, which in turn induces protease activity.Pepsin, present in thegastric juice, cleaves proteins into many smaller chains. These small peptidechains stimulate receptor cells in theantrumof the stomach to producemoregastrin.Gastrinalso has the ability to stimulate limited secretionin the intestine and pancreas. Inadditon,gastrinstimulates an increasein gastric motility and helps keep the sphincter of the esophagus tightlyclosed. When partially digested food, orchyme, reaches the duodenum, its physicalpresence initiates peristalsis over the entire intestine. In addition, theacid in thechymecauses receptor cells to release the hormone secretin.Secretinhas many functions. It slows down stomach motion and decreasesgastric juice andgastrinproduction. This effectively keeps the foodin the stomach for a longer period. At the same time,secretin preparesthe intestine for neutralization of the acidicchymeby stimulating therelease of alkaline bicarbonate ions in pancreatic juice. Other receptor cells in the duodenum sense free amino acids and fats. These cells release a hormone calledcholecystokininor CCK. CCK stimulatesthe release of a pancreatic juice rich in digestive enzymes while slowingdown the motion in the stomach. The more fat and protein there is ina meal, the longer will digestion occur. A re-view of the regulatory activitiesof gastrointestinal hormones is given in the accompanying table. Note that CCK andsecretinact synergistically. CCK stimulates secretionof a pancreatic juice rich in digestive enzymes, but it accentuatessecretin'sstimulation of a pan-creatic juice rich in bicarbonate. Secretinstimulates se-cretion of a juice high in bicarbonate, but it accentuatesCCK's secretion of a juice high in enzymes. The control of secretions in digestion is a complex interaction of severalfactors. One's emotional state, as well as physical health, have a greatinfluence on digestion. Age and diet are also critical factors. When a childis born, it does not have sufficient enzyme production to digest many foodsother than milk. As we age, most adults stop producing rennin and losethe ability to fully digest milk. Humans have the most varied diet of all animals. This is reflected in the complex interactions of enzymes and hormonesthat control our digestion.

Question:

Write a program to find the remainder of the quotient of two integer numbers. Test the program with sample data.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G10-0231.htm

Solution:

The remainder is given by R = N - INT(N/D)\textasteriskcenteredD where N = numerator, D = denominator, and integer quotient is INT(N/D), (This statement will work properly only on those systems where INT(X) function determines the largest integer number that does not exceed X. For a few systems where INT(X) just truncates fractional parts of X, the statement would need some modification.) 5 PRINT "NUMERATOR", "DENOM", "REMAINDER", "INT UOT" 1\O READ N,D 15 REM FIND REMAINDER WHEN 'N' IS DIVIDED BY ' DJ 2\O LET R = N - INT(N/D)\textasteriskcenteredD 3\O PRINT N,D,R, INT(N/D) 4\O GO TO 1\O 5\O DATA 93, 12, 1\O\O 25, 365, 52, 365, 7 6\O END Sample Output: NUMERATOR DENOM REMAINDER INT QUOT 93 12 9 7 100 25 4 0 365 52 1 7 365 7 1 52

Question:

The electric field outside of an infinite cylindrically symmetrical charge distribution is equivalent to the field due to an infinite line charge of equal linear charge density.

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/Users/wenhuchen/Documents/Crawler/Physics/D17-0563.htm

Solution:

First, we compute the field due to a cylindrical shell of charge. We note that the field must be radial since no other direction is preferred. We erect as a Gaussian surface, a cylinder, of radius r (r > R) and height l, concentric to the shell. (See figure (a)). By Gauss' law (since E is constant along any cylinder concentric to the axis): \int E^\ding{217} \textbullet dA^\ding{217} = 4\piq, 2\pirlE = 4\pi(s\piRl\sigma) where \sigma = q/area is the surface change density of the shell. We note that since the field is radial, the flux through the top and bottom of the Gaussian cylinder is zero. Thus: E = 4\piR\sigma/r = 2 [{2\piR (q/2\piRl)}/r] = [2(q/l)]/r = 2\lambda/r where \lambda = q/l is the linear charge density. This formula holds for a cylindrically symmetrical charge distribution since we can consider such a configuration as the superposition of many cylindrical shells of charge density \lambda_i , i = 1, 2, 3,... E = (2\lambda_1/r) + (2\lambda_2/r) + (2\lambda_3/r) +... = 2/r (\lambda_1 + \lambda_2 + \lambda_3 +...) where \lambda' = \lambda_1 + \lambda_2 + \lambda_3 + ... = q_1/l + q_2/l + q_3/l + ... = (q total)/l. Now, we compute the field due to an infinite line charge. Figure (b) depicts an infinite line of charge of linear charge density \lambda, with a cylindrical Gaussian surface of radius r and height l, surrounding it. Again, we see as with the cylindrical shell, that the field must be radial since no other direction is preferred. Thus, the flux through the top and bottom of the Gaussian surface is zero. Ap-plying Gauss's law: \int E^\ding{217} \textbullet dA^\ding{217} = 2\pirl E = 4\pil\lambda = 4\piq E = 2\lambda/r. The same result is thus obtained.

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Question:

Which molecule of each of the following pairs would exhibit a higher degree of polarity.HClandHBr, H_2O and H_2S;BrCl and IF?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E17-0641.htm

Solution:

Polarity indicates that there is an uneven sharing of electrons between 2 atoms. This creates a charge distribution in the molecule where one atom is partially positive and the other is partially negative. The degree of polarity is measured by finding the difference in the abilities of the two atoms to attract electrons. This tendency to accept electrons is called theelectronegativity. The greater theelectronegativitydifference, the greater the degree of polarity. From the table ofelectronegativityvalues, thefollowing electronegativitiescan be obtained: Compounds Electronegativity Difference (1)HCl HBr 3.0 - 2.1 = .9 2.8 - 2.1 = .7 (2)H_2O H_2S 3.5 - 2.1 = 1.4 2.5 - 2.1 = .4 (3)BrCl IF 3.0 - 2.8 = .2 4.0 - 2.5 = 1.5 In pair (1) (HClandHBr),HClhas the largerelectronegativity. Hence,HClhas a greater degree of polarity thanHBr. For the same reasons, H_2O in pair (2) and IF in pair (3) have the greater degrees of polarity.

Question:

If an E. coli cell contains 10^6protein molecules, each of mol. wt. 40,000, calculate the total length of all the polypeptide chains of a single E. coli cell, assuming they are entirely \alpha-helical.

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Solution:

When a protein exists in the \alpha-helical configuration, each amino acid contributes 1.5 \AA to its length. To find the total length of all the polypeptide chains present in one E. coli cell, first calculate the number of amino acids present and then multiply this number by 1.5 \AA. If there are 10^6 protein molecules, there are (10^6 molecules) / (6.02 × 10^23 molecules/mole) or 1.66 × 10^-18 moles present. The weight of the proteins becomes 1.66 × 10^-18 moles × 40,000 g/mole or 6.64 × 10^-14 g. The average molecular weight of an amino acid is 120, therefore the average weight of one amino acid residue is (120g/mole) / (6.02×10^23residues/mole) or 1.99 × 10-22g/residue. One can solve for the number of residues present by divid-ing the total weight of the protein chains by the weight of an amino acid residue. no. of ammo acid residue = (6.64 × 10^-14 g) / (1.99 × 10^-22 g/residue) = 3.33 × 10^8 residues. It is now possible to solve for the length of the protein chain. length of chain = no. of residues × 1.5 \AA = 3.33 × 10^8 residues × 1.5 \AA/residue = 5.0 × 10^8 \AA = 5.0 × 10^8 \AA × 1 cm/10^8 \AA = 5.0 cm.

Question:

Describe the structure and function of the nuclear membrane.

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/Users/wenhuchen/Documents/Crawler/Biology/F02-0041.htm

Solution:

The nuclear membrane actually consists of two leaflets of membrane, one facing the nucleus and the other facing the cytoplasm. The structure of each of these two leaflets is fundamentally similar to the structureof the plasma membrane, with slight variations. However, the twoleaflets differ from each other in their lipid and protein compositions. The nuclear membrane is observed under the electron microscope to be continuousat some points with the membranes of the endoplasmic reticulum. Nuclear pores are the unique feature of the nuclear membrane. They are openings which occur at intervals along the nuclear membrane, andappear as roughly circular areas where the two membrane leaflets cometogether, fuse, and become perforated. Many thousands of pores maybe scattered across a nuclear surface. The nuclear pores provide a means for nuclear-cytoplasmic communication. Substances pass into and out of the nucleus via these openings. Evidence that the pores are the actual passageways for substancesthrough the nuclear membrane comes from electron micrographswhich show substances passing through them. The mechanismsby which molecules pass through the pores are not known. At present, we know that the pores are not simply holes in the membrane. This knowledge comes from the observation that in some cells, small moleculesand ions pass through the nuclear membrane at rates much lowerthan expected if the pores were holes through which diffusion occursfreely.

Question:

What is the major difference between themethodof swimming in a fish such as a tuna, and that of a dolphin?

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/Users/wenhuchen/Documents/Crawler/Biology/F19-0489.htm

Solution:

The major difference between these two organisms with regard to swimming is in the movement of the tail fin. In the tuna, the tail fin is oriented dorso-ventrally. Swimming is accomplished by the side to side movement of the fin by muscle layers in the side of the body. In the dolphin , the tail fin is oriented laterally. Swimming here is accomplished by the movement of the findorso-ventrally. In addition, the dolphin's hind limbs , or pelvic fins, which are generally used for the maintenance of balance while swimming in other fish, are only vestigial. The forelimbs, which have been modified into swimming paddles, take over the role of ba-lance . In the tuna, the pelvic fins are intact. In both organisms, steering is accomplished by contractions of muscles in the body wall.

Question:

Geologists on a plane are attempting to locate the exact position of the iron reserves in a region by measuring the variation of g in the area, since a large mass of iron will exert an appreciable additional gravitational attraction on a body in the vicinity. They hover above selected spots and observe the movement of a mass suspended from a light spring. If the system has a natural period of 2 s, and the smallest movement of the mass which can be detected is 10^-6 m, what is the minimum change in g which they can observe?

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Solution:

At a point at which the acceleration due to gravity is g, the mass, when in equilibrium, has two forces acting on it: the weight mg down and the restoring force F =kx(i.e. Hooke's law) up, and these must be equal. Thus g = (k/m)x. At another point, where local conditions vary, the stretching of the spring will be x +dxif the value of the acceleration due to gravity is g + dg (wheredxand dg are very small changes in x and in g, respectively). \therefore g +gd= (k/m)(x +dx) or dg = (k/m)(dx). But, if the mass-spring system is allowed to oscillate, its period is given by T = 1/f = (2\pi)/(2\pif) = (2\pi)/(\omega) = 2\pi\surd(m/k) \therefore k/m = (4\pi^2)/(T^2) and dg = [(4\pi^2)/(T^2)](dx). If the smallest value ofdxobservable is 10^-6 m, and we are given that T = 2 sec, the smallest value dg detectable is thus dg = [(4\pi^2)/(4s^2)] × 10^-6 m = 9.87 ×10^-6 m\bullets^-2.

Question:

How many electrons must be added to a spherical conductor (radius 10 cm) to produce a field of 2 × 10^\rule{1em}{1pt}3 N/couljust above the surface?

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/Users/wenhuchen/Documents/Crawler/Physics/D18-0592.htm

Solution:

The electric field E^\ding{217} at a point in space, due to a spherical conductor having a total charge Q is given by E = KQ / R^2, where K is a constant having a value of 9 × 10_9 (N \bulletm^2 / c^2 ) and R is the distance from the center of the sphere to the point at which we wish to calculate E^\ding{217}. The charge needed to produce a field of 2 x 10^\rule{1em}{1pt}3 N/coul. at the surface of the sphere (R = 10 cm = .1 m) is then Q = R^2E / K Q = [{(.1m)^2(2 ×10\rule{1em}{1pt}3N/C)} / {(9 × 10^9 N \bullet m^2) / (C^2)}] =2.22 × 10^\rule{1em}{1pt}15 C (The radius was converted to meters in order to make it compatible with the MKS system being used.) Since one electron has a charge of 1.6 × 10^\rule{1em}{1pt}19coul., n electrons will produce a charge of 2.22 × 10^\rule{1em}{1pt}15coul. Setting up the following proportion, (1electron / 1.6 × 10^\rule{1em}{1pt}19coul) = (n electrons / 2.22 × 10\rule{1em}{1pt}15coul) We may solve for n N = (2.22 × 10^\rule{1em}{1pt}15coul) / (1.6 × 10^\rule{1em}{1pt}19coul) = 1.39 × 10^4.

Question:

In the Bohr-atom model an electron of mass 9.11 × 10^-31 kg revolves in a circular orbit about the nucleus. It completes an orbit of radius 0.53 × 10^-10 m in 1.51 × 10^-16 sec. What is the angular momentum H of the electron in this orbit?

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Solution:

H\ding{217}= r\ding{217}× p\ding{217}where p\ding{217}is the linear momentum of the electron, and \ding{217} \ding{217} \ding{217} \ding{217} r\ding{217}is the vector from the proton to the electron. Because the orbit of the \ding{217} electron is circular, r\ding{217}is perpendicular to p\ding{217}. Hence \ding{217} \ding{217} \vertH\ding{217}\vert = H = rp = mrv \ding{217} But v =\cyrchar\cyromegar, where \cyrchar\cyromega is the angular velocity of the electron. The frequency of revolution is f = 1/T, where T is the period , or the time it takes for the election to make one revolution in its orbit. In T units of time, then, the electron goes through an angle of 2\pi radians. Hence \cyrchar\cyromega = 2\pi/T Therefore, H = mr^2\cyrchar\cyromega = (m)(2\pi/T)(r^2) = H = (9.11 × 10^-31 kg){(2\pi rad)/(1.51 × 10^-6 sec)} (0.53 × 10^-10m)^2 = 1.06 x 10^-34 kg-m^2 /sec.

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Question:

What is the pH of 0.500 M NaHSO_3? TheK_eqfor NaHSO_3 is 1.26 × 10^-2.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E12-0429.htm

Solution:

The pH is defined as - log [H^+], where [H^+] is the concentration of H^+ ions. NaHSO_3 ionizes as shown in the equation: NaHSO_3 \rightleftarrows Na^+ + HSO_3^- Thus for each NaHSO_3 that ionizes, one H^+ and one HSO_3^- are formed. Therefore, [H^+] = [HSO_3^-]. The equilibrium constant (K_eq) for NaHSO_3 is 1.26 × 10^-2. The equation forK_eqis K_eq = {[H^+] [SO_3-]} / [NaHSO_3] = 1.26 × 10-2 One knows that [NaHSO_3] = 0.500 and that for each H^+ and NaSO_3^- formed, one NaHSO_3 is ionized. Therefore, at equilibrium, the concentration of NaHSO_3 is .500 - [H^+]. One can now solve for [H^+], using the equation for the equilibrium constant. K_eq = {[H^+] [HSO_3^-]} / [NaHSO_3] = 1.26 × 10^-2 and because [H^+] = [SO_3-], you have K_eq = {[H^+] [H^+]} / {.500 - [H+]} = 1.26 × 10-^2 [H+]^2 = {.500 - [H+]} × 1.26 × 10^-2 [H+]^2 = (6.3 × 10^-3) - 1.26 × 10^-2 [H^+] Rewriting, [H^+]^2 + 1.26 × 10^-2 [H^+] - 6.3 × 10^-3 = 0. Using the quadratic formula: ax^2 +bx+ c = 0a = 1 b = 1.26 × 10^-2 c = - 6.3 × 10^-3 x = [-b \pm \surd(b^2 - 4ac)] / 2a x = (-1.26 × 10^-2 \pm \surd{(1.26 × 10^-2)^2 -[4×1× (-6.3 × 10^-3)]}) / {2 × 1} x = [-1.26 × 10^-2 \pm \surd{1.59 × 10^-4 + (2.52 × 10^-2)}] / 2 x = [(-1.26 × 10^-2) / 2] \pm [\surd(2.52 × 10^-2) / 2] x = -6.3 × 10^-3 \pm [(1.59 × 10^-1) / 2] x = -6.3 × 10^-3 \pm 7.97 ×10^-2 x = -6.3 × 10^-3 + 7.97 ×10^-2 = 7.34 ×10^-2or x = -6.3 × 10^-3 + 7.97 ×10^-2 = 8.60 ×10^-2 x must equal 7.34 × 10^-2 because [H^+] cannot be negative. One can now findpH. pH = - log [H^+][H^+] = 0.734 pH = - log 0.734 pH = 1.13.

Question:

A corporate executive who is slowly embezzling money from thecompany's pension fund has instructed his partner in crime, an investment banker, to send all correspondence in code. The computer operators on his staff type the corres-pondenceon cards, each card containing 60 characters. To decode the message, you must first divide up the stringinto 5 blocks containing 12 characters each, with every twelfthcharacter used to build a word. In addition, each letter isto be replaced with the letter six ahead of it in the alphabet. Thus, A becomes G, B becomes H, and U becomesA, V becomes B, and so on. Write a procedure inPL/Ito read the characters from cards, decode the characters of each card, and print the 12 outputwords, four per line. The program should terminate whena card contains all consonants (consider Y to be a con-sonant) .

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Solution:

To complete this program, we make use of the INDEX, VARYING andSUBSTR functions, three built-in routines useful for handling characters. We shall go through the program, explaining the implementationof each of these functions in the program. In the declaration section, we use the VARYING attribute to adjust thelength of the string. Based on the number of characters in the string, eachelement in the array having less than 5 characters will not be padded withblanks. IfWORD(12) were 'QOR', it would be stored just as is. Without the VARYING (VAR for short) attribute,WORD(12) would contain 'QOR '. After this we have a section which determines if the mes-sage is complete. Here we use the INDEX function. This is a powerful tool which searchesa character string to determine whether a specified character or combinationof characters is contained there. The five IF statements check tosee if any vowels are contained in the strings. If not found, a value of 0 isreturned to the program, and the next step is execut-ed. If it is found, INDEX returns a single integer indicating the number of the character wherethe match begins. This section causes the program to terminate if thecard contains all consonants. The statement in which characters are replaced deserves special attention. First, the double bar symbol stands for the concatenation operation. Its function is to connect two character strings to form a single, longerone. In this statement,WORD(J) is attached to a lengthy expression, be-ginning with the substring function SUBSTR. Let us digress for a moment to explain the SUBSTR func-tion. The generalform of this function is SUBSTR (a, n) where a is the name of a characterstring variable, and n is any numerical expression that produces apositive integer. It examines the string in a, reaches the nth character, andpulls from the string all characters, including the nth one, to the right. Say we have stored in WORD the characters 'WDPSO'. If we write NEWWORD =SUBSTR(WORD,3), the variable NEWWORD will contain thesubstring 'PSO'. We can even insert a third argument into the SUBSTR function, which has the same restric-tions as n, to indicate the lastcharacter we wish to remove. Now, the section of the statement following the \vert\vert symbol cam be explained. Character string ALP is the string which contains the master code. The INDEX function determines the number of the character to be moved, while the inner SUBSTR function finds the character in the string. The integer 6 is added to the number returned by the INDEX function to assigna letter six ahead of it in the alphabet. This is done to each characterin sequence. The output is given in tabular form with four words per row. DECODE:PROCOPTIONS(MAIN); DCL MESSAGECHAR(60), WORD(12) CHAR(5) VARYING, ALP CHAR (32) INIT ('ABCDEFGHIJKLMNOPQRSTUVWXYZABCDEF '); /\textasteriskcentered READ STRING TO DETERMINE IF MESSAGE IS COMPLETE \textasteriskcentered/ LOOP: GETDATA(MESSAGE); IFINDEX(MESSAGE,'A')=0 THEN IFINDEX(MESSAGE,'E')=0 THEN IFINDEX(MESSAGE,'I')=0 THEN IFINDEX(MESSAGE,'O')=0 THEN IF INDEX (MESSAGE,'U')=0 THEN GO TO FINISH; /\textasteriskcentered INITIALIZE ARRAY AND COUNTER \textasteriskcentered/ WORD=' ' ; K = 0; /\textasteriskcentered NEST WORD LOOP INSIDE LETTER LOOP \textasteriskcentered/ LETTERS: DO I = 1 TO 5; WORDS: DO J = 1 TO 12; /\textasteriskcentered NOW INCREMENT COUNTER, DETERMINE REPLACEMENT FOR LETTER, AND ADD REPLACEMENT TO WORD \textasteriskcentered/ K=K+1; WORD (J) = WORD (J) \vert\vert SUBSTR (ALP, INDEX (ALP, SUBSTR (MESSAGE,K, 1)) + 6,1) ; END WORDS; END LETTERS; /\textasteriskcentered PRINT DECODED WORDS \textasteriskcentered/ PRINTIT: DO I = 1 TO 12 BY 4; PUT SKIP DATA [WORD (I) ,WORD(I+1) ,WORD(I+2) ,WORD (I+3)] ; END PRINTIT; GO TO LOOP; FINISH:ENDDECODE;

Question:

The long solenoid In Fig. 1 Is wound with n = 1000 turns per meter, and the current in its windings is increased at the rate of 100 amp/sec. The cross-sectional area A of the solenoid is 4 cm^2 = 4 × 10^-4 m^2 . (a)Find the rate of change of magnetic flux inside the solenoid. (b)What is the induced electric field E at a distance d = 10 cm from the axis of the solenoid?

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Solution:

(a)The magnetic field B inside the solenoid, due to current I flowing in it, is given by B = \mu_0 n I. Then, the magnetic flux \varphi_B inside the coil becomes \varphi_B = BA = \mu_0 nA I and the rate of change of this flux is [(d\varphi_B )/(dt)] = A [(db)/(dt)] = \mu_0 nA [(dI)/(dt)] = (4\pi × 10^-4 (w/a-m) [1000 (turns/m)] (4 × 10^-2 m^2) [100 (amp/sec)] = 16\pi × 10^-7 w/sec (b)A long solenoid can be thought of as a long cylindrical current with the magnetic field outside becoming negligible in comparison to the field inside (at least far away from the edges). Therefore, the problem reduces to that of finding the electric field Induced by a time varying uniform magnetic field B while it is Increasing (see Fig. 2). For this, we shall employ Faraday's law around the circle with radius d: \oint E^\ding{217}.dl^\ding{217} = - (d\varphi_B /dt ) The direction and shape of the electric field can be obtained by remembering that, it will tend to set up a new flux to oppose the increase in magnetic flux through the coll. The electric field is circular and it is constant around the path we defined for the above integral. Therefore it can be taken outside the integral sign. E^\ding{217} \oint dl^\ding{217} = El = E 2\pid = - (d\varphi_B)/(dt) orE = - [1/(2\pid)] [(d\varphi_B)/dt)] = 8 × 10^-5 (w/m \bullet sec) = 8 × 10^-5 (v/m).

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Question:

Write a FORTRAN program which uses the modified Euler method to simulate the growth of an isolated species from time t = 0 to t =t_f, if the number of individuals at time t is N(t), N(0) = N_0 and the rate of population change (per unit of time), N, is given by: Ṅ = [K + E(t) - N/M)N](1) where K and M are positive constants and E(t) is a periodic forcing input resulting from seasonal cycles, given by: Case 1: E(t) is sinusoidal with amplitude = A, period =2\pi/\omega, \omega and phase angle \varphi . Case 2: E(t) is a pulse train of width w, height, h and period Pw, P, h, w > 0, and starting at time t = t_1 .

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Solution:

Equation (1) is a nonlinear differential equation, for which a numerical method is most appropriate. Since t appears explicitly, the EULER (not MEULER!) subroutine is used. Case 1: E(t) = A sin (\omegat + \varphi) . \omega Substituting into equation (1), Ṅ = (K + A sin (\omegat + \varphi) - N/M)N \omega The program uses P0P0 for N(0) = N_0; omega for\omega, PHI for \varphi : \omega REAL T/0.0/,M,K,N COMMON K,A,OMEGA.,PHI,M READ,N, TFIN,ACCUR, K,A, OMEGA, PHI,M, POPO CALL EULER(N,TFIN,T,POPO,ACCUR) STOP END FUNCTION F(T,W) COMMON K,A,OMEGA,PHI,M REAL M,K F = (K+A\textasteriskcenteredS IN (0MEGA\textasteriskcenteredT+PHI) - W/M)\textasteriskcenteredW RETURN END Case 2: E(t) is a pulse train. The derivation of the TRAIN sub-program to evaluate the pulse train function was given earlier. The program uses POPO for N_0 : REAL T/0.0/,M,K COMMON K,M,H,T1,W,P READ,N,TFIN,ACCUR,K,M,POPO,A,T1,W,P,H CALL EULER (N, TF1N, T, POPO,ACCUR) STOP END FUNCTION F(T,Z) COMMON K,M,H,T1,W,P REAL M,K F = (K+TRAIN(T) - Z/M)\textasteriskcenteredZ RETURN END FUNCTION TRAIN (T) COMMON K,M,H,T1,W,P TIME = T IF (T.GE. (T1 + P\textasteriskcenteredW)) TIME = AM0D((T-T1)/(P\textasteriskcenteredW))\textasteriskcenteredP\textasteriskcenteredW + T1 IF ((TIME.LT.T1).OR. (TIME.GE. (T1+W)))TRAIN = 0 IF ((TIME. GE.T1) .AND. (TIME.LT. (T1 + W)))TRAIN = H RETURN END

Question:

High-level languages specifically provide us with subroutines tolessen repetitious coding. How does a simple subroutine ofFORTRAN compare to a basic assembler language subroutine? What specifically does assembler languageuse forrepetitious procedures?

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Solution:

Although both are referred to as subroutines, the subroutine of the high-levellanguage and that of assembler show no noteworthy similarities. We will not be concerned with the functions of the two subroutines that are usedas examples, for this is not relevant to our discussion. Our attention willbe primarily focused on the manner in which these subroutines are called , the passing of parameters, and ending procedure. Let us consider the subroutine of FORTRAN This subroutine is initiated by a 'CALL' statement in the main program. This CALL statement has parameters that are passed to the subroutine. In general, it is only through the CALL statement that a subroutine can be accessed. In addition the subroutines may be said to be completely depen-dent on the operands passed to it by the main program. Now on inspecting the subroutine of the IBM/360-370 computers, we find something completely different. The sub-routine is not called but is branched to as if the computer was branching to any other part of the main program. There are no operands passed in this branching process and hence, the subroutine is not dependent on the main program. In fact, the subroutine of the 360-370 computer could quite easily be mistaken for just another part of the main program. The subroutine of the high-level language requires an 'END' statement whereas that of assembler does not. When the end of the high- level subroutine is found, control auto-matically returns to the instruction that comes immediately after the call of the subroutine. With the assembler subroutine, the computer has to be directed (branch) to this instruction . The high-level subroutine is generally placed outside of the main program but, the subroutine of the as-sembler language is almost like another part of the main program. The following example will clearly illustrate the difference between the two subroutines: Example 1A FORTRAN: CALLLARGER (DATA1,DATA2,MAXI) \textbullet \textbullet \textbullet C THIS SUBROUTINE FINDS THE LARGER OF TWO NUMBERS SUBROUTINE LARGER (FIRST, SECOND, RESULT) INTEGER FIRST, SECOND, RESULT IF (FIRST. GT. SECOND) RESULT = FIRST IF (SECOND. GT. FIRST) RESULT = SECOND RETURN END Example 1B (Basic Assembler Language Subroutine) Assembly: BAL11, LCNTRT 1 \textbullet \textbullet \textbullet \textasteriskcenteredLINECOUNTING AND PRINTING SUBROUTINE LCNTRTLCPLINECNT, = P '50' BLLCNTRT2 OUTPRTOUT, HDGLINE ZAPLINECNT, = P '0' MV1WORKAREA, C '0' LCNTRT2PUTPRTOUT, WORKAREA APLINECNT, = P '1' BR11 For FORTRAN: The CALL statement initiates the subroutine, and passes operands DATA1, DATA2 and MAXI to it. This data is accepted by FIRST, SECOND and RESULT respectively. Note, the data is 'accepted' in the order in which it is passed, so DATA1 is accepted by FIRST, DATA2 by SECOND and MAXI by RESULT. In-cidentally, 'LARGER' is the name of this routine , and its end is indicated by the END statement. At this point the computer goes back to the instruction following CALL LARGER (DATA1, DATA2, MAXI) statement. For Assembler: The BAL statement tells the computer to branch to the label LCNTRT1, but before doing this, it must store the ad-dress of the next instruction in register 11. Going to LCNTRT1, the assembler processes the instructions until it comes to the BR 11 instruction. Here, the computer branches to the address previously stored in register 11. This is the address of the instruction following the BAL 11, LCNTRT1 instruction. So, as you can see, these two subroutines have almost nothing in common . The subroutine of a high-level language is comparable to that of an assembler MACRO. The calling procedure and operand passing are quite similar, as the following example illustrates. Example 2 SUMWDS5,WORDA, WORDB \textbullet \textbullet \textbullet MACRO SUMWDS,ER1, EW1, EW2 SRER1, ER2 AER1, EW1 AER1, EW2 MEND In the main program the name of the macro (SUMWDS) calls the macro. The operands following the name are passed to the macro and accepted exactly like those illustrated in example 1A. The word MACRO indicates the start of the macro. The end is indicated by MEND. At this point control goes back, in the main program, to the instruction immedi-ately following the call (SUMWDS 5, WORDA,WORDB).

Question:

Write the micro-instructions for the execute cycles of the following register reference instructions. Instruction Symbol Hexadecimal Code Definition CIR 7080 Circulate right E and AC CIL 7040 Circulate left E and AC SPA 7010 Skip if AC is positive SNA 7008 Skip if AC is negative SZA 7004 Skip if AC is zero SZE 7002 Skip if E is zero

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Solution:

Each register-reference instruction is speci-fied when the expression q_7I ̅c_2t_3 [MBR]_i is true (logical 1) where the i_th bit of the MBR is the high bit specified by each instruction. All the instructions can be exe-cuted during one timing signal t_3 . CIR:The circulate right instruction is explained with the use of Figure 1. The content of each bit is shifted to the next higher bit, E is shifted to bit 0 of the AC, and bit 15 of the AC is lost. q_7I ̅c_2 [MBR] _8 t_3 :n\leftarrow15 AC_n \leftarrow [AC]_n-1;for 1 \leq n \leq 15 A_C0\leftarrow [E] E\leftarrow0 F\leftarrow0 CIL:This is just the opposite of the CIR instruction, with E getting the bit shifted off the end. q_7I ̅c_2t_3 [MBR] _9 :n\leftarrow0 AC_n\leftarrow [AC]_n+1;for 0 \leq n \leq 14 AC_15\leftarrow E E\leftarrowAC_0 F\leftarrow0 SPA:The PC is incremented only if the AC is positive. The AC is positive when bit 0 is 0 and the AC is negative when bit 0 is 1 (2's complement notation). Note that when the AC is zero it is also positive (zero is treated as a positive number). q_7I ̅c_2t_3 [MBR]_11 [AC]_0 :PC\leftarrow[PC]+1 F\leftarrow0 orq_7I ̅c_2t_3 [MBR]_11 [AC]_0 :F\leftarrow0 The SNA and SZA micro-instructions are written in a similar manner. SNA : q_7I ̅c_2t_3 [MBR]_12 [AC]_0 :PC\leftarrow[PC]+1 F\leftarrow0 orq_7I ̅c_2t_3 [MBR]_12 [AC]_0 :F\leftarrow0 SZA:The AC is zero when A_0A_1 . . . A_15 = 0. q_7I ̅c_2t_3 [MBR]_13 [A_0+A_1+ . . . +A_15] :PC\leftarrow[PC]+1 F\leftarrow0 orq_7I ̅c_2t_3 [MBR]_13 [A_0+A_1+ . . . +A_15] :F\leftarrow0 SZE:The PC is incremented if E = 0. q_7I ̅c_2t_3 [MBR]_14 E ̅ :PC\leftarrow[PC]+1 F\leftarrow0 orq_7I ̅c_2t_3 [MBR]_14 E :F\leftarrow0

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Question:

A man and a woman are heterozygous for tongue rolling, and have three sons, The three sons marry women who are not tongue rollers. Assuming that each of the three sons has a different genotype, show by diagram what proportion of their children might have the ability to roll their ton-gues.

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Solution:

Let us again use the letter R to represent the dominant tongue rolling allele, and r to represent the recessive allele. Only three possible genotypes could result from the cross of the heterozygous parents: These alleles are:RR, Rr and rr. Let one of the sons be RR and therefore a tongue roller, the second son Rr and also a tongue roller, and the third son rr (homozygous recessive) and so unable to roll his tongue. All three sons have married woman who are not tongue rollers and who are therefore rr. Let us now determine the proportion of the offspring who will be able to roll their tongues. F_1Rr100% Rr: tongue rollers 50% Rr: tongue rollers F_1Rr;rr50%rr: non-tongue rollers F_1rr100% rr: non-tongue rollers As a result of the three crosses, we can see that of the total offspring from the three marriages, half will be tongue rollers and half will be non-tongue rollers. The tongue rollers consist of all the children of the first son plus half the children of the second son. The non-tongue rolling children consist of all the offspring of the third son's marriage plus half the offspring from the second son's marriage.

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Question:

Using a specific case, show that the effect of a 10\textdegreeK rise in temperature will have a greater effect on the rate constant, k, at low temperatures than it does at high temperatures.

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Solution:

The best way to demonstrate this fact is to use the natural logarithm form of Arrhenius' equation, which gives an indication of the effect of temperature on reaction rate. The expression may be written lnk= lnA - E_a /2.303 RT, where k is the rate constant, A = the Arrhenius constant, E_a = activation energy, R = universalgas constant (8.314 J mol^-1 deg^-1) and T = temperature in Kelvin (Celsius plus 273). For a low temperature, consider room temperature (T = 300\textdegreeK). For a first-order reaction, A might be 1.0 × 10^14sec^-1 and E might be 80 k J/mole. Thus, at T = 300\textdegreeK, you have log k = log (1 × 10^14) - 80,000 / (2.303) (8.314) (300) = .07. Thus, k = 1.2 sec^-1. If you increase the temperature 10\textdegreeK to 310\textdegreeK, you find by exactly the same method of calculation, that k = 3.3 sec^-1. Thus, at a temperature of 10\textdegreeK higher (in a low temperature range), k was increased nearly three times. Now consider a higher temperature range, say T = 900\textdegreeK. For this, log k = (1 × 10^14) - 80,000/(2.303) (8.314)(900) = 9.358. Solving log k = 9.358 k = 2.28 × 10^9. Again, let us increase the temperature by 10\textdegreeK to 910\textdegreeK. Using the same type of calculations, you find that k becomes 2.56 × 10^9. The percent change = {(2.56 × 10^9 - 2.28 × 10^9)/ (2.28 × 10^9)} × 100= 12.28% . Thus, at the higher temperatures, the rate constant was increased only 12% as compared to 300% at the low tem-peratures .

Question:

A variable resistor in series with a 2-V cell and an ammeter is adjusted to give a full-scale deflection on the meter, which occurs for a current of 1 mA. What resistance placed in series in the circuit will re-duce the meter readings by 1/f? The meter is calibrated to measure resistance on this basis, but the emf of the cell drops by 5% and the variable resistor is readjusted so that the full-scale deflection again corresponds to the zero of the resistance scale. What percentage error is now given on a resistor which has a true resistance of 3800 \Omega?

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Solution:

The total resistance in the circuit when the meter is giving full-scale deflection is R = (\epsilon/I) = [(2 V)/(10^-3 A)] = 2000 \Omega, for we are given that full scale deflection on the ammeter occurs for a current of 1 mA = 1 × 10^-3 A. If an unknown resistance X is added to the circuit and produces a meter reading of (1/f) mA, then R + X = [(2 V)/{(1/f) × (10^-3 A)}] = 2000 f \Omega. \thereforeX = (2000f - 2000)\Omega = 2000(f - 1)\Omega.(1) The emf of the cell now drops to 95/100 (= 95%) of 2 V = 1.9 V. Alternatively since we are given that the emf of the cell drops by 5%, then the emf loss is 5/100 of 2 V = .1 V. The new emf is now 2V - .1 V = 1.9 V. For full-scale deflection the resistance in the circuit is now R' = [(1.9 V)/(10^-3 A)] = 1900 \Omega, and if a further resistance of 3800 \Omega is inserted in the circuit, the current is 1.9 V/(1900 + 3800)\Omega =1/3 mA. But from the meter calibration, when the current drops to one-third of its value (i.e. the meter read-ing is reduced by 1/f = 1/3, or f = 3) the inserted resistance should have a value given by (1) X = [2000 (3 - 1)]\Omega = 4000 \Omega. The error in the reading is thus (4000 - 3800)\Omega = 200 \Omega, and the percentage error is (200/3800) × 100% = 5.3% .

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Question:

Explain three ways in which a person's basal metabolic rate maybe determined.

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Solution:

A person's body is said to be at the basal metabolic rate when it is usingenergy at a rate just sufficient to maintain the life of its cells. A person'sbasal metabolic rate (BMR) usually approximates his metabolic rateimmediately after awakening, when he is not physically active and onlythe necessary metabolic processes to maintain life are occurring. The BMR varies with age, height, weight and sex, as well as with certain abnormalitiessuch as thyroid disease. The thyroid gland produces the hormonethyroxine, which regulates the rate of cellular metabolism. An abnormalBMR may indicate abnormalthyroxinelevels.BMR has been measuredby placing a person in a chamber and measuring the amount of heatgiven off (since heat is a by-product of metabolism), but such is an awkwardand difficult method. The BMR is more easily determined by indirectlymeasuring the amount of oxygen used in a given period of time. Since any oxygen used will serve as the final electron acceptor in respiration, the amount of oxygen used is proportional to the metabolic rate. Another common method entails measuring the amount ofthyroxine presentin a person's blood. As mentioned previously, a high BMR could indicatean excess ofthyroxinewhile a low BMR could indicate a deficiency. Eachthyroxinemolecule contains iodine and when released, is boundto a protein in the blood. This protein is the only protein in the blood whichcontains iodine. Thus, by measuring the amount of protein-bound iodinein a blood sample, the amount ofthyroxinemay be determined. This, in turn, gives an indication of the rate of metabolism. BMR may be increased through administration of thyroid pills, whichcontainthyroxineproduced by animals. The increased concentrationof this hormone in the blood increases the metabolic rate. The underlying reason for this is thatthyroxineacts to uncouple oxidative phosphor-ylation; i.e. it dissociates the formation of ATP from the electron transportchain in the cell. Much of the energy released during the oxidationof NADH will be lost as heat rather than trapped in the form of ATP.The ATP/ADP ratiodeclines,and the relatively heightened level of ADP acts as a positive or stimulatoryallostericmodulator for the TCA cycle(through its action on an enzyme -isocitratedehydro-genase) . Thus theTCA cycle will function actively producing a continuous supply of NADH, the oxidation of which will fail to be coupled to ATP formation. Oxygen, the final electron acceptor, will be consumed rapidly, and a higherBMR measure-ment will result. To decrease the BMR,thyroxine-suppressing drugs may be administered, or part of the thyroid gland may be surgically removed. Previously, radioactive iodine was injected in order to kill the cells of the thyroidgland. However, this treatment has recently been found to be extremelycarcinogenic (cancer-causing).

Question:

Write a subprogram in FORTRAN to add two numbers in any baseK. Use the arrays L(I) and M(I) for I = 2,3,4...30 to store thedigits of the numbers, saving L(l) and M(l) for the sign. Also letN(I) hold the answer, with N(l) for the sign.

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Solution:

First you must decide which way the digits are to be added. In other words, you could enter the sum of 4379 + 2512 as either: L(2)L(3)L(4)L(5) 4379 +M(2)M(3)M(4)M(5) 2512 oras L(30)L(29)L(28)L(27) 9734 +M(30)M(29)M(28)M(27) 2152 We will use the latter example, because it leaves us room for an over-flowon the final digit. Hence, we need a decrementing loop. Also noticethat the maximum number of digits we can have is 28 because of thisoverflow problem. SUBROUTINEADD(L,M,N,K) INTEGER L( L( 3\O), M(3\O), N(3\O), K 3\O), M(3\O), N(3\O), K NN = 3\O NN = 3\O DO 2\O J = 2, NN DO 2\O J = 2, NN I = NN - J + 1 I = NN - J + 1 15N(I) = L(I) + M(I) N( I) = L(I) + M(I) IF (N(I). LT.K) GO TO 20 N(I) = N(I) - K N(I - 1) = L(I - 1) + M(I - 1) + 1 I = I - 1 GO TO 15 2\OCONTINUE

Question:

Write a computer program in Basic that plays the computer game of LIFE.

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Solution:

The game of life can be played either manually on a checkerboard square or on a CRT terminal. The game may be interpreted as a dynamic example of the laws of genetics. The rules of the game are as follows: 1) The game is played on a chessboard (imagined infinite) with counters. An initial distribution of counters is as-sumed. The pattern of the initial distribution changes ac-cording to the number of survivals, deaths and births. 2) Survivals: Every counter with two or more neighboring counters survives for the next generation. Neighbors are defined by the diagram 7 6 5 8 C 4 1 2 3 We see that counter C has eight potential cells on which neighbors may sprout. 3) Deaths: Each counter with four or more neighbors dies (is removed) because of over-population. Every counter with one or no neighbors dies from isolation Fig. 2 In Fig. 2 the circled counters die from isolation and are removed. The squared counter dies from lack of resources. 4) Births: Each empty cell adjacent to three and only three neighbors is a birth cell. A counter is placed on it on the next move. Fig. 3 In Fig. 3 the circled cell will have a counter placed on it in the next generation (move). Finally, note that births, deaths and survivals consti-tute a move. The game continues until a stable population is reached (no more births or deaths) or until the popula-tion dies out. An oscillating population is also possible. As an example, consider the following part game: InitialXX InitialXX Distribution :XXX X X Generation 1:XX XXXX XX X Generation 2:XX XXX XX XX Generation 3:XX XXX X XX Generation 4:XXX XXX X The program follows: 10PRINT "LIFE" PRINT "LIFE" 20PRINT "ENTER YOUR PATTERN. TYPE `DONE' WHEN FINISHED. " 30X1 = 1: Y1 = 1: X2 = 24: Y2 = 70 40DIM A (24, 70), B$ (24) 50C = 1 60INPUT B$ (C) 70IF B$ (C) = "DONE" THEN B$ (C) = " " : GO TO 120 80IF LEFT $ (B$ (C), 1) = " . " THEN B$ (C) = " " + RIGHT $ (B$ (C), LEN (B$ (C) ) - 1) 100C = C + 1 110GOTO 60 120C = C - 1: L = 0 130FOR X = 1 TO C - 1 140IF LEN (B$ (X) ) > L THEN L = LEN (B$ (X) ) 145REM SEE END OF PROGRAM 150NEXT X 160X1 = 11 - C/2 170Y1 = 33 - L/2 180FOR X = 1 TO C 190FOR Y = 1 TO LEN (B$ (X) ) 200IF MID$ (B$ (X), Y 1) <> " " THEN A (X1 + X, Y1 + Y) = 1: P = P + 1 220NEXT Y 230NEXT X 240PRINT: PRINT: PRINT 250PRINT "GENERATION: " ; G, "POPULATION: " ; P ; 260IF I9 THEN PRINT "INVALID"; 270X3 = 24: Y3 = 70: X4 = 1: Y4 = 1: P = 0 280G = G + 1 290FOR X = 1 TO X1 - 1: PRINT: NEXT X 300FOR X = X1 TO X2 310FOR Y = Y1 TO Y2 320IF A (X,Y) = 2 THEN A (X,Y) = 0: GOTO 400 330IF A (X,Y) = 3 THEN A (X,Y) = 1: GOTO 350 340IF A (X,Y) <> 1 THEN 400 350PRINT TAB (Y); " \textasteriskcentered " ; 360IF X < X3 THEN X3 = X 370IF X > X4 THEN X4 = X 380IF Y < Y3 THEN Y3 = Y 390IF Y > Y4 THEN Y4 = Y 400NEXT Y 410NEXT X 420FOR X = X2 + 1 TO 24: PRINT: NEXT X 430X1 = X3: X2 = X4: Y1 = Y3: Y2 = Y4 440IF X1 < 3 THEN X1 = 3: 19 = - 1 450IFX2 > 22 THEN X2 = 22: I9 = - 1 460IF Y1 < 3 THEN Y1 = 3: I9 = - 1 470IF Y2 < 68 THEN Y2 = 68: I9 = - 1 480P = 0 490FOR X = X1 - 1 TO X2 + 1 500FOR Y = Y1 - 1 TO Y2 + 1 510C = 0 520FOR I = X - 1 TO X + 1 530FOR J = Y - 1 TO Y + 1 540IF A (I,J) = 1 OR A (I,J) = 2 THEN C = C + 1 550NEXT J 560NEXT I 570IF A (X,Y) = 0 THEN 610 580IF C < 3 OR C > 4 THEN A (X,Y) = 2: GOTO 600 590P = P + 1 600GOTO 620 610IF C = 3 THEN A (X,Y) = 3: P = P + 1 620NEXT Y 630NEXT X 640X1 = X1 - 1: Y1 = Y1 - 1: X2 = X2 + 1: Y2 = Y2 + 1 650GOTO 250 660END If your computer does not accept conditional execution, change it into few simple statements with the same meaning. For example: 140 IF LEN (B$ (X)) > L THEN L = LEN (B$ (X)) can be changed into: 140IF LEN (B$ (X)) > L THEN 142 141GOTO 150 142L = LEN (B$ (X) )

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Question:

Explain briefly the background and applications of the PL/I programminglanguage.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G13-0312.htm

Solution:

PL/I (Programming Language One) was developed by the International Business Machines (IBM) corporation for its System/360 seriesof computers. PL/I is a multipurpose pro-gramming language-it is designedto be used by both scientific and commercial programmers and togive these programmers lan-guage features powerful enough to handle alltheir problems. PL/I is constructed so that the individual programmer canmove and operate easily at his own level of experience, whether this levelis low or high. The language has many automatic and default conditionsto assist the beginning programmer, while the specialist with extensiveexperience may detail each step of his complicated program. Being a high-level language, PL/Iletsthe user describe complicatedprocedures in simple, easy to understand terms. This virtually freesthe programmer from having to learn the machine organization. Although the notation of PL/I is not identical to con-ventional algebraicnotation, it is close enough to be easily read by anyone knowing algebra. All these features make PL/I one of the most powerful and flexible languages in Computer Science today.

Question:

Compare the function of the swimmerets in the male and female crayfish in sexual reproduction.

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/Users/wenhuchen/Documents/Crawler/Biology/F12-0313.htm

Solution:

Swimmerets are the abdominal appendages of the crayfish. In many crustaceans these function as swimming legs. (The thoracic appendages generally function as feeding appendages, pincers, and walking legs.) Cray-fish generally do not swim by means of the swimmerets. The swimmerets of both male and female crayfish aremodi fied for reproduction and are calledpleopods. The male reproductive system consists of paired con-nected testes in the eighth thoracic segment. The testes lead to a sperm duct. At its terminal end, the sperm duct becomes a muscular ejaculatory duct, which opens to the outside on the ventral surface of the male on the base of the leg of the last thoracic segment. The first two pairs ofpleopodsare modified to aid in transfer of sperm to the female. The ovaries are located in the sixth thoracic segment. The oviduct opens to the outside on the ventral base of the last thoracic appendage. Most female crayfish possess a median seminal receptacle. This is a pouch that opens to the outside on the ventral surface of the last, or second to last, thoracic segment. In copulation, the tips of the malepleopodsare inserted into the seminal receptacle of the female cray-fish. Sperm flow along grooves in the pleopods and are deposited in the seminal receptacle. The sperm may be stored in the pouch until the female lays eggs. Fertilization occurs when eggs are laid. The female cray-fish lies on its back and curls the abdomen forward. By beating thepleopods, a water current is created, and this drives the eggs into a chamber created by the curved ventral abdominal surface. The eggs remain attached to the abdomen while they are brooded. A cementing material is associated with the egg membrane, and thepleopodsare modified to hold the eggs in place. Male and female swimmerets are considerably differ-ent in decapods, making it easy to distinguish the two sexes. Almost all female decapods brood the eggs and havepleopodswhich are modified to hold the eggs. Female crayfish never use thepleopodsfor swimming.

Question:

Write a FORTRAN program to compute the coefficient of rank correlation for the following data: X Y 1. .3 10 2. .6 15 3. .9 30 4. 1.2 35 5. 1.5 25 6. 1.8 30 7. 2.1 50 8. 2.4 45

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G19-0487.htm

Solution:

The correlation coefficient r is an appropriate measure of association between two variables when cardinal ordering is in-appropriate. Sometimes scores can only be ranked in order without any information concerning the distance between two elements in the ranking. For example, two judges in a beauty contest may give dif-ferent rankings. The correlation between them will depend on ordinal properties (first, second, third, etc.), not on cardinal properties (1, 1.5, 2.3, etc.). To find the coefficient of rank correlation between X and Y, first order Y's from smallest to largest, provided that X's are initially given in increasing order. X Rank of X Y Rank of Y .3 X_1 = 1 10 Y_1 = 1 .6 X_2 = 2 15 Y_2 = 2 .9 X_3 = 3 20 y_3 = 3 1.2 X_4 = 4 35 Y_4 = 6 1.5 X_5 = 5 25 Y_5 = 4 1.8 X_6 = 6 30 Y_6 = 5 2.1 X_7 = 7 50 Y_7 = 8 2.4 X_8 = 8 45 Y_8 = 7 The correlation coefficient r is equal to r = [\sum(x-x) (y-y)] / [(\sum(x-x)^2)^1/2 (\sum(y-y)^2)^1/2] This formula can be modified into a rank correlation coefficient. Omitting the details of the transformation: r_s= 1 - {[6^n\sum_i_=1 d^2] / [n(n^2-1)]} where = d^2 = (x_i-y_i)^2, the difference in ranks between thei-thx score and i-thy score squared. The 6 in the numerator arises from the sum ^n\sum_i_=1 i^2 = [n(n+1)(2n+1)] / 6 which is computed during the transformation. Observe that if the rankings of x_i and y_i are identical, r_s = 1 - {[6 (0)] / [n(n^2-1)]} = 1. For the given data x_i-y_i (x_i-y_i)^2 0 0 0 0 0 0 -2 4 1 1 +1 1 -1 1 1 1 \sum(x_i-y_i)^2 = 8n = 8 and r_s= 1 - {[6 (8)] / [8(63)]} = .873 The first task of the FORTRAN program for computing the rank cor-relation coefficient is the ordering of the giveny'sin ascending order. This can be accomplished using the subroutine "SORT", developed in the chapter named "DATA STRUCTURES". Assuming this is already done, the program looks as follows: INTEGER YFIN(N), XFIN(N) DIMENSION Y(N),YFIN(N), XFIN(N), M(N) READ, N READ, Y(I) CALL SORT (Y(J)) I = 1 J = 1 10IF(Y(J) - Y(I)) 2,3,2 2K = 0 6IF(Y(J) - Y(I+1). EQ.O) GO TO 5 K = K + 1 I= I + 1 GO TO 6 5YFIN(J) = I I = I - K GO TO 7 3YFIN(J) = I 7I = I + 1 J = J + 1 IF(J.EQ.N+1) GO TO 11 GO TO 10 11M(N) = 0 DO 12 L = 1,N M(L) = (XFIN(L) - YFIN (L)) \textasteriskcentered\textasteriskcentered2 M(N) = M(N) + M(L) 12CONTINUE RCC = 1 - 6\textasteriskcenteredM(N) / (N\textasteriskcentered(N\textasteriskcentered\textasteriskcentered2 - 1)) PRINT, RCC STOP END

Question:

An observation balloon has a volume of 300 m^3 and is filled with hydrogen of density 0.1 g/liter. The basket and passengers have a total mass of 350 kg. Find the initial acceleration when the balloon is released, assum-ing that the air resistance is zero when the velocity is zero. The density of air is 1.3 g/liter and the upward force on the balloon is equal to the weight of air dis-placed by the balloon.

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Solution:

The weight of air that the balloon displaces equals the upward force U on it, as a consequence ofArchimides' principle. This law states that a body, wholly or partly immersed in a fluid (either a liquid or a gas), is buoyed up with a force equal to the weight of the fluid displaced by the body. This force is then u =w_air=\rho_airg v where\rho_airis the density of the air, g is the acceleration due to gravity, and V is the volume of the balloon. We have \rho_air = (1.3 g/liter) (10^3 liter/m^3) (10^-3 kg/g) = 1.3 kg/m^3 Then U = (1.3 kg/m^3) (9.8 m/sec^2) (300 m^3) = 3822 N The mass that must be moved consists of the basket and passengers (350 kg) and that of the balloon. The massm_bof the balloon is m_b =\rho_hv where\rho_his the density of the hydrogen gas \rho_h= (0.1 g/liter) (10^3 liter/m^3) (10^-3 kg/g) = 0.1 kg/m^3 Thereforem_b= (0.1 kg/m^3) (300 m^3) = 30 kg and the total mass m is m = 350 kg +m_b= 350 kg + 30 kg = 380 kg. Due to the gravitational force acting on this mass, there is a force W acting downward and equal to the weight of the total mass. W = mg = (380 kg)(9.8 m/sec) = 3724 N From Newton's second law, the sum of the forces acting on a body equals the product of its mass m and its acceleration. \sumF = ma. The total initial force is equal to the sum of the weight of the balloon and the upward buoyant force. Taking the upward direction as positive, U - W = ma 3822 N - 3724 N = (380 kg) a The initial acceleration is a = (98 N)/(380 kg) = 0.258 m/sec^2. a = (98 N)/(380 kg) = 0.258 m/sec^2.

Question:

One of the holes on a golf course runs due west. When playing on it recently, a golfer sliced his tee shot badly and landed in thick rough 120 yd. WNW of the tee. The ball was in such a bad lie that he was forced to blast it SSW onto the fairway, where it came to rest 75 yd. from him. A chip shot onto the green, which carried 64 yd., took the ball to a point 6 ft. past the hole on a direct line from hole to tee. He sank the putt. What is the length of this hole? (Assume the golf course to be flat.)

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Solution:

Since the course is flat, all displacements are in the one horizontal plane. Since we know that the hole is due west of the tee, we only need to calculate its easterly component which is the sum of the easterly components of the ball's displacements (see figure). We take east to be the positive abscissa of the axes shown, and the direction angles \textphi of all displacement vectors will be measured counter-clockwise from the positive east-axis. Since we know that WNW means 22(1/2)\textdegree west of north-west, or \textphi = 157(1/2)\textdegree, and that SSW means 22(1/2)\textdegree south of southwest, or \textphi = 247(1/2)\textdegree; \sum easterly components = 120 cos 157(1/2)\textdegree yd + 75 cos 247(1/2)\textdegree yd + 64cos \texttheta yd + 2 yd = - 110.9 yd - 28.7 yd + 64 cos \texttheta yd + 2 yd = - 137.6 yd + 64 cos \texttheta yd We can solve for \texttheta by noting that the sum of the northerly components of displacement must equal zero: \sum northerly components = 120 sin 157(1/2)\textdegree yd + 75 sin 247(1/2)\textdegree yd + 64 sin \texttheta yd = 0 Sin \texttheta = {(- 120 sin 157(1/2)\textdegree - 75 sin 247(1/2)\textdegree}/64 = {(- 45.9 + 69.3)/64} = (23.4/64) Thus:\texttheta = 158.6\textdegree cos \texttheta = - 0.93125 Finally, inserting this into the equation for the sum of the easterly components: \sum easterly components = - 137.6 yd + 64(- 0.93125)yd = - 137.6 yd - 59.6 yd = - 197.2 yd Thus, the hole is 197.2 yd due west of the tee.

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Question:

Two glass plates are nearly in contact and make a small angle \texttheta with each other. Show that the fringes produced by interference in the air film have a spacing equal to (\lambda / 2\texttheta) if the light is incident normally and has wavelength \lambda.

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/Users/wenhuchen/Documents/Crawler/Physics/D37-1084.htm

Solution:

Let \texttheta be the angle between the two glass plates and let x_1 and x_2 be the distances from the edge of the wedge of two consecutive fringes, as shown in the figure. If d_1 and d_2 denote the air gaps for these two fringes, then 2d_1 = n\lambda(1) and 2d_2 = (n+1)\lambda(2) where \lambda is the wavelength and n is an integer. Subtracting (1) from (2) 2(d_2 - d_1) = \lambda.(3) From the figure, tan\texttheta = (d_1 / x_1) = (d_2 / x_2) /. Assuming \texttheta to be very small, the approximation tan\texttheta \approx \texttheta can be made. Therefore. d_2 - d_1 = \texttheta(x_2 - x_1) = \textthetax(4) where x = (x_2 - x_1) Substituting for d_2 - d_1 into equation (3), 2\textthetax = \lambda Then, x = (\lambda / 2\texttheta).

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Question:

Suppose genes A and B are 14.5 map units apart. Another gene, C, linked to these, is found to cross with gene B, 7 percent of the time. Is this data sufficient to determine the exact order of the three genes? If not, what other information is needed to order the genes?

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/Users/wenhuchen/Documents/Crawler/Biology/F25-0676.htm

Solution:

A good way to order genes is to do it visually. We are told that genes A and B are separated by a map distance of 14.5 units. Translating this into a genetic map, we get: We also know that C recombines with B, 7 percent of the time, which means that the map distance between B and C is 7 map units. On the genetic map, this fact becomes: Combining the three genes on a genetic map, we can have: Or we can have: It is important to realize that the genes are probably not ordered CAB because the distance from B to C is less than the distance from B to A. Thus both I and II are equally probable if we depend only on the data given. To decide which one of the two is correct, we would have to know the map distance between A and C. If this distance is greater than 14.5 map units, (the distance between A and B) then we know that I is correct. If the distance between A and C is less than 14.5 map units (i.e., 7.5 m.u.), then we know that II is correct. In this way, the exact sequence of the three genes can be determined.

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Question:

When 10.00 g of phosphorus was reacted with oxygen, it produced 17.77 g of a phosphorus oxide. This phosphorus oxide was found to have a molecular weight of approximately 220 in the vapor state. Determine its molecular formula.

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Solution:

The molecular formula of a substance indicates the relative number of atoms in a molecule of the substance. Therefore, to solve this problem, you must first calculate the ratios of the gram-atoms to each other, the empirical formula, and then extrapolate to the molecular formula via the molecular weight. The number of gram-atoms of phosphorus (P) is (wt. in grams of P) / (atom weight) = (10) / (30.97) = 0.323 gram - atoms P. For oxygen we have (7.77g) / (16.00) = 0.484 gram-atoms The weight in grams of oxygen is 7.77 because the final product weighs 17.77 g and the phosphorus weighs. 10.00 g. Since, the only other element is oxygen, its weight must be the difference. The ratio of the gram-atoms of P and O is respect-ively 1 : 1.5 or 2 : 3. Therefore, the empirical formula of the oxide is P_2O_3. To calculate the molecular formula, we must use the stated molecular weight of 220. We must look for a formula that totals to this molecular weightANDmaintains the 2 : 3 ratio of P : O as expressed in the empirical formula. With some arithmetic, we find that the only formula that meets these two requirements is P_4O_6 . 4 : 6 is the same as 2 : 3. The atomic weight of P and O is respectively 30.97 and 16. We have four P atoms for a total of 123.88 and we have 6 O atoms for a total of 96. Now add: 123.88 + 96 = 219.88, which is approximately 220. Another method for determining the molecular formula is to divide the molecular weight of the molecule, 220, by the weight of 1 P_2O_3, 110. no. of P_2O_3 = (220g) / (110g/mole of P_2O_3) = 2 mole of P_2O_3. The formula is, therefore 2 × P_2O_3 or P_4O_6.

Question:

As shown in the figure, a block of mass .5 slugs moves on a level frictionless surface, connected by a light flexible cord passing over a small frictionless pulley to a second hanging block of mass .25 slugs. What is the acceleration of the system, and what is the tension in the cord connec-ting the two blocks?

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Solution:

In order to find the system's acceleration, we must relate the net force on the system to the acceleration via Newton's Second Law. First we isolate the rope, and calculate its acceleration. By the second law, T_1 - T_2 = m_ropea where T_1 and T-_2 are in opposite directions. In this problem, we assume m_rope = 0, and T_1 = T_2 Hence, the rope acts only to transmit the force of tension to the block. Applying Newton's Second Law to the horizontal (x) and vertical (y) directions of motion of the block on the table, we obtain T = m_1a_(x)1 N - w_1 = a(y)1 where m_1 is the mass of the block on the table, and a_(x)1 is its horizontal acceleration. Noting that a_(y)1 = 0, since the block doesn't accelerate vertically, we find T = m_1a_(x)1(1) N = w1(2) We next apply the third law to the hanging block of mass m_2, and m_2g - T = m_2a_(y)2(3) where a_(y)2is the vertical acceleration of block 2. Now, since the 2-block system moves as a unit, a_(x)1 = a_(y)2 = a, and, using (1), (2) and (3) T = m_1a(4) N = w1(5) m_2g - T = m_2a(6) Substituting (4) in (6), and solving for a m_2g - m_1a= m_2a a= m_2-g/(m_1 + m_2) From (4),T= m_1m_2g/(m_1 +m_2) Substituting the given data in these equations a = [{(.25 sl)(32 f/s^2)} / {.75 sl}] = 10.7 f/s^2 T = (m_1m_2g) / (m_1 + m-_2) = m_1a = (.50 sl)(10.7 f/s^2) T = 5.4 lb

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Question:

Discuss the mechanism by which the photoreceptors are stimulated by light. How are rods and cones distributedin the retina?

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Solution:

Photoreceptors are sensory cells that are sensitive to light. In the human retina, they are called rods and cones according to their shapes. Both types of cells contain light-sensitive molecules called visual pigments , whose primary function is to absorb light. The rods contain rhodopsin (visual purple) which is composed of achromophore(a variant of vitamin A) and a protein (opsin) . The cones containiodopsinwhich is made up of the samechromophoreas inrhodopsinbut with a different protein . Light from the outside enters the eye and stimu-lates the rods or cones , thus triggering the emission of nerve impulses by the receptor cells . Light does not directly provide the necessary energy to set off the impulse . The energy comes from the chemical bonds in therhodopsinor iodopsin molecule. In this respect, the phenomenon of vision is basically different from the phenomenon of photosynthesis, in which light supplies the energy to drive the series of chemical reactions. When light strikes the visual pigments, it acts upon the chromophore , which then splits away from theopsin. This splitting occurs because light changes the molecular configuration of thechromophorein such a way that it no longer can bind to theopsin.Simultaneosly, impulses are triggered and these travel to the brain (via the optic nerve) where they are interpreted. Rhodopsinis sensitive to a very small amount of light. Rod cells, which contain this kind of visual pigment, are used to detect objects in poor il-lumination such as in night vision. They are not responsible for color vision , but are important in the perception of shades of gray, and brightness . Their acuity - ability to distinguish one point in space from another nearby point - is very poor. Rods are most numerous in the peripheral retina, that is, that part closest to the lens, and are absent from the very center of the retina (the fovea). On the other hand, cones operate only at high levels of illumination and are used for day vision. The primary function of the cones is to perceive colors. Their visual acuity is very high, and because they are concentrated in the center of the retina, it is that part which we use for fine, detailed vision.

Question:

A standard 48-candle lamp is placed 36 in. from the screen of a photometer and produces there the same illuminance as a lamp of unknown intensity placed 45 in. away. What is the luminous intensity of the unknown lamp?

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Solution:

Illuminance E is equal to the ratio of the luminous intensity I to the square of the distance R from the source. Since the illuminance of the two lamps is equal, E_1 = E2 where the subscripts 1 and 2 refer to the lamps at dist-ances of 45 in. and 36 in. respectively. Then I_1 / R_1 = I_2 / R2 and substituting values I_1/(45 in)^2 = (48 candles) / (36 in)2 I_1 = {(45 in)2(48 candles)} / (36 in)2= 75 candles Note that the distances can be expressed in any unit as long as they are consistent .

Question:

In the human body, the enzymephosphoglucomutase catalyzes the conversion of glucose-1-phosphate into glucose-6- phosphate: glucose-1-phosphate\rightleftarrowsglucose-6-phosphate. At 38\textdegreeC, the equilibrium constant, k, for this reaction is approximately 20. Calculate the free energy change, ∆G\textdegree, for the equilibrium conversion. Calculate the free energy change ∆G for thenonequilibriumsituation in which [glucose-1- phosphate] = 0.001 M and [glucose-6-phosphate] = 0.050 M.

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Solution:

Thenonequilibriumfree energy change ∆G is related to the standard free energy change ∆G\textdegree by ∆G = ∆G\textdegree + 2.303 RT log k, where R is the gas constant and T the absolute temperature. At equilibrium, ∆G = 0, hence ∆G\textdegree = - 2.303 RT log k. Thus, for the equilibrium conversion with k = 20 and T = 38\textdegreeC = 311\textdegreeK. ∆G\textdegree = - 2.303 RT log k = - 2.303 × 1.987 cal/mole-deg × 311\textdegreeK × log 20 = - 1850 cal/mole. The equilibrium constant for the conversion glucose-1-phosphate \rightleftarrowsglucose-6-phosphate isk = [glucose-6-phosphate]/[glucose-1-phosphate] In the case where [glucose-6-phosphate] = 0.050 M and [glucose-1-phosphate] = 0.001 M k_n = [glucose-6-phosphate]/[glucose-1-phosphate] = [(0.050 M)/(0.001 M)] = 50, where the subscript "n" has been added to distinguish this ratio from the equilibrium constant. Then ∆G = ∆G\textdegree + 2.303 RT logk_n = - 1850 cal/mole + (2.303) (1.987 cal/mole-deg) (311\textdegreeK) (log 50) = - 1850 cal/mole + 2420 cal/mole = + 570 cal/mole.

Question:

Explain the general meaning of assembler and its basic Explain the general meaning of assembler and its basic func-tions.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G09-0214.htm

Solution:

An assembler is a program that accepts a symbolic language and produces its binary machine language equiva-lent. The input symbolic program is called the source pro-gram and the resulting binary program is called the object program. The assembler is a program that operates on char-acter strings and produces an equivalent binary interpre-tation. An assembler must know the arrangment of the char-acters in memory so it can distinguish between symbols and lines. The symbolic program constitutes the input data for the assembler. Now we shall see how a two pass assembler translates the symbolic program of the previous problem. First Pass: A two pass assembler scans the entire sym-bolic program twice. During the first pass, it generates a table that correlates all user defined address symbols with their binary equivalent value. The binary translation of the program is done during the second pass. In order to keep track of the location of instructions, the assembler uses a memory word called location counter (LC). The ORG pseudo instruction initializes the LC to the value of the first location; in case ORG is missing LC is set to 0 initially. The tasks performed by the assembler during the first pass are described in the flow chart of Fig. 1. LC is initially set to 0. A line of symbolic code is scanned to see if it has a label (by presence of comma).If the line of code has no label, the assembler checks in the instruction field. If it contains ORG, it sets LC to the number followed by ORG. If it is END pseudo instruc-tion, the assembler terminates the first pass and goes to the second pass: If the line of code has a label, it is stored in the address symbol table together with its binary equivalent number specified by LC. Label LC MIN 100 SUB 107 DIFF 108 Fig. 2 On the first pass, the assembler encounters ORG in the example program (previous problem) and sets the LC to 100. It then scans the next line. If it has neither label nor END, LC is incremented by 1 to 101 & proceeds until HLT, when LC has value of 105. Now LC is incremented by 1 to 106. Upon scanning, it finds label MIN, then SUB and DIFF. The address symbol table is shown in Fig. 2. On encounter-ing the last instruction, END, the assembler goes to sec-ond pass. Second Pass: - Machine instructions are translated during the second pass by means of table-lookup procedures. A table lookup procedure is a search of table entries to determine whether a specific item matches one of the items stored in the table. The assembler uses four tables. 1) Pseudo-instruction table. 2) MRI table. 3) Non-MRI table. 4) Address-symbol table. The way the assembler proceeds in the second pass is described in the flow chart of Fig. 3. LC is initially set to 0. Labels are neglected during the second pass, so the assembler goes immediately to the instruction field. It first checks the pseudo-instruction table. For a match with ORG, the assembler sets LC to num-bers followed by ORG. A match with END terminates the translation process. An operand pseudo instruction causes a conversion of the operand into binary. This operand(e.g. DEC 50 from the previous problem) converted into its binary value is placed in the memory location specified by con-tents of LC i.e. at location 106. The LC is incremented by 1 and the assembler continues to analyze the next line of code. If the symbol encountered is not a pseudo-instruction, the assembler refers to the MRI table. If the symbol is not found in this table, the assembler refers to the non- MRI table e.g. in a scan of the second line, the symbol is CMA with LC = 101. The assembler stores the bit code of CMA into a memory location specified by LC i.e. at place 101. Again LC is incremented by 1 and a new line is analyzed. When a symbol is found in the MRI table, the assembler ex-tracts its equivalent code and inserts it in, say, bits 2-4 of the memory location specified by LC. The correspond-ing second symbol is an address. This is converted to bi-nary by searching the address symbol table. The first bit of instruction is set to 0 or 1, depending on whether the letter I is absent or present. For example the first line from the previous problem is LDA SUB. LDA is an MRI and its equivalent 3 bit code, say, 010 is set in bit position 2-4, SUB is searched in the address symbol table and is found to be at location 107. I is not present, therefore, LDA SUB has the HEX code 2107; its equivalent binary is stored in memory location specified by LC = 100. Therefore memory location 100 contains 2107. One important task of an assembler is to check for possible errors in the symbolic program. This was just the basics of how an assembler performs translation. A practical assembler is much more complicated.

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Question:

A projector with a lens of 40 cm focal length illuminates an object slide 4.0 cm long and throws an image upon a screen 20 m from the lens. What is the size of the pro-jected image?

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/Users/wenhuchen/Documents/Crawler/Physics/D28-0882.htm

Solution:

We use the lens formula 1/p = (1/f) - (1/q) where p is the object distance, f is the focal length of the lens, and q is the image distance. Upon substitution of the given values in the formula, we have 1/p = (1/40cm) - (1/20m) In order to do the subtraction, we must change 20 m to centimeters. Since 1 m = 10^2cm, 20 m = 2000 cm and 1/p = (1/40cm) - (1/2000cm) 1/p = (2000cm - 40cm) / 8000cm2= 1960 / 80000cm p = 80000cm / 1960 = 40.8cm This is the object distance. The magnification M of the lens is defined as M = q/p = (2000cm / 40.8cm) = 49. Given, then, an object 4 cm long, the size of its projected imageis Image length = M x object length = 49 × 4cm\approx 200cm

Question:

As the explosive TNT burns, it releases about 4.2 x 106 joules/kg. The molecular weight of TNT is 227. Find the energy released by one molecule of TNT in the process. What is the fractional change in mass that occurs as one kg of TNT explodes?

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/Users/wenhuchen/Documents/Crawler/Physics/D31-0923.htm

Solution:

The molecular weight is expressed in terms of atomic mass units. Oneamuequals 1.66 × 10^-27 kg. Therefore, the mass of one molecule is 1.66 × 10^-27 × 227 kg .The amount of energy released by one molecule is, energy release = 4.2 × 10^6 × 1.66 × 10^-27 × 227 joule = 1.58 × 10^-18 joule = 9.8eV. This energy should be compared with the quantity 7.6 × 10^10 joules involved in the fission of a uranium atom into zirconium and neodymium. We caneasily calculate the fraction of mass converted into energy in an explosion of TNT. The energy released by 1 kg is 4.2 × 10^6 J; thus the mass change is ∆m = E/c^2 = [4.2 × 10^6 /(3 × 10^8)^2 = 4.7 × 10^-11kg. The fractional change in mass is ∆m/m = [(4.7× 10^-11kg)/1] = 4.7 × 10^-11.

Question:

The radiation from an x-ray tube operated at 50 kV is diffracted by a cubic KC1 crystal of molecular mass 74.6 and density 1.99 × 10^3 kg\textbulletm^-3. Calculate (a) the short- wavelength limit of the spectrum from the tube, and (b) the glancing angle for first-order reflection from the princi-pal planes of the crystal for that wavelength.

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/Users/wenhuchen/Documents/Crawler/Physics/D33-1006.htm

Solution:

(a) When an electron passes through a potential difference V, it acquires energyeV. If all this energy is used in producing one quantum of x-radiation, thenhf=eVor f =eV/h, where f is the photon frequency. Since \lambda = c/f, where c is the speed of light and \lambda is the photon wavelength \lambda = (c/f) = (ch/eV). The electron may have lost some of its acquired energy before producing the quantum of radiation. The f-value calculated above is thus the maximum possible frequency of the x-radiation emitted and \lambda the corresponding short-wavelength limit of the emitted spectrum. The value is \lambda = [(3.0 × 10^8 m\textbullets^-1 × 6.6 × 10^-34 J\textbullets) / (1.6 × 10^-19 C × 5 × 10^4 V)] = 0.248 \AA. (b) In order to apply Bragg's law, we must find the separation of the principal planes of the KC1 crystal. We know that the atomic mass of KC1 is equal to the sum of the atomic masses of K(35.1) andCl(39.1). Its value is then 74.6amu. Since 1amuis the gram mass of 1 mole of a substance, 1 mole of KC1 has a mass of 74.6 g or 74.6 × 10^-3 kg. If the density of KC1 is 1.99 × 10^3 kg/m^3, 1 mole of KC1 has a volume V = [(74.6 × 10^-3 kg) / (1.99 × 10^3 kg/m^3)] = 3.75 × 10^-5 m^3 Since there are 6 × 10^23 molecules in 1 mole, the volume v occupied by 1 molecule is v = [v / (6 × 10^23)] = [(3.75 × 10^-5 m^3) / (6 × 10^23)] v = 6.25 × 10^-29 m^3 . Each molecule of KC1 is composed of 2 atoms. The volume, v' per atom is Then v' = (v/2) = 3.13 × 10^-29 m^3 . If we assume each atom to be centered in a cube of side d, then v' = d^3 = 3.13 × 10^-29 m^3 or d = 3.150 × 10^-10 m Thus the linear separation of atoms, i.e., the separation of the principal planes, is d = 3.150 × 10^-10 m = 3.150 \AA. Bragg's law relates the glancing angle to d by the equation 2d sin \texttheta =m\lambda, where m is the order number. Giving \lambda the value calculated above, we obtain sin \texttheta = (m\lambda/2d) = [{(1) (0.248 \AA)} / {(2) (3.150 \AA)}] = 0.0394. \texttheta = 2.25\textdegree.

Question:

The friendly neighborhood bank is planning to start a lottery toattract new customers. A new account receives a number containingthree digits. The digits are be-tween 000 and 999 andwe will assume that all 1,000 combinations have been givenout. Prizes are awarded according to the following scheme: 3 numbersalike$100 gift certificate 2 numbersalike$10 gift certificate What will be the cost of the lottery to the bank?

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G18-0449.htm

Solution:

There are 1000 numbers between 000 and 999. Each occurrence ofa number composed of the same digits (like 000,111...) contributes $100 to the total cost, while each occurrence of a number with two similar digits(like 005, 511, 363...) adds $10 to the cost. The total cost is composedof the cost of all "triplets\textquotedblright added to the cost of all "doublets". 10PRINT "GAMBLING BANKS" 20S1 = 0 30S2 = 0 40FOR I = 0 TO 9 50FOR J = 0 TO 9 60FOR K = 0 TO 9 70IF I = J THEN 100 80IF J = K THEN 110 85IF I = K THEN 110 90GO TO 130 100IF J = K THEN 120 110S2 = S2 + 10 115GO TO 130 120S1 = S1 + 100 130IF 100I + 10J + K<>999 THEN 150 140S3 = S1 + S2 150NEXT K 160NEXT J 170NEXT I 180PRINT "COST OF FIRST KIND"; S1; "PLUS COST OF 190PRINT "SECOND KIND"; S2; "EQUALS TOTAL" 200PRINT "COST OF LOTTERY" S3 210END

Question:

A spinning target in a shooting gallery is dynamically equiv-- alent to two thin disks of equal mass, m, connected by a light bar. The radius of the disks is a = 400 mm and the bar has a length 2a = 800 mm. The target is initially rotating about its' axis of symmetry with a rate \omega_0 = 40 rpm. A bul-let of mass m_0 = (m / 100) and traveling with a velocity v_0 = 100 m/s relative to the target, hits the target and is embedded in point C. Determine (a) the angular velocity of then target immediately after collision, (b) the precession of the ensu-ing motion, (c) the rates of precession and spin of the ensuing motion.

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/Users/wenhuchen/Documents/Crawler/Physics/D37-1075.htm

Solution:

We need the principal axes of the target to solve this problem. With the simplifying assumption that the bullet does not change the principal axes, note that the x, y, z axes shown are the principal axes. I = I_z = (1 / 2)ma^2. Find I' = I_x = I_y by use of the parallel-axis theorem. The moment of inertia of the disk with respect to an axis through it, parallel to the x or y axis in the figure, is (1 / 4)mr^2 . Since the mass of each disk is [(1/2)m] and the distance from the parallel axis to the axis in the figure is a, the moment of inertia of one disk is I' = I_x = I_y = (1/4)[(1/2)m]a^2 + [(1/2)m]a^2 . Due to symmetry, it is possible to multiply by two to find the Moment of Inertia of the total system. I' = I_x = I_y = 2[(1/4)[(1/2)m]a^2 + [(1/2)m]a^2] = (5/4)ma^2 The total angular momentum of the target-bullet system is conserved during collision and remains constant during the ensuing motion. Taking moments about the coordinate origin G yields H^\ding{217}_G = -a\^{\j} × m_0v_0kˆ + l\cyrchar\cyromega_0kˆ H^\ding{217}_G = -m_0v_0\^{\i} + l\cyrchar\cyromega_0kˆ(1) Now determine the angular velocity after impact by invoking the conservation of H_G. Using H^\ding{217} = K_x\cyrchar\cyromega_x\^{\i} + l_y\cyrchar\cyromega_y\^{\j}+ l_z\cyrchar\cyromega_zkˆ, the result is -m_0v_0a = l'\cyrchar\cyromega_x = (5/4)ma2\cyrchar\cyromega_x \cyrchar\cyromega_x = (4/5)[(m_0v_0) / (ma)]; 0 = I'\cyrchar\cyromega_y , \cyrchar\cyromega_y = 0; I\cyrchar\cyromega_0 = I\cyrchar\cyromegaz \cyrchar\cyromega_z = \cyrchar\cyromega_0. \cyrchar\cyromega^\ding{217} = -(4 / 5) [(m_0v_0) / (ma)]\^{\i} + \cyrchar\cyromega_0kˆ.(2) While this equation is correct it is not very helpful since the body axes x, y, z are moving in space in an unknown way. For the target considered, \cyrchar\cyromega_0 = 40 rpm = 4.19 (rad / sec), )(m_0 / m) = (1 / 100), a = 0.400 m, v_0 = 100 (m / sec); we find \cyrchar\cyromega_x = -2 (rad /sec), \cyrchar\cyromega_y = 0, \cyrchar\cyromega_z = 4.19 (rad / sec). This gives \cyrchar\cyromega = \surd[(\cyrchar\cyromega^2_x) + (\cyrchar\cyromega^2_z)] = 4.64 (rad / sec) = 44.3 rpm with angle \Upsilon = arctan [(-\cyrchar\cyromega_x) / (\cyrchar\cyromega_z)] = 25.5\textdegree to the z-axis. (Figure 2). The angle \texttheta shown in fig. 2 is between the precession axis and the z-axis. (H_G is fixed in space; the target wobbles around it.) \texttheta = arctan[(m_0v_0a) / (I\cyrchar\cyromega_0)] = arctan [(2 m_0v_0)( ma\cyrchar\cyromega_0)] \texttheta = 50.0\textdegree . Figure 3 shows \cyrchar\cyromega^\ding{217}, the space and body cones and \varphi ׂ and \Psi ׂ which, added vectorially, must equal \cyrchar\cyromega^\ding{217}. Using the law of sines, \mid[(\cyrchar\cyromega) / {sin(\pi - \texttheta)}]\mid = \mid[(\varphi ׂ ) / (sin \Upsilon )]\mid = \mid[(\Psi ׂ ) / {sin(\texttheta -\Upsilon)}]\mid; (\cyrchar\cyromega / sin \texttheta) = (\varphi ׂ / sin \Upsilon)= (\Psi ׂ / sin \texttheta -\Upsilon) . This gives \varphi ׂ = 24.9 rpm , \Psi ׂ = 24.0 rpm.

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Question:

A man stands at the center of a turntable, holding his arms extended horizontally with a 10-lb weight in each hand. He is set rotating about a vertical axis with an angular velocity of one revolution in 2 sec. Find his new angular velocity if he drops his hands to his sides. The moment of inertia of the man may be assumed constant and equal to 4 slug-ft^2. The original distance of the weights from the axis is 3 ft, and their final distance is 6 in.

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/Users/wenhuchen/Documents/Crawler/Physics/D04-0188.htm

Solution:

If friction in the turntable is neglected, no external torques act about a vertical axis and the angular momentum about this axis is constant. That is, I\omega = (I\omega)_0 = I_0\omega_0, where I and \omega are the final moment of inertia and angular velocity, and I_o and \omega_0 are the initial values of these quantities. I =I_man+I_weights The moment of inertia of a weight at a distance r from the axis of rotation is given by I = mr^2 Therefore I = 4 + 2 (10/32)(1/2)^2 = 4,16 slug \bullet ft^2, I_0 = 4 + 2 [10/32](3)^2 = 9.63 slug\bulletft^2, \omega_0 = 2\pif_0 = (2\pi) [(1/2) rev/sec) = \pirad/sec where f_0 is the original frequency of rotation \omega = \omega_0(I_0/I) = 2.13\pirad/sec. That is, the angular velocity is more than doubled.

Question:

Suppose thecodonsfor amino acids consisted of only two bases rather than three. Would there be a sufficient number ofcodonsfor all twenty amino acids? Show how you obtain your answer.

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/Users/wenhuchen/Documents/Crawler/Biology/F24-0617.htm

Solution:

There are two ways to approach this question. One way is to use a mathematical principle known as permuta-tion and the other is to do it by common sense. The latter method will be discussed first. In this question, we are told that acodonconsists of two bases only. We know there are four different kinds of bases in DNA, namely adenine, guanine, cytosine and thymine. To get the total number of possible two-basecodons, we will have to pick from the four bases and put them into two positions on thecodon. For the first position, we can have any one of the four bases. That means there are four possible ways of filling the first position. For each of these four possible first positions, we again can put any one of the four bases into the second position. The only restriction we face is the number of different bases we have, which is four. So the total number ofcodonsconsisting of two bases is 4 × 4 or 16 codons. The other way to solve this problem is to use per-mutation. The general formula used in permutation isnPr= n^r where n is the total number of objects and r is the number of times permuted. nPr is read as "n permuted r times," and n^r as "n to the power of r." Applying this formula: n is the number of different bases (that is, 4) and r is the number of bases in acodon(which is 2) . Substi-tuting in the numerical data, we have the total possible number ofcodonsmade up of 2 bases. (n^r) = 4^2 =16 In either case, we arrive at the same answer. There will be 16 differentcodonsif eachcodoncontains two bases. However, there are 20 different kinds of amino acids, thus 16codonswill be insufficient to code for all 20 amino acids.

Question:

Calculate the wavelength of a car whose mass is 2 × 10^3 kg andwhose speed is 30 m/s. Determine whether the quantum aspectsof particle motion will be observable.

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/Users/wenhuchen/Documents/Crawler/Physics/D33-0967.htm

Solution:

The de Broglie wavelength of the auto,\lambda_auto, is given by \lambda_auto= (h/mv) = [(6.6 × 10^-34 J s)/{2 × 10^3 kg)(3 × 10^1 m/s)}] = 1.1 × 10^-38 m Note that because the speed of the car is much less than the speed of light, this calculation is clearly non-relativistic. This number is very small. Since the di-mensions of all man-sized objects with which the automobile interactsare very much larger than 1.1 × 10^-38 m, we do not expect to observewave phenomena (i.e., dif-fraction , interference) for cars , and classicalmechan-ics gives a very satisfactory description of the car's motion. This is true for the motion of most man-sized objects the quantum featuresof man-sized or larger objects are usually unimportant and are difficult to observe.

Question:

2KClO_3 (s) \rightarrow 2KCl (s) + 3O_2 (g). He collects the O_2 gas by water displacement at 20\textdegreeC. He observes a pressure of 753 mm Hg. Assuming the pressure of water vapor is 17.5 mm Hg at 20\textdegreeC and he started with 1.28 g of potassium chlorate (KClO_3) what volume of gas is produced? (R = .0821 l-atm/mole \textdegreeK.)

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/Users/wenhuchen/Documents/Crawler/Chemistry/E05-0213.htm

Solution:

You are asked to find the volume of gas produced and can do so using the equation of state, PV =nRT, where P = pressure, V = volume, n = number of moles, R = universal gas constant and T = temperature in Kelvin (Celsius plus 273\textdegree). In the problem you are told the pressure at which the gas is collected, which must be modified due to the presence of water vapor and the temperature. If you knew n, you could substitute for the values in the equation of state and solve for V. Your procedure will be to find the number of moles of gas produced. This can be determined from the reaction equation given and the use ofstoichiometry. If you knew how many moles of KClO_3 you started with, you would know the number of moles of O_2 produced, since 3 moles of O_2 are generated for every 2molesof KClO_3. You can deter-mine how many moles of KClO_3 you started with. You are told that the chemist has 1.28 g. Since the molecular weight of KClO_3 is 122.55 g, you have [(1.28) / (122.55)] = 1.045 × 10^-2 moles of KClO_3. Therefore, from reaction there are (3/2) (1.045 × 10^-2) = 1.57 × 10^-2 moles of O_2 gas produced. Now you go back to PV =nRT. Recall that the gas was collected by water displacement. Thus, it is saturated by water vapor. Hence, the pressure of O_2 is only 753 mm Hg - 17.5 mm Hg, or 735.5 mm where 17.5 mm Hg is the water vapor pressure. Since 1atm= 760 mm, (735.5 mm) [(1atm) / (760 mm)] = .968 atm. Now, substitute and solve for the volume V of O_2 produced. Rewriting and substituting, V= (nRT/P) = [(.0157) (.0821) (293\textdegreeK)/(.968)] = .390 liters or 390 ml.

Question:

Excluding hexagonal unit cells, when counting the number of points inside a cell, a point on an edge is 1/4 inside the cell; and a lattice point at a corner is 1/8 inside the cell. Justify these fractions. Calculate, also, the net number of lattice points in the following unit cells: simple cubic, body-centered cubic, face-centered cubic, and tetragonal.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E06-0218.htm

Solution:

To solve this problem properly you need to know the definition of a unit cell and lattice point. You need, also, to know the actual structures of the unit cells given. You proceed as follows: A unit cell is that small fraction of a space lattice, which sets the pattern for the whole lattice. In other words, it is the smallest portion of the space lattice (which is just the pattern of points which describes the arrangement of atoms or molecules in the crystal), which, when moved repeatedly a distance equal to its own dimensions along the various directions, generates the whole space lattice equivalent to the original lattice. The lattice is, of course, just points that denote atoms or molecules; thus, the term lattice point. The accompanying illustrations give you the structures in question. From this, you can obtain justification for the fractions and find the net number of lattice points. For example, take the simple cubic structure. In a space lattice, each atom is shared with three other unit structures as seen in figure d. Note; This point is shared by 8 unit cells, if you place another 4 cubes on the side. Thus, each point is only 1/8 in the cell. This same sort of procedure and reasoning can be used to justify all fractions in all unit cells given. Calculations now become easy. Simple cubic: Have 8 lattice points. Each is only 1/8 inside cell. Thus, you have net of (1/8) (8) = 1. Face-centered cubic: 8 lattice points at corners, each only 1/8 inside cell. Have a point on each face. Each point is only 1/2 inside cell. Have six points. Thus, net = (8) (1/8) + (6)(1/2) = 4. Body-centered: Again have (8) (1/8) from lattice points at corners. 1 point in center, which is entirely in cell. Thus net becomes (8) (1/8) + (1)(1) =2. Tetragonal: Only have a net of (8) (1/8) from lattice points at corners.

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Question:

In an electric shaver, the blade moves back and forth over a distance of 2 mm, with a frequency of 60 cps. This constitutes a simple harmonic motion with amplitude R = 1 mm. (a) Find the maximum accelera-tion. (b) Find the maximum velocity, (c) Calculate the velocity and acceleration at point C, a distance x = 0.5 mm from 0, the center point (see diagram).

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/Users/wenhuchen/Documents/Crawler/Physics/D09-0384.htm

Solution:

For simple harmonic motion, the force acting on the blade must be proportional to its dis-placement x, but opposite in direction. The limits of the oscillation are equidistant from the equi-librium position. We can then write F = -kx = ma = m[(d^2x)/(dt^2)](1) where k is a constant, m is the mass of the blades, and the negative sign indicates that the force acts in the opposite direction to the displacement. Equation (1) can be written [(d^2x)/(dt^2)] = -(k/m)x(2) [(d^2x)/(dt^2)] = -(k/m)x(2) which requires a solution where the second derivative of x(t) is proportional to -x(t). The sinusoidal functions have such a property. Assuming that at t = 0 sec, the displacement is at its maximum ampli-tude R, we can write x = R cos \omegat(3) where \omega is a constant. Differentiating twice, dx/dt = - \omega R sin \omegat(4) d^2x/dt^2 = - \omega^2 R cos \omegat(5) Substituting equations (3) and (5) into (2), we find - \omega^2 R cos \omegat= -(k/m) R cos \omegat \omega2= k/m Therefore, if the value for \omega^2 is chosen to be k/m, then x = R cos \omegat is a solution of the equation obtained for the simple harmonic oscillation of the blade. From this solution, we see that \omega must be the angular frequency of the motion since the function cos \omegat repeats itself after a time 2\pi/\omega. The velocity and acceleration are dx/dt and d^2x/dt^2, respectively as found in equations (4) and (5). (a) Looking at the acceleration function a = -\omega^2 R cos \omegat(6) we note that the maximum value is attained when cos \omegat is one or \omegat = n\pi where n is an integer. This occurs when the displacement is a maximum. The angular frequency for the blade is \omega = 2\pif = (2\pi)(60 sec^-1) = 377 sec^-1 Therefore, the maximum acceleration is a_max= \omega^2R = (377 sec^-1)^2 × 1 mm = 1.42 × 10^5 mm/sec^2 = 1.42 × 10^2 m/sec2 (b) To find the maximum velocity, see that v = - \omegaR sin \omegat(7) has a maximum whenever sin \omegat = 1. This happens when-ever the blade is at its equilibrium position at which point \omegat = nr/2,where n is an integer. The maximum velocity is calculated to be v_max = \omegaR = 377 sec^-1 × 10^-3 m = 0.377 m/sec (c) To express the velocity function in terms of the displacement x, we square equation (7). Then, v^2 = \omega^2R^2 sin^2 \omegat = \omega^2R^2 (1 - cos^2 \omegat) = \omega^2(R^2 - R^2 cos^2 \omegat) = \omega^2(R^2 - x^2) Expressing this in terms of v_max = \omegaR, we write v^2 = \omega^2R^2[1 - (x^2/R^2)] = v^2_max [1 - (x^2/R^2)] At x = 0.5 mm v = v_max \surd[1 - (x^2/R^2)] = 0.377 m/sec \surd[1 - (0.5 mm/1.0 mm)^2] v = 0.377 \surd[{1 - (1/2)^2} m/sec] = 0.377 × 0.865 m/sec = 0.326 m/sec The acceleration function of equation (6) can be expressed as a = -\omega^2x and at x = 0.5 mm = (5)(10^-4m), the acceleration is a = \omega^2x = (377 sec^-1)^2 × (5 × 10^-4m) = 71 m/sec^2

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Question:

There are N_0 = 10^25 uranium-238 nuclei in a sample. (a) What is the number of uranium nuclei remaining in the sample after 10^8 years if the decay rate per nucleus, \lambda, is 5 × 10\Elzbar18sec\Elzbar1? (b) What is the half-life of uranium?

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/Users/wenhuchen/Documents/Crawler/Physics/D34-1032.htm

Solution:

(a) It is known that each nuclear disintegration takes place independently of any of the others. The radio-active decay rate therefore involves only the instantane-ous number N(t) of the decaying nuclei present ; [dN(t)] /dt] =\Elzbar \lambda N (t)(1) where the constant of proportionality \lambda, is known as the decay rate per nucleus. The solution of (1) is known to be N(t) = N_0- e\Elzbar\lambdat(2) where N_0 is the number of nuclei at t = 0. There are approximately 3.1 × 10^7 sec in one year, therefore 10^8 years equals (10^8 years) × (3.1 × 10^7 sec/year) = 3.1 × 10^15 sec. The number of uranium atoms remaining at the end of this period is N = 10^25 × e^\Elzbar (5 × 10\Elzbar18sec\Elzbar1) × (3.1 × 10^15 sec) = 10^25 × e^\Elzbar0.0155 = 10^25 × 0.984 = 9.84 × 10^24 which is practically the original number . (b) The half-life, T, is defined as the time after which the number of radioactive nuclei has decreased to half its original value. From equation(2) (1/2) N_0 = N_0 e\Elzbar\lambdaT, or (1/2) = e\Elzbar\lambdaT givingIn (1/2) =\Elzbar In 2 =\Elzbar\lambdaT, T = [(In 2) / \lambda] = [(0.7) / (5 × 10\Elzbar18sec\Elzbar1) ] = 1.4 × 10^17 sec. = [(1.4 × 10^17 sec ) / (3.1 × 10^7 sec/year )] = 4.5 × 10^9 years .

Question:

What is the magnetic field 10 cm from a long straight wire carrying a current of 10 amp?

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Solution:

The magnitude of the current in the CGS system of units is 10 amp = (10 A)3 × 10^9 statA/A = 3 × 10 statA The magnetic field, B^\ding{217} , due to the current I in the wire can be found by applying Ampere's law to the circular path shown in the figure, \oint B^\ding{217} \bullet dl^\ding{217} = (4\pi/c) I. From the symmetry of the configuration, the lines of B^\ding{217} form con-centric circles around the wire, then B^\ding{217} \bullet dl = B dl.B at a dis-tance r from the wire is B \int dl = (4\pi/c) I orB = (4\piI)/(c 2\pir) B = (2I/cr) = [2 × (3 × 10^10 statA)]/ [(3 × 10^10 cm/sec) × 10cm] = 0.2 gauss For comparison, this is approximately equal to the magnitude of the Earth's magnetic field at the surface of the Earth and at middle latitudes.

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Question:

Write a Fortran program to calculate the odds of winning in a game of craps.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G23-0556.htm

Solution:

Craps is played with a pair of dice. The rules are as follows: 1) If, on his first throw a player throws a 7 or 11 he wins the bet. 2) If, on his first throw, a player throws a 2, 3, or 12 he loses the bet. 3) If, on his first throw, a player throws any of the re-maining points he continues to throw: a) If he obtains the same score as on the first throw before throwing a 7, he wins. b) If he throws a 7 before obtaining the same score as on the first throw, he loses. The program is written using a function to generate ran-dom numbers arising out of a uniform distribution on ]0,1[. Once the random number generator has been created, an-other function subprogram uses it to obtain two random num-bers between zero and 1. These represent the throw of the dice as follows: X\in [0, 1/6] (1/6, 2/6](2/6, 3/6] (3/6, 4/6] (4/6, 5/6] (5/6. 1] V (x) 123456 Thus X = .45820000 corresponds to a spot value of 3. A single throw of the dice is obtained by adding the two func-tional values of X. Finally, one sequence of throws result-ing in either a win or a loss will be done by a third func-tion subprogram. In the program, note the following: 1)The number of games is G. 2)The seed for the random number generator is KSEED. 10READ(5,100)G, KSEED, IOUT WRITE(6,200)G, KSEED X = XRAND(KSEED) CGENERATE G RANDOM PLAYS KSUM = 0 DO 15 I = 1,G K = KSCORE(O) KSUM = KSUM + K IF (IOUT. EQ. 1) WRITE(6,300)I, K 15CONTINUE CCALCULATE FINAL PERCENTAGE OF WINS XNUM = FLOAT(KSUM) XDENOM = FLOAT(G) FRACT = XNUM/XDENOM WRITE(6,400)G, FRACT 100FORMAT(3110) 200FORMAT(41H1 COMPUTER SIMULATION OF 1A GAME OF CRAPS/,25H0 SIMULATION 2BASED UPON, I5, 13H RANDOM PLAYS/, 39H0 KSEED =, I10//) 300FORMAT(10X, 2HI =, I5, 10X, 7HKSCORE = 1I1) 400FORMAT(3 2H0 THE PERCENTAGE OF WINS 1AFTER, I5, 10H PLAYS IS, F8.6) GO TO 10 END FUNCTION KSCORE(KSEED) CTHIS SUBPROGRAM GENERATES THE OUTCOME OF CONE PLAY OF CRAPS. KSCORE = 0 IMPLIES A CLOSS AND KSCORE = 1 IMPLIES A WIN. K = KTOSS(KSEED) IF ((K.EQ.7).OR.(K.EQ.11)) GO TO 2 IF((K.EQ.2).OR.(K.EQ.3).OR.(K.EQ.12)) GO TO 3 CIF K DOES NOT EQUAL 2, 3, 12, 7 OR 11. CTHE PLAY CONTINUES G = K 10K = KTOSS(KSEED) IF(K.EQ.G) GO TO 2 IF(K.EQ.7) GO TO 3 GO TO 10 2KSCORE = 1 RETURN 3KSCORE = 0 RETURN END FUNCTION KTOSS(KSEED) CTHIS FUNCTION GENERATES THE OUTCOME OF A CRANDOM TOSS OF TWO DICE KTOSS = 0 DO 10 I = 1,2 X = XRAND(KSEED) IF(X.GT. 0.1666667) GO TO 1 KTOSS = KTOSS + 1 GO TO 10 IF(X.GT. 0.33333333) GO TO 2 KTOSS = KTOSS + 2 GO TO 10 2IF(X.GT. 0.5) GO TO 3 KTOSS = KTOSS + 3 GO TO 10 3IF(X.GT. 0.6666667) GO TO 4 KTOSS = KTOSS + 4 GO TO 10 4IF(X.GT. 0.8333333) GO TO 5 KTOSS = KTOSS + 5 GO TO 10 5KTOSS = KTOSS + 6 10CONTINUE RETURN END FUNCTION XRAND(KX) CTHIS FUNCTION GENERATES A UNIFORMLY CDISTRIBUTED RANDOM NUMBER BETWEEN ZERO CAND ONE. IF(KX. GT. 0) IX = KX IY = 65539 \textasteriskcenteredIX IF(IY.LT.0) IY = IY + 2147483647 + 1 XRAND = .4656613E - 9\textasteriskcenteredFLOAT(IY) IX = IY RETURN END

Question:

What is mutualism? Give three examples of this type of relationship.

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Solution:

Mutualism,(or symbiosis) like parasitism and commensalism, occurswidely through most of the principal plant and animal groups and includesan astonishing diversity of physiological and behavioral adaptations. In themutualistictype of relationship, both species benefit fromeach other. Some of the most advanced and ecologically important examplesoccur among the plants. Nitrogen - fixing bacteria of the genus Rhizobiumlivein special nodules in the roots of legumes. In exchange for protectionand shelter, the bacteria provide the legumes with substantial amountsof nitrates which aid in their growth. Another example of plant mutualism is the lichens. They are actually"compound" organisms consisting of highly modified fungi that harborblue-green and green algae among theirhyphae(filaments). Together the two components form a compact and highly efficient unit. In general, the fungus absorbs water and nutrients and forms most of the supportingstructure, while the algae provides nutrients for both componentsvia photosynthesis. In a different form of mutualism, many kinds of ants depend partly orwholly upon aphids and scale insects for their food supply. They milk aphidsby stroking them with theirforelegs and antennae. The aphids respondby excreting droplets of honeydew, which is simply partly digestedplant sap. In return for this sugar-rich food, the ants protect their symbiontsfrom parasitic wasps, predatory beetles and other natural enemies. In man, certain bacteria that synthesize vitamin K live mutualisticallyin the human intestine which provides them with nutrients anda favorable environment.

Question:

Describe the sequence of succession leading to a climax communityfrom a bare rock surface.

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Solution:

Ecological successions are common in most regions, including thosewhich develop on bare rocks. The first pioneer plants may be lichens(algae and fungi living symbiotically) which grow during the brief periodswhen the rock surface is wet and lies dormant during periods whenthe surface is dry. The lichens release acids and other substances thatcorrode the rock on which they grow. Dust particles and bits of dead lichenmay collect in the tiny crevices formed and consequently pave the wayfor other pioneer communities. Mosses can gain anchorage in these crevicesand grow in clumps. These clumps allow the formation of a thickeninglayer of soil composed of humus rotted from dead mosses, lichenparts, and also grains of sand and silt caught from the wind or from therunoff waters of a rainstorm. A few fern spores or seeds of grasses andannual herbs may land on the mat of soil and germinate. These may befollowed by perennial herbs. As more and more such plants survive andgrow, they catch and hold still more mineral and organic materials, andthe new soil layer becomes thicker. Later, shrubs and even trees may startto grow in the soil that now covers what once was a bare rock surface. This type of succession is referred to as primary succession in thatthe region involved has never supported life forms before. Secondary succession is best understood by considering what happensto a farm field when it is abandoned. In the very first year, the abandonedfield is covered with annual weeds. By the second summer the soilis completely covered with short plants, and the perennial herbs have seizedthe available space. Grasses start to form a turf. In the next few years, the turf of herbs thickens, but also woody plants, such as brambles andthorny shrubs, become established. The shrubbery grows until the fieldis blocked by an almost impenetrable tangle of thorny plants. Trees growup through them, changing the field in a decade or two into a rough woodforest. Other communities of trees replace these trees in the succession. The stages from ploughed fields to forests have been witnessedmany times and the successions of vegetation before the climax community is reached are better known asseralstages.

Question:

Write a program which reads in words and prints out only those which end with letter "K".

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G15-0388.htm

Solution:

If we could somehow separate the last character from each word, it would be possible then to check that character and, if it were "K", output the corresponding word. Unfortunately, there is no straightforward way to perform the separation. However, SNOBOL provides a very convenient way of separating the first letter from the word. In order to apply this method, it is necessary to reverse the order of letters in the word. The format - free flow-chart of the solution is shown In the figure. There is no primitive (built-in) function in SNOBOL IV compiler which reverses a string; therefore, it has to be defined. SNOBOL IV provides the programmer with this capability. A function is defined by executing the primitive function DEFINE to specify the function name, formal arguments, local variables, and the entry point of the function. The entry point is the label of the first of a set of statements constituting the procedure for the function. Thus: DEFINE ( 'REVERSE (W)', 'A') defines a function REVERSE having one formal argument, W, and entry point A. As was explained in the previous problem LEN(I) is a primitive function which returns a pattern that matches any string of the length specified by the integer I. Therefore, function LEN (1) will return the first matched string of length one, which is the first letter of the input word. Thus the statements ASTRING = LEN (1) .LETTER BW STRING =: F (RETURN) REVERSE = LETTER REVERSE : (B) form a procedure that reverses the string W. The first statement of this procedure (labelled) A, which indicates the entry point of the function REVERSE) indicates that both STRING and LETTER will be assigned the first one-character string found (matched). The second statement assigns the first letter of the input word under variable name W to STRING and consequently, to LETTER, and deletes that letter. The third statement concatenates values of LETTER and REVERSE (which is now just a variable with initial value 0) assigns the result to REVERSE and sends the execution back to the second statement. The next letter is taken now and the procedure repeats. When all the characters of the input word are reversed, the second statement fails and returns the execution to the main program, after hitting the RETURN label in the transfer field of the second statement. The full solution for this problem looks as follows: STARTWORD = INPUT: F(END) DEFINE( 'REVERSE(W)', 'A') WORD 1 = REVERSE(WORD) WORD 1 LEN (1) . K K'K':F(START) OUTPUT = REVERSE (WORD1)(START) ASTRING = LEN (1) . LETTER BW STRING =: F(RETURN) REVERSE = LETTER REVERSE: (B) END The statement W0RD1 = REVERSE (WORD) calls the function REVERSE, performs it for the part of the input stored in variable WORD, and assigns the final (reversed) value to the variable WORD1. The next statement, W0RD1 LEN(1) . K takes the first letter of the reversed word (which is the last letter of the original word) and assigns it to variable K. The statement K 'K' : F(START), then, checks the variable K for occurrence of letter K, and transfers the execution to the first statement if it fails. The first state-ment reads the next word, and the procedure repeats. However, if the check succeeds, the computer executes the next statement, OUTPUT = REVERSE(W0RD1):(START), which reverses the word for the second time to give it its original form, outputs the word, and sends the control to the first statement. The program terminates when the last word of the input is processed. Note that a function has to be defined before it is called, while the procedure statements may appear later in the program. Also note, that the entry point label in the DEFINE statement could be omitted. In that case, the entry point to the procedure is taken to be the same as the function name. For example: DEFINE ( 'REVERSE (W) ') could have the procedure: REVERSESTRING = LEN(l). LETTER BWSTRING =: F (RETURN) REVERSE = LETTER REVERSE : (B)

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Question:

Birth is associated with a number of rather drastic but important changes that occur in a short period of time. Discuss these changes.

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Solution:

Several important changes must occur in the newborn baby in a very short period of time. During the time of its development in the mother, the fetus obtained all its oxygen and eliminated its CO_2 through the placenta. Its lungs were collapsed. Shortly after birth, when the placenta is no longer present, the carbon dioxide level in the blood of the baby rises to the point at which it stimulates the respiratory center in the medulla of the brain to ini-tiate breathing. The lungs fill with air for the first time and normal breathing follows. A second important change occurs in the heart and blood vessels. In the fetal heart there exists a hole between the right and left atria called the foramenovale, allowing blood from both sides of the heart to mix. In addition, in the fetus there exists a duct between the pulmonary artery and the aorta. This duct, termed theductusarteriosus, effectively cuts down the blood supply to the lungs by diverting it into the aorta. At the time of birth, theductusarteriosusbecomes constricted and eventually is completely blocked. This allows more blood to enter the lungs, the flow increasing by approximately seven times. Also at this time the flaps around the opening between the atria fuse together, thus effectively separating the two atria and allowing the heart to function as two separate pumps. While in the mother's womb the heart serves to pump oxygenated blood, obtained from the mother, to the rest of the fetal body. When the fetus is born, however, the heart must serve the function of pumping deoxygenated blood to the lungs as well as oxygenated blood to the rest of the body. Should the foramenovalefail to close, deoxygenated and oxygenated blood would be able to mix and the infant would be deprived of its full oxygen supply. When this happens, the newborn is termed a "blue baby,\textquotedblright because that is the color of the infant due to the deficiency of oxygen in blood. The foramenovalemust be surgically closed.

Question:

Ethanol (C_2H_5OH, molecular weight = 46 g/mole) unlike most ingested substances, is absorbed directly by the stomach lining. If 44 g of pure ethanol (4 oz of whiskey or 5.5 oz of a martini) is consumed, the resulting concentration of ethanol in the blood is 0.080 g ethanol/100 ml blood. What percent of the ingested ethanol is in the blood? Assume that the total blood volume of an adult is 7.0 liters.

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Solution:

We must first calculate the total mass of alcohol in the blood and then determine to what percent of the ingested mass of alcohol this corresponds. The total mass of ethanol in the blood is equal to the product of the concentration of ethanol and the total blood volume, or 0.080 g ethanol/100 ml blood × 7.0 liters blood = 0.080 g ethanol/100 ml blood × 7000 ml blood = 5.6 g ethanol. Note: 7 liters = 7000 ml, since 1 l = 1000 ml. This corresponds to (5.6 g / 44g ) × 100% = 13% of the ingested ethanol.

Question:

A 0.10 M solution ofHClis prepared. What species of ions arepresent at equilibrium, and what will be their equi-librium concentrations?

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Solution:

Two processes are occurring simultaneously: the reaction ofHCl withH_2O (dissociation ofHCl) and theautoionizationof H_2O. HClreacts with H_2O according to the equation HCl+ H_2O \rightleftarrows H_3O^+ +Cl^-. For every mole ofHClthat dissociates, one mole ofCl^- and one moleof H_3O^+ are produced. The initial con-centration ofHClis 0.10 M. Thus, if we assume thatHCldissociates completely, [H_3O^+] = 0.10 Mand[Cl^-] = 0.10 M. Waterautoionizesaccording to the equation H_2O + H_2O \rightleftarrows H_3O^+ + OH^-. The water constant for this process is K_w= 10^-14 moles^2/liter^2 = [H_3O^+][OH^-]. Hence, [OH^- ]= [(10^-14 moles^2/liter^2)/(H_3O^+)] Since [H_3O^+] was determined to be 0.10 M = 10^-1 M = 10^-1 moles/liter, [OH^-] = [(10^-14 moles^2/liter^2)/(H_3O^+)] = [(10^-14 moles^2/liter^2)/(10^-1 moles/liter)] = 10^-13 moles/liter = 10^-13 M. Hence, atequlibrium, H_3O^+, OH^-, andCl^- are present in the concentrations [H_3O^+] = 0.10 M, [OH^-] = 10^-13 M, [Cl^-] = 0.10 M.

Question:

The thyroid gland is located in the neck and secretes several hormones, the principal one being thyroxine. Trace the formation of thyroxine. What functions does it serve in the body? What happens when there is a decreased or increased amount of thyroxine in the body?

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Solution:

The thyroid gland is a two-lobed gland which manifests a remarkably powerful active transport mechanism for uptaking iodide ions from the blood. As blood flows through the gland, iodide is actively trans-ported into the cells. Once within the cell, the iodide is converted to an active form of iodine. This iodine combines with an amino acid called tyrosine. Two molecules of iodinated tyrosine then combine to form thyroxine . Following its formation, the thyroxine becomes bound to a polysaccharide-protein material called thryroglobulin. The normal thyroid gland may store several weeks supply of thyroxine in this 'bound' form. An enzymatic splitting of the thyroxine from the thyroglobulin occurs when a specific hormone is released into the blood. This hormone, produced by the pituitary gland, is known as thyroid-stimulating hormone (TSH). TSH stimulates certain major rate-limiting steps in thyroxine secretion, and thereby alters its rate of release. A variety of bodily defects, either dietary, hereditary, or disease-induced, may decrease the amount of thyroxine released into the blood. The most popular of these defects is one which results from dietary iodine deficiency. The thyroid gland enlarges, in the continued presence of TSH from the pitui-tary, to form a goiter. This is a futile attempt to synthe-size thyroid hormones, for iodine levels are low. Normally, thyroid hormones act via a negative feedback loop on the pituitary to decrease stimulation of the thyroid. In goiter, the feedback loop cannot be in operation - hence continual stimulation of the thyroid and the inevitable protuberance on the neck. Formerly, the principal source of iodine came from seafood. As a result, goiter was prevalent amongst in land areas far removed from the sea. Today, the incidence of goiter has been drastically reduced by adding iodine to table salt. Thyroxine serves to stimulate oxidative metabolism in cells; it increases the oxygen consumption and heat production of most body tissues, a notable exception being the brain. Thyroxine is also necessary for normal growth, the most likely explanation being that thyroxine promotes the effects of growth hormone on protein synthesis. The absence of thyroxine significantly reduces the ability of growth hormone to stimulate amino acid uptake and RNA synthesis. Thyroxine also plays a crucial role in the closely related area of organ development, particularly that of the central nervous system. If there is an insufficient amount of thyroxine, a condition referred to as hypothyroidism results. Symptoms of hypothyroidism stem from the fact that there is a reduction in the rate of oxidative energy-releasing reactions within the body cells. Usually the patient shows puffy skin, sluggishness, and lowered vitality. Hypothyroidism in children, a condition known as cretinism, can result in mental retardation, dwarfism, and permanent sexual immaturity. Sometimes the thyroid gland produces too much thyroxine, a condition known as hyperthyroidism. This condition produces symptoms such as an abnormally high body temperature, profuse sweating, high blood pressure, loss of weight, irritability, and muscular weakness. It also produces one very characteristic symptom that may not be predicted because it lacks an obvious connection to a high metabolic rate. This symptom is exophthalmia, a condition where the eyeballs protrude in a startling manner. Hyperthyroidism has been treated by partial removal or by partial radiation destruction of the gland. More recently, several drugs that inhibit thyroid activity have been discovered, and their use is supplanting the former surgical procedures.

Question:

What hormone is responsible for metamorphosis in amphibians?

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Solution:

In amphibians, the thyroid hormones control metamorphosis. This is seen by the fact that very young larvae that have had their thyroids removed did not undergo metamorphosis unless they were immersed in or injected with thyroxine solutions. In addition, normal larvae with thyroids metamorphosize precociously when they have thyroxine added to their water or administered by injection . The matemorphosis includes regressive changes (ex: resorption of the tadpole tail, gills, and teeth) as well as constructive changes (ex: the development of limbs, tongue, and middle ear). The pituitary gland initiates metamorphosis by releasing thyroid stimulating hormone (TSH), which is carried by the bloodstream to the thyroid. The thyroid is induced to release thyroxine into the circulation. Metamorphosis involves extensive changes at the biochemical level. A new set of enzymes appears in the liver, a new type of hemoglobin protein is synthesized; and a novel protein is formed in the retina. Structures that are resorbed contain high levels of proteases, nucleases and other digestive enzymes. These changes are all inducible with a single substance - thyroxine.

Question:

In acting as a reducing agent a piece of metal M, weighing 16.00g,gives up 2.25 × 10^23 electrons. What is theweight ofone equivalent of the metal?

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Solution:

One equivalent of a reducing agent is defined as that mass of the substancethat releases the Avogadro number of electrons. Avogadro's numberis 6.02 × 10^23 , thus one can find the number of equivalents in 16.00 g of the metal by dividing 2.25 × 10^23 by 6.02 × 10^23. no. of equiv =(2.25 × 10^23 electrons) / (6.02 × 1023electrons / equiv) Thus, .374 equiv weigh 16.00 g, one can find the weight of one equivalentby dividing 16.0 g by 0.374equiv. weightof 1 equiv = (16.0 g) / (0.374equiv) = 43.78 g / equiv .

Question:

Write a FORTRAN program to set up a direct-access file with one record for each of 1,869 students in a school. Assume that the students have been assigned identification numbers from 1 to 1,869. Allow some extra space for expansion of student population as well as for expansion of items within a student record.

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Solution:

This problem introduces the DEFINE FILE statement along with the associated WRITE statement. This statement takes the form DEFINE FILE N1 (NREC, NRECL, E, INT) where N1 is a unit number (for restrictions on N1 see your computer installation manual), NREC is the number of records, NRECL is the record length in bytes, E stands for format control, and INT is an associated variable which is the index to the next sequential record in the file following a READ or WRITE operation. The form of the WRITE statement is WRITE (UNIT, INDEX, FMT) LIST UNIT is an integer variable, or integer constant four spaces long, which has a value between 1 and 99 and which corresponds to the file to be written. This file must have been previously defined with a DEFINE FILE statement. The INDEX is an integer expression whose value indicates the index of the record to be written. FMT is a statement number which refers to the FORMAT statement to be used. Finally, LIST defines the data to be written, and usually contains one or more FORTRAN variable names separated by commas. INTEGER STUDNO, NAME (5) DEFINE FILE 12(2000, 150, E, INT) DO 22 I = 1, 1869 READ 3, STUDNO, NAME 22WRITE (12 , STUDNO, 4) NAME 3FORMAT (I4, 5A4) 4FORMAT (5A4) STOP END

Question:

A student wanted to produce a sample of lactic acid. He carried out the following synthesis: CH_3CH_2CO_2H \rightarrow CH_3CHCICO_2H \rightarrow CH_3CHOHCO_2H. He obtained a product that appeared to be lactic acid, and yet, it was optically inactive. Does this mean the product was not truly lactic acid?

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Solution:

A carbon atom that has four different atoms or groups attached to it is termed an asymmetric carbon atom or a chiral center. These four groups can give two spatial arrangements, whose relationship to each other is that of an object and its mirror image. For example, see the accompanying figure. The two arrangements are not identi-cal, because they are not superimposable. In such cases, these molecules,i.e. those which differ only in the positioning of groups around asymmetric carbon atoms, are called optical isomers. Such isomers can rotate plane polarized light. Plane polarized light vibrates in only one plane. When substances can rotate plane polarized light, they are optically active. Each of the two optical isomers rotates the light in a different direction. The dextrorotary isomer rotates it to the right, while the levorotatory rotates it to the left. If both isomers are present in equal amounts, the light will not be rotated at all. For one rotates it to the left and the other rotates it to the right, which means there is a cancellation of the rotations. When plane-polarized light is not rotated, the solution is optically inactive. With this information, one can now answer the question. The student formed CH_3CHOHCO_2H, which he believed was lactic acid. The solution was optically inactive, and yet, it is known that lactic acid is optically active. The only possible explanation for this is that the student formed equal amounts of the two possible optical isomers. As mentioned, this results in optical inactivity. There is every reason to believe that formed in equal amounts. The star (\textasteriskcentered) indicates the asymmetric carbon. Thus, the student did form lactic acid. It is just that a mixture of the two optical isomers in equal proportions is present.

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Question:

What are the results of fertilization?

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Solution:

There are four results of fertilization. First, fusion of the two haploid sex cells produces a zygote with a diploid number of chromosomes. For higher organisms in which the prominent generation is the diploid one, fertilization allows the diploid state to be restored. Second, fertilization results in the initia-tion of cleavage of the zygote. It sets off cleavage by stimulating the zygote to undergo a series of rapid cell divisions, leading to the formation of the embryo. Third, fertilization results in sex determination. It is at fertilization that the genetic sex of a zygote is determined. Hormonal factors, which are dependent upon the genotypic sex, regulate the development of the reproduc-tive system during the embryonic period and of the secon-dary sex characteristics after birth, completing the sex differentiation of an individual. In humans, the male genetic sex is determined by the presence of the XY chromosomal pair, whereas the female is determined by the XX chromosomal pair. The fourth result of fertilization is the rendering of species variation. Because half of its chromosomes have a maternal source and the other half a paternal source, the zygote contains a new, unique combination of chromosomes, and thus a new set of genetic information. Hence fertilization provides for the genetic diversity of a species.

Question:

Write a FORTRAN program to analyze a population according to these three genetic criteria: 1) Hair color 2) Eye color 3) Skin color The program should read a deck of cards, each of which contains for one individual the integers, 1, 2, or 3 to identify the three criteria. Output should consist of the number of individuals found to have the corresponding set of genetic factors.

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Solution:

We will need a three-dimensional array to store hair, eye, and skin data. These will be in the form of five-digit integers; e.g., black hair = 01111, blue eyes = 02100, green skin = 03999. As each card is read, the program should add 1 to the appropriate cell of the array. In the output, we want to print the number of individuals that have the same set of genetic factors. Note that we are not in-terested in the individual cases, but only in the number of matches found in the population. The variable IEND appears at the end of each card; the program terminates when a card is encountered where IEND is nonzero. DIMENSION KOUNT (3,3,3) DO 10 I = 1,3/{_\ast}INITIALIZING ARRAY{_\ast}/ DO 20 J = 1,3 DO 30 K = 1,3 KOUNT (I, J,K) = 0 30CONTINUE 20CONTINUE 10CONTINUE 40READ (2,100) IHAIR, IEYES, ISKIN, IEND 100FORMAT (415) IF (IEND.NE.0) GO TO 50 KOUNT(IHAIR, IEYES, ISKIN) = KOUNT (IHAIR, IEYES, 1ISKIN) + 1 GO TO 40 50CONTINUE DO 60 I = 1,3 DO 61 J = 1,3 DO 62 K = 1,3 WRITE (6,100) KOUNT (I,J,K) 62CONTINUE 61CONTINUE 60CONTINUE STOP END

Question:

Why does a child with a certain type of heredity have PKU? Explain in detail.

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/Users/wenhuchen/Documents/Crawler/Biology/F24-0638.htm

Solution:

To understand PKU, or phenylketonuria, we have to trace metabolic pathways. When we digest protein foods, they are broken down into their component amino acids, one of which is phenylalanine. Eventually, these amino acids are taken up by the cells. Within the cells, some enzymes cause the phenylalanine to be combined with amino acids to form structural proteins that become part of the cell protoplasm. If phenylalanine is not used this way by the body cells, it remains in the blood and as it passes through the liver, it may be acted on by an enzyme which converts it into tyrosine, another amino acid. Still another enzyme will convert some of the, phenylalanine into phenylpyruvicacid. Each arrow in the diagram indicates a metabolic pathway mediated by an enzyme. Each enzyme is produced by a gene and a mutation in any one of these genes can break the chains of conversions. In PKU, the gene which codes forphenylalanine H-monooxygenase. the enzyme responsible for conversion of phenylalanine to tyrosine, is altered so that it no longer codes for that protein. Individuals homozygous for this recessive gene cannot produce the mono-oxygenase. Since the conversion of phenylalanine to tyrosine is blocked, there is an increase in the levels of the other by products of phenylalanine metabolism, the important one here being phenylpyruvate. Phenylpyruvate accumulates in blood, at levels from 50 to 100 times the normal values, and is excreted in large amounts in the urine. In childhood, excess circulating phenylpyruvate impairs normal brain development, causing severe mental retardation. Since the disease results from a genetic defect, it can be inherited and passed from generation to generation.

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Question:

A particular type of x-ray equipment discharges a capacitor to obtain its electrical energy. The capacitor of capacitance 0.25\muFis charged to 100 kV and discharged to 40 kV during a 0.1-s exposure. Calculate the quantity of charge supplied to the x-ray tube, the average current through it, and the energy dissipated.

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/Users/wenhuchen/Documents/Crawler/Physics/D20-0690.htm

Solution:

The charge possessed by the capacitor initially is Q_1 = CV_1 = 0.25 × 10^-6 F × 10^5 V = 0.025 C. At the end of the exposure the capacitor is left with charge Q_2 = CV_2 = 0.25 × 10^-6 F × 4 × 10^4 V = 0.01 C. The charge supplied to the x-ray set is thus Q_1 - Q_2 = 0.015 C. This charge is supplied in \cyrchar\cyrt = 0.1 s. The average current flowing is thus i i = [Q_1 - Q_2 /(\cyrchar\cyrt)] = [(0.015 C)/(0.1s)]= 150 = [Q_1 - Q_2 /(\cyrchar\cyrt)] = [(0.015 C)/(0.1s)]= 150 mA . The energy dissipated in this time is the difference in the energy stored in the capacitor before and after discharge: W = (1/2) CV2_1 - (1/2) CV^2 _2 = (1/2) × 0.25 × 10^-6 F (10^1 0 - 16 × 10^8 ) V^2 = 1050 J.

Question:

Calculate the energy (in ergs) of a single quantum of each of the following: (a) 0.5 \AA X-ray, (b) a radio wave of 900 kilocycles/sec. 10^-8 cm = 1 \AA, h = 6.626 × 10^-27 erg-sec, c = 2.99 × 10^10 cm/sec and 900 kilocycles/sec = 9 × 10^5 sec^-1.

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Solution:

The energy (E) of a single quantum can be calculated using the formula E =hѵ, where h = Planck's constant and ѵ = frequency. Sometimes the frequency is not known, but the wavelength, \lambda, is. \lambda and ѵ are related via the expression\lambdaѵ= c, where c = speed of light. Thus, if one solves for ѵ and then for E. Proceed as follows: (a)It is given that the wavelength of the X-ray = .5 \AA or .5 \AA [{(10^-8)/(\AA)} cm] = .5 × 10^-8 cm, \lambda = .5 × 10^-8 cm. and ѵ = (c/\lambda) = [(2.99 × 10^10 cm/s) / (.5 × 10^-8 cm)] = 5.98 × 10^18 sec^-1. Solving for E: E =hѵ= (6.626 × 10^-27 erg-sec) (5.98 × 10^18 sec-1) = 4.0 × 10^-8 erg. (b)Use E =hѵ. One is given the frequency as 900 kilocycles/sec = 9 × 10^5 sec^-1. Thus, E = (6.626 × 10^-27 erg-sec) (9 × 10^5 sec^-1) = 5.96 × 10^-21 erg.

Question:

Explain how the elements of a structure are initialized. In addition, list the initial values of each of the elements of each ofthe structures given below. a)DCL1 A, 2 BFLOAT(3) INIT(2\textasteriskcentered5E0), 2 C, 3 DFIXED(4,2) INIT(4), 3 ECHAR(2) INIT(BA), 2 F LIKE C; b)DCL1A(4), 2 B(3) FLOAT(3) INIT((3)2\textbullet5E0) , 2 C (2) 3D(2) FIXED(4,2) INIT(0,3), 3 ECHAR(2) INIT(BA) 2 F LIKE C;

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G13-0322.htm

Solution:

Each structure consists of a Major Structure con-taining a few Minor Structures.Each Minor Structure consists of other Minor Structures and Elements.The Elements are the items of a structure which can have arithmeticor character- string attributes. Hence, it is the Elements only whichcan have initial values. The rules regarding the assignment of initial valuesto the elements of a structure are the same as that for the assignmentof ordinary variables in a program. Moreover, each Item of a structure, whether Major, Minor or an Element, can be an array. In this case, rules regard-ing initialization of arrayelements apply to the arrays in the structure. a) The various elements of the structure and their initial values are listed below. Elements ofInitial Value stored theStructurein the memory A \textbullet B0 \textbullet 250E + 01 A \textbullet C \textbullet D04 \textbullet 00 A \textbullet C \textbullet EBA A \textbullet F \textbullet D04 \textbullet 00 A \textbullet F \textbullet EBA b) Here, the Item A of the structure is an array of 4ele-ments.The followingtable gives the initial values for the elements of the structure correspondingto the element A(1) of the array. The rest of the elements of thestructure cor-responding to the elements A(2), A(3) and A(4) of the arraycan be obtained by simply substituting for A(1) in the Table. Elements ofInitial Value stored theStructurein the memory A (1) \textbulletB (1)0 \textbullet 250E + 01 A (1) \textbulletB (2)0 \textbullet 250E + 01 A (1) \textbulletB (3)0 \textbullet 250E + 01 A (1) \textbulletC (1) \textbulletD (1)00 \textbullet 00 A (1) \textbulletC (1) \textbulletD (2)03 \textbullet 00 A (1) \textbulletC (2) \textbulletD (1)00 \textbullet 00 A (1) \textbulletC (2) \textbulletD (2)03 \textbullet 00 A (1) \textbulletC (1) \textbulletEBA A (1) \textbulletC (2) \textbulletEBA A (1) \textbulletF\textbullet D (1)00\textbullet00 A (1) \textbulletF\textbullet D (2)03\textbullet00 A (1) \textbulletF\textbullet EBA Note in the above table that when the LIKE C attribute is used for F, the dimensionsof the array C are not copied. So F becomes a one element array, i.e., a simple variable having a minor structure of 3D (2) and 3E. If F wasrequired to be an array of 2 elements like C, the declaration statementwould have to be modified to 2F(2) LIKE C insteadof2F LIKE Cas given.

Question:

Calculate the approximate freezing point of a solution of 162g ofHBrin 500 g H_2O, assuming that the acid is 90% Ionized.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E07-0244.htm

Solution:

The freezing point of a solution is acolligativeproperty, which means that it depends upon the number of solute particles present in the solution. The depression in the freezing point is related to the number of particles by the equation: freezing pt. depression =K_f×molalityof solute. Wherek_fis the freezing point constant. The freezing point constant is defined as the number of degrees the freezing point is lowered when 1 mole of solute is added to 1 kg of solvent.k_ffor H_2O is 1.86\textdegree. One now must determine themolalityof the solution. Themolalityof unionizedHBr is found by first determining the number of moles present. (MW of HBr = 81.) The number of moles present equals the weight divided by the molecular weight. no. of moles = (162g)n / (81g / mole) = 2 moles One can now find themolalityby dividing the number of moles by the number of kilograms of solvent (1000 g = 1kg) molality= (2 moles / 0.5 kg) = 4 m Themolalityof the unionizedHBris 4.0 m, but one is given that the compound is 90% ionized. This means that the total number of particles in the solution will increase. One must therefore calculate the effective molality .HBrionizes as shown: HBR \leftrightarrows H^+ + Br^- OneHBrforms, one H^+ and one Br^-. Because the HBR is 90% ionized, only 10% of the originalHBris left and .90 ×molalityof H^+ and Br^- ions are formed. The effectivemolalityof the solution is the sum of the concentrations of the three components. HBr.10 × 4m = .4m H^+.90 × 4m = 3.6m Br^-.90 × 4m =3.6m effectivemolality= 7.6 m One can now find the amount that the freezing point has lowered. freezing pt. depression =K_f× eff.molality freezing pt. depression = 1.86\textdegree × 7.36 = 13.7\textdegree The freezing point of water is lowered by 13,7\textdegree. It was originally 0\textdegreeC, it is now -13.7\textdegreeC.

Question:

What are the basic differences between acomputer designed for scientific use and the one designed for business use? Also give examples of computer languages designed for scientific use and business use, and compare them .

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G01-0014.htm

Solution:

The scientific computer is designed to be extremely fast and accurate . It will be a number-cruncher where the compu-tations within CPU are very fast, and the I/O to external de-vices is minimal (or none). For example , the embedded systems used in robotics or space applications may not even have any disk or data transfers to disks. These systems generally use the real-time executives that need not carry out file management functions. Supercomputers and high-performance multiple- processor systems are used for scientific applications. A business computer is designed to move vast amounts of data amongst primary memory, secondary storage, I/O peripherals, and communication devices. There may not be too much number crunching, and hence no need for a very fast CPU or multiple processors. In such systems , the disk access is fast, and the secondary storage capacities are high . FORTRAN (derived fromFORmulaTRANslation) is a well-known example of scientific language. It can support various types of numbers (integer, real, complex) in single or double precision. It can accurately and efficiently evaluate complex arithmetic expressions. COBOL (derived fromCOmmonBusiness Oriented Language) is a good example of business language. It can create and access many different types of files, and support complex structured data (records with fields and subfields up to the depth of 49). It has statements for moving large chunks of data, searching through tables and files, sorting, and merging . In short, it has powerful I/O facilities. Language such as Pascal or C may be used for both scien-tific and business applications. These languages can process the mathematical expressions efficiently and also support complex data structures. These languages have some additional features that are desirable for clarity and reliability . Such features include structured programming constructs, strong typing of variables (e.g. as characters, numbers, etc.), block structure with notions of local and global variables.

Question:

main \textasteriskcenteredthis program prints the message\textasteriskcentered \textasteriskcenteredHi from C on the screen\textasteriskcentered printf ("Hi from C\textbackslashn") }; };

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G17-0422.htm

Solution:

1st error: main is a function and the format (of main) is: main ( ) 2nd error: The executable part of the function is always to be enclosed in braces { }. These braces are analogous to the 'begin-end' of Pascal, 'DO-END' of PL/I, etc. In the given program, the terminal brace, }, is present, but the initial is missing. The initial brace, { , has to be placed in the beginning of the second line, before the first \textasteriskcentered. 3rd error: C recognizes everything enclosed by /\textasteriskcentered and \textasteriskcentered/ as a comment. There are two distinct ways by which the above given informa-tion i.e. \textasteriskcentered this program prints the message\textasteriskcentered \textasteriskcenteredHi from C on the screen\textasteriskcentered can be represented. 1st method: /\textasteriskcentered this program prints the message\textasteriskcentered/ /\textasteriskcenteredHi from C on the screen\textasteriskcentered/ 2nd method: /\textasteriskcentered this program prints the message Hi from C on the screen\textasteriskcentered/ Both the above mentioned methods are correct. 4th error: In the statement printf ("Hi from C\textbackslashn") semicolon ( ;) at the end of line has not been given. This semicolon serves as a statement terminator to signal the com-piler where an instruction ends. 5th error: Semicolon placed after the end of main i.e. }; is wrong. Right brace just defines the end of function main in this example and is never followed by a semicolon. Therefore the correct program segment is: main ( ) { /\textasteriskcentered this program prints the message Hi from C on the screen\textasteriskcentered/ printf ("Hi from c\textbackslashn"); }

Question:

In what ways do mosses and liverworts resemble and differ from each other?

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Solution:

Mosses and liverworts are two classes of bryophytes. Being in the same phylum, they resemble each other in several respects. First, they share an ability to live and reproduce on land. Second, they have evolved similar structures to help them survive on land: an epidermis to prevent excessive evaporation, pores on the surface to effect gaseous exchange, rootlike projec-tions called rhizoids for anchorage, and small green leaflets to manufacture food. Third, the two classes have a similar life cycle in which the gametophyte is the dominant generation and thesporophyteis partially dependent upon the gametophyte for water, minerals, and anchorage . Fourth, both produce their sex cells inmulticellularsex organs (calledanantheridiumin the male and anarchegoniumin the female) and have flagel-lated sperm requiring a moist medium for transport. Last, the embryos of both develop protected within the archegonia. However, the mosses and liverworts have certain structural differences which separate them into these two distinct classes. First, the moss plants have an erect stem supporting spirally-arranged leaves. The more primitive liverworts are simply flat, sometimes branched,ribbonlike structures that lie on the ground, attached to the soil by numerous rhizoids . Second, the rhizoids of the liverworts are specialized, unicellular projections extending downward from the leaf-like plant body (thallus). The rhizoids of the mosses, on the other hand, are composed of filaments of cells extending from the base of the stem. Third, the liverworts'thallus, unlike the leaflets of the moss, is more than one cell layer thick and bear scales , frequently brown or red, on its lower, ground-facing surface. On the opposite, upper surface are small cups, known asgemmaecups. These form oval structures, thegemmae, which when detached from the parent plant give rise asexually to new game-tophyte individuals. Finally, whereas in the mosses, the sex organs are borne at the tip of the stem, in the liverworts they are embedded in deep, lengthwise depres-sions or furrows in the dorsal surface of thethallus.

Question:

Write a program which, using a recursive function, determines and prints out the steps necessary to move N discs from one needle to another in the Tower of Hanoi game.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G15-0392.htm

Solution:

A Tower of Hanoi is a game where N discs of decreasing size are stacked on a needle. There are two empty needles nearby. The aim is to move all the discs from the original needle to a second needle using, when necessary, the third needle as temporary storage. Only one disc can be moved at a time, and at no time may a larger disc rest upon a smaller disc. A recursive procedure has the property that the function Itself is called in the procedure. While convenient, recursive procedures may lead to computational inefficiencies. Nevertheless, recursion is frequently the most natural way of expressing a function, and may considerably simplify programming. The solution of this problem is as follows: \textasteriskcenteredTHE TOWER OF HANOL GAME DEFINE(`HANOI(N,NS,ND,NI)'):(HANOI.END) HANOIEQ(N,O):S(RETURN HANOI(N- 1,NS,NI,ND) OUTPUT= 'MOVE DISC 'N' FROM 'NS' TO 'ND HANOI(N- 1,NI,ND,NS)(RETURN) HANOI.END TESTHANOI(5, 'A' , 'C' ,'B') END In this case N is equal to 5. On entry to the function HANOI, the value of N is compared with zero. If N is zero, no discs are moved, and the function returns. If N is not zero, HANOI is called recursively to move N - 1 discs from the starting needle (NS) to the intermediate storage needle (NI) Having done that, the command to move the Nth disc from the starting needle to the destination needle (ND) is printed. Finally, HANOI is called a second time to move the N - 1 discs from intermediate storage to the destination needle. The output for this program is as follows MOVE DISC 1 FROM A TO C MOVE DISC 2 FROM A TO B MOVE DISC 1 FROM C TO B MOVE DISC 3 FROM A TO C MOVE DISC 1 FROM B TO A MOVE DISC 2 FROM B TO C MOVE DISC 1 FROM A TO C MOVE DISC 4 FROM A TO B MOVE DISC 1 FROM C TO B MOVE DISC 2 FROM C TO A MOVE DISC 1 FROM B TO A MOVE DISC 3 FROM C TO B MOVE DISC 1 FROM A TO C MOVE DISC 2 FROM A TO B MOVE DISC 1 FROM C TO B MOVE DISC 5 FROM A TO C MOVE DISC 1 FROM B TO A MOVE DISC 2 FROM B TO C MOVE DISC 1 FROM A TO C MOVE DISC 3 FROM B TO A MOVE DISC 1 FROM C TO B MOVE DISC 2 FROM C TO A MOVE DISC 1 FROM B TO A MOVE DISC 4 FROM B TO C MOVE DISC 1 FROM A TO C MOVE DISC 2 FROM A TO B MOVE DISC 1 FROM C TO B MOVE DISC 3 FROM A TO C MOVE DISC 1 FROM B TO A MOVE DISC 2 FROM B TO C MOVE DISC 1 FROM A TO C

Question:

A cylindrical glass tube of length L is half submerged in mercury, as shown in Figure a. The tube is closed by a finger and withdrawn. Part of the mercury flows out. What length of mercury column remains in the tube? (Figs. b and c). Take the atmospheric pressure P_A to be H cm Hg.

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Solution:

When the tube is submerged in mercury with both ends open, mercury rises in the tube until the levels of mercury in the tube and in the container are the same (Fig. a). We expect no change in the level of the mercury inside the tube if we close the top of the tube since the air trapped inside was previously in equilibrium with the atmospheric pressure P_a. Closing the top does not change the state of the air inside so that the air pressure on the mercury inside the tube is still P_a (Fig. b). When the tube is in air with its top still closed, the mercury flows out until the pressure at the bottom of the tube equals air pressure (Fig. c). If P is the pressure of the air in the tube and h is the length of the remaining mercury, we can write P_A = P + [(weight of the mercury)/(cross-sectional area)] = P + (W_Hg/A) = P + [(\rho_Hg A\bullet h)/A] = P + h\rho_Hg(1) where \rho_Hg is is the specific weight of mercury. We can find P by using Boyle's law for the air trapped in the tube. Since the temperature remains constant, we have, (P_a) × (Volume in Fig. b) = (P) × (Volume in Fig. c) orP_a (1/2)LA = P (L - h)A givingP = [LP_a] / [2(L - H)].(2) Substituting (2) in (1), we get P_a = P_a{L/[2(L - H)]}+ h\rho_Hg.(3) We are told that P_a is equal to the weight of H cm long mercury column with unit cross-sectional area, P_a = H\rho_Hg Hence, (3) becomes H\rho_Hg = H\rho_Hg{L/[2(L - H)]} + h\rho_Hg orH = H{L/[2(L - H)]} + h.(4) If we solve (4) for h, we get 2H(L - h) = HL + 2h(L - h) 2HL - 2Hh = HL - 2hL + 2h^2 = 0 2h^2 - 2h(H + L) + LH = 0. The roots of this equation are, by the quadratic formula, h = {H + L\pm\surd[(H + L)^2 - 2LH]} / 2 = [H + L \pm \surd(H^2 + L^2)]/2 Since P < P_a, we see that h\rho_Hg < H\rho_Hg, or h < H. The root [H + L + \surd(H^2 + L^2)]/2 is greater than H, therefore it can not be the physical choice. Hence h = [H + L - \surd(H^2 + L^2)]/2.

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Question:

Write a program to read in a man's earning for the day, cal-culate his salary for the week and print it.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G09-0201.htm

Solution:

APWKDAYS,= P'0'INITIALIZE COUNTER READREADCARD CARDIN,CRDEOFREADS IN EARNINGS FOR DAY GOES TO LABEL 'CRDEOF' WHEN NO MORE DATA IS FOUND. PACKWKERN, CRDERNCONVERT EBCDIC DATA TO PACK DECIMAL FOR CALCULATIONS APWKSUM, WKERNADDS DAY'S EARNINGS TO SALARY APWKDAYS , = P '1'ADDS 1 TO NUMBER OF DAYS WORKED CPWKDAYS, = P '6'MAN WORKS 6 DAYS PER WEEK BESALFND BREAD\textasteriskcenteredif the program reaches this statement it means the pre-vious test has failed, it tells the computer that days worked does not yet equal 6, therefore another card must be read. \textasteriskcenteredAT THIS POINT SALARY FOR WEEK IS FOUND SALFNDZAPWKDAYS, = P'0'RESET WKDAYS TO ZERO MVC PRTSUM, WKSUMPREPARE TO PRINT SALARY FOR THE WEEK PRINTLINPRTSAL, 132PRINT SALARY \textbullet \textbullet \textbullet CRDEOF . . .END OF DATA; DO END OF FILE PROCEDURES \textasteriskcenteredTHE DATA DEFINITION FOR THE INPUT AREA CARDINDS\varphiCL80 \varphi \varphi CRDERNDSCL4 DSCL76 \textasteriskcenteredTHE DATA DEFINITION FOR THE PRINT AREA. PRTSALDS\varphiCL132 \varphi \varphi PRTSUMDSCL6 6 DS126C" DS126C" \textasteriskcenteredDATA DEFINITION FOR WORK AREAS WKERNDSPL3 WKSUMDSPL4 WKDAYSDSPL2 READCARD CARDIN, CRDEOF: The first operand of this instruc-tion specifies the name of the data definition for this area. The data definition indicates how the data should be read in. On a card, the data must be punched exactly in the format specified by the data definition. You should also observe that there is a data definition for the print area. PACK WKERN, CRDERN: The data is read in EBCDIC form and cannot be operated on in this form. Hence, we convert it to packed decimal which is an operable form. ZAP WKDAYS, = P'O': The man has worked for six days, so after the sixth day we reset WKDAYS by zeroing it. MUC PRTSUM, WKSUM: Here the salary WKSUM is moved into the print area prior to printing. It is generally a good idea to edit (convert back to EBCDIC) data in packed decimal be-fore printing it. PRTLIN PRTSAL, 132:PRTSAL is the name of the data defini-tion for the print area, and 132 is the length' of the print area. You should note that the general format of a data definition is NAME DS\varphiCLX. The\varphiindicates \varphi \varphi that the symbols that follow are sub-area of the field. X is the length of the field, 80 for card, 132(or 133) for a print line.

Question:

Distinguish between a soap and a detergent. How does a soap clean? Explain the function of phosphate in a detergent.

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Solution:

Soaps are usually potassium or sodium salts of fatty acids (long carbon chain carboxylic acids of formula R-COOH, where R is a long alkyl group). They are produced commercially by heating a fat with an aqueous sodium or potassium hydroxide solution. The structure of soap explains its cleansing action. Soap is a long hydrocarbon (non-polar) group with an ionic group at one end. (See figure A.) The ionic group is water-soluble and the hydrocarbon group is oil soluble. Dirt particles are oil-containing. The hydrocarbon end of the soap is attached to and dissolves in dirt particles, leaving the ionic end exposed to water. The dirt particle is dispersed in the water as a stable emulsion and removed from the article being cleaned. It is known that the cleansing action of soap is reduced by hard water, which contains calcium or magnesium ions. To avoid this problems, soaps have been replaced by detergents; they are more resistant to precipitation by hard water. The active ingredients, or surfactants, of detergent possess the same principal features as soaps. Detergents have a long hydrocarbon chain and an ionic group, usually a sulfonyl group. One example is the sodium salt of alkylbenzene sulfonate, (shown in Fig. B). In addition to surfactants, detergents possess builders. The most widely used builder is Na_5P_3O_8. The phosphate in the detergent functions as a water softener by tying up hard-water ions. It aids the surfactant by dispersing the suspended dirt. It can also act as a buffer, thereby maintaining the pH of the solution at the optimum point for effective washing action.

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Question:

For PCI_5\rightleftarrows PCl_3 + CI_2 , K = 33.3 at 760\textdegreeK . Suppose 1.00g of PCI_5is injected into a 500-ml evacuated flask at 760\textdegreeK and it is allowed to come into equilibrium. What percent of the PCI_5 will decompose?M.W. of PCI_5 = 208.235.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E09-0321.htm

Solution:

To find the percent of decomposition of PCI_5 , you need to know its initial concentration and final concentration (i.e., equili-brium concentration). Initially, there is 1.00g of PCI_5in the 500ml flask. Concen-tration = moles/liter. There are 1000ml in 1 liter, so that 500ml is .5 liters. The molecular weight of PCI_5 = 208.235 grams/mole. A mole is defined as [{grams (mass)} / {molecular wt}]. Thus, number of moles of PCI_5 = 1g / (208.235g/mole ) = .0048 moles . Therefore, the initial concentration of PCI_5 = [(.0048moles) / (.5 liter)] = .0096 M . To find its concentration at equilibrium, use the equilibrium constant expression, which states that K, the equilibrium constant, is equal to the ratio of the concentrations of products to reactants, each raised to the power of its coefficient in the equation. Here, you have K = { [CI_2 ] [PCI_3 ] } / [PCI_5 ] . But, you are told K = 33.3. You can equate to obtain 33.3 = { [CI_2 ] [PCI_3 ] } / [PCI_5 ] . Let x = concentration of [CL_2 ] at equilibrium. If this is the case, then x = [PCI_3 ] also, since from the chemical equation it is shown that they are formed inequimolaramounts. If x moles/liter of each CL_2 and PCI_3form, and the only source is PCI_5 , then the [PCI_5 ] at equilibrium is the initial concentration minus x (moles / liter) = .0096\Elzbar x . Substituting these values into the equilibrium constant expression, you obtain 33.3 = { (x)(x)} / (.0096\Elzbar x) . Solving for x, using the quadratic equation, you obtain x = 9.597 × 10-^3= [PCI_3] = [CL_2] at equilibrium. [PCI_5 ] = .0096\Elzbar .009597 = 3 × 10-^6 . This means per-cent dissociation = 100 × [ (Initial\Elzbar final) concentration ] / [initial concentration] = 100 × (.009\Elzbar 3 × 10-^6) / .0096 = 99.9 % dissociated.

Question:

When an adult mouse or bird is castrated,itssexual behaviorvirtually disappears. Compare the effects of castrationof highly developed primates with the less developedanimals.

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Solution:

In simple behavior, animals respond to a particular stimulus in the environmentwith an appropriate response. However, few animals respond soautomatically; stereotyped behavior can be modified. Sometimes the animaldoes not respond; sometimes it responds differently than expected. Its actions often depend on other control-ling factors, such as hormone levelsand previous experi-ence. This question involves the use of these twofactors in modifying sexual behavior. Hormones modify behavior by affecting the level of an animal's motivationalstate. They alter the animal's physiology, which can then alter behavioralpatterns. For example, the breeding term in birds is initiated by physiologicalchanges in the reproductive organs which produce the sex hormones. These sex hormones cause the development of special breedingplumage which subsequent-ly affects reproductive behavior. The malehormones also induce courtship displays like bowing and cooing uponsight of the female. The female responds to these court-ing displays bythe release of its own reproductive hormones which affect, egg production. Eventually nest building occurs and copulation results. In each stepof this reproductive behavior, hormones lead to behavior acts which causefurther release of other hormones di-recting further behavior. However, as animals have evolved, there is less hormone-directed behavior, and more learned responses. Learning is the change in behavior asa result of ex-perience. Only higher animals with larger and more complexbrains demonstrate the process of learning. Learning affects man'sbehavior, very little of which is stereotyped. In man, hormones play aminor role in modifying behavior; learning plays a major role. When a mouse is castrated (testes removed), it can no longer produceany male reproductive hormones. As in the bird, these hormones arenecessary for the reproductive behavior of the mouse. Removal of the hormonesthus causes the sexual behavior of the mouse to decline and eventuallydisappear. The mouse ap-parently cannot learnitssexual behavior. In higher primates such as man, castration does not affect the abilityto perform the sexual act, provided the male is sexually mature and experienced. This is because the behavior has been learned through experience.

Question:

Which tissue of plants most resembles, in function, theskin ofanimals?

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/Users/wenhuchen/Documents/Crawler/Biology/F02-0057.htm

Solution:

Surface tissues form the protective outer covering of the plant body. The epidermis is the prin-cipal surface tissue of roots, stems, and leaves. It resemblesthe skin of animals in having a protective function. This tissue canbe one cell thick, although some plants that live in very dry habitats, whereprotein from water loss is critical, have a thick epidermis con-sisting ofmany layers of cells. Most epidermal cells are relatively flat and contain avery large vacuole with only a thin layer of cytoplasm. Their outer walls arethicker than their inner walls. Epidermal cells often secrete a waxy, water-resistantsubstance, calledcutin, on their outer surface. The thick outercell wall and the cuticle aid in protection of the plant against loss of water. Some epidermal cells on the surface of the leaves are specialized asguard cells. These regulate the size of small holes in the epidermis, the stomata, through which gases can move into or out of the leaf interior. Epidermal cells of roots have no cuticle which may in-terfere with their functionof absorption. They are, how-ever, characterized by hair-like outgrowths, called root hairs, that increase the absorptive surface for the intakeof water and dissolved minerals from the soil. Peridermis the surface tissue that constitutes the corky outer bark oftree trunks or the outer cork layer of large roots. Cork cells are dead whenthey are mature, and their cell walls contain another waterproof material,suberin, for additional protection.

Question:

Which of the following, if any, are not alcohols derived from the methane series of hydrocarbons: C_6H_5OH, C_17H_33OH, C_4H_8OH, C_9H_19OH?

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Solution:

Alcohols are derived from molecules whose hydrogen atoms have been replaced by one or more hydroxyl (-OH) groups. The simplest alcohols are derived from thealkanesor the methane series and contain only one hydroxyl group per molecule. These have the general formula ROH, where R is an alkyl group of composition, C_nH_2n+1. Thus, the alcohols to be derived from the methane series will follow this formula. (a) C_6H_5OH has 6 carbons. Therefore, its hydrogen content should be 2n+1= 2(6) + 1 = 13. Because there exist only 5hydrogens, excluding the H from OH, it cannot be derived from the methane series. Compound (b), C_17H_33OH, also does not fit the general formula (n= 17, 2n + 1 = 35), and thus, is not derived from the methane series. If it did fit, it would possess 35hydrogensinstead of 33. Compound (c), C_4H_8OH, does not fit the general formula (n = 4, 2n+ 1 = 9) and, thus, is not derived from the methane series. If it did fit, the alkyl group it would posses 9 H instead of 8. Compound (d), C_9H_19OH, does fit the general formula (n = 9, 2n + 1 = 19), and is derived from the methane series.

Question:

Describe the structure and functions of the plasma membrane.

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/Users/wenhuchen/Documents/Crawler/Biology/F02-0035.htm

Solution:

Each cell is surrounded by a selective membrane, a complex elastic covering that separates the cell protoplasm from the external environment. The structure of this covering, called the plasma membrane, has been under major investigation for many years. Studies of membrane permeability, electron microscopy, and biochemical analysis have enabled biologists to better understand the structure and composition of the plasma membrane. The plasma membrane contains about 40 percent lipid and 60 percent protein by weight, with considerable variation from cell type to cell type. The different types and amounts of lipids and proteins present determine to a great extent the characteristics of different membranes. As seen in electron micrographs, all membranes appear to have a similar fundamental structure. The plasma membrane is revealed by electron microscopy to resemble a railroad track in cross-section - two dark lines bordering a central lighter line. The membranes of cellular organelles also display this charac-teristic. The two dark lines were suggested to corres-pond to two layers of protein and the light middle layer to lipid. It was soon revealed that the lipid actually exists in two layers. The lipid molecules of the plasma membrane are polar, with the two ends of each molecule having different electric properties. One end is hydrophobic ("fear of water"), which means it tends to be insoluble in water. The other end is hydrophilic ("love of water"), which means it has an affinity for water (see Figure 1). The lipid molecules arrange themselves in two layers in the plasma membrane so that the hydrophobic ends are near each other, and the hydrophilic ends face outside toward the water and are stabilized by water molecules (see Figure 2). In this bilayer, individual lipid molecules can move lateral-ly, so that the bilayer is actually fluid and flexible. Protein molecules of the plasma membrane may be arranged in various sites but embedded to different degrees in relationship to the bilayer. Some of them may be par-tially embedded in the lipid bilayer, some may be present only on the outer surfaces, and still others may span the enitire lipid bilayer from one surface to the other (see Fig. 3). The different arrangements of proteins are determined by the different structural, conformational, and electrical characteristics of various membrane proteins. Like the lipid bilayer, the protein molecules tend to orient them-selves in the most stable way possible. The proteins are usually naturally folded into a globular form which enables them to move laterally within the plane of the membrane at different rates. Certain proteins can actually move across the membrane. Thus membrane proteins are not static but dynamic. The functions of the plasma membrane are highly specific and directly related to its structure, which is in turn, dependent on the specific types and amounts of proteins and lipids present. The discriminating permea-bility of the membrane is its primary function. It allows certain substances to enter or leave the cell, and prevents other substances from crossing it. Whether or not a molecule can cross a membrane depends on its size after hydration, electric charge, shape, chemical proper-ties, and its relative solubility in lipid as compared to that in water. This selective permeability of the plasma membrane gives the cell the ability to keep its interior environment both chemically and physically dif-ferent from the exterior environment. The plasma membrane is also found to be particularly important in cell adherence. Because of the specificity of protein molecules on the membrane surface, cells can recognize each other and bind together through some inter-action of their surface proteins. Surface proteins are believed to provide communication and linkage between cells in division so that cells divide in an organized plane, rather than in random directions giving rise to an amorphous mass of cells as in cancer. Surface proteins of the plasma membrane are also proposed to recognize foreign substance due to their remarkable specificity; they can bind with the foreign substance and inactivate it. Membrane proteins are further suggested to interact with hormones, or convey hormonal messages to the nucleus so that a physiological change can be effected. The plasma membrane also is involved in the conduction of impulse in nerve cells. The axon of nerve cells transmits impulse by a temporary redistribution of ions inside and outside the cell, with a subsequent change in the distribution of charges on the two surfaces of the membrane.

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Question:

(A) Explain the purpose of a coliform bacteria count in natural water. (B) What is meant by hard or refractory organic compounds?

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Solution:

(a) One of the most important problems associated with polluted water is the possible presence of disease-causing bacteria. A test for the presence of bacteria is made by measuring the amount of common coli form bacteria in water. (Coliform bacteria are harmless and reside in the large intentines of humans.) Coliform bacteria are present in plentiful quantities in feces, which become part of sewage wastes. Since coliform bacteria do not thrive inahaquatic environment, they give an in-dication of how recently and to what extent sewage pol-lution has occurred by their count (which is proportional to the amount of disease-causing bacteria present). If the count is high, human sewage contamination has taken place, possibly accompanied by the presence of dangerous bacteria. (b) Hard or refractory organic compounds are nondegradable substances that decay very slowly in water. Degradation refers to the breaking down of substances into simpler forms. Examples of nondegradable substances are certain pesticides and detergents.

Question:

In rabbits, spotted coat (S) is dominant to solid color (s), and black (B) is dominant to brown (b). In a large population, brown spotted rabbits are mated to solid black ones and all the offspring are black spotted. What are the genotypes of the parents? What would be the appearance of the F_2 if two of these F_1 black spotted rabbits were mated? Illustrate your answer with a diagram.

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Solution:

The first thing to note about this problem is that the cross is dihybrid, and so deals with two separate traits. We can determine the genotypes of the parents by comparing their phenotypes with those of the F_1 offspring. The brown spotted parents have two possible genotypes, bbSS and bbSs. The solid black parents have two possible genotypes, BBss and Bbss. There are four possible matings then: 1)bbSS×BBss, 2)bbSS×Bbss, 3)bbSs×BBss,and 4)bbSs×Bbss. We can see that crosses 2, 3, and 4 could produce off-spring with either or both of the homozygous recessive traits, brown and solid. But we are told that we are dealing with a large sample size and that all the off-spring were black spotted. Therefore the genotypes in 1) must be the parental genotypes. Looking at the cross: BS Bs bS Bs BS BBSS BBSs BbSS BbSs BS BBSs BBss BbSs Bbss bS BbSS BbSs bbSS bbSs bs BbSs Bbss bbSs bbss Phenotypically: BBSS BBSs BbSS BbSs BBSs9 black, spotted BbSs BbSS BbSs BbSs BBss Bbss3 black, solid Bbss bbSS bbSs3 brown, spotted bbSs bbss1 brown; solid That is, a9:3:3:1ratio blackblackbrownbrown spottedsolidspottedsolid Note that whenever we come across a 9 : 3 : 3 : 1 ratio in a dihybrid cross, it indicates that the cross involved dominant-recessive characteristics, and was between two heterozygotes.

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Question:

Calculate the pH_I values of glycine, alanine, serine and threonine from their pK^1 values. The pk' values for the ionizing groups of some amino acids (25\textdegreeC) pK_1' \alpha-COOH pK_2' \alpha-NH'_3 pK_R' R-group Glycine 2.34 9.6 Alanine 2.34 9.96 Leucine 2.36 9.60 Serine 2.21 9.15 Threonine 2.63 10.43 Glutamine 2.17 9.13 Aspartic acid 2.09 9.82 3.85 Glutamic acid 2.19 9.67 4.25 Histidine 1.82 9.17 6.0 Cysteine 1.71 10.78 8.33 Tyrosine 2.20 9.11 10.07 Lysine 2.18 8.95 10.53 Arginine 2.17 9.04 12.48

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Solution:

pH_I is defined as the isoelectric pH. At the isoelectric point, there is no net electrical charge on the molecule and the molecule will not move in an electric field. pH_I can be determined using the following equation. pH_I = (1/2)(pK'_1 + pK'_2) One can now solve for the pH_I values of the various amino acids mentioned. glycine: pH_I = (1/2)(2.34 + 9.6) = 5.97 alanine: pHI= (1/2)(2.34 + 9.69) = 6.02 serine:pH_I = (1/2)(2.21 + 9.15) = 5.68 threonine : pH_I = (1/2)(2.63 + 10.43) = 6.53

Question:

What is the approximate number of molecules in a drop of water which weighs 0.09 g?

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Solution:

The number of molecules in a mole is defined to be 6.02 × 10^23 molecules. Thus, to find the number of molecules in a drop of water, one must know the number of moles making up the drop. This is done by dividing the weight of the drop by the molecular weight of H_2O. (MW of H_20 = 18) number of moles ={0.09g / 18 (g/mole)} = .005 moles. The number of molecules present is now found by multiplying the number of moles by Avogadro' number (6.02 × 10^23). no. of molecules = .005 moles × 6.02 × 10^23 molecules/mole = 3.01 × 10^21 molecules.

Question:

Show how you would use the Full Adder for adding and subtracting two 4 bit numbers.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G05-0100.htm

Solution:

To add two 4 bit numbers, A = A_4 A_3 A_2 A_1 and B = B_4 B_3 B_2 B_1, take four full adders F_4, F_3, F_2 and F_1 and connect at the input of the four full adders A_4 B_4, A_3 B_3, A_2 B_2, and A_1 B_1 to F_4, F_3, F_2 and F_1 respectively. Also, the output carry from the previous full adder is connected to the input carry of the next full adder. The complete circuit is shown in Figure 1. For implementing subtraction using Full Adders, we use 2's complement technique, i.e. take a complement of the subtrahend, add it to the minuend, and add 1 in the least significant position of the result, i.e. Difference.For example 8100010001's complement method -5-01011010 3001110010 +1 0011 or810001000 -810000111 01111 +1 0000 The subtraction circuit is shown in Figure 2. In Figure 3 a circuit realizing adding and subtracting functions is shown depending on the control line state. If 1 is placed on the control line, the two numbers are added, and if a 0 is placed on the control line, the two numbers are subtracted.

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Question:

Show that (v^2) does not equalv^2.

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/Users/wenhuchen/Documents/Crawler/Physics/D14-0497.htm

Solution:

Consider four particles with the following velocities: 1,2,3, and 4 cm/sec. The square of the average of v is (v^2) = [(1 + 2 + 3 + 4)/4]^2 cm^2/s2= (10/4)^2 cm^2/s^2 = 6.25 (cm/sec)^2 whereas the average of v^2 is v^2 = {[(1)^2 + (2)^2 + (3)^3 + (4)^4]/4} cm^2/s^2 = [(1 + 4 + 9 + 16)/4] cm^2/s^2 = 7.5 (cm/sec)^2 so that there is a substantial difference between the two methods of averaging. If the individual velocities are +1, -2, -3 and +4 cm/sec, there will, of course, be no change in the value ofv^2, but the average velocity v will be zero.

Question:

What is meant by the term "social" and "solitary" as applied to insects?

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Solution:

"Solitary" insects are those capable of many functions including sexual reproduction, food gathering and other types of non-reproductive behavior. "Social" insects are insects living in aggregations in which there is extensive division of labor and internal societal organization. Colonial organization in insect societies is similar to that in a coelenterate colony, except that the insects are not anatomically joined, as are the members of anObeliacolony. The individual insects of a society are incapable of independent survival, and both their structure and function are specialized for their particular role in the society. Division of labor in insect societies is based upon major biological dif-ferences between individuals. Exactly how different castes are determined is not fully understood, and the process is not the same in different insect species. Ants, termites, certain wasp and bee species are social insects. Insect behavior is based primarily on instinct, and is subject to relatively little modification, or learning. The individual has no choice of role in this system. Insect societies are not analogous to human societies. Most adult humans are capable of reproduction. Insect societies set apart several individuals for a solely re-productive role, while the other individuals are sterile and function in maintaining the colony. Human behavior is based-primarily on learning, and division of labor is not chiefly based on biological differences. Humans can survive for extended periods apart from society, and are able to change their role within a society.

Question:

Develop a pseudocoded program that converts the nodes of the following undirected graph into an adjacency matrix.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G07-0165.htm

Solution:

A graph is a collection of nodes and branches in which each branch links one node to another. The particular graph above is called an undirected graph because there is no particular direction associated with any of the branches. As you can see, the nodes are numbered. Two nodes that are connected by branches may be represented by pairs; for this graph, we notice the pairs: (1, 2), (1, 3), (1, 4), (2, 3), and (2, 5). One way of representing the graph is to use a list of the branches as defined by the pairs of nodes. We will look at the representation of the graph through an adjacency matrix. We define an adjacency matrix as one whose (i, j)th element is nonzero if nodes i and j are joined by a branch. Other-wise, the (i, j)th element is zero. For this graph, we have the adjacency matrix below: Notice that the nonzero elements are found at both the (i, j)th elements and the (j, i) th elements. This is because the graph is undirected: nodes may be accessed along branches from either side. For this simple graph, we store nodes in a 5 by 5 array. Other data structures may be used to represent graphs; a chained list could be used if the nodes are to be changed often. The pseudocode is presented below. After the array is filled with zeros, the (i, j)th elements and the (j, i)th elements are changed to ones. C TO CONVERT FROM PAIRS TO THE ADJACENCY MATRIX, C ASSUME THE BRANCHES ARE GIVEN AS PAIRS A(K) , B(K) integer A, B, I, J, K, m (5, 5) INPUT (A (1), B (1)), (A (2), B (2)) , ... , (A (5),B (5)) do for I \leftarrow 1 to 5 do for J \leftarrow 1 to 5 m (I, J) \leftarrow 0 end do for end do for C NOW CHANGE ELEMENTS TO NONZERO FOR PAIRS A (K), B (K) C FOR EACH PAIR K indicates the branch number and A and B the nodes of that branch. do for K \leftarrow 1 to 5 m (A (K), B (K)) \leftarrow 1 m (B (K), A (K)) \leftarrow 1 end do for output m (K, K) end program

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Question:

Calculated the resultant force, on the charge q_3 in the figure.

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Solution:

The force exerted by q_1 on q_3 is F_1 = q_1q_3 / R^21 = {(50) (\rule{1em}{1pt}5)} / (10)^2 = \rule{1em}{1pt}2.5 dyne The negative sign denotes an attractive force. The force exerted on q_3 by q_2 is F_2 = q_2q_3 / R^22 = {(20) (\rule{1em}{1pt}5)} / (30)^2 = \rule{1em}{1pt}100 / 900 = \rule{1em}{1pt}0.111 dyne Since q_2 is positive and q_3 is negative this force is attractive and is directed to the right toward q_2. The resultant force on q_3 is F_R = F_1 \rule{1em}{1pt} F_2 = \rule{1em}{1pt}2.5 \rule{1em}{1pt} (\rule{1em}{1pt} 0.111) = \rule{1em}{1pt}2.389 dyne and is directed to the left.

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Question:

What is modular programming?Name the six functional Name the six functional modulesof a structured program.

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Solution:

Modular programming is a term which describes a disciplined meansof writing computer programs. By writing program segments which performspecific tasks, you will find that it is easier to debug your program. This concept is a step to take some of the "art" out of computer programmingand to replace It with more logical, intelligent, and efficient techniques. There are six types of modules needed in almost every type of program. The table below gives an overview of these modules. FUNCTIONAL MODULE DESCRIPTION 1) INPUT Performs the input of data necessary to accomplish the task 2) INPUT DATA VALIDATION Checks the validity of the incoming data and diagnoses what type of error if any, has occurred 3) PROCESSING Completes the basic task requested of the algorithm 4) OUTPUT Performs output of information needed by the user 5) ERROR SPECIFICATION Analyzes problematical points in the execution of the program and takes appropriate action 6) CLOSING Finalizes program execution The above modules should stand alone in their meaning. That is, If definedthoroughly enough, the module should be readily adaptable to anotherprogram with only minor changes in variable names. In addition, eachmodule should have only one point of entry and one point of exit. Multiple returns and branches, if used excessively, tend to bewilder the readerof your program. The term robustness is used in programming to mean the degree of completenessa program offers. A robust algorithm takes care of almost all possibilitieswith swift and precise action. The idea of modular programming, if followed carefully, yields robustness.

Question:

A small pebble lies at the bottom of a tank of water 24 feet deep. Determine the size of a piece of cardboard which, when floating on the surface of the water, directly above the pebble, totally obscures the latter from view.

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Solution:

We can consider the pebble as a point source of light. If the cardboard is big enough for its purpose, then all rays of light from the pebble which would be refracted into the air at the surface must be blocked off by the cardboard, and all rays striking the surface of the water outside the cardboard must be totally internally reflected (see figure). The cardboard must obviously be circular, and, if its center is directly above the pebble, a ray of light striking the edge of the cardboard must do so at an angle \textphi \geq the critical angle. By Snell's Law, n_w sin \textphi = n_a sin \texttheta where \textphi is the angle of incidence, \texttheta is the angle of refraction, n_a is the index of refraction of air (n_a = 1) and n_w is that of water. Then, n_w sin \textphi = 1 because \texttheta = 90\textdegree to satisfy the requirement that \textphi be a critical angle. Therefore, using the figure n_w [r/\surd(r^2 + d^2)] = 1 r^2 + d^2 = n^2_wr^2orr^2 = [d^2/(n^2_w - 1)] r = [24 ft/\surd{(16/9) - 1}] = (72 ft/\surd7) = 27.2 ft.

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Question:

Find the pH of 0.10 MHOAcsolution that has 0.20 MNaOAc dissolved in it. The dissociation constant ofHOAcis 1.75 × 10^-5.

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Solution:

To find the pH of this solution, calculate the H_3O^+ concentration; pH is defined as - log [H_3O^+], This necessitates considering the dissociation of acetic acid,HOAc, since the only source of H_3O^+ is from this dissociation. The dissociation reaction may be written HOAc + H_2O \rightleftarrows H_3O^+ +OAc^-. To find the H_3O^+ concentration, the equilibrium constant expression must be written. This expression measures the ratio of the concentrations of products to reactants, each raised to the power of their respective coefficients in the chemical equation. Thus, K_a = {[OAc^-] [H_3O^+]} / [HOAc]. (Note: water is excluded since it is considered a constant.) Let x = [H_3O^+] at equilibrium. The [OAc^-] at equi-librium will be from two sources: the dissociation ofHOAcand the presence ofNaOAc, which exists as ions Na+ andOAc^-. From the dissociation, the concentration is x, since the reaction indicates that H_3O^+ andOAc^- are formed inequimolar amounts. Since one starts with 0.2 MNaOAc, one has a concentration of 0.2 M forOAc^-. The total is x + 0.2. If one starts with 0.1 MHOAcand x M of H_3O^+ orOAc^- form, then, at equilibrium, one has 0,1 - x left ofHOAc, Substituting these values into the equi-librium constant expression, and noting that K_a = 1.75 × 10^-5 (given), K_a = {x (0.2 + x)} / (0.1 - x) \approx (0.20 x) / (0.1) =1.7 × 10^-5. The acetic acid is weak; its dissociation to H_3O^+ is low, thus, one can make the above approximations. Solving for x; x = 8.5 × 10^-6 M = [H_3O^+]. pH = - log [H_3O^+] = - log [8.5 × 10^-6] = 5.07.

Question:

At the position x_0 = 1 unit on a screen, the wave function for an electron beam has the value +1 unit, and in an interval\Deltax and around x_0 = 1 there are observed 100 light flashes per minute. What is the intensity of flashes at x_0 = 2, 3, and 4 units whereihas the values +4, +2, and -2 units, respectively?

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Solution:

In this problem, one can imagine that we are sending photons into an electron cloud, and recording photon-electron collisions (light flashes) in order to find the intensity ofiat various points. \verti(x_0)\vert^2 represents the probability density for the electron's wave function and, the probability of finding the electron in an interval\Deltaxabout x_0 is \verti(x_0)\vert^2\Deltax. But the intensity of light flashing at x_0 is proportional to \verti(x_0)\vert^2\Deltax . This is due to the fact that a light flash, as was said above, cor-responds to a collision between a photon and an electron. The photon rebounds and reaches the observer. The number of flashes per unit time at a point therefore, indicates the probability of finding an electron at the point. Introducing a constant of proportionality A, we can write this relation as I(x_0) = A \verti(x_0)\vert^2\Deltax Therefore, I(x_0 = 1) = A \verti(x_0 = 1)\vert^2\Deltax = A \vert + 1\vert^2\Deltax =A\Deltax = 100 flashes/min so that A\Deltax= 100 flashes/min ThenI(x_0 = 2) = A \vert + 4\vert^2\Deltax = 16A\Deltax = 1600 flashes/min I(x_0 = 3) = A \vert+2\vert^2\Deltax = 4A\Deltax = 400 flashes/min I(x_0 = 4) = A \vert- 2\vert^2\Deltax = 4A\Deltax = 400 flashes/min Notice that even thoughi(x_0) differs in sign for the last two cases, the intensity is the same because the intensity is proportional to the square of the wave function.

Question:

For a condensation polymerization of a hydroxyacid in which 99% of the acid groups are used up, calculate the average number of monomer units in the polymer molecules, (b) the probability that a given molecule will have the number of residues given by this value, and (c) the weight fraction having this particular number of monomer units.

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Solution:

(a) The average number of monomer units is equal to X_n = [1 / (1 - p)] whereX_n is the average number of units and p is the extent of the reaction. X n = [1 / ( 1 - 0.99)] = 100 (b) To determine the number of molecules that have a particular number of monomers examine the number of molecules of various molecular weights. To obtain the distribution of molecular weights, consider the possi-bilities \pi_i of finding molecules of various degrees of polymerization (those containing i number of monomers) in a partially polymerized sample. The probability \pi_1 of finding an unreacted A (or B) group in the mixture is (1 \rule{1em}{1pt} p). The probability \pi_2 that an AB has reacted with another AB to form ABAB is p(1 - p) , since the probability of two independent events is the product of then two in-dependent probabilities for having a bond p and not having a bond (1 - p). The probability \pi_3 ABABAB, that is, two successive bonds, is p^2(1 - p). To generalize, consider \pi_1 = (1 - p) \pi_1 = p(1 - p) \pi_1 = P^2(1 - p) \pi_1 = P^i-1(1 - p) Here, p = 0.99 and i = 100 as shown in the previous section. \pi_100 = 0 .99(100 - 1)(1 - 0.99) = 0.99^99 (0.01) = 3.7 × 10^-3 (c) The weight fraction, W_i, of polymers containing this particular number of monomer units in an i-mer {a polymer containing i units) is equal to W_i = i p^i-1 (1 \rule{1em}{1pt} p)^2 W_i = 100(0.99^99) (1 - 0.99)^2 = 3.70 × 10^-3 monomers.

Question:

A manufacturer can produce X units of an item at a cost given by the following equation: Cost = C = 15 + 0.001X^2 The relationship between the number of units and price is given by P = 35 - . 1X (the producer is a monopolist). How many units should the monopolist produce in order to maximize his total return?

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G18-0455.htm

Solution:

The maximization problem may be written as follows: Max {X.(P-C)} X where P = 35 - O.1X C = 15 + 0.001X^2 , X \geq 0 When X, the quantity produced is zero, there is zero pro-fit. Similarly, when the price of X equals the cost of X, profits are zero. This occurs when P = C or, 35 - 0.1X = 15 + 0.001X^2 Solving for X gives X = 100. Thus profits are zero when X = 0 and X = 100. Assuming profit is a continuous function of output Rolle's theorem may be applied to conclude that the maximum profit point lies in the interval (0,100). Numerical approximation of this maximum is performed on the computer using the GOLDEN SECTION method. To illustrate the method, suppose the graph of profit versus production looks as follows: In the interval [0,100] there is only one maximum. Suppose we pick two points, X_1 and X_2 in that interval and find their ordinates on the curve. If f(X2) > f(X1), then the interval of search may be reduced from [0,100] to [X_1,100] and the interval [0,X_1] can be eliminated. The original interval [X_L X_u] is equal to 1 (Fig. 2). Let \tau equal the ratio of the long subinterval to the total interval, i.e. (X_2 - X_L) /( X_u - X_L) = \tau Since and X1 and X2 have been located symmetrically about the center of the interval, the distance between X_1 and X_u is also \tau. From Fig. 2, the interval [X_L X1] will be elimi-nated after the first iteration since f(X_2) > f(X_1). In order for X2 to be properly located within the interval [X_1,X_u] for the next iteration, the following must be true: (X_u - X_2) / (X_u - X_1) = \tau But, from Fig. 2, X_u - X_2 = 1 - \tau and X_u - X_1 = \tau. Thus, [(1 - \tau) / \tau] = \tau,or \tau^2 + \tau - 1 = 0.(1) Solving (1) for \tau we obtain \tau = [(\surd5 - 1) / 2] = 0.618. Thus, according to the GOLDEN SECTION rule, X_2 should be 0.618 from X_L and X_1, should be 0.618 from X_u . For the given problem, the selection of points in the search interval would proceed as follows: X_1 = 100 - 0.618 × 100 = 100 - 61.8 = 38.2; X_2 = 0.618 × 100 = 61.8. From Fig. 1, f(X_2) > f(X_1) so the interval [0,38.2] is eliminated. The next point is located at 0.618 × (length of interval) = 0.618 × (61.8) = 38.2 from the left end of the interval, or X_3 = 38.2 + 38.2 = 76.4. Since f(X_2) > f(X_3), [76.4,100] is eliminated. X_4 is located at 0.618 × (length of interval) = 0.618 × (38.2) = 23.6 from the right end or 76.4 - 23.6 = 52.8. After three iterations the interval within which the optimum is located has been reduced from [0,100] to [52.8]. These values are now assigned to X_L and X_u. The process is continued until the length of the interval is less than some preassigned \epsilon > 0. The average of f(X) at final X_L and X u is then the required approximation to the optimum. The most efficient method of locating the points is the GOLDEN SECTION technique.

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Question:

The total resistance of a wire wound rheostat when "cold" is 300 ohms. In use it experiences a rise in temperature. How does the current which it draws on a 100 volt line after its temperature has risen 50\textdegree C, compare with that drawn at the start when it is "cold" ?

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/Users/wenhuchen/Documents/Crawler/Physics/D20-0664.htm

Solution:

This problem involves a combination of Ohm's Law and the dependence of resistance upon temperature. I = (V/R) (Ohm's Law) But R = R_0 (1 +\alphat) where \alpha for metals is .0038/\textdegreeC (approx). That is the resistance of the metal increases at a rate directly proportional to the increase in temperature. When "cold" R_1 = 300 ohms. \thereforeI_1 = (V/R_1 ) = (100/300) = .333 ampere But when "hot" R_2 = 300[1 + .0038 (50)] orR_2 - 300 = 300 (.0038) (50) = 3 × 3.8 × 5 × 10\textdegree = 57 \thereforeR_2 = 357 AndI_2 = (V/R_2 ) = (100/357) = .280 ampere ThusI_2 = (.280/.333) I_1 = .84 I1(approx).

Question:

Mrs. Jones has some money which she wants to invest in a number of activities (investment programs) in such a way that the total return is maximized. Assume that she has $8,000 for allocation and that the investments can only be integral multiples of $1000. Three investment programs are available. The return function for each program is tabulated below: Return Functions h_i (x) X(0000) 0 1 2 3 4 5 6 7 8 h_1(x) h_2(x) h_3(x) 0 0 0 5 5 4 15 15 26 40 40 40 80 60 45 90 70 50 95 73 51 98 74 52 100 75 53 Using the principles of dynamic programming, how would the optimal investment in each program be determined so as to maximize total return?

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Solution:

The problem can be solved recursively, i.e., the optimal solution for onestage is used as input for the next stage. Step 1: Assume that program 3 is the only program. Then the optimal returnfrom investing X = 0,1,2,...,8 inh_3(x) is given by the last row of the table above. In particular,h_3(8) = 53 isthe optimal return. Step 2: Now assume only programs 2 and 3 are available and that d_2 (x), d_3 (y) can be invested in programs 2 and 3, where x = 0,1,...,8 y = 0,1,2,...,8 and x + y = 0,1,2,...,8 Thus, with 8 we can invest 0 in 2 and 8 in 3, with 5 3 in 2 and 2 in 3 and so on. Find the optimum of all these choices for each x in 0,1,2,....8. The functional equation is f_2 (x) = max[g_2y + f_3(x)] y = 0,1,... ,8 - x x = 0,1,...,8 Step 3: The final stage is the same as the original prob-lem. We now assume all three programs are available. We examine the results of investing z in program 1 and 8 - z units in programs 2 and 3 (the optimal amounts for each 8 - z (z = 0,1,2, ..,8) have already been found in Step 2). The functional equation for the last stage is f_i ( x) = max [g_i(z) + f_i+1 (x - z) x = 0,...,8 Z = 0,1,.....,X d_i(x) = value of z that yieldsf_i(x) where f_i(x) is the optimal return from investing x units in programsi,i+ 1,..,3andd_i(x) is the optimal amount to invest in programiwhen x units are available to in-vest in programsi,i+ 1,..., 3 fori= 1,2,3.

Question:

A beam of electrons of kinetic energy T = 5 MeV impinges on a single slit of width 1 micron (10^-6 m), causing a diffraction pat-tern (see figure). Calculate the width of the central maximum at a plane 3 m behind the plane of the slit.

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Solution:

According to the de Broglie wave theory of matter every particle has wave-like characteristics. Electrons then should display dif-fraction effects when passing through a narrow slit, as do light waves. The wavelength associated with the moving electron is the de Broglie wavelength. \lambda = (h/p)(1) where p is the momentum of the electron and h is Planck's constant. The p appearing in (1) is the relativistic momentum of the electron. This quantity is related to the energy E of the electron by E^2 = p^2c^2 + m^2_0 c^4 where m_0 is the rest mass of the electron. The energy E is equal to the kinetic energy of the electron, T, plus the rest energy, m_0c^2 . E = T + m_0c^2 Hence, (T + m_0c^2)^2 = p^2c^2 + m^2_0c^2 T^2 + 2m_0c^2T = p^2c^2 Therefore p^2 = [(T^2 + 2m_0c^2T)/c^2] p = [\surd{T(T + 2m_0c^2)}/c] Now,m_0c^2 = (9.31 × 10^-31 kg)(9 × 10^16 m^2/s^2) = 83.79 × 10^-15 Joules because 1Joule = 6.24 × 10^18 eV m_0c^2 = .5 × 10^6 eV Hence, p = [\surd{(5 × 10^6 eV)(5 × 10^6 eV + 1 × 10^6 eV)} / (3 × 10^8 m/s)] p = [\surd{(30.0 × 10^12 eV^2)} / (3 × 10^8 m/s)] p = [(5.48 × 10^6 eV) / (3 × 10^8 m/s)] p = 1.83 × 10^-2 (eV\textbullets/m) whence,\lambda = (h/p) = [(6.63 × 10^-34 J\textbullets)/{1.83 × 10^-2 (eV\textbullets/m)}] since 1Joule = 6.24 × 10^18 eV \lambda = [{(6.63 × 10^-34)(6.24 × 10^18)m} / (1.83 × 10^-2)] = 22.6 × 10^-14 m = 2.26 × 10^-13 m The diffraction pattern due to the electron waves may be treated as if it were caused by the diffraction of light waves. We may then use the formula for the maxima produced by diffracting light rays to analyze the maxima produced by electron waves. With respect to the diffraction pattern (see the figure), we have d = 10^-6 m and L = 3 m. x_1 = (\lambdaL/d) = [(2.26 × 10^-13 m × 3m)/(10^-6 m)] = 6.8 × 10^-7 m The width of the central maximum will be 2x_1 = 1.4 × 10^-6 m, which would be difficult to observe. If the electron beam were of lower energy, 50,000 eV for example, the central maximum would have a width of 3.2 × 10^-5 m (0.032 mm), which is more easily measured.

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Question:

What is meant by the statement "ontogeny recapitulates phylogeny"? With the advent of knowledge in genetics, how has this theory of recapitulation been modified and what significance does its present form have?

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/Users/wenhuchen/Documents/Crawler/Biology/F28-0745.htm

Solution:

The idea of recapitulation was first proposed by Ernst Haeckel in 1866, in the principle that "ontogeny recapitulates phylogeny.\textquotedblright Haeckle claimed that embryos, in the course of development, recapitulate, or repeat, the evolutionary history of their ancestors in some abbreviated form. Thus, if mammals evolved from reptiles, which evolved from amphibians, which in turn evolved from fish, then a mammalian embryo should first resemble a fish, then an amphibian, then a reptile, and finally a mammal. Through research and experimental findings, later workers modified Haeckle's idea and developed the modern theory of recapitulation. This theory states that the embryos of the higher animals resemble the embryos of lower forms, not the adults, as Haeckle had believed. An embryo passes through developmental stages similar to some of those which lower animal forms pass through. Ontogeny does not repeat the adult stages of phylogeny, but it does repeat, in an altered form, some of the ontogenic characteristics of ancestral forms. Inherent in the theory of recapitulation is the existence of an evolutionary relationship between animal groups. There is also the implication that a species arises from some previous species and may give rise to new ones. Because embryonic development must primarily be controlled by genes, ontogeny can be rightfully thought of as an expression of the inherited genetic composition of an organism. Differences in ontogeny can be related, then, to differences in genetic composition. The modern theory of recapitulation thus also holds that a given species expresses a slightly altered version of the ontogenic program that it inherited genetically from the previous species, which in turn acquired its developmental program genetically from its ancestor. According to this, a species will closely resemble, both genetically and ontogenically, its recent ancestors and less closely its more distant ones. In other words, the more remote the ancestry, the fewer the similarities. It is also observed that in remote relationships, the similarities are restric-ted to the very early stages of embryonic development. Beyond those early stages evolutionary changes have grossly altered the direction of development. Thus, the ontogenies of fish and mammals resemble each other only in their earliest stages, while the ontogenies of reptiles and mammals show a close resemblance through later stages, the reptiles being a more recent ancestor of the mammals than the fish (see Figure). The theory of recapitulation, as we can see, is significant for three reasons. First, it confirms that evolution does exist; second, it shows that evolution is a genetic occurrence; and third, taken with due caution, it may indicate the evolutionary hierarchy of some animals.

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Question:

It is known that 1 ton of TNT can produce 4.6 × 10^9 J of energy. A chemist wanted to obtain this much energy from acetylene (C_2 H_2) and oxygen (O_2) . How much C_2 H_2 and O_2 would be required? The ∆H\textdegree = - 2510 KJ for the reaction between C_2 H_2 and O_2.

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Solution:

The equation for this reaction is 2 C_2 H_2 + 5O_2 \rightarrow 4CO_2 + 2H_2O, ∆H\textdegree = - 2510 KJ. In a chemical reaction, energy is usually liberated or absorbed The ∆H\textdegree enthalpy, expresses this energy liberation or absorption quantitatively. The negative sign denotes the release of energy. Two moles of acetylene and 5 moles of oxygen react to form 2510 KJ or 2510 × 10^3 J of energy. Similarly, 2x moles of acetylene and 5x moles of oxygen produce 2.510 × 10^6 × J of energy. In this case, the energy released must equal 4.6 × 10^9 J. Therefore, 2.510 × 10^6 x = 4.6 × 10^9 , or x = 1.83 × 10^3. From the equation for the reaction one knows that 2 moles of C_2 H_2 and 5 moles of O_2 react to produce 2510 KJ. The calculations already done show that to produce as much energy as 1 ton of TNT there must be 1.83 × 10^3 × 2 moles of C_2 H_2 and 1.83 × 10^3 × 5 moles of O_2 to form 4.6 × 10^9 J. To solve for the amount of each needed, multiply the number of moles by the molecular weight of the substance. (MW of C_2 H_2 = 26). wt of C_2 H_2= (1.83 × 10^3 × 2)moles × 26 g/moles = 9.52 × 10^4 g = 9.52 × 10^4 g × 1.10 × 10^-6 tons/g = .1044 tons. (MW of O_2 = 32) wt of O_2= (1.83 × 10^3 × 5) moles × 32 g/mole = 5.86 × 10^4 g = 5.86 × 10^4 g × 1.10 × 10^-6 tons/g = 0.646 tons.

Question:

What are the rules regarding the specification of a procedure?

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Solution:

The rules for the specification of a procedure are as follows. 1. The procedure block may appear in any part of the PL/I program bodyin which a Declare statement is allowed. 2. Each procedure block must start with a procedure statement whichcontains at least one name for the proce-dure .E.g., GRAVY: BREAD: PROCEDURE; In the above, BREAD and GRAVY are names of the procedure. 3. Each procedure block must end with an END procedure-name statement.E.g., END BREAD; 4. When a procedure statement is encountered by the program controlduring execution, the program control by-passes the whole procedureblock up to the end statement of the procedure. 5. Program control will execute the procedure only when the procedureis invoked by a procedure reference. A procedure reference maybe a function reference or a sub-routine reference.E.g., CALL BREAD; Y = U\textasteriskcenteredSHIP(X,Y); In the above, BREAD and SHIP are procedures which are ex-plicitly invokedas part of a program. The procedures will then be executed.

Question:

Distinguish between a cistron, a muton, and a recon. Which of these is the largest and which is the smallest?

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/Users/wenhuchen/Documents/Crawler/Biology/F24-0621.htm

Solution:

A cistron is defined as the genetic unit of biochemical function. One cistron corresponds to one functional unit (one gene coding for a polypeptide chain, protein, or enzyme) on a chromosome. A cistron can be defined by means of complementation tests. Complementation refers to the production of a normal phenotype through the combined activities of two chromosomes, each of which alone is incapable, due a recessive mutation, of producing a normal phenotype. If two homologous chromosomes contain mutations in different cistrons, the normal gene of one can compensate for the mutated gene of the other, and vice-versa, when the two chromosomes are in combination with each other in the organism (see figure 1) . If each chromosome of the pair contains a mutation within the same cistron, neither can compensate for the other, and a mutant phenotype will result. A cistron is thus the unit within which two chro-mosomes can have mutations and not complement each other when they occur in combination. It follows then that a cistron is the smallest unit capable of producing a gene product, and is equivalent to a gene. A muton is defined as the smallest unit which, if altered, gives rise to a mutant phenotype. A muton may thus be as small as a single nucleotide, since a change in a single nucleotide can cause a mutant phenotype (think, for example, of sickle-cell anemia). Of course, some single -point mutations may not result in a mutant phenotype. If a given codon were changed via a single base substitution, for example, from CCA to CCG, that codon would still specify proline, due to the degeneracy of the genetic code. A recon is defined as the smallest unit that is interchangeable, but not divisible, by recombination. In other words, recons are the smallest units between which crossing over can occur. This is known to be equivalent to a single nucleotide. This is because the breakage point in crossing over is between any two given nucleotides. Given two crossovers, it is thus possible for a single nucleotide to be exchanged (see figure 2). FIGURE 2.Double crossover between two strands of DNA, the base sequences of which are shown. Since a cistron consists of a sequence of mutons or recons, it is the largest of the three. A recon may be as small as a single nucleotide, as may be a muton.

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Question:

A 2.0-kg ball traveling with a speed of 22 m/sec over-takes a 4.0-kg ball traveling in the same direction as the first, with a speed of 10 m/sec. If after the collision the balls separate with a relative speed of 9.6 m/sec, find the speed of each ball.

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Solution:

No external forces act on the system. Since this is a collision problem, the principle of con-servation of momentum can be used. Letting u_1 and u_2 be the initial velocities and v_1 and v_2 the final velocities of the two masses m_1 = 4.0 kg and m2= 2.0 kg, we find m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2(1) (4.0 kg) (10 m/sec) + (2.0kg)(22 m/sec) = (4.0 kg)v_1 +(2.0 kg) v_2 (2.0 kg)v1+ (1.0 kg)v_2 = 42 kg-m/sec or2.0 v_1 + v_2 = 42 m/sec(2) The difference in speed of the two masses after collision is given as 9.6 m/sec. Since the two masses initially move along the same axis, they must continue to do so after the collision, assuming their center of mass lies along this axis of motion. Since, physically, the 2.0 kg ball cannot pass through the 4.0 kg ball, the 4.0 kg ball continues to have greater velocity in the initial direction of motion. Taking this initial direction as positive, v_1 - v_2 = 9.6 m/sec Substituting v_2 = v_1 + 9.6 m/sec in equation (2), we find the final velocities to be v_1 = 17.2 m/sec v_2 = 7.6 m/sec Both balls continue to move in the initial direction but with their speeds changed.

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Question:

One of the two most common ingredients in medication designed for the relief of excess stomach acidity is aluminum hydroxide (Al (OH)_3, formula weight = 78 g/mole). If a patient suffering from a duodenal ulcer displays a hydrochloric acid (HCI, formula weight = 36.5 g/mole), concentration of 80 × 10^-3 M in his gastric juice and he produces 3 liters of gastric juice per day, how much medication containing 2.6 g Al(OH)_3 per 100 ml of solution must he consume per day to neutralize the acid?

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Solution:

Al (OH)_3 neutralizes HCI according to the following reaction: Al (OH)_3 + 3HCI \rightarrow AlCl_3 + 3H_2O. When all the HCI has been neutralized, the number of equivalents of Al (OH)_3 is equal to the number of equi-valents of HCI. But since the number of equivalents is equal to the product of the normality, N, and the volume, V, this condition may be written as N_AI _(OH)3 V_AI _(OH)3 = N_HCI V_HCI. We must solve for V_AI(OH)3 v = [(N_HCI V_HCI) / (N_AI(OH)3)] Themolarityof the Al (OH)_3 solution is equal to the number of moles of Al (OH)_3 divided by the volume in liters. To determine the number of moles corresponding to 2.6 g Al (OH)_3, we divide 2.6 g by the formula weight of Al (OH)_3. We then divide this by 100 ml = 100 ml × 1 liter/1000 ml = 0.100 liter to obtain themolarity. Hence, concentration of Al (OH)_3 = [(2.6 g)/(100 ml)] = [(2.6 g/78 g/mole)/(0.100 liter)] =0.33 M. Since HCI has oneionizableproton, its normality is equal to its molarity, or N_HCI = 80 × 10^-3 M = 80 × 10^-3 N. Al (OH)_3 contains three hydroxyl groups, hence, its normality is equal to three times itsmolarity, or N_Al_(OH)3 = 3 × 0.33 M = 1.0 N. The required volume of medication is then V_AI _(OH)3 = [(N_HCI V_HCI)/(N_AI(OH)3)] = [(80 × 10^-3 N × 3 liters)/(1.0 N)] = 0.240 liter = 0.240 liter × 1000 ml/liter = 240 ml.

Question:

A combinational system is to receive a decimal value encoded according to code I. It also receives a control signal, J, that affects the output of the system. When J is TRUE, the input is converted to its equivalent in code II. When J is FALSE the output is chosen from code III. Design the combinational system for implementing this process. Consider both sum of products (SOP) and product of sums (POS) forms. Minterm Code I ABCD Code II WXYZ Code III WXYZ 1 2 4 8 9 10 12 13 14 15 0001 0010 0100 1000 1001 1010 1100 1101 1110 1111 0000 0001 0010 0011 0100 0101 0110 0111 1000 1001 0011 0100 0101 0110 0111 1000 1001 1010 1011 1100

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Solution:

A five variable K-map is used to minimize each output. Notice that there are don't care minterms, namely, 0,3,5,6,7, and 11. Half of the K-map is used for J = 1 and the other half for J = 0. A K-map is drawn for each output W,X,Y, and Z. It is seen from the W K-map that W =JAB + BC +JAC and in the POS form: W = (A) (J+A+ C) (A+ B + C) (J+ D + B) . Similarly: X =JAD+ ABD +JCD+JABC+ JABCD+JABC X = (J+A) (A+D) (J+A+B+C) (J+A+C+D) (J+B+C+D) (J+A+B+C) Y =JCD + J BC+ ABCD+JB CD Y = (J+C) (C+D) (B+C) (J+B+C+D) (J+B+C+D) Z = JBD + J ABD+ JBCD+JBD +JBD Z = (J+B+D) (J+A+D+B) (J+B+C+D) (J+B+D) (J+B+D) To decide which method would be cheapest to implement, SOP or POS, we will look at the number of gates required for implementation. The table of fig. 1 shows the number of gates needed to implement the SOP and the POS form of each Variable. It is seen from fig. 1 that W and Y are best expressed in the SOP form because the SOP form takes the least amount of gates. X and Z can be expressed in either SOP or POS form as both require the same number of gates. We will express X in POS form and Z in SOP form. Figure 2 shows the circuit implementation with W,Y, and Z expressed in the SOP form and X in the POS form.

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Question:

(a) You desire to plate 5g of Ag from a solution of Ag^+ . Determine the number of amperes required over a period of 1000 seconds. (b) With a current of 0.5 ampere, how much time is required to plate out 10g of gold from an Au^3+ solution?

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Solution:

Both of these problems can be solved using the equation q = I(t_2 - t_1) = nF, where q is the charge in coulombs, I the current in amperes, (t_2 - t_1) the difference in time, n the number of equivalents oxidized or reduced, and F Faraday's constant, 96,490 coulombs. For part (a), the time span , t_2 - t_1 , and the weight of material to be reduced is known. The number of equivalents to be reduced (or oxidized) equals the weight of material divided by the gram-equivalent weight. The atomic weight of Ag = 107.88 (g / mole) and the value of n = (5g of Ag^+) / (107.88 g/equiv) . Substituting the known values into the equation I = (nF)/(t_2 - t_1) = [(5/107.88)(96.490)] / (1000) = 4.45 amps. For part (b), the time span, t_2 - t_1, is unknown. The atomic weight of Au = (97 g/mole. Using the same equation n = (10g Ag^+++) / (65.7 g/equiv.); Au^3+ requires 3 electrons to be reduced and the atomic weight must be divided by 3. Therefore, (t_2 - t_1) = (nF) / (I) = [(10g/65.7) (96,490)] / [(0.5 coulombs/sec)] = 29,400 seconds.

Question:

Develop a sorting program that arranges the elements of the array A (1:N) in ascending order. Use your knowledge of binary trees and heaps to complete the problem.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G07-0161.htm

Solution:

A sort of this type is referred to as a heap sort. While not as conceptually accessible as the insertion sort or bubble sort, the heap sort is most efficient, having a run time proportional to nlog_2n. A binary tree is based on principles similar to those which are used in the creation of a family tree. The tree consists of nodes (the uncircled numbers in Fig. 1) and branches (the lines connecting individual nodes). At the top of the tree, we have the root node. Two is the maximum number of branches that can originate from any single node. Each node is labeled (the cirled integers in Fig. 1). The notation for the labels is NODE (I), where I is the en-circled integer. In this example, NODE (1) = 6, NODE (3) = 5, and NODE (8) = 1. Remaining consistent with the idea of a family tree, we may say that any node from which branches originate is a father. The father may have right and left sons, which are the elements which emanate from the father. You can probably guess that each node can have ancestors and descendants. The root node, however, may be considered Adam, since it has no ancestors. Notice the method of labeling in the tree. The root node is always N0DE (1). After that, the rule is: If a node has label I, its left son is always 2I, and its right son is always 2I + 1. Another restriction on the tree is that there are no holes. In other words, if NODE (I) exists, then so do nodes 1, 2, ... , I - 1. To translate a binary tree into a heap, remember the following caveat: A binary tree is a heap only if for any node I, NODE (each ancestor of I) \geq NODE (I). Now, to calcu-late fathers and sons, we recognize the following: FATHER (A (I)) is B (FLOOR (I / 2)) LEFTSON (A (I)) is B (2I) RIGHTSON (A (I)) is B (2I + 1) First, we must make B (N) into a heap, and then we must sort the heap in ascending order. Notice that nodes 1 - 8 are already in order. Node 9, however, is out of order because its value, 8, is larger than the value in node 4, which is 3. We must move the unordered value up the tree until the order is restored. This is accomplished by swap-ping values of father and son until the condition NODE (each ancestor of I) \geq NODE (I) is satisfied. This is the swapping algorithm, written in PL / 1, which makes A (N) into a heap: / \textasteriskcentered TRANSLATION OF BINARY TREE INTO HEAP \textasteriskcentered / HEAP_IT: DO I = 1 TO N BY 1; / \textasteriskcentered MAKE A (1:1) INTO A HEAP, ASSUMING \textasteriskcentered / / \textasteriskcentered A (1:I - 1) IS ALREADY HEAPED \textasteriskcentered / DO WHILE (J > 1); FATHER_J = FLOOR (J / 2EO); IF A (FATHER_J) > = A (J) THEN GO TO TERM; SWAP_TEMP = A (FATHER_J); A (FATHER_J) = A (J); A (J) = SWAP_TEMP; J = FATHER_J; END; TERM: ; END HEAP_IT; Consider the number of times the DO WHILE loop is executed. Each time the loop is traversed, the unordered value is moved up one level in the tree. Hence, the maximum number of executions possible is the number of levels minus one. On reflection, you can see that the level is exactly FLOOR (LOG2 (I)) +1. For a given value of I, the maximum number of loops is LOG2 (I). times. The total number of times the inner loop body is executed is LOG2 (1) + LOG2 (2) + ... + LOG2 (N) , a sum which has as an upper bound N\textasteriskcenteredLOG2 (N). Now, the heap A satisfies the relation A (FATHER (I)) \geq A (I). To output a sort in ascending order, we want A to satisfy the relation A(I) \leq A (I + 1). In other words, the largest value must be put in the last node of the tree. This will destroy the heap relation, such that only the value A (I) will disrupt the heap. / \textasteriskcentered SORT HEAP A (1: N). \textasteriskcentered / I = N; -SWAP: DO WHILE (I > 1); TEMP = A (1); A(1) = A(I); A(I) = TEMP; I = I - 1; / \textasteriskcentered MAKE A (1: l) INTO A HEAP, WITH THE \textasteriskcentered / / \textasteriskcentered VALUE A (1) THE ONLY DISRUPTER OF THE HEAP \textasteriskcentered / J = 1; / \textasteriskcentered K is J's \textasteriskcentered / K = 2\textasteriskcenteredJ; / \textasteriskcentered LARGER SON \textasteriskcentered / DO WHILE (K < = I); If K < I THEN IF A (K)< A (K + 1) THEN K = K + 1; If A (J)> = A (K) THEN GO TO DUMMY_END; A_TEMP = A (J); A (J) = A (K); A (K) = A_TEMP; J = K; K = 2\textasteriskcenteredJ; END; DUMMY_END: ; END SWAP;

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Question:

A women feeds her new cat daily with canned cat food. As she opens the cans with a noisy electric can opener, she calls the cat. The cat comes in response, to its name, and the woman feeds it. Why does the cat eventually come running when it hears the electric can opener?

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Solution:

Many animals can learn to associate two or more stimuli with the same reward or punishment. This behavior, called associative learning, was first demon-strated by Pavlov, a Russian psychologist. When Pavlov placed meat powder in a dog's mouth, the dog salivated. Pavlov then added a new stimulus, the ticking of a metro-nome, at the same time the meat was given. After a number of such pairings, the dog was only permitted to hear the metronome, without any presentation of meat. It still salivated. The dog associated the ticking of the metronome with the presentation of food, and hence salivated even when no food was given. Associative learning is called classic reflex conditioning, when the response is a reflex, such as salivation. The cat and can opener example is a variation of Pavlov's experiment. When the woman calls the cat by its name, it responds by running to her. The cat is rewarded or positively reinforced with food. A short time before the cat is fed, it hears the sound of the can opener. After a while the woman stops calling the cat. However, it still comes running when it hears the sound of the can opener. The cat has associated the sound of the can opener with the presentation of food, and it responds to that sound by running to the woman even when she is silent. One would expect this associative learning to continue as long as the cat is still given food (reward). If the cat is no longer fed after responding to the sound of the can opener, it will eventually stop responding to the can opener. The lack of reward will extinguish the behavior.

Question:

Define the following terms, making clear the distinctions between them:a) Bitb) Bytec) Word.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G01-0004.htm

Solution:

a) The bit is the most elementary data unit that corresponds to a binary digit in computer's memory. Generally the binary 1 is represented as high voltage and binary 0 is represented as low or negative voltage in a digital computer's registers and memory. However, this also depends on the nature of media. For example, on magnetic disks and tape reels, the bit is stored as traces of magnetism. The bit is also typically used to measure the rate at which data is transmitted over communica-tion media (e.g. twisted copper wire, coaxial cable, air, etc). A kilobit per second is one thousand bits per second, a megabit per second is one million bits per second. As of 1990, we do not yet speak of gigabits per second since there are no medium over which that high a communication rate can be achieved. b) The byte is a group of 8 bits. A byte may represent ASCII (American Standard Code for Information Interchange) char-acters or numbers in the computer. Byte is the basic unit for expressing storage capacities in computers. A kilobyte is a thousand bytes, a megabyte is a million bytes, and a gigabyte is one billion bytes. c) The word is less well defined. Its definition is depen-dent upon the hardware architecture of the computer. In general, the term "word" is used to indicate the length of registers and addressable memory locations. If the computer has an 8-bit pro-cessor (CPU), such as INTEL 8080A, then the word is 8 bits. If the computer has a 16-bit processor, such as PDP-11, then the word is 16 bits. Finally, if the computer has a 32-bit proces-sor, then the word is 32 bits. The term "word", other than indi-cating the smallest addressable units in a computer's memory, is not a very useful measure since it is not well defined. Perhaps the best definition is that a word is 16 bits (two bytes), a double word is 32 bits (four bytes).

Question:

Design a program that can guess any integer from 100 to 999999999 by asking questions of the user about the sums of certain digits. Use BASIC.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G21-0509.htm

Solution:

Let us choose 718999 as our number to be "guessed" . The algorithm we shall use requires several pieces of information . First, you must type in the number of digits, which in this case is 6. Then you must type in the sums of adjacent digits starting from the left, until you reach the last digit. Finally, the sum of the second and the last digits from the left are needed if the number is even; the sum of the first and last digits, if the number is odd. At this point, the array T has stored the following quantities: T (0) = 8T (3) =18 T (1) = 9T (4) = 18 T (2) = 17T (5) = 16 Now we hit the loop in which S is computed. Let us write down these steps as they will be executed: S = 0 + (9 \textasteriskcentered 1) = 9 S = 9 + (17 \textasteriskcentered (-1)) = -8 S = -8 + (18 \textasteriskcentered 1) = 10 S = 10 + (-18) = -8 S = -8 + 10 = 2 S is then divided by 2 and stored in T(6). The final loop calculates each digit, multiplies by the appropriate factor of 10, and adds the totals to give the answer. 5 DIM T(15) 7 PRINT 9 PRINT "WRITE ANY NUMBER FROM 3 TO 9 DIGITS ON A PIECE OF PAPER" 10 PRINT "WHEN READY, TYPE A ZERO AND RETURN" 11 INPUT W 12 IF W < > 0 THEN 65 13 LET S = 0 14 LET E = 0 15 LET H = -1 16 MATT = ZER 17 PRINT "TYPE IN THE NUMBER OF DIGITS IN YOUR NUMBER" 18 PRINT 19 INPUT N 20 LET G = N/2 21 IF G <> INT(N/2) THEN 23 22 LET E = 1 23 FOR J = 0 TO N - 2 24 LET K = J + 1 25 PRINT 26 PRINT "WHAT IS SUM OF DIGIT" K "AND DIGIT" K + 1 "?" 27 PRINT 28 INPUT T(J) 29 NEXT J 30 PRINT "WHAT IS SUM OF DIGIT" E + 1 "AND LAST DIGIT?" 31 PRINT 32 INPUT T(K) 36 FOR J = E TO N - 1 38 LET H = -1 \textasteriskcentered H 40 LET S = S + T(J) \textasteriskcentered H 41 NEXT J 42 LET S = S/2 43 LET T(K + 1) = S 44 LET G = 1 45 LET S = 0 46 FOR J = 0 TO N - 1 50 LET L = N - J - 1 52 LET T(L) = T(L) - T(L + 1) 53 LET S = S + T(L) - T(L + 1) 54 LET G = 10 \textasteriskcentered G 56 NEXT J 58 PRINT 60 PRINT "I BET YOUR NUMBER IS" S 62 GO TO 7 65 END

Question:

Explain why blood is such an important tissue in many animals. Discuss the major functions of blood.

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/Users/wenhuchen/Documents/Crawler/Biology/F14-0344.htm

Solution:

All cells, in order to survive, must obtain the necessary raw materials for metabolism, and have a means for the removal of waste products. In small plants and animals living in an aquatic environment, these needs are provided for by simple diffusion. The cells of such organisms are very near the external watery medium, and so nutrients and wastes do not have a large distance to travel. However, as the size of the organism increases, more and more cells become further removed from the media bathing the peripheral cells. Diffusion cannot provide sufficient means for transport. In the absence of a specialized transport system, the limit on the size of an aerobic organism would be about a millimeter, since the diffusion of oxygen and nutrients over great distances would be too slow to meet the metabolic needs of all the cells of the organism. In addition, without internal transport, or-ganisms are restricted to watery environments, since the movement to land requires an efficient system for material exchange in non-aqueous surroundings. Therefore, larger animals have developed a system of internal transport, the circulatory system. This system, consisting of an exten-sive network of various vessels, provides each cell with an opportunity to exchange materials by diffusion. Blood is the vital tissue in the circulatory system, transporting nutrients and oxygen to all the cells and removing carbon dioxide and other wastes from them. Blood also serves other important functions. It transports hormones, the secretions of the endocrine glands, which affect organs sensitive to them. Blood also acts to regulate the acidity and alkalinity of the cells via control of their salt and water content. In addition, the blood acts to regulate the body temperature by cooling certain organs and tissues when an excess of heat is produced (such as in exercising muscle) and warming tissues where heat loss is great (such as in the skin). Some components of the blood act as a defense against bacteria, viruses and other pathogenic (disease-causing) organisms. The blood also has a self-preservation system called a clotting mechanism so that loss of blood due to vessel rupture is reduced.

Question:

Iodine pentoxide (I_2O_5) is a very important reagent because it can oxidize carbon monoxide (CO), which is a major pollutant of the atmosphere, to carbon dioxide (CO_2) in the absence of water. The reaction involved is I_2O_5 + 5CO \rightarrow I_2 + 5CO_2 How many kg of I_2O_5 is needed to completely clear 140,000 kg of CO from the air?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E24-0848.htm

Solution:

From the stoichiometry of this equation, 1 mole of I_2O_5 completely reacts with 5 moles of CO. To solve this problem, calculate the number of moles of CO that will re-act by dividing its weight by its molecular weight of 28 g/mole. Determine, also, how many moles of I_2O_5 will completely react with this, and then calculate the weight of I_2O_5 that will completely react by multiplying the number of moles of I_2O_5 by its molecular weight (MW of I_2O_5 = 334 g/mole). Thus, number moles CO = (140,000,000 g CO) / (28 g/mole) = 5.00 × 10^6 mole. This means that one fifth of this number of moles will be the moles of I_2O_5 required. Recall, there will exist 5 moles of CO for every mole of I_2O_5. Therefore, moles I_2O_5 = (1/5) (5 × 10^6) = 1 × 10^6 moles. This means that (1 × 10^6 moles) (334 g/mole) = 3.34 × 10^8 grams of I_2O5is needed to completely clear 140,000 kg of CO from the air.

Question:

A galvanometer has a resistance of 5 ohms with 100 divisions on its face. Maximum deflection of the meter occurs when the current through it is 2ma. What series resistance should be used with the galvanometer in order to employ it as a voltmeter of range 0 to 200 volts? Voltmeter consisting of a galvanometer and resistor in series

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/Users/wenhuchen/Documents/Crawler/Physics/D20-0671.htm

Solution:

The maximum voltage we need to read is 200 volts. We therefore want maximum deflection to occur when the voltage across the voltmeter is 200 volts. This occurs when the current through the galvanometer is 2ma or 0.002 amp. To accomplish this, we can put a resistor R in series with the galvanometer. The total resistance in the circuit, R + r (see figure), must then be, by Ohm's Law, R + r = (V/I) = [(200 volts)/(0.002 amp)] = 100,000 ohms andR = 100,000 - 5 = 99,995 ohms Since the voltage across the terminals is proportional to the current flowing through the galvanometer, each division of the meter corresponds to [(200 volts)/(100 divisions)] = 2 volts/division.

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Question:

The executioner in charge of the lethal gas chamber at the state penitentiary adds excess dilute H_2 SO_4 to 196 g (about (1/2) lb) ofNaCN. What volume of HCN gas is formed at STP?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E05-0207.htm

Solution:

The equation for this reaction is H_2 SO_4 + 2NaCN \rightarrow 2HCN + Na_2SO_4 This means that for eachNaCNreacted, one HCN is formed. After one knows the number of moles of HCN formed, one can find the volume that the gas occupies by multiplying the number of moles by 22.4 liters (volume of one mole of gas at STP). STP (Standard Temperature and Pressure) is defined as 273\textdegreeK and 1 atm. The number of moles ofNaCNreacted can be found by dividing 196 g by the molecular weight ofNaCN. (MW ofNaCN= 49.) number of moles = [(196 g)/(49 g \bullet mole)] = 4.0 moles From thestoichiometryof the reaction, we see there will be 4.0 moles of HCN formed. The volume of the gas can now be found. volume = 4.0 moles × 22.4 liters/mole = 89.6 liters.

Question:

Describe the structure of a centriole . What hypothetical function does it serve ?

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/Users/wenhuchen/Documents/Crawler/Biology/F02-0043.htm

Solution:

Centrioles are small bodies located just outside the nucleus of most animal cells and some plant cells in a specialized region of cytoplasm that has been known to play a role in cell division. Centrioles usually occur as a pair in each cell oriented at right angles to each other. Each centriole of the pair is composed of nine groups of tubules arranged longitudinally in a ring to form a hollow cylinder. Each group is a triplet composed of three closely associated tubular elements called micro-tubules. (See Figures.) The space immediately surrounding a pair of centrioles is called the centrosome. The centrosome appears to be clear or empty in the light microscope. Centrioles do not occur in most higher plant cells, although they are found in some algae and fungi and in a few reproductive cells of higher plants. The centrioles seem to play some part in directing the orderly distri-bution of genetic material during cell division. At the beginning of cell division the centrioles replicate, and the two pairs of centrioles that result move to opposite poles of the dividing cell. Under the electron microscope each pair is seen to send out spindle fibers, structures involved in separating and moving chromosomes to the oppo-site ends of the cell. This observation leads to the hypothesis that centrioles are needed in the formation of the spindle fibers. However, the existence of cells with-out centrioles yet capable of spindle formation seems to refute this hypothesis.

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Question:

What unusual feature of the cycads makes them are remarkable group among the seed plants?

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/Users/wenhuchen/Documents/Crawler/Biology/F08-0195.htm

Solution:

Cycads are remarkable among the seed plants in that they form motile, swimming sperm within their pollen tubes. Of all the living seed plants, only the cycads and gingko, another gymnosperm, possess swimming sperm cells. The sperm formed within the pollen tubes of other seed plants are represented solely by nuclei. The motile sperm swim by means of flagella, and those of the cycad have thousands of flagella per cell. Each of these flagella have the 9 + 2 pattern of axial filaments characteristic of eukaryotes, and are attached to a spiral band that encircles the anterior end of the sperm cell. The sperm cells, in addition, are remarkably large, measuring as much a 400\mumin one species, and may be seen with the naked eye. They are the largest motile male gametes known among the higher plants. When the sperm become active and move about in the pollen tube, the tube bursts and releases the sperm which then swim briefly in the fluid of the fertilization chamber. The sperm that will fertilize the egg enters the egg cytoplasm and loses its flagella. The presence of swimming sperm is a primitive condition, characteristic of lower aquatic plants and early terrestrial plants, such as ferns and mosses, which require water for fertilization. Swimming sperm is of no advantage to a seed plant, and virtually all have been eliminated through natural selection as the plants have become more adapted to terrestrial life. For this reason, the cycad is considered to be a primitivegymnosperm which provides a connecting link between the ferns and lycopodsand the more advanced seed plants.

Question:

Find the force acting on the bottom of an aquarium having a base 1 foot by 2 feet and containing water to a depth of one foot.

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/Users/wenhuchen/Documents/Crawler/Physics/D10-0408.htm

Solution:

Pressure, p, is given by height times density (when density is weight per volume). The density of the water is 62.4 lb/ft^3. Therefore, p = hw = 1 ft × 62.4 lb/ft^3 = 62.4 (lb/ft^2 ) Force = pressure × Area of bottom, therefore, F =pA = 62.4 (lb/ft^3 ) × (1 ft × 2 ft) = 124.8 lb. Note that the shape of the vessel is not considered in this solution. It would be the same even if the sides were sloping outward or inward.

Question:

Calculate the magnitude of the gravitational red shift for a photon traveling from the sun to the earth.

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/Users/wenhuchen/Documents/Crawler/Physics/D32-0955.htm

Solution:

Because of Einstein's Mass-Energy Relation, any object having energy E has an equivalent mass m = E/c^2 where c is the speed of light. If the "object" is a photon, E = hѵ and the equivalent mass is m = hѵ /c^2(1) where h is Planck's constant. Now, we may ask, what happens if a photon is in a gravitational field? In traveling between 2 points in this field the potential energy of the photon changes. Hence, so must its total energy, E. We may write E' = E + (\textphi' - \textphi)(2) where \textphi' - \textphi is the change in potential energy of the photon, and E' is its new energy as a result of moving in the gravitational field. Rewriting(2) E' - E = \textphi' - \textphi or hѵ' - hѵ = \textphi' - \textphi whence, h( ѵ' - ѵ) = \textphi' - \textphi Hence, we expect the photon to undergo a frequency shift of ѵ' - ѵ = [(1/h)(\textphi' - \textphi)](3) as a result of a change in the potential energy of the photon. If the photon travels from the surface of the sun to the surface of the earth, the primed variables will refer to the surface of the earth. Similarly, the unprimed variables will refer to the sun's sur-face. In this case, the fractional change in frequency of the photon is (ѵ' - ѵ) / ѵ =(\textphi' - \textphi) / hѵ(4) Now, (see figure) \textphi' = [(- GM_em') / R_e] - [(GM_sm') / D](5) and \textphi' = [(- GM_sm) / R_s] - [(GM_sm) / D] whereG = 6.67 × 10^-11 [(N.m^2)/ kg^2] . From (1) m' = hѵ' / c^2andm = hѵ / c^2(6) Substituting (6) in (5) \textphi' = [(- GM_e hѵ') / (c^2 R_e)] - [(GM_s hѵ ') / (c^2 D)] \textphi' = [(- GM_s hѵ) / (c^2 R_s)] - [(GM_e hѵ) / (c^2 D)] Rewriting these equations \textphi' = [(- Ghѵ' / c^2 ){(Me/ R_e) + (M_s / D)}] \textphi = [(- Ghѵ / c^2 ){(Ms/ R_s) + (M_e / D)}] and = (\textphi' - \textphi) / hѵ = [(- Gѵ' / c^2ѵ ){(Me/ R_e) + (Ms+ D)}] + [(Gѵ/ c^2) {(Ms/ R_s) + (M_e + D)}] But (ѵ' - ѵ) / ѵ =(\textphi' - \textphi) / hѵ And (ѵ' - ѵ) / ѵ = [{(- Gѵ') / (c^2ѵ )} {(M_e / R_e) + (Ms/ D)}] + [(G/ c^2) {(Ms/ R_s) + (M_e / D)}] Or (ѵ' / ѵ) - 1 + [{(Gѵ') / (c^2ѵ )} {(Me/ R_e) + (Ms/ D)}] =[(G/c^2) {(Ms/ R_s) + (M_e / D)}] (ѵ' / ѵ) {1 + [( G/ c^2){(Me/ R_e) + (M_s / D)}] } = [(G / c^2) {(Ms/ R_s) + (Me/ D)}] + 1 (ѵ' / ѵ)= [(G/ c^2){(Ms/ R_s) + (Me+ D)} + 1] / [( G / c^2 ) {(M_e/R_e) + (Ms+ D)} +1] Hence , (ѵ' / ѵ) = [G {(Ms/ R_s) + ( M_e / D)} + c^2] / [G {(Me/ R_e) + ( M_s / D)} + c^2](7) Before solving for ѵ' / ѵ , we may make some approximations in (7). First, the data needed are M-_s = 1.99 × 1030kgR_e = 6.37 × 10^6m M_e = 5.98 × 10^24 kgR_s = 3 6.96 × 10^8m D = 1.49 × 10^11m Now, note that [M-_s / R_s] >> [M_e / D] and may be neglected in (7).Furthermore, [(GM_e) / R_e] + (M-s/ D) << c^2 and may also be neglected in (7), whence ѵ' / ѵ \approx {G {(Ms/ R_s) + c^2} / c^2 = 1 + (G / c^2) (Ms/ R_s) ѵ' / ѵ \approx 1+ [{(6.67 × 10^-11 n.m2/kg^2)(1.99 × 1030kg)} / {(9 × 10^16 m^2/s^2)(6.96 × 10^8 m)}] ѵ' / ѵ \approx 1 + [{(1.32733 × 10^20) / (6.264 × 10^25)} ѵ' / ѵ \approx 1.0000021(8) Hence the fractional change in the frequency of a photon traveling from the sun to the earth is (ѵ' - ѵ) / ѵ = (1.0000021 ѵ - ѵ) / ѵ = 0.0000021

Question:

A particle of mass m is attached to the end of a string and moves in a circle of radius r on a frictionless horizontal table. The string passes through a frictionless hole in the table and, initially, the other end is fixed. (a) If the string is pulled so that the radius of the circular orbit decreases, how does the angular velocity change if it is \omega_O when r = r_O ? (b) What work is done when the particle is pulled slowly in from a radius r_O to a radius rO/ 2?

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/Users/wenhuchen/Documents/Crawler/Physics/D37-1070.htm

Solution:

a) In this problem, no external torques act on the particle, therefore, angular momentum is conserved. Angular momentum, L, equals r^\ding{217} × mv^\ding{217} and, for this problem, L = mr^2\omega since the motion is circular and r\omega^\ding{217} = v^\ding{217}. If the strings length is altered from r_O to r, the angular rotation changes from \omega_O to \omega. However, L = L_O (angular momentum is conserved) and it follows that However, L = L_O (angular momentum is conserved) and it follows that mr 2 \omega= mrO^2\omega_O. = mrO \omega_O. Rearranging the above expression and dividing by m yields Rearranging the above expression and dividing by m yields \omega = (r_O ^2\omega_O ) / r^2 = [(rO) / r]_2\omegaO \omega = (r_O ^2\omega_O ) / r^2 = [(rO) / r]_2\omegaO b) The work done shortening the string is equal to the integral of the tension in the string times an infinitesimal distance integrated over the total length change. Mathemati-cally, this is written as W =^(r)O/2\int_(r)O T \bullet dr W =^(r)O/2\int_(r)O T \bullet dr where W is work and T is tension. where W is work and T is tension. The tension in the string is merely the centripetal force in this The tension in the string is merely the centripetal force in this problem. T = mv^2 / r = mr\omega^2 Integrating,W = ^(r)O/2\int_(r)Omr\omega^2dr Using the expression for \omega from part a), W = ^(r)O/2\int(r)O[(r_0 / r)^2 / \omega_0]^2 dr W = mrO^4 \omegaO2(r)O/2\int(r)O(dr / r^3)dr W = mrO^4 \omegaO2[ \rule{1em}{1pt}(1 / 2)(1 / r^2)](r)O/2(r)O W = mrO^4 \omegaO2[{\rule{1em}{1pt}(1 / 2)} {(2 /r_O)^2 \rule{1em}{1pt} (1 / r_0)^2}] W = mrO^4 \omegaO^2[\rule{1em}{1pt}(3 / 2)]r_0^\rule{1em}{1pt}2 W = \rule{1em}{1pt}(3 / 2) mrO^2 \omegaO^2 The negative sign indicates that the work is done on the particle, which is as it should be.

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Question:

How much area, in square meters, will one liter of paint cover if it is brushed out to a uniform thickness of 100 microns?

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Solution:

Because one is asked to give the final area in square meters, one should first convert the volume and thickness to meter units. One liter is equal to 1,000 cc. Since 1 m = 100 cm, one can convert centimeters to meter units by cubing both sides of the equality: 1m^3 = (100cm)^3 1m^3 = 1.0 × 10^6 cc 1m^3 / 1.0 × 10^6 = 1cc 10\Elzbar6m^3 = 1cc Therefore, 1000 cc or 1 liter is equal to 10\Elzbar6m^3 × 1,000 or 10\Elzbar3m^3. There are 106microns in 1 m. Thus, 1 micron = 10\Elzbar6m and 100 microns = 10\Elzbar4 m. The area of the film is equal to the volume divided by the thickness. Therefore, area = (10\Elzbar3m^3 ) / (10\Elzbar4m) = 10 m^2 .

Question:

When changes are occurring in the ectoderm during the process of neurulation, differentiation of the pharynx is also occurring. What are the steps involved?

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/Users/wenhuchen/Documents/Crawler/Biology/F23-0600.htm

Solution:

One of the principal events occurring in the region of the archenteron at the time of neurulation is the rapid expansion of the anterior end to form the region known as the pharynx. Lateral expansion of the pharynx produces pouches which come in contact with the ectoderm. The two tissues form, respectively, pharyngeal pouches and pharyngeal clefts. Shortly after contact between the ectoderm and the endoderm of the pharynx, the point of contact breaks through to create pharyngeal clefts. The tissue now remaining between the two pharyngeal clefts is known as a pharyngeal arch. This structure is covered on the exterior by ectoderm and on the interior by endoderm. The arch is composed of loosely organized mesodermal cells called mesenchyme. In amphibians, the gills develop from the pharyngeal arches. In the human and other higher vertebrates, the pouches give rise to other structures or disappear. The first pair of pouches becomes the cavities of the middle ear and the connection with the pharynx, the Eustachian tube. The second pair of pouches becomes a pair of tonsils, while parts of the third and fourth pairs of pouches become the thymus gland, and other parts of them become the parathyroids. The fifth pair of pouches forms part of the thyroid. The bulk of the thyroid develops from a separate outgrowth on the floor of the pharynx. The cavity of the mouth arises as a shallow pocket of ectoderm that grows in to meet the anterior end of the pharynx. The membrane between the two ruptures and disappears during the fifth week of development.

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Question:

An electric iron with a resistance of 24 ohms is connected to a 120-volt line. What power is being used?

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Solution:

The formula for power is P = (E^2/R) . This formula has two equivalent forms. From P = (E^2/R)with E = IR (Ohm's law) we get P = {(IR)^2/R} = I^2R(1) = I(IR) = EI(2)(again by Ohm's law) Using the original form {P = (E^2/R)} we arrive at solution 1: Solution 1: E = 120 volts,R = 24 ohms P = (E^2/R) = {[120^2 (volts)^2]/[24 ohms]} = 600 watts For both the other solutions we need to know the current I. I = (E/R) = (120 volts/24 ohms) = 5 amp Solution 2: Here we use form (1) of the equation. P = I^2R = 5^2 (amp)^2 × 24 ohms = 600 watts Solution 3: Here we use form (2) of the equation. P = El = 120 volts × 5 amp = 600 watts.

Question:

Explain why it has been necessary to postulate the existence of a cyclic form of glucose?

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Solution:

Glucose (C_6H_12O_6), is one of the isomeric aldohexoses - a carbohydrate or sugar. Glucose exists in two forms, one of which is the mirror image of the other. Its configurational formulas may be written At one time, it was believed that these glucose isomers existed in this linear form. Now, however, it is believed that cyclic forms exist, as indicated in the accompanying diagram. This has been postulated for the following reasons: It has been found that when you take a solution of pure (+) - glucose or a solution of pure (-) - glucose form, a pair of optical isomers are obtained, i.e., ones with the ability to rotate plane- polarized light. For this to occur, in a solution of pure (+) or (-) glucose, another asymmetric carbon must be present. An asymmetric carbon is one with 4 different atoms or molecules attached. Asymmetric carbons can generate optical isomers. In the formation of cyclic structures, the carbon atom of the aldehyde group (C = O) must be the additional asymmetric carbon. This would account for the two optical isomers formed. In other words, that glucose exists in cyclic forms was deduced from the fact that only in such a structure is an additional asymmetric carbon generated, which would account for the optical somers produced.

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Question:

What are the basic flags incorporated into a condition code register?State the purpose of each flag.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G05-0108.htm

Solution:

Each flag in the condition code register designatesa certain conditioninside the CPU. The flags are the basis for computer decision- makingin conditional instructions such as "Jump to location 31 if the Accumu-lator is Zero." When the computer reads such an instruc-tion it will checkthe flag in the condition code register which designates a zero accumulator. If the flag is 1 the accumulator is zero and the computer jumpsto loca-tion 31. If the flag is 0 the accumulator is not zero and the computerreads the next instruction. As can be seen, the more flags a computerhas in its condition code register, the stronger the computer will bein its decision-making. The basic flags incorporated into a condition register are: 1.Carry:This flag is set to 1 if the last operation generated a carry fromthe most significant bit. 2.Zero:This flag is set to 1 if the last operation resulted in a zero. This flag is often used in loop control and searching for a certain data value. 3.Overflow:This flag is set to 1 if the last opera-tion produced a 2's complementoverflow. 4.Sign:This flag is set to 1 if the most significant bit of the result of thelast operation was 1, desig-nating a negative two's complement number. 5.Parity:This flag is set to 1 if the result of the last operation contains aneven number of l's (sometimes called Even Parity). 6.Half-Carry:This flag is set to 1 if the last opera-tion generated a carryfrom the lower half word to the upper half word. 7.Interrupt Enable: This flag is set to 1 if an inter-rupt is allowed, 0 if not. A computer may have several interrupt enable flags if it has several interruptinputs or levels.

Question:

Explain the subtraction function using the complementing system as in decimal and in binary number system.

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Solution:

To understand the complementing system in computers and why this system is used, consider the following example: 12 + 7 = 19 - 12 + 7 = - 5 Each of these is defined as an addition between two operands (numbers), but because one of the numbers is actually negative in the latter case, it is really a subtraction. This means that, even though we instructed the computer to do an addition, it would first have to examine the numbers and, if one of them is negative, do a subtraction instead. This makes the job of the computer harder, because it has to examine the numbers, make a decision on what to do, and then do it. For this reason the complementing system was developed, so that addition could work for both positive and negative numbers. For decimal numbers there are two complementing systems - 10's and 9's. 10's complement system: Consider the example 5-3 = 2 \equiv5 + 10 - 3 - 10 = 2 \equiv5 + (10 - 3) - 10 = 2 \equiv5 +7 - 10 = 2 \equiv12 - 10 = 2 e., for - 3, take its 10's complement, 10 - 3 = 7. Now add 7 to 5 = 12 and drop the last 1. This gives12 = 2, the necessary answer. 9's complement method: Here we subtract the negative number from 9. Perform addition and the last digit is again added back to the result. 5 - 3 = 2 1st step9 - 3 = 6 2nd step5 + 6 = 11 3rd step11= 1 + 1 = 2 One of the methods for constructing a 1's complement for a binary number is the interchange of zeros and ones, e.g., decimalsubtraction 5101 - 3- 011 2010 = 2 Now the 1's complement of 011 = 100. Thus, This is equivalent to the 9's complement in decimal. Similarly, there is a 2's complement method equivalent to the 10's complement method. Here we take the complement of a binary number and add one to it. 011 = 1001st step complement + 1 1012nd step add 001 to it Now 5101 -3\equiv+100 21010drop out last bit as in 10's complement method. Now consider the case where the result of the subtraction is negative. For example, 6 - 7 - 1(decimal subtraction) or, 6 = 0110 in binary, and, 7 = 0111 in binary. 1's complement of 7 in binary = 1000. Hence, As no carry is generated in the last stage, it means the result is negative. Hence, take the complement 1110 \rightarrow 0001 Hence, the answer is: - 0001, i. e, - 1 (binary). Also, 2's complement of 7 in binary = 1000 +1 1001 Hence, the answer = - 0001, i. e., - 1. Hence, the answer = - 0001, i. e., - 1.

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Question:

The property of skeletal muscle contraction in which the mechanical response to one or more successive stimuli is added to the first is known as summation. What is the underlying explanation for this phenomenon?

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/Users/wenhuchen/Documents/Crawler/Biology/F19-0473.htm

Solution:

A possible explanation of this phenomenon, based on the role of calcium in excitation-contraction coupling, is that the amount of calcium released from the sarcoplasmic reticulum during a single action poten-tial is sufficient to inhibit only some of the troponin-tropomyosin in the muscle. Multiple stimulation would then release more calcium so that more troponin-tropomyosin would be inhibited, allowing for further contraction. However, the truth is that more than enough calcium is released by the first action potential to inhibit all the troponin-tropomyosin, so this proposal must be discarded. The explanation of summation involves the passive elastic properties of the muscle. Tension is transmitted from the cross bridges through the thick and thin fila-ments, across the Z lines, and eventually through the extracellular connective tissue and the tendons to the bone. All these structures have a certain amount of elasticity, analogous to a spring that is placed between the contractile components of the muscle and the external object. In the muscle, the contractile elements in their fully active state begin to stretch the passive elastic structures immediately following calcium release. Only when the elastic structures are all taut, can increasing contraction by the muscle occur. Summation occurs because a second stimulus is given, very close in time to the first, while the elastic structures are still a bit taut and not yet slack. Under this condition, the active state of the contractile proteins is maintained and the result is contractions that are stronger than any single simple twitch. Should sustained stimulation occur, the elastic elements would never have time to relax at all, and it is at this point that maximal force by the muscle fibers is attained; the individual contractions are indistinquishably fused into a single sustained contraction known as tetanus, (see accompanying figure) If stimula-tion of the muscle continues at this frequency, the ulti-mate result will be fatigue and possibly complete cessation of activity due to exhaustion of nutrients. It is not surprising to note that cardiac muscle has an extremely long refractory period, allowing the elastic components to relax and thus avoiding tetanus, which would result in death due to loss of pumping action of the heart.

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Question:

For a condensation polymerization of a hydroxyacid with a residue weight of 200, it is found that 99% of the acid groups are used up. Calculate a) the number average mo-lecular weight and b) the weight average molecular weights.

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Solution:

a) The number average molecular weight is the average weight of all particles present, where each unit is given equal weight. Thus, the number average molecular weight is equal to the individual residue weight times the average number of residues. This number can be calcu-lated from the fact that 99% of the acid groups have re-acted to form the polymer chain. When monomers join to form a polymer, the last monomer is left unreacted at the end of the chain. If 99% of the acid reacts then 1% is left unreacted, so that 1% of the average number of monomers is equal to 1. Let the number average of the monomers equalX_n . Then, .01X_n = 1 X_n = 100 One can now solve for the number-average molecular weight, M_n. number-average mol.wt = mol.wt of a monomer xX_n = 200 g/mole × 100 = 2.0 × 10^4 g/mole. b) The weight average molecular weight, M_w, weights molecules proportionally to their molecular weight in the averaging process. It is found from some complex derivations that M_w = M_n (1 + p), where p is the extent of the reaction. In this problem, p is .99. From this one can solve for M_w M_w= (2.0 × 10^4 g/mole)(1 + 0.99) = 3.98 × 10^4 g/mole.

Question:

At standard conditions, 1 liter of oxygen gas weighs almost 1.44 g, whereas 1 liter of hydrogen weighs only .09 g. Which gas diffuses faster? Calculate how much faster.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E03-0097.htm

Solution:

This question deals with diffusion of gases, which is governed by Graham's law. A gas that has a high density diffuses more slowly than one with a lower density. The rates of diffusion of 2 gases are inversely proportional to the square roots of their densities, as shown by the equation r_1 / r_2 = \surdd_2 / \surdd_1 ,where r_1, r_2 = rates of diffusion of the gases, and d_1, d_2 are their respective densities. Therefore, by Graham's law, we know that H_2 diffuses faster because of its lower density. (H_2's density = M / V = (.09 g) / (1 l) =.09 g/ l while O_2's density = 1.44 / 1 = 1.44 g/liter.) To find out how much faster H_2 diffuses, plug the necessary factors into the equation r_(H)2 / r_(O)2 = \surd(d_(O)2) / \surd(d_(H)2).Rewriting, r_(H)2 = r_(O)2 [\surd(d_(O)2) / \surd(d_(H)2)] = r_(O)2[\surd(1.44) / \surd(.09) ] r_(H)2 = 4r_(O)2 Thus, the calculated comparison is in agreement with the predicted comparison that H_2 diffuses more quickly. To be precise, the rate of diffusion of H_2 is 4 times that of the rate of diffusion of O_2.

Question:

Solve f(x) = x2 - e^x = 0. (a) graphically (b) using the Newton-Raphson method of iteration (program in BASIC)

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G21-0515.htm

Solution:

Observe that f(x) is a continuous function and that f(-1) = 1 - e^-1 = 0.632 > 0 and f(0) = -1 < 0 so that f(x) has a root between -1 and 0. By graphing f(x) we can get a good pre-liminary estimate of where the root lies. Values plotted are given in the Figure. Note that there is a root near x = -0.7. We can use this root as an input estimate to the Newton-Raphson method. This is an iterative method in which we solve the equation x_n+1 = x_n -f(x_n) / f'(x_n), wheref(x_n) = x^2_n -e^(x)n and f'(x_n) = 2x_n -e^(x)n . We make an initial guess x_0 (in this case, x_0 = -0.7), and evaluate the function. Next, we compute the error as an absolute value. This value is compared to the convergence condition, a previously establish-ed number that we have chosen. If our error is less than the conver-gence condition, then we write the solution x_n+1 . If not, we return to the start of the loop and continue. A sample program solution is given below: PROGRAM EXPLANATION 0010 INPUT X,L Input initial guess x_0 and limiting no. of iterations 0020 F = (X\uparrow2 - EXP (X)) Evaluate f(x) 0030 D = (2\textasteriskcenteredX - EXP(x)) Evaluate D = f'(x) 0040 A = F/D Evaluate f(x)/f'(x) 0050 E = ABS (A) E = \vertA\vert = \vert x_n+1 - x_n \vert = error 0060 IF E < 0.000005 THEN 110 Convergence condition 0070 X = X - A x_n+1 = x_n - f(x_n)/f'(x_n) 0080 N = N + 1 Increment loop counter 0090 IF N = L THEN 110 Check for limit on itera-tions 0100 GO TO 20 loop back 0110 PRINT X,N,A output results 0120 END end of program

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Question:

Assuming that the density of water is .9971 (g / cm^3)at 25\textdegreeC and that of ice at 0\textdegree is .917 (g / cm^3), what percent of a water jug at 25\textdegreeC should be left empty so that, if the water freezes, it will just fill the jug?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E07-0234.htm

Solution:

Density = [(mass) / (volume)]. When water freezes, the mass is constant, but its density decreases.This means, therefore, that an in-crease in volume occurs. To solve this problem, you must determine the change in volume of water from 25\textdegree to 0\textdegree. Because you are given the densities of the substance at each temperature, you can say for 1 mole of H_2O (MW = 18) (mass / volume) = 18 [(g / mole) / (y) = (0.997g / cm^3) for water at 25\textdegreeC and (mass / volume) = 18 [(g / mole) / x] = (0.997g / cm^3) for ice at 0\textdegreec Rewriting, you have .997 y = 18 and .917 x = 18. Both expressions equal 18, which means .997 y = .917 x, or (x / y) = [(.997) / (.917) = 1.087. This means, then, that at 0\textdegreeC the volume is 1.087 times greater than the original at 25\textdegreeC. If you let B equal the fraction of water to be put in the jug, then you can say 1.087 B = 1, or B = .920. In other words, only 92% of the jug should be filled at 25\textdegree to obtain a completely filled jug at 0\textdegree. This means that 8 % is empty space.

Question:

If 5 joules of work are done in moving 0.025 coulomb of positive charge from point A to point A', what is the difference in potential of the points A and A'?

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/Users/wenhuchen/Documents/Crawler/Physics/D17-0553.htm

Solution:

To solve this problem we use our formula for the definition of the volt, E = W/Q . The work is 5 joules and the charge Q is 0.025 coulomb. Then the potential difference E = (5 joules)/(0.025 coulomb) = 200 volts

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Question:

If Young's Modulus for steel is 19 × 10^11 dynes/cm^2, how much force will be required to stretch a sample of wire 1 sq mm in cross section by 0.2% of its original length?

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/Users/wenhuchen/Documents/Crawler/Physics/D08-0348.htm

Solution:

The problem is recognized as one involving Young's Modulus, the defining equation for which is Y = (Stress)/(Strain) = (F/A)/(∆x/x) We are given that ∆x = .10x (or ∆x/x = 1/10) and Y is given as 19 × 10^11 dynes/cm^2. Moreover A = 1 mm^2 = 1 mm^2 × 1cm^2/100 mm^2 = .01cm^2. F/A = Y (∆/x/x)andF = YA(∆x/x) F = 19 × 10^11 dyne/cm^2 × 10^-2 cm^2 × 10^-1 = 19 × 10^8 dynes.

Question:

The extremes of temperature in New York, over a period of 50 years, differ by 116 Fahrenheit degrees. Express this 50 years, differ by 116 Fahrenheit degrees. Express this range range in Celsius degrees. in Celsius degrees.

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/Users/wenhuchen/Documents/Crawler/Physics/D12-0442.htm

Solution:

Fahrenheit and Celsius temperature scales are related by C\textdegree = (5/9) (F\textdegree \rule{1em}{1pt} 32). Since in this example, only a change in temperature is being converted from one linear scale to the other, we have \DeltaC\textdegree = (5/9) \DeltaF\textdegree Substituting, we get C\textdegree = (5/9) x 116 \textdegreeF = 64.5 C\textdegree

Question:

For earlier computer systems, the storage of data in main memory was done by means of magnetic cores (donut-like structures whose diameter is a fraction of an inch). To represent these cores, we use the binary digits (Bits) 0 and 1. In the interpretation of bits, what are the rules used by the IBM/360-370 computers to make distinction be-tween their fixed point integer and packed decimal repre-sentations?

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G09-0196.htm

Solution:

The IBM/360-370 processors have many ways of in-terpreting groups of bits stored in memory. When storing fixed point integers, these computers use two forms; a short form and a long form. In the short form, two bytes (16 bits) are used to store the integer; for the long form, four bytes (32 bits) are used. In either case, if the processor determines that it has encountered an integer, the sign of this integer is indicated by using the leftmost bit of the 2 or 4 byte field. If the integer is positive this bit is zero, if it is negative the bit is one. The computers store packed decimal numbers by placing two decimal digits in each byte (8 bits) field. This is done for all the bytes of the field with the exception of the rightmost. For this byte, the rightmost four bits in-dicate the sign of the number. If the number is positive, the sign is an 'F' or a 'C', if negative, it is a 'D'. The length of a packed decimal field can be between 1 and 16 bytes. Assuming this, the field can be a minimum of 1 digit and a sign or a maximum of 31 digits and a sign.

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Question:

List the arithmetic operators used in BASIC and state their priorities of execution in a statement. Also give a list and briefly explain the functions of the most popular statements.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G10-0222.htm

Solution:

Arithmetic operations in BASIC are represented by the following symbols: +-ADDITION --SUBTRACTION AND NEGATION \textasteriskcentered-MULTIPLICATION /-DIVISION \textasteriskcentered\textasteriskcentered or \uparrow-EXPONENTIATION Exponentiation has the highest priority of execution; multiplication and division are executed next; addition and subtraction are done last. If 2 or more equally "powerful" operators are present in the statement, their execution goes from left to right. Operations in parenthesis are always done first. The most popular BASIC statements are: LET variable = expression (The arithmetic assignment statement) INPUT variable list (Statement for data entry from terminal into a program during the execution phase) READ variable, variable, etc. (Statement for data entry from the DATA statement) DATA constant, constant, etc. (Information supplier statement for READ statement) PRINT variable, constant, "junk", etc. (Primary output statement) REM comments (Comments, insertion statement) GO TO n ("Sends" the computer to the statementlabelledn) More complicated statements will be explained in later problems.

Question:

When carbon dioxide combines with water, carbonic acid is produced and rapidly dissociates to the bicarbonate ion and free hydrogen ions: CO_2 + H_2O\rightarrowH_2CO3 H_2CO_3\rightarrowH^+ + HCO_3^\rule{1em}{1pt} When carbon dioxide is carried by the blood, why doesn't the blood become acidic owing to the presence of free H^+?

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Solution:

Not only does the blood transport oxygen from the lungs to the tissues (via hemoglobin), it also transports carbon dioxide in the reverse direction, from the tissues to the lungs. Some of the carbon dioxide (\sim 7%) is carried as gas dissolved in the plasma. A slightly larger proportion (\sim 23%) is carried by hemoglobin; the carbon dioxide-hemoglobin complex is called carbaminohemoglobin. The loosely bound CO_2 is combined with the terminal amino groups of all four polypeptide chains; each hemoglobin molecule having four polypeptide chains, can bind four CO_2 molecules. Since CO_2 does not combine with hemoglobin at the same site as oxygen, hemoglobin can transport both oxygen and carbon dioxide at the same time. However, most of the CO_2 (70%) is transported as the bicarbonate ion, HCO_3^-. Carbon dioxide enters red blood cells where it reacts with water to form carbonic acid: This reaction can only occur because of an enzyme called carbonic anhydrase. The carbonic acid quickly dissociates into hydrogen and bicarbonate ions: H_2CO_3\rightarrowH^+ + HCO_3^- The presence of free H^+ ions should drop the pH of the blood significantly. Since cells can only live within a narrow range of pHs, this change would be physiologically harmful. But the blood exhibits only very slight variations (\pm.04 pH units) around the normal pH of 7.4. What accounts for this phenomenon? The pH would not drop if the H^+ ions could be combined with some substance, so as to remove them as free ions from solution. The H^+ ions combine with an ionized form of hemoglobin (Hb^-) to form acid hemoglobin (HHb) while the HCO_3^- leaves the red blood cell and enters the plasma. HCO_3^- + H^+ + Hb^-\rightarrowHCO_3^- + HHb Acid hemoglobin dissociates to a much lesser degree than does carbonic acid, so that most H^+ ions remain bound, and therefore effectively removed from solution. Acid hemoglobin acts as a buffer in the blood. A buffer is a substance which prevents large changes in pH. When there is an excess of H^+ ions(when the pH drops), hemoglobin binds H^+ ions to remove them from solution, thus restoring the pH. Hb^-+H^+\rightarrowHHb When the blood becomes alkaline because of an excess of hydroxide ions (OH^-) or a decrease of H^+, the acid hemo-globin dissociates, releasing H^+ and restoring the pH: OH^- + HHb \rightarrow OH^- + H^+ +Hb^- \rightarrow H_2O + Hb^- Hemoglobin and some plasma proteins act to prevent the large deviations from the normal pH which would occur in carbon dioxide transport. A summary of CO_2 transport is provided in the following illustration: The reverse process occurs in the lungs. Carbaminohemoglobin dissociates to release CO_2. Acid hemoglobin dissociates to produce H^+ ions which combine with bicar-bonate ions to form carbonic acid. This acid, then dissoci-ates to form water and CO_2, which is expired from the lungs.

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Question:

In question 17, a linked list for 10 students was created. Write or modify the program. a) Delete an element if the name (member) of that element coincides with the given name. b) Given name is to be read in. c) The program should indicate whether it was successful or not in deleting that particular element. Note: If element is not deleted means that particular name was not found in the linked list.

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Solution:

Modifications: These are all additions to be done. #define TRUE 1 #define False 0 main ( ) int not _ found; char name [MAX_NAME]; struct students \textasteriskcenteredprevious _ptr; printf ("Enter the student name to be deleted \textbackslashn"); scanf ("%s\textquotedblright, name); not _ found = TRUE; /\textasteriskcenteredThis segment checks 1st element \textasteriskcentered/ if (equal _ string (start \rightarrow name, name) == 0) { not _ found = FALSE; start = start \rightarrow next; } else {ptr= start \rightarrow next; previous _ptr= start; while(ptr&& not _ found) {if(equal _ string (ptr\rightarrow name, name) = =0) {not _ found = FALSE; previous _ptr\rightarrownext =ptr\rightarrow next; } previous _ptr=ptr; ptr=ptr\rightarrow next; }/\textasteriskcenteredend while \textasteriskcentered/ }/\textasteriskcenteredend else \textasteriskcentered/ if(not _ found) printf("STUDENT NOT FOUND IN THE LIST\textbackslashn "); else printf("STUDENT DELETED FROM THE LIST \textbackslashn"); }/\textasteriskcenteredend main \textasteriskcentered/ The logic followed in this program is to 1) Locate whether the given name is in the first element of the list. If so delete the first element and point the pointer start to the second element. 2) Check the other elements. If the given name coincides with the name in the element then delete it from the list. ROUTINE : equal _ string ( ): This is a routine in UNIX systems which examines two strings character by character until the NULL value is found, or strings are found to be equal. If strings are of equal length and contain the same sequence of letters, value 0 is returned. If strings are unequal and con-tents are not the same then value -1 is returned.

Question:

Calculate the number of electrons that can be removed from metallic cesium with the energy required to remove one electron from an isolated cesium atom. Assume the following: the ionization potential of Cs = 3.89eV, h = 6.63 × 10^-34 J sec, and Cs metal is sensitive to red light of 700 nm.

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Solution:

This problem entails the understanding of the photoelectric effect. The photoelectric effect states that when one shines light of a given frequency on a metallic material, electrons will be emitted. The equation: E =hv(where E = energy, v = frequency, and h = Planck's constant) determines the amount of energy produced by a given frequency of light. The ionization potential is the amount of energy needed to cause the electrons to be ejected from the metal. To solve this problem, there-fore , one must calculate how much energy is present in light with a wavelength of 700 nm and then compare this to the ionization potential of a Cs atom. It is given that the wavelength of the light is 700 nm. The frequency of this light is equal to speed of light/wavelength = (3.00 × 10^8 m/s) / 700 × 10^-9 m. Thus, E = (6.63 × 10^-34 J sec) [(3.00 × 10^8 m/s) / (700 × 10^-9 m)] = 2.84 × 10^-19 J/photon 1eV= 1.60 × 10^-19 J. The ionization energy is 3.89eVor (3.89eV) (1.60 × 10^-19 J/eV) = 6.22 × 10^-19 J. Because Cs is sensitive to light with a wavelength of 700 nm, one photon of this light flashed on the metal Cs will cause 1 electron to be emitted from the Cs. Therefore, the number of electrons emitted when 6.22 × 10^-19 J, (the ionization potential of an isolated Cs atom), is added to the metallic system is found by dividing 6.22 × 10^-19 J by 2.84 × 10^-19 J no. of electrons emitted = [(6 22 × 10^-19 J) / (2.84 × 10^-19 J)] = 2.2 electrons Therefore, the maximum number of electrons that can be emitted is 2.

Question:

Find the Q value for the disintegration _60Nd^144 \ding{217} _2He^4 + _58Ce^140 .

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Solution:

Q represents the amount of energy released as a result of the conversion of mass to energy. To find the amount of mass converted, find the difference in the amount of mass of the reactants and the products. From the tables of isotope masses, in atomic mass units, _2He^4 = 4.00387_60Nd^144 = 143.95556 _58Ce^140 = 139.94977 / 143.95364 Products =143.95364 \Deltam=0.00192 The atomic mass unit is 1 / 12 of the mass of a C^12 atom and is equal to 1.66 × 10^-27 kg. To find \Deltam in kg, it has to be multiplied by one amu. Q = \Deltamc^2 = 2.87 × 10^-18 joule = (0.00192amu) × (1.66 × 10^-27 kg/amu) × (3.0 × 10^8 m/ sec^2) Since very small energies are involved, when dealing with atoms, energies are expressed in electron volts. Con-verting joules to MeV, Q = (2.87 × 10^-16 joule ) / {1.6 ×10^-16(joule / eV)} = 1.79 eV.

Question:

Metallic sodium violently liberates H_2 gas according to the Reaction 2Na(s) + 2H_2 0(l) \rightarrow 2NaOH + H_2 (g) If you collect the gas at 25\textdegreeC under 758 mm Hg of press-ure, and it occupies 2.24 liters, (a) find the number of moles of hydrogen produced, (b) find the number of moles of water decomposed and (c) find the percent error in (a) if the gaseous hydrogen had been collected dry, assuming the vapor pressure of water is 23.8 mm at 25 C. (R = .0821 l-atm/mole \textdegreeK.) (R = .0821 l-atm/mole \textdegreeK.)

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Solution:

(a)You want to determine the number of moles of hydrogen gas liberated given its pressure, volume and temperature. Therefore, you make use of the equation of state, which indicates that PV = nRT, where P = pressure, V = volume, R = universal gas constant, T = temperature in Kelvin (Celsius plus 273\textdegree), and n = moles. Thus, to find n, substitute into this equation and solve for n. In 1 atm, there is 760 mm of pressure. You use atmospheres as the units of pressure, since R is expressed in atmospheres. Hence, P = 758 mm = [(758)/(760)] atms = .997 atm. We are given R in the problem statement and that T = 298\textdegreeK and V = 2.24 l. Substituting, n = [{(.997) (2.24)} / {(.0821) (298\textdegreeK)}] = .0913 mole H_2 . Now that you know how many moles of H_2 are liberated, you also know how much water decomposed via the stoichiometry of the reaction. In (b),therefore, you make use of the equation 2Na(s) + 2H_2 O(l) \rightarrow 2NaOH + H_2 (g), which tells you that for every mole of H_2 (g) produced you have 2 moles of H_2O decomposed. You calculated a release of .0913 moles of H_2. Therefore, 2 × .0913 = .183 mole of water must have decomposed. For (c) you must realize that if the hydrogen gas were wet (not dried) , the pressure given is not the true pressure of the H_2. For in this pressure is included the vapor pressure of water. Thus, the actual pressure of H_2 is only 758 - 23.8 = 734.2, where 23.8 was the given vapor pressure of water at 25\textdegreeC. The % error is, then [(23.8)/(734.2)]× 100 = 3.24 %.

Question:

What are some of the cues used by animal in migrating?

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Solution:

Some animals have the extraordinary ability to find their way home aftermaking long-distance journeys. They must somehow "remember" the wayhome. Several cues which such animals use in their migrations have beenestablished. One cue used by many animals is scent. Salmon are known to be bornin fresh-water streams and shortly thereafter journey to the sea. After theymature, they swim back upstream to spawn. But they almost invariablyswim back up the same stream in which they were born. It has beendetermined that salmon have an acute sense of smell, and can rememberthe odor due to chemical composition of the stream in which theywere hatched. Salmon whose olfactory tissues have been incapacitatedcannot locate their native stream. Birds are known to undertake long-distance migratory journeys. Each spring many birds flyNorthto obtain food and to rear their young. As winter approaches, food becomes scarce and they flySouthagain. One species, the golden plover, migrates from northern Canada to southernSouth America. Birds make these long journeys by relying on celestialclues. At migration time, if a caged bird is allowed to see the sun, itwill attempt to fly in the direction of its migratory route. If it is overcast, themovements are undirected. Some birds fly at night and use the relative positionsof the stars. The physiological state of such birds can be altered byexposing them to different artificial day lengths. Thus some birds will enterthe spring migratory condition and others the fall migratory condition. When exposed to the artificial night sky of a planetarium, the "spring" birds tryto fly North while the "fall" birds try to fly South. In studying the homing behavior of pigeons and other birds, it was discoveredthat birds are able to sense the earth's magnetic field. When tinymagnets were attached to homing pigeons in order to cancel the effectsof the earth's magnetic field, the pigeons became disoriented and couldnot make their way home. Experiments are being done to see if othercues are also used, such as barometric pressure and infra-sound (low-frequencysound waves).

Question:

A galvanometer of resistance 20 \Omega gives a full-scale deflection when a current of 1 mA passes through it. What modification must be made to the instrument so that it will give full-scale deflection for (a) a current of 0.5 A, and (b) a potential difference of 500 V?

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/Users/wenhuchen/Documents/Crawler/Physics/D20-0670.htm

Solution:

If a galvanometer has a resistance R of 20 \Omega and gives full-scale deflection for a current I of 1 mA, then the voltage drop across it under these circumstances is given by Ohm's Law as V = IR = 10^-3 A × 20 \Omega = 0.02 V. (a) If we want a full-scale deflection of the galvanometer for a current of .5 A, we must arrange things so that only 1 mA of this current passes through the galvanometer. Referring to figure (a), a shunt resistor R must be added to the galvanometer. This resistor must take 499 mA, allowing only 1 mA through the galvanometer. But since r and R are in parallel, the potential difference across each is the same. Thus, if R is the resistance of the shunt, then, by Ohm's Law, 1 mA × 20 \Omega = 499 mA × R (b) The galvanometer's full-scale deflection occurs when a voltage of .02 V is across the galvano-meter resistance. Hence, to convert to a voltmeter reading up to 500 V, one must add a series resistor to the galvanometer. Only 0.02 V is dropped across the galvanometer for the maximum current of 1 mA. Thus 499.98 V must be dropped across the resistor of resistance R. The same current flows through both resistor and galvanometer. Hence, by Ohm's Law R = [(499.98 V)/(10^-3 A)] = 499, 980 \Omega .

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Question:

A ball is projected horizontally with a velocity v_0 of 8 ft/sec. Find its position and velocity after 1/4 sec (see the figure).

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/Users/wenhuchen/Documents/Crawler/Physics/D03-0089.htm

Solution:

Since the acceleration of gravity, g, is constant, we may use the equations for constant acceleration to find the velocity (v_y) and position (y) of a particle undergoing free fall motion v_y = ^v0_y - gt y = y_0 + ^v0_y - (1/2) gt^2 y = y_0 + ^v0_y - (1/2) gt^2 Here, y_0 and ^v0_y are the initial y position and velocity of the particle. In this case, the departure angle Is zero. The initial vertical velocity component is therefore zero. The horizontal velocity component equals the initial velocity and is constant. Since no hori-zontal force acts on the flying object, it is not accelerated in the horizontal direction. Therefore, v_y = -gty = (1/2) gt^2 v_x = ^v0_xx = ^v0_x t and, at t = 1/4 sec,y = (- 1/2)(32 ft/sec^2)(1/16 sec^2) = - 1 ft. v_y = (-32 ft/sec^2)(1/4 sec)x = (8 ft/sec) (1/4 sec) = 2 ft. v_x = 8 ft/sec

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Question:

An ac source of internal resistance 9000 \Omega is to supply current to a load of resistance 10 \Omega. How should the source be matched to the load, and what is then the ratio of the currents passing through load and source?

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/Users/wenhuchen/Documents/Crawler/Physics/D23-0775.htm

Solution:

The matching may be done by means of a transformer (see the figure). If the numbers of turns on primary and secondary windings are N_1 and N_2, then the internal resistance of the source,R_1,and load re-sistance, , , R_2, are related by the equation R_1 = (N_1/N_2)^2R_2 . \therefore (N_1/N_2)^2 = (R_1/R_2) = (9000/10) = 900 or (N_1/N_2) = 30. Therefore, a transformer with a turns ratio of 30 : 1 must be employed. The transformer lowers the voltage in the ratio 30 : 1, but correspondingly increases the current in the same ratio, for (V_1/V_2) = (N_1/N_2)and(I_1/I_2) = {(1)/(N_1/N_2)}

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Question:

Each of the following is not a correct FORTRAN statement. Give reasons in each case.

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Solution:

a) The only acceptable characters in a variable name in any FORTRAN statement are the twenty-six letters of the alphabet and the ten digits 0 to 9 (some systems also accept the $ included in the variable name). Algebraic signs cannot be included, and therefore M + 1 is an invalid variable. b)For FORTRAN statements involving arithmetic operations, we use the following symbols: -for subtraction +for addition {_\ast}{_\ast}for exponentiation {_\ast}for multiplication /for division No arithmetic operations are allowed on the left side of the "=" sign. c) In the WRITE statement the first number (3) indicates the device that is to be used to print the values obtained at the end of the program. (Typewriter, printer). The second number (13) is the number of the corresponding FORMAT statement (will be discussed in later chapters). The listing of the variables should not be followed by a period. This is an error. d) FORTRAN has no symbol \textquotedblright×" (presumably we mean multiplication, in which case the correct FORTRAN statement is K = 2 {_\ast} I . (Z^Y)^W or (Z)^(Y)W. These are not always equal. For example, if Z = 2, Y = 3, W = 4: (2^3)^4 \not = (2)^(3)4.

Question:

Design an experiment which demonstrates that both positive and nega-tive ions move in electrolytic conduction.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E16-0558.htm

Solution:

The accompanying figure illustrates the set up of an ex-periment which will demonstrate the migration of both positive and nega-tive ions in electrolytic conductivity. A U- tube is initially half filled with a deep purple solution of Cu(MnO_4)_2 in water. The color of the blue hydrated Cu^2+ ions is masked by the purple of the MnO^-_4 ions. A colorless aqueous solution of HNO_3 floats on top of the Cu(MnO_4)_2 solution in each arm of the U- tube. If an electric field is established for a period of time across the solution by 2 electrodes, the blue color, characteristic of hydrated Cu^2+ ions, moves into the region marked A, suggesting a migration of positive ions toward the negative electrode. At the same time, the purple color, characteristic of MnO^-_4 , moves into the B region, indicating a migration of nega-tive ions toward the positive electrode. Thus, there is empirical evi-dence of the migration of both positive and negative ions.

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Question:

In plants, which tissue is responsible for new growth?

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Solution:

Meristematictissue is responsible for new growth in plants. There areseveral regions in a growing plant wheremeristematictissues are found. The apicalmeristemnear the tips of roots, the stem, and branches isimportant in the growth and differentiation in these areas of a plant. The meristemsin the buds of a stem are responsible for outgrowths of the stem. Anothermeristem-atictissue is found in the cambium; this tissue, calledthe vascular cambium, is where thickening of the stem occurs. The cells of themeristematictissue are often small, thin-walled, moreor less rounded, each with a large nucleus and few or no vacuoles. Furthermore,meristematiccells generally are closely packed together, andthus few intercellular spaces are found. These cells are embryonic, relativelyundifferentiated cells that are capable of rapid cell division. The resultantnew cells grow and transform into specific types of plant tissue. Themeristemof the root, for example, gives rise to cells that eventually dif-ferentiateinto all of the cell types present in the root. A plant embryo is composed entirely ofmeristematiccells and consequentlyis capable of rapid growth. As the plant develops, most of themeristemundergoes changes and differentiates into specific tissues. Some of themeristemcontinues to function in certain parts of the adult plant, and thus provides for continued growth. This gives some plants suchas the perennials, the unique ability to con-tinue growth throughout theirlives.

Question:

Air at pressure 1.47 lb/in.^2 is pumped into a tank whose volume is 42.5 ft^3. What volume of air must be pumped in to make the gage read 55.3 lb/in.^2 if the temperature is raised from 70 to 80\textdegreeF in the process?

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Solution:

P_i = 14.7 lb/in.^2 Gage pressure is the pressure present minus the air pressure. Therefore, when the gage reads 55.3 lb/in.^2, the actual pressure P_2 is P^2 = 14.7 lb/in.^2 + 55.3 lb/in.^2 = 70.0 lb/in.^2 T_1 = 70\textdegreeF = 294\textdegreeK T_2 = 80\textdegreeF = 320\textdegreeK V_2 = 42.5 ft^3 The volume of air V_1 pumped in can be found from the ideal gas law.(P_1V_1)/(n_1T_1) = (P_2V_2)/(n_2T_2) Since the number of moles of air is constant, this reduces to (P_1V_1)/T_1 = (P_1V_1)/T_1 [(14.7 lb/in.^2)V_1] / [294\textdegreeK] = [(70.0 lb/in.^2)(42.5 ft^3)] / [320\textdegreeK] V_1 = [(70.0) (42.5) (294)] / [(14.7)(320)] ft^3 = 186 ft^3

Question:

Calculate the voltage of the cell Fe; Fe^+2 \vert\vert H^+ ; H_2 if the iron half cell is at standard conditions but the H^+ ion concentrations is .001M.

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Solution:

The voltage (E) of a cell is found using the Nernst Equation because it involves the use of concentration factors. It is stated E = E\textdegree - (RT / NF) In Q, where R is 8.314 joules per degree, F is 96,500 coulombs, N is the number of moles of electrons transferred and Q is the concentration term. T = 25\textdegree C, by definition of standard conditions, in this equa-tion (or 298\textdegree K). But to solve the problem we must first obtain E\textdegree. This is done by writing down the appropriate half-reactions. Oxidation is the loss of electrons and reduction is the gain of electrons. ReactionTypeE\textdegree_red Fe \rightarrow Fe^+2 + 2e^-Oxidation+ .44 2H^+ + 2e-\rightarrow H_2Reduction0 Next, take the algebraic sum of the E\textdegree_redand E\textdegreeoxidwhich gives E\textdegree. E\textdegree = + .44 + 0 = + .44. Now set up the concentration term. 1nQ = 1n {[H^+]2/ [Fe^+2]} = 1n {[.001]^2 / [1]} Standard conditions always means a concentration of 1 M. Substituting all these terms into the Nernst equation, one calculates E to be E = + .44 - (.59 / 2)In (10^-6 / 1) = .85 volt.

Question:

In order to determine their density, drops of blood are placed in a mixture ofxyleneof density 0.867 g\bulletcm^-3, and bromobenzene of density 1.497 g\bulletcm^-3, the mixture being altered until the drops do not rise or sink. The mixture then contains 72% ofxyleneand 28% ofbromobenzeneby volume. What is the density of the blood?

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Solution:

Using the definition of density density = mass/ volume every 72 cm^3 ofxylenehave a mass of 72 cm^3 × 0.867 g\bulletcm^-3 = 62.424 g, and every 28 cm^3 ofbromobenzenehave a mass of 28 cm^3 × 1.497 gm\bulletcm^-3_= 41.916 g. Thus, 100 cm^3 of the mixture have a mass of (62.424 + 41.916)g = 104.340 g. Thus the density of the mixture is 1.0434 g\bulletcm^-3. But blood neither rises nor sinks in this mixture, showing that the blood has no net force acting on it. Thus the weight of any drop of blood is exactly equal to the up thrust acting on it. But, by Archimedes' principle, the up thrust is the weight of an equal volume of mixture. Hence the blood and the mixture have the same densities; thus the density of blood is 1.0434 g\bulletcm^-3.

Question:

Air consists of a mixture of gas molecules which are constantly moving. Compute the kinetic energy K of a molecule that is moving with a speed of 500 m/s. Assume that the mass of this particle is 4.6 × 10^-26 kg.

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Solution:

The mass of the gas molecule, m = 4.6 × 10-^26 kg, and its speed v = 5 × 10^2 m/s, are the known observables. Using the equation: K = 1/2 mv^2 K = (1/2) (4.6 × 10^-26 kg) (5.0 × 10^2 m/s)^2 = 5.75 × 10^-21 J.

Question:

A chemist dissolves 300 g of urea in 1000 g of water. Urea is NH_2CONH_2. Assuming the solution obeys Raoult's'law, determine the following: a) The vapor pressure of the solvent at 0\textdegree and 100\textdegreeC and b) The boiling and freezing point of the solution. The vapor pressure of pure water is 4.6 mm and 760 mm at 0\textdegreeC and 100\textdegreeC, respectively and the K_f. = (1.86 C / mole) and K_b = (.52 C / mole).

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Solution:

To solve (a), you must employ Raoult's law, which states P_H(2)O= P\textdegree_H(2)O X_H(2)O, where P_H(2)O = the partial pressure of water (the solvent). P\textdegree_H(2)O = the vapor pressure of water when pure, and X_H(2)O is the mole fraction of H_2O in the solution. The mole fraction is equal to (N_H(2)O) / [(N_solute) + (N_H(2)O)],where N_H(2)O = moles of H_2O, and N_solute = mole of solute.You are told the P\textdegree_H(2)O for H_2O at the temperatures in question. Thus, to find the vapor pressures of the solution, you must calculate X_H(2)O and substitute into Raoult's law. To calculate the concentration of urea, remember moles = (grams) / (molecular weight). Since you have 300 grams of urea and the molecular weight of urea is 60.06 g, you have (300 / 60.06) or 4.995 moles of urea. Similarly, for water (MW = 18). you have (1000 g) / [(18 g / mole)] = 55.55 moles of water. Thus, the mole fraction of water = (55.55)/(4.995 + 55.55) = .917. Therefore, at 0\textdegree, [P_H(2)O = P\textdegree_H(2)O X_H(2)O = (4.6 mm)(.917) = 4.2 mm]at 760\textdegree, [P_H(2)O = P\textdegree_H(2)O X_H(2)O = (760)(.917) = 697 mm]. To solve (b), you use the equations ∆T_f = K_fm, where ∆T_f = freezing point depression, Kf = molal freezing pt. depression constant, and m = molality and ∆T_b = K_bm where ∆T_b = boiling point elevation, Kb = molal boiling point depression constant. If you find ∆T_f, you can calculate the freezing point, since the normal freezing point is decreased by this amount to give you the new freezing point. Boiling point works in the same way, except that you add to the normal boiling point. To calculate ∆T_f and ∆T_b, you must know molality. Molality = moles solvent per 1 kilogram solvent. You have 4.995 moles of urea and 1000 g of water or solvent, so that (4.995 mole) / (1 kg) \approx 5molal solution, Thus, ∆T_f = (1.86)(5) = 9.3\textdegreeC and ∆T_b = (.52)(5) = 2.6\textdegreeC. Therefore, the freezing point is now 0.0\textdegreeC - 9.3\textdegreeC = -9.3\textdegreeC and the boiling point is = 100\textdegreeC + 2.6\textdegreeC = 102.6\textdegreeC.

Question:

The density of alcohol is 0.8 g/ml. What is the weight of50 ml. of alcohol?

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Solution:

Density is defined as weight per unit volume. density= (weight) / (volume) = g / ml . Thus, one can solve for the weight of the alcohol by multiplying the density bythe volume. weight= density × volume weight= 0.8 g/ml × 50 ml = 40 g.

Question:

How should 5 capacitors, each of capacitance 1 \muF, be connected so as to produce a total capacitance of 3/7 \muF? Series Connection of C_1 and C_2

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Solution:

If all capacitors are joined in parallel, the resultant capacitance is 5 \muF. (For the resultant capacitance of a set of capacitors connected in parallel equals the sum of the individual capacitances). If the capacitors are connected in series, the resultant capacitance is 1/5 \muF (for the resultant capacitance of a set of capacitors connected in series equals the reciprocal of the sum of the reciprocals of the in-dividual capacitances). The connection is thus more complicated. Suppose that n capacitors are connected in parallel and 5 - n in series. The resultant capacit-ances are thus such that C1 = (1 + 1 + . . .) \muF = n \muF n times and(1/C_2 ) = [(1/1) + (1/1) + . . .] = (5 - n) \muF^-1 (5 - n) times orC_2 = (1/5 - n) \muF. If C_1 and C_2 are connected in parallel, then C_r = C_1 + C_2 = [n + (1/5 - n)] \muF = (3/7) \muF. 5n - n^2 + 1 = [(15 - 3n)/(7)] 35n - 7n^2 + 1 = 15 - 3n or7n^2 - 38n + 8 = 0 By the quadratic formula, n = [{- (- 38) \pm \surd{(- 38)^2 - 4 (7) (8)}} / {2 (7)}] There is thus no integral solution for n. But if C_1 and C_2 are connected in series (see figure), then (1/C_r ) = (1/C_1 ) + (1/C_2 ) = [(1/n) + 5 - n] \muF^-1 = (7/3) \muF^-1 . \therefore3 + 15n - 3n^2 = 7nor3n^2 - 8n - 3 = 0. \therefore(3n + 1) (n - 3) = 0. This has an integral solution for n, n = 3. Thus the required capacitance is given if 3 capacitors are connected in parallel, and the combination is connected in series with the other two.

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Question:

Find K_b andpK_bfor the acetate ion CH_3COO^-, The ionization constant of CH_3COOH is K_a = 1.75 × 10^-5;K_w= 1.00 × 10^-14.

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Solution:

This problem is concerned with conjugate acid- base pairs. In the general reaction, HA+H_2O\rightleftarrowsH_3O^++A^- acidbaseacidbase , HA is the conjugate acid of A^-, or conversely, A^- is the conjugate base of HA. H_2O has acted as a weak base. The acid and base ionization constants for a conjugate pair, as illustrated above, are K_a and K_b. K_a K_b =K_wwhich meanspK_a+pK_b=pK_w. Thus, if given the ionization constant for an acid or base, one can calculate the ionization constant for its conjugate base or acid. Also, remember that the K andpKvalues are related by the general equation, pK= - log K. Therefore, the values of K_b andpK_bare K_b =K_w/ K_a = 1.00 × 10^-14 / 1.75 × 10^-5 = 5.71 × 10^-10. pK_b= - log K_b = - log 5.71 × 10^-10 = - [- 10 + log 5.71] = 9.244.

Question:

Water flows into a water tank of large cross-sectional area at a rate of 10^-4 m^3/s ,but flows out from a hole of area 1 cm^2, which has been punched through the base. How high does the water rise in the tank?

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Solution:

When the water reaches its maximum height in the tank the pressure head is great enough to produce an outflow exactly equal to the inflow. Equil-ibrium is then reached and the water level in the tank stays constant. Since the cross-sectional area of the tank is large in comparison with the area of the hole, the water in the tank may be considered to have zero velocity. Further, the air above the tank and outside the holes are each at atmospheric pressure. Apply Bernoulli's theorem, p_1 + \rhogy_1 + (1/2) \Rhov^2_1 = p_2 + (1/2)\rhogy_2 + (1/2) \rhov p_1 + \rhogy_1 + (1/2) \Rhov^2_1 = p_2 + (1/2)\rhogy_2 + (1/2) \rhov 2 2 with point 1 at the surface of the water at a height h above the hole and point 2 the hole itself. Then p_1 + \rhogh + 0 = p_-a + 0 + (1/2) \rhov^2, where v is the velocity of efflux from the hole. Hence, v = \surd(2 gh). But at equilibrium v is the rate of influx divided by the area of the hole. That is, v = (10^-4 m^3/s)/(10^-4 m^2) = 1 m/s. Therefore the maximum height of water in the tank is h = v^2/2g = (1^2 m^2/s^2)/(2 × 9.8 m/s^2) =5.1 cm.

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Question:

List the parts of the human respiratory system. How is each adapted for its particular function?

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Solution:

The respiratory system in man and other air-breathing vertebrates includes the lungs and the tubes by which air reaches them. Normally air enters the human respiratory system by way of the external nares or nostrils, but it may also enter by way of the mouth. The nostrils, which contain small hairs to filter in-coming air, lead into the nasal cavities, which are separated from the mouth below by the palate. The nasal cavities contain the sense organs of smell, and are lined with mucus-secreting epithelium which moistens the incoming air. Air passes from the nasal cavities via the internal nares into the pharynx, then through the glottis and into the larynx. The larynx is often called the "Adam's apple," and is more prominent in men than women. Stretched across the larynx are the vocal cords. The opening to the larynx, called the glottis, is always open except in swallowing, when a flap-like structure (the epiglottis) covers it. Leading from the larynx to the chest region is a long cylindrical tube called the trachea, or windpipe. In a dissection, the trachea can be distinguished from the espophagus by its cartilaginous C-shaped rings which serve to hold the tracheal tube open. In the middle of the chest, the trachea bifurcates into bronchi which lead to the lungs. In the lungs, each bronchus branches, forming smaller and smaller tubes called bronchioles. The smaller bronchioles terminate in clusters of cup-shaped cavities, the air sacs. In the walls of the smaller bronchioles and the air sacs are the alveoli, which are moist structures supplied with a rich network of capillaries. Molecules of oxygen and carbon dioxide diffuse readily through the thin, moist walls of the alveoli. The total alveolar surface area across which gases may diffuse has been estimated to be greater than 100 square meters. Each lung, as well as the cavity of the chest in which the lung rests, is covered by a thin sheet of smooth epithelium, the pleura. The pleura is kept moist, en-abling the lungs to move without much friction during breathing. The pleura actually consists of two layers of membranes which are continuous with each other at the point at which the bronchus enters the lung, called the hilus (roof). Thus, the pleura is more correctly a sac than a single sheet covering the lungs. The chest cavity is closed and has no communication with the outside. It is bounded by the chest wall, which contains the ribs on its top, sides and back, and the sternum anteriorly. The bottom of the chest wall is cov-ered by a strong, dome-shaped sheet of skeletal muscle, the diaphragm. The diaphragm separates the chest region (thorax) from the abdominal region, and plays a crucial role in breathing by contracting and relaxing, changing the intrathoracic pressure.

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Question:

The solar constant, the power due to radiation from the sun falling on the earth's atmosphere, is 1.35 kW\textbulletm^\rule{1em}{1pt}2. What are the magnitudes of E^\ding{217} and B^\ding{217} for the electro-magnetic waves emitted from the sun at the position of the earth?

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/Users/wenhuchen/Documents/Crawler/Physics/D25-0810.htm

Solution:

Starting with the electromagnetic waves at the earth, it is possible to determine E^\ding{217} and B^\ding{217} by two methods, (a) ThePoyntingvector S^\ding{217} = (1/\mu_0) E^\ding{217} × B^\ding{217} gives the energy flow across any section of the field per unit area per unit time. Here, E^\ding{217} and B^\ding{217} are the instantaneous electric field and magnetic induction, respectively, at a point of space, and \mu_0 is the permeability of free space. If we approximate the sun as a point source of light, then we realize that it radiates electromagnetic waves in all directions uniformly. However, the distance between earth and sun is very large, and we may approximate the electromagnetic waves arriving at the surface of the earth as plane waves. For this type of wave, E^\ding{217} and B^\ding{217} are perpendicular. Thus \mid S^\ding{217} \mid = \mid (1/\mu_0) E^\ding{217} × B^\ding{217} \mid = EH = 1.35 × 10^3 W\textbulletm^\rule{1em}{1pt}2 where we have used the fact that \mid H^\ding{217} \mid = \mid (B^\ding{217}/\mu_0) \mid in vacuum. (H is the magnetic field intensity.) But in the electromagnetic field in vacuum, \epsilon_0E^2 = \mu_0H^2, or E \surd(\epsilon_0/\mu_0) = H. Then E × \surd(\epsilon_0/\mu_0) E = EH = 1.35 × 10^3 W\textbulletm^\rule{1em}{1pt}2 orE^2 = \surd(\mu_0/\epsilon_0) × 1.35 × 10^3 W\textbulletm^\rule{1em}{1pt}2 = 377 \Omega × 1.35 × 10^3 W\textbulletm^\rule{1em}{1pt}2. E = \surd(5.09 × 10^5) V\textbulletm^\rule{1em}{1pt}1 = 0.71 × 10^3 V\textbulletm^\rule{1em}{1pt}1. Similarly, B = \mu_0H = [{\mu_0 (1.35 × 10^3 W\textbulletm^\rule{1em}{1pt}2)}/E] B = {[(4\pi × 10^\rule{1em}{1pt}7 Weber \textbulletA^\rule{1em}{1pt}1 \textbulletm^\rule{1em}{1pt}1) (1.35 × 10^3 W \textbulletm^\rule{1em}{1pt}2)]/(.71 × 10^3 V \textbulletm^\rule{1em}{1pt}1)} B = 2.39 × 10^\rule{1em}{1pt}6 {(Weber \textbulletW\textbulletA^\rule{1em}{1pt}1 \textbulletm^\rule{1em}{1pt}2)/V} But 1 W = 1 J\textbullets^\rule{1em}{1pt}1 and 1 V = 1 J\textbulletC^\rule{1em}{1pt}1 whence B = 2.39 × 10^\rule{1em}{1pt}6 {(Weber \textbullet J \textbullet s^\rule{1em}{1pt}1 \textbulletA^\rule{1em}{1pt}1 \textbullet m^\rule{1em}{1pt}2)/(J \textbullet C^\rule{1em}{1pt}1)} B = 2.39 × 10^\rule{1em}{1pt}6 Weber \textbullet m^\rule{1em}{1pt}2. (b) The electromagnetic energy density (or, energy per unit volume) in an electromagnetic field in vacuum is \mu_0H^2 = \epsilon_0E^2. The energy falling on 1 m^2 of the earth's atmosphere in 1 s is the energy initially con-tained in a cylinder 1 m^2 in cross section and 3 x 10^8 m in length; for all this energy travels to the end of the cylinder in the space of 1 s. Hence the energy density near the earth is \mu_0H^2 = \epsilon_0E^2 = {(1.35 × 10^3 W\textbulletm^\rule{1em}{1pt}2)/(3 × 10^8 m\textbullets^\rule{1em}{1pt}1)} Here, \epsilon_0 is the permittivity of free space. E^2 = {(1.35 × 10^3 W\textbulletm^\rule{1em}{1pt}2)/(8.85 × 10^\rule{1em}{1pt}12 C^2 \textbulletN^\rule{1em}{1pt}1 \textbulletm^\rule{1em}{1pt}2 × 3 × 10^8 m\textbullets^\rule{1em}{1pt}1)} E^2 = {(1.35 × 10^7)/26.55} {W/(C^2 \textbulletN^\rule{1em}{1pt}1 \textbulletm\textbullets^\rule{1em}{1pt}1)} But1 W = 1 J\textbullets^\rule{1em}{1pt}1 = 1 N\textbulletm\textbullets^\rule{1em}{1pt}1 E^2 = 5.085 × 10^5 N/C N^\rule{1em}{1pt}1 = 5.085 × 10^5 N^2/C^2 orE = .71 × 10^3 N\textbulletC^\rule{1em}{1pt}1 = .71 × 10^3 V\textbulletm^\rule{1em}{1pt}1 Also\mu_0H^2 = (B^2/\mu_0) = {(1.35 × 10^3 W\textbulletm^\rule{1em}{1pt}2)/(3 × 10^8 m\textbullets^\rule{1em}{1pt}1)} orB^2 = {(4\pi × 10^\rule{1em}{1pt}7 N\textbulletA^\rule{1em}{1pt}2 × 1.35 × 10^3 W\textbulletm^\rule{1em}{1pt}2)/(3 × 10^8 m\textbullets^\rule{1em}{1pt}1)} B = 2.36 × 10^\rule{1em}{1pt}6 Wb\textbulletm^\rule{1em}{1pt}2.

Question:

A cable drum of inner and outer radii r and R is lying on rough ground, the cable being wound round the inner cylinder and being pulled off from the bottom at an angle \texttheta to the horizontal. An inquiring student stroll-ing by notes that when the cable is pulled by a workman, with \texttheta a small angle, the drum rolls without slipping toward the workman. Whereas if \texttheta is large, the drum rolls without slipping in the opposite direction. He works out a value for the critical angle \texttheta_0 which separates the two types of motion. What is the value of \texttheta_0?

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/Users/wenhuchen/Documents/Crawler/Physics/D04-0204.htm

Solution:

We define the drum's acceleration to be positive when the drum moves toward the workman. Hence the mathematical condition that the drum roll toward or away from the worker is that a (the acceleration) be positive or negative, respectively. The critical con-dition distinguishing the 2 types of motion is that a = 0. Therefore, to find the critical angle \texttheta_0, we find a as a function of \texttheta, set it equal to zero, and solve for \texttheta. We can use Newton's Second taw to relate the net force on the drum to its acceleration. (We do this for the vertical and horizontal directions separately.) Figure (1) shows the drum with the forces acting on it. The force applied by the workman acts tangentially to the inner cylinder at such a position that the angle between this tangent, and the horizontal is \texttheta. It follows that the angle between the corresponding radius and the vertical is \texttheta also, and that this radius is at right angles to the tangential force F^\ding{217}. The other forces acting are the weight Mg^\ding{217} of the cable drum, the normal forces exerted by the ground on the drum at the two points of contact, which combine into a resultant N^\ding{217} passing through the center of gra-vity, and the frictional forces at the same points of contact which combine to form a single resultant force f^\ding{217} . There is no movement in the vertical direction. Hence N = Mg - F sin \texttheta.(1) The forces in the horizontal direction produce an acceleration a. Thus F cos \texttheta - f = Ma.(2) Further, the moments of the forces about the center of mass, O produce a rotational acceleration about that point. The only forces whose lines of action do not pass through the center of gravity are F and f. Hence Fr - fR = l\alpha(3) where I is the moment of inertia of the drum about its center of mass. At the points at which the drum touches the ground no slipping occurs. Therefore instantaneously these points are at rest. But all points of the drum have an acceleration a forward and in addition the points of contact, due to the rotation about the center of mass, have a further linear acceleration R\alpha forward. Thus a + \alphaR= 0 or\alpha= - a/R(4) We wish to eliminate f from (2). Solving (3) for f (Ff - I\alpha)/R= f(5) Substituting (4) in (5) f = (Ff - Ia/R)/R f = [(Fr)/R] + [(Ia)/R2](6) Inserting (6) in (2) F cos \texttheta - [(Fr)/R] - [(Ia)/R^2 ] = Ma Solving for a [F cos \texttheta - {(Fr)/R}]/[M + (I/R^2)] = a ora = [F(cos \texttheta - r/R)]/[M + (I/R^2 )](7) Since we do not know F, we solve (1) for F and insert this in (7) F = (Mg - N)/(sin \texttheta) anda = [{(Mg - N)/(sin \texttheta)} {cos \texttheta - (r/R)}] / [M + (I/R^2)](8) The critical value of \texttheta, (\texttheta_0), is found by setting (8) equal to 0, whence a = 0 cos \texttheta_0 - (r/R) = 0 cos \texttheta_0 = (r/R)(9) If cos \texttheta > r/R, the drum rolls towards the workman, and vice versa. This result could be obtained more easily by con-sidering rotation about A, the line of the drum in-stantaneously at rest. The only force that does not pass through A is F^\ding{217}, the applied force. If the line of action of F^\ding{217} cuts the ground to the left of A, the moment of F^\ding{217} about A causes the drum to roll to the right. If the line of action of F^\ding{217} cuts the ground to the right of A, the moment of F^\ding{217} about A causes the drum to move to the left. If the line of action of F^\ding{217} passes through A, the drum is stationary and \texttheta has the critical value \texttheta_0. Figure (2) shows this situation. Since the line of action of F^\ding{217} is tangential to the inner cylinder, OB and AB are at right angles and \angle AOB is \texttheta_0. \therefore cos \texttheta_0 = r/R .

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Question:

Why does succession occur? Why can the first colonists not simplyseize an area and hold it against subsequent intruders?

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/Users/wenhuchen/Documents/Crawler/Biology/F30-0779.htm

Solution:

Succession occurs independently of the climate. Climate may be a majorfactor in determining what types of species will follow one another, butsuccession itself results from other changes. The most important of theseare the modifications of the physical environment produced by the communityitself. Mostsuccessionalcommunities tend to alter the area in whichthey occur in such a way as to make it less favorable for themselves andmore favorable for other communities. In effect, each community in thesuccession sows the seeds of its own destruction. Consider the alterations initiated by pioneer com-munities on land. Usually these communities will produce a layer of decay matter or litter on thesurface of the soil. The accumulation affects the runoff of rainwater, thesoil temperature, and the formation of humus (organic decomposition products). The humus, in turn contributes to the soil development and thus altersthe availability of nutrients, the water, the pH and aeration of the soil, and the types of soil organisms that will be present. But the organismsthat are characteristic of the pioneer communities that producedthese changes may not prosper under the new conditions, and theymay be replaced by invading competitors that do better in an area withthe new type of soil. In another manner, plants sometimes foreclose futurereproduction by the process of their own growth. In Australia, for example, the Eucalyptus trees of the open, sunny savannas provide shadein which young tress from the nearby rain forests can sprout and grow. When these intruders reach their full size, they cast too much shade forthe Eucalyptus to reproduce themselves. The Eucalyptus finallydieout asthe rain forest takes over the land completely. Not allsuccessionalstages prepare the way for their own decline andfall. In the eastern United States, mixed forests of pine and oak modifythe soil in a way that makes it more favorable for the growth of their ownseedlings than for those of their competitors. Succession in such casesoccurs simply because slower growing trees rise to dominance at a latertime and alteration of the environment therefore may have nothing to dowith re-placement. An orderly sequence of communities in a succession usually occurs becauseas the species specialized for each stage persists and changes theenvironment, an excellent habitat for the next set of species is provid-ed. In fact, ecologists recognize two broad classes of species involvedin this ordered kind of succession. Opportunistic species are able todisperse widely, and they grow and breed rapidly. The opportunistic speciesinclude the pioneers and the species involved in the early successionstages. They have adopted the strategy of finding and utilizing emptyspace before other species preempt it. Stable species, on the other hand, specialize in competitive superiority. Forest trees of most kinds belongto this category. They grow and disperse more slowly, but in their encounterswith opportunistic species, they are able to grow successfully atthe expense of the other species and remain for longer periods of time. Forest trees commonly make up a large percentage of the land climax communities.

Question:

What percent of the AB particles are dissociated by water if the freezing point of a(.0100 m) AB solution is -0.0193\textdegreeC? The freezing point lowering constant of water is (-1.86\textdegreeC / mole).

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/Users/wenhuchen/Documents/Crawler/Chemistry/E07-0245.htm

Solution:

To answer this question, you must determine the number of particles dissociated and divide that number by the initial number of particles available for dissociation. To accomplish this, you must be able to relate quantitatively the freezing point depression with the concentration of particles in solution. From the equation, ∆T_f=k_fm, where ∆T_f= the depression of the freezing point,k_f= the freezing point depression constant, and m =molalityof the solution, one can determine the concentration of the particles. From the data given, the total concentration of particles, m = (∆T_f/k_f) = (-.0193\textdegreec) / (-1.86\textdegree c/m) = .0104m. With this in mind, let x be the concentration of AB particles that dissociate.As such,you have xmolalfor A^+ and xmolalfor B^-, since AB \rightarrow A^+ + B^-. You are told that you are start with amolalsolution of.0100. Therefore, the number of particlesundissociatedis .01 - x. As such, the total concentration of particles is x + x + (.0100 - x) , which is equal to .0100 + x. Before, however, you said it was .0104 m from the equation ∆T_f=k_fm. It follows then, that you can equate the two. Thus, .0104 m = (.0100 + x)m, Solving for x, you obtain, x = .0004 m. This number, x = .0004 m, represents, themolalcon-centrations of each A^+ and B^-. Therefore, it must be the number of particles that dissociated. You started with .0100 m concentration of particles, Therefore, the percent that dissociated is (.0004) / (.0100) = (.04) = 4%

Question:

A small sports arena is designed by an architect in the form of a dome with radius of curvature R = 115 ft mounted on a cylindrical base 75 ft in radius and 30 ft in height. The dome acts as a spherical mirror with a focal length f = 1/2R = 57.5 ft. The top of the dome is the vertex of the mirror. It is of interest to calculate the location of the focal point with respect to the ground surface of the aren

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/Users/wenhuchen/Documents/Crawler/Physics/D26-0829.htm

Solution:

From the diagram y^2 + (R - x)^2 = R^2 (R - x)^2 = R^2 - y^2 = (115^2 - 75^2)ft^2 = 87^2ft^2 R - x = 87 ft x = (115 - 87)ft x = 28 ft The distance from the vertex of the dome to the ground sur-face is x + 30 = 58 ft, the same as the focal length. Thus the focal point of the mirror lies on the ground surface at the center of the arena. As a result of this there will be a tendency for spec-tator noise to be focused at the center of the ground sur-face, and the noise there is liable to be deafening. There exists a hockey arena that has been designed this way (accidentally), and in the center ice region the noise is of such intensity that the players can not hear the whistles of the officials.

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Question:

Obtain polar solutions for the following polar equations, in which the ordered pair (R, T) is interpreted as R = length of the vector going from the origin to the point R, and T = angle between polar axis and that vector: (a) R = 1 - 2 sin T (b) R = 2 + 2 sin T (c) R = 1 + 2cosT - sin^2T

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G10-0239.htm

Solution:

Although it is not absolutely necessary, we can make use of the defined statement function. The value of T will be incremented by 15 degrees each time by establishing a loop. Conversion to radians is accomplished with k = \pi/180. The output will appear in tabular form, with values rounded to the nearest hundredth: 1\O REM SOLVING POLAR EQUATIONS IN 15 - 2\O REM DEGREE STEPS 3\O LET K = 3.14159/18\O.\O 4\O PRINT "ANGLE", "1-2SIN(T)", \textquotedblright2+2SIN(T)", "1+2COS(T) - SIN(T)_\uparrow2" 5\O DEF FNR(X) = INT(X\textasteriskcentered1\O\O)/1\O\O 6\O DEF FNA(X) = 1 - 2\textasteriskcenteredSIN(K\textasteriskcenteredX) 7\O DEF FNB(X) = 2 + 2\textasteriskcenteredSIN(K\textasteriskcenteredX) 8\O DEF FNC(X) = 1 + 2\textasteriskcenteredCOS(K\textasteriskcenteredX) - SIN(K\textasteriskcenteredX)_\uparrow2 9\O FOR T = \O TO 3\O\O STEP 15 1\O\O PRINT T, FNR(FNA(T)), FNR(FNB(T)), FNR(FNC(T)) 11\O NEXTT 12\O END [Note: a variety of polar equations can be evaluated with this program by making minor adjustments.]

Question:

Translate the following Boolean expressions into Venn diagrams: a) A + B(Inclusive OR) b) A \bullet B(AND) c) \backsim A(NOT)

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G03-0038.htm

Solution:

Venn diagrams offer a pictorial representation of logical relations. When translating from Boolean algebra, sometimes the nota-tion changes. We will give the diagrams and the alternate notations: a) The expression A + B is an OR function, also known as the union of two sets. Elements can be in A, or in B, or in both A and B. In the Venn diagram of Fig. 1, the shaded area is the area of interest. A + B can also be written as AUB, using set notation. b) The AND function of Boolean algebra is given by A \bullet B. This may also be considered to be the intersection of sets A and B, written as A\capB . By \cap definition, an intersection is the set of all elements contained in both A and B simultaneously. Using a Venn diagram, we represent it as shown in Fig. 2. Again, the shaded area is the part described by the relation A \textbullet B (or A\capB). \cap c) The NOT function, \textasciitilde A, is also known as the complement of a set A. The complement of a set A includes all elements in the universe that are not in A. We can write the complement as \={A}, and our Venn diagram is shown in Fig. 3. The shaded area represents \={A}.

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Question:

Differentiate between herbaceous and woody plants and between annual, biennial and perennial plants. How do the stem of woody plants increase in diameter?

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/Users/wenhuchen/Documents/Crawler/Biology/F10-0248.htm

Solution:

Plants are generally differentiated into two types based on the nature of their stems. Herbaceous plants have a supple, green, rather thin stem, woody plants have a thick, tough stem or trunk, covered with a layer of cork. This cork is derived from the cork cambium, a layer of meristematic cells in the outer cortex of woody stems. Herbaceous plants are typically annuals or biennials. Annual plants start from seed, develop flower, and produce seeds within a single growing season, and die before the following winter. Biennial plants characteristically have two-season growing cycles. During the first season, while the plant is growing, food is stored in the root. Then the top of the plant dies and is replaced in the second growing season by a second top which produces seeds. Carrots and beets are examples of biennials. Woody plants, exemplified by a wide variety of trees and shrubs, are usually perennial plants, which live longer than two years. Perennials have been found that live hundreds or even thousands of years. They produce seeds yearly. The stems of woody plants resemble herbaceous ones during their first year of growth, with each containing an epidermis, a cortex, an endodermis, a pericycle, vascular bundles, and a pith. But by the end of the first growing season, a cambium layer has taken a form which extends as a continuous ring between the primary xylem and phloem, pushing the phloem to the periphery of the stem and keeping the xylem toward the pith (see figure). This cambium layer, called the vascular cam-bium, is a region of rapid cell division, that is, the cambium is meristemic. In each successive year, this ring of meristem divides to form two types of cells; those inside the ring differentiate into the secondary xylem elements, and those outside the ring become the secondary phloem cells. The yearly deposits of xylem form the annual rings. Addition of a new ring to the old ones each year causes the stem of a woody plant to increase in diameter. Monocotyledons and some dicotyledons have herbaceous stems, while all gymnosperms and some dicotyledons have woody stems.

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Question:

A rectangular cistern 6 ft x 8 ft is filled to a depth of 2 ft with water. On top of the water is a layer of oil 3 ft deep. The specific gravity of the oil is.6. what is the absolute pressure at the bottom, and what is the total thrust exerted on the bottom of the cistern?

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/Users/wenhuchen/Documents/Crawler/Physics/D10-0413.htm

Solution:

The total force F acting at the bottom of the cistern is the sum of the weights of the water Wg, the oil W_0, and the air in the atmosphere above the cistern [F_atm]. Since Density (d) = Mass/Volume and Pressure = Force/Area Then F = M_wg + M_0g + F_atm = d_wv_wg + d_0v_0g + P_atm A where V_w and V_0 are the volumes of water and oil re-spectively, and A is the cross sectional area of the cistern. V_w = h_2A V_0 = h_1A ThereforeF = (d_wh_2g + d_0h_1g + P_atm)A d_w = (62.4 slug)/(32 ft^3) By definition of specific gravity (or relative density) of oil S_0 = d_0/d_w Since we are given that S_0 = .6 we have d_0 = .6 d_w = (.6) [(62.4 slug)/(32 ft^3)] Also P_atm = 14.7 lb/in^2 = 14.7 lb/in^2 × 144 in^2/1 ft^2 h_1 = 3 ftandh_2 = 2 ft Therefore F = [{(62.4 slug)/(32 ft^3)} (2 ft) (32 ft/sec^2) + .6{(62.4 slug)/(32 ft^3)} (32 it/sec^2) + (14.7) (144)lb/ft^2]A orF = (2357 lb/ft^2) A P_bottom = F/A = 2357 lb/ft^2 SinceA = 6 ft × 8 ft F = (2357 lb/ft^2) (6 × 8 ft^2) = 113,000 lb.

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Question:

For the formation of hydrogen iodide, H_2 (g) + I_2 (g) \rightleftarrows 2HI(g) , the value of the equilibrium constant at 700K is 54.7. If one mole of H_2 and one mole of I_2 are the only materials initially present, what will be the equilibrium concentrations of H_2 ,I_2 , and HI .

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/Users/wenhuchen/Documents/Crawler/Chemistry/E09-0314.htm

Solution:

This problem is an application of the expression for the equilibrium constant K = [HI]^2 / {[H_2] [I_2]} , except that the concentrations are unknown. However, as we shall see, we do not need to know the concentrations. Let x be the number of moles of H_2 that disappeared at equili-brium. Since one mole of I_2 is consumed and two moles of HI are formed per mole of H_2 consumed, then, at equilibrium, x moles of I_2 will have reacted, 2x moles of HI will have formed, and 1\Elzbar x moles of H_2 and 1 \rule{1em}{1pt} x moles of I_2 will remain. If we let the volume of the vessel in which the reaction takes place be v, then [H_2 ] = (1\Elzbar x)/v , [I_2] = (1\Elzbar x)/v , and [HI] = 2x/v . The expression for the equilibrium constant is then k= 54.7 = [HI]^2 / {[H_2 ] [I_2]} = [2x/v]^2/ { [(1\Elzbar x)/v] [(1\Elzbar x)/v] } = [4x^2/ (1\Elzbar x)^2 ] [v^2 / v^2 ] = 4x^2/ (1\Elzbar 2x + x^2 ) . or 54.7\Elzbar 2(54.7)x + 54.7x^2 = 4x^2 50.7x^2\Elzbar 109.4x + 54.7 = 0 . Using the quadratic formula, x = { 109.4 \pm \surd[ (\Elzbar109.4)^2\Elzbar 4(50.7)(54.7)] } / { 2(50.7) }] or x = 0.79 , x = 1.37 . Since x, the number of moles of H_2 or I_2 that reacted, cannot be greater than one, the answer x = 1.37 must be a nonphysical entity, and we retain only the answer x = 0.79. Hence, at equilibrium, [H_2] = 1\Elzbar x = 1\Elzbar 0.79 = 0.21 mole [I_2]= 1\Elzbar x = 1\Elzbar 0.79 = 0.21 mole [HI]= 2x = 2(0.79) = 1.58 moles .

Question:

What is an anhydro bond? Give examples of these bonds in common biological molecules.

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/Users/wenhuchen/Documents/Crawler/Biology/F01-0017.htm

Solution:

In biology, the most important covalent bonds joining molecules together are anhydro bonds. These are formed by removing a molecule of water to join two other molecules together. In other words, an -OH group is removed from one molecule and an -H group from the other to leave the two molecules bonded to each other usually by an oxygen or nitrogen atom. Examples of anhydro bonds in common biological molecules include: glycoside bonds, peptide bonds, and ester bonds (see figure). Glycoside bonds are the anhydro bonds of carbohydrates and are formed by removing an H from an alcohol group (-OH) on one sugar and an OH from the other sugar. Peptide bonds are the anhydro bonds of proteins and are formed by removing an -OH from the carboxyl group (-COOH) of one amino acid and an H from the amino group (-NH_2) of another amino acid. Ester bonds are an-hydro bonds of fats, formed by removing an OH from the carboxyl group of a fatty acid and an H from an alcohol group of glycerol. Other ester bonds that have great biological significance are phosphate esters [formed by removing an OH from phosphoric acid (HO-PO_3H_2) and an H from a sugar] and thioesters (formed by removing an OH from the carboxyl group of an acid and an H from a -SH group).

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Question:

Soon after the Mendelian laws became firmly established, numerous exceptions to Mendel's second law, the Law of Independent Segregation, were demonstrated by experiments. Parental non-allelic gene combinations were found to occur with much greater frequencies in offspring than were the non-parental combinations. How can this be explained?

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Solution:

The Law of Independent Segregation states that when two or more pairs of genes are involved in a cross, the members of one pair segregate independently of the members of all other pairs. This law was substantiated by Mendel's experiments with sea plants. However, in Mendel's time, the physical nature of genes was not known, nor was it known how they are carried. When it was learned that chromosomes are the bearers of genes, the reasons why Mendel's law was both supported by some experiments and negated by others became obvious. It was seen that the chromosomes are relatively few in number. For example, Orosophila have only four pairs of chromosomes. Man has twenty-three pairs. In compa-rison, however, the number of genes possessed by each species is very large, often in the thousands. Since there are so many more genes than chromosomes, and the chromosomes carry the genes, it follows that there must be many genes on each chromosome. And since it is now known that it is whole chromosomes which segregate independently during meiosis, gene separation can only be independent if the genes in question are on different chromosomes. Genes located on the same chromosomes are physically forced to move together during meiosis. Such genes are said to show linkage. When genetic experiments are performed using genes that are linked, very different ratios from the expected Mendelian ratios are obtained. Genes that were linked together on parental chromosomes tend to remain together in the gametes, and so occur in conjuction with one another more frequently in the offspring than they would if they had segregated independently. Exactly how linkage produces these variant ratios will be dealt with in later problems. At the present we can see that linkage explains how exceptions to the Law of Independent Segregation could possibly occur.

Question:

Calculate the number of parity bits required in order to code aninformation consisting of one binary bit on each input line, intothe Hamming code, if each input information has: a) 8 bits, and, b) 4 bits.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G06-0131.htm

Solution:

The number of parity bits K, required to be added to n information linesis calculated as follows: with K parity bit, the total number of combinationsof 0's and 1's that one can have is 2^K . Out of these 2^K combinations, one must reserve one combina-tion to denote a no-error conditionwhen the lines are received at the receiver. The other (2^K - 1) combinationscan be used to identify each one of the incoming lines, so thatone can point out which incoming line has developed an error. Now, the number of incoming lines = n + k i.e., sum of the n information-bits lines plus the k parity-bits lines that were addedat the transmitter. Hence, one must satisfy the condition that 2k\geqn + k + 1 . (a) As there are 8 information bits or lines,n =8 . \therefore2k\geq 8 + k+1 . i.e.,2k\geq9 + k Now, k can be found by a trial and error method. Let k = 5 therefore,2k= 2^5 = 32. And 9 + k = 9 + 5 = 14. Hence, 32 \geq14 . The condition is satisfied, but maybe a more "economical" number can be found. Letk = 4 then2k= 2^4 = 16 . And, 9 + k = 9 + 4 = 13. 16 \geq13 . Again the condition is satisfied. But if k = 3, the condition will no longer be satisfied. Hence, k = 4 is the correct answer, i.e., 4 parity bits must be added. (b) If there are 4 information bits or lines, n = 4. Again, k is selected by trial anderror method until the condition is just satisfied (i.e., the least possible k). Let k = 3 \therefore 2^k = 2^3 = 8. And, n + k + 1 = 4 + k + 1 = 5 + k = 5 + 3 = 8 8 \geq8 . thecondition is satisfied, while k = 2 leads to the incorrect comparision4 \geq 7 . Thus, with n = 4 information bits, one must have k = 3 paritybits.

Question:

Mitosis is only one of four stages of the complete cell life cycle. List the other three and indicate the amount of RNA synthesis in each of the four stages. Explain why RNA output is greater in some stages than in others.

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/Users/wenhuchen/Documents/Crawler/Biology/F24-0626.htm

Solution:

The length of the cell cycle under optimal conditions varies from 18 - 24 hours for most cells. Cell cycles are usually divided into four phases: m (mitosis), G_1 (period prior to DNA synthesis), S (period of DNA syn-thesis) and G_2 (period between DNA synthesis and mitosis). M is the shortest stage, usually lasting about one hour. There is no synthesis of either DNA or RNA in this stage. The absence of RNA synthesis may be the result of the tightly coiled and condensed nature of the DNA at this point, making it impossible to transcribe. Recall that for transcription to occur, DNA must first unfold. During G_1, there is production of mRNA and the cell increases in size. Next, in the S phase, the DNA of the cell becomes doubled because the genes are duplicating. Duplicating genes cannot produce RNA, but since all the genes do not duplicate at once, there is still some output of RNA, although in reduced quantity. Finally, full output of RNA is resumed in G_2 . RNA production is greater in G_1 and G_2 because during these two stages the DNA is relaxed (uncondensed) and is not involved in duplication. The relaxed nature of the DNA favors RNA synthesis, which can only occur when the double stranded DNA separates into two individual strands.

Question:

A plane has an airspeed of 120 mi/hr. What should be the plane's heading if it is to travel due north, relative to the earth, in a wind blowing with a velocity of 50 mi/hr in an easterly direction?

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/Users/wenhuchen/Documents/Crawler/Physics/D01-0021.htm

Solution:

The figure shows the situation. The plane has a velocity relative to the air, V\ding{217}_PA, of 120 mi/hr in a direction of \texttheta degrees west of north. The air has a velocity relative to the earth of V\ding{217}_AE (50 mi/hr east). We require the plane to travel with speed v_pa due north. From the figure tan \texttheta = (V\ding{217}_AE/V\ding{217}_PE) = [V_AE/(V_PA^2 - V_AE^2)^1/2] Here, we have used the Pythagorean theorem. tan \texttheta = [(50 mi/hr)/({(120 mi/hr)^2 - (50 mi/hr)^2})^1/2] = [(50 mi/hr)/(109.09 mi/hr)] = tan^-1 (.4584) \approx 24\textdegree38'.

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Question:

How many kinds of neurons are there? What functions are performed by each type?

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Solution:

Neurons are classified on the basis of their functions into sensory, motor orinterneurons. Sensory neurons, or afferent neurons, are either receptors that receive impulses or connectors of receptors that conduct information to the central nervous system (brain and spinal cord). Motor neurons , or efferent neurons, conduct information away from the central nervous system to the effectors, for example, muscles and glands. Interneurons , which connect two or more neurons, usually lie wholly within the central nervous system. There-fore, they have both axonal and dendritic ends in the central nervous system. In contrast, the sensory and motor neurons generally have one of their endings in the central nervous system and the other close to the periphery of the body.

Question:

At pH values of 4 to 9, natural amino acids exist as polar or zwitter ions: H_2N^+CHZCO^-_2. At a pH of 12, what would be the predominant ionic type? What would it be at a pH of 2? Can the uncharged molecular form, H_2NCHZCO_2 H ever predominate at any pH?

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Solution:

To answer these questions, you need to con-sider the chemical composition of amino acids. Amino acids are compounds whose molecules possess both the Amino (-NH_2) and the carboxy (-CO_2H) functional groups. An amino group is basic - if it is an electron donor. The carboxylic group is acidic; it releases protons (H^+). In a zwitter ion, as indicated, the carboxylic group has donated its proton to become COO^- \equiv CO_2 and NH_2 has received a proton to become (NH^+_3). The presence of such an ion is dependent on pH, which is a measure of the H^+ or hydronium ion (H_3O^+) concentration. Thus, pH gives an indication of the acidity or basicity of a solution. The lower the pH, the more acidic the solution, and, as such, the greater the concentration of H^+ ions. The higher the pH, the more basic the solution, and, the lower the con-centration of H^+ ions. One can now proceed as follows: pH = 12. At this pH, which is high, the H^+ con-centration is low; the solution is basic. Thus, the amino group (NH_2) will probably not be protonated. The carboxylic group, which is acidic, can and does react in the basic solution to go to COO^-. Thus, at pH = 12, H_2NCHZCO_2^- pre-dominates. At pH = 2, the reverse is true. The solution is very acidic, so that NH_3 will be protonated to NH^+_3. CO_2H, the acid portion, has no base to react with. Thus, it does not ionize to CO_2^-. Thus, H^+_3HCHZCO_2H. H_2NCHZCO_2H can never predominate for the molecule contains both an acid and base groups; they interact with each other to form zwitter ions.

Question:

During low tide, tough, shapeless organisms can be seen attached to rocks near the low tide line. These organisms can be seen to eject water from their bodies when they contract. What are these organisms?

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/Users/wenhuchen/Documents/Crawler/Biology/F13-0325.htm

Solution:

These sessile marine organisms are known as the tunicates, or sea squirts, which belong to the subphylum Urochordata. There is nothing in the external appearance of the adult that suggests it is a member of the chordates. The adult body is covered by a tough, leathery tunic, composed principally of a kind of cellulose. Within the tunic, the greater portion of the body is comprised of a pharynx perforated by numerous gill slits. The tunic has two openings: the incurrent and excurrent siphens. Water and nutrients enter the body via the incurrent siphon. After entering the pharynx and having the food particles digested, the waste products are carried by the water to the excurrent siphon where they are released from the body. During reproduction, the gametes leave the animal via the excurrent siphon. Tunicates are filter- feeders, removing food particles from the stream of water passing through the pharynx. The food particles are trapped in mucus secreted by part of the pharynx called the endostyle, and are carried down by ciliated cells into the esophogus and digestive gland. The flow of water through the tunicate is maintained by the contraction of the body wall. Whereas the adult tunicate is a sessile organism, the tunicate larva is motile, and shows a greater re-semblance to the other chordates. They possess a well- developed dorsal hollow nerve cord, a notochord beneath it in the tail region, and pharyngeal gill slits. When the larvae settle down and undergo metamorphosis into the adult form, the notochord is completely lost. Most of the nerve cord is also lost, the rest remaining only as a ganglion or "brain" above the pharynx.

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Question:

What is the explanation for the following trends inlattice energies ? NaF - 260 Kcal / moleNaCl- 186 Kcal / mole NaCl - 186 Kcal / moleKCl- 169 Kcal / mole NaBr - 177 Kcal / moleCsCl- 156 Kcal / mole

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/Users/wenhuchen/Documents/Crawler/Chemistry/E17-0633.htm

Solution:

Lattice energy is an indication of the stability of the species. In the first column, since the sodium (Na) ion is common to all species , one should concentrate on F^-,Cl^-, and Br^- ions. Is there any change in any property when these ions are compared? Notice that in going from F^- toCl^- to Br^-, there is an increase in ionic size. The same situation exists in the second column in going from Na+to K^+ to Cs^+, with Cl - being common to all. How does this relate to lattice energy? The energy derived from two ions in a compound stems from the mutual attraction of the oppositely charged ions. The closer the ions, the greater the attraction and the larger the lattice energy. In each column in this problem , as the size of one ion increases, the ions in the lattice become further apart. As such, there is a decrease in attraction between ad-jacent ions and, thus, a lowering of the magnitude of the lattice energy.

Question:

Pauli exclusion principle andHund's rule.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E18-0646.htm

Solution:

An atom is described by four quantum numbers: the principal quantum number N, the angular momentum quantum number l, the magnetic quantum number m, and the spin quantum number s. According to the Pauli ex-clusion principle, no two electrons in an atom can have the same four quantum numbers. If two electrons did, they would have identical fingerprints, a situation forbidden by nature. Each orbital can accommodate a maximum of two electrons. With this in mind,Hund' s rule states that once an electron is in an orbital, a second electron will not enter into that same orbital if there exist other orbitals in thatsubshellthat contain zero electrons. In other words, all orbitals in asubshellmust contain one electron, before a second one can enter,Hund'srule also states that single electrons in their separate orbitals of a givensubshellwill have the same spin quantum number.

Question:

You have the reaction H_2O(l) \rightarrowH_2O (g) , For both states, 1 mole of water is at 100\textdegreeC and 1 atm pressure. The volume of 1 mole of water = 18 ml, ∆H = 9710 cal/mole and there are 24.2 cal/liter-atm. Calculate the work done in this conversion and the value of ∆E.

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Solution:

In the conversion from liquid water to steam, the pressure and temperature remain constant, work (w) = p (pressure) times ∆V(change in volume). You are given the pressure. To find the volume, note ∆V = volume of gas - volume of liquid. Thus, to find the work done in the conversion, find the volume that the steam occupies. This can be done by using the equation of state, which indicates PV = n RT, where P = pressure, V = volume, n = moles, R = universal gas constant and T = temperature in Kelvin (Celsius plus 273\textdegree). You know all values, except V. Substituting and re-writing, V= nRT/P = [(1) (.0821) (373)] / [1] = 30.62 liters. This is the volume of the steam. The volume of the liquid = 18 ml or .018 liters. Thus, ∆V = 30.62 - .018 = 30.602 l and w = work = (1 atm)(30.602 liter) = 30.602 liter-atm. 24.2 cal. exist per liter-atm. Thus, w = work = 24.2 cal/liter-atm × 30.602 liter-atm = 740.56 cal. To find ∆E, remember that ∆E = ∆H - P∆V, where ∆E = change of energy, ∆H = change of enthalpy and P∆V = work, which was calculated. You are given ∆H, therefore ∆E = 9710 cal - 740.56 cal = 8969.44 cal.

Question:

About how many molecules are there in 1 cm^3 of air and what is their average distance apart?

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Solution:

The number of molecules in 1 cm^3 can be calculated from the ideal gas equation: N =pV/kT The pressure of the air is approximately p = 10^6 dyne cm^-2. The temperature of the air is approximately 300\textdegree K. If V = 1 cm^3, N = (10^6 dyne/cm^2 × 1 cm^3)/(1.38 × 10^-16 erg/\textdegreeK × 300\textdegree K) N = (10^6 dyne/cm^2 × 1 cm^3)/(1.38 × 10^-6 dyne-cm/\textdegreeK × 300\textdegreeK) = 2.5 × 10^19. In 1 cm^3 of air there are approximately 2.5 × 10^19 molecules. Imagine the 1 cm^3 to be divided up into little cubes of side a, each of which contains a molecule. Then the volume of each cube is a^3. Hence, in 1 cm^3, there are 1 cm^3/a^3 cubes. Since there is 1 molecule in each cube, the number of cubes must equal the number of molecules in 1 cm^3. (1 cm^3)/(a^3)= 2.5 × 10^19 a3= [(1 cm^3)/(2.5 × 10^19)]^= 4 × 10^-20 cm^3 a = 3.4 × 10^-7 cm This is the average distance apart of the molecules and is about 20 times the size of an oxygen or nitrogen molecule.

Question:

An automobile travelling at 10 miles per hour produces 0.33 1b of CO gas per mile. How many moles of CO are produced per mile?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E02-0048.htm

Solution:

The number of moles of a substance is equal to the quotient of the mass (in grams) of that substance and its molecular weight (in g/mole), or moles = [{mass (g)}/{molecular weight (g/mole)}] The mass of CO is 0.33 1b = 0.33 1b × 454 g/1b = 150 g. The molecular weight of CO is the sum of the atomic weight of C and the atomic weight of O, or molecular weight (CO) = atomic weight (C) + atomic weight (O) = 12 g/mole + 16 g/mole = 28 g/mole. Hence, Moles = {mass (g)}/{molecular weight (g/mole)}] = [(150 g)/(28 g/mole)] = 5.4 moles per mile.

Question:

What type of mirror is required to form an image, on a wall 3 m from the mirror, of the filament of a headlight lamp 10 cm in front of the mirror? What is the height of the image if the height of the object Is 5 mm?

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/Users/wenhuchen/Documents/Crawler/Physics/D27-0851.htm

Solution:

The image I has to be real since it appears on the wall. Therefore, we must use a concave mirror and the object 0 must be placed beyond the focal point F for a real image (see the figure). We rule out a convex mirror since one cannot obtain images that can be shown on a screen with a convex mirror. If p and q are the object and image distances from the mirror respect-ively, the first mirror equation is (1/p) + (1/q) = (1/f) where f = (R/2) is the focal distance of the mirror. Therefore, we have p = 10 cm,q = 300 cm. (1/10 cm) + (1/300 cm) = (2/R), and R = 19.4 cm. The radius R has a positive sign, as it is required for a concave mirror. The optical magnification m is m = - (q/p) where the negative sign is required because we want m to come out negative to signify an inverted image. The image height is obtained as follows. h_1 = h_0m = - h_0 (q/p) = -0.5mm × (300 cm/10 cm) = - 15 cm. The image is therefore 30 times bigger than the object and is inverted.

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Question:

Determine ∆S\textdegree, ∆H\textdegree, and ∆G\textdegree for the following reaction at 25\textdegreeC. CO (g) + Cl_2 (g) \rightarrow COCl_2 (g) Use the following table. Substance S\textdegree ∆H\textdegree_f CO Cl_2 COCl_2 47.3 53.3 69.1 -26.4 0 -53.3 S\textdegree is expressed in cal/mole \textdegreeK and ∆H\textdegree_fis in kcal/mole. ∆H\textdegree_fis in kcal/mole.

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Solution:

The change in entropy (∆S\textdegree) or randomness of a reaction is obtained by subtracting the entropy (S\textdegree) of the reactants from the S\textdegree of the products. If more than one mole of a compound reacts or is formed in a reaction, the S\textdegree should be multiplied by the appropriatestoichio-metric coefficient. Here, one mole of CO reacts with one mole of CI_2 to form one mole of COCI_2. ∆S\textdegree = S\textdegree of COCI_2 - (S\textdegree of CO + S\textdegree of Cl_2) ∆S\textdegree = 69.1 - (47.3 + 53.3) = - 31.5cal/mole - \textdegree K The change in enthalpy (∆H\textdegree) of a reaction is determined by subtracting the enthalpy of formation (∆H\textdegree_f) of the reactants from the ∆H\textdegree_fof the products. ∆H\textdegree_fof any pure element is always 0. When more than one mole of a com-pound either reacts or is formed in a reaction, the ∆H\textdegree_fmust be multiplied by the correspondingstoichiometriccoefficient when calculating ∆H\textdegree. ∆H\textdegree = ∆H\textdegree_fof COCI_2 - [∆H\textdegree_fof CO +H\textdegree_fof Cl_2 ] ∆H\textdegree = -53.3 kcal - (-26.4 kcal + 0) = -26.9 kcal. The change in free energy (∆G\textdegree) can be obtained if one remembers how ∆G\textdegree is related to ∆S\textdegree, T and ∆H\textdegree. Namely, ∆G\textdegree = ∆H\textdegree - T∆S\textdegree, where ∆G\textdegree is the change in free energy, ∆H\textdegree is the change in enthalpy, ∆S\textdegree is the change in entropy and T is the absolute temperature. In this problem, one has already obtained ∆H\textdegree and ∆S\textdegree, and one can quickly obtain T. The absolute temperature can be found by adding 273 to the temperature in \textdegreeC. T = 25 + 273 = 298\textdegreeK In the previous part of this problem, one calculated ∆S\textdegree in cal/mole - \textdegreeK and ∆H\textdegree in kcal/mole. To use ∆S\textdegree and ∆H\textdegree in the formula to find ∆G\textdegree, one must have them both in the same units. Thus, one should change the ∆S\textdegree term from cal to kcal by multiplying it by the conversion factor 1. Kcal/1000 cal. One can now obtain ∆G\textdegree. ∆G\textdegree = ∆H\textdegree - T∆S\textdegree ∆G\textdegree = -26.9 kcal - (298K) (-31.5 cal/\textdegreeK) (1 kcal/1000 cal) = -26.9 + 9.4 = -17.5 kcal.

Question:

A mechanical device moves a charge of 1.5 coulombs for a distance of 20 cm through a uniform electric field of 2 × 10^3 newtons/coulomb. What is theemfof the device?

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Solution:

The electromotive force oremf\epsilon is the work done by transporting a unit charge through an opposing electric field E. This input of work becomes available as an electric potential (as in a battery). In order to move the charge q, a force of magnitude F =Eqmust act opposite to the electric force on the charge. An amount of work W = F^\ding{217} \textbullet d^\ding{217} =Eqd will be supplied over a distance d. Theemfis this work per unit charge, therefore \epsilon = W/q = Ed = (2 × 10^3 N/coul) (0.2m) = 400 volts.

Question:

Write a BASIC program to alphabetize a list of names. Assume that the name is fifteen letters or less.

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Solution:

A flowchart giving the program logic is given below. Consider the operation of this program. Statements 2\O to 8\O provide for the entering of the names in arbitrary order. Next a pair of nested loops is set up. The outer one runs from statement 9\O to statement 18\O, and the inner one runs from 11\O to 15\O. In statement 1\O\O, we set S$ = "ZZZZZZZZZZZZZZZ" Since we assume all names are 15 letters or less, S$ repre-sents the largest possible numerical representation of a name. Now consider the inner loop. The variable with the smallest numerical value corresponds to the first name of the alphabetical list. Thus after the inner loop is completely cycled, J1 will be the number of the name that is alphabetized to be first. That is, A$(J1) will be the first name in the alphabetical list. After control leaves the inner loop, A$(J1) is printed. Next, we set A$ (J$) = "ZZZZZZZZZZZZZZZ" and cycle the loop again. The first name in the alphabetical list has now been moved to the end. Now, when the inner loop is cycled again, the second name in the alphabetical list will be printed. Hence, when the outer loop goes through a complete cycle, the complete list Will be printed in alpha-betical order. 1\OREM ALPHABETIZING PROGRAM 2\ODIM A$ (1\O\O) 3\OPRINT "ENTER TOTAL NUMBER" 4\OINPUT N 5\OPRINT "ENTER NAMES" 6\OFOR I = 1 TO N 7\OINPUT A$ (I) 8\ONEXT I 9\OFOR I = 1 TO N 1\O\OLET S$ = "ZZZZZZZZZZZZZZZ" 11\OFOR J = 1 TO N 12\OIF S$ < = A$ (J) THEN 15\O 13\OLET S$ = A$ (J) 14\OLET J1 = J 15\ONEXT J 16\OPRINT A$ (J1) 17\OLET A$ (J1) = "ZZZZZZZZZZZZZZZ" 18\ONEXT I 999END

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Question:

At 986\textdegree C, K = 1.60 for the reaction, H_2(g) + CO_2(g) \rightleftarrows H_2O(g) + CO(g). If you inject one mole each of H_2 , CO_2 , H_2 O, and CO simultaneously in a 20-liter box at time t = 0 and allow them to equilibrate at 986\textdegreeC, what will be the final concentrations of all the species? What would happen to these concentrations if additionalH_2 was injected and a new equilibrium was established?

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Solution:

Final concentrations of the species can be found by using the equilibrium constant expression. This expression equates K, the equi-librium constant, to the concentration ratio of products to reactants, each raised to the power of its coefficient in the chemical equation. Thus, for this reaction, you can say K = {[H_2 O] [CO] } /{[H_2 ] [CO_2 ]} = 1.60. Initially, there was 1 mole of each component in the 20 liter container. Since, concentration = moles/liter, all had an initial concentration of (1mole) / (20 liter) = .05 M . At equilibrium, let x = the number of moles/liter of H_2 that have re-acted. Thus, its concentration, at equilibrium, becomes .05 - x. If x moles/liter of H_2 react, the same number of moles/liter of C0_2 must react also, since they react inequimolaramounts; this is seen from the chemical reaction. Thus, [CO_2] = .05 - x, at equilibrium. These x moles/liter have been converted to products. Thus, [H_2 O] = [CO] = .05 + x, at equilibrium. The two products have the same concentration, since, again, the reaction shows they are formed inequimolaramounts. As such, you can substitute these values to obtain {(.05 + x)(.05 + x)} / {(.05 - x)(.05 - x) } = 1.60 or (.05 + x) / (.05 - x) = \surd(1.60) . Solving for x, you obtain x = .00585. Thus, the concentrations become [H_2 ] = [CO_2 ] = .0442 and [H_2 O] = [CO ] = .0558. If more H_2 is injected, the equilibrium is subjected to a stress, one component's concentration has been increased, and according to LeChatelier'sprinciple, the system will act to relieve the stress by shifting the equilibrium. To do this, more H_2 reacts with CO_2 , thus, decreasing their concentrations, to produce more H_2 O and CO, thereby, increasing their concentrations.

Question:

The maximum safe concentration of fluoride ion in drinking water is 2.0ppm(parts per million). Express this concentration in milligram %.

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Solution:

Milligram percent is the number of milligrams of solute in 100ml. of liquid. It is the given that there are 2 g. of fluoride in 1.0 × 10^6 g H_2O. Because 1 ml of H_2O weighs 1 g, there are 2 g of fluoride in 1 × 10^6 ml of H_2O. Solving for mg %: milligram % = [(100 ml × 2g × 1000 mg/g) / (1 ×10^6 ml)] = .20 mg %. The concentrations of substances in body fluids such as urine or blood are rather low, and for this reason they are generally expressed in milligram %.

Question:

If the dot under a question mark has a mass of 1 × 10^-6 g , and you assume it is carbon, how many atoms are required to make such a dot?

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Solution:

Two facts must be known to answer this question. You must determine the number of moles in the carbon dot. You must also remember the number of atoms in a mole of any substance, 6.02 × 10^23 atoms/mole (Avogadro's number). A mole is defined as the weight in grams of a sub-stance divided by the atomic weight (or molecular weight). The atomic weight of carbon is 12 g/mole. Therefore, in the dot you have (1 × 10^-6 g) / (12 g/mole) moles of carbon. Therefore, the number of atoms in such a dot is the number of moles × Avogadro's number, {(1 × 10^-6 g) / (12 g/mole) } 6.02 × 10^23 atoms/mole = 5 × 10^16 atoms.

Question:

A 10,000-ohm, 1-watt resistor has been connected up with two capacitors of capacitance 0.2 and 0.5 microfarads (\muF). We propose to plug this into the 120-volt, 60- cycle outlet (see Figures). Will the 1-watt resistor get too hot?

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/Users/wenhuchen/Documents/Crawler/Physics/D24-0797.htm

Solution:

(i) The angular frequency \omega = 2\pif = 2\pi × 60 sec^\rule{1em}{1pt}1 = 377 sec^\rule{1em}{1pt}1. At times we will work with the reciprocal of impedance, or admittance. Admittance of (i) C_2 = i\omegaC_2= (377) (2 × 10^\rule{1em}{1pt}7) i An actual network (a) ready to be connected to a source of electromotive force, and (b) the circuit diagram. = 0.754 × 10^\rule{1em}{1pt}4i ohm^\rule{1em}{1pt}1 (ii) Admittance of the resistor = (1/R) = 10^\rule{1em}{1pt}4 ohm^\rule{1em}{1pt}1 (iii) Admittance of = (i) + (ii) = 10^\rule{1em}{1pt}4 (1 + 0.754i) ohm^\rule{1em}{1pt}1 (iv) Impedance of = [1/{10^\rule{1em}{1pt}4 (1 + 0.754i)}] = [10^4/(1 + 0.754i)] [(1 - 0.754i)/(1 - 0.754i)] = [10^4 (1 - 0.754i)/(1^2 + 0.754^2)] = (6360 - 4800i) ohms (v) Impedance of C_1 = 1/(i\omegaC) = \rule{1em}{1pt} [i/(\omegaC)] = \rule{1em}{1pt} [i/{(377) (5 × 10^\rule{1em}{1pt}7)}] = \rule{1em}{1pt} 5300 i ohms Here we used the fact that 1/i = \rule{1em}{1pt} i. (vi) Impedance of the entire circuit = (iv) + (v) = (6360 \rule{1em}{1pt} 10,100i) ohms (vii) By Ohm's law I_1 (V/Z)= [(120 v)/{(6360 - 10100i)\Omega}] = [(120 v)/{(6360 - 10100)\Omega}] [(6360 + 10100i)/( 6360 + 10100i)}] = [120 (6360 + 10,100i)]/[( 6360)^2 + (10,100i)^2] = (5.37 + 8.53i) × 10^\rule{1em}{1pt}3 amp In polar form I_1 = [(5.37)^2 + (8.53)^2]^1/2 × 10^\rule{1em}{1pt}3/tan^\rule{1em}{1pt}1 8.53/5.37 = 10.0 × 10^\rule{1em}{1pt}3/1.01 rad. for if a number A is expressed in rectangular form, A = x + jy, then in polar form A = \surd(x^2 + y^2) tan^\rule{1em}{1pt}1 y/x. \surd(x^2 + y^2) is called the modulus of A \textbullet tan^\rule{1em}{1pt}1 y/x = \textphi is called the phase factor. The 120-volt source is the root mean square voltage (rms) (i.e., the square root of the square of the mean voltage). Since we have used this rms voltage in Ohm's law above, we obtain rms current. That is, the modulus of the complex number I_1, which is 10.OmA, is the rms current in amperes. An ac milliammeter inserted in series with the line would read 10mA. This current has a phase angle \textphi =1.01 radians with respect to the line voltage. The average power delivered to the entire circuit is then: (viii)P= VI cos \textphi = (120 volts) (0.010 amp) cos 1.01 = 0.64 watt In this circuit the resistor is the only dissipative element, so this must be the average power dissipated in it. Just as a check, we can find the voltage V_2 across the resistor: (ix) V_1 = I_1 [1/(i\omegaC_1)] = I_1 [\rule{1em}{1pt}i/(\omegaC_1)] = (5.37 + 8.53i)(\rule{1em}{1pt}5300i)10^\rule{1em}{1pt}3 = (45.2 - 28. 4i) volts (x) V_2 = 120 - V_1 = (74.8 + 28.4i) volts The current I_2 in R will be in phase with V_2, of course, so the average power in R will be P= (V_2^2/R) = [{(74.8)^2 + (28.4)^2}/(10^4)] = 0.64 watt which checks. Thus the rating of the resistor isn't exceeded, for what that assurance is worth. Actually, whether the re-sistor will get too hot depends not only on the average power dissipated in it but also on how easily it can get rid of the heat. The power rating of a resistor is only a rough guide.

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Question:

What is the nature of a "transforming agent"? What importance may this phenomenon have on our understanding of the chemical basis of inheritance?

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/Users/wenhuchen/Documents/Crawler/Biology/F24-0609.htm

Solution:

The concept of transformation arises from the experiments performed by Griffith in 1928. It was observed that when injected into mice, some strains of pneumococcus bacteria caused pneumonia and usually death. Other strains of the bacteria, were relatively harmless. The infective form always had a capsule (a complicated polysaccharide coating). The non-infective form did not have a virulent capsule. The encapsulated strain was called the "smooth strain" because the colonies looked smooth on a culture plate, and the harmless, unencapsulated strain was called the "rough strain" because of the rough appearance of its colonies. In his famous experiment, Griffith injected one group of mice with the virulent smooth strain and another with the harmless rough strain. As expected, mice from the former group died while the latter group survived. Griffith then injected a third group of mice with the heat-killed smooth strain. This group lived, showing that bacteria killed by heat were no longer virulent. However, when a fourth group of mice was injected simul-taneously with both the harmless rough strain and the heat- killed smooth bacteria, the mice died. The disease- causing organisms were of the smooth type bacteria. This process, by which something from the heat killed bacteria converted rough bacteria into smooth bacteria, is known as transformation. Griffith felt that protein from the dead bacteria might be the active transforming agent. But Griffith's interpretation of his experiment was later shown to be incorrect by Avery, MacLeod, and McCarthy in 1944. They made an extract from heat-killed smooth cells, and purified the extract by removing any substance that did not cause transformation of rough bacteria into smooth bacteria. Eventually they determined that DNA was the essential transforming agent. Moreover, when this extracted DNA was added to other types of rough strains, the bacteria formed from transformation were always identical to the bacteria that donated the DNA. This indicated that he-reditary information must be carried by DNA. Transformation suggests that in bacteria, genetic traits can be passed via DNA alone, from one bacterium to another.

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Question:

Two small conducting balls, each of mass 0.25 g, are hanging from the roof on insulating threads of length 50 cm so that they just touch. A charge, which they share equally, is given to them and each takes up a position such that the thread by which it hangs makes an angle of 45\textdegree with the vertical. What is the charge on each?

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/Users/wenhuchen/Documents/Crawler/Physics/D18-0587.htm

Solution:

There are three forces acting on each ball, the weight mg^\ding{217} acting downward, the Coulombian repulsive force F^\ding{217} acting horizontally, and the tension T^\ding{217} in the supporting thread. Since the ball is in equilibrium, horizontal and vertical components must separately balance (see figure). Thus mg = T cos 45\textdegreeandF = T sin 45\textdegree or, dividing these 2 equations: F = mg tan 45\textdegree = mg. But, by Coulomb's Law F = [q^2 / 4\pi\epsilon_0r^2 ]= q^2 / [4\pi\epsilon_0(2h sin 45\textdegree)^2 ], where q is the charge on each ball, and r is the separation of the 2 balls. \thereforeq2 = 4\pi\epsilon_0(2h sin 45\textdegree)^2 mg \thereforeq2 = 4\pi\epsilon_0(2h sin 45\textdegree)^2 mg = [{2 x 0.5 m × (1\surd2) ]^2 × 2.5 ×10^\rule{1em}{1pt}4 kg × 9.8 m\bullets^\rule{1em}{1pt}2} /(9 ×10^9N\bulletm^2\bulletC^\rule{1em}{1pt}2)] = 13.6 × 10^\rule{1em}{1pt}14C^2. \thereforeq = 3.7 × 10^\rule{1em}{1pt}7 C.

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Question:

A ball is thrown with an initial velocity, v_0, of 160 ft/sec, directed at an angle, \texttheta_O, of 53 with the ground. (a) Find the x- and y-components of v_0. (b) Find the position of the ball and the magnitude and direction of its velocity when t = 2 sec. (c) At the highest point of the ball's path, what is the ball's alti-tude (h) and how much time has elapsed? (d) What is the ball's range d? (See figure).

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/Users/wenhuchen/Documents/Crawler/Physics/D03-0090.htm

Solution:

(a) Using the figure v 0_x = v_0 cos \texttheta_0; ^v0_y = y_0 sin \texttheta_0 v 0_x = 160 ft/sec \bullet cos 53\textdegree = 160 ft/sec (3/5) = 96 ft/sec v 0_y = 160 ft/sec \bullet sin 53\textdegree = 160 ft/sec (4/5) = 128 ft/sec (b) The acceleration due to gravity is constant. Furthermore, there is no force acting on the projectile in the x-direction, and its ac-celeration in the x-direction is therefore zero. Hence a_x(t) = 0a_y(t) = -g v_x(t) = ^v0_xv_y(t) = ^v0_x -gt x(t) = x_0 + ^v0_x ty(t) = y_0t ^v0_y t - (1/2) gt^2 Here, x_0,y_0 are the initial coordinates of the projectile, and ^v0_x, ^v0_y are the initial x and y components of the ball's velocity, Taking the origin (0) as shown in the figure, we have, at t = 2 sec v_x = 96ft/sec x= (96 ft/sec) (2sec) = 192 ft. v_y- = 128 ft/sec - (32 ft/sec^2)(2 sec) = 64 ft/sec y= (128 ft/sec) (2 sec) -(1/2)(32 ft/sec^2)(4 sec^2) y= 256 ft - 64 ft = 192 ft. The magnitude of the ball's velocity is v = (v^2_x + v^2_y)^1/2 v = [ (64 ft/sec)^2 + (96 ft/sec)^2]^1/2 v = 115.4 ft/sec. The direction of the velocity relative to the x-axis is tan \texttheta = (v_y)/(v_x) = 64/96 = 2/3 \texttheta = 34\textdegree (c) At the highest point of the path, the ball has no vertical velocity. Then, by our kinematics equations, v_y = 0 = v_0 - gt t= (^v0_y - ^0)/g = (128 ft/sec)/(32 ft/sec^2) = 4 sec It takes 4 sec. for the ball to reach its maximum height. It has traveled a vertical distance, y_max = ^v0_y t - (1/2) gt^2 = (128 ft/sec) (4 sec) - (1/2)(32 ft/sec^2)(4 sec)^2 = 512 ft - 256 ft = 256 ft. (d) It takes the ball as much time to fall as it does to rise. Hence, the entire trajectory requires 8 sec. By the kinematics equations, we find its horizontal position at the end of its trajectory, x(t) = ^v0_x t = 96 ft/sec \textbullet 8 sec = 768 ft. This is the range of the ball.

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Question:

Which animal groups dominated the Mesozoic Era? What factors may have contributed to the extinction of the dinosaurs?

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/Users/wenhuchen/Documents/Crawler/Biology/F28-0737.htm

Solution:

The Mesozoic Era began some 230 million years ago and was characterizedby a wide variety of reptiles. In fact, the Mesozoic Era is commonlyreferred to as the "Age of Reptiles"; common reptiles of this era werethe primitive lizards; snakes, turtles, alligators, crocodiles, pterosaurs (flyingreptiles) and later in the era, the dinosaurs. All of these, and also themammals which came later, evolved from an importantpaleozoic groupcalled the stem or root reptiles (cotylosaurs). By the end of the Mesozoic Era the vast majority of the reptiles, of whichthe dinosaurs were the prominent group, dramatically disappeared. No totally satisfactory explanation has been established for their disappearance. One possible reason may be the climatic changes brought aboutby the Rocky Mountain Revolution. As the climate became increasinglycold and dry, many of the plants died out. The dinosaurs, manyof which were herbivorous, may have disappeared because of the lackof food. Some of the dinosaurs did not favor a dry environment, and theysubsequently declined in number as the swamps dried up. Another reasonmay have been theemerganceof the mammals. The smaller, warm-bloodedmammals were probably better able to compete for food, escapefrom enemies, and adapt to a colder environment than the larger, cold-blooded reptiles. The early mammals fed on the dinosaurs' eggs, andthe late dinosaurs, it is believed, resorted to using reptilian eggs for foodas the supply of food from the environment became scarce. The fatalityof the reptilian eggs may be another factor contributing to the extinctionof the dinosaurs, it is generally agreed that the demise of the dinosaurswas most probably the result of a combination of factors, rather thanany single one.

Question:

Define the terms random access, direct access, and sequential access. Give examples of each.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G01-0013.htm

Solution:

Direct or sequential access is dictated by character-istics of storage media . For example, tapes and printers allow only sequential access, whereas disks allow either direct or sequential access. In the case of semiconductor memories, the random rather than "direct'' is used, as in the example of Random Access Memory (RAM). Random access or sequential access may also be dictated by the way the data is organized (and not by the characteristic of the media). For example , data may be organized either for random or for sequential access on a magnetic disk. A record is a logi-cal piece of data that describes something meaningful, such as the report header line, the report detail line, etc. If data is organized in such a way that a program can calculate the posi-tion of the n-threcord and access it without processing any other records, then the access is random. If, a program, in order to access the n-threcord, must read the previous records (up to n-1 depending upon the data organization), then the ac-cess is said to be sequential .

Question:

Compare the methods of obtaining oxygen in the earthworm, the frog, the fish and man.

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/Users/wenhuchen/Documents/Crawler/Biology/F16-0409.htm

Solution:

The earthworm belongs to the group of invertebrates called the Annelids, or roundworms. The earthworm has no specialized respiratory system. Respira-tion takes place by diffusion of gases through the body integument or skin. The skin is kept moist by the secretions of mucous glands directly underneath the skin. The thin cuticle is quite permeable to both oxygen and carbon dioxide. The red blood cells contain dissolved hemoglobin, which aids in the transportation of oxygen to various parts of the body. In most adult amphibians (ex. frogs) the moist skin, which is supplied by a large capillary network, functions as a respiratory organ in addition to the lungs. When submerged, the frog utilizes the dissolved oxygen in the water, which enters its body via diffusion through the skin. Similarly, carbon dioxide produced by the frog diffuses into the water. When on land, respiration is accomplished by the skin and the lungs. The lower part of the throat is covered with a flap of skin that can be lowered and raised. When lowered, air comes into the mouth through the nostrils; when raised, the nostrils are closed with a valve-like flap of skin and the air is forced down into the lungs. Unlike man, the frog does not have a diaphragm to facilitate breathing. The larval frog has a method of respiration entirely different from that of the adult; it is an aquatic animal and breathes via gills. Fish and other complex water animals have gills, as their main respiratory organ. The gills are made of finely divided structures that are very thin, and richly supplied with blood capillaries. Diffusion of gases occurs readily across the thin surface of the gills. Gills are principally a respiratory adaptation to an aquatic environment. Gas exchange in man takes place in the lungs. Lungs can be thought of as organs that serve the same purpose in terrestrial animals as gills do in aquatic animals. The lungs are located internally in order to prevent dessication and possible damage. It should be noted that as an animal becomes larger and more complex, and more metabolic energy is needed, the surface area for gas exchange likewise increases. This facilitates the uptake of oxygen in addition to the release of carbon dioxide. The lamellar, filamentous surface of the gills, and the extensive branching of the bronchioles into alveoli in the lungs are adaptations towards increased surface area for gas exchange.

Question:

Meats possess large amount of connective tissue that can be tenderized by the action of proteolytic enzymes. Can such enzymes harm the stomach lining when the food is eaten?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E23-0833.htm

Solution:

This question deals with proteolytic enzymes, which may be defined as protein degradation catalysts. Two factors that affect enzymic activity are pH, and temperature. Whether these enzymes damage the stomach or not will depend, to a great extent, on the conditions in the stomach. The accompanying figure shows how pH can alter enzyme activity of pepsin and trypsin. From the diagram, you see that these enzymes do not function well at extremely low pH. Due to the presence of HCl in the stomach as a hydrolysis agent, the acidity of the stomach is high, which means an environment of low pH exists. It becomes doubtful that these proteolytic enzymes would have much activity at the low pH of the stomach. This means that little, if any, damage would be done to the stomach. Heating alters enzymic activity, also. It is likely that most of the enzyme would be inactivated by the cooking of the meat.

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Question:

Determine the mass of water to which 293 g ofNaCl (formula weight = 58.5 g/mole) is added to obtain a 0.25 molalsolution.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E08-0278.htm

Solution:

To solve this problem we use the relationship molality= [(number of moles of solute)/(kg of solvent)] or, molallty= (moles ofNaCl/kg of water). The number of moles ofNaClin 293 g is determined by dividing 293 g by the formula weight ofNaCl, or molesNaCl= massNaCl/formula weightNaCl= 293 g/58.5 g/mole \cong 5.0 molesNaCl. Then, solving for the kg of water, kg of water = (moles ofNaCl/molality) = (5.0 molesNaCl/ 0.25molal) = 20 kg H_2O.

Question:

Give a list of the character set used inPL/I. Also define the term'variable', and explain what is meant by an 'assign-ment statement'.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G13-0313.htm

Solution:

Altogether, 60 different characters constitute the PL/I character set, allor only some of them may appear in a specific PL/I program. In PL/I thereis a choice of the 26 capital letters A through Z, the 10 digits 0 through9, and also 24 special characters. The graphics and names by whichthese 24 special characters are represented are as follows: =Equal or assignment symbol +Plus symbol -Minussymbol \textasteriskcenteredAsterikor multiply symbol /Slash or divide symbol (Left parenthesis symbol :Colon symbol \lnotNOT symbol &AND symbol \vertOR symbol >Greater than symbol

Question:

Describe the development of a mammalian egg from cleavage up to the development of the fetal membranes.

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/Users/wenhuchen/Documents/Crawler/Biology/F23-0595.htm

Solution:

Cleavage in a mammalian egg takes place as the egg moves slowly down the oviduct, driven along by cilia and muscular contraction of the wall of the duct. The journey takes about four days in the case of cells, the morula, which divides and rearranges into a hollow ball of cells, the blastocyst. The blastocyst subdi-vides into an inner cell mass from which the embryo devel-ops, and an enveloping layer of cells, the trophoblast. The cavity of the blastocyst may be compared to the blastocoel, but the embryo is not a blastula, for its cells are diffe-rentiated into two types. The cells ot the inner cell mass differentiate further into a thin layer of flat cells, the hypoblast, which is located on the interior surface of the mass adjacent to the blastocoel, and which represents the endoderm. The remaining cells of the inner cell mass become the epiblast. The cells of the hypoblast, spreading along the inner surface of the trophoblast, eventually surround the cavity of the blastocyst, forming a "yolk sac," which is filled with fluid, not yolk. As the hypoblast spreads out, the inner cell mass also spreads and becomes a disc shaped plate of cells similar to the blastodisc of bird and reptilian eggs. The blastodisc becomes delimitated from the rest of the embryo. Gastrulation begins with the formation of a primitive streak and Hensen's node in which cells migrate downward, laterally and anteriorly between the epiblast and hypoblast. Those cells which remain between these two layers comprise the mesoderm. Those cells which join the hypoblast become endoderm. A crevice appears between the cells of the inner cell mass, which then enlarges to become the amniotic cavity. The cavity of the blastocyst becomes filled with mesodermal cells, and is comparable to the subgerminal cavity in the bird. The embryonic disc comes to lie as a plate between the two cavities, connected to the trophoblast at the posterior end by a group of extraembryonic mesoderm cells, the body stalk or allantoic stalk. The non- functional endodermal part of the allantois, which develops as a tube from the yolk sac, is rudimentary and never reaches the trophoblast. Thus, after two weeks of development the human embryo is a flat, two-layered disc of cells about 250 microns across, connected by a stalk to the trophoblast.

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Question:

Write a FORTRAN program to solve linear congruences of the form ax = b (mod m).

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G22-0552.htm

Solution:

Two integers are said to be congruent modulo an integer m if their difference is divisible by m. Thus a \equiv b (mod a) means (a-b)/m is an integer. For example, 3 \equiv 8 (mod 5) since [(3 - 8) / 5] = (-5/5) = -1 (an integer). Congruences are similar to equations and linear congruences are similar to linear equations. The linear congruence ax \equiv b(mod m) has a solution if and only if (a, m) (the greatest common divider of a and m) divides b,i.e., (a, m)/b. Furthermore if x0 is such a solution, then x \equiv x0 (mod m/(a, m)) is also a solution. For example, solve the linear congruence 8x \equiv 4 (mod 6).(1) By definition, (1) means (8x - 4) / 6 = 3 integer When x = 2, 8x - 4 = 12, and six divides twelve. Now, (8,6) = 2. Hence any x satisfying x \equiv 2 (mod 6/(8,6)), or x \equiv 2 (mod 3) is also a solution of 8x \equiv 4 (mod 6). Thus, 5,8,-1,-4,.. .etc. are solutions. To prove the above (first part) note that ax \equiv b (mod m) means (ax - b) / m = r (an integer), or ax - b = rm or ax - rm = b.(2) Equation (2) is a linear Diophantine equation and is solvable for r and x if and only if (a, m) divides b. The strategy in the presented computer program is as follows: 1) Read in values of a, b and m (< 10000). 2) Call a subroutine that solves the linear Diophantine equation described above. 3) Return to main program. The program terminates when the value of M read in is less than, or equal to 0 . 5BEAD (5,10) IA,M, IB 10FORMAT (3I4) IF (M.LE.O) GO TO 99 20CALL GLDE(IA,M,IB,M1,N1,NA,INDEX) IF (INDEX) 30,30,22 22WRITE (6,25) 25FORMAT (1HO,45HNO VALUE OF X SATISFIES 1THE LINEAR CONGRUENCE) GO TO 40 30WRITE (6,35) IX 35FORMAT (1HO.3HX =, I6, 32H SATISFIES THE LINEAR CONGRUENCE) 40WRITE (6,44), IA,IB,M 44FORMAT (1H , I6, 5HX (=),I16, 4H(MOD,I6,1H)) GO TO 5 99STOP END SUBROUTINE GLDE(M,N,K,M1,N1,NA,INDEX) CTHIS SUBROUTINE RECEIVES INTEGERS FROM CMAIN PROGRAM. FIRST, ANOTHER SUBROUTINE LDE IS CALLED, THAT FINDS THE GCD NA OF (N,M), CAND DETERMINES VALUES M1 AND N1 THAT SATISFY THE LINEAD CDIOPHANTINE EQUATION M\textasteriskcenteredM1 + N \textasteriskcentered N1 = K. CTHE VALUE OF M1 RETURNED HAS THE LEAST CPOSSIBLE ABSOLUTE VALUE. INDEX IS SET CTO 0 IF A SOLUTION EXISTS, TO / IF NOT, CAND TO -1 IF THE VALUE RECEIVED FOR CK IS, SAY, 30000. USE K = 30000 IF A CSOLUTION TO THE EQUATION CM\textasteriskcenteredM1 + N\textasteriskcenteredN1 = NA (GCD OF M AND N) CIS DESIRED. NOTE: THE NUMBER 30000 CIS SOMEWHAT ARBITRARILY CHOSEN. IT IS CASSUMED THAT ONE WOULD NEVER WISH CTO SOLVE THE EQUATION MX + NY = 30000. CALL LDE (M,N,M1,N1,NA) IF (K - 30000) 22,21,22 21INDEX = -1 K = NA GO TO 27 22MULT = K/NA IF (K - MULT \textasteriskcentered NA) 23,25,23 23INDEX = 1 RETURN 25Ml = M1 \textasteriskcentered MULT N1 = N1 \textasteriskcentered MULT 27MTEST = M1 ND = N/NA MD = M/NA MX = M1 30MX = MX + ND IF (IABS(MX) - LABS (M1)) 35,40,40 35M1 = MX N1 = N1 - MD GO TO 30 40IF (MTEST - M1) 65,45,65 45MX = M1 50MX = MX - ND IF (IABS(MX) - IABS(M1)) 55,65,65 55M1 = MX N1 = N1 + MD GO TO 50 65INDEX = 0 RETURN END SUBROUTINE LDE(M,N,M1,N1,NA) MO = 1 NO = 0 M1 = 0 N1 = 1 MA = IABS(M) NA = IABS(N) MS = M/MA NS = N/NA 25IQUOT = MA/NA IREM = MA - NA \textasteriskcentered IQUOT IF (IREM) 35,35,30 30M2 = MO - IQUOT \textasteriskcentered M1 N2 = NO - IQUOT \textasteriskcentered N1 MO = M1 NO = N1 M1 = M2 N1 = N2 MA = NA NA = IREM GO TO 25 35M1 = MS \textasteriskcentered M1 N1 = NS \textasteriskcentered N1 RETURN END

Question:

Write a procedure in PL/I that gives information about undergraduate alumni for a university's records. Your input will be a deck of 80-column cards, each card specified to the fol-lowing format: Columns Contents Specifications 1 - 15 Last name Characters 16 - 30 First name Characters 31 - 34 Year entered Fixed-point 35 - 38 Year graduated Fixed-point 39 - 40 Empty 41 - 60 GPI for each year Fixed-point 61 - 80 Empty Your output should contain the alumnus' name, years of attendance, and final GPI. Input is terminated by a card with a blank last name field.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G13-0327.htm

Solution:

This is a simple data processing routine which contains some interesting format specifications. In the first PUT EDIT statement, the options PAGE and LINE (3) are executed prior to the transmission of data. The A specifi-cations allow 15 spaces for each of the headings. X speci-fications insert blanks, and the COL(40) option jumps the preceding heading to the fortieth column of the page. The GET statement acquires the first and last names as characters, but unlike the list-directed and data-directed formats, it does not require that the characters be in quotes. Now let us look at a sample card from the input stream to illustrate the possible defaults that could occur: The specification 4F(5,3) denotes usage of format for the four yearly averages. These rules will clarify the way in which data is taken into this program. First, if an explicit decimal point appears in the in-put stream, it overrides the instructions specified in the F format item, as is the case of 2.9. If there is no explicit decimal point, the decimal point is placed implicitly by the format statement. Hence the digits 3127 are read in as 3.127. If an explicit decimal point appears, the value may appear anywhere within the field width, as shown by the last two values 3.612 and 3.7. The rest of the program is apparent, and is presented below: LIST:PROC OPTIONS(MAIN); DCL(LAST,FIRST) CHAR(15), (ENTER,GRAD) FIXED, (AVER(4),SUM) FIXED DECIMAL(4,3); /\textasteriskcenteredWRITE LIST HEADINGS\textasteriskcentered/ PUT EDIT('LAST NAME', 1 FIRST NAME \textasteriskcentered, 'YEAR ENTERED', 'YEAR GRADUATED', 'FINAL AVERAGE') [PAGE,LINE(3),A(15),X(1),A(15),COL(40),A(15), X(2),A(15),X(2),A(15)]; PUT SKIP (1); NEW: GET EDIT (LAST, FIRST, ENTER, GRAD, AVER) [A(15),A(15),2F(4),X(2),4F(5,3)]; IF LAST=' ' THEN GO TO FINISHED? SUM = 0; DOI = 1 TO 4; SUM = SUM + AVER(I); END; PUT EDIT(LAST, FIRST, ENTER, GRAD, SUM/4) [SKIP,A(15),X(1),A(15),COL(40), 2(X(4),F (4),X(9)),X(5),F(5,3)]; GET SKIP; GO TO NEW; FINISHED:END LIST;

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Question:

Discuss the action of enzymes in intestinal digestion.

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/Users/wenhuchen/Documents/Crawler/Biology/F17-0432.htm

Solution:

Digestion and absorption of ingested food takes place mainly in the small intestine. The partially digested food that enters the upper part of the small intestine (the duodenum) from the stomach is very acidic, due to the high acid content of the stomach. This acidity stimulates special receptor cells in the intestinal lining. These cells, in turn, produce a hormone, cholecystokinin (CCK) that enhances the release of bile from the gall Digestion of protein. Pepsin in the stomach hydrolyzes peptide bonds at the amino end of tyrosine (Tyr) and phenylalanine (Phe). Then the food moves into the intestine, where trypsin and chymotrypsin from the pancreas hydrolyze bondsadjacent to lysine and arginine and to tyrosine, leucine, and phenylalanine, respectively (chymotrypsin also hydrolyzes bonds adjacent to tryptophan and methionine when they are present). Pepsin, trypsin, and chymotrypsin hydrolyze only internal bonds, not bonds attaching terminal amino acids to the chains. Terminal bonds at the amino end of chains may be split by aminopeptidase and those at the carboxyl end by carboxpeptidase. Bonds between pairs of amino acids are split by dipeptidases whereupon digestion is completed. bladder. CCK also stimulates the release of a pancreatic juice rich in digestive enzymes (proteases, lipase, nucle-ases and amylase). Secretin, also produced by the intes-tinal mucosa, causes release of a pancreatic juice rich in sodium bicarbonate, which neutralizes the stomach acid. Proteases are enzymes that hydrolyze proteins, which are long chains of amino acids joined together by peptide bonds. Generally, short chains of amino acids are called polypeptides. The intestinal proteases are usually called peptidases because they break the peptide bonds of peptide chains. There are many different kinds of peptidases (see figure). Pepsin, produced in the stomach, is an endopeptidase; it cleaves only those peptide bonds that are within the peptide chain, splitting the large chain into many shorter chains. Trypsin and chymotrypsin are the endopeptidases produced by the pancreas. Unlike pepsin, which requires the high acidity of the stomach in order to function, these intestinal enzymes prefer an alkaline environment. Like pepsin of the stomach, these enzymes will only break the peptide bonds between certain amino acids. Trypsin will split the bond linking the amino acids lysine or arginine. Chymotrypsin cleaves the peptide bonds next to phenylalanine, tyrosine, or tryptophan. The second category of peptidases are the exopeptidases. Exopeptidases only attack the ends of the peptide. The many peptides produced as the result of endopeptidase ac-tivity provide numerous sites for attack by the exopeptidases. The two ends of a peptide chain are chemically different. One end is called the amino end(-NH2) while the other is the carboxylic end (-COOH). Thus, there are two kinds of exopeptidases: aminopeptidase and carboxypeptidase. These enzymes sure secreted by glands in the intestinal lining and the pancreas, respectively. When digestion of a protein is completed by these enzymes, all that remains are the free amino acids, which are taken up by the cells lining the villi. Lipase, produced by the pancreas, digests the lipids in our food, with the help of the bile. Two other pancreatic enzymes, ribonuclease and deoxyribonuclease, break up any RNA and DNA from the nuclei of the cells we ingest. The products of their digestion are quickly absorbed through the Villi. Pancreatic amylase breaks up any remaining starch into maltose. In addition to aminopeptidase, the intestinal glands secrete enzymes which degrade sugars. Maltase splits maltose into two glucose molecules; sucrase splits sucrose into glu-cose and fructose; and lactase splits lactose into glucose and galactose molecules.

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Question:

What are the chief characteristics of theprotozoans?

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/Users/wenhuchen/Documents/Crawler/Biology/F11-0265.htm

Solution:

Theprotozoansare a heterogeneous assemblage of a large number of species, which are almost exclusively microscopic organisms. Theprotozoansare grouped into the phylum Protozoa within the Animal Kingdom, although this classification remains controversial. Some biologists still believe that theprotozoanshave more in common with the KingdomProtista. Protozoa live either singly or in colonies. These organisms are usually said to be unicellular. There-fore they contain no tissue or organs, which are defined as aggregations of differentiated cells. Instead of organs, they have functionally equivalentsubcellularstructures called organelles. These organelles do show a great deal of functional differentiation for the purposes of locomotion, food procurement, sensory recep-tion, response, protection and water regulation. Certainprotozoans have interesting plant-like characteristics in both structure and physiology. Reproduction among theprotozoansis variable. An individual may divide into two, usually equal halves, after which each grows to the original size and form. This form of reproduction is called binary fission and can be seen in the flagellates, among the ciliates, and in organisms such as the amoeba. Multiple fission, orsporulation, where the nucleus divides repeatedly and the cytoplasm becomes differentiated simultaneously around each nucleus resulting in the production of a number of offspring, is also seen among theprotozoans. Other types of reproduction characteristics of theprotozoansareplasmotomy, which isthe cytoplasmic division of a multi-nucleate protozoan without nuclear division, resulting in smaller multinucleate products. Budding is another reproductive process by which a new individual arises as an outgrowth from the parent organism differentiating before or after it becomes free. All thereproductorymechanisms thus far mentioned illustrate asexual means of reproduction. Sexual reproduction may also occur by the fusion of two cells, called gametes, to form a new individual, or by the temporary contact and nuclear exchange (conjugation) of two protozoans (for example, two paramecia). The result of conjugation may be "hybrid vigor", defined as the superior qualities of a hybrid organism over either of its parental lines. Some species have both sexual and asexual stages in their life cycles. With regard to their ecology,protozoansare found in a great variety of habitats, including the sea, fresh water, soil, and the bodies of other organisms. Someprotozoansare free-living; meaning that they are free- moving or free-floating, whereas others have sessile organisms. Some live in or upon other organisms in either acommensalistic,mutualistic, or parasitic relationship. The mechanisms for the acquisition of nutrition is also variable among theprotozoans. Some areholozoic, meaning that solid foods such as bacteria, yeasts, algae,protozoans, and small metazoans or multicellular organ-isms, are ingested. Others may besaprozoic, wherein dissolved nutrients are absorbed directly;holophytic, wherein manufacture of food takes place by photosynthesis; ormixotrophic, which use both the saprozoic andholophyticmethods. It should be pointed out that the unicellular level of organization is the only characteristic by which the phylum Protozoa can be described. In all other respects, such as symmetry and specialization of organelles, the phylum displays extreme diversity.

Question:

A rectangular coil of 300 turns has a length of 25.0 cm and a width of 15.0 cm. The coil rotates with a constant angular speed of 1800 rev/min in a uniform field of induction 0.365 weber/m^2 . (a) What EMF is induced in a quarter revolution after the plane of the coil is perpendicular to the field? (b) What is the EMF for a rota-tion of 180 o from the zero position? (c) What is the EMF for a full rotation?

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/Users/wenhuchen/Documents/Crawler/Physics/D22-0754.htm

Solution:

The flux through a coil of N turns, is, because B^\ding{217} is constant. \varphi = B^\ding{217} \textbullet A^\ding{217} = BA cos \texttheta where \texttheta is the angle between B^\ding{217} and A^\ding{217}. The angular velocity \omega, for constant speed, is \omega = (\texttheta/t) where \omega is the angular velocity of the coil. By definition, \omega = 2\pin, where n is the frequency of rotation of the coil. According to Faraday's law of induction, the EMF induced in a coil due to a changing flux is given by \epsilon = [(-Nd\varphi)/(dt)] The negative sign indicates that the EMF is induced in such a direction as to oppose the change in flux which induced it. Then \epsilon = -N (d/dt) (BA cos \omegat) = \omegaNBA sin \omegat = \omegaNBA sin \texttheta = 2\pin NBA sin \texttheta Therefore, \epsilon = 2\pi[(1800/60) sec^-1 ] (300) [0.365 (W/m^2 )] [25cm × 15cm × (1cm^2 /10^-4 m^2 )] × sin \texttheta or\epsilon = 4.6 × 10^12 × sin \texttheta volts . (a)When\texttheta = 90 o sin \texttheta = 1therefore\epsilon = 4.6 × 10^12 v. (b)When\texttheta = 180 o sin \texttheta = 0therefore\epsilon = 0 (c)When\texttheta = 360 o sin \texttheta = 1therefore\epsilon = 4.6 × 10^12 v.

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Question:

Suppose a hole is punched into the bottom of the tank of the previous problem. If the water flows out at a constant rate, Q_1, modify the program of the previous problem to simulate this system.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G12-0311.htm

Solution:

Since Q_1 is the rate of flow out of the tank, it tends to decrease V; therefore we give it a negative sign. V is the accumula-tion of the net flow, Q + (-Q_1) : From this, the equation can be written: V̇ = Q - Q_1(1) Since both Q and Q_1 are constant, Q - Q_1 is constant too and equation (1) can be written: V = Q_2 where Q_2 = Q - Q_1 Hence, the programmer needs only to substitute the value of Q_2 for Q in his input data.

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Question:

Find Antilog_10 0.8762 - 2.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E33-0966.htm

Solution:

Let N = Antilogy_10 0.8762 - 2. The following relationship between log and antilog exists: log_10x = a is the equivalent of x = antilog_10a. Therefore, log_10 N = 0.8762 \rule{1em}{1pt} 2. The characteristic is \rule{1em}{1pt}2. The mantissa is 0.8762. The number that cor-responds to this mantissa is 7.52. This number is found from a table of common logarithms, base 10. Therefore, N= 7.52 × 10^\rule{1em}{1pt}2 = 7.52 × (1/10^2) = 7.52 × (1/10^2) = 7.52 × (1/100) = 7.52 × (1/100) = 7.52(.01) N= 0.0752 . Therefore, N = Antilogy_10 0.8762 - 2 = 0.0752.

Question:

Given pairs (x, y) on data cards, write a PL/I program to test x, y and decide in which quadrant they lie. Also find the distance from the origin. If the distance from originis less is less than 10 classify as small; if distance is less than 20 classify as medium and all other cases classify as large. Your program should not to have any GOTO except for the one GOTO needed for repetitive input.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G13-0339.htm

Solution:

This is an exercise in IF THEN and IF ELSE con-trol statements. The program reads in the values of X & Y from each data card. To obtain the information about the quadrants consider the figure. Test for X and Y Quadrant the point lies X>0,Y>01st quadrant X>0,Y<0IVth quadrant X<0,Y<0Illrd quadrant X<0,Y>0Ilnd quadrant X = 0Y axis Y = 0X axis X = 0,Y = 0Origin Distance from origin to the point with coordinates (X,Y) is given by Distance = \surd(X^2 + Y^2) The program is as follows. TABLE:PROCEDURE OPTIONS(MAIN); /\textasteriskcentered THIS PROGRAM GIVES THE LOCATION OF THE POINT AND THE DISTANCE FROM THE ORIGIN \textasteriskcentered/ DCL(X,Y,DIST) FIXED REAL DEC(6,2); PUT LIST('X CO-ORD',1Y CO-ORD',1QUADRANT','DISTANCE', 'CLASSIFICATION'); /\textasteriskcentered THIS PROVIDES A HEADING FOR EACH COLUMN \textasteriskcentered/ PUT SKIP(2); REPEAT:GET LIST(X,Y); IF X =1000,Y=1000 THEN STOP; /\textasteriskcentered THE LAST CARD HAS THIS VALUE OF X,Y THIS IS A WAY TO END LOOPING AND EXIT \textasteriskcentered/ PUT LIST(X,Y); Note that in the following, the brackets { } have been used to delineate the THEN clause and the ELSE clause. IF conditionTHEN expression 1; ELSE expression 2; It should be borne in mind that the use of the { } brackets is limited only for purposes of explanation of this example. The { } brackets are not a standard PL/I feature. IF X = 0 THEN {IF Y = 0THEN PUT LIST('ORIGIN'); ELSE PUT LIST('Y AXIS') ;} ELSE {IF Y = 0THEN PUT LIST('X AXIS'); ELSE; } IF X>0 THEN {IF Y>0THEN PUT LIST('FIRST QUADRANT'); ELSE PUT LIST('THIRD QUADRANT');} ELSE{IF Y>0THEN PUT LIST('SECOND QUADRANT'); ELSE PUT LIST('FOURTH QUADRANT');} DIST = SQRT(X\textasteriskcenteredX+Y\textasteriskcenteredY); PUT LIST(D); IF D<10 THEN {PUT LIST('SMALL') ;} ELSE {IF D<20THEN PUT LIST('MEDIUM'); ELSE PUT LIST('LARGE');} PUT SKIP(2); GO TO REPEAT; END TABLE;

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Question:

Consider the gas thermometer illustrated below. At 0\textdegreeC, the volume of the gas is 1.25 liters. Assuming that the cross-sectional area of the graduated arm is 1 cm^2, what is the distance (in cm.) from the 0\textdegreeC reading and a reading at 35\textdegreeC?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E02-0035.htm

Solution:

Since volume, V, and absolute temperature, T, are the only two variables being considered, we can approach this problem by means of Charles' Law, V = kT, where k is the Charles' law constant, to be determined for our system. We can determine k by. using the initial values of 0\textdegreeC (0\textdegreeC = 273.15\textdegreeK) and 1.25 liters (= 1250 ml = 1250 cm^3). Then, k = (V/T) = [(1250 cm^3 )/(273,15\textdegreeK)] = 4.576 cm^3 -\textdegreeK^-1 The volume at 35\textdegreeC (308.15\textdegreeK) can now be determined: V = kT = 4.576 cm^3 -\textdegreeK^-1 × 308.15\textdegreeK = 1410.1 cm^3 = 1.410 l. The difference between this volume and the initial volume is 1410.1 cm^3 - 1250 cm^3 = 160.1 cm^3. This volume of liquid will be displaced in the graduated arm. The difference in height of the liquid in the graduated arm, can now be calculated from the volume displaced and the cross-sectional area: height = (volume/area) = [(160.1 cm^3 )/(1 cm^2 )] = 160.1 cm.

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Question:

What scale readings would you predict when a uniform 120-lb plank 6.0 ft long is placed on two balances as shown in the figure, with 1.0 ft extending beyond the left support and 2.0 ft extending beyond the right support?

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/Users/wenhuchen/Documents/Crawler/Physics/D02-0030.htm

Solution:

From the first condition for equilibrium, the forces upward must equal the forces downward, F_A + F_B - 120 lb = 0 The plank is uniform, meaning that the center of mass is at the center of the beam, three feet from each end. This is the point at which the 120 lb gravitational force can be considered to act. Torque about a point is defined as the tendency of a force to cause rotation about the point. The magnitude of the torque is given by the product of the magnitude of the force and the perpendicular distance of the line of action of the force (the line along which the force acts) from the point of rotation. The direction of the torque can be found using the right hand rule. Place the fingers of the right hand in the direction of the distance vector. Rotate the distance vector into the direction of the force vector. If this rotation is in the clockwise direction, the torque is negative. For counterclockwise rotation, the torque is positive. For equilibrium, the sum of all the torques about any point in the body must equal zero. To apply this second condition for equilibrium, we may choose to write torques about an axis through A, noting that the center of mass of the plank is 2.0 ft from A. -120 lb × 2.0 ft + F_B(3.0 ft) = 0 or F_B = 80 lb Substitution of 80 lb for F_B in the first equation gives F_A = 40 lb. Alternatively, we may write a second torque equation, this time about an axis through B. + 120 lb × 1.0 ft - F_A(3.0 ft) = 0 or F_A = 40 lb.

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Question:

What trends in the evolution of sexual reproduction are evident among the green algae? Have these trends been continued in higher plants?

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/Users/wenhuchen/Documents/Crawler/Biology/F06-0174.htm

Solution:

Among the green algae, the major lines of evolutionary development in sexual reproduction are: (1) the change from isogamy to heterogamy to oogamy, (2) the division of labor between the cells of a thallus so that vegetative and reproductive cells per-form separate functions, (3) the development of con-jugation processes which enable gene exchange through recombination, and (4) the presence of a diploid multicellular organism in the life cycle. The unicellular green algae, Chlamydomonas, and the colonial genus Gonium have the simplest type of life cycles. The haploid adult organism is capable of asexual reproduction and of sexual reproduction. Usually Chlamydomonas reproduces asexually by zoospores. Under certain conditions, Chlamydomonas reproduces sexually. The mature cell divides mitotically several times to produce gametes. These gametes are all alike and they cannot be separated into male and female types. This condition is termed isogamy. In Gonium, individual cells are released from the colony and function as gametes. These gametes are also alike. Two gametes of one species fuse to form a zygote. The zygote divides meiotically to yield either individual haploid cells in Chlamydomonas, or a colony of haploid cells in Gonium. The zygote is the only diploid stage in these life cycles. In these two groups, the cells are not specialized and any cell is capable of either asexual or sexual reproduc-tion. Pandorina is a genus of colonial forms slightly more complex than Gonium. Sexual reproduction is heterogamous with small male gametes fusing with larger female gametes. Both types of gametes are free swimming, and fertilization (fusion of gametes) occurs outside the parent colony. Pleodorina is an even more advanced genus. In sexual reproduction, the female gametes are not flagel-lated and remain embedded in the parent colony. A flagellated male gamete swims to the nonmotile egg cell and fertilization occurs. This type of heterogamy is termed oogamy. Volvox, a colonial genus, has an oogamous sexual reproductive life cycle, and also shows cellular specialization. Most of the cells of the large spherical colonies are exclusively vegetative, incapable of sexual reproduction. A few cells are specialized for repro-duction. The female cell is large and non-motile and is produced by a specialized cell - an oogonium. The male cells are small and free-swimming, and are produced by sperm producing cells termed antheridia. Fertiliza-tion occurs inside the parent colony. The change from isogamy to oogamy and the speciali-zation of vegetative and reproductive cells are trends that have been continued among higher plants. All of the algae discussed so far have a dominant haploid stage and the only diploid stage is the one-celled zygote. Ulva, or sea lettuce, is a multicellular algae whose life cycle includes both multicellular haploid and multicellular diploid stages. Its life cycle exhibits alternation of generations. This alternation of genera-tions is a major evolutionary event in the development of plants. In the Ulva, reproductive cells differ from vege-tative cells, but any vegetative cell is capable of becoming a reproductive cell. Ulothrix is a multicellular filamentous algae which exhibits a type of isogamous sexual reproduction termed heterothallism. Isogametes are of two different mating types termed plus and minus. A plus gamete can only fuse with a minus type, and not with another plus gamete. Heterothallic isogamy is fairly common in algae and fungi, and may be considered an evolutionary stage between homothallic isogamy (where any gamete may fertilize any other gamete) and heterogamy. The sexual reproductive cells of Ulothrix differ from the usual vegetative cells and from zoospores. The zygote is the only diploid stage. Spirogyra is a filamentous green algae with a rather odd sexual reproductive pattern. The gametes are not released from the parent thallus. Instead two filaments come to lie side by side and protuberances develop on cells next to one another. The protuberances enlarge, fuse, and the cell wall disintegrates between the two cells. One cell then travels through this conjugation tube and the two cells fuse to form a zygote. Any cell in a filament is capable of uniting with a cell in a neighboring filament. The two fusing cells are similar (isogamy). Although in higher plants, specialized tubes develop to allow fertilization (pollen tubes), this is not analogous to the conjugation tube of Spyrogyra. In Spirogyra, both haploid cells develop protuberances that will form the conjugation tube. In higher plants only the haploid male gametes form pollen tubes. These enable the male nucleus to unite with the nucleus of the female egg cell. In all the life cycles discussed, both sexual and asexual reproduction occur. In filamentous algae, asexual reproduction occurs by fragmentation or by zoospores.

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Question:

Write a FORTRAN program that accepts two positive integers and out-puts a message indicating whether the integers between the two limits are perfect, abundant, or deficient.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G22-0541.htm

Solution:

The proper divisors of a positive integer include all divisors except the integer itself. For instance, the proper divisors of 12 are 1,2,3,4, and 6. We use three terms to classify integers according to the sums of their divisors. An integer is perfect if the sum of its proper divisors is equal to that integer. There exist only 6 perfect numbers less than 10 billion, all of which are even. One of them is integer 6. Its proper divisors are 1,2 and 3, the sum of which is equal to 1 + 2 + 3= 6. The rest of the integers may be either abundant or deficient. An integer is abundant if the sum of its proper divisors is greater than that integer. Conversely, an integer is deficient if the sum of its proper divisors is less than an integer itself. Perfect numbers are related toMersenneprimes, which take the form 2^K -1. We can say that if 2^K - 1 is prime, then n = 2^K - 1(2K - 1) is perfect. Note that we refer only to even perfect numbers. No odd perfect numbers have yet been demonstrated. The program uses a subprogram called NSIGMA, named for the \sigma-function. If q is a prime, then\sigma(q) = q + 1. In general, \sigma \sigma \sigma \sigma(q)^K = [(q^K+1 -1) / (q -1)] for q = 1, 2, 3,... \sigma \sigma An array, NPRIM, is used to store all primes between 2 and 181. Although this example uses a DATA statement for the primes, you could also use a subprogram for the Sieve of Eratosthenes to generate primes. The main program reads in upper and lower bounds (HI and LO, respectively), an increment term (I), and a case number (K), which indicates the following according to the values assigned to it: K = -2 output deficient numbers only K = -1 output deficient and perfect numbers only K = 0 output all numbers K = 1 output abundant and perfect numbers only K = 2 output abundant numbers only Since the highest prime number stored in NPRIM is 181, the upper bound cannot exceed (181)2 = 32761, otherwise some of the divisors may end up missing. The program looks as follows: INTEGER HI READ (5,100) LO, HI, I, K 100FORMAT (4(I5)) DO 60 N = LO, HI, I CUSE NSIGM& TO CHECK FACTORS IF (NSICMA (N) - 2\textasteriskcenteredN) 15, 20, 25 15IF (K) 30, 30, 60 20IF (IABS(K) - 1) 40, 40, 60 25IF (K) 60, 50, 50 30WRITE (6,101) N 101FORMAT (1H, I5, 13H IS DEFICIENT) GO TO 60 40WRITE (6, 102) N 102FORMAT (1H ,I5, 11H IS PERFECT) GO TO 60 50WRITE (6, 103) N 103FORMAT (1H , I5, 12H IS ABUNDANT) 60CONTINUE STOP END CFUNCTION SUBPROGRAM NSIGMA FUNCTION NSIGMA DIMENSION NPRIM (42) CFILL ARRAY WITH PRIMES FROM 2 TO 181 DATA NPRIM/2,3,5,7,11,13,17,19,23,29,31,37,41, 143,47,53,59,61,67,71,73,79,83,89,97,101,103,107,109,113, 2127,131,137,139,149,151,157,163,167,173,179,181/ INT = N RN = INT NROOT = SQRT(RN) NSIGMA = 1 MSIGMA = 1 DO 40 I = 1,42 NF = 0 NP = NPR(I) IF (NP - NRT) 32, 32, 50 32KINT = INT/NP IF (INT-KINT\textasteriskcenteredNP) 36, 34, 36 34INT = KINT NF = 1 WSIGMA = NSIGMA\textasteriskcenteredNP + MSIGMA GO TO 32 36IF (NF) 40, 40 , 38 38MSIGMA = NSIGMA RN = INT NROOT = SQRT (RN) 40CONTINUE 50IF (INT-1) 60, 60, 56 56NSIGMA = NS IGMA\textasteriskcenteredINT + NSIGMA 60RETURN END

Question:

Suppose that three devices are connected in parallel to a 12-V battery. Let the resistances of the devices be R_1 = 2\Omega, R_2 = 3\Omega, and R_3 = 4\Omega, What current is supplied by the battery and what is the current in each device?

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/Users/wenhuchen/Documents/Crawler/Physics/D20-0676.htm

Solution:

Since the resistors are in parallel, we have for the equivalent resistance R, (1/R) = (1/R_1 ) + (1/R_2 ) + (1/R_3 ) = (1/2\Omega) + (1/3\Omega) + (1/4\Omega) = (13/12)\Omega^-1 The equivalent resistance, therefore, is R = (12/13)\Omega = 0.92\Omega. The voltage across each device is 12 volts since they are in parallel to the battery. Therefore, using Ohm's law and the figure, I_1 = (V/R_1) = [(12 V)/(2\Omega)] = 6 amp I_2 = (V/R_2) = [(12 V)/(3\Omega)] = 4 amp I_3 = (V/R_3) = [(12 V)/(4\Omega)] = 3 amp The current supplied by the battery is found by applying Kirchoff's node equation at point A. Hence I = I_1 + I_2 + I_3 = (6 + 4 + 3) amp I = 13 amp.

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Question:

What is the acceleration of a point on the rim of a flywheel 0.90 m in diameter, turning at the rate of 1200 rev/min?

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/Users/wenhuchen/Documents/Crawler/Physics/D04-0154.htm

Solution:

For uniform circular motion, the acceleration of a particle at

Question:

Lithopone white paint is made by heating a mixture ofBaS and ZnSO_4 according to the equation BaS + ZnSO_4 \ding{217} BaSO_4 +ZnS What is the weight ratio of products BaSO_4 andZnSin the paint ?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E27-0904.htm

Solution:

From thestoichiometryof this equation, 1 mole ofBaS+ 1 mole of ZnSO_4 \ding{217} 1 mole of BaSO_4 + 1 mole ofZnS. Assuming that 1 mole of each product is contained in this paint, their weights would be equivalent to their molecular weights. The molecular weight of BaSO_4 is 233 g / mole; for ZnS it is 97 g / mole. The weight ratio is the ratio of weight of one substance compared to the weight of another substance. Thus, the weight ratio of BaSO_4 toZnSin lithopone is BaSO_4 :ZnS= 233 : 97 or 2.4 : 1. The initial assumption that only 1 mole of each product is contained in the paint is therefore not necessary. The weight ratio would be the same no matter how many moles of products were formed, since the products are formed ineguimolaramounts.

Question:

What are the properties of a colloidal system? Why does it afford a particularly good medium for chemical reactions? Give some examples of colloidal solutions.

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/Users/wenhuchen/Documents/Crawler/Biology/F01-0028.htm

Solution:

A colloid is a stable suspension of particles that do not settle out of solution upon standing. The particles or dispersed phase, of a colloid are between 10^-1 and 10^-4 microns in size. The medium which holds these particles is called the dispersing phase. Collodial systems display properties which are unique to them, such as the Tyndall effect and Brownian movement. The Tyndall effect is due to colloidal particles scattering light at all angles from the direction of a primary beam. This scattering of light can be observed, for example, when a beam of light enters a darkened room through a small opening. The particles of dust suspended in the air in the room cannot be seen if they are colloidal in size, but the light scattered by them, however, can be seen, and thus the particles appear as bright points in the beam. This phenomenon can only be observed at right angles to the light source. Brownian movement is the random motion of colloidal particles in their dispersing medium. To observe this, an optical microscope should be focused on a colloidal solution. Due to the light-scattering properties just discussed, the dispersed substance will then be observed not as particles with definite outlines, but as small, sparkling specks traveling a random zig-zag path. The reason colloidal solutions provide a particu-larly good medium for chemical reactions is another special property of colloids. Due to their physical characteristics, they have an enormous total surface area. All surfaces contain energy which can contribute to a chemical reaction. This is best exemplified by absorption (the process which accounts for the fact that solid substances can hold appreciable quantities of gases and liquids on their surface). The capacity for absorption and surface area are approximately proportional. The particles suspended in the dispersing phase can absorb ions, molecules and atoms of other substances. Many colloids have a preference for certain substances which permits the use of absorption in such processes as: removing undesirable colors and odors from certain materials, se-paration of mixtures, concentration of ores and various purification methods. Examples of various colloidal systems are given in the following table: dispersed substance dispersing substance general name examples gas liquid foam whipped cream, beer froth gas solid solid foam pumice, marshmallow liquid gas liquid aerosol fog, clouds liquid liquid emulsion mayonnaise, milk liquid solid solid emulsion cheese, butter solid gas solid aerosol smokes, dust solid liquid sol most paints, jellies solid solid solid sol many alloys, black diamonds

Question:

Obtain the 9's and 10's complements of the following decimalnumbers: (a) 13579(b) 90090(c) 09900(d) 10000(e) 0000

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G02-0030.htm

Solution:

The nine's and ten's complements of a given decimal number X are foundby subtracting it from a sequence of nine's and from the power of tengreater than X respectively. (a) The nine's complement of 13579 is: 99999 - 13579 86420 The ten's complement is: 100000 - 13579 86421 (b) The nine's complement of 90090 is: 99999 - 90090 09909 Ten's complement is: 100000 - 90090 09910 From (a) and (b) note that Ten's complement (X) = Nine's complement (X) + 1 (c) The nine's complement of 09900 is: 99999 - 09900 90099 The ten's complement is 90099 + 1 =90100 . (d) When numbers are represented in decimal form, the number ten is calledthe base or radix of the decimal system. Then, forming the radix complementof a decimal number means finding its ten's complement representationand the radix minus one complement is equal to the nine's complement. The radix complement of 10000 is: 100000 - 10000 90000 The radix minus one complement is 89999. (e) The radix minus one complement of 0000 is: 9999 - 0000 9999 Hence, the radix complement is 10000. Radix complement representationsof numbers are useful for subtracting two numbers. For example, sup-pose (c) is to be subtracted from (b). 90090 - 09900 80190 In performing this subtraction, we had to borrow twice from the next column. To eliminate the need for borrowing, the subtraction may be performedas follows: i) Form the nines complement of (c) (09900): 90099 ii) Add 1 to obtain the ten's complement: 90099 + 1 = 90100 iii) Add (b) (90090) to the result: 90100 + 90090 = 180190 iv) Subtract100000 from iii) to obtain the final answer 180190 - 100000 = 80190.

Question:

a)odd(x)b)eoln(f)c)eof(f) d)abs(x)e) trunk (x)f) round (x) g)ord(x)h)pred(x)i)succ(x)

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G16-0398.htm

Solution:

a) The actual parameter of this function is an integer. ODD(x) returns BOOLEAN value 'true' if integer x is odd, and 'false' otherwise, ex : odd(8) = false; odd(9) = true b)eoln(f) returns 'true' if end of a line in the file f has been reached , and 'false' otherwise. c)Similarly ,returns 'true' if end of file f has been reached. d)abs(x) computes the absolute value of x. ex:abs(-5) = abs(5) = 5 e)Trunc(x) - truncates the real number x. ex:trunk(7.98) 7,trunk(3.005) = 3 f) Round(x) - round the real number x to the nearest integer. ex:Round(7.98) = 8,Round (3.005) = 3. Note that for non-negative values of x, round (x) = trunk(x + 0.5), for negativevalues Round (x) = trunk(x - 0.5) g)ord(x) gives ordinal number of the element x in an enumer-ated type. Suppose we have following declaration: TYPE DAYS = (sun,mon,tue, wed,thu,fri, sat); Theordvalues of such user-defined type are ordered starting with 0. Therefore, ord (sun) = 0;ord(mon) = 1;...ord(sat) = 6. Theordfunction can be applied to standard types such as Boolean and char, but the values will be different for different machines. h)pred(x) returns the value preceding x in an enumerated type. Thus,pred(6) = 5,pred(tue) =mon pred ( true) = false,pred(sun) = undefined i )Similarly ,succ(x) returns the value following x in an enumerated type ex : succ(7) = 8 ;succ(tue) = wed succ (false) = true,succ(ture) = undefined.

Question:

If the cost of electricity is 5 cents per kilowatt-hour, what does it cost to run a motor 2 hours if the motor draws 5 amperes from a 120-volt line?

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/Users/wenhuchen/Documents/Crawler/Physics/D20-0650.htm

Solution:

E = 120 volts,I = 5 amp,t = 2 hours Work = Power × Time = El × t W =Elt= 120 volts × 5 amp × 2 hr = 1200 watt-hours = 1.2kw-hr This is the work done by the motor in the given time period. Multiplying this by the cost per hour we have; Cost = 1.2kw-hr × 5 \textcent /kw-hr = 6 \textcent .

Question:

When a piece of magnesium ribbon weighing 0.32 g is burnedin oxygen, the resultant oxide weighs 0.53 g. Whatis thepercentage composition of the oxide?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E04-0152.htm

Solution:

To find the percent composition of magnesium and oxygen in the oxide, one must first know the weight of the magnesium and of the oxygen makingup the compound. The percent composition of magnesium is equalto the weight of the magnesium divided by the weight of the compoundmultiplied by 100. The percent composition of the oxygen is foundin the same manner by substituting the weight of oxygen for the weightof the magnesium. One is told (in this problem) that 0.32 g of magnesium is used, and that theweight of the oxide formed is 0.53 g. The weight of oxygen in the oxide isthe difference in the weight of the oxide and the magnesium. weightof oxygen = 0.53 g - 0.32 g = 0.21 g. The oxygen in the oxide weighs 0.21 g. The percent composition of theoxide can now be determined. percentMg in oxide = {(0.32g) / (0.53g)} ×100 = 60 % percentO in oxide = {(0.21) / (0.53)} × 100 = 40. %. The oxide is therefore 60 % Mg and 40 % O .

Question:

Explain what a procedure is, and whatare the different types ofprocedures.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G13-0353.htm

Solution:

If a certain set of calculations or a certain set of expressions to be executedoccur often in a program, it becomes convenient to use a procedurefor the given set. PL/Iprovidesthis very important facility. A procedure that is contained in some other procedure is called an iternalprocedure and a procedure that is not included within another procedureis said to be an external procedure. A reference in the program to the name of a procedure is called a procedure-reference. A procedure can be a function procedure or a subroutine procedure. A function procedure is like a built-in function and initiates an evaluationof the .procedure. The result value produced will be used for furthercalculations in the pro-gram.

Question:

Write a FORTRAN subroutine and an accompanying calling programto perform conversions from base 10 to base M, whereM \leq 36.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G02-0037.htm

Solution:

We initialize an array, NB, with all the possible characters needed forthe different bases. The subprogram runs through the division process forconverting between bases (developed earlier), each time calculating a quotientIQUOT and a remainder IREM. The first dividend is the absolute valueof N_10, placed in location NDIVID. The remainders, integers in base 10 between 0 and n - 1 inclusive, are identified with a corresponding alphanumericcharacter from NB, as shown by the state-ment NA(J) = NB(IREM + 1). Then we test the quotient: if it is not zero, it becomes the newdividend NDIVID, and the loop continues. As written, the subroutine uses remainders to identify with alpha- numericcharacters. Suppose the remainder is 5 (IREM = 5 when J = 14). Then the statementNA(J) = NB(IREM + 1) causes the contents of NB(6) tobe placed in NA(14). The rightmost digits are placed at the higher positionsin array NA. The program looks as follows: DIMENSIONNA(16) 10READ(2,100) M,N 100FORMAT (12,18) IF(M - 1) 15,15,20 15GO TO 66 20CALL C0NVRT (M,N,NA) WRITE(3,101) N, (NA(I), I = 1,16),M 101FORMAT(1Hb, 16, 13Hb(BASE 10) =, 16A1,7Hb,BASE b,2,1H)) GO TO 10 END SUBROUTINECONVRT(M,N,NA) DIMENSIONNA(16), NB(36) DATA NBLANK, MINUS/' ', '-'/ DATA NB/'0', '1','2','3', '4', '5', '6', '7', '8', '9', 'A', 'B', 1'C', 'D', 'E', 'F', 'G', 'H', 'I', 'J', 'L', 'M', 'N', 'O', 'P', 'Q', 2'R', 'S', 'T', 'U', 'V', 'W', 'X', 'Y', 'Z', 'S'/ NDIVID =IABS(N) J = 16 25IQUOT = NDIVID/M IREM = NDIVID - M\textasteriskcenteredIQUOT NA (J) =NB(IREM + 1) J = J - 1 IF (IQUOT) 30,45,30 30NDIVID = IQUOT GO TO 25 45IF (N) 50,55,55 50NA(J) = MINUS IF (J - 1) 65,65,52 52J = J - 1 55DO 60 I = 1,J NA(I) = NBLANK 60CONTINUE 65RETURN 66END

Question:

Why do some viruses contain only one type of protein in their protective coat?

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/Users/wenhuchen/Documents/Crawler/Biology/F05-0149.htm

Solution:

The nucleic acid in a virus particle is surrounded by a protein coat, called a capsid, made up of protein subunits called capsomeres. A complete virus particle is called a virion and it may be covered by an envelope. Those that do not have envelopes are termed naked virions, while those virions that do have envelopes are termed enveloped. A typical naked and an enveloped virion are shown in Fig.1. Some of the protein coats are complex, containing numerous layers and various different proteins. Some also include lipid and carbohydrate molecules. Others, however, contain only one or a few types of protein molecules in their coats. The use of a large number of identical proteins in the protective coat of viruses is often mandatory. The extremely small viruses have a limited nucleic acid content which restricts the total number of amino acidsthat can be used to synthesize its protein. For example, there are approximately 6000 nucleotides in a TMV RNA chain. The TMV RNA can therefore code for a protein 2000 amino acids in length. (Three nucleotides are needed to code for one amino acid.) Since each amino acid weighs about 125 daltons (a dalton is a unit of weight equal to the weight of a single hydrogen atom), this protein molecule of 2000 amino acids should weigh 2.5 × 10^5 daltons. However, since we know the protein coat of TMV weighs 3.5 × 10^7 dal- tons, about 140 (3.5 × 10^7 \div 2.5 × 10^5) of these protein molecules would be needed. The virus does not use all its nucleic acids to code for a single protein. Since only a very small part of the RNA chain codes for the coat protein, the proteins consist of fewer amino acids. Therefore, many smaller identical protein molecules are used to construct the protein coat. Since the number of identical protein molecules used to form the TMV protein coat is about 2150, each protein molecule consists of about 128 amino acids {[(3.5 × 10^7 daltons) / (2150 proteins)] × [(1 amino acid) / (125 daltons)] = (128 amino acids) / (proteins )} The protein subunits either possess helical symmetry or cubical (or quasi-cubical) symmetry. The TMV shell con-sists of identical protein molecules helically arranged around a central RNA molecule (see Fig. 2) . The Adenovirus is an icosahedral virion, exhibiting quasi-cubical sym-metry (see Fig. 3). An icosahedron is a twenty-sided geometric figure.

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Question:

The type of wall socket commonly found in the house is capable of delivering a current of 5 amperes. If this current flows through a cop-per wire with a diameter of 0.1 cm, what is the drift velocity v_d of the electrons?

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/Users/wenhuchen/Documents/Crawler/Physics/D19-0636.htm

Solution:

The drift velocity (v_d) is the velocity of the electrons in the wire due to the accelerating field E^\ding{217}. But, v_d is constant because the electrons are constantly colliding with the copper atoms of the wire. Hence, the net effect of the collisions and accelerating field is to pro-pel the electrons at constant velocity v_d. Our aim is to relate the number of electrons per unit volume of the wire to the current carried by the wire. We will then find a relation for v_d. We shall need to know n_0, the number of free electrons per cm^3 . The copper atom is known to have one loosely bound electron, so we shall assume that one free electron is provided by each copper atom,n_0 is therefore equal to the number of copper atoms in 1cm^3. The atomic mass of copper is 64 which is the gram mass of Avogadro's number of copper atoms. Since the density of copper is 9.0 gm/cm^3,1 cm^3 of copper has a mass of 9.0 gms. Therefore, (6 × 10^23 atoms / 64 gm) = (n atoms in 1cm^3 / 9gm) n = [(9gm) (6 × 10^23 atoms)] / (64gm) n = 8.4 × 10^22 Hence, there are 8.4 × 10^22 atoms in 1cm3ofcopper, and n0= 8.4 × 10^22 (atoms / cm^3 ). The current is 5 amp = 5 amp× (3×10^9 stata amp / 1 amp) or i = 1.5 × 1010statamps The cross- sectional area of the wire in the figure is A = \pir^2 = \pi × (1 / 2 cm)2 The charge on the electron is 4.80 × 10^\rule{1em}{1pt}10 statcoulomb. Then the net charge passing a point of the wire in a time t is the amount of charge in a cylinder (see figure) of length v_dt, where v_d is the drift velocity of the charges, and base area A. This is equal to the number of electrons in the cylinder times the charge of an electron. Q = n_0 AvdteandL = Q / t = n_0 Avde Hence v_d = i /n_0 Ae = (1.5 × 1010satata amp) / [(4.80 × 10^\rule{1em}{1pt}10 satata coul) {(8.4 × 1022(1 / cm^3)} (7.854 ×10\rule{1em}{1pt}3cm^2)] Drift velocity, v_d = 4.74 X 10^\rule{1em}{1pt}2 cm/sec If the wire is 1 meter long, the time taken for an electron to drift from one end to the other is 100 / (4.74 X 10^\rule{1em}{1pt}2 )sec = 2.11 X 10^3 sec = 35 minutes Notice that we chose to work in CGS units throughout. In this particular example we could have obtained the correct answer if we had left i in amps and expressed e in coulombs. To be consistent, we would then have had to express n_0in atoms per m^3, A in m^2 and v_d in m/sec. We would, however, have obtained the correct answer if we had left n_0, A and v_d in CGS units, but this cannot be relied upon without careful consideration of the problem. The student is advised never to mix systems of units and then he can be certain of not running trouble.

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Question:

If a particular character in a certain species of animal were always transmitted from the mother to the offspring, but never from the father to the offspring, what could you conclude about its mode of inheritance?

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/Users/wenhuchen/Documents/Crawler/Biology/F25-0670.htm

Solution:

The inheritance pattern outlined in this question cannot be explained by a mutation. Such a mutation would have to have occurred somewhere in the genes of the normal chromosomal complement of the egg during meiosis, and so coded for a trait not found in theunmutated genes of the sperm. If the trait was indeed caused by such a mutation, we would have to propose a hypothesis that the same mutation happens to all mothers of that species, and that every mother, but not the fathers, are affected . Such a hypothesis is extremely unlikely since mutation rates are by definition very low. This inheritance also cannot be accounted for by the absence of certain genes in the father. If this were true, the first generation offspring would inherit that from their mothers but not their fathers. However in the matings of the offspring of the first generation, the special mode of inheritance would no longer be observed. This is because both the male and female offspring of the first generation will have equal chance of possessing the gene responsible for that trait from their mother. Thus, second generationoffspringswill not necessarily inherit the characteristic from their mothers as opposed to the fathers. Let us apply our knowledge of gamete formation to the problem. We know that spermatogenesis gives rise to sperm which contain little or no cytoplasm.Oogenesisproduces large eggs with enormous amounts of cytoplasm . A small but significant amount of DNA may be present in the cytoplasm . This DNA may carry genes that are not carried on the DNA of the chromosomes. Such genes are called extrachromosomal genes. A male is unable to transmit the traits of such genes, even if he possesses them in his body cells, because his gametes, the sperm, are not formed with any substantial amount of cytoplasm. The female, because of the large amounts of cytoplasm incorporated in the formation of her gamete, the egg, transmit the gene. Thus, we see that an extrachromosomal trait could only be transmitted by the mother. It can be expressed in both male and female offspring , but only the females will be able to transmit the trait to the next generation .

Question:

Why does population growth follow a logistic (S-shaped) form until it reaches the carrying capacity of the environment?

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/Users/wenhuchen/Documents/Crawler/Biology/F30-0768.htm

Solution:

Since nearly all mature individuals in a population can produce offspring, the rate at which an unrestricted population increases is a function of the generation time, which is the period required for an organism to reproduce. When a population initially grows, the size of the population will double every generation time. This kind of growth rate is referred to as exponential growth (Fig. I). A population of animals has a potential through reproduction to increase at a given rate which is termed the intrinsic rate of increase. The intrinsic rate of increase is equal to the difference between the average birth rate and average death rate of a population. It varies enormously from species to species. In a typical experiment, 40 individual paramecia were placed in a small tube of water and each day there-after fresh food was added. Therefore, a constant but limited daily input of food to the system was provided. Under these conditions, the paramecia reproduced quickly at first, with their number increasing exponentially until there were about 4000 animals in the tube. Then the rate of growth leveled off, and the population remained at a steady state. The rate of replacement was finally about equal to the death rate, and the population became balanced. It seems likely that the balance reached by experimental populations is equivalent to the general balance existing in nature. If animals breed prolifically when there is plenty of space and food, but cease to do so when crowded, then it is reasonable to suggest that cessation may be a result of increasing intraspecific (between the members of one species) competition for the resources of the habitat. Breeding may also become suppressed by some effect of the environment which inhibits the population more strongly as crowding increases. The growth of a population in a confined space with a limited input of energy is described by an "S"-shaped graph (see Fig. II). The number of individuals increases rapidly, but in time there is a leveling off of the rate of growth of the population so that the growth rate eventually becomes zero. This suppressing effect of competition gets stronger as the animals get more numerous, until it is strong enough to bring the rate of increase down to zero and an upper limit is reached where the environment in which they live cannot support any more of them. The limiting size is called the carrying capacity of the environment. At this population size, the death rate equals the birth rate, and therefore the rate of increase in the number of individuals is zero. Temporary deviations from zero growth will probably occur causing the population to grow for a short time, or to decline, but the average value over long periods of time will be zero. The actual value of the carrying capacity of the environment for a specific species is determined by the interaction of several factors, including the total energy flow in the ecosystem, the trophic level to which the species belongs, and the size and metabolic rate of the individuals. Eventually, energy from food will be a limiting factor for any population.

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Question:

Simplify the quotient [{2x^0} / {(2x)^0}].

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/Users/wenhuchen/Documents/Crawler/Chemistry/E33-0949.htm

Solution:

The following two laws of exponents can be used to simplify the given quotient: 1) a^0 = 1 where a is any non-zero real number, and 2) (ab)^n = a^nb^n where a and b are any two numbers. In the given quotient, notice that the exponent in the numerator applies only to the letter x. However, the exponent in the denominator applies to both the number 2 and the letter x; that is, the exponent in the denominator applies to the entire term (2x). Using the first law, the numerator can be rewritten as: 2x^0 = 2(1) = 2 Using the second law with n = 0, the denominator can be rewritten as: (2x)^0= 2^0 x^0 Using the first law again to further simplify the denominator: (2x)^0= 2^0 x^0 = (1) (1) = 1 Therefore,[2x^0 / (2x)^0]=2/1 = 2. Therefore,[2x^0 / (2x)^0]=

Question:

Differentiate between acids, bases and salts. Give examples of each.

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/Users/wenhuchen/Documents/Crawler/Biology/F01-0018.htm

Solution:

There are essentially 2 widely used definitions of acids and bases: the Lowry-Bronsted definition and the Lewis definition. In the Lowry- Bronsted definition, an acid is a compound with the capacity to donate a proton, and a base is a compound with the capacity to accept a proton. In the Lewis definition, an acid has the ability to accept an electron pair and a base the ability to donate an electron pair. Salts are a group of chemical substances which generally consist of positive and negative ions arranged to maximize attractive forces and minimize repulsive forces. Salts can be either inorganic or organic. For example, sodium chloride, NaCl, is an inorganic salt which is actually best represented with its charges Na^+ Cl^-; sodium acetate, CH_3COONa or CH_3COO^-Na^+ is an organic salt. Some common acids important to the biological system are acetic acid (CH_3 COOH), carbonic acid (H_2 CO_3), phosphoric acid (H_3 PO_4), and water. Amino acids, the building blocks of protein, are compounds that contain an acidic group (-COOH). Some common bases are ammonia (NH_3), pyridine (C_5H_5N), purineand water. The nitro-genous bases important in the structure of DNA and RNA carry the purine or pyridine functional group. Water has the ability to act both as an acid [(H_2 O-H^+\rightarrowOHH^-] and as a base (H_2 O + H^+ \rightarrowH_3 O^+) depending on the H conditions of the reaction, and is thus said to exhibit amphiprotic behavior.

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Question:

A free harmonic oscillator (FHO) is a system whose behavior can be described by a second-order, linear differential equation of the form: ӱ = -Ay(t)(1) where A is a positive constant. Two FHO systems are a spring-masssystem and an LC electric circuit: FHO y(t) Ẏ=dy/dt A Spring-mass LC circuit x-displacement q-charge in coulombs v =velocity i =current in amperes K/M 1/LC Given the initial conditions y(0) = y0and ẏ(0)=z0, write a Given the initial conditions y(0) = y0and ẏ(0)=z0, write a FORTRAN program that uses the modified Euler method to simulate this system from t = 0 to t = t_f .

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G24-0586.htm

Solution:

Since equation (1) is of the form d^2y/dx^2 = g_1(y), we can use the EULER1 subroutine developed in a previous problem. First, rewrite equation (1) as a system of simultaneous first-order equations: ẏ = dy/dt = z(2) \.{z} = dz/dt = -Ay(3) with initial conditions y(0) = y_0 and z(0) = y'(0) = z_0. Now use the modified Euler method to find solutions to both equations (2) and (3): REAL T/0.0/,TFIN,N,ACCUR,Y0,Z0,A COMMON A READ,N,TFIN,ACCUR,Y0,Z0, A CALL EULER1(N,TFIN,T,Y0,Z0,ACCUR) STOP END FUNCTION G1(W) REAL G1,W,A COMMON A G1 = -A\textasteriskcenteredW RETURN END

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Question:

Calculate the density of the tightly coiled tropocollagen molecule, which may be considered to be a cylinder 2,800 \AA long and 14 \AA in diameter. It contains three poly-peptide chains of 1,000 amino acid residues each.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E22-0805.htm

Solution:

Tropocollagen is a triple helix made up of 3 polypeptide strands as shown in the figure. It is found in tendons. Density is defined as mass divided by volume and is usually in the form of (g / cm^3). Because the tropo-collagen is considered to be a cylinder, its volume is equal to the height times the area of the cross-section. The diameter of the cross-section is 14 \AA or 1.4 × 10^-7 cm. Its radius is 7.0 × 10^-8 cm and its area (\pir^2, where r = radius) equals (7.0 × 10^-8 cm)^2 \pi or 1.54 × 10^-14 cm^2. Solving for its volume: volume = 1.54 × 10^-11 cm^2 × 2.800 × 10^-5 cm (height) = 4.31 × 10^-19 cm^3 The mass of the tropocollagen is found by first determining the number of moles of amino acid residues present. The average molecular weight of amino acid residues is 120. Therefore, the mass of the collagen is equal to 120 g/mole × no. of moles of amino acid residues. Collagen is made up of 3 polypeptide strands each con-taining 1000 amino acid residues. Therefore 3000 amino acid residues are present. 3000 amino acid residues is equal to (3000 residues) / (6.02 × 10^23 residues / mole) = 4.98 × 10^-21 moles The mass of the collagen molecule is, then, mass = 4.98 × 10^-21 moles × 120 (g / mole) = 5.98 × 10^-19 g The density is equal to mass / volume. density = (5.98 × 10^-19) / (4.31 × 10^-19) = 1.39 g/ cm^3.

Question:

A chemist dilutes two 10-ml samples of waste water to 300 ml with aerated water. One sample is analyzed for dissolved oxygen at 20\textdegreeC.immediately, but the second sample is incubated for 5 days before being analyzed for dissolved oxygen. The results are (DO)^20\textdegreec_0 days = 7.9 mg/liter and (DO)^20\textdegreeC5 days= 1.0 mg/liter. What is BOD_5 of the waste water?

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Solution:

BOD stands for Biochemical Oxygen Demand. It measures the potential oxygen requirement of sewage and of sewage polluted waters. (DO)20 \textdegreeCstands for the concentra-tion of dissolved oxygen at 20\textdegreeC. The numerical value of BOD can be found by measuring the dissolved oxygen content (DO) of the sample before and after the 5 day incubation period. BOD is the difference between these dissolved oxygen concentrations (after accounting for corrections of dilution and any additional organic material that might be present in the diluting water). Proceed as follows: The 10 ml sample was diluted to 300 ml or 0.3 l. The DO concentrations at day 0 and day 5 are multiplied by 0.3 l to determine the amount of oxygen used on those days (7.9 mg/ l) (0.3 l) = 2.37 mgDay 0 (1.0 mg/ l) (0.3 l) = 0.3 mgDay 5 These amounts are subtracted to determine the BOD of the sample. 2.37 - 0.3 = 2.07 mg The original sample was 10 ml; the density of water is 1 g/ml. Thus, 10 ml weighs 10 g. The BOD of this sample is 2.07 mg/10 g. BOD's are usually expressed in ppm. Con-verting 2.07 mg/10 g to ppm one contains 207 ppm. For a diversified warm-water biota, including game fish, the DO concentrations should be at least 5 mg/ l or 5 ppm. A body of water is considered polluted whentheDO concentration drops below the level necessary to sustain a normal biota for that water. The BOD range for untreated municipal sewage is 100-400 ppm.

Question:

Suppose that a mass of 8 grams is attached to a spring that requires a force of 1000 dynes to extend it to a length 5 cm greater than its natural length. What is the period of the simple harmonic motion of such a system?

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Solution:

An interesting property of springs is that the length that they stretch is directly proportional to the applied force. The magnitude of this force is: F =kx where k is the force constant. The force constant is k = F/x = (1000 dynes)/(5 cm) = 200 dynes/cm Therefore, the period is by definition: \cyrchar\cyrt = 2\pi\surd(m/k) = 2\pi\surd[8g/(200 dynes/cm)] = 2\pi\surd[8g/(200 {g - cm}/{cm - sec^2})] = 2\pi ×\surd(4/100 sec^2) = 2\pi ×0.2 sec = 1.26 sec and the frequency is by definition: v = 1/\cyrchar\cyrt = 1/1.26 sec = 0.8 Hz.

Question:

Show how Le Chatelier'sprinciple for oxidation-reduction reactions corresponds to the Nernst equation.

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Solution:

Le Chatelier's principle predicts that increasing the con-centration of a reactant favors its tendency to react and decreasing the concentration of a reactant diminishes its tendency to react. Similarly, decreasing the concentration of a product favors the tendency towards formation of that product. The Nernst equation relates E, the poten-tial for a reaction or half- reaction at non-standard conditions, to E\textdegree, the standard potential for that reaction or half-reaction at unit acti-vities. Namely, E = E\textdegree -(.591 / n)log Q, where n = the number of electrons transferred in the reaction and Q is the equilibrium expression. Q gives the ratio of products, to reactants, each raised to the power of their coefficients in the chemical equation. Notice, that both Le Chatelier's principle and the Nernst equation re-late what happens to reaction tendencies when concentrations are changed. For example, in the Nernst equation, if the concentrations are such that yield a positive potential, E, then the reaction is favorable in the direction written. The difference between the Nernst equation and Le Chatelier's principle is that the latter relates reaction tendencies with a change of concentration qualitatively, while the former discusses it quantitatively .

Question:

In a chemical reaction requiring two atoms of phosphorus for five atoms of oxygen, how many grams of oxygen are required by 3.10 g of phosphorus?

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Solution:

Because the relationship between the phosphorus and oxygen is given in atoms, the relationship also holds for moles. There must be 2 moles of phosphorus for every 5 mole of oxygen. This is true because there is a set number of atoms in any one mole. Therefore, one must first find the number of moles of phosphorus present. From this, one can find the number of moles of oxygen present. From the number of moles of oxygen, one can find the weight by multiplying by the molecular weight. The number of moles of phosphorus present is found by dividing The number of grams by the molecular weight (MW = 31). no. of moles = [(3.10 g)/(31 g/mole)] = .10 moles Because there must be five moles of oxygen for every two moles of phosphorus, the following ratio can be used to determine the number of moles of oxygen. Let x = number of moles of O. [(2 moles P) / (5 moles 0)] = [(.10 moles P) / (x)] x = [(5 moles O × .10 moles P) / (2 moles P)] = .25 moles O The weight of the oxygen is then found by multiplying the number of moles by the molecular weight (MW = 16). no. of grams of O = 16 g/mole × .25 mole = 4.00 g O.

Question:

A firm is considering investing $75,000 in a new venture. Its economists have projected the following returns over the next five years. YearReturn 120,000 230,000 335,000 440,000 550,000 Total$175,000 Describe how the discounted cash flow interest rate would be found, given a current interest rate of 10%.

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Solution:

From the given table we observe that the gross return from the investment is $175,000 - $75,000 = $100,000. However, this is assuming that the money invested has no alternative uses, e.g. it could not be invested in a bank at 10% interest. Taking the time value of money into account, to obtain $20,000 after 1 year we would have to invest $20,000 = X + .IX now. Thus, X = (20,000 / 1.1) = $18182 Similarly, $30,000 two years from now is worth $30,000 = (X+.1X)(1+.1) = X(1+.1) (1+.1) = X(1.1)2 Or, X = [$30,000 / (1.1)^2] = $24,794 Applying the same analysis to the remaining returns: Year Return Present Value 1 2 3 4 5 20,000 30.000 35,000 40,000 50,000 20,000/1.1= 18,182 30,000/(1.1)2= 24,794 35,000/(1.1)3= 26,296 40,000/(1.1)4= 27,321 50,000/(1.1)^5= 31, 047 Adding the present values of returns over the five years gives the present value of returns from the investment. PV = $127,640. The net present value of the investment when the current interest rate is 10% is therefore NPV = PV - I = $127,640 - $75,000 = $52,640. This means that the $100,000 return over five years is actually worth $52,640 to the firm now. Note that as the interest rate changes continuously from 0% to 100%, the NPV changes continuously from $100,000 (the gross return becomes the net present value when the interest rate is 0%) to $-59,063. Thus, there must be an interest rate at which the NPV equals zero. This interest rate value is called the DISCOUNTED CASH FLOW interest rate. A graph of the NVP versus interest rate might look as shown in Fig. 1. To actually compute the DCF we must find a root of the equation f (r) = 0 where NVP = f (r) . We cannot use the Newton-Raphson method because the functional relationship is not in polynomial form. The bisection method, however, may be applied. The procedure is as follows: We know that f(.10) = 52,640 and f(1.00) = -$59,063. Compute the value of f at the mid-point of the interval [10, 1.00], f(.55). If f(.55) \geq 0 the root must lie in the interval [.55, 1.00] (since f(.10) is positive and f(1.00) is negative). If f(.55)<0, then the root lies in the interval [.10, .55]. In either case the interval of search has been reduced by 1/2. Repeat the above procedure with the new interval i.e. take the mid-point and evaluate f at this point. A flowchart for computing the DCF is as shown in Fig. 2

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Question:

In what ways are studies of twins useful insupplying information about the relative importance of inheritance and environment in the determination of a given trait?

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Solution:

Although both environment and heredity are involved in the development of any trait, a change in some aspect of the environment may alter one character relative-ly little as compared to its effect on another character. An individual's blood type, for example, seems fairly impervious to practically all environmental effects, while the phenotype of a diabetic can be radically changed by a mere alteration in diet. Moreover, not all characters are determined by simple genetic effects which have easily observable relationships with simple environmental changes. Some traits , such as intelligence in animals, are probably determined by complex genetic-environmental interactions, the results of which are then viewed as a single trait. Performing studies using twins is one way of deter-mining the relative contributions of genetic composition and environment to the expression of a given trait. There are two kinds of twins: identical or monozygotic twins, which arise from a single fertilized egg, and fraternal ordizygotictwins, which result from the fertilization of two separate eggs. From this definition, we can see that monozygotic twins are genetically the same in every respect, while dizygotictwins may differ genetically for any character. By comparing both kinds of twins with regard to a particular trait, we can evaluate the roles of environment and heredity on the development of that character . On the one hand, we have genetically identical individuals of the same age and sex, raised in a single uterine environment?andon the other , we have genetically dissimilar individuals of the same age, though not necessarily of the same sex, raised in a common uterine environment. Presumably the difference between these two kinds of twins is only in the extent of their genetic similarity. Using twins for genetic studies, then, provides built-in controls for both the effect of environment and the effect of heredity on the expression of a given trait. For example, if phenotypic similarities for a particular character are greater among identical twins than among fraternal twins, we can ascribe this to the genetic similarity of the identical pair and the genetic dissimi-larity of the fraternal pair. On the other hand, if both identical and fraternal pairs show the same extent of phenotypic differences for a particular character, we can assume that genetic similarity or dissimilarity plays less of a role than the differences which may occur in the post-uterine environment of the twins. As a concrete illustration of this, consider a pair of monozygotic twins reared in different postnatal environments and a pair ofdizygotictwins reared in a common postnatal environment. If the monozygotic twins were found to exhibit a wide range of dissimilarities and thedizygotictwins a significant range of similarities , the importance of the environment as a phenotypic modifier of the genetic composition of an individual can readily be appreciated.

Question:

Why is the Hardy-Weinberg population only anidealistic modeland not representative of every natural population?

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Solution:

A Hardy-Weinberg population, like all statistical samples, exists onlywhen the number of individuals involved is significantly large. In additionto a large population size, there must be no selection for or againstany specific characteristic, no occurrence of genetic mutations, no movementof individuals into or out of the population, and complete randommating. The Hardy-Weinberg Principle fails to explain the genetic compositionof the population if any of the above conditions is not met. Natural populations rarely meet more than one, if any, of the conditions. While it is true that many populations in the natural environment are large, manyare small and relatively isolated on islands, mountain slopes and otherphysical or geographical barriers. These isolated populations may containhomo-geneous individuals as a result of continued inbreeding over longperiods of time or through the fixation of genes caused by genetic drift. The condition of no selection for or against any specific trait is also readilyviolated in most natural populations. Selection by natural forces includingwind, cold, heat, disease, and predators is clearly evident, and theindividuals that survive tend to have certain common adaptive traits. Mating of individuals in a population or species is rarely random. The more colorful male birds of some species, for example, are more effectivein attracting and stimulating female birds. It is also not unusual to observethat in some species, bigger individuals tend to mate with each other, and the same is true of smaller individuals. Mutations in the natural populationhave been shown to occur constantly, contributing a great numberof changes to a species in its evolutionary development. Finally, whereassome populations are indeed isolated and have no free migration ofindividuals, many populations do allow the "come and go" of their members. Those mig-rating into the population may be able to mate with thenative members, adding new genes to the population as a result. Those leaving the population are in effect, removing genes from the population.

Question:

How many coulombs are required to deposit 0.10 gram of zinc, the atomic mass of zinc being 65.38 and its valence 2?

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Solution:

In order to do this problem, we must realize that for every 2 electrons transported through the circuit, one zinc atom willdedeposited. We therefore know that exactly half as many zinc molecules (or moles of zinc molecules) will be deposited as transported electrons (or moles of transported electrons). We may calculate the number of moles of zinc we need to deposit by divid-ing the mass of material we want by the atomic mass of zinc. Equating this with one half the number of moles of electrons transported (obtainable by dividing the amount of transported charge by 96,500 coulombs/mole) we have [0.10 gm / (65.38 gm / mole)] = (1/2) [Q in coulombs / (96,500 coulomb / mole)] Solving for Q we have. Q = [0.10 gm / (96,500 coulomb / mole)] / [32.69 gm / mole] =295 coulomb, approximately.

Question:

What is meant by the statement that the atoms of the chlorophyll molecule constitute a "resonating system"? Of what importance is this in the process of photosynthesis?

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Solution:

If we examine closely the structure of the chlorophyll molecule (see Figure) we will see that its atoms constitute a conjugated system, with double and single bonds alternating around the ring. Such a conjugated system provides for many possible resonance structures and is essentially a resonating system in which the pi electrons are spread, or delocalized, about the carbon atoms of the ring. In other words, a pi electron is no longer associated with a single atom or bond but with the conjugated system as a whole. This property, called resonance, gives the chlorophyll molecule considerable stability. As a conse-quence of resonance stabilization, only a small amount of energy is required to raise the pi electrons of the carbon atoms to a higher energy level. The chlorophyll molecule absorbs energy by having its pi electrons excited and "pumped" to a new, more energetic orbital (energy level). The energy of excitation comes from the sun in the form of visible light energy, and is converted to chemical energy in the light reactions of the photosynthetic process. The absorbed energy is always in a definite, specific quantity, known as a photon. In order to raise an electron to a given new energy level, the energy of the photon must just equal the difference in energy content of the electron's old and new orbitals. Unless a photon can raise the energy of the electron by just the right amount for a defined energy level, that photon will not be absorbed. The adaptive importance of the chlorophyll molecule lies therefore in its unique capacity to function in capturing the energy of visible light with a high degree of efficiency.

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Question:

It has been estimated that each square meter of the earth's surface supports 1 × 10^7 g of air above it. If air is 20% oxygen (O_2, molecular weight = 32 g/mole) by weight, approximately how many moles of O_2 are there above each square meter of the earth?

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Solution:

This problem is solved by first calculating what weight of oxygen is present in 1 × 10^7 g of air and then dividing by the molecular weight of oxygen to con-vert this mass to moles. Using the definition of weight percent, weight % of O_2 = [weight of O_2 / total weight] × 100% 20% = [weight of O_2 / 1 × 10^7 g] × 100% 20 = [weight of O_2 / 1 × 10^7 g] × 100 Solving for the weight of O_2, weight of O_2 = (20/100) × 1 × 10^7 g = 2 × 10^6 g. The number of moles of O_2 is equal to the weight of O_2 divided by the molecular weight, or moles of O_2 = weight of O_2 / molecular weight= (2 × 10^6 g) / (32 g/mole) \cong 6 × 10^4 moles. Therefore, each square meter of the earth's surface supports 6 × 10^4 moles of O_2.

Question:

A certain solution contains 5 % FeSO_4. How many pounds of Fe could be obtained from 1 ton of this solution?

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Solution:

If a solution contains 5 % FeSO_4, this means that 5 % of the total weight of the solution is FeSO_4. In this problem, 5 % of a ton is FeSO_4. To determine the weight of Fe in this amount of FeSO_4, one must calculate the weight percent of Fe in FeSO_4. This is done by dividing the molecular weight of Fe by the molecular weight of FeSO_4 and multiplying the quotient by 100. To solve this problem: (1) Determine the percent weight of FeSO_4, (2) Determine the weight of 5 % of a ton (3) Determine how much of this weight is Fe. (1) The molecular weight of a compound is determined by adding togetherthe weight contributed by the elements of which it is composed. MW of Fe = 55.8, MW of S = 32, MW of O = 16. Thus , for FeSO_4, 1 atom of Fe1 × 55.8 = 55.8 1 atom of S1 × 32= 32 atoms of O4 × 16=64 molecularweight ofFeSO_4= 151.8 percent weight of Fe in FeSO-_4 = ( weight of Fe / weight of FeSO_4) × 100 = (55.8 × 151.8) × 100 = 37 % 37 % of FeSO_4 , by weight is Fe. (2) Determining 5 % of a ton. 1 ton = 2000 lbs. 5% of a ton = .05 × 2000 lbs = 100 lbs. Thus, 5 % of a ton is 100 lbs. (3) Determining the weight of Fe in 100 lbs. 37 % of this is Fe. weight of Fe = .37 x 100 lbs = 37 lbs.

Question:

A chemist performs a neutralization reaction. She finds that 1 g of C_6H_10C_4, an acid, requires 0.768 g of KOH for completeneutralization. Determine the number of neutralizableprotons in this molecule.

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Solution:

To solve this problem, first determine how many equivalents of the acidor base are involved. Once this is known, the number ofneutralizable protonswill also be known. One equivalent is defined as the mass of the substance, if it is an acid, needed to furnish one mole of H_3O^+ or, if it is a base, needed to furnish one mole of OH^-. The number of grams/equiv. for KOH is 56.1, since only 1 mole of OH^- can be furnished. The number of equi-valents is [(0.768 g)/(56.1 g/equiv)] = 0.0137 equiv of KOH. The number of equivalents of base must equal that of the acid. The numberof equivalents of acid = 0.0137. The number of grams/equiv is [(1.00 g of C_6H_10C_4)/(0.0137 equiv.)]= 73.0 g/equiv. One mole of C_6H_10C_4 weighs 146.1 g, since a mole = weight in grams/molecular weight. Therefore, the number of moles/equiv is [(73.0 g/equiv)/(146.1 g/mole)] = 0.5 moles/equiv. It follows that each mole of acid furnishestwo H_3O^+. As such, there are twoneutralizableprotons in C_6H_10C_4.

Question:

What is the pH of a neutral solution at 50\textdegreeC? pK_W= 13.26 at 50\textdegreeC.

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Solution:

A neutral solution is defined as [H^+] = [OH^-]; an acid solution has [H^+] > [OH^-], and a basic solution has [H^+]<[OH^-]. For a solution at 25\textdegree, pK_W= 14, pK_W,indicates the amount of dissociation of water. To find the neutral pH, one lets pH =pOH= x. Since pH+pOH= 2x = 14 =pK_W, x = 7. However, the solution in question is at 50\textdegree. At 50\textdegreepK_W= 13.26. Therefore, to find the neutral pH, pH+pOH= 2x = K_W = 13.26 x = 6.63 = neutralpH.

Question:

If the frequency of an oscillating source on the surface of a pool of water is 3 Hz, what is the speed of the wave if the wavelength is observed to be 0.5 m?

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Solution:

An example of an oscillating wave, is a sinusoidal wave. Three important properties of an oscillating wave are its velocity of propagation, frequency, and wavelength. Its frequency f is the number of cycles per unit time at which any point oscillates or, expressed in another way, the number of waves that pass a given point per unit time. The wavelength \lambda is the distance between two adjacent crests of the wave. For a relation between these quantities, note that the time t required for the wave to make one oscillation is 1/f. During this time, the wave moves a distance d = \lambda. From d =vt we have\lambda = v \textbullet (1/f) = (v/f) orv =f\lambda Substituting the known values, v = (3 Hz) (5 × 10^\rule{1em}{1pt}1 m) = 1.5 m/sec (Note: 1 Hz = sec^\rule{1em}{1pt}1)

Question:

When estimating the total number of Calories required by a personin a day, we consider such factors as sex, amount of activity, and the temperature in which the person lives. Why arethese factors necessary?

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Solution:

One must first clarify the difference between the terms "calorie" and "Calorie."One standard calorie is defined as the heat necessary toraise thetemperature of one gram of water by 1\textdegreeC from 14\textdegreeC to 15\textdegreeC. However, thenutritionist's Calorie is equal to 1000 of these small calories, or what a physicistwould call a kilocalorie. The energy yield of food measured in Calories is the amount of heatenergy that can be extracted from a given amount of food. When one mole(180 g) of glucose is burned (oxidized) in a calorimeter, 690 Calories aregiven off. Fats have a higher caloric value than carbo-hydrates, yielding about9 Calories per gram. This gives a value of 1620 Calories for 180 g offats. A minimum amount of Calories each day is needed to supply the energyrequirements of the body. If we ingest more food than we really needthe excess energy is usually stored as fat. Each person's Caloric requirementdiffers, however, according to sex, amount of activity and environmentaltemperature. A man usually requires more Calories per day than does a woman. A woman's metabolism is usually lower than that of a man, so that women generallyhave to eat less per body weight than men. A lower metabolic rateindicates a lower Caloric need since less energy is required to drive thereactions of the body. The Caloric need of a person also varies with the amount of physicalactivity. More Calories are needed for greater amounts of activity. A highly active person requires more energy for body processes than does asedentary person. A jogger needs ATP for muscle contrac-tion during running; his cardiac and respiratory rates increase to speed the delivery of O2 to his cells and the removal of CO2 from his cells - these processes involvefurther ATP need. A sedentary person needs just enough Calories tosustain normal metabolic activity. The temperature of a person's surroundings also affects Caloric requirements. A person working in a cold environ-ment needs energy, not onlyfor his regular metabolic needs, but also for heat production to maintainhis normal body temperature. A person in a warmer climate has nosuch difficulties.

Question:

What makes the blood flow throughout the body? In what part of the circulatory system does the blood flow most slowly? Why?

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Solution:

Blood flow is the quantity of blood that passes a given point in the circulation in a given period of time. It is usually expressed in milliliters or liters per minute. Blood flow through a blood vessel is determined by two factors: (1) the pressure difference that drives the blood through the vessel; and (2) the vascular resistance, or impedance to blood flow. These relations can be expressed mathematically: Flow = (\Delta Pressure / Resistance) . As the pressure difference increases, flow increases; as the resistance increases, flow decreases. The pressure difference between the two ends of the vessel causes blood to flow from the high pressure end to the low pressure end. This pressure dif-ference or gradient is the essential factor which causes blood to flow. Although the heart acts to pump the blood, it is the pres-sure difference (about 140 nun Hg in the aorta and near 0 in the vena cava at the right atrium) that is critical for blood flow. The overall blood flow in the circulation of an adult is about 5 liters per minute. This is called the cardiac output because it is the amount pumped by each ventricle of the heart per unit time. The cardiac output is determined by multiplying the heart rate and the volume of blood pumped by each ventricle during each beat (stroke volume): cardiac output = heart rate × stroke volume. Normally the heart rate is about 70 beats per minute and the stroke volume is about 70 milli-liters per beat: CO = 70 (beats / min) × 70 (mls/ beat) = 4900 (mls/ min). The normal cardiac output is thus about 5 liters per minute. The rate of flow throughout the circulation is not constant. The rate is rapid in the arteries (about 30 centimeters per second) but falls as it moves through the arterioles. The rate is slowest in the capillaries but then increases again in thevenulesand veins. There is a physical explanation for this phenomenon. The velocity of blood flow in each type of blood vessel is inversely proportional to its total cross-sectional area. If any fluid passes from one tube to another of larger radius, the rate of flow is less in the larger tube. The rate is fastest in the aorta since it has the smallest total cross-sectional area of any vessel type (2.5 cm^2) The rate decreases in the arteries (20 cm^2) and decreases further in the arterioles (40 cm^2). The rate of flow decreases drastically in the capillaries which have a total cross-sectional area of 2500 cm^2 . Although the individual branches of the capillaries have a diameter much less than the aorta, their total cross-sectional area is much greater; therefore the rate in the capillaries is 1/1000 the rate in the aorta. The rate of flow increases in thevenules (250 cm^2) and increases further in the veins (80 cm^2). Finally, the rate increases rapidly in the vena cava where the cross-sectional area is 8 cm^2 .

Question:

0.001 mol ofNaOHis added to 100 ml. of a solution that is 0.5 M HC_2H_3O_2 and 0.5 M NaC_2H_3O_2. Determine the pH of this Solution; theK_dissof acetic acid is 1.8 × 10^-5.

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Solution:

This problem requires an understanding of the concept of buffer solution. In some processes, a solution of constant pH is desired. This constancy is maintained by the buffering action of an acid-base equilibrium. A buffer contains both an acid and a base and responds to the addition of either H_3O^+ or OH^- to maintain thepH.In this problem, the 0.5 M solutions of HC_2H_3O_2 and NaC_2H_3O_2 act as a buffer that must respond to the addition of base,NaOH. One has equal numbers of moles of HC_2H_3O_2 and NaC_2H_3O_2 in the buffer, which means that the {[HC_2H_3O_2] / [C_2H_3O_2^-]} ratio is one. In solution, one has HC_2H_3O_2 + H_2O \rightleftarrows H_3O^+ + C_2H_3O_2^- , which indicates that each mole/liter of HC_2H_3O_2 must produce the same number of mole/liter of C_2H_3O_2^-, For this reaction, k = {[H_3O^+] [C_2H_3O_2^-]} / [HC_2H_3O_2]or[H_3O^+] = {k [HC_2H_3O_2]} / [C_2H_3O_2^-] , which is an expression for the concentration ofhydroniumions in the buffer. Recalling that pH = - log [H_3O^+] and thatpK= - log K, by taking the negative log of both sides of the above equation, one can write - log [H_3O^+] = - log K - log {[HC_2H_3O_2] / [C_2H_3O_2^-]} , pH =pK- log {[HC_2H_3O_2] / [C_2H_3O_2^-]} For this solution, it is given that K = 1.8 × 10^-5 and that {[HC_2H_3O_2] / [C_2H_3O_2^-]} is equal to one. Thus, pH =pK- log {[HC_2H_3O_2] / [C_2H_3O_2^-]} pH = (- log 1.8 × 10^-5) - log (1) pH = 4.74 - 0 = 4.74 One adds 0.001 moles ofNaOH. This base will convert an equal number of moles of the acid, [HC_2H_3O_2] to [C_2H_3O_2^-], by a neutralization reaction. If one started with 0.5 M HC_2H_3O_2 in 100 ml, one had 0.05 mole of it. Upon addition ofNaOH, however, 0.001molesof it is converted to C_2H_3O_2^-. This means one has 0.05 - 0.001 = 0.049 moles of HC_2H_3O_2 left. There were 0.05 moles of NaC_2H_3O_2 to start. NaC_2H_3O_2 exists as the ions Na+ and C_2H_3O_2^-. There were 0.001 moles of C_2H_3O_2^- produced upon addition ofNaOH, which means one has a total of 0.001 + 0.05 = 0.051 moles of C_2H_3O_2^-. The volume of the solution remained at 100 ml. Therefore, these mole amounts are 0.49 M concentration of HC_2H_3O_2 and 0.51 M concentration of C_2H_3O_2^-. To find the new pH, one need only substitute these values into pH =pK- log {[HC_2H_3O_2] / [C_2H_3O_2^-]} pH = - log (1.8 × 10^-5) - log (.49/.51) = 4.74 + 0.017 = 4.76.

Question:

A spotlight equipped with a 32-candle bulb concentrates the beam on a vertical area of 125 ft^2at a distance of 100 ft. What is the luminous intensity of the spot of light in the beam direction?

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Solution:

In a lamp, electrical power is supplied and radiation is emitted. The radiant energy emitted per unit of time is called the radiant flux. Only a fraction of this lies within the wavelength interval (400m\mu to 700 m\mu) which can produce a visual sensation in the human eye. The part of the radiant flux which affects the eye is called the luminous flux F and is measured in lumens. The luminous intensity I of a source is defined as theflux it emits flux it emits per unit solid angle (I = F/\cyrchar\cyromega). The solid angle w is a measure of the size of the cone of radia-tion and is defined in steradians as \cyrchar\cyromega = A / R2 where R is any distance and A is the area of a spherical surface surrounding the source which is irradiated by the light source at this radius R. Assuming that the spotlight radiates in all di-rections, the total area illuminated at a distance R is equal to the area of a sphere of radius R or 4\piR^2. Therefore the total solid angle \cyrchar\cyromega_total = 4\piR2/ R2= 4\pi sterad . and the total flux emitted by the bulb is given by F_total = I \cyrchar\cyromega_total = (32 candles) (4\pi steradians) = 400 lu This total flux is concentrated into a beam with a small solid angle through the use of a reflector and lens. This solid angle is \cyrchar\cyromega = A / R2= 125ft^2 / (100ft)2= 0.0125 sterad. The intensity of this beam is then I = Ftotal/ \cyrchar\cyromega = 400lu/ 0.0125stread= 32,000 candles

Question:

Using a virus, how can one transformE.colibacteria unable to utilize galactos (gal-mutants) into those that can utilize galactose (gal^+).

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Solution:

E.coliare usually able to utilize the mono-saccharide galactose as a coarbon and energy source. However, mutations arise which affect an enzyme necessary for galac-tose utilization. TheseE.colimutants are unable to utilize this sugar, and are called gal^-mutants. To transform some of these gal^-mutants into the wild type, gal^+, one can use a lysogenic virus such as the \lambda (lambda) bacteriophage. One can infect a culture of prototrophicE.coli(gal^+) with the lysogenic phages. The viruses will inject their DNA into the bacterial cells. In the host, the viral DNA molecule changes from a linear structure to a circular one. At a specific attachment site on the bacterial chromosome, the viral DNA pairs with the bacterial chromosome and integrates into it after recombination. The viral DNA, now called a prophage, remains incorporated within theE.colichromosome, until conditions are favorable for the excision of the prophage and its induction to a vegetative virus which replicates and lyses the cell. The gene necessary for galactose utilization is very near the incorporated prophage. In the case of the \lambda phage, its proximity to the bacterial gal^+ gene allows rare errors to occur in which excision of the prophage includes the gal^+ gene. The \lambda prophage may coil in such a way that the recombination event leading to excision of the circular viral DNA includes the gal gene. To remain approxi-mately the same size, the circular viral DNA leaves behind some of its own genes, which were replaced by the substituted bacterial region. Since it now lacks some necessary genes, the virus, now called a transducing particle or \lambda gal, is considered to be defective. These transducing particles can lyse the bacterial cell, but they cannot establish a lysogenic relatonship with, or cause lysis of, subsequent bacterial cells. The excision of the prophage which has the incorporated gal^+ gene can be illustrated as follows: One can then harvest these transducing particles and add them to a culture of gal^-E.colimutants. These phages attach to the mutant bacteria and inject their DNA (containing the gal^+ gene). The gal^+ region of the viral DNA may then recombine with the gal^- region of the bacterial chromosome. The gal^+ gene may be incorporated into theE.colithrough the recombination process, transforming them to the wild type, which are able to utilize galactose. The whole viral DNA rarely becomes incorporated, since the transducing phage lacks a complete genome. The viral chromosome, containing the gal^- gene, cannot be inserted into the bacterial chromosome and so is lost. This process by which a virus transfers genes from one bacterial cell to another is called transduction. In specialized transduction, as opposed to generalized transduction, only one gene is transferred. Transduction thus serves as a mechanism for recombination in bacteria.

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Question:

A light horizontal bar is 4.0 ft long. A 3.0-lb force acts vertically upward on it 1.0 ft from the right- hand end. Find the torque about each end.

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/Users/wenhuchen/Documents/Crawler/Physics/D02-0028.htm

Solution:

Since the force is perpendicular to the bar, the moment arms are measured along the bar. About the right-hand end L_r = 3.0 lb × 1.0 ft =3.0 lb-ftclockwise About the right-hand end L_1 = 3.0 lb × 3.0 ft = 9.0 lb-ftcounterclockwise The torques produced by this single force about the two axes differ in both magnitude and direction. This causes the bar to twist through an angle \texttheta which is proportional to the torque.

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Question:

How do the algae differ from the fungi in their method of obtaining food?

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Solution:

Algae areautotrophs, while the fungi areheterotrophs. Certain algae contain the green pigment chlorophyll, others have the accessory pigments. These pigments enable them to utilize solar energy to synthesize organic compounds from CO_2 and H_2O. Fungi, lacking photo synthetic pigments, must obtain organic compounds direct-ly. The eucaryotic algae (all algae except the blue-green) contain chlorophyll in chloroplasts,subcellularorganelles where photosynthesis occurs. Solar energy is trapped by the chlorophyll and converted to ATP, the ATP is used by the chloroplast to convert carbon dioxide and water to organic compounds, and oxygen is liberated. Algae obtain water and CO_2 from their environment, as well as the other minerals needed for their organic con-stituents. The CO2and minerals are dissolved in the aquatic environment, and are absorbed through the cell wall and cell membrane of the algae. Fungi live where organic compounds are present. Some fungi are animal or plant parasites but most are saprophytes. Mostsarophyticfungi excrete digestive enzymes into their environment, and break down food materialextracellularly. The products of digestion are then absorbed through the cell wall and cell membrane by structures called thehautoria. Saprophytic fungi de-compose vastquatitiesof dead organic material. Some cause spoilage of foodstuffs (bread, fruit, vegetables) and deterioration of leather goods, paper, fabrics, and lumber. Parasitic fungi may carry out extracellular digestion or they may directly absorb organic materials produced by the body of their host. However, many fungi are benefical to man.

Question:

a notochord, a dorsal hollow nerve cord, a backbone, hinged jaws, lungs and large, fleshy paired pelvic and pectoral fins. A marine biologist concluded that the animal very closely represented the direct ancestor of the land vertebrates. What kind of animal was this?

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/Users/wenhuchen/Documents/Crawler/Biology/F13-0333.htm

Solution:

By assigning the observed characteristics to a class of organisms in a progressively specialized manner, we may deduce the identification of this marine animal. The presence of a notochord and a dorsal hollow nerve cord indicates that the animal belongs to the phylum Chordata. The presence of a backbone and hinged jaws leads us to reason that the animal is a vertebrate; more specifically, it is a bony fish (Osteichthyes). The existence of lungs and large, fleshy fins further classifies the animal as a lobe-finned fish. Since the animal was thought to represent the ancestors of the land vertebrates, our conclusion is confirmed. The animal is a lobe-finned fish, otherwise known as a coelacanth.

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Question:

Define a BASIC function which takes on the value \surd(x^2 - y^2) if x^2 \geq y^2, but \surd(y^2 - x^2) if y^2 > x^2.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G10-0237.htm

Solution:

A multiline function which takes into account the sign of x^2 - y^2 is needed for solving this problem. The multiline function starts with a defining statement. It does not include an assignment statement (unlike the DEF statement of a single line function), just the function name and a list of "dummy" variables. BASIC statements defining the calculation of the function follow. Also, there must be a function ending command. It is the same for all multiline functions and consists of a statement number followed by the word FNEND. Note that the multiline function can be placed anywhere in the program. As in the case of a single line function, it is good program-ming practice to place all functions at the beginning (or end) where they can be kept track of by the programmer. See the program below: 1\O\O DEF FNA(X,Y) 11\O LET X1 = X_\uparrow2 - Y_\uparrow2 12\O IF X1 > = \O then 14\O 13\O LET X1 = -X1 14\O LET FNA = X1_\uparrow\O.5 15\O FNEND

Question:

Suppose the force applied to a mass M by a spring stretched in the x direction isF_x= -Cx, where C is a constant. Consider a non-inertial frame with the acceleration a_0^\ding{217} = a_0x˄ in the x direction. Derive a relation between the displacement of the spring (x) and the acceleration of the non-inertial frame (a_0) relative to the earth.

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/Users/wenhuchen/Documents/Crawler/Physics/D04-0213.htm

Solution:

If we wish to analyze this problem from the point of view of an observer in a non-inertial reference frame, we cannot use Newton's Laws in their usual form. (A non-inertial frame is one which accelerates with respect to the fixed stars.) However, if we modify these laws, we may apply them in accelerated frames. The modi-fied form of Newton's Second Law is F^\ding{217} - F'^\ding{217} = Ma"^\ding{217}(1) where a"^\ding{217} is the acceleration of the system as examined with respect to the non-inertial frame, F^\ding{217} is the sum of all real forces acting on the mass M, and -F'^\ding{217} is the sum of all fictitious forces acting. (Real forces are gravitational, elastic, etc. while fictitious forces arise only because we insist upon doing the problem in an accelerated frame. Examples of the latter are Carioles forces, centrifugal forces, etc.) In our case, M is at rest in the accelerated frame, so that a"^\ding{217} = 0. F^\ding{217} is the real force acting on M and is supplied by the spring. Hence, F = -Cx(2) The minus sign indicates that F is a restoring force. No matter how we displace the mass, the spring tends to restore M to its initial position. -F', the fictitious force, is defined as -F'^\ding{217} = - Ma'^\ding{217}, where a'^\ding{217} is the acceleration of the non-inertial frame with respect to the earth (a_0x˄ in this problem). Therefore, -F' = - Ma_0(3) Inserting (3) and (2) in (1) we find -Cx- Ma_0 = 0(4) Solving (4) fora_O, we obtain a_O = - (Cx/M)(5) which is the desired relation. The non-inertial frame might be an aircraft or an automobile. We see that Eq. (5) describes the operation of an accelerometer in which a mass M is attached to a spring and constrained to move in the direction of the acceleration. The displacementx of the mass measures the accelerationa_Oof the non-inertial reference frame.

Question:

Taking the wave properties of the electron into account, find the connection between the errors \Deltap_x and \Deltax when measuring the momentum and coordinates of an electron, if \Deltax is determined by the width d of the slit through which the electron beam passes (Figure below).

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/Users/wenhuchen/Documents/Crawler/Physics/D33-0991.htm

Solution:

As the electron is passing through the slit, we lose sight of it. Therefore its position at the slit has an x-axis uncertainty \Deltax = d. As a result of it, the momentum of the electron after it passes through the slit can have an x-component \Deltap_x = p sin \alpha . In order to observe the wave characteristics of the electron, we should be able to display its diffraction pattern. Therefore, the slit should have a size consistent with the single slit condition for observing the first fringes of the diffraction pattern sin \alpha \geq (\lambda/d) where \lambda. is the wave length (de Broglie wave length) of the electron. \lambda = (h/p). Now we calculate \Deltax \Deltap_x , to obtain the uncertainty relation for the electron; \Deltax \Deltap_x \approx dp sin \alpha \geq dp (\lambda/d) = dp (h/pd) \geq h.

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Question:

One of the methods for the synthesis of dilute acetic acid from ethanol is the use ofacetobacter. A dilute solution of ethanol is allowed to trickle down overbeechwoodshavings that have been inoculated with a culture of the bacteria. Air is forced through the vat countercurrent to the alcohol flow. The reaction can be written as C_2H_5OH (aq) + O_2 CH_3COOH (aq) + H_2O (l) acetic acid Calculate the Gibbs Free Energy, ∆G\textdegree, for this reaction, given the following Gibbs Free Energies of Formation: Ethanol = - 43.39 Kcal/mole, H_2O (l) = - 56.69 Kcal/mole, and Acetic acid = - 95.38 Kcal/mole.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E28-0910.htm

Solution:

The ∆G\textdegree of a reaction is a measure of the free energy, i.e., the energy available to do work. It can be calculated from the free energy of formation, ∆G\textdegree_f, of the reactants and products. ∆G\textdegree_fis a measure of the free energy needed to form a compound from its elements in their standard states. By definition, the ∆G\textdegree_fof an element is equal to 0. The ∆G\textdegree of the reaction C_2H_5OH + O_2 \rightarrow CH_3COOH + H_2O can be found by taking the sum of the ∆G\textdegree_f's of the products minus the sum of the ∆G\textdegree_f's of the reactants, where products and reactants are each multiplied by their molar amount, indicated by the coefficients in the chemical equation. In other words, ∆G_reaction= \sum∆G_formationof products- \sum∆G_formationof reactants For the reaction in this problem, ∆G_reaction= [(∆G\textdegree_f_ CH(3)COOH + ∆G\textdegree_fH(2)O) - (∆G\textdegree_fC(2)H(5)OH+ ∆G\textdegree_fO(2))] ∆G\textdegree_fO(2)is zero, because O_2 is an element in its standard state. Substituting the ∆G\textdegree_f's given and solving one obtains: ∆G = [(- 95.38 - 56.69) - (- 43.39 + 0)] = - 108.7 Kcal/mole.

Question:

Sea shells are mostly calcium carbonate. When heated, calcium carbonate decomposes into calcium oxide (lime) ecomposes into calcium oxide (lime) and carbon dioxide, (a) Write the balanced equation for the reaction, (b) Calculate the number of kilograms of calcium carbonate needed to produce 10 kg of lime.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E27-0897.htm

Solution:

Calcium carbonate decomposes as follows: CaCO_3 \ding{217} CaO + CO2 From the stoichiometry of this equation, 1 mole of CaCO_3 into 1 mole of CaO and 1 mole of CO_2. If the number of moles of lime is known, the number of moles of CaCO_3 can be determined. The number of moles of lime is its weight divided by its molecular weight (56 g / mole) or number of moles of lime = [(10,000 g) / (56 g / mole)] = 179 moles. Since 179 moles of lime are produced, 179 moles of CaCO_3 are consumed. The weight of 179 moles of CaCO_3 is the weight of one mole of CaCO_3 (or the molecular weight) . Thus, the weight of CaCO_3 = (179 moles) (100 g / mole) = 17,900 g or 17.9 kg of CaCO_3 consumed.

Question:

Compare the functions of the seminal vesicles and the seminal receptacles in the earthworm. m.

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/Users/wenhuchen/Documents/Crawler/Biology/F12-0297.htm

Solution:

The seminal vesicles are glands that secrete a thick fluid which nourishes the sperm cells. The mix-ture consisting of sperm cells and fluid from the seminal vesicles is called semen. In the earthworm, there are three pairs of seminal vesicles. The testes, which produce the sperm, are housed within the seminal vesicles. There are two pairs of testes within the three pairs of seminal vesicles. Sperm are discharged from the testes into the vesicles, where they are stored and nourished. During copulation, the sperm enter the funnel-shaped openings of the vas deferens, and pass to the outside. Forming a part of the female reproductive system, but completely separated from the female oviducts, are the seminal receptacles. These storage chambers are simple pairs of sacs, opening onto the ventral surface of the segment containing them. The number of seminal receptacles ranges from one to five pairs, each pair located in a separate segment. During copulation, sperm are deposited in the receptacles and stored for later use. It should be noted that an earthworm contains both seminal It should be noted that an earthworm contains both seminal vesicles and receptacles, therefore it is called a hermaphroditic organism. However, they do not self- fertilize.

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Question:

In Figure 1 assume that B = 2.0 T,1 = 10cm, and v = 1 .0m /s. Calculate (a) the induced electric field observed by S' and (b) the emf induced in the loop.

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/Users/wenhuchen/Documents/Crawler/Physics/D37-1078.htm

Solution:

(a) The electric field, which is apparent only to observer S', is associated with the moving magnet and is given in magnitude by E = vB = (1.0 m/s)(2.0 T) = 2.0 (V/m). (b) Observer S would calculate the induced (motional) emf from \epsilon = Blv = (2.0 T)(1.0 × 10^-1m)(1.0 m/s) = 0.20 V. Observer S' would not regard the emf as motional and would use the relationship \epsilon = El = (2.0 V/m) (1.0 x 10^-1m) = 0.20 V. As must be the case, both observers agree as to the numerical value of the emf.

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Question:

What is the angular momentum of a long thin rod that rotates as shown in the figure if it makes three revolutions per second? Assume that the mass of the rod is 2 kg and its length is 6 m.

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/Users/wenhuchen/Documents/Crawler/Physics/D06-0343.htm

Solution:

The angular momentum of an object about a fixed axis is L = I\cyrchar\cyromega where I is the moment of inertia of the object about that axis and \cyrchar\cyromega is the angular velocity of the object. To calculate L for a long thin rod, we must first find I. By definition, I = \int r^2dm(1) where dm is a mass element of the rod, and r is the distance of dm from the axis of rotation of the rod, z. The integral is evaluated over the total mass of the rod. If \textphi is the mass density of the rod, then by definition \textphi = dm/dv anddm = \textphi dv where dv is a volume element of the rod. If the rod is very thin, we may consider it to be one dimensional. Hence, using cylindrical coordinates (r,\texttheta, z), dv = dranddm = \textphi dr. Using (1)I =^l/2\int- l/2\textphi r^2 dr Where l is the length of the rod. I = (\textphir^3)/3^l/2\vert- l/2= (\textphil^3)/12. But \textphi = M/ l I = (Ml^2)/(12) ThereforeL = [(Ml^2)/(12)]\cyrchar\cyromega Using the given data L = [(2kg) (36 m^2) (3 rev/sec)] / [12] = (18 m^2\bulletkg\bulletrev)/(sec) But(1 rev)/(sec) = 2\pi [(rads)/(sec)] L = (2\pi)(18) [ {m^2\bulletkg\bulletrad} / sec] L = 113.04 [ {m^2\bulletkg\bulletrad} / sec] L = 1.13 × 10^2[(kg \bulletm)/(sec^2)]m \bullet sec \bullet rad L = 1.13 × 10^2 Joules \bullet sec.

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Question:

A suitcase is dragged 30 m along a floor by a force F = 10 newtons inclined at an angle 30\textdegree to the floor. How much work is done on the suitcase?

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/Users/wenhuchen/Documents/Crawler/Physics/D05-0255.htm

Solution:

Work is defined as the scalar product of the force acting on an object, and the distance through which the object moves while the force is being applied. W = F^\ding{217} \textbullet d^\ding{217} where F^\ding{217} is the force and d^\ding{217} is the distance. (See the figure above.) Note that the force and distance are vectors while work is a scalar, hence the "scalar product" nomenclature for the dot. W = F^\ding{217} \textbullet d^\ding{217} = Fdcos \texttheta = Fd cos 30\textdegree = (10N) (30 m)[\surd3/2] = 150 \surd3 N\bulletm

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Question:

Classify each of the following as a member of the methane series, the ethylene series, or the acetylene series: C_12H_26, C_9H_16, C7H_14 , C_26H_54.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E20-0732.htm

Solution:

Before beginning this problem, one should first know the general formulas for each of the series. For thealkanes(methane series), the general formula is C_nH_2n+2' where n is the number of carbon atoms and 2n + 2 is the number ofhydrogens. Molecules of the ethylene series, also called thealkeneseries, have two adjacent carbon atoms joined to one another by a double bond. Any member in this series has the general formula C_nH_2n. The acetylene series, commonly called thealkyneseries, has two adjacent carbon atoms joined to one another by a triple bond. The general formula for this series is C_nH_2n-2. with this in mind one can write: C_12H_26:C_nH2n+2:alkaneseries C_9H_16:C_nH_2n-2: acetylene series C_7H14:C_nH2n: ethylene series C_26H_54:C_nH2n+2:alkaneseries

Question:

Of the total number of atoms in the universe, approximately 93 % are hydrogen (H, atomic weight = 1.0 g/mole) and 7 % are helium (He, atomic weight = 4.0 g/mole). What percentage of the universe by weight is hydrogen?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E04-0155.htm

Solution:

The percentage by weight is equal to the total weight of H atoms divided by the total weight of the universe. Suppose that there are N atoms in the universe. Then the number of hydrogen atoms is 93 % × N = 0.93 N and the number of helium atoms is 7 % × N = 0.07 N. The mass of hydrogen atoms is mass H = number H × mass H = 0.93 N × 1.0 g/mole and the mass of helium atoms is mass He = number He × mass He = 0.07 N × 4.0 g/mole. Then, on a per mole basis, % g/mole = {(g/mole of H) / (total g/mole) } × 100% = {(0.93N × 1 g/mole) / (0.93 N × 1.0g/mole + 0.07 N × 4.0g/mole)} ×100% On a weight basis, % H by weight ={(0.93N × 1 g/mole ) / (0.93 N × 1.0g/mole + 0.07 N × 4.0g/mole)} × (1mole / 1mole) ×100% = {(0.93 N g) / (0.93N g + 0.28N g )} × 100% = (0.93 N g / 1.21N g) × 100% = (0.93 g / 1.21g ) × 100% = 77% .

Question:

If the vapor pressure of ethyl alcohol, C_2H_5OH, is 0.132 atm at 34.9\textdegreeC, and 0.526 atm at 63.5\textdegreeC what do you predict it will be at 19.0\textdegreeC?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E07-0261.htm

Solution:

Equilibrium vapor pressure is the pressure exerted by a vapor when the vapor is in equilibrium with its liquid. The magnitude of the equilibrium vapor pressure depends (1) on the nature of the liquid and (2) on its temperature. The vapor pressure is related to temperature by the following equation log P = [(-∆H) / (19.15 T)] + c where p is the vapor pressure, T is the absolute temperature, ∆H is the heat required to transform one mole of liquid to the ideal gas state, and C is a constant dependent on the liquid and units used for expressing pressure. One is given P and T for two trials, thus AH for ethyl alcohol can be found. Forp = 0.132 atm T = 34.9\textdegreeC + 273 = 307.9\textdegreeK a)log 0.132 = [(-∆H) / {(19.15)(307.9\textdegreeK)}] + C For p = 0.526 atm T = 63.5\textdegreeC + 273 = 336.5\textdegreeK log p = -1.263 P = .0545 atm.

Question:

Discuss the etiology, epidemiology and prophylaxis of tuberculosis .

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/Users/wenhuchen/Documents/Crawler/Biology/F05-0159.htm

Solution:

It is first necessary to define the terms used in the question before one can answer it. Etiology refers to the cause of a disease. Epidemiology deals with the cause and control of epidemics. An epidemic is the unusual prevalence or sudden appearance of a disease in a community. Prophylaxis is the preventive treatment used to protect against disease. Many of the important infections of man are air-borne and cause diseases of the respiratory tract. Tuberculosis is an endemic respiratory disease ; it is peculiar to a locality or a people. Although evidence of itsexistence has been found in Egyptian mummies 3000 years old, it is still a leadingcause of death. In the nineteenth century, Robert Koch isolated the causative agent of human tuberculosis -Mycobacteriumtuberculosis. Koch proved that pure cultures of these bacilli would produce the infection in experimental animals, and he later recovered the bacilli form these animals . There are several types of tubercle bacilli (the common name): the human, bovine, avian and other varieties. One strain is almost exclusively a human parasite, and is responsible for over ninety percent of all cases of tuberculosis. The less common bovine strain can cause tuberculosis in man through ingestion of in-fected beef or milk from an infected cow. It is characteristic of tuberculosis and other res-piratory diseases to occur in epidemic form, attacking many people within a short time. Their incidence usually increases during fall and winter, when many people frequently remain indoors. The tubercle bacillus is transmitted by association with infected people through the secretions of the nose and throat, which is spread in droplets by coughing and sneezing. It may be transmitted by articles which have been used by an infected person, such as eating and drink-ing utensils and handkerchiefs. The lungs are the most commonly affected tissue , often exhibiting tubercles or small nodules. These are areas of destroyed lung tissue in which the tubercle bacilli grow. Symptoms of tuber-culosis include pleurisy (inflamationof the pleural mem-brane which line the chest and cover the lungs), chest pains, coughing, fatigue and loss of weight. Treatment consists of bed rest, nourishing diet, and sometimes chemotherapy. The most effective drug, INH (isonicotinicacid hydrazide ) is used both for chemotherapy andchemopropylaxis. For actual prevention, the drug is administered before infection to highly susceptible individuals. High death rates in developing countries are due to substandard housing , overcrowding, and malnutrition. Prophylaxis for a community is best achieved when living and working conditions are not overcrowded, diet is ade-quate, and proper sanitation is available.

Question:

Analyze the following flowchart and ask yourself these questions: a) Is the flowchart a structured one? b) Can you write pseudocode to structure it? c) What difficulties or inconveniences could arise when coding?

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Solution:

a) This flowchart is somewhat fictitious, and it does not satisfy the one-entry one-exit rule of modules in structured program-ming. We will assume that this flowchart represents a program segment, since we are not shown where the four nodes (written as circles enclosing numbers) lead. The problem lies in the module for tasks A-F. It has two points of entry, an unstructured means of transferring program control. b) Recalling the three logical structures of programming -- sequencing, decision-making, and looping \textemdash you should recognize that the flow-chart can be rendered into proper pseudocode easily. We need the IF-THEN-ELSE construct, along with some kind of nesting of decisions. The pseudocode is rather straightforward: if (P) then if (Q) then tasks G-N else tasks A-F else if (R) then tasks O-Z else tasks A-F end if then else c) Notice that tasks A through F are referenced twice in the program. If these tasks take up many lines of instructions, it would be tedious to write out all of these instructions twice. As an alternative, you may want to call a subroutine to perform these tasks. By doing this, you need write out tasks A through F only once.

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Question:

A photon rocket is being propelled through space. The propellantconsists of photons. The mass of the rocket is not constantthroughout the motion, for it continual-ly loses mass inthe form of photons. Use the principles of conservation of momentumand conservation of energy to show that only about5 per cent of the mass of the photon rocket remains afterthe rocket has reached a speed of 99.5 per cent of the speedof light.

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Solution:

Beforetakeoffthetotal momentum of the photon-rocket system is zerofor the rocket is at rest. Since no external force F acts on the system, then F = 0 =dP/ dt whereP is the total momentum of the system. Hence P remains constant intime. Consequently, after attaining final speed, the total momentum muststill be zero; therefore, the momentum of the photons must equal the finalspaceship momentum. (These momentum vectors are oppositely directedto give zero total momentum.) When the rocket has reached its finalspeed, its rest mass will be a fraction f of the original total rest mass m_0. The fraction (1 - f) of the original rest mass has been ejected in the form ofphotons. totalejected photon momentum = p finalrocket momentum = {fm_0(v)} / \surd{1 - (v^2 / c^2) = \Upsilon(fm_0)v where \Upsilon = 1 /\surd{1 - (v^2 / c^2)is the relativtistic correction of the restmass due to the rocket velocity. Therefore, by conservation of momentum, p = \Upsilonfm_0 Since the rocket speed is 99.5 per cent of the speed of light, v may be replacedby c with negligible error. p = \Upsilonfm_0c The total mass-energy of the rocket system must remain constant intime, for no energy enters or leaves the closed ejected photons and rocketsystem. Then totalinitial energy = m_0c^2 finalrocket energy = \Upsilon(fm_0)c^2 totalejected photon energy = pc sinceinitial energy = final energy m_0c^2 = \Upsilonfm_0c^2 + pc Substituting for p from the previous equation gives m_0c^2 = \Upsilonfm_0c^2 + \Upsilonfm_0c^2 = 2\Upsilonfm_0C^2 Solving for f, f = {1 / (2\Upsilon)} At a speed of 99.5 per cent c, the value of \Upsilon is 1 / \surd {1- (0.995c /c )^2} = 1.0 × 10^1 Therefore, f = 1 / {(2)(1 × 10^1)} = 0.05 = 5 per cent Therefore, only 5 per cent of the original mass of the rocket remains when the rocket has achieved a speed of 99.5 per cent c.

Question:

A household cleaning solution has a hydronium ion concentration of 10^-11 M. What is the pH of the solution? Is the solution acidic or basic?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E26-0878.htm

Solution:

The pH of any solution by definition is determined by using the following equation. pH = - log [H^+] Thus, if hydronium concentration = 10^-11 M, pH = - log 10^-11 or pH = 11. The solution is basic because it has a pH greater than 7.

Question:

Explain how the codon for phenylalanine was discovered.

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/Users/wenhuchen/Documents/Crawler/Biology/F24-0618.htm

Solution:

It has been known for some time now that the genetic code is a triplet code - three bases in a messenger RNA molecule code for one amino acid. Each triplet is known as a codon and the entire set of codons comprising the RNA molecule constitute the genetic code for a polypeptide chain. The translation of the genetic code resulted from recent research the aim of which was to determine the exact sequence of the bases on the mRNA molecule which specifies a particular amino acid. An important breakthrough in this research came when a method of artificially synthesizing mRNA was developed. Ribose sugar, phosphate ions, and various bases, when combined under the proper conditions, result in the formation of a synthetic RNA molecule. By varying the kinds of bases used in producing this RNA, it has been possible to determine which base sequences result in which amino acids. The synthetic mRNA is placed with ribosomes, amino acids, ATP, and tRNA. The resulting polypeptide chain is analyzed for its component amino acids. The codon for the amino acid phenylalanine was the first to be discovered using the above procedure. The only base used in the synthesis of the mRNA, was uracil, and therefore the molecule had a series of UUU codons. This mRNA was added to ribosomes, amino acids, ATP and tRNA, and the resulting polypeptide consisted only of phenylalanine. Thus, it was deduced that the codon UUU is the sequence of bases which causes phenylalanine to be placed in the polypeptide chain.

Question:

What is the height of the Coulomb barrier for an \alpha particle and a Pb^206 nucleus? Use nuclear radius R \cong 1.4 A^1/3 × 10\Elzbar^13cm where A is the mass number.

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/Users/wenhuchen/Documents/Crawler/Physics/D34-1019.htm

Solution:

Within the nucleus there is an attractive nuclear force which is short range. This force is much larger than the repulsive Coulomb force between the protons, but it is effective only within the nuclear radius R. Therefore \alpha particles within a nucleus are in a potential well consisting of a nuclear potential (negative since the force is attractive) for radius r < R and a Coulomb potential (positive since the force is repulsive) for r > R. This is shown in the diagram. For the alpha particle to escape the nucleus, it must acquire an energy greater than the Coulomb potential barrier \epsilon_C. The maximum repulsion between two nuclei occur when they are just touching. If they were any closer, the nuclear force would overcome the Coulomb repulsion force and the nuclei would attract and fall into each other. The maximum Coulomb potential \epsilon_Coccurs when the nuclei repulsion is at a maximum. Therefore, the Coulomb barrier \epsilon_Cis just the electro-static potential energy between the two nuclei when the distance between their centers is equal to the sum of their radii (that is, when they are just in "contact"). R_Pb \cong 1.4 × (206)^1/3 × 10\Elzbar13cmR_\alpha \cong 1.4 ×(4)^1/3 × 10\Elzbar13cm \cong 1.4 × 5.91 ×10\Elzbar13cm\cong 1.4 × 1.59 ×10\Elzbar13cm \cong 8.2 ×10\Elzbar13cm\cong 2.2 ×10\Elzbar13cm Electrostatic potential energy is PE = (q_1 q_2 ) / r where q_1 and q_2 are the charges and r the-distance between them. The charge on each of the nuclei is equal to the number of protons Z they contain, multiplied by the charge e on a proton. Therefore \epsilon_C = [{(Ze)_Pb × (Ze)_\alpha } / (R_Pb + R_\alpha ) ] \cong [(82e × 2e) / {(8.2 ×10\Elzbar13cm) + (2.2 ×10\Elzbar13cm)}] \cong [(164e^2 ) / (10.4 × 10\Elzbar13cm) ] Now, the square of the electronic charge can be expressed as e^2 = 1.44 × 10\Elzbar13MeV-cm. Thus, \epsilon_C \cong [{164 × (1.44 ×10\Elzbar13MeV-cm)} / (10.4 ×10\Elzbar13cm)] \cong 22. 7 MeV This energy is large compared to the kinetic energy of \alpha particles emitted from radioactive nuclei.

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Question:

A pendulum with a bob of mass M is raised to height H and released. At the bottom of its swing, it picks up a piece of putty whose mass is m. To what height h will the combination (M + m) rise?

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Solution:

There are three phases to the problem\textemdashthe fall of M, the collision of M and m, and the rise of M + m. The first and last phases involve energy conservation and the second phase involves momentum conservation. (1) Fall:(PE)_initial = (KE)_final MgH = (1/2) Mv^2 from which v = \surd(2gH) (2) Collision: P_initial = P_final Mv = (M + m)v' (3) Rise:(KE)_initial = (PE)_final (1/2)(M + m)v'^2 = (M + m)gh from which v' = \surd(2gh) Substituting Eqs. 1 and 3 into Eq. 2 gives M\surd(2gH) = (M + m) \surd(2gh) Canceling \surd2g and squaring, we have M^2H = (m + M)^2h so that the final height is h = [M/(m + M)]^2 H

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Question:

Discuss how the carbons of a glucose molecule may be converted to the carbons of a fat molecule.

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Solution:

A fat molecule is actually a molecule of triacylglycerol (once called triglyceride), a compound formed when the three hydroxyl groups of glycerol are bonded to three fatty acids. A reaction of triacylgly-cerol synthesis can be written as follows: R represents the hydrocarbon chain of the fatty acid, and will usually vary in its length and number of double or triple bonds. Thus, it is the fatty acid chain of fats which determine the nature of the fat. Triacylglycerols that are solid at room temperature are called "fats" and those which are liquid are called "oils." Glucose is degraded in glycolysis to fructose -1,6- diphosphate, which is split into two three-carbon sugars, glyceraldehyde -3-phosphate (G-3-P) and dihydroxyacetone phosphate (DHAP). DHAP is reduced to \alpha-glycerophosphate (glycerol-3-phosphate) with oxidation of NADH. \alpha- glycer-ophosphate is dephosphorylated to glycerol with concomit-ant production of ATP. The synthesis of fatty acids requires molecules of acetyl coenzyme A as precursors of all the carbon atoms in the fatty acid chain. Chain growth during fatty acid synthesis starts at the carboxyl (-COOH) group of acetyl -CoA and proceeds by successive addition of acetyl residues at the carboxyl end of the growing chain. Acetyl -CoA is formed from oxidation of pyruvate, the breakdown product of glucose in glycolysis. Thus, both the glycerol and fatty acid carbon atoms of a fat molecule can originate from those of glucose molecules. Although this synthesis of fats from glycerol and fatty acids can occur in vitro (Latin, in glass), a different pathway is used in vivo (Latin, in the living). All naturally occurring triacylgly-cerols are formed by reaction of dihydroxyacetone phosphate and fatty acyl coenzyme A (a fatty acid "activated" by coenzyme A). Both these reactants can originate from glucose, as explained earlier.

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Question:

Methane is burnt in oxygen to produce carbon dioxide and water vapor according to the following reaction, CH_4 (g) + 2O_2 (g) \rightarrow CO_2 (g) + 2H_2O (g). The final pressure of the gaseous products was 6.25torr. Calculate the partial pressure of the water vapor.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E03-0080.htm

Solution:

The partial pressure of each of the components in a mixture of gaseous substances will be equal to the product of the mole fraction of that component in the gas and the total pressure of the system. Given the total vapor pressure of the gases, one can calculate the mole fraction of the water vapor. Let P_A = partial pressure of the water vapor, N_T = total number of moles of gases produced, N_A = number of moles of water vapor, and P_T = total pressure. This law can now be expressed in these terms: P_A = [N_A / N_T] P_T . From thestoichiometryof the equation, the total number of moles of products is 3. Out of these 2 moles are water vapor. Thus, substituting, P_A = 2/3 (6.25torr) = 4.16torrof water vapor.

Question:

A block of mass m slides down a plane inclined at an angle \texttheta, the surface of which has coefficient of sliding friction \mu. The mass collides with a spring, having force constant k, after it has slid a distance d from rest. The spring will then reduce the speed of the block, which, will come to a momentary halt, when the spring has been compressed a distance X. The block will then be pushed back up the incline and begin a frictionally damped harmonic oscillation. Calculate X.

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Solution:

We will use the law of energy conservation. The kinetic energy of the block at the moment it collides with the spring is equal to the sum of its loss of gravitational potential energy and the work done on it by friction. Since the frictional force, f, acts in a direction opposite to the motion, the work due to friction is negative. Then (1/2) mv^2 = mgh - \muNd where f = \muN, and h = d sin \texttheta is the vertical height through which the block falls, (see figure). N = mg cos \texttheta is the force the plane exerts, normal to its surface, on the block to cancel the component of the block's weight perpendicular to the plane. Thus: (1/2) mv^2 = mg d sin \texttheta - \mu mg d cos \texttheta = mg d(sin \texttheta - \mu cos \texttheta) Once the spring has brought the block to rest, the spring's potential energy is equal to the sum of the block's kinetic energy, the further gravitational po-tential energy the block loses as it falls through height h' = X sin \texttheta, and the additional (negative) work done by friction on the block (see figure). (1/2) kX^2 = (1/2) mv^2 + mgh'-\mu NX (1/2) kX^2 = mgd(sin \texttheta - \mu cos \texttheta) + (mg sin \texttheta)X - (\mu mg cos \texttheta)X (1/2) kX^2 - mg (sin \texttheta - \mu cos \texttheta)X - mgd (sin \texttheta - \mu cos \texttheta) = 0 From the quadratic formula: x = [- b \pm \surd(b^2 - 4ac)] / 2awith a = (1/2) k, b = - mg (sin \texttheta - \mu cos \texttheta), and c = -mg d (sin \texttheta - \mu cos \texttheta) X = [mg(sin \texttheta - \mu cos \texttheta) \pm \surd{m^2g^2 (sin \texttheta - \mu cos \texttheta)^2 - 4(1/2 k)}] / [2{(1/2)k}] [-mgd (sin \texttheta - \mu cos \texttheta)] / [denominator] X = [mg(sin \texttheta - \mu cos \texttheta) \pm \surd{m^2g^2 (sin \texttheta - \mu cos \texttheta)^2 + 2mgdK(sin \texttheta - \mu cos \texttheta)}] / (k) This yields two answers for X. If we had been given numerical values for m, d, \mu, \texttheta, and k, we would find that one of the solutions for X would be negative and unacceptable. The other solution would be correct for the problem at hand.

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Question:

Solid ammonium nitrate,amitol, is much used as an explosive; it requires a detonator explosive to cause its decomposition to hot gases. NH_4NO_3 \ding{217} N_2O + 2H_2O Assume 900 g NH_4NC_3 almost instantaneously decomposed. What total volume of gas in liters forms at 1092\textdegreeC?

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Solution:

From thestoichiometryof the equation, 1 mole of NH_4NO3produces a total of 3 moles of products (1 mole of N_2O + 2 moles of H_2O). The number of moles of NH_4NO_3 900 g (MW = 80 g/mole) is [900 g / (80 g / mole)] = 11.25 moles of NH_4NO_3. There are, thus, a total of 3 × 11.25 moles = 33.75 moles of products. Using the relationship that 1 mole of any ideal gas has a volume of 22.4 l at STP (Standard Temperature and Pressure), the total volume of gaseous products is (22.4 l / mole) (33.75moles) = 756 l at STP. To obtain the volume at 1092\textdegreeC, Charles' Law is used (V1/ V_2) = (T_1 / T_2) where V_1 = 756 l, T_1 = 273\textdegreeK (the STP temperature), and T_2 = 1092\textdegree + 273\textdegree = 1365\textdegreeK. Solving for V_2, V_2 = (V_1T_2 )/ T_1 = [{(756 l) (1365oK)} / (273oK)] = 3780 l.

Question:

An interferometer illuminated with red light from cadmium (\lambda = 6438 A) is used to measure the distance between two points. Calculate this distance, D, if 120 minima pass the reference mark as the mirror is moved from one of thepoints to the other.

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Solution:

The fringes observed represent the interference of light rays. Consider 2 rays of light, initially in phase. (That is, they interfere to produce areas of high light intensity, or maxima). If the phase of one ray is varied until a minima is observed, we will find that the phase difference of the 2 rays is now\lambda/2 where \lambda is the wavelength of the light. Hence, 2 minima are separated by \lambda. Therefore, D =N\lambda= 120(6.438 × 10^-5 cm) = 0.00773 cm.

Question:

A boy is sledding on a snowy slope and looks very weary as he drags his sled up again after each run down. A helpful physics student who is passing by, and who knows that the coefficient of kinetic friction between a sled and snow is around 0.10, points out to the boy that he is exerting pull on the tow rope at an incorrect angle to the ground for minimum effort. At what angle to the slope should the pull be exerted?

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Solution:

There are four forces acting on the sled, as shown in the diagram; the weight W\ding{217}, the normal force N\ding{217} exerted by the slope, the frictional force F_f\ding{217} exerted by the snow down the slope opposing the motion, and the upward pull P\ding{217} exerted by the boy at an angle \texttheta to the slope. The sled is moving with uniform velocity up the hill and thus the forces are in equilibrium, and F_f = \mu_kN, where \mu_k is the coefficient of kinetic friction. Let us now resolve all the forces into components parallel and perpendicular to the surface of the slope. We will take forces going up the slope as positive in the parallel direction, and forces pointing upward out of the slope as positive in the perpendicular direction. Imposing the conditions of equilibrium, we have: \sumF_\parallel = P cos \texttheta - W sin \alpha - \mu_kN = 0(1) \sumF_\perp = N + P sin \texttheta - W cos \alpha = 0(2) where \alpha is the angle of the slope's incline. We can thus solve for P. From equation (2) N = W cos \alpha - P sin \texttheta From equation (1) P cos \texttheta = W sin \alpha + \mu_kN = W sin \alpha + \mu_k(W cos \alpha - P sin \texttheta) P(cos \texttheta + \mu_k sin \texttheta) = W sin \alpha + \mu_kW cos \alpha P = (W sin \alpha+ \mu_kW cos \alpha)/(cos \texttheta + \mu_k sin \texttheta) We see now that P is a function of \texttheta alone since W, \alpha, and \mu_k are constants. Thus, to find the angle \texttheta at which the boy must exert the minimum force, we use the calculus to find the value of \texttheta at which a minimum occurs for P. These minima occur when the derivative of P with respect to \texttheta is zero. Since the numerator is constant with respect to \texttheta: (dP/d\texttheta) = - [(W sin \alpha + \mu_kW cos \alpha)/{(cos \texttheta + \mu_k sin \texttheta)^2}](- sin \texttheta + \mu_k cos \texttheta) = 0 orsin \texttheta = \mu_k cos \texttheta tan \texttheta = \mu_k = 0.10 \texttheta = 5.7\textdegree Thus the boy should drag his sled up in such a way that the rope makes an angle of 5.7\textdegree with the slope. It might have seemed at first sight that a force parallel to the slope would be most efficient. It is now clear that this is not so. Any component of the pull P\ding{217} at right angles to the slope decreases the normal force N\ding{217} and thus the frictional force F\ding{217}. The best compromise between maximum forward force and least frictional force is achieved at the angle 5.7\textdegree.

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Question:

Modify the automobile suspension program of parti) of the previous problem so that it is possible to find a value for D - the damping coef-ficient, that will not allow the displacement of the mass x(t) to exceed 1.6, given K = 400.0, M = 2.0, E(t) = h = 1.0. Start with D = 6.0 and increase it by one unit whenever it is found that x(t) > 1.6, until the desired damping is found.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G12-0308.htm

Solution:

The modification is made when the value of XCOR is found with desired accuracy. We test to see if XCOR \leq 1.6. If so, the program proceeds as in parti) of the previous problem. If not, D is increased by 1 and variables are reinitialized so that the algorithm may begin again using the new value of D. The program follows: REAL K/400.0/,M/2.0/,N,T/0.0/,XCOR/0.0/, VCOR/0.0/, H/1. 0/, D/ 6.0/ F(Y,Z) = (H - D{_\ast}Y - K{_\ast}Z)/M READ N, TFIN, ACCUR DT = (TFIN - T)/N 50 50X = XCOR V = VC0R A = F(V,X) PRINT T,X,D T = T + DT IF (T.GT.TFIN) STOP XPRED = X + V{_\ast}DT VPRED = V + A{_\ast}DT 100XDOT = VPRED VDOT = F(V,XPRED) XCOR = X + 0.5{_\ast}(V + XDOT){_\ast}DT VCOR = V + 0.5{_\ast}(A + VDOT){_\ast}DT XDIF = ABS(XCOR - XPRED) VDIF = ABS(VCOR - VPRED) IF (XDIF .LE.ACCUR.AND.VDIF.LE.ACCUR) GO TO 110 XPRED = XCOR VPRED = VC0R GO TO 100 110IF (XCOR.LE.1.6) GO TO 50 D = D + 1.0 T = 0.0 XCOR = 0.0 VCOR =0.0 GO TO 50 END

Question:

A chemist has a saturated solution of CaSO_4 in .0100M Na_2 SO_4 . The K_sp of CaSO_4is 2.4 × 10-^5 . calculate the concentration of. Ca^2+ ion in this saturated solution.

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Solution:

The K_sp is the solubility constant of a substance. For the general reaction, AB_(s)\rightarrow A+ + B- ,K_sp = [A+] [B-]. The K_sp = the product of the concentrations of the ions. In this problem, K_sp = [Ca++] [SO_4-] . You are given that K_sp = 2.4 × 10-^5 . Equating, you obtain 2.4 × 10-^5 = [Ca++] [SO_4-]. To solve this problem, you must determine the value of [Ca++]. Let x = the moles/liter of CaSO_4 . Since CaSO_4 \rightarrow Ca++ + SO_4- , this x also is the concentra tion of Ca++ . The concentration of SO_4- , however, has two sources. SO_4- comes from CaSO_4 AND NaSO_4 . The concentration of Na_2 SO_4 was given as .01. Therefore, the total concentration of SO_4- is x + .01, where x represents the amount of SO_4 contributed by CaSO_4 . Thus, [x] [.01 + x] = 2.4 × 10-^5 . Solving for x you obtain, x = 2.0 × 10-^3 = [Ca++] .

Question:

Contrast the major features of cartilaginous and bony fish.

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Solution:

The cartilaginous fish of the class Chondrichthyes are distinguished by, as their name implies, their cartilaginous skeletons. No bone is present in this group. The presence of a cartilaginous skeleton in this class must be regarded. as a specializa-tion because there is evidence that the ancestors of the group, the placoderms, had bony skeletons. Included in the class Chondrichthyes are the sharks, skates, and rays. The sharks are streamlined, fast-swimming, voracious feeders. One kind, the whale shark, is the largest known fish. It may attain a length of 50 feet or more. Rays are sluggish, flattened bottom feeders. Chondrichthyes have neither swim bladders nor lungs. Osmoregulation in some groups is unusual, in-volving retention of high concentrations of urea in the body fluids. Fertilization is internal, and the eggs have tough, leathery shells. Most species are predatory but a few are plankton feeders. Like the cartilaginous fish, the bony fish, belonging to the class Osteichthyes, arose from the placoderms. In most of them, the adult skeleton is composed largely of bone. Unlike cartilaginous fish, fertilization is external in most bony fish. The gills of bony fish are covered by a hinged bony plate, the operculum, on each side of the phcirynx. In most bony and cartilaginous fish the body is covered with scales. Unlike the cartilaginous fish, most bony fish have a swim bladder, which arose as a modification of the lungs of the earliest Osteichthyes. There are still some bony fish today that possess lungs instead of swim bladders. By modifying the gaseous content of the swim bladder, the fish can modify its buoyancy. Cartilaginous fish have no swim bladder and are denser than the sea water around them. They would sink to the bottom if it were not for the movements of the pelvic and tail fins which serve to maintain buoyancy in these fish. The early Osteichthyes gave rise to three main lines of bony fishes: the lobe-finned fish (Order Crossopterygii), the lungfish (Order Dipnoi), and the telecosts (Order Tele- ostei) which form the dominant group of marine and aquatic animals today). The lungfish and lobe-finned fish have functional lungs as well as gills. Teleosts posses a swim bladder, but no lungs.

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Question:

Define and draw a contrast between compilers and interpreters .

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Solution:

A compiler is a translator program that converts the entire program into the machine language equivalent. Once translated, the program can be run over and over unless logi-cally modified. Compilers treat the program written in a high- level language as data, and may use several passes (processing of the entire program) to produce the correct and efficient machine language equivalent. Therefore, if frequent modifica-tions of the program source code is required, the compiler is not very efficient. An interpreter is a translator program that interprets the statements one by one (on the fly so to speak) and directs the computer to do what the high-level statement intends. The trans-lation takes place every time the program is run. The interpret-ers slow down the execution of a program , but tend to be highly interactive, where frequent modifications and reruns of a source language program is easy. Thus, during the development phase that we can regard as the experimental phase, it is best to interpret a high-level language (HLL) program. After the program is thoroughly tested and debugged, it may then be compiled into an object code ready to be executed. Note that some languages, such as C, allow both interpreta-tion and compilation . In some cases, it is possible to have both an interpreter and a compiler for a language such as BASIC or Lisp on the same computer system . Therefore, whether a high level language program should be interpreted or compiled becomes a matter of convenience.

Question:

What are the difficulties in the classification of tissuesof higherplants?

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Solution:

The characteristics of plant cells themselves make it difficult for botaniststo agree on any one classi-fication system. The different types of cellsintergradeand a given cell can change from one type to another duringthe course of its life. As a result, the tissues formed from such cells alsointergradeand can share functional and structural characteristics. Some plant tissues con-tain cells of one type while others consist of a varietyof cell types. Plant tissues cannot be fully characterized or distinguishedon the basis of any one single factor such as location, function, structure or evolutionary heritage. Plant tissues can be divided intotwo major categories:meristematictissues, which are composed of immaturecells and are regions of active cell division; and permanent tissues, which are composed of more mature, differentiated cells. Permanent tissues can be subdivided into three classes of tissues - surface, fundamental and vascular. However, the classification of plant tissuesin-to categories based purely on their maturity runs into some difficulties. Some permanent tissues may change tomeristematicactivity undercertain conditions. There-fore, this classification is not absolutely reliable.

Question:

Determine the molecular weight of a gas if 4.50 g of it occupies 4.0 liters at 950 torr Hg and 182\textdegreeC. R = .082 liter - atm/mole - \textdegreeK.

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Solution:

One can find the molecular weight of this gas, once one knows the number of moles present in 4.50 g. The molecular weight is equal to 4.50g divided by the number of moles present MW = [(4.50 g)/(no. of moles)] The number of moles of the gas can be found by using the Ideal Gas Law. This law is stated PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant (0.082 liter-atm/mole-\textdegreeK) and T is the absolute temperature. One must convert the pressure in torr to atm to use the gas constant. There are 760 torr in 1 atm. Thus, torr are converted to atm by multiplying the number of torr by 1 atm/760 torr. P (in atm.) = 950 torr × 1 atm/760 torr = 1.25 atm. To use the Ideal Gas Law, the temperature must be in \textdegreeK. Here, it is given in \textdegreeC. Temperature in \textdegreeC can be converted to \textdegreeK by adding 273 to the temperature in \textdegreeC. T = 182 + 273 = 455\textdegreeK One can now use the Ideal Gas Law to solve for n. n = (PV/RT)P = 1.25 atm. V = 4.0 liters R = 0.082 liter-atm/mole-\textdegreeK T = 455\textdegreeK n = [(1.25 atm × 4.0 liters)/{0.082 (liter-atm / mole-\textdegreeK) × 455\textdegreeK}] = 0.13 moles In 4.5 g of this gas, there are 0.13 moles. The molecular weigth can now be found. MW = [(4.50 g)/0.13 moles)] = 34.7.

Question:

Most cells fall within the range of 5-20 microns. What factors keep the size of the cell within these boun-daries?

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Solution:

Normally, a cell cannot attain a size greater than 20 microns due to the limits imposed by the size inter-relationships of its components. For example, there is a relationship between the amount of nuclear material and the size of the cell; this limits cells to a size where there is an optimal proportion of nuclear material to the rest of the cell. A cell can have only as much material as its number of genes can handle. And since the nucleus controls the entire cell, there is a limit to how far away parts can be located from the nucleus. However, some larger cells have solved these problems by having more than one nucleus. Because materials must be transported throughout the cell and among the organelles, the relationship between the amount of cytoplasm and the amount of cellular organelles is also important in limiting cell size. If there is too much cytoplasm, transportation and communication between various areas of the cell would not be efficient. This transport problem is slightly minimized by processes such as cytoplasmic streaming. The effect of the amount of membrane surface area as compared to cell volume is the most important factor in limiting cell size. The intake of nutrients, the exchange of gases, and the excretion of wastes are all dependent on the surface area available for diffusion of these materials. How this is affected by increases in cell size can be seen through the following mathematical relationships: volume (sphere) = (4/3)\pir^3 (r = radius) surface area = 4\pir^2 proportionally: (volume / surface area) = [{(4/3)\pir^3 } / {4\pir^2}] = (1/3) r. It can be seen that the volume of the cell increases with the cube of the radius while the surface area in-creases only with the square of the radius. In other words, the surface area increases at a rate 1/3 r slower than the volume. Thus, if a cell's size increases, its surface-area may not increase proportionally enough to satisfy the needs of the larger cell. There is also a limit as to how small a cell can be. The minimum size can be predicted by determining how many macromolecules a cell must have in order for it to keep up its metabolism. Mathematical analysis of the cellular content of the PPL organism (the smallest known single- celled organism) will supply the answer. It is estimated that there are 1600 macromolecules (nucleic acids, proteins, etc.) in the PPL organism. Assuming that the cell must perform approximately 500 reactions in order to survive, we can determine that (1600 macromolecules / 500 reactions) are required on the average, or 3 to 4 macromolecules are required for each reaction. Taking into account the average size of the macromolecules necessary, the minimum size that any cell could be can be approximated.

Question:

At a particular time, N_0 atoms of a radioactive substance with a disintegration constant \lambda_1 are separated chemically from all other members of the radioactive series. At what time thereafter is the number of radioactive atoms of the daughter product, with disintegration constant \lambda_2 , a maximum?

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Solution:

The figure shows the situation described. Att = 0, we isolate N_0 atoms of a radioactive ma-terial. This is the parent substance. As time progresses the parent undergoes decay I, and becomes daughter product A. Similarly, daughter A decays into daughter B. Our problem is to find out at what time the number of atoms of daughter A is a maximum. Let us first look at decay I. It is observed experimentally that the rate of decay of any radio-active element is proportional to the number of atoms available for decay at a particular time t. Hence, [{dN(t)} / dt ] \alpha N (t) This can be written as an equation if we insert a constant of proportionality, \lambda. In the case of radio-active decay, \lambda is called the disintegration constant. Hence, [{dN(t)} / dt ] = -\lambdaN (t) where the minus sign indicates that the number of radioactive atoms of the parent element, N(t), is decreasing with time. Applying this equation to the decay of the parent to daughter product A, we obtain [{dN_1 (t)} / dt ] = - \lambda_1 N_1 (t)(1) Where \lambda_1 is the disintegration constant for decay I, and N_1(t) is the number of parent atoms at any time t . Now, daughter product A gains radioactive atoms from decay process I, but also loses atoms due to its own decay. Hence, the time rate of change of atoms of daughter product A is given by [{dN_2 (t)} / dt ] = -[{dN_1 (t)} / dt ] - {\lambda_2 N_2 (t)}(2) where \lambda_2 is the disintegration constant for decay II , and N_2 (t) is the number of atoms of daughter A at time t. (Note that we use - dN1(t)/dt in (2) be-cause the number of atoms gained by daughter A from decay I is greater than zero, but {dN_1 (t)}/dt < 0, from (1).) Using (1) in (2) [{dN_2 (t)} / dt ] = {\lambda_1N_1 (t)} - {\lambda_2 N_2 (t)}(3) Solving the differential equation in (1) will give us N_1 (t) [{dN_1 (t)} / {N_1 (t)}] = - \lambda_1dt or ^N1(t)\int_N0 [{dN_1(t)} / {N_1(t)}] = -\lambda_1 ^t\int_0 dt Where we assume that N_1(t) = N_0att = 0 N_1(t) = N_1(t)att = t . Then l_n N_1 (t)N(1) (t)▏N0= -\lambda_1t and l_n {N_1 (t) / N_0} = -\lambda_1t Taking the expotential of both sides of this equation, we obtain N_1 (t) = N_0- e ^-\lambda(1)t(4) Inserting (4) in (3) {dN_2(t)} / dt = (\lambda1N_0 e^-\lambda(1)t ) - {\lambda_2 N_2 (t)} or [{dN_2(t)} / dt] +{\lambda_2 N_2 (t)} = (\lambda1N_0 e^-\lambda(1)t ) Multiplying both sides by e^-\lambda(2)t e^-\lambda(2)t [{dN_2(t)} / dt] + \lambda_2e^-\lambda(2)t N_2(t) = \lambda_1 N_0 e[\lambda(2)-\lambda(1)] t or d/dt {e^-\lambda(2)t N_2(t)} = \lambda_1 N0e[\lambda(2) - \lambda(1)] t Solving this differential equation d {e^-\lambda(2)t N_2(t)} = \lambda_1 N0e[\lambda(2) - \lambda(1)] tdt and \intd {e^-\lambda(2)t N_2(t)} = \lambda_1 N_0 \int e[\lambda(2) - \lambda(1)] tdt {e^-\lambda(2)t N_2(t)} = (\lambda_1 N_0/ \lambda2-\lambda_1) \inte[\lambda(2) - \lambda(1)] t(\lambda1-\lambda_2) dt {e^-\lambda(2)t N_2(t)} = [(\lambda_1 N_0) / (\lambda2-_ \lambda_1)] e[\lambda(2) - \lambda(1)] t+ c(5) where C is a constant of integration. Noting that, at t = 0, there are no daughter atoms of type A, we can write N_2(0) = 0 andN_2(0) = (\lambda_1 N_0/ \lambda2-_ \lambda_1) + c = 0 Hence,C = -(\lambda_1 N_0/ \lambda2-\lambda_1)(6) Inserting (6) in (5) {e^-\lambda(2)t N_2(t)} = (\lambda_1 N_0/ \lambda2-\lambda_1) {e[\lambda(2) - \lambda(1)] t- 1} Solving for N_2(t) N_2(t) = (\lambda_1 N_0/ \lambda_2 -\lambda_1) {e-\lambda(1)t-e- \lambda(2)t)} This gives us the number of daughter atoms of type A as a function of time. This is a maximum when dN_2(t) / dt = 0 ThendN_2(t) / dt = (\lambda_1 N_0/ \lambda2-\lambda_1) [-\lambda_1 e-\lambda(1)t+ \lambda_2 e- \lambda(2)t] = 0 This equation is zero when \lambda_1 e-\lambda(1)t= \lambda_2 e- \lambda(2)t or\lambda1/ \lambda2= e-[\lambda(2) - \lambda(1)] t Solving for t ln(\lambda1/ \lambda2)= (\lambda_2 - \lambda_1)t = (\lambda1- \lambda2)t ort = 1 / (\lambda1- \lambda2) ln(\lambda_1 / \lambda2)(7) Hence, N_2 (t) is a maximum at the time given by (7).

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Question:

Given that the earth's mean radius = 6.37 × 10^6 m, normal atmosphere pressure = 1.013 × 10^5 N/m_2, and the gravitational acceleration = 9.8 m/sec^2, what is the mass of the homosphere?

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Solution:

The homosphere is that part of the atmosphere that extends from the surface of the earth up to 80 kilo-meters. The constituent gases in the homosphere are well mixed. The proportions of the principal constituents at high altitudes do not differ to a great extent from those found at sea level. To find the mass of the homosphere, calculate the total mass of the atmosphere. The homosphere is equal to 99.99% of the atmosphere and, hence, their masses are approximately equal. The mass of the atmosphere can be derived from the atmospheric pressure. Since Pressure = (Force / Area) and Force = mass × gravity Combining these two equations one obtains Pressure = (mass × gravity) / Area Since one is solving for mass the equation can be rearranged to mass = (Pressure × Area) / gravity One is given the atmospheric pressure and the force of gravity. The surface area of a sphere is equal to 4\pir^2. Substituting and solving for the mass of the homosphere: (N = kg - m/s^2) mass = (1.013 × 10^5 kg-m/m^2s^2) (4\pi) (6.37 × 10^6 m)^2 / (9.8 m/sec^2) = 5.27 × 10^18 kg.

Question:

Suppose the collision in the figure is completely inelastic and that the masses and velocities have the values shown. What is the velocity of the 2 mass system after the collision? What is the kinetic energy before and after the collision?

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/Users/wenhuchen/Documents/Crawler/Physics/D06-0321.htm

Solution:

Since we wish to relate the final velocity of the system to its initial velocity, we will use conservation of momentum The total momentum before the collision is equal to the total momentum after the collision, if no forces external to the system act. Hence m_Av^\ding{217}_Ai + m_Bv^\ding{217}_Bi = (m_A + m_B)v^\ding{217} where v^\ding{217}_Ai, v^\ding{217}_Bi are the initial velocities of m_A and m_B, and v^\ding{217} is the final velocity of the combined masses. Solv-ing for v, v^\ding{217} = (m_Av^\ding{217}_Ai + m_Bv^\ding{217}_Bi) / (m_A + m_B) Changing the vectors to magnitudes, and noting that v^\ding{217}_Ai and v^\ding{217}_Bi are in the opposite directions v = (m_A v_Ai - m_Bv_Bi) / (mA+ m_B) where v_Ai is in the positive x direction. Therefore, v = [(5 kg)(2 m/s) - (3 kg)( 2 m/s)] / [8 kg] v = .5 m/s Since v_2 is positive, the system moves to the right after the collision. The kinetic energy of body A before the collision is (1/2) m_Av^2 _Ai = (1/2)(5 kg)(4 m^2/s^2) = 10 joules and that of body B is (1/2) m_Bv^2_Bi = (1/2)(3 kg)(4 m^2/s^2) = 6 joules The total kinetic energy before collision is there-fore 16 joules. Note that the kinetic energy of body B is positive, although its velocity vBi and its momentum mv_Bi are both negative. The kinetic energy after the collision is (1/2) (m_A + m_B)v^2 = (1/2) (8 kg) (.25 m^2/s^2) = 1 joule Hence, far from remaining constant, the final kinetic energy is only 1/16 of the original, and 15/16 is "lost" in the collision. If the bodies couple together like two freight cars, most of this energy is converted to heat through the production of elastic waves which are eventually absorbed. If there is a spring between the bodies and the bodies are locked together when their velocities be-come equal, the energy is trapped as potential energy in the compressed spring. If all these forms of energy are taken into account, the total energy of the system is conserved although its kinetic energy is not. However, momentum is always conserved in a collision, whether or not the collision is elastic.

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Question:

The cortex of the adrenal gland is known to produce over 20 hormones, but their study can be simplified by classifying them into three categories: glucocorticoids, mineralocorticoids, and sex hormones. Explain the function of and give examples of each of these three groups of hormones. What will result from a hypo - functional adrenal cortex?

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Solution:

The glucocorticoids include corticosterone, cortisone, and hydrocortisone (cortisol). These hormones serve to stimulate the conversion of amino acids into carbo-hydrates (a process known as gluconeogenesis), and the for-mation of glycogen by the liver. They also stimulate forma-tion of reserve glycogen in the tissues, such as in muscles. The glucocorticoids also participate in lipid and protein tabolism. Glucocorticoids are also essential for coping with stress and acting as anti-inflammatory agents. Glucocorti-coid secretion is controlled by the anterior pituitary hormone ACTH (adrenocorticotrophic hormone). Aldosterone, the major mineralocorticoid, stimulates the cells of the distal convoluted tubules of the kidneys to decrease reabsorption of potassium and increase reabsorption of sodium. This in turn leads to an increased reabsorption of chloride and water. These hormones, together with such hormones, together with such hormones as insulin and glucagon, are important regulators of the ionic environment of the internal fluid. The adrenal sex hormones consist mainly of male sex hormones (androgens) and lesser amounts of female sex hormones (estrogens and progesterone) . Normally, the sex hormones released from the adrenal cortex are insignificant due to the low concentration of secretion. However, in cases of excess secretion, masculinizing or feminizing effects appear. The most common syndrome of this sort is virilism of the female. Should there be an insufficient supply of cortical hormones, a condition known as Addison's disease would result. This disease is characterized by an excessive excretion of sodium ions, and hence water, due to a lack of mineralocorticoids. Accompanying this is a decreased blood glucose level, due to a deficient supply of glucocorticoids. The effect of a decreased androgen supply cannot be observed immediately. Injections of adrenal cortical hormones promptly relieves these symptoms. Hormonal production in the adrenal cortex is directly controlled by the anterior pituitary hormone called adrenocorticotrophic hormone (ACTH).

Question:

Describe the major trends in the evolution of the lungs.

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/Users/wenhuchen/Documents/Crawler/Biology/F16-0406.htm

Solution:

When the vertebrates adopted a land habitat, the gill slits and the gills disappeared in the adults, and their physiological role was taken over by a pharyngeal outgrowth. This outgrowth already existed in some fish (i.e. the lung fish, the lobe-finned fish, and some bony fish), and is called the swim bladder. In these fish, this bladder is used primarily as a device to facilitate float-ing and maintain buoyancy, though it may also function as a respiratory organ. Swim bladders can occur in pairs, but they usually occur as a single structure. In some fish the swim bladder can be seen to have lost its pharnygeal connection. At the anterior end of the swim bladder are specialized cells which transfer oxygen from the blood into the bladder. This unique ability is un-paralleled anywhere in the animal kingdom. In addition to these anterior cells, there are also specialized cells at the posterior end of the bladder, which function in the transference of oxygen from the bladder to the blood. Because of the pumping action of these two groups of cells, the fish can vary its buoyancy to maintain a certain depth in the water without muscular effort. While the swim bladder has been found to serve as an organ for the production of sound, this is relatively rare and is found in only a few fish. Some stages in the evolution of the lungs. It is believed that the tetrapods evolved from the group of fish called the lobe-finned fish. These fish, which were once thought to be extinct but were rediscovered in 1939, have a swim bladder which can be used as a lung. In contrast to the bladder of other fish, the swim bladder of the lobe finned fish is equipped with a pulmonary artery. These fish can force air into their swim bladder, which functions as a lung. In the event that their pond would dry up, the fish would survive by breathing air directly, and with the aid of their lobe fins they would be able to traverse land to other streams. The presence of a "lung" and lobe fins facilitated the evolution of these creatures into the first tetrapods, the amphibians. The lungs of the tetrapods are homologous (arise embryologically in the same way) to the lungs of the lobe- finned and lung fish. The lungs of the mud puppies (the most primitive amphibians) are two long simple sacs, the outside of which is covered by capillaries. In frogs and toads, the surface area of the lung is increased by folds on the inside of the lung sacs. The method of breathing in frogs and toads is much different from that of humans, because of the absence of a diaphragm and rib muscles. Valves in the nostrils and muscles in the throat are in-corporated into the respiratory anatomy of these amphibians. Following the amphibian stage of lung development, the general evolutionary trend was toward a greater subdivision of the lung into smaller and smaller sacs. In this respect, the structure of the lung became increas-ingly complex in reptiles, birds and mammals. In some liz-ards, for example, the lungs have supplementary air sacs which, when inflated, are thought to be a protective device to frighten predators. The respiratory system of birds shows some remarkable peculiarities. Birds possess large, thin-walled air sacs scattered throughout the body. These are situated amongst the muscles and within the interior of bones. Bronchial tubes connect these air sacs to the lung. The alveoli are spongy masses of minute air pockets where gas exchange takes place, and occur along the sides of the smaller bronchial tubes called bronchioles. The functioning of the system is not completely understood, but it appears that air passes through the lungs, and then into and out of the air sacs. The air sacs therefore act as bellows, so that the lungs are completely ventilated at each breath. This system is lacking in other vertebrates, where considerable amounts of used air remains behind in the lungs. The air sacs also act to lower the specific gravity of the bird, which is advantageous for flight.

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Question:

What effect do bond angles have on bond strain? What is the influence of bond strain on bond energy?

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Solution:

All molecules have an optimum configuration of their atoms in space. Bond angles are a function of this configuration. For example, carbon atoms that have four bonds assume a tetrahedral arrangement with bond angles of 109.5\textdegree. When the arrangement of the atoms is forced to differ from this configuration by external forces, the bond angles must, therefore, adjust and are said to be strained. The bond energy is the amount of energy necessary to break a bond between two atoms. When the bonds of a molecule are strained, less energy is needed to break the bonds. Hence, the greater the bond strain the lower the bond energy.

Question:

List and compare the tissues that support and hold together theother tissues of the body.

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/Users/wenhuchen/Documents/Crawler/Biology/F02-0050.htm

Solution:

Connective tissue functions to support and hold together structures ofthe body. They are classified into four groups by structure and/or function: bone, car-tilage, blood, and fibrous connective tissue. The cells of thesetissues characteristicallyscretea large amount ofnoncellular material, called matrix. The nature and func-tion of each kind of connective tissueis determined largely by the nature of its matrix. Connective tissue cellsare actually quiteseperatefrom each other, for most of the connectivetissue volume is made up of matrix. The cells between them functionindirectly, by secreting a matrix which performs the actual functionsof connection or support or both. Blood consists of red blood cells, white blood cells, and platelets in aliquid matrix called the plasma. Blood has its major function in transportingalmost any substance that is needed, anywhere in the body. The fibrous connective tissues have a thick matrix composed of interlacingprotein fibers secreted by and sur-rounding the connective tissuecells. These fibers are of three types:collagenousfibers, which are flexiblebut resist stretching and give considerable strength to the tissues containingthem; elastic fibers which can easily be stretched, but return to theirnormal length like a rubber band when released, and reticular fibers whichbranch and interlace to form complex networks. These fibrous tissuesoccur throughout the body, and hold skin to the muscle, keep glandsin position, and bind together many other structures. Tendons and ligamentsare specialized types of fibrous connective tissue. Tendons are notelastic but are flexible, cable-like cords that connect muscles to each otheror to bones. Ligaments are semi-elastic and connect bones to bones. The supporting skeleton of vertebrates is composed of the connectivetissues cartilage and bone. Cartilage cells secrete a hard rubberymatrix around themselves. Cartilage can support great weight, yet itis flexible and somewhat elastic. Cartilage is found in the human body at thetip of the nose, in the ear flaps, the larynx and trachea,intervertebral discs, surfaces of skeletal joints and ends of ribs, to name a few places. Bone has a hard, relatively rigid matrix. This matrix contains many collagenousfibers and water, both of which prevent the bone from being overlybrittle. Bone is impreg-nated with calcium and phosphorus salts. These give bone its hardness. Bone cells that secrete the body matrix containingthe calcium salts are widelyseperatedand are located in specializedspaces in the matrix. Bone is not a solid struc-ture, for most boneshave a large marrow cavity in their centers. Also, extending through thematrix areHaversiancanals, through which blood vessels and nerve fibersrun in order to supply the bone cells.

Question:

Distinguish between infection and infestation and between virulence andpathogenicity.

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Solution:

Although these terms are used interchangeably, the meanings of these words are rather distinct. The term infection implies an interaction between twoliving organisms . The host and the parasitic microorganism compete for superiority over the other.If the micro-organism prevails, disease results.If the host is dominant, immunity or increased resistance to the disease may de-velop . Infestation indicates the presence of animal (non- microbial) parasites in or on the hosts' body. Lice, fleas and flat-worms are infesting organisms . They may transmit an infection (e.g. a louse carries typhus - a disease caused by microorganisms calledRickettsia) to man. Parasitism is a type of antagonism in which one organism, the parasite , lives at the expense of the other the host. Infection is a type of parasitism . Pathogenicity refers to the ability of a parasite to gain entrance to a host and produce disease. (A pathogen is any organism capable of producing disease.) The degree ofpathogenicity, or ability to cause infection , is called virulence. The virulence of a microorganism is not only determined by its inherent properties but also by the host's ability to resist the infection. A pathogen may be virulent for one host andnonvirulentfor another . For example, streptococci, although found in the throats of some healthy individuals, can be pathogenic under differ-ent conditions or in other individuals. Resistance is the ability of an organism to repel in-fection. Immunity is resistance (usually to one type of microorganism) developed through exposure to the pathogen by natural or artificial means. Lack of resistance is called susceptibility.

Question:

What are the three chief characteristics of phylum Chordata?

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Solution:

All chordates display the following three characteristics: first, the central nervous system is a hollow tube containing a single continuous cavity, and is situated on the dorsal side of the body. A second characteristic feature is the presence of clefts in the wall of the throat region, usually referred to as gill slits, originally utilized perhaps as a food catching device. The third characteristic is the presence of a notochord, a rod lying dorsal to the intestine, extending from the anterior to posterior end, and serving as a skeletal support. In the Vertebrata, one of the three subphyla of the chordates, the notochord is partially or wholly replaced by the skull and vertebral column. It should be noted that the dorsal hollow nerve cord, the notochord and the gill slits need only be present at some time in. the life of an organism for it to be considered a chordate. In a tunicate, for example, which belongs to the Chordata subphylum, Urochordata, the dorsal hollow nerve cord and the notochord are confined to the tail in the larval stage, and disappear in the adult stage. In the subphylum, Cephalochordata, all chordate characteristics are retained in the adult.

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Question:

After each collision in a solid, the valence electrons re-bound in a random direction. However, when an electric field is present, there is a general drift of the electrons superimposed on this random motion, as shown in the figure. Show that the average drift velocity vdmay be written v_d = at / 2 Where a is the acceleration produced by the electric field and t is the average time interval between collisions. If the acceleration is 7 x 10^10 m/s^2 , what is the numerical value of the drift velocity for free electrons in copper metal? How long does it take these electrons to drift 1 m in the presence of the electric field? The mean free time between collisions with the crystal lattice for an electron in a crystal is 3 / 3 ×10^\rule{1em}{1pt}14sec. The presence of an electric field \epsilon in a metal crystal produce a general drift of electron which is superimposed upon their random motion.

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Solution:

We assume that each time an electron suffers a collision within the crystal, it rebounds in a random fashion and on the average has no component of motion para-llel to the field. So the initial velocity v_0 in the field direction must be zero. The electric force accelerates the electron until its next collision with an ion. The final velocity acquired during the time interval t between collisions is v. The distance traveled in the direction of the force can be found from the kinematics equations for constant acceleration, or x = x_0 + v_0t + 1/2 at2 But v_0 = 0, and x \rule{1em}{1pt} x_0 = d, the distance travelled by the electron in the direction of E. Hence, d = x \rule{1em}{1pt} x_0 = 1/2 at^2 Since the average drift speed during this time interval is given by the distance travelled by the electron divided by the time t, we have v_d = d / t Because d = 1/2 at^2 , v_d = (1/2 at^2 ) / E = 1 / 2 at The average drift velocity of free electrons in copper is, therefore vd= [(7× 1010m / s^2) (3.3 × 10 ^\rule{1em}{1pt}14s)] / 2 = 1.2 × 10^\rule{1em}{1pt}3 m /s This number is very small compared to the average speed 1.21 × 10^6 m/s associated with the random electron motion. The time t required for electrons to drift a distance of 1m is, t = 1m/ (1.2 × 10^\rule{1em}{1pt}3m/ s) = 8.3 × 102s

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Question:

How long will it take to form a thickness of 4 cm of ice on the surface of a lake when the air temperature is -6\textdegreeC? The thermal conductivity K of ice is 4 × 10^-3 cal/s-cm -\textdegreeC and its density is \rho = 0.92 g/cm^3.

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Solution:

Let the thickness of the ice layer at any instant be represented by y. The next layer of thickness dy forms through the transfer of heat dQ from water to air through the ice as shown in the figure. Let this transfer take place in dt seconds. The heat lost by the freezing water, is dQ = \rhoA dy L where A is the area of ice formed, l is its density, and L is the latent heat of water. This is equal to the heat transmitted through the layer of ice already formed, dQ = [-K A (T_2 - T_1)dt]/y where T_1 and T_2 are the temperatures of the water (near the ice) and air, respectively and k is a constant. Hence \rhoA dy L = {[-K A [(T_2 - T_1)]/y}dt ordt = [\rhoL/{k(T_2 - T_1)}]ydy Integrating the above equation from y = 0 to y = 4 cm, we get t \int_0dt = [\rhoL/{k(T_2 - T_1)] ^4\int_0 ydy = [\rhoL/{k(T_2 - T_1)}][(1/2)(y^2)]^4_0cm where t is the time it takes for the ice to grow 4 cm thick. t = [{-\rhoL×8 cm^2}/{k(T_2 - T_1)}] = (8 cm^2 × 0.92 gr/cm^3 × 80 cal/gr)/(4 × 10^-3 cal/s \textdegreeC deg\bulletcm × 6 \textdegreeC deg) = 24.53 × 10^3 s = 409 min = 6 hr 49 min.

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Question:

If a 50 g bullet traveling at 40,000 cm/sec is stopped by a steel plate, how much heat is generated, on the assumption that the energy is completely converted into heat?

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Solution:

We are told that all of the bullet's energy is converted into heat, Q. Since the bullet has only kinetic energy Q = (1/2) mv^2 where m is the bullet's mass, and v is its speed. Hence, the amount of heat energy produced is Q = (1/2) (50 g) (4 × 10^4 cm/s)^2 Q = 25 × 16 × 10^8 ergs Q = 4 × 10^10 ergs

Question:

A 1000 lb weight Is suspended from the wooden boom (see the figures) which has a weight of 200 lb. Calculate the tension in the supporting cable, and the compression in the boom. This compression is the force exerted by the boom on the wall and the force exerted by the boom at the point of connection of the two cables.

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Solution:

Figure (B) shows all the forces acting on the boom. Note that that the force R\ding{217} exerted on the boom by the wall cannot be assumed to act along the boom if the weight of the boom is not neglected. This force R\ding{217} has been broken up into x and y components, as shown. Since R\ding{217} is unknown, we find the tension T\ding{217} in the cable by taking moments about point 0 at the left end of the boom. About this point, R\ding{217} does not contribute to the net torque. Though the length of the boom is not given, all distances can be expressed as some fraction of this length L, so that L appears on both sides of the moment equation and cancels. The second condition of equilibrium, applied about 0,yields (200)(L/2) + 1000L = T(sin 40\textdegree)L = (0.6428T)L 1100 = 0.6428T T = 1711 lbs. The horizontal component of R\ding{217} can be obtained using the first condition of equilibrium. \sumF_x = 0 = R_X - T cos 40\textdegree R_x = T cos 40\textdegree = (1711) (0.7660) R_x = 1311 lbs. For the vertical component of R\ding{217}, \sumF_y = 0 = R_y + T sin 40\textdegree - 1000 - 200 R_y = 1000 + 200 - T sin 40\textdegree = 200 - (1711)(0.6428) R_y = 1200 - 1100 = 100 lbs. From vector summation, we know R = \surd[R_x^2 + R_y^2] = \surd[(1311)^2 + (100)^2] = 1314 lbs. The angel is found from tan \texttheta= (R_y)/(R_x) = 100/1311 = 0.0763 \texttheta= 4.33\textdegree

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Question:

If the "average" alpha particle speed is 1.3 × 10^7 m/s within the nucleus of ^238 _92U, how many collisions will the particle make with the barrier each second? Also determine how many years are required before the average alpha particle escapes from the uranium nucleus. One year is approximately 3.1 × 10^7 s. Only one out of 10^38 collisions with the barrier results in the escape of an alpha particle.

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Solution:

As a result of the very strong nuclear attrac-tion, an \alpha particle in the nucleus is like a particle in a deep well. The attractive barrier that confines the \alpha particle to the nucleus is actually a "potential well,\textquotedblright as shown in the figure. As long as the particle doesn't have sufficient kinetic energy, according to classical physics it cannot get out of this potential well. However, quan-tum physics states that even if the particle is not energetic enough, there is a finite probability of escape from the well. In fact, it was quantum physics that explained the radioactive emission of certain particles from nuclei, in which tremendous nuclear forces bind the nuclear matter together. We can assume that the diameter, 2R_0 , of the nucleus is representative of the distance L the particle travels between collisions with the walls of the well. Therefore, the distance L is L = 2R_0 = (2)(9 × 10\Elzbar15m) = 1.8 × 10\Elzbar14m The number of times the alpha particle encounters the bar-rier each second, N, can be calculated since we know the average speed of the particle between collisions. If a distance of L meters must be traversed to make one colli-sion, then the total number of collisions taking place in one second is N =v/ L = (1.3 × 10^7m/s) / (1.8 × 10\Elzbar14m) = 7.2 × 10^20 collisions/s We are told that 10^38 collisions must occur on the average before an alpha particle tunnels through the barrier. The time t required for an \alpha particle to escape is t = (1 × 10^38 collisions ) / (7.2 × 10^20 collisions /s) = 1.4 × 10^17 s The average time for escape, t, may be written as t = (1.4 × 10^17s) / (3.1 × 10^7s/year) = 4.5 × 10^9years.

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Question:

Explain why the capillaries, rather than the veins or arteries, are the only vessels where exchange of nutrients, gases, and wastes can take place.

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/Users/wenhuchen/Documents/Crawler/Biology/F15-0372.htm

Solution:

The circulatory system of humans and other vertebrates consists basically of a heart and three types of blood vessels: arteries, veins and capillaries. An artery is a blood vessel that carries blood away from the heart. The walls of the arteries are the thickest of all the blood vessels. They are composed of three layers: (1) an outer layer of connective tissue, containing elastic fibers which give the arteries their characteristic elasticity; (2) a middle layer of smooth muscle which can regulate the size of the arterial lumen (opening); and (3) an inner layer of connective tissue lined with endothelial cells. The largest artery, the aorta, has a wall about three millimeters thick. A vein is a blood vessel that carries blood back toward the heart. The walls of veins are thinner, and the lumens larger than those of the arteries. Although the same three layers are present, the outer layer has fewer elastic fibers and the middle layer of smooth muscle is much thinner. The walls of veins are therefore much less rigid and easily change shape when muscles press against them. Unlike arteries, veins have valves which prevent the backflow of blood and thereby maintain the direction of flow to the heart. Capillaries are the blood vessels that connect the arteries with the veins. The capillaries have walls composed of endothelium only one cell thick. It is the thinness of the capillary walls which allows for the diffusion of oxygen and nutrients (also hormones) from the blood into the tissues, and for carbon dioxide and nitrogenous wastes to be picked up from the tissues by the blood. The walls of arteries and veins are thick and impermeable, and are thus unsuit-able for this exchange process. Much of the exchange, however, is not across the endothelial cells of the capillary but rather between the cells through pores. The exchanged substances do not enter the tissue cells directly but first enter the interstitial fluid which lies in the spaces between the cells. They then dif-fuse through the plasma membranes of the neighboring cells. Capillaries usually form an extensive network, creating a large surface area for exchange. The arterioles are the last small branches of the arterial system which release blood into the capil-laries . The venules, the smallest veins, then collect blood from the capillaries. A capillary network is shown in Figure 2. (The amount of branching is actually much more extensive).

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Question:

What are the bacterial factors influencing virulence?

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/Users/wenhuchen/Documents/Crawler/Biology/F05-0140.htm

Solution:

Virulence is the degree of ability in micro-organisms to produce disease . Some organisms are more virulent than others. There are several factors influencing virulence. One of these is the production of toxins which are poisonous substances produced by some microorganisms. Both the ability to produce the toxin and the potency of the toxin affect the organism's capability to produce disease. Toxins which are secreted into the surrounding medium during cellular growth are calledexotoxins, and toxins retained in the cell during growth and released upon cell death and lysis are calledendotoxins.Exotoxinsare released into the surrounding medium , e.g., into a can of vegetables containingClostridiumbotulinum (thebacterium causing botulism) or into damaged tissue infected with Clostridium tetani(the bacterium causing tetanus). Both these types of bacteria exhibit difficulty in penetrating the host. However, these bacteria are virulent because of their toxins. When the spatial configuration of the amino acids in a toxin molecule is altered, the toxicity is lost, and the resulting substance is called atoxoid. Both toxins andtoxoidsare able to stimulate the production of anti-toxins, substances made by the host which are capable of neutralizing the toxin.Endotoxinsare liberated only when the microorganism disintegrate and are generally less toxic than exotoxins . They do not formtoxoidsand are usuallypyroqenic,inducinq fever in the host (someexo-toxinsproduce fever also).Exotoxinsare usually associated with gram-positive bacteria whileendotoxinsar associated with the gram-negative ones. Another factor influencing virulence is the ability of bacteria to enter the host and penetrate host tissue. Specific bacterial enzymes are involved . One enzyme that is produced by some of the Clostridia and cocci ishyaluronidase. This enzyme facilitates the spread of the pathogen by aiding itspenetranceinto host tissues. It hydrolyzeshyaluronicacid, an essential tissue "cement", and thus increases the permeability of tissue spaces to both the pathogen and its toxic products. The virulence of pathogens is also influenced by their ability to resist destruction by the host. In certain bacteria, this resistance is due to the presence of a non-toxic polysaccharide material which forms a capsule surrounding the bacterial cell. For example,pneumococciare Streptococcus pnemoniaevirulent when capsulated but are oftenavirulent when not capsulated. Capsulatedpnemococciare resistant to phagocytosis (the ingestion and destruction of microorganisms by host phagocytes ), while those without capsules are ingested by leucocytes and destroyed .

Question:

Explain how visual displays are important in communi-cating an animal's "mood."

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/Users/wenhuchen/Documents/Crawler/Biology/F31-0796.htm

Solution:

For animals having a well-developed visual sense, visual displays serve as signals to convey their potential behavior. Such displays can convey an animal's readiness to mate, to attack, or to retreat. During territorial contests and sexual encounters, cichlids, which are African fresh water fish, transmit their moods by displaying distinctive coloration or patterns. (See Figure 1) The particular coloration suggests the behavior that is likely to ensue. Another example of a visual display is the courtship behavior of ducks. The odd movements of male ducks in spring signal to the female that they are eager to mate. These visual displays also synchronize the sexual physio-logy of both male and female and make the female more receptive to the male's advances. Each species has its own display pattern, ensuring that a female will only mate with a male of its own species. Visual displays occur in other than reproductive behavior. They can be seen in the everyday behavior of cats and dogs. The greeting display, which involves the erection of the tail in cats and the wagging of the tail in dogs, conveys the animal's receptive mood. When dogs and cats are displaying antagonistic behavior, the teeth are bared, the hairs on the neck and back are raised, the ears are laid back, and the body raised. This conveys Figure 1. African fresh-water fish (Hemichromis fasciatus) can change its body coloring rapidly. The color changes are displays that express sight different moods. Figure 2. Agonistic displays of Black-headed Gull. (A) The first response given by a male on his territory when another male approaches is the "Long Call," in which the body is tilted downward, the wing butts are lifted, the head is thrust foward, and a characteristic call is given. (B) If the other male continues to approach, the defending bird may move to meet the intruder at the boundary of his territory, where he gives the "Upright" display, lifting his still-folded wings, stretching his neck up-ward, and pointing his bill downward. (C) If the intruder performs counterthreat displays, the defender may then adopt the "Choking" posture, tilting his body head down in an almost vertical position and moving his head in a series of quick up-and-down jerks. (D) "Facing Away" is an appeasement display in which a gull turns his head so that the other bird cannot see the beak and eyes or the black facial mask. the animal's readiness to attack. A defeated or submissive animal displays appeasement behavior: the fur is sleeked back, the body lowered with legs bent, the tail is down or tucked between the legs, and the head turned away from the victor. Such a display shows the animal's willingness to retreat. In general, attack behavior usually involves directing the face toward the antagonist and making the body appear as large as possible through hair-raising and posturing. The antagonistic displays of a black- head gull are shown in figure 2. Retreat or appeasement be-havior involves turning the face away and making the body appear small and vulnerable. Such behavior sometimes serves to prevent actual fighting between animals, as one of the contestants usually backs down. For instance, a defeated rat will roll over on his back, which usually causes the aggressor to stop his attack.

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Question:

A charge is placed on a string of infinite length so that the linear charge density on the string is n coulombs/m. Find the electric field due to this charge distribution.

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/Users/wenhuchen/Documents/Crawler/Physics/D17-0555.htm

Solution:

This problem can be solved using Gauss's Law. We construct as our Gaussian surface a cylinder, whose axis of symmetry coincides with the string, of length L, and of radius R. It can be seen from symmetry that there can be no component of the electric field parallel to the string, since all contributions to the field in that direction will cancel (see Fig. 2.). Since the field vector can be expressed in terms of components that are either perpendicular or parallel to the string, the resulting field will be radial. It can also be seen from symmetry that the magnitude of the field will be uniform over the surface of the cylinder, excluding the circular top and bottom. The flux through these portions is zero, however, since the field lines do not pass through their area. The flux through the cylinder is therefore: \Phi_E = \intE^\ding{217} \bullet dS^\ding{217} = E \int cos 0\textdegree dS = E \int dS = ES = E \textbullet 2\piRL Since the field is constant in magnitude it can be factored outside of the integral sign. The field lines coincide with the surface vector elements of the cylinder, which is the same as saying that they make an angle of zero degrees with each other. The corresponding term cos 0\textdegree in the integral reduces to 1, leaving only dS in the integrand which reduces to S, the total surface area of the cylinder (excluding the top and bottom). The charge on length L of string is nL, thus by Gauss's Law: \Phi_E = 4\pik_Eq = 4\pi (1/4\pi\epsilon_0) q = q/\epsilon_o E \bullet 2\piRL = nL/\epsilon_o E = (n/2\pi\epsilon_oR) = 2K_E (n/R) where k_E = 1/4\pi\epsilon_o .

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Question:

A box is dragged up and down a concrete slope of 15\textdegree to the horizontal. To get the box started up the slope, it is necessary to exert six times the force needed to get it started down the slope. If the force is always parallel to the slope, what is the coefficient of static friction between the box and the concrete?

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Solution:

When the box is about to slide down the slope, the forces acting on it are as shown in (figure (a)). The weight of the box W acts vertically downward, the frictional force which attempts to prevent the motion acts up the slope, and the concrete exerts a normal force at right angles to the slope. when the box is just on the point of moving, F = \mu_sN, where \mu_s is the coefficient of static friction required. Let us resolve the force W\ding{217} into its components along, and at right angles to, the slope. Since the angle between the slope and the horizontal is 15\textdegree, this is also the angle between the normal to the slope and the normal to the horizontal (i.e., the vertical). Thus W\ding{217} has components W cos 15\textdegree at right angles to the plane and W sin 15\textdegree down the plane. The box is just in equilibrium. From the conditions for equilibrium, we know that N = W cos 15\textdegree. andP + W sin 15\textdegree = F = \mu_sN = \mu_sW cos 15\textdegree. \thereforeP = \mu_sW cos 15 - W sin 15\textdegree. Figure (b) shows that, when the box is about to slide up the slope, the situation is very similar. The box is in equilibrium once more, so that N = W cos 15\textdegree andP' = W sin 15\textdegree + F = W sin 15\textdegree + \mu_sN = W sin 15\textdegree + \mu_sW cos 15\textdegree. But we know that the force P' = 6P. \thereforeW sin 15\textdegree + \mu_sW cos 15\textdegree = 6(\mu_sW cos 15\textdegree - W sin 15\textdegree). \therefore5\mu_sW cos 15\textdegree = 7 W sin 15\textdegree. \therefore\mu_s = (7/5) tan 15\textdegree = (7/5) × 0.268 = 0.375.

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Question:

A sample of air held in a graduated cylinder over water has a volume of 88.3 ml at a temperature of 18.5\textdegreeC and a pressure of 741 mm (see figure). What would the volume of the air be if it were dry and at the same temperature and pressure?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E03-0085.htm

Solution:

This problem is an application of Dalton's Law which states that the total pressure of a mixture, P_total, of gases is the sum of the individual partial pressures, P_1, P_2. Stated mathematically P_total = P_1 + P_2 + P_3 ... The mixture of gases in this problem is located in the cylinder and is made up of dry air and water vapor. At 18.5\textdegreeC the vapor pressure of water is 16 mm. The pressures are related by the equation P_total = Pdry air+ Pwater vapor Using this equation, we can calculate Pdry air. Having obtained this value, the problem becomes a direct pressure-volume relationship. As the volume increases, the pressure decreases and vice versa. These factors are related by the equation P_1V_1 = P_2V_2 where P_1 and V_1 represent pressure and volume in the presence of water vapor, and P_2 and V_2 represent the same quantities for dry air. By Dalton's Law: P_dry air= P_total - P_water vapor = 741 - 16 = 725 mm when water vapor is present. If there is no water vapor present and the dry air were to exert the entire pressure of 741 mm then, we predict that the dry air volume will be smaller. Next substitute the values into the pressure-volume equation and solve for unknown volume V_2 V_2 = (P_1V_1) / P_2 = [725(88.3)] / 741 = 86.4 ml Therefore, as predicted the volume has decreased.

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Question:

SO^2-_3 + CrO^2-_4 \rightarrow SO^2-_4 + Cr (OH)_3

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/Users/wenhuchen/Documents/Crawler/Chemistry/E16-0566.htm

Solution:

Three rules can be used to balance oxidation- reduction reactions: oxygen by adding water. (3) Balance atoms (of hydrogen) by adding hydrogen to the appropriate side. These 3 rules will balance the redox equation. You proceed as follows: Reduction: CrO^2-_4 \rightarrow Cr(OH)_3. Add 2OH^-, to the right side so that charge is balanced. You obtain CrO^2-_4 \rightarrow Cr(OH)3+ 2OH^- . Balance oxygens by adding one water molecule to left side. Thus, H_2O + CrO^2-_4 \rightarrow Cr(OH)3+ 2OH^-. Balance H's by adding three H's to right side. You have (3 / 2)H_2 + H_2O + CrO^2-_4 \rightarrow Cr(OH)3+ 2OH^- Oxidation: SO^2-_3 \rightarrowSO^2-_4 . Charges are already balanced. To balance oxygen, add water to left side. As such, H_2O + SO^2-_3 \rightarrow SO^2-_4 . Now balance hydrogens to obtain H_2O + SO^2-_3 \rightarrow SO^2-_4+ H_2. In summary, the balanced half-reactions are oxid:H_2O + SO^2-_3 \rightarrow SO^2-_4+ H_2 red:(3 / 2)H_2 + H_2O + CrO^2-_4 \rightarrow Cr(OH)3+ 2OH^- So that no free H's appear in overall reaction, multiply the oxidation reaction by 3 and red by 2. You obtain oxid:3H_2O + 3SO^2-_3 \rightarrow 3SO^2-_4+ 3H_2 red:3H_2 + 2H_2O + 2CrO^2-_4 \rightarrow 2Cr(OH)3+ 4OH^- overall (oxid + red): 5H_2O + 3SO^2-_3+ 2CrO^2-_4\rightarrow3SO^2-_4 + 2Cr(OH)_3+ 4OH^-. Notice: The H_2's dropped out.

Question:

Consider a unit cell of sodium chloride in the accompanying figure, (a) What fraction of a Cl^- at each corner of the cube is within the unit cell? (b) What fraction of a Cl^- appears in each face? (c) What fraction of each Na^+ is with-in the unit cell? (Note that one complete Na^+ is in the center.) (d) Add all the fractions for each ion within the unit cell. What is the ratio of Na^+ to Cl^-?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E06-0228.htm

Solution:

The following rules may be used to determine the number of unit particles (atoms) associated with each type of cubic unit cell: 1) An atom at a corner contributes 1/8 of its volume to each of 8 adjacent cubes. 2) An atom on an edge of a cube contributes 1/4 of its volume to each of 4 adjacent cubes. 3) An atom in the face of a cube contributes 1/2 of its volume to each of 2 adjacent cubes. 4) An atom completely within a cube contributes all of its volume to the unit cell. (a) According to rule 1 an atom at a corner contributes 1/8 of its volume to the unit cell. (b) From rule 3 one can see that an atom on the face of the unit cell contributes 1/2 of its volume to the cell. (c) From rule 2 one can see that atoms on the edge of the unit cell contribute 1/4 of their volume to the unit cell. Therefore the sodium ions on the edge of the cell contribute 1/4 of their volume and the one in the center of the cell contributes its entire volume. (d) There are 14 Cl^- ions contributing to the cell. 8 are at corners and 6 are on faces - using parts a + b of this problem one can calculate the number of atoms that the Cl^- ions contribute to the unit cell. total number of Cl^- ions = 8 corners × 1/8 atom/corner + 6 faces × 1/2 atom/face = 4 atoms of Cl^- The Na^+ ions can be dealt with in a similar manner. There are 13 Na^+ ions, 12 on edges and 1 in the center - using part c of this problem one can find the number of Na^+ ions contributing to the unit cell. total number of Na^+ ions = 12 edges × 1/4 atom/edge + 1 atom in center = 4 atoms of Na^+ There are 4 atoms of Cl^- and 4 atoms of Na^+ contributing to the unit cell. The ratio will therefore be -Na^+ : Cl^- is 4 : 4 or 1 : 1. Na^+ and Cl^- contribute equally to the unit cell.

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Question:

The first step in the radioactive decay of ^238_92U is ^238_92U = ^234_90Th + ^4_2He. Calculate the energy released in this reaction. The exact masses of ^238_92U, ^234_90Th, and ^4_2He are 238.0508, 234.0437 and 4.0026amu, respectively. 1.0073amu= 1.673 × 10^-24 g.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E19-0714.htm

Solution:

The energy released in this process can be determined from the change in mass that occurs. Energy and mass are related in the following equation, ∆E = ∆mc^2 where, ∆E is the change in energy, Am the change in mass and c the speed of light (3.0 × 10^10 cm/sec). ∆m is found by subtracting the mass of ^238_92U from the sum of the masses of ^238_90Th and ^4_2He. ∆m= (234.0437amu+ 4.0026amu) - 238.0508amu = - 0.0045amu. Energy is expressed in ergs (g - cm^2/sec^2), therefore, ∆m must be converted to grams before solving for ∆E. ∆m= (- 0.0045amu) [(1.673 × 10^-24 g) / (1.0073amu)] = - 7.47 × 10^-27 g. Solving for ∆E: ∆E= (- 7.47 × 10^-27 g) (3.0 × 10^10 cm/sec)^2 = - 6.72 × 10^-6 g (cm^2/sec^2) = - 6.72 × 10^-6 erg Therefore, 6.72 × 10^-6 ergs are released.

Question:

A solute of formula AB is slightly dissociated into A+ and B-. In this system, there is a dynamic equilibrium such that A+ + B-\rightleftarrows AB. Explain what happens if more acid is introduced into this system.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E09-0325.htm

Solution:

An acid is a species which, when added to a solvent (such as H_2O), dissociates into protons (H+) and anions. In this particular case, the proton is represented as A+. When more acid is added to this general solvent system, more A+ is introduced. The increased concentration of A+ places a stress on the equilibrium and the result is a shift in this equilibrium. According to LeCh\^{a}telier'sprinciple, an equilibrium system will readjust to reduce a stress if one is applied. Thus, the equilibrium A+ + B- \rightleftarrowsAB will readjust to relieve the stress of the in-creased A+ concentration. The stress is relieved by the reaction of A+ with B- to produce more AB. The concentration of B- will decrease as compared to its concentration prior to the addition of the acid. Also, the concentration of the product AB will increase with the addition of the acid.

Question:

What is meant by "coincident current" in a computer's core memory?How does this relate to binary states?

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G05-0093.htm

Solution:

A single unit of core can take on a value of 1 or 0, depending upon whether the direction of its mag-netization is clockwise or counterclockwise. If all the cores were strung on a single wire each time current was passed down the wire, all the cores would become magne-tized in the same direction. This would render the use of core inefficient. To avoid this possibility, the method of coincident cur-rent is used. Three wires are strung through each core to mediate control of the direction of magnetization. One entire plane of cores resembles a grid with cores at each intersection. Two of the three wires are current wires, while the third is called a sense wire. Let us present an example of the procedure necessary to have a single core represent the binary digit 1. Each of the current wires sends only one-half the necessary cur-rent to make the core become magnetized in a clockwise direction. The core located where the current wires inter-sect will become magnetized in a clockwise direction; all other cores along the lines remain unaffected. The sense wire is a feedback device which can determine at any time if a particular core is representing a binary 1 or a binary 0.

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Question:

At two points on a horizontal tube of varying circular cross section carrying water, the radii are 1 cm and 0.4 cm and the pressure difference between these points is 4.9 cm of water. How much liquid flows through the tube per second?

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Solution:

Since the tube is horizontal there is no pressure difference along the tube due to hydrostatic effects because the static pressure due to the weight of the fluid plays no part in the problem. Thus Bernoulli's equation is P_1 + (1/2) \rhov^2_1 = p_2 + (1/2) \rhov^2_2(1) where \rho is the density of the liquid, v its velocity, p its pressure, and the subscripts 1 and 2 refer to any two points along the tube. In time t, those particles of the fluid which were originally at point 1(see figure), move a distance v_1t. The total volume of fluid which moves past point 1 in time t is there-fore A_1V_1t. Its rate of flow per unit time is then A_1V_1. Similarly, the rate of flow past point 2 is A_2V_2. Assuming the fluid is incompressible, these two rates of flow must be equal. We have A_1V_1 = A_2V_2(2) This is the equation of continuity which states that the quantity of an incompressible liquid which flows through a tube per second is constant at all points. The pressure difference is given as 4.9 cm of water. This equals the pressure produced by the weight of 4.9 cm of water or P_1 - P_2 = \rhog x 4.9 cm(3) From equations (1) and (3), v^2_2 - v^2_1 = [2(p_1 - p_2)]/\rho = [2(\rhog) (4.9 cm)]/\rho = (2g)(4.9 cm) = (2)(980 cm/sec^2) (4.9 cm) andv^2_2 - v^2_1 = 98^2 cm^2/sec^2(4) Using equation (2), v_1/v_2 =A_2/A_1 = (\pi × 0.4^2 cm^2)/(\pi × 1^2 cm^2) = 0.16. Substituting v^2_1 = 0.16^2 v^2_2 in equation (4) yields v^2_2(1 - 0.16^2) = 98^2 cm^2 \bullet s^-2 orv_2 = \surd[(98^2 cm^2 \bullet s^-2)/(0.9744)] The quantity of water flowing through the tube per second is thus A_1V_1 = A_2V_2 = \pi × (0.4 cm)^2 × \surd[(98^2 cm^2 \bullet s^-2)/(0.9744)] = 50 cm^3 \bullet s^-1.

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Question:

An engine used to pump water out of a mine shaft raises the water 150 ft and discharges it on the surface with a speed of water 150 ft and discharges it on the surface with a speed of 20 mph. It removes 2 slugs per second from the mine. One- 20 mph. It removes 2 slugs per second from the mine. One- fifth of the work it does is used in overcoming frictional fifth of the work it does is used in overcoming frictional forces. What is the horsepower of the engine? forces. What is the horsepower of the engine?

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Solution:

During the process of removal from the mine, the water gains both

Question:

When an object is placed 20 in. from a certain lens, its virtual image is formed 10 in. from the lens. Determine the focal length and character of the lens.

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/Users/wenhuchen/Documents/Crawler/Physics/D28-0884.htm

Solution:

Since the image is virtual, on the same side of the lens as the object, its distance from the lens, q, is negative. Substitution in the general equation for lenses yields 1/p + 1/q = 1/f (1/20in) + [1/(-10in)] = 1/f 1/f = (- 10 in + 20 in) / {(20in)×(-10)} = -(10/200in) f = -20in. The negative sign for the focal length indicates that the lens is diverging. Diverging lenses are concave.

Question:

Will precipitation occur if .01 mole of Ba+^2 is added to a liter of solution containing .05 mole of SO_4 -^2 ?K_spfor BaSO_4 = 1 × 10-^10 .

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Solution:

K_spis the solubility product constant . It is defined as the minimum product of the concentration of the ion needed to bring about the formation of a precipitate (i.e., a sold compound ) . The general form of theK_sp equation for compound A_nB_mis : K_sp= [A]^n [B]^m Therefore, for BaSO_4 theK_spequtionis : K_sp= [Ba+^2 ] [SO_4 -^2 ] = 1 × 10-^10 . To find out if precipitation occurs, substitute in the given concentration values. If the calculated value is greater than the given value forK_sp, then precipitation occurs . If it is lower , then no precipitation occurs . Thus , K_sp= [Ba+^2 ] [SO_4 -^2 ] = [1 × 10-^10 ] [5 × 10-^2 ] = 5 × 10-^4 . Since this value is greater than the givenk_spvalue , precipitation occurs .

Question:

The terms defecation, excretion and secretion are sometimes confused. What are meant by these terms?

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/Users/wenhuchen/Documents/Crawler/Biology/F18-0458.htm

Solution:

Defecation refers to the elimination of wastes and undigested food, collectively called feces, from the anus. Undigested food materials have never entered any of the body cells and have not taken part in cellular meta-bolism; hence they are not metabolic wastes. Excretion refers to the removal of metabolic wastes from the cells and bloodstream. The excretion of wastes by the kidneys involves an expenditure of energy by the cells of the kidney, but the act of defecation requires no such effort by the cells lin-ing the large intestine. Secretion refers to the release from a cell of some substance which is utilized either lo-cally or elsewhere in some body processes; for example, the salivary glands secrete saliva, which is used in the mouth in the first step of chemical digestion. Secretion also involves cellular activity and requires the expenditure of energy by the secreting cell.

Question:

A mass m hangs at the end of a rope which is attached to a support fixed on a trolley (as shown in the figures). Find the angle \alpha it makes with the vertical, and its tension T when the trolley 1) moves with a uniform speed on horizontal tracks, 2) moves with a constant acceleration on horizontal tracks.

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Solution:

1) When the trolley moves with a uniform speed, the only external force acting on m is the gravitational force mg\ding{217}. It is balanced by the \ding{217} tension T\ding{217}in the rope, since m is in equilibrium (Fig. a). Hence \alpha = 0. 2) \ding{217} When the trolley has acceleration a, the effect of the acceleration is transmitted to the mass through the rope. We see (Fig. b), that the magnitudes of the gravitational and horizontal accelerations determine the magnitude of the angle \alpha. The tension in the rope provides the upward force -mg\ding{217}to hold the mass (since m is in vertical equilibrium) as well as \ding{217} the external force ma acting on the mass as a result of the motion of the trolley, (Fig. c). Hence, T sin \alpha = ma(1) T cos \alpha = mg(2) Dividing (1) by (2), we get [T sin \alpha] / [T cos \alpha]= (ma/mg) tan \alpha= (a/g) Therefore, \alpha is given by \alpha = tan^-1 (a/g).

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Question:

Write equations to show how Ca(HCO_3)_2 is involved in (a) stalagmite formation in caves, (b) formation of scale in tubes or pipes carrying hot water, (c) the reaction with soap (C_17H_35COONa), and (d) reaction with Na_2CO_3 in water softening.

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Solution:

(a) To answer this, one must know the composition of stalagmites, Which are rocks composed of CaCO_3 and are formed in underground caverns. Thus, one knows that somehow the reaction Ca(HCO_3)_2 CaCO3 occurs. Since Ca^+2, CO_3^-2 are not reduced or oxidized, no electrons are exchanged. One needs to balance the reaction with respect to mass. Ca (HCO_3)_2 \ding{217} CaCO_3 + H_2O + CO2 This equation has both charge and mass balanced. (b) The formation of scales in pipes is the result of CaCO_3 build up from the decomposition of bicarbonates. The reaction for this process is Ca (HCO_3)_2 CaCO_3 + H_2O + CO_2 (c) The reaction of Ca(HCO_3)_2 with soap (C_17H_35COONa) occurs when the Na^+ on the soap molecule is replaced by the Ca^++. Ca (HCO_3)_2 + 2C_17H_35COONa \rightarrow (C_17H_35COO)_2 Ca + H_2O + CO2 (d) Permanent hardness in water results from chlorides or sulfates of metals, such as calcium or magnesium. Boiling does not remove these salts; a substance called water softener must be added to the water to form a precipitate with the salts. Water containing Ca(HCO_3)_2 can be softened by adding the proper amount of Na_2CO_3. The reaction is Ca (HCO_3)_2 + Na_2CO_3 \ding{217} CaCO_3(s) + 2NaHCO_3.

Question:

A chemist dissolves methyl alcohol in n-octane. He adds this to excessmethylmagnesiumiodide, which is dissolved in the high-boiling solvent, n-butyl ether. A gas is evolved, which is collected at STP. It occupies a volume of 1.04 cc. What is the gas and how many grams of methyl alcohol were present?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E21-0767.htm

Solution:

Grignard reagents, which have the general formula ofRMgX(R is an alkyl or aryl group, Mg is magnesium, and X is a halogen atom) , are extremely re-active. In the presence of any acid stronger than analkane, thegrignardreagent is displaced to form RH. This happens because the grignardreagent may be con-sidered the magnesium salt of a weak acid RH. Thus, a stronger acid displaces RH from its salt. Alcohols are stronger acids thanalkanes. Therefore, when methyl alcohol (CH_3OH) is added to methyl magnesium iodide (CH_3MgI), the following reaction occurs: CH_3OH + CH_3MgI \rightarrow CH_4+ H_3COMgI. At S.T.P. (0\textdegreeC and 1atm), CH_4 exists as a gas. To find the amount of CH_3OH present, remember that at STP a mole of gas occupies 22.4 liters. The gas produced here occupies only 1.04 cc or 0.00104 liters (1000 cc = 1 liter). Let x = moles of methane gas produced. Thus, you can set up the proportion (22.4 liters / 1 mole) = (.00104 liters / x mole) . Solving, x = 4.64 × 10-^5 moles. From thestoichiometryof the equation, 1 mole of CH_3OH produces 1 mole of CH_4 . Thus, you have 4.64 × 10-^5 moles of CH_3OH. The molecular weight (MW) of CH_3OH is 32 g/mole. By definition, 1 mole equals the weight in grams divided by MW. Substituting and solving for weight in grams, 4.64 × 10-^5 moles = (grams of CH_3OH) / 32 g/mole) ; grams of CH_3OH present = 32 (4.64 × 10-^5 ) = 1.48 × 10-^3 grams.

Question:

What is meant byphotoperiodism?How would you determine whether a particular flowering plant is a "long-night," a "short-night," or an "indeterminate" plant?

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Solution:

Photoperiodismis defined as the biological response to a change in the proportions of light and darkness in a 24-hour daily cycle. Photoperiodism has now been shown to control a wide range of biological activities, including the induction of flowering in flowering plants, the stimulation of germination in seeds of certain species, and the initiation of mating in certain insects, birds, fish and mammals. Based on the phenomenon ofphotoperiodism, flowering plants can be classified as either long-night (short-day), short-night (long-day), or indeterminate (night-neutral). Experimentally, we can determine the photoperiodic classi-fication of a given flowering plant by varying the length of darkness per day to which the plant is exposed and obser-ving the effect of flowering - whether it occurs or whether it is inhibited. If the plant can produce flowers only when subject to a dark period of about nine hours or more per day, the plant is said to be a long-night plant. Long-night plants, such as asters, cosmos, chrysanthemums, dahlias, poinsettias and potatoes, normally flower in the early spring, late summer, or fall. If, on the other hand, the plant can be made to flower only when exposed to a period of darkness less than nine hours per day, it is said to be a short- night plant. Short-night plants, such as beets, clover, coreopsis, delphinium and gladiolus, normally flower in the late spring and early summer. If the flowering of a plant is unaffected by the amount of darkness per day, it is neither a long-night nor short-night plant, but rather an indeterminate plant. Ex-amples of indeterminate plants are carnations, cotton, dandelions, sunflowers, tomatoes and corn. It must be emphasized that the critical length of daily darkness for flowering depends on the individual species of plant, and the nine-hour period of darkness as a criterion for classifying flowering plants is at best an approximate figure. The cocklebur, for instance, is a long-night plant, at least 8(1/2) hours of darkness per 24-hour cycle in order to flower.

Question:

Find the ∆H\textdegree, heat of reaction, for the hydrolysis of urea to CO_2 and NH_3. Although the reaction does not pro-ceed at a detectable rate in the absence of the enzyme, urease, ∆H may still be computed from the following thermodynamic data: Component ∆H\textdegree_f (Kcal/mole Urea (aq) - 76.30 CO_2 (aq) - 98.69 H_2O (l) - 68.32 NH_3 (aq) - 19.32 The hydrolysis reaction is H_2O (l) + H_2N - CO - NH_2 \rightarrow CO_2(aq) + 2NH_3 (aq) (urea)

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Solution:

The heat of reaction, ∆H\textdegree, may be found from the heats of formation, ∆H\textdegree_f, of the reactants and products. Heat of formation may be defined as the heat absorbed or evolved in the synthesis of one mole of a compound from its elements, all components being in their standard states. The heat of reaction is the sum of the heats of formation of the products minus the sum of the heats of formation of the reactants. In other words, ∆H_reaction = \sum∆Hformation of products- \sum∆Hformation of reactants Therefore, for the hydrolysis of urea to CO_2 and NH_3 ∆H_reaction = [(∆H\textdegree_f,_CO2 + 2∆H\textdegree_f,_NH3) - (∆H\textdegree_f,H2N-CO-NH2 + ∆H\textdegree_f H2O)] Given these values, one can substitute to obtain: ∆H_reaction = [(-98.69 - 2 × 19.32) - (- 76.30 - 68.32)] From this equation one obtains ∆H_reaction = 7.29 Kcal/mole.

Question:

What is meant by the "vital capacity" of a person? In what conditions is it increased or decreased?

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/Users/wenhuchen/Documents/Crawler/Biology/F16-0395.htm

Solution:

During a single normal breath, the volume of air entering or leaving the lungs is called the tidal vol-ume. Under conditions of rest, this volume is approximately 500 ml. on average. The volume of air that can be in-spired over and above the resting tidal volume is called the inspiratory reserve volume, and amounts to about 3000 ml. of air. Similarly, the volume of air that can be expired below the resting tidal volume is called the expiratory reserve volume, and amounts to approximately 1000 ml. of air. Even after forced maximum expiration, some air (about 1000 ml.) still remains in the lungs, and is termed the residual volume. The vital capacity is the sum of the tidal volume and the inspiratory and expiratory reserve volumes. The vital capacity then represents the maximum amount of air that can be moved in and out during a single breath. The average vital capacity varies with sex, being 4.5 liters for the young adult male, and about 3.2 for the young adult female. During heavy work or exercise, a person uses part of both the inspira-tory and expiratory reserves, but rarely uses more than 50% of his total vital capacity. This is because deeper breaths than this would require exhaustive activities of the inspiratory and expiratory muscles. Vital capacity is higher in an individual who is tall and thin than in one who is obese. A well developed athlete may have a vital capacity up to 55% above average. In some diseases of the heart and lungs, the vital capacity may be reduced considerably.

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Question:

An x-ray tube with a copper target is found to be emitting lines other than those due to copper. TheK_\alphaline of copper is known to have a wavelength of 1.5405 \AA, and the other two K_\alphalines observed have wavelengths of 0.7092 \AA and 1.6578 \AA. Identify the impurities.

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Solution:

According to Moseley's equation forK_\alpharadiation, 1/\lambda = R(Z - 1)^2 [(1/1^2) - (1/2^2)]. Thus, if \lambda is the wavelength ofCuK_\alpharadiation and \lambda_1 and \lambda_2 the wavelengths of the two unknownK_\alpharadiations, then \lambda_1/\lambda = (z - 1)^2 / (z_1 - 1)^2 = 0.7092 \AA / 1.5405 \AA . But for copper Z = 29. Therefore Z_1 - 1 = 28\surd(1.5405 / 0.7092) = 41. Z_1 = 42, and we know that the impurity is molybdenum. Similarly, Z_2 - 1 = 28\surd(1.5405 / 1.6578) = 27. Z_2 = 28, and we know that the impurity is nickel.

Question:

Two sources separated by 10 m vibrate according to the equations y_1 = 0.03 sin \pit and y_2 = 0.01 sin \pit. They send out simple waves of velocity 1.5 m/sec. What is the equation of motion of a particle 6 m from the first source and 4 m from the second?

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Solution:

The equation of a wave travelling along the x \rule{1em}{1pt} axis is y = A sin (\omegat \pm Kx) where \omega is the angular frequency of the wave A is its am-plitude and K is its wave number. The factor \pmKx accounts for the direction of travel of the wave; \rule{1em}{1pt}Kx indicates that the wave travels towards increasing values of x, and vice-versa. Note that this equation assumes that, at x = 0, t = 0, the y displacement is zero. Since \omega = 2\pif, where f is the frequency of the wave, and k = 2\pi/\lambda, where \lambda is its wavelength, we obtain, y = A sin (\omegat \pm Kx) y = A sin {2\pift \pm (2\pix/\lambda)} But \lambda = (v/f), where v is the wavespeed. Therefore, y = A sin 2\pi [ft \pm (fx/v)] y = A sin 2\pi f[t \pm (x/v)] We suppose that source 1 sends out waves in the +x direction, with amplitude a_1, and frequency f_1. y_1 = a_1 sin 2\pif_1 [t - (x_1/v)](1) and that source 2 sends out waves in the \rule{1em}{1pt}x direction, with amplitude a_2 and frequency f_2. y_2 = a_2 sin 2\pif_2 [t + (x_2/v)](2) In these equations x_1 and x_2 measure position from sources 1 and 2 as origin, respectively. Comparing equations (1) and (2) with the equations given in the question, we find that, 2\pif_1 = \pi and 2\pif_2 =\pi or f_1 = f_2 = 1/2 Furthermore, v = 1.5 m/sec a_1 = 0.03 ma_2 = 0.01 m x_1 = 6 mx_2 = \rule{1em}{1pt}4 m Thus, y_1 = (.03 m) sin \pi(t \rule{1em}{1pt} 4) y_1 = (.03 m) [sin \pit cos 4\pi \rule{1em}{1pt} cos \pit sin 4\pi] y_1 = (.03 m) sin \pit y_2 = (.01 m) sin \pi (t \rule{1em}{1pt} 8/3) y_2 = (.01 m) [sin \pit cos (8\pi/3) \rule{1em}{1pt} cos\pi t sin (8\pi/3)] y_2 = (.01 m) [sin \pit (\rule{1em}{1pt}1/2) \rule{1em}{1pt} cos \pit(\surd3/2)] y_2 = \rule{1em}{1pt} .005 m sin \pit \rule{1em}{1pt}.00866 cos \pit The resultant wave motion of the particle is, y = y_1 + y_2 = (.03 m) sin \pit \rule{1em}{1pt}(.005 m) sin \pit \rule{1em}{1pt}(.00866 m)cos \pit = (.025m) sin \pit \rule{1em}{1pt}.00866 cos \pit(3) We will write this in the form, y = A sin (\pit +\textphi) = A sin \pit cos \textphi+ A cos \pit sin \textphi(4) Comparing (4) with (3), A cos \textphi = .025 m A sin \textphi = \rule{1em}{1pt}.00866 m A^2 (sin^2 \textphi + cos^2 \textphi) = A^2 = (.025 m)^2 + (.00866 m)^2 A = .0264 m and, tan \textphi= (\rule{1em}{1pt} .00866/.025) = \rule{1em}{1pt} .346 \rule{1em}{1pt}tan \textphi = .346 But, \rule{1em}{1pt}tan \textphi = tan (\rule{1em}{1pt}\textphi) and tan (\rule{1em}{1pt}\textphi) = .346 \rule{1em}{1pt}\textphi= tan^\rule{1em}{1pt}1 (.346) \rule{1em}{1pt}\textphi= 19,1\textdegree

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Question:

Ethanol and methyl ether are isomers, having the molecular formula C_2H_6O. Based on structure, which one would be expected to have the higher vapor pressure? The higher boiling point? The greater solubility in water? Explain why in each case.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E20-0746.htm

Solution:

The ethers are a class of compounds that contain two hydrocarbon groups bound to an atom of oxygen. The general formula for ethers is ROR' where R and R' are alkyl groups and may or may not be the same Both alcohols and ethers derived from alkanes have the same general formula C_nH_2n+2O (ethanol is isomeric with methyl ether). Unlike alcohols, ethers do not form associated molecules by means of hydrogen bonds. Thus, the boiling point of an ether is considerably below the boiling point of its isomeric alcohol. Methyl ether is a gas, bp -23\textdegreeC, whereas ethanol is a liquid, bp + 78\textdegreeC. Therefore, it is reasonable to expect that methyl ether has a higher vapor pressure than ethanol. Weak hydrogen bonding between water molecules and ether molecules leads to slight water solubility. The functional group that is characteristic of ethers, the C O C group, is chemically inactive compared to the hydroxyl group of alcohols. Thus, ethanol has the greater solubility.

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Question:

The computer is a PDP-8. The program counter (PC) is at 400. The contents of the various memory locations are as follows: Show what will be stored in the AC in each case if the content of location 400 is:

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Solution:

a) The instruction is 1152. This instruction is in octal. Writing it out in binary, we get 001001101010. The first bit on the left hand side is bit #0. The right--most bit is bit #11. Bit #4 is 0. This means the ad-dress is on Page 0. Hence, the last 7 bits give the address = 152. Bit #5 is also 0. This means that it is direct ad-dressing Hence, location 152 contains the operand. Thus, the AC gets the value of location 152, which is 1050. b) The instruction is 1352, in octal. Hence, in bi-nary the instruction is:001011101010. Bit #4 is 1. This means that the address is on the same page as the in-struction. But, the instruction is in location 400, i.e., on Page 2. Hence, the address is also on Page 2. Hence, the address is given by the first five bits of the PC concatenated with the last seven bits of the instruction. The program counter contains 0400, i.e., 000100000000. And, the instruction is: 001011101010. Hence, the add-ress is 000101101010, i.e., 0552. Now, the contents of location 552 are 320. Therefore 320 gets stored in the AC. c) The instruction 1552 means: 001 101 101 010. Bit #4 is 0, therefore address is on Page 0. Thus, the ad-dress is: 00000 1101 010 = 000 001 101 010 = 0 152 And, the Bit #3 is 1. This means it is Indirect Addressing. Hence, the location 0152 on Page 0 contains not the operand itself, but another address where the operand can be found. The content 1050 of location 152 is therefore an add-ress. And hence, the content of location 1050, viz., 5 is the operand which goes to the AC. Therefore it indicates the same page as the instruction at 400, i.e., at 000 100000000. Hence, the address is: 000101101010 = 000 101 101 010 = 0552. Now, the #3 bit is 1. Therefore it means indirect address-ing, and the content of 552 is an address of the op-erand. The content of 552 is: 320. Hence, 320 is the address of the operand. Now, the content of 320 is 3620. Hence, 3620 is the operand which goes to the AC. Hence the AC contains the following:

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Question:

The radioactive decay constant for radium is 1.36 × 10^-11. How many disintegrations per second occur in 100 g of radium?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E19-0720.htm

Solution:

The number of disintegrations per second of a given amount of a particular element can be determined by using the following equation. D =\lambdaN where D is the number of disintegrations per second, \lambda the decay constant and N the number of atoms present. In this problem one is given \lambda and must determine N before solving for D. One is told that 100 g of radium is present, the number of moles present is determined by dividing 100 g by the molecular weight of radium, 226 g/mole. no. of moles = [(100 g) / (226 g/mole)] = 0.442 moles There are 6.02 × 10^23 particles per mole, thus, one can calculate the number of atoms (N) in 0.442 moles. N= (0.442moles) (6.02 × 10^23 atoms/mole) = 2.66 × 10^23 atoms Solving for D: D=\lambdaN = (1.36 × 10^-11) (2.66 × 10^23) = 3.62 × 10^12dis/sec.

Question:

Explain with illustrations how data are read in (acquired) , writtenout (disposed of) and referred to in array structures.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G13-0326.htm

Solution:

A matrix (i.e. a two dimensional array) is de-clared by the statement DCL X [9, 9] FIXEDDECIMAL(4); Now if a problem requires that the values of the main diagon-al of the matrixbe read in, there are two techniques for doing it.i) Use of a single getstatement of the form given below: GET LIST(X(1,1), X(2,2), X(3,3), X(4,4), X(5,5), X(6,6), X(7,7) , X(8,8) , X(9,9)) , or, ii) the iterative DO group illustrated below: DO I = 1 TO 9 BY 1; GET LIST X [I, I] ; END; The first method has the disadvantage that a long data specification is required. The second method is short and flexible. One of the ways to acquire data for one dimensionalarraysis as follows: GETLIST(N); DO I = 1 TO N BY 1; GETLIST(A(I)); END; If it is a two dimensional array then the following program segment can be used: GETLIST(M,N); DO I = 1 TO M; DO J = 1 TO N; GET LIST(X(I,J)); END; END; This method can be generalized to a multidimensional array: Take the upperbounds of all the dimensions. Then set DO loops with the most frequentlyvarying dimension at the in-nermost position and the least frequentlyvarying dimension at the outermost position. There is another technique provided by the PL/I machine, called repetitivedata specification, which reads in an en-tire array in just one GET statement. This is shown below: GETLIST(N, {X(I) DO I = 1 TO N}) ; Because of the left to right order, the variable N will be acquired beforeexecution of the DO-loop, where it is used as the upper limiting value. If it is a two dimensional array, for exampleX(10, 4) the statement willbe as follows: GETLIST((X (I, J) DO J = 1 TO 4) DO 1 = 1 TO 10); The repetitive specification consists of the following: a)thedata item to be used during execution of the DO portion (e.g., X(I) , Y(I,J)); b)thecontrol information which determines the number of times the dataitem is to be used for reading/writing values of the elements. (e.g. DO 1 = 1 TO N BY 1); c)apair of parentheses that enclose the repetitive specification. Similarly, to write out (dispose of) array elements, the repetitive dataspecification can also be used e.g., DECLARE (A,B)(10) FLOAT DECIMAL(5); PUTLIST[{B(I), A(I) DO I=1 TO 10}] SKIP; The value ofB(1) is printed out first, followed by A(1) , and then a line is skipped. After testing the limiting con-dition, iteration continues and outputs B(2), A(2) and so on. e.g.,DCL (A(100,4), B (100)) FIXED BIT(8); PUT DATA( (B(J) , (A(J,K) DO K = 1 TO 4) DO J=1 TO 100)); The put statement produces an output stream as follows: B(1) =XXXA(1,1) =XXXA(1,2) =XXXA(1,3) =XXXA(1,4) =XXX B(2) =XXXA(2,1) =XXXA(2,2) =XXXA(2,3) =XXXA(2,4) =XXX \textbullet\textbullet\textbullet\textbullet\textbullet \textbullet\textbullet\textbullet\textbullet\textbullet \textbullet\textbullet\textbullet\textbullet\textbullet To refer to an element of array the following is done: a) A subscripted reference, also termed a subscripted name, is used. b) The subscript list must contain as many subscripts as there are dimensionsdeclared for the array. c) A subscripted reference consists of the name of the array followedby a list of one or more subscripts enclosed in parentheses. d) The integer value resulting from a subscript evalu-ation must fall withinthe bounds declared by the correspond-ing boundary specification. To understand clearly the above rules of referring to an element of anarray, consider the example given below: DECLARE A(10,10) FIXED DECIMAL(10,2), B(-1:5,-5:0,10:98) The subscript references A(3,5) A(1,2×5 - 2) B(2,- 3,40) B[- (- 6 + 3),0,90] areall well formed and valid. Elements of the array A are referred by the arrayname A. Since A is a two dimensional array, the subscript reference hastwo subscripts and the in-teger value of the subscript is within the boundsdeclared for the array. So is the case for a reference made to arrayB. Now observe the following: The subscript references A(1,2,3) A(5) A(12,- 3) arenot valid. In the first case,A(1,2,3), there are three subscripts, alluding toa three dimensional array. But ar-ray A is declared as 2 dimensional. The same goes for the second case, but here, the subscript reference has onlyone dimension instead of the 2 which are required. In the third case, thereare two subscripts and array A is also of two di-mensions. However, boththe dimensions of array A have a lower bound of 1 and a higher boundof 10. Hence, 12, the first subscript element is greater than the higherbound, and - 3, the second subscript element is less than the lower bound1. This is not permissible.

Question:

A small ball swings in a horizontal circle at the end of a cord of length l1which forms an angle \texttheta_1 with the vertical. The cord is slowly shortened by pulling it through a hole in its support until the free length is l_2 and the ball is moving at an angle \texttheta_2 from the vertical, a) Derive a relation between l_1, l_2, \texttheta_1 and \texttheta_2. b) If l_1 = 600 mm, \texttheta_1 = 30\textdegree and, after shortening, \texttheta_2 = 60\textdegree, determine l_2.

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Solution:

In the system shown,l_n is the cord length, \texttheta_n is the cone angle cut out by the cord as it revolves, r_n is the distance from the mass to the axis of rotation, and v_n is the velocity of the mass. To find the relationship between l_1, l_2, \texttheta_1 and \texttheta_2 when the cord is shortened from l_1 to l_2, one should first analyze a free body diagram of the particle. Since the particle is in a circular motion, the inward component of T must supply the centripetal force, m v^2_n / r_n. T_n sin\texttheta_n = m v^2_n / r_n(1) Also, the vertical component of T must balance the weight, T_n cos \texttheta_n = m g(2) Dividing (1) by (2) yields: Tan \texttheta_n = (m v^2_n / m g r_n ) = (v_n^2 / g r_n )(3) One more relationship is needed to solve the problem. Since no external torques act on the system, angular momentum L is conserved. L_1 = L_2 m r_1 v_1 = m r_2 v_2 r_1 v_1 = r_2 v_2.(4) If one rearranges (3), g r_n^3 tan \texttheta_n = rn^2 v_n ^2.(5) From (4), it follows that (r_1 v_1)^2 = (r_2 v_2)^2.(6) Combining (5) and (6), g r_1^3 tan \texttheta_1 = g r2^3 tan \texttheta2 .(7) Finally, from geometry, r_n = l_n sin \texttheta_n(8) Using this relation in (7) and dividing both sides by g pro-duces the desired result. (l_1 sin \texttheta_1)^3 tan \texttheta_1 = (l_2 sin \texttheta_2)^3 tan \texttheta_2. In this problem, l_1 = .6m, \texttheta_1 = 30\textdegree, \texttheta_2= 60\textdegree Solving for l_2 yields l_2 = 3\surd[{(l_1 sin \texttheta_1)^3(tan \texttheta_1)} / (sin \texttheta_2 ^3 tan \texttheta_2)] l_2 = [(l_1 sin \texttheta_1) / (sin\texttheta_2)]3\surd[(tan \texttheta_1) / (tan \texttheta_2)], and substituting in the numbers yields l_2 = [{.6m (sin 30)} / (sin 60\textdegree )] 3\surd[(tan 30\textdegree) / (tan 60\textdegree)] = [{.6m (.5)} / (.87)] 3\surd[(.58) / (1.73)] = .24 m.

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Question:

A satellite of mass 3 ×10^3 kg moves with a speed of 8 × 10^3 m/s in an orbit of radius 7 × 10^6 m. What is the angular momentum of the satellite as it revolves about the earth?

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Solution:

The angular momentum of an object about a point 0 is defined as L\ding{217}= r\ding{217}× p\ding{217} \ding{217} \ding{217} \ding{217} where p\ding{217}is the linear momentum of the object and r\ding{217}is the distance \ding{217} \ding{217} vector from 0 to the object. In this problem, r\ding{217}and p\ding{217}are perpendicular, \ding{217} \ding{217} hence the magnitude of L\ding{217}is \ding{217} L = rp = r(mv) = mvr. The known observables are mass, m = 3 × 10^3 kg; speed v = 8 × 10^3 m/s; and orbit radius, r = 7 × 10^6 m; therefore, L = mvr = (3 × 10^3 kg) (8 × 10^3 m/s) (7 × 10^6 m) = 1.68 × 10^14 Js.

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Question:

Plant cells are able to withstand much wider fluctuation in the osmotic pressure of the surrounding medium than animal cells. Explain why.

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Solution:

A plant cell is able to withstand much greater fluctuations in the makeup of the surrounding fluids than can an animal cell. This can be best under-stood by examining the role played by the plants' cell wall, a structure absent in animal cells. First, take the case in which the cell is placed in a hypotonic medium. This means that the surrounding fluid has a lower concentration of solute molecules and a higher concentration of solvent (water) than the cell. Thus, the cell has a higher osmotic pressure than the medium, which means that the net movement of water should be into the cell. This is exactly what happens in an animal cell (see Figure 1). An animal cell will take in water by osmosis causing it to swell. It will take in enough water so that the osmotic pressure of the cell is equal to the osmotic pressure of the medium, thus they will be isotonic to each other. However, if the original differ-ence in pressure was great, the animal cell may have to take in more water than its membrane can allow. Then the animal cell would burst. This is called lysis. If the original cell was a red blood cell, this is hemolysis. A plant cell placed in a hypotonic medium would also have water enter into it, causing it to swell, (see Figure 2). However, an upper limit as to how much water can enter is imposed by the cell wall. As the cell swells, its plasma membrane exerts pressure on the cell wall which is called turgor pressure. The wall exerts an equal and opposing pressure on the swollen membrane. When the pressure exerted by the cell wall is so great that further in-crease in cell size is not possible, water will cease to enter the cell. Thus, plant cells will only absorb a certain amount of water, even in an extremely dilute medium. Figure 1 A typical animal cell in varying environments. Now consider a cell placed in a hyperosmotic medium. In this case, the medium has a greater osmotic pressure than the cell. Thus, the net movement of water will be out of the cell and into the medium. Since the cell is losing water it will shrink. In an animal cell this phenomenon is called crenation (see Figure 1). In a plant cell, the shrinkage will cause the plasma membrane to pull away from the cell wall. The cell is said to be plasmolyzed, and the phenomenon is plasmolysis (see Figure 2).

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Question:

Look at the followingpseudocodedprogram. Can you think of an instance in which the program would get stuck in an infinite loop ? Modify the program to deal with this possibility. integer N, NSUM NSUM \leftarrow 0 input N do while N \not = 0 NSUM \leftarrow NSUM + N N \leftarrow N - 1 end do while output NSUM end program

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Solution:

A problem arises if N is a negative number. Say we input for N the value -6. On the initial pass through the loop, NSUM takes on a value of -6. But when the next statement is encountered, N becomes -7. The loop will continue to decrement the value of N so that N will never equal 0. Hence, the loop is Infinite as written. Probably the programmer wants to sum the digits between -6 and 0. To accomplish this, we can include an IF-THEN-ELSE clause in the DO- WHILE construct: do while N \not = 0 NSUM \leftarrow NSUM + N if N < 0 then N \leftarrow N + 1 else N \leftarrow N - 1 end if end do while Now the program can handle both positive and negative values of N.

Question:

Live bacteria, all of which were red in color, were placed under an ultra-violet lamp. After several days, groups of white bacteria began to appear among the red. What conclusions, if any, can be made at this stage? What further experiments should be performed?

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/Users/wenhuchen/Documents/Crawler/Biology/F24-0635.htm

Solution:

At this stage, we can guess that the white bacteria probably arose as the result of a mutation. We know that exposure to ultraviolet radiation usually causes an increase in the mutation rate, i.e., it is said to induce mutations. We also know that the white bacteria arose from among the red, and in the same culture medium. If the original culture was indeed genetically pure (all the individuals having come from the same cell and thus having identical genotypes), there is no reason for any of the bacteria to exhibit different behavior or charac-teristics while remaining in the same environment. A change in phenotype could only be caused by a change in the genetic structure, in this case, a genetic mutation. In order to be absolutely sure that the change in color is indeed caused by a genetic mutation, we can test the permanence of the change. If there has been a change in genetic structure, it must necessarily be passed on to offspring of the cell with the original mutation. We can remove the white cells from the culture and grow them in a separate, identical medium. If they reproduce true- to-type, and produce a colony of only white bacteria, we can be sure that a mutation has indeed occurred. Of course there is always a possibility that back mutation will occur, with a reversion to red color, but this is very rare. It is important to note that while we know that UV radiation is a potential mutagen (agent inducing mutation), we cannot be entirely sure that all of the mutations resulted from UV rays. Some could have arisen as a result of a spontaneous change. Spontaneous muta-tions are chance occurrences, which are not caused by a specific external agent. The facts of the experiment, however, do fit well with what we know of the mechanisms of UV-induced mutations. Ultraviolet rays are theorized to affect the pyrimidines in DNA. They do this by (1) hydrolysing cytosine to a product that may cause mispairing, and (2) joining thymine molecules together to form dimers which block normal transcription from occur-ring. If such changes did occur, they would affect the production of substances that is coded for genetically. In this case, the gene product could have been a specific enzyme necessary for the production of red color. Its absence would account for the reversion to white color in the mutants. Of course an equally probable cause would be the production of an enzyme responsible for the whitecolor morphology in the mutants.

Question:

A boat travels directly upstream in a river, moving with constant but unknown speed v with respect to the water. At the start of this trip upstream, a bottle is dropped over the side. After 15 minutes the boat turns around and heads downstream. It catches up with the bottle when the bottle has drifted one mile downstream from the point at which it was dropped into the water. What is the current in the stream?

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/Users/wenhuchen/Documents/Crawler/Physics/D04-0212.htm

Solution:

Consider a coordinate system at rest with respect to the water. Then the water is at rest and it is the banks which appear to move upstream. The bottle is at rest with respect to the water. From the point of view of this coordinate system, it is just as though the boat were moving at speed v in a perfectly still pond. We can see that the return trip downstream must also take 15 minutes. Once it is known that the round trip takes half an hour, we can see that the current in the river must be 2 miles per hour since the bottle moves one mile in a half hour.

Question:

What is the available energy for the \alpha decay of Po^210 ?

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/Users/wenhuchen/Documents/Crawler/Physics/D34-1023.htm

Solution:

Some unstable nuclei, especially those with mass numbers above 200, spontaneously emit helium nuclei (\alpha particles). Emission of an \alpha particle by a nucleus decreases the original nuclear charge particles Z (protons) by two and decreases the original mass number A (protons plus neutrons) by four. N^A _Z \rightarrowNA\Elzbar4Z\Elzbar2+ He^4 _2 If N^A _Z has a greater mass than the combined mass of NA\Elzbar4Z\Elzbar2and He^4 _2 it is unstable and can decay by the emission of an \alpha particle. The available energy for the \alpha-decay process is given by the mass available for mass- energy conversion. Alpha-particle emission from Po^210 leaves Pb^206 . Therefore, the pertinent masses are: m(Po^210) = 209.98287 AMUm(Pb^206) = 205.97447 AMU m(He^4 ) =4.00260 AMU m(Pb^206) + m(He^4 ) = 209.97707 AMU The available energy\epsilon_ais: \epsilon_a= [m(Po^210)\Elzbar m(Pb^206)\Elzbar m(He^4 )] × c^2 \epsilon_a= (209.98287 AMU\Elzbar 209.97707 AMU) × c^2 = (0.00580 AMU) × (931.481MeV/AMU) = 5.40MeV Actually, this decay energy is shared by the \alpha particle and the Pb^206 nucleus (because the linearmo-mentaof the two fragments must be equal and opposite). Consequently, the \alpha particle emitted by Po^210 has a kinetic energy of 5.30 MeVand the recoil Pb^206 nucleus has a kinetic energy of 0.10MeV.

Question:

An equilibrium solution of the complex ion Ag(NH_3)_2^+ contains 0.30 M of NH_4^+ and 0.15 M of the actual ^+complex ion. To obtain a concentration of Ag^+ equal to 1.0 × 10^-6 M, what must the pH of this solution be?K_dissof Ag(NH_3)_2^+ = 6.0 × 10^-8, K_b of NH_3 = 1.8 × 10^-5 and K_W = 1 × 10^-14.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E10-0382.htm

Solution:

To find the pH required, calculate the [H^+] (pH = - log [H^+]). To find [H^+], the [OH^-] is needed. One is given K_W, which is the constant for the autodissociationof H_2O. K_W is defined as [H^+][OH^-] for H_2O \rightleftarrows H^+ + OH^-. If [OH^-] was known, one could solve for [H^+]. To find [OH^-], consider the K_b value given. When bases, such as NH_3, are placed in water, these bases dissociate into positive and negative ions. The K_b, the equilibrium constant, measures the ratio of the concentration of products to reactants, each raised to the power of its coefficient in the chemical reaction. When NH_3 is placed in water, following dissociation exists: NH_3 + H_2O \rightleftarrows NH_4^+ + OH^- .Thus, K_b = {[NH_4^+][OH^-]} / [NH_3] . Note: H_2O is omitted; it is assumed to be a constant. There-fore, if [NH_4^+]/[NH_3] was known, one could find [OH^-]. To find [NH_4^+]/[NH_3], calculate [NH_3] since [NH_4^+ ] is given. To calculate [NH_3], use the equilibrium constant expression for the complex ion dissociation. This constant,K_diss_' is defined in the same way as K_b (i.e., ratio of products to reactants). The problem provides all the information needed to find [NH_3]. Proceed as follows: The complex ion dissociation is similar in nature to other equilibrium dissociations. Ag (NH_3)_2^+ \rightleftarrows 2NH_3 + Ag^+.As such, K_diss= {[Ag^+][NH_3]^2} / [Ag(NH_3)_2^+] . As given, [Ag(NH_3)_2^+] = 0.15 M, one wants [Ag^+] = 1.0 × 10^-6 M at equilibrium. Thus, substituting these values, one finds that K_diss= 6.0 × 10^-8 = {(1.0 × 10^-6)[NH_3]^2}/(0.15) Solving for [NH_3], one obtains [NH_3] = \surd[{(6.0 × 10^-8)(0.15)}/(1.0 × 10^-6)] = 9.5 × 10^-2 . One is told K_b = 1.8 × 10^-5 . By definition, K_b = {[NH_4^+][OH^-]}/[NH_3] , [NH_3] has been found and [NH_4^+] is given. Thus, by sub-stitution one can solve for [OH^-]. 1.8 × 10^-5 = {(0.30) [OH^-]}/(9.5 × 10^-2)]or [OH^-] = [{(1.8 × 10^-5) (9.5 × 10^-2)}/(0.30)] = 5.7 × 10^-6 . Recalling that K_W = 1.0 × 10^-14 = [OH^-][H^+], one has 1. 0 × 10^-14 = [H^+][5.7 × 10^-6] Solving,[H^+] = [(1.0 × 10^-14)/(5.7 × 10^-6)] = 1.75 × 10^-9 SincepH = - log [H^+], pH = - log [1.75 × 10^-9] = 8.76 . Thus, bring the pH up to 8.76 to obtain an [Ag^+] = 1.0 × 10^-6 M.

Question:

snake, lizard, salamander , turtle or alligator.

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/Users/wenhuchen/Documents/Crawler/Biology/F13-0339.htm

Solution:

The classReptiliais divided into four orders. Turtles are reptiles belongingto the orderChelonia; crocodiles and alligators belong to the OrderCrocodilia; lizards and snakes are in the OrderSquamata; and the tuatarabelong to the OrderRynchocephalia, which has only one surviving species. Of the five animals mentioned in the question, only the salamanderis not a reptile. The salamander is an amphibian in the Order Urodelia.By anatomical observation, it is easily deduced that the salamanderis not of the same class as these other animals. The smooth andmoist porous skin of the amphibious salamander contrasts sharply withthe coarse dry skin of the reptilian animals. This epidermal variation is basedupon the different environmental adaptations that these animals underwent. The skin of the amphibian is adapted for a semi-aquatic environmentwhereas that of the reptile is adapted for a strictly terrestrial environment.

Question:

Explain how density dependent controls of population occur.

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/Users/wenhuchen/Documents/Crawler/Biology/F30-0769.htm

Solution:

An important characteristic of a population is its density, which is thenumber of individuals per unit area. A more useful term to ecologists is ecologicdensity, which is the number of individuals per habitable unit area. As the ecologic density of a given population begins to increase, thereare regulatory factors that tend to oppose the growth. These regulatingmechanisms operate to maintain a population at its optimal size withina given environment. This overall process of regulation is known as thedensity dependent effect. Predation is an example of a density dependent reg-ulator. As the densityof a prey species rises, the hunting patterns of predators often changeso as to increase predation on that particular population of prey. Consequently, the prey population decreases; the predators then are left withless of a food resource and their density subsequently declines. The effectof this is often a series of density fluctuations until an equilib-rium is reachedbetween the predator and prey populations. Thus, in a stable predator-preysystem, the two populations are actually regulated by each other. Emigration of individuals from the parent population is another form ofdensity dependent control. As the population density increases, a larger numberof animals tend to move outward in search of new sources of food. Emigration is a distinctive behavior pattern acting to disperse part of theovercrowded population. Competition is also a density dependent control. As the population densityincreases, the competition for limited resources becomes more intense. Consequently, the deleterious results of unsuccessful competition suchas starvation and injury become more and more effective in limiting thepopulation size. Physiological as well as behavioral mechanisms have evolved that helpto regulate population growth. It has been observed that an increase inpopulation density is accompanied by a marked depression in inflammatoryresponse and antibody formation. This form of inhibited immuneresponse allows for an increase in susceptibility to infection and parasitism. Observation of laboratory mice has shown that as population densityincreases, aggressive behavior increases, reproduction rate falls, sexualmatu-rity is impaired, and the growth rate becomes suppressed. These effects are attributable to changes in the endocrine system. It appearsthat the endocrine system can help regulate and limit population sizethrough control of both reproductive and aggressive behavior. Although these regulatory mechanisms have been demonstrated with labo-ratorymice, it is not clear to what extent they operate in other species.

Question:

What is the weight of 1.0 liter of carbon monoxide (CO)at STP?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E04-0158.htm

Solution:

At STP (Standard Temperature and Pressure, 0\textdegreeC and 760torr) ,a moleof any gas has a volume of 22.4 liters. This means that the gram molecularweight of a gas is contained in 22.4 liters. In this problem, one is lookingfor the weight of 1 liter of CO. The molecular weight of CO is 28 g. Because there are 28 g of CO in 22.4 liters of gas,28g must be divided by 22.4tofind the weight of one liter. weightof one liter = gram molecular weight / 22.4 weightof 1 liter of CO = 28g / 22.4 = 1.25 g.

Question:

Why would you not expect conjugation among a group of paramecia that had descended from a single individual through repeated fission?

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/Users/wenhuchen/Documents/Crawler/Biology/F11-0273.htm

Solution:

Conjugation occurs among certainprotozoans, ciliates in particular. It is a method of exchanging genes between organisms. In paramecium conjugation may take place whenparameciumsof plus and minus strains are present in the same pond or culture. Conjugation will not however take place with two organisms of similar mating types. During conjugation two animals come together and adhere to each other in the oral region. Next, the micronuclei divide by meiosis. Then all but two of the resulting haploid micronuclei in each cell disintegrate; the macronucleus usually also disintegrates. One of the two nuclei in each cell remains stationary and functions as the female nucleus. The other, the male nucleus, moves into the conjugant cell and fuses with the female nucleus in that cell in the process of fertilization. Therefore, each cell acts as both male and female, donating one nucleus and receiving another, and when the two cells part, each has a new diploid nucleus. This nucleus then under-goes more divisions, and some of the new nuclei become macronuclei. Following somecytoplasmiccleavages, the normal nuclear number of the cell both micro and macro, is restored. As stated, conjugation involves the mixing of genes between organisms, and requires two different mating types. In a group of paramecia descended from a single individual through repeated fission, all individuals are identical genetically, and conjugation could not occur. The exchange of nuclear material through conjugation between these organisms would not result in greater genetic diversity and would, therefore, be pointless.

Question:

The masses of _1H^1 and _2He^4 atoms are 1.00813 amu and 4.00386 amu, respectively. How much hydrogen must be converted to helium in the sun per second if the solar constant is 1.35 kW\textbulletm^2 and the earth is 1.5 × 10^8 km from the sun?

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/Users/wenhuchen/Documents/Crawler/Physics/D36-1046.htm

Solution:

The energy falling on a 1- m^2 area at the distance of the earth from the sun, in 1 second, is 1.35 × 10^3 J. The total energy radiated from the sun is thus E = (1.35 × 10^3 J\textbullets^-1 )4\pi r^2 where 4\pir^2is the area of a sphere centered at the sun, with radius r equal to the separation distance of the earth and sun. E = 1.35 × 10^3 J\textbulletm^-2 × 4\pi × 1.5^2 × 10^22 m^2 = 3.82 × 10^26 J If four atoms of hydrogen combine to give one atom of helium, energy is released. This follows because the rest mass energy of 4 hydrogen atoms is greater than that of 1 helium atom. From Einstein's mass-energy relation, this excess rest mass energy must be released as energy every time 4_1H^1 atoms combine to form 1 _2He^4 atom. The energy released is then E' = [(4 × 1.00813 - 4.00386)amu]c^2 But 1 amu= 1.66 × 10^-27kg, whence E' = [(4.03252 - 4.00386)amu] (1.66 × 10^-27 kg/amu)× (9 × 10^16 m^2/s^2) E' = (.02866)( 14.94 × 10^-11 J) E' = 4.27 × 10^-12 J Hence, for every 4 _1H^1 atoms used, an amount of energy equal to E' is released. The number of hydrogen atoms converted per second to produce the energy radiated by the sun is thus n = 4 × (E/E '), and the mass of hydrogen converted per second is M = n × (mass of 1 _1H^1 atom) Noting that 1 _1H^1 atom has a mass m = 1.00813 amu we findM = n × m = 4Em / E' M ={(4) (3.82 × 10^26 J) (1.00813 amu)} / (4.27 x 10^-12 J) Using the fact that 1 amu = 1.66 × 10^-27kg M = {(4) (3.82 × 10^26 J) (1.00813 amu) (1.66 × 10^-27kg / amu)} / (4.27 x 10^-12 J) M = 5.99 × 10^11 kg

Question:

A man standing on the roof of a building 30 meters high throws a ball vertically downward with an initial velocity of 500 cm sec^-1 as it leaves his hand (see the figure). The acceleration due to gravity is 9.8 m sec^-2. (a) What is the velocity of the ball after it has been falling for 0.5 second? (b) Where is the ball after 1.5 second? (c) What is the velocity of the ball as it strikes the ground?

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/Users/wenhuchen/Documents/Crawler/Physics/D03-0080.htm

Solution:

We will use the MKS system of units because most of the given quantities are expressed in these units. Then the initial velocity must be expressed as 5 meters per se-cond since 1 cm/s = .01 m/s. Place the origin at the top of the building. Then x = 0 when t = 0. Let the positive direction of x be downward. The initial velocity is down-ward and therefore positive, so v_0 = +5 m sec^-1. The acceleration is downward and therefore positive, The acceleration is downward and therefore positive, so a = +9.8 m sec^-2 so a = +9.8 m sec^-2 . (a) In this part of the problem one is given v_0, a,and t, and must deduce v. Since the acceleration is con-stant, we may use the kinetmatics equations for constant acceleration, or v = v_0 + at = [+5 (m/sec)] + [+9.8 (m/sec2)] (+0.5 sec) = 5 (m/sec) + 4.9 (m/sec) = +9.9 m/sec. After 0.5 second the velocity is 9.9 m sec^-1 downward. (b) In this part of the problem one is given v_0, a, and t, and must calculate x. By definition of velocity , v = dx/dt . Therefore, ^x\int_x0 dx = ^t\int_t=0 v dt where x_0 is the initial position of the ball. Using the formula for v given in the previous part, we have x - x_0 = ^t\int_0 (v_0 + at) dt or x = x_0 + v_0t + (1/2)at^2. Because x_0 = 0 x = v_0t + (1/2)at^2 = {+5 (m/sec)} (+1.5 sec) +[(1/2) {+9.8 (m/sec^2)}] (+1.5 sec)^2 = 7.5 m + {4.9 (m/sec^2)} (2.25 sec)^2 = 7.5m + 11.025 m = 18.525 m. After 1.5 second the ball is 18.525 meters below the roof or 11.475 meters above the ground. (c) When the body strikes the ground x = +30 m. So one is given v_0, a, and x, and asked to calculate v. The correct equation, then, should not contain t as a variable. The equation to be used is, since a= constant, v^2 = v^2_0 + 2ax = {+5 (m/sec)}^2 + 2 {+9.8 (m/sec^2)} (+30 m) = 25 (m^2/sec^2 )+588 (m^2/sec^2 ) v^2 = 613 (m^2/sec^2 ) v= 24.76 m/sec. When it strikes the ground, the ball has a velocity of 24.76 m sec^-1.

Question:

Predict the total spin for each of the following electronic configurations:(a) 1s^2 2s^1;(b) 1s^2 2s^2 2p^3; and (c) 1s^2 2s^2 2p^6 3s^2 3p^6 3d^5 4s^2.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E18-0652.htm

Solution:

One determines the total spin of an atom by calculating the sum of all the s quantum numbers of the electrons. There are four quantum numbers needed to describe each electron. They are the principal quantum number n, the azimuthal quantum number l, the magnetic quantum number m and the spin quantum number s. In each orbital of the atom two electrons can be accommodated. These electrons will have opposing spins of (+1/2) and (-1/2). The total spin is calculated by determining the total number of unpaired electrons and multiplying this number by (1/2). It is assumed that all unpaired electrons will have the same spin. Solving for the total spins: Figure A shows the total number of electrons that can occupy each orbital. (a)1s^2 2s^2. Filled s orbitals each contain 2 electrons. One writes this electronic configuration diagrammatically as shown in figure B. Because there are no unpaired electrons the total spin for this atom is 0. (b)1s^2 2s^2 2p^3. Here the two s orbitals are filled but the p orbital is only half filled. There are three subshells in the p orbital and there will be one electron occupying each of these. The configuration of lowest energy is assumed when each of these 3 electrons has the same spin. This configuration is written diagrammatically as shown in figure C. This configuration contains 3 unpaired electrons. The total spin is, thus 3 × (1/2) or (3/2). (c)1s^2 2s^2 2p^6 3s^2 3p^6 3d^5 4s^2. From figure A one sees that the only unfilled orbital in this configuration is the 3d. Therefore, unpaired electrons can only occur in this shell. From figure A, one sees that the 3d orbital can contain 10 electrons. Because each subshell contains 2 electrons, there are 5 subshells in this orbital. In this configuration there are 5 electrons occupying the 3d orbital, therefore, each subshell con-tains 1 electron and there are 5 unpaired electrons in the configuration. The total spin of this configuration is, thus, 5 × (1/2) or (5/2).

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Question:

When the atria contract, only 30% of the ventricular space becomes filled. How does the additional 70% become filled? Describe the process of the cardiac cycle in general.

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/Users/wenhuchen/Documents/Crawler/Biology/F15-0377.htm

Solution:

The cardiac cycle, is the period from the end .of one con-traction to the end of the next contraction. Each heartbeat consists of a period of contraction or systole, followed by a period of relaxation, or diastole. Since the normal heart rate is 70 beats per minute, each complete beat (systole and diastole) lasts .85 second. The atria and the ventricles do not contract simultaneously. The atria contract first, withatrialsystole lasting .15 seconds. The ventricles then con-tract, with ventricular systole lasting .30 second. Diastole therefore lasts .40 second. The cardiac cycle begins withatrialsystole. The wave-like contraction of the atria is stimulated by the impulse of thesino-atrialor S-A node. The contraction forces into the ventricles only 30% of the blood that will fill them. How the other 70% enters the ventricles will be explained shortly. During contraction of the two atria, the tricuspid and bicuspid valves are opened. There is a very brief pause before ven-tricular systole begins due to the delay in transmission of the cardiac impulse at the atrio-ventricular or A-V node. The impulse is then transmitted to the A-V bundle and bundle branches and then very rapidly to the Purkinje fibers, which stimulate the ventricles to contract as a unit. The contraction of the ventricular muscles causes a rapid increase in pressure in the ventricles. This increased ventricular pressure immediately closes the tricuspid and bicuspid valves, preventing backflow of the blood into the atria. It is important to realize that the valves work passively. There is no nervous stimulation which directly regu-lates their opening and closing; it is only the difference in pressure due to the relative amount of blood in the atria and ventricles (or ventricles and arteries) which controls this. As ventricular systole progresses, the ventricular pressure further increases (due to contraction). At this point, no blood is en-tering the ventricles, since the tri- and bicuspid valves are closed, and no blood is leaving, since the semilunarvalves are closed. When the-ventricular pressure becomes greater than the pressure in the arteries (the pulmonary artery and the aorta), thesemilunarvalves open and blood is forced into these vessels. After the ventricles complete their contraction, ventricular diastole begins. As the ventricles relax, the pressure decreases. When the ventricular pressure is less than arterial pressure, thesemilunarvalves shut, preventing backflow. Since the pressure in the ventricles is still higher than in the atria, the tri- and bi-cuspid valves are still closed, and no blood enters or leaves the ventricles. But some blood is entering the now relaxed atria. During further relaxation of the ventricles, the ventricular pres-sure continues to decrease until it falls below theatrialpressure. At this point, the tri- and bicuspid valves open, allowing blood to rapidly flow into the ventricles. The atria do not contract to force this blood into the ventricles. It is a passive flow due to the fact that theatrialpressure is greater than the ventricular pressure. This flow allows for 70% of the ventricular filling beforeatrialsystole. Thus, the major amount of ventricular fil-ling occurs during diastole, notatrialsystole, as one might expect.

Question:

Find the entropy rise ∆S of an ideal gas of N molecules occupying a volume V_1, when it expands to a volume V_2 under constant pressure.

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/Users/wenhuchen/Documents/Crawler/Physics/D15-0524.htm

Solution:

IfdQis the heat added to the system,dWis the work done by the system anddUis the change in the internal energy of the system, the first law of thermo-dynamics states that dQ =dU+dW, orTdS=dU+PdV,(1) where S is the entropy, and P, V, T are, respectively, the pressure, the volume and the temperature of the gas. The internal energy of an ideal gas (whose molecules have only translational motion) is U = (3/2) NkT NkT where k is Boltzmann's constant. Since for an ideal gas PV =NkT,(2) its internal energy can be expressed as U = (3/2)PV. If P is constant, we get dU= (3/2)PdV.(3) Substituting (3) in (1), TdS = (3/2)PdV+PdV= (5/2)PdV ordS= (5/2)(P/T)dv. From (2), we see that P/T =Nk/V.Therefore dS = (5/2)Nk(dV/V)(4) Integrating (4) between initial and final states as the gas expands, we get the entropy rise ∆S = (5/2)Nk^v2\int_v1 (dV/V) = (5/2)Nk1n (V_2/V_1).

Question:

a) a block diagram,b) a timing diagram, c) a Karnaugh Map,d) an excitation equation, e) a state table,andf) a state diagram.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G04-0081.htm

Solution:

The block diagram is shown in figure 1. The slave gets the inverted clock pulse. The input to the slave is taken from the x'-output of the master. b) The timing diagram is shown in figure 2. Normally a D flip-flop behaves simply as a delay line. However, here we have fed x' of the master as the input to the slave. Hence, our master-slave combination gives an inversion plus a delay (see D and Y). c) The Karnaugh Map is drawn as shown in figure 3. d) From the Karnaugh Map, we write down the excitation equation as: y\rightarrow= D. \rightarrow e) The state table can be prepared as shown in figure 4: Dy y\rightarrow \rightarrow REMARKS 00 01 0 0 Y\rightarrowIS SAME AS D \rightarrow 10 11 1 1 Y\rightarrowIS SAME AS D \rightarrow f) The state diagram is also plotted with the help of the Karnaugh Map as shown in figure 5.

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Question:

Calculate B at the center o\pm a circular loop of wire (point P in the diagram).

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/Users/wenhuchen/Documents/Crawler/Physics/D21-0704.htm

Solution:

For this problem we apply the Biot - Savart Law which gives us the magnitude and direction of the magnetic field at a certain point due to a current carrying wirer dB^\ding{217} = [(\mu_0 i)/(4\pi)] [(d^\ding{217}l × r^\ding{217})/(r^3)] where r^\ding{217} is a displacement vector from a current element to P, and dl^\ding{217} is the length of this current element in the direction of current flow. Since the contribution from any current element will be in the same direction (into the page), we may add (inte-grate) these contributions directly neglecting the vector nature. The magnitude, dB, is given by dB = [(\mu_0 i)/(4\pi)] [(dl sin \texttheta)/r_2] where \texttheta is the angle between dl^\ding{217} and r^\ding{217}. Since we are dealing with a circular current loop, \texttheta is 90\textdegree and sin \texttheta = 1. The magnetic field strength at point P is found by integrating dB over the entire loop B = \int dB = \int [(\mu_0 i)/(4\pi)] [(dl)/r^2] But [(\mu_0 i)/(4\pi)] and r are constants. Therefore, B = [(\mu_0 i)/(4\pir^2)] \int dl. The integral of dl is just 2\pir, the circumference. Therefore B = [(\mu_0 i)/(2r)] (directed into the page).

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Question:

What is an organizer and how does it influence develop-ment ?

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/Users/wenhuchen/Documents/Crawler/Biology/F23-0598.htm

Solution:

An organizer is a chemical secreted by certain cells in the embryo, which diffuses into and influences surrounding cells. The presence of organizers is the whole basis for the concept of embryonic induction. In the frog and salamander, it was found that the dorsal lip of the blastopore was very important in organizing the development of the body parts . When a portion of this dorsal lip was transplanted to the ventral region of another embryo, it acted as an organizer. The cells surrounding it formed into a separate head, and part of the body. The result was a Siamese type of twin salamander. This demonstrated that the cells in one part of the body of the embryo had the power to form body parts characteristic of other regions, but they needed the stimulation of the organizer from the dorsal lip in order to function. Hans Spemann did pioneering work on organizers. The optic vesicle in the embryo, he noted, stimulates the adjacent ectoderm to form the lens of the eye. When the optic vesicle is removed from the head and inserted at any point under the skin of the embryo's body, it will organize the overlying ectoderm into a lens . This principle of embryonic induction also takes place in many other areas of the embryo and plays a role in the development of such tissues as the nervous system and the limbs.

Question:

An approach to controlling industrial pollution is to recover waste products in usable form. In the reaction, 2H_2S + SO_2 \rightarrow 3S + 2H_2O waste H_2S and SO_2 are recovered as elemental sulfur liquid. How much sulfur can be recovered if 650,000 kg of SO_2 are deposited into the air?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E24-0847.htm

Solution:

From the stoichiometry of this equation, 2molesof H_2S react with 1 mole of SO_2 to produce 3 moles of S and 2 moles of H_2O. To solve this problem, calculate how many moles of SO_2 are polluted into the air by dividing its weight by its molecular weight (MW SO_2 = 64 g/mole). From the stoichiometry of the equation, determine how many moles of S would be produced, and then calculate the weight of S by multiplying the number of moles of S times its molecular weight (MW S = 32 g/mole). Thus, number moles of SO_2= (650,000,000 g SO_2) / (64 g/mole) = 1.02 × 10^7 moles From the stoichiometry of the equation 1 mole SO_2 \rightarrow 3 mole S or 1.02 × 10^7 mole SO_2 \rightarrow 3.06 × 10^7 mole S. weight of S = (3.06 x 10^7 mole S) (32 g/mole) = 97.9 x 107 g = 979,000 kg.

Question:

Write a BASIC program to obtain unit fractions. Any proper fraction can be expressed as the sum of unit fractions, as in the example (17/21) = (1/2) + (1/4) + (1/17) + (1/1428)

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Solution:

For this problem, you can enter any two numbers to represent the numerator and denominator of the fraction you choose. Then, by factoring the fraction continually, the program will print out the unit fraction's denominators. If the fraction has too many factors (i.e., more than 29), the program will print out a message telling you to try again. The flowchart will illustrate the procedure. Based on the flowchart, the program looks as follows: 10 REM UNIT FRACTIONS BY FACTORING 11 DIM M(29) 15 PRINT "TYPE N AND D, REMEMBERING THAT D MUST BE GREATER THAN N". 16 PRINT 18 INPUT N,D 19 FOR J = 0 TO 29 20 LET M(J) = 0 21 NEXT J 22 IF D < = N THEN 14 23 LET K = 0 24 LET P = INT(D/N) + 1 35 LET M(K) = P 38 LET K = K + 1 40 IF K < 30 THEN 50 44 PRINT "TOO MANY FACTORS: TRY AGAIN" 45 PRINT 48 GO TO 14 50 LET R = N\textasteriskcenteredP - D 55 IF R = 0 THEN 75 56 LET G = D/10 58 LET D = P\textasteriskcenteredD 60 LET N = R 62 LET W = D/N 64 LET Y = INT(W) 66 IF Y < > W THEN 24 70 LET M(K) = W 75 PRINT 76 PRINT "DENOMINATORS OF UNIT FRACTIONS ARE:" 77 PRINT 80 LET K0 84 IF M(K) = 0 THEN 96 86 PRINT 88 PRINT M(K) 92 LET K = K + 1 94 GO TO 84 96 PRINT"--------------------------" 97 PRINT 98 GO TO 15 99 END

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Question:

A Venturi meter is a device by means of which the velocity of flow of a fluid can be measured. The diagram illustrates the operation of one type of such a meter. If the value of A is 10 times that of a, and colored water is used in the U-tube, what is the vel-ocity v_1 of flow of water when the difference in levels h is 6 in.? See Fig.

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Solution:

The operation of the meter, according to diagrams A and B is based upon Bernoulli's Principle. The indicator tube shows that the pressure at the constriction (P_2) is less than p_1. The pressure P_1 due to the flowing fluid in the wide tube must balance both p_2 as well as the pressure due to the weight of water in the column, hence it must be greater than p_2. By the law of continuity Av_1 = av_2 This result follows from the assumed incompressibility of the fluid. In the same time interval ∆t, the same volume of fluid which enters the constriction must leave it. Or (see figure C). (A∆x_1)/∆t = (a∆x_2)/∆t where A and a are the cross-sections indicated in figure B. By definition of velocity v_1 = (∆x_1)/(∆t) and (1/2) = (∆x_2)/(∆t). ThenAv_1 = av_2 We are given that A = 10a, then 10 av_1 = av_2 10 v_1= v_2(1) By Bernoulli's Principle (1/2) dv^2_1 + P_1 = (1/2) dv^2_2 + P_2 where d is the density of the fluid. This principle follows from the work energy theorem. Consider a volume of fluid (fig. C) which flows into the constriction. The total work done on this element of fluid, is equal to the work done on it by the fluid behind it in pushing it through the constriction minus the work it does on the fluid in front of it when it pushes this fluid forward. This total work is equal to the change in kinetic energy of this element of fluid. Or \intF_1^\ding{217} \bullet dx_1^\ding{217} - \intF_2^\ding{217} \bullet dx_2^\ding{217} = F_1∆x_1 - F_2∆x_2 = (1/2) mv^2_2 - (1/2) mv^2_1(2) Here F_1 is the constant force exerted on the element of fluid, by the fluid to the left of it, F_2 is the force it exerts on the fluid to the right of it, and v_1 and v_2 are the velocity of the fluid element before it enters the constriction, and after it enters the constriction, respectively. M is the mass of the fluid element. Equation (2) may be written as F_1∆x_1 + (1/2) mv^2_1 = F_2∆x_2 (1/2) mv^2_2(3) Using the fact that the volume of fluid element pushed to the right equals the volume of fluid element pushed to the left (i.e., A∆x_1 = a∆x_2) then F_1∆x_1 + (1/2) mv^2_1 = F_2(A/a)∆x_1 (1/2) mv^2_2 Dividing both sides by the volume A∆x_1 (F_1/A) + (1/2) (m/A∆x_1) v^2_1 = (F_2/a) + (1/2) (m/A∆x_1) v^2_2 orP_1 + (1/2) dv^2_1 = P_2 + (1/2) dv^2_2 since density = (Mass)/(volume)and Pressure = (Force)/(Area). d is the density of water. Then P_1 - P_2 = (1/2) d(v^2_2 - v^2_1) = (1/2)d (100v^2_1 - v^2_1) = (1/2)d (100v^2_1) (approx) where we used equation (1) But P_1 - P_2 = (F_1 - F_2)/A_x Where A_x is the cross sectional area of the indicator tube, and the difference of these forces (due to the fluid in motion) must equal the weight of water in the column of height h. Then F_1 - F_2 = mg = d(Ah)g where m is the mass of water in the column, d is the density of water, and Ah is the volume of the water. Upon division of both sides by A_x, P_1 - P_2 = (F_1 - F_2)/A_x = dhg Therefore hdg = (1/2)d (100)v^2_1 v^2_1 = [(2g)/(100)] h = [2(32)]/[100] h = .64h v_1 = .8\surdh Substituting h = 6 in. = .5 ft: v_1 = .8(.7 ft)_s = .56 ft/sec

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Question:

Do you believe the various races of man living today constitute one species or several? Why? What characteristics distinguish the present races of man?

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Solution:

The populations of man are now more diverse than ever in history. From all of the stereotyping that we are accustomed to, we all know each race's distinct physical and cultural features. But because the populations of large cities consist of samplings from almost every race, and interracial matings , the human races are slowly losing their characteristic distinctions. It would be improper to consider each race a separate species since individuals of any race are capable of mating with any other individual of another race and of producing fertile offspring. This is the major characteristic in deciding whether organisms belong to the same species. Thus, the various races of man living today constitute only one species. Genetically, a race is a subdivision of a species that consists of a population with a characteristic combi-nation of gene frequencies. Races are populations that differ from one another not in any single feature, but in having different frequencies of many genes that affect body proportions, skull shape, degree of skin pigmentation, tex-ture of hair, abundance of body hair, form of eyelids, thickness of lips, frequencies of various blood groups, tasting abilities and many other anatomical and physiological traits. While each race has its fundamental characteristics, no race is "pure" because the evolutionary history of human beings has been one of continuous intermixture of races as peoples migrated, invaded and conquered their neighbors or were conquered by them. The worldwide inter-breeding typical of mankind is a characteristic which is absent in other higher primates. The daily or lifetime range of other individual prima-tes or primate social groups is relatively small. For example, most apes need to be close to a source of water and to the same food supply that they are accustomed to, thus their diet is not very diverse. They also need a form of shelter for protection when they sleep at night. In addition, most apes remain in the small close-knit social group that they were born into during their entire lifetime. Because of the non-migrating feature of ages, there is only little genetic diversity within the various species. Widespread migration and interbreeding in humans are thus primarily responsible for the genetic diversity of races. Almost every combination of genes has the possibility of occurring in mankind and this might be related to the evolutionary advances that have taken place in our history as well as those that will take place in the future. Mankind is evolving even today, but the rate is extremely slow. Usually an organism will tend to evolve or adapt in response to a change or stress in its environment. Living in a large city, with all of its pollution, technology, tension, and stress, any prove to be a selection pressure for individuals who can better function in this type of situation. It is hard to imagine what the races of human beings will be like in the future.

Question:

What hormones are involved with the changes that occur in a pubescent female and male?

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Solution:

Puberty begins in the female when the hypothalamus stimulates the anterior pituitary to release increased amounts of FSH (follicle stimulating hormone) and LH (luteinizing hormone). These hormones cause the ovaries to mature, which then begin secreting estrogen and progesterone, the female sex hormones. These hormones, particularly estrogen, are responsible for the development of the female secondary sexual characteristics. These characteristics include the growth of pubic hair, an increase in the size of the uterus and vagina, a broadening of the hips and development of the breasts, a change in voice quality, and the onset of the menstrual cycles. Before the onset of puberty in the male, no sperm and very little male sex hormone are produced by the testes. The onset of puberty begins, as in the female, when the hypothalamus stimulates the anterior pituitary to release increased amounts of FSH and LH. In the male, FSH stimulates maturation of the seminiferous tubules which produce the sperm. LH is responsible for the maturation of the interstitial cells of the testes. It also induces them to begin secretion of testosterone, the male sex hormone. When enough testosterone accumulates, it brings about the whole spectrum of secondary sexual characteristics normally associated with puberty. These include growth of facial and pubic hair, deepening of the voice, maturation of the seminal vesicles and the prostate gland, broadening of the shoulders, and the development of the muscles. If the testes were removed before puberty, the secondary sexual characteristics would fail to develop. If they were removed after puberty, there would be some retro-gression of the adultsexual characteristics, but they would not disappear entirely.

Question:

Calculate the wavelength of the first line in the visible (Balmer) series of the hydrogen spectrum, for which the electron transition is from orbit 3 to orbit 2.

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/Users/wenhuchen/Documents/Crawler/Physics/D31-0922.htm

Solution:

The total energy of the nth orbit of the hydrogen atom is given by E_n = - (1/2) K (e^2/r_n) wherer_n= n^2 r_1, r_1 being the first Bohr radius. If an electron jumps from a higher orbit to a lower orbit, it will emit photons (electromagnetic waves) because it has to decrease its energy to match the binding energy of the new orbit. Therefore, the energy carried away by radiation is equal to the difference between the two energy levels E^ph = E(n)1- E_(n)1 = - (1/2) k (e^2/r_1)[(1/n^2_1) - (1/n^2_1)]. where n_1 and n_2 denote the initial and final energy levels. If the emitted electromagnetic wave has a wave length, \lambda, its energy ishc/\lambda. We have, then, for the wavelength : 1/\lambda = (1/2) (ke^2/hcr_1)[(1/n^2_1) - (1/n^2_1)] Substituting for the constants, we get 1/\lambda = R[(1/n^2_1) - (1/n^2_1) Where R = (ke^2/2hcr_1) = [{(9 × 10^9 m/farads) × (1.6 ×^ 10^-19coul)^2} / {(2) × (5.29×10^11m) × (6.63×10^-34J-sec) × (3×10^8m/sec)}] = 1.097 × 10^7 m^-1 = 1.097 × 10^7 m^-1 × 10^-10 m/A = 1.097 × 10^-3 A-1 Forn_1 = 3 and n_2 = 2, the wavelength is 1/\lambda = 1.097 × 10^-3 A^-1[(1/2^2) - (1/3^2)] = 1.52 ×10^-4 A-1 \lambda = 6560A

Question:

If it takes 30.3 KJ/mole of heat to meltNaCl, calculate the melting point ofNaCl, assuming the entropy increase is 28.2 J mol^-1 deg^-1.

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Solution:

To solve this problem, relate free energy, enthalpy, and entropy. The formula, ∆G = ∆H - T∆S, does this. ∆G = free energy, ∆H = enthalpy, T = tem-perature in Kelvin and ∆S = entropy. Let T = the melting point of NaCl . Entropy is a measure of randomness. Re-actions are favored that increase the randomness of the system. Reactions will also be favored that proceed to a lower energy state. Above the melting point, randomness increases, but so does energy. However, randomness pre-dominates, and so that solid melts, when the temperature is greater than the melting point. The situation below the melting point is exactly the opposite; here the energy consideration predominates. Only at the melting point, or when the solid and liquid are in equilibrium, will ∆H and T∆S be equal. As such, at the melting point ∆G = 0. You have, therefore, ∆G = 0 = ∆H - T_MP ∆S or ∆H = T_MP ∆S or T_MP = ∆H/∆S. Because it takes 30.3 KJ/mole to meltNaCl, this must be the heat content or enthalpy. The change of entropy was given as 28.2 J mol^-1 deg^-1. You have therefore, T_MP = [(∆H^1)/(∆S)] = [(30, 300 J/mole)/(28.2 J mole^-1 deg^-1 )] = 1070\textdegreeK.

Question:

Give a brief introduction to the operations and applications of the SNOBOL IV language.

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Solution:

Developed at Bell Telephone Laboratories, Inc. in 1962, SNOBOL IV is a computer programming language containing many features not commonly found in other programming languages. The basic data element of SNOBOL IV is a string of characters, such as this line of printing. The language has operations for joining and separating strings, for testing their contents, for making replacements in them, etc. SNOBOL IV provides numerical capabilities with both integers and real numbers. Conversions among integers, real numbers, and strings represen-ting integers or real numbers are performed automatically as required. SNOBOL IV provides great flexibility of function usage. Functions can be defined and redefined during program execution. Function calls can be made recursively with no special program notation. These and many other important features make this language highly applicable to such areas as compilation techniques, machine simula-tion, symbolic math, text preparation, natural language translation, linguistics, music analysis, and many others. SNOBOL IV has been implemented on several different computers, in-cluding the IBM 360, UNIVAC 1108, GE 635, CDC 3600, CDC 6000, PDP-10, SICMA 5,6,7, ATLAS 2, and RCA SPECTRA 70 series.

Question:

The accompanying figure shows an energy-level diagram. Make a comparison of the energy of the n = 5 and n = 4 transition for an electron with a nucleus of Z = 3 and the energy of the transition from n = 2 to n = 1 for an electron with a nucleus of Z = 2.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E18-0692.htm

Solution:

When an atom is in the ground state, it is at its lowest energy level, n = 1. If energy is added, the electrons become "excited" and move to a higher energy level; n = 2, 3, 4 . . .. After a time, the electron falls back to lower energy states and releases the energy needed to move it to the higher energy levels. The equation E_b - E_a = Z^2 (e^2/2a_0) [(1/n_a ^2) - (1/n_b ^2)], where n = the (quantum) energy level, E = energy, e = charge on electron, a_0 = Bohr radius, and Z = atomic number, measures this energy release between energy levels. You are asked to compare the energy releases between two different atoms and energy levels. You do not need to know the actual energy released. Because e^2/2a_0 is a constant, it need not be evaluated. (It will cancel out when you compare). Therefore, you have for n = 5 to n = 4 E_b - E_a = (e^2/2a_0) [(1/n_a ^2) - (1/n_b ^2)] Z^2 = (e^2/2a_0) [(1/16) - (1/25)] × 9 For n = 2 to n = 1, you have E_b - E_a = (e^2/2a_0) [(1/n_a ^2) - (1/n_b ^2)] Z^2 = (e^2/2a_0) [(1/1) - (1/4)] × 4 By comparison, in the form of a ratio, one has {(e^2/2a_0) [(1/16) - (1/25)] × 9} / {(e^2/2a_0) [(1/1) - (1/4)] × 4} = {[(1/16) - (1/25)] × 9} / {[(1/1) - (1/4)] × 4} = [(.2025) / (3)] [(.2025) / (3)] × 100% = 6.8%. Thus, the energy involved in the n = 5 to n = 4 transition is 6.8% as large as compared to the energy emitted in the n = 2 to n = 1 transition.

Question:

Two particles of equal mass move initially on paths parallel to the x axis and collide. After the collision one of the particles is observed to have a particular value v_y(1) of the y component of the velocity. What is the y component of the velocity of the other particle after the collision? (Recall that each component x, y, or z of the total linear momentum is conserved separately).

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Solution:

As shown in the figure, before the collision the particles were moving along the x axis, so that the total y component of the momentum is zero. By momentum conservation the total y component of momentum must also be zero after the collision, so that M [ v_y(1) + v_y(2) ] = 0 where M is the mass of the particles, whence v_y(2) = - v_y(1) The Velocities (a) before and (b) after collision. We cannot calculate v_y(1) itself without specifying the initial trajectories and the details of the forces during the collision process.

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Question:

If you hydrolyze a peptide bond in a protein molecule, what type of compounds are formed?

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Solution:

When proteins are boiled in dilute acids or bases they are hydrolyzed, i.e. degraded or broken down, to amino acids. Amino acids are compounds whose molecules possess both the amino (- NH_2) and the carboxyl (- CO_2H) functional groups. The general formula may be written as The union of amino acids, which result in the protein, is due to bonds that come about as a result of the elimi-nation of a hydrogen from the - NH_2 group and an OH from the - CO_2H group. This linkage is termed a peptide bond. It is depicted as When this substance is hydrolyzed, the following reaction occurs. In acid: (back) the amino and carboxylic group. Note: NH_2 is basic, so that in acid, you actually get - NH_3^+. Thus, you obtain a compound that possesses an amino and a carboxylic group, which is, as you recall, an amino acid. With base, the same result is obtained, but, you immediately obtain NH_2 and not NH_3^+. The carboxylic portion, which is acidic, reacts with the OH^- from base to yield COO^-.

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Question:

The specific rotation [(\alpha)^25\textdegree_D] of L-alanine is + 1.8. The specific rotation [(\alpha)^25\textdegree_D] of L-alanine is + 1.8. Calculate the observed rotation of a 1.10 M solution in a 2.5-dm polarimeter tube at 25\textdegreeC.

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Solution:

All amino acids with the exception of glycine show optical activity, i.e., they can rotate the plane of plane-polarized light when examined in a polarimeter. Optical activity is shown by compounds capable of existing in two forms, whose structures are nonsuperimposable mirror images of each other. Optical activity is expressed quantitatively as the specific rotation [(\alpha)^25\textdegree_D] : [(\alpha)^25\textdegree_D] = (observed rotation ×100) / [optical path length (dm) × concentration (g/100 ml)] With this information, one can find the observed rotation of L-alanine after converting the concentration from (moles/liter) to (g/100 ml). The molecular weight of alanine is 89. [1.10 moles] / [1 liter (= 1000ml)] × 89 g/mole = (97.9 g) / (1000ml) = (9.79 g) / (100ml) Solving for the observed rotation: observed rotation = [(\alpha)^25\textdegree_D × optical path length (dm) × concentration (g/ml)] / [100] observed rotation =[+1.8 × 2.5dm × 9.79] / [100] = 0.44\textdegree.

Question:

Draw a typical bacterial growth curve and label the different phases. Discuss the factors for the existence of each phase.

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/Users/wenhuchen/Documents/Crawler/Biology/F05-0133.htm

Solution:

If one inoculates a flask of nutrient broth with a given number of bacteria and follows the rate of growth during the incubation period, there is found a series of different growth rates. A plot of the logarithms of the number of cells versus time is used to illustrate the growth curve, which is shown in the following diagram: There is an initial period of no growth, followed by one of rapid growth, and then there is a leveling off. The last period is one of decline in the population. The curved portions designates the transitional period between phases. These represent periods where the bacteria enter a new phase. Following the addition of inoculum into the new medium, the bacterial population does not increase but each individual bacterium increases in size. During this period, called the lag phase, the bacteria are physiologically active and are adapting to the new environment, but the bacterial population number remains constant. In addition, the bacteria may alter the environment for their means (eg. exretion of CO_2 to lower pH, so that more favorable growth conditions are achieved.) At the end of the lag phase, the bacteria start to divide. The lag phase is followed by the logarithmic or exponen-tial phase where the cells divide at a constant and maximal rate according to their generation time. Most of the bacteria during this phase are uniform in terms of metabolic activity, unlike the other phases. After the log phase, growth begins to level off in the stationary phase. The population remains constant because of a cessation of reproduction or an equalization of growth and death rates. The decreased growth is usually due to the exhaustion of nutrients or the production of toxic or inhi-bitory products. The final phase is called the death phase. Here, the bacteria die faster than they are being produced, if any reproduction is occurring at all. Depletion of nutrients and accumulation of inhibitory products, such as acid, cause the increased death rate. The number of bacteria increases exponentially in the log phase, and the number of bacteria decreases exponentially in the death phase. Some species die rapidly, so that few living cells remain after about three days, while other species die after several months.

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Question:

Write a program in BASIC to compute the n-thpower of a square n × n matrix A.

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Solution:

A concept of identity matrix will be used in the solution of this problem. An identity matrix in BASIC is the matrix in which all the elements in the leading diagonal are 1 and all other elements are zero. A 3 × 3 identity matrix, for example, looks as follows: \mid 100 \mid \mid 010 \mid \mid 001 \mid An identity matrix multiplied by another matrix M (of the same dimensions) produces a matrix identical to M. This property is used in the solution in the following way: Matrix B is made into an identity matrix. Then matrix A is multiplied by B to produce another matrix C equal to A. After that, matrix C is assigned into B, and the multiplication of B and A is repeated. The new value of C is again assigned into B and B × A is executed again. This procedure repeats n times, where n is the desired power of A (as well as the matrix dimensions in this case). The final value of the matrix C is equal to (A)^n . Note, that this is not the only possible solution to the problem. The program looks as follows: 10DIM A (N, N), B (N, N), C (N, N) 20INPUT N 30MAT B = IDN 40FOR 1 = 1 TO N 50MAT C = B\textasteriskcenteredA 60MAT B = C 70NEXTI 80MAT PRINT C: 90END

Question:

In an isolated mountain village, the gene frequencies of A, B and O blood alleles are 0.95, 0.04, and 0.01, respectively. If the total population is 424 calculate the number of individuals with O, A, B, and AB type blood.

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Solution:

Multiple allelic systems establish equilibrium in the same way as single-pair alleles, and the same equilibrium principles can be applied, though naturally the system is more complex. In the inheritance of blood groups, three genes, A, A^B, and a are involved. Letting p, q, and r, respectively, represent their frequencies, we can say that the total of these frequencies equals 1 (p + q + r = 1). We can represent the total possible gene combinations (genotypes) in the population by means of the same type of expansion used with single-pair allelic systems; in this case, however, three alleles are involved. Thus: (p+q+r)^2 = p^2 + 2pq + q^2 + 2pr + r^2 + 2qr where: p = frequency A q = frequency A^B r = frequency a The following table summarizes what we know of the geno-types of each blood group, (see previous chapter for an explanation of the inheritance of blood groups) and the corresponding allelic and genotypic frequencies. blood type genotypes genotypic frequency O aa r^2 A AA, Aa p^2, 2pr B A^BA^B, A^Ba q^2, 2qr AB AAb 2pq In the given population, p = 0.95, q = 0.04, and r = 0.01. Knowing this we can calculate, the blood type frequencies using the results of the expansion. The frequency of type O blood in the population is: r^2 = (0.01)^2 = 0.0001 The frequency of type A blood is the sum of the frequencies of the two genotypes comprising this phenotypic blood group: p^2 + 2 pr = (0.95)^2 + 2(0.95) (0.01) = .9025 + .019 = 0.9215 The frequency of type B blood is the sum of the frequencies of the genotypes comprising the group: q^2 + 2 qr= (0.04)^2 + 2(0.04) (0.01) = (.0016) + .0008 = .0024 The frequency of type AB blood is simply: 2 pq = 2(0.95) (0.04) = .076 We can see that the sum of the frequencies is 1: frequency O= r^2=.0001 A = p^2 + 2 pr=.9215 . B = q^2 + 2 qr=.0024 AB = 2 pq=.076 1.000 1.000 Converting the frequencies to actual numbers of individuals in the population: O = (r^2) (424) = (.0001) (424) \sim 0 (.0424) A = (p^2 + 2pr) (424) = (.9215) (424)= 391 B = (q^2 + 2qr) (424) = (.0024) (424)=1 AB = (2pq) (424)= (.076) (424)=32 424

Question:

Write a general purpose BASIC program which prints a table of values for any function.

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Solution:

The solution uses the notion of string variables and user defined functions. This program also makes use of the STOP statement, which, in BASIC, essentially terminates the execution of the program In the same fashion as the END statement. A term with $ following a single letter is used to record character data. 1\O REM PRINTS A TABLE OF VALUES FOR ANY FUNCTION 2\O DEF FNA(X) = X 3\O PRINT "TO DEFINE A NEW FUNCTION TYPE Y; ELSE TYPE N." 4\O INPUT A$ 5\O IF A$ = "N" THEN 11\O 6\O PRINT "TYPE YOUR FUNCTION AS A FUNCTION OF X, AFTER YOU TYPE'' 7\O PRINT "2\O DEF FNA(X) =" 8\O PRINT "THEN HIT THE RETURN KEY AND FINALLY"; 9\O PRINT "TYPE THE COMMAND RUN" 1\O\O STOP 11\O PRINT "WHAT ARE THE MINIMUM, MAXIMUM, AND INCREASE IN X" 12\O INPUT X1, X2, D 13\O FOR I = X1 TO X2 STEP D 14\O PRINT "X; I; "F(X) ="; FNA (I) 15\O NEXT I 16\O END

Question:

A researcher notices that the frequency of a note emitted by an automobile horn appears to drop from 284 cycles \bullet s^-1 to 266 cycles \bullet s^-1 as the automobile passes him. From this observation he is able to calculate the speed of the car, knowing that the speed of sound in air is 1100 ft \bullet s^-1. What value does he obtain for the speed?

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/Users/wenhuchen/Documents/Crawler/Physics/D26-0841.htm

Solution:

This is an example illustrating the Doppler effect. When there is no movement of the surrounding medium the relation between the frequency as heard by a moving observer and that emitted by a moving source is f_L /(u \pmv_L) =f_S/( u \mpv_S) wheref_Lis the frequency heard by the listener,f_Sthe frequency emitted by the moving source,v_Lthe velocity of the listener,v_Sthe velocity of the source, and u the velocity of sound (=1100 ft \bullet s^-1). The upper signs (+ left side of equation, - right side) correspond to the source and observer moving along the line joining the two and approaching each other and the lower signs (- left,+ right) correspond to source and observer receding from one another. In this case the frequencies heard by the stationary listener (v_L= 0) will bef_L=uf_S/(u \mpv_S). As the auto-mobile approaches the observer he records a frequency of 284 cycles \textbullet s^-1 and as the automobile moves away from him, he records 266 cycles \bullet s^-1 Thus 284 s^-1 =uf_S/(u -v_S)(1) and266 s^-1 =uf_S/(u +v_S)(2) Dividing (1) by (2) (u +v_S)/(u -v_S) = 284/266 266 (u +v_S) = 284 (u -v_S) (266 + 284)v_S= (284 - 266)u orv_S/u = 18/550 \thereforev_S= (18/550) × 1100 ft \bullet s^-1 = 36 ft \bullet s^-1 = 36 ft \bullet s^-1 = 36 ft \bullet s^-1 × (1 mile/5280 ft) × (60/1 min) × (60 min/ 1 hr) = 24.5 mph

Question:

Consider gases confined by a liquid, as shown in the diagram below. Find an expression for the pressures P_1, P_2, and P_3 in terms of the density of the liquid, \rho (g/ml), the heights h_1 and h_3 (mm), and the baro-metric pressure P_atm (mm Hg).

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Solution:

The device shown above is called a eudiometer. It is used to compare the pressures of several gases. The pressure of the confined gas is equal to the baro-metric pressure plus the pressure needed to depress the column of liquid (for P_1) or to the barometric pressure minus the pressure needed to support the column of liquid (for P_3). The pressure of the liquid column is given by pressure of liquid = height or depth of column (mm) × \rho (g/ml) (ml/cm^3) (cm/10mm) Hence: P_1 = barometric pressure + pressure needed to depress liquid = P_atm + h_1 × \rho (ml / cm^3 ) cm/10mm = P_atm + 0.1 h_1 (g/cm^2) \rho P_2 = barometric pressure + pressure needed to depress liquid (or - pressure needed to elevate liquid) = P_atm + 0 × \rho × m1/cm^3 × cm/l10mm = P_atm P_3 = barometric pressure - pressure needed to elevate liquid = P_atm - h_3 × \rho × ml/cm^3 × cm/10mm = P_atm - 0.1 h_3 (g/cm^2 ) = P_atm - 0.1 h_3 (g/cm^2 ) \rho

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Question:

The half-life of radon is 3.80 days. After how many days will only one-sixteenth of a radon sample remain?

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Solution:

A half-life of 3.80 days means that every 3.80 days, half the amount of radon present decays. Since one- sixteenth is a power of one-half, this problem can be solved by counting . After 3.80 days, one-half the original sample remains. In the next 3.80 days, one-half of this decays so that after 7.60 days one-fourth the original amount remains. After 11.4 days, one-eighth the original amount remains, and after four half-lives (15.2 days), one-sixteenth the original amount remains. As an alternate solution, the formula for decaying matter can be used. N/N_0 = e(\Elzbar\lambdat)(1/2) \lambda is an experimental constant which can be determined from the half life. Therefore, for N/N_0 = 1/2, 1n(N/N_0 ) =(\Elzbar\lambdat)(1/2) \lambda = [{1n(N_0/N )} / {t _(1/2) }] =[(1n 2) / (t _(1/2) )] = 0.693 / 3.80days = 0.182/days. We want N/N_0 = 1/16 = e(\Elzbar\lambdat)= e\Elzbar0.182t. \Elzbar0.182t = 1n 1/16 =\Elzbar2.77 t = 15.2 days.

Question:

Consider an airplane trip which takes place in four stages. Each stage is represented by a vector as follows (see figure). A to B AB = 120 mi\textphi_1 = 30\textdegree B to C BC = 50 mi\textphi_2 = 60\textdegree C to D CD = 700 mi\textphi_3 = 210\textdegree D to E DE = 400 mi\textphi_4 = 90\textdegree The angle describing these vectors is with respect to the positive x-axis. Find the resultant displacement vector.

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Solution:

First we can calculate the x- and -y components. (AB)_x = AB cos \textphi_1 = 120 cos 30 = 60 \surd3 mi (BC)_x = BC cos \textphi_2 = 50 cos 60 = 25 mi (CD)_x = CD cos \textphi_3 = 700 cos 210 = - 350 \surd3 mi (DE)_x = DE cos \textphi_4 = 400 cos 90 = 0 (AB)_y = AB sin \textphi_1 = 120 sin 30 = 60 mi (BC)_y = BC sin \textphi_2 = 50 sin 60 = 25 \surd3 mi (CD)_y = CD sin \textphi_3 = 700 sin 210 = - 350 mi (DE)_y = DE sin \textphi_4 = 400 sin 90 = 400 mi These components are summed to find the x and y components of the resultant. x-component y-component AB 104 mi 60 mi BC 25 mi 43 mi CD - 606 mi - 350 mi DE 0 mi 400 mi Resultant AE - 477 mi 153 mi The magnitude of the resultant is therefore, by the Pythagorean theorem: AE^2 = (- 477)^2 + (153)^2 AE = 501 mi and its direction is given by the angle \textphi where sin \textphi = (153/501) = 0.305 cos \textphi = (- 477/501) = - 0.952 \textphi = 17.8\textdegree The same thing can be found by making a graph (see figure). The resultant vector, AE, is drawn from the starting point A to the end point E of the trip.

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Question:

A river flows due north. Which side of the bank should be the most worn?

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Solution:

This will depend on the hemisphere in which the river exists. The force which presses the river against the bank is the Carioles force: F_c- ^\ding{217} + 2mv^\ding{217} × \omega^\ding{217} In the northern hemisphere, we see from the diagram that v^\ding{217} × \omega^\ding{217} is to the east, whereas in the southern hemisphere v^\ding{217} × \omega^\ding{217} is to the west. Thus the river will tend to wear down its right bank in the northern hemisphere and left bank in the southern hemisphere.

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Question:

Compare the processes of cyclic andnoncyclic photophosphorylation.

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Solution:

There are two types ofphotophosphorylation: cyclic andnoncyclic. In cyclicphotophosphorylation, the flow of electrons from light-excited chlorophyll molecules to electron acceptors proceeds in a cyclic fashion through electron carriers back to the original chlorophyll.PhotosystemI, sensitive to longer wavelengths of light thanphotosystemII, functions in cyclicphotophosphor-ylation. No oxygen is liberated, since water is not split; no NADP^+ is reduced, since it does not receive electrons. ATP is formed during the electron flow, and light energy is thus converted into chemical energy in the ATP molecules. However, since NADPH is not formed, cyclicphotophosphorylationis not adequate to bring about CO_2 reduction and sugar formation, processes which require the energy of NADPH . Noncyclic photophosphorylation, on the other hand, produces both ATP and NADPH molecules necessary for the dark reactions of photosynthesis. In this pro-cess, electrons from excited chlorophyll a molecules are trapped by NADP in the formation of NADPH and do not cycle back to chlorophyll a. Electrons are ejected from chlorophyll b to be donated to chlorophyll a through a series of electron carriers. To restore chlorophyll b to its ground level, water is split into protons, elec-trons and oxygen. The electrons are picked up by chlor-ophyll b, thehydrogensare used to form NADPH, and oxygen escapes to the atmosphere in its molecular form. BothphotosystemsI and II are involved in the process of noncyclic photophosphorylation.

Question:

In the Cephalochordate, one of the three chordate subphyla, all three chordate characteristics are highly developed and retained in the adult. Outline the features of an adult Cephalochordate.

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Solution:

The cephalochordates are represented by the lancelets, most notablyAmphioxus. These small, marine chordates have been the subject of much study because of the almost diagrammatic way in which they illustrate the basic chordate characteristics. In Amphioxus, the notochord extends from the tip of the head to the tip of the tail, and aids the animal in burrowing through the mud. There are numerous gill slits in the pharynx. The dorsal hollow nerve cord exterids along the entire length of the body. Water enters the mouth (the opening into the pharynx), and passes through the gill slits into a chamber known as the atrium and leaves the body through the atriopore. Food is removed from the water in the pharynx with the aid of secretions from the endostyle in the ventral portion of the pharynx. The food is then sent posteriorly into the instestine; undigested food is eventually eli-minated via the anus. Metabolic wastes are excreted by segmentally arranged, ciliated protonephridia that open into the atrium. The muscles of the body are V shaped, segmentally arranged structures. A continuous dorsal fin, supported by fin rays, expands into a tail fin and then continues ventrally to the atriopore as the ventral fin. Although superficially similar to fishes, the lancelets are more primitive for they lack paired fins and jaws . Anterior to the atriopore, however, are two lateral folds that may have been the forerunners of the paired pectoral and pelvic fins of fishes. It hardly seems possible that any form living today is in the direct line of vertebrate ancestry, but the lancelet most closely represents the ancestral vertebrates. Its body plan, although relatively simple, resembles that of vertebrates, and the larva of the lamprey, one of the lowest vertebrates, is in many ways similar to the lancelets.

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Question:

V \leftarrow 3 4 2 -1. Find the result of the following operations.

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Solution:

a) The symbol \vert, when written as a monadic function, as \vertV, specifies the absolute value of V. V is a vector, which consists of the elements 3 4 2 -1. The absolute value of V is 3 4 2 1 i.e., \vertV is equal to 3 4 2 1. b) The symbol \Gamma represents a maximum function. When used with the Reduction Operator /, the maximum function selects the largest from among all the elements of the vec-tor, V in this case. Thus, if V consists of 3 4 2 -1, then \Gamma/V is equivalent to the following: 3\Gamma4\Gamma2\Gamma - 1 i.e.,3\Gamma4\Gamma2 i.e.,3\Gamma4 i.e.,4 In the above, using the right to left rule of APL for eval-uating an expression, the elements of V are compared two by two from the right. The maximum is selected at each com-parison. Thus, finally when the whole string has been checked out, the largest element is obtained. c) The symbol L represents a minimum function. When used with the Reduction Operator / , the minimum function selects the least from among all the elements of the vec-tor (V in this case). Thus, if V consists of 3 4 2 -1, then L/V is equivalent to the following: 3L4L2L - 1 i.e.,3L4L-1 i.e.,3L-1 i.e.,-1 In the above, using the right to left rule, the elements of V are compared two by two, and the minimum is selected.

Question:

How does communication serve in predator avoidance?

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Solution:

Both recognition and avoidance of predators can be considered forms of communication between species. One method by which animals avoid detection is called cryptic appearance. In this method, the animal uses its color or shape to resemble objects in the environment in which it lives. Many insects are shaped to resemble inedible objects, such as twigs. Some animals can even change their colors to resemble the surface on which they are temporarily resting. Warning appearance is another form of inter-species communication. It is used by some animals to warn potential predators that they are dangerous or poisonous, thus serving to prevent attack. The bold red-and-black bands of the venomous coral snake and the black-and- yellow stripes of the hornet are familiar examples. Mimicry is another method used to avoid attack by predators. It is used by animals that are not naturally protected; that is, they do not actually possess any characteristics harmful to a predator. The mimic, however, resembles some other species that does have dangerous characteristics and warning signals. In the form of mimicry called Batesian mimicry, the predator is unable to differentiate between the dangerous model and the harmless mimic. It thus leaves both alone. For example, the Monarch butterfly is distasteful to birds, who therefore avoid it. It serves as the model for the Viceroy butterfly, the edible mimic who closely resembles the Monarch in color, shape and behavior. Having the same warning appearance as the Monarch the Viceroy butterfly is protected. (See Figure) A second form of mimicry, called Mullerian mimicry, involves the evolution of two or more inedible or unpleasant species to resemble each other. Since each species serves as both model and mimic, greater protec-tion is afforded to them because their repellent qualities are more frequently advertised.

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Question:

Write the micro-instruction for the execute cycles of the following I/O instructions : Instruction Symbol Hexadecimal Code Definition INP F800 Input character to lower eight OUT F400 Output character from lower eight bits of AC SKI F200 Skip on input flag SKO F100 Skip on output flag ION F080 Interrupt on IOF F040 Interrupt off

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Solution:

The I/O instructions are specified when the control signal q_7Ic_2 [MBR]_i is true (logic 1), where the i^th bit of the MBR is the high bit specified by each instruction. All of the I/O instructions can be executed during one timing signal t_3. INP:This instruction transfers the eight bit word from the INPR to the AC. After transferring the word, the FGI flag must be cleared so the teletype can insert another word in the INPR, if it needs to. q_7Ic_2t_3 [MBR]_4 :AC_8-15 \leftarrow [INPR] FGI\leftarrow 0 F\leftarrow 0 The first micro-instruction transfers the data to the AC. The second micro-instruction clears the FGI. The third micro-instruction sets the fetch cycle control signal. OUT:This instruction transfers the word from the AC to the OUTR and clears the FGO flag so that the OUTR cannot be changed while the teletype is printing. q_7Ic_2t_3 [MBR]_5 :OUTR\leftarrow [AC]_8-15 FGO\leftarrow 0 F\leftarrow 0 SKI:This instruction increments the PC when the FGI flag is set. q_7Ic_2t_3 [MBR]_6 [FGI] :PC \leftarrow [PC]+1 F\leftarrow 0 orq_7Ic_2t_3 [MBR]_6 [FGI] :F\leftarrow 0 SKO:This instruction increments the PC when the FGO flag is set. q_7Ic_2t_3 [MBR]_7 [FGO] :PC \leftarrow [PC]+1 F\leftarrow 0 orq_7Ic_2t_3 [MBR]_7 [FGO] :F\leftarrow 0 ION:This instruction turns the interrupt on. The inter-rupt is on when the IEN flag is set. q_7Ic_2t_3 [MBR]_8 :IEN\leftarrow1 F\leftarrow 0 IOF:This instruction turns the interrupt off. q_7Ic_2t_3 [MBR]_9 :IEN\leftarrow0 F\leftarrow 0

Question:

In what ways are the bodies of earthworms (i.e.Lumbricus) and marine worms (i.e.Nereis) adapted to their habitats?

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/Users/wenhuchen/Documents/Crawler/Biology/F12-0295.htm

Solution:

Earthworms andNereisare both classified in the phylumAnnelida, the segmented worms. At a finer taxonomic level, earthworms are in the classOligochaeta(few bristles) andNereisin the classPolychaeta(many bristles). These two classes of annelids are adapted to live different lifestyles. Earthworms live in a moist terrestrial habitat. They are commonly seen burrowing through damp soil.Nereis, like manypolychaetesare marine worms which swim freely in the sea. Often they are found burrowing in sand and mud near the shore, or living in tubes formed by secretions from their body wall. The obvious difference in lifestyles between the two worms, the earthworm being a soil burrower,Nereis being a free swimmer, will be important in analyzing their adaptive body structures. The earthworm has four pairs of bristles or chaetae (also spelled setae) on each body segment except the first. These bristles enable the earthworm to grasp the ground or adhere to the walls of its burrow while in motion.Nereisalso has chaetae which cover the paired paddles or parapodia , which appear on each of their body segments. Theparapodia are the structures which giveNereisthe ability to swim. Adaptations are also apparent in the methods of nutrient procurement in the different habitats of the earthworm andNereis. Earthworms feed on decomposing matter as they sweep along the ground. The anterior segment bears no special appendages. In contrast, Nereis is a predator and the anterior end of the body commonly bears tentacles, bristles,palps, and antennae. The earthworm, living a subterranean life, has no well-developed sense organs.Nereis, an active swimmer, has two pairs of eyes, and organs sensitive to touch and to chemicals in the water. Fertilization is external in both the earthworm andNereis. However, Nereis , which relies on a watery medium for the union of egg and sperm, has no special protective device for the developing young. Earthworms produce special cocoons, and deposit eggs and sperm in them to facilitate union and so that the young can develop in a sheltered environment. Earthworms are hermaphroditic, while inNereis, sexes are separate.

Question:

Develop a FORTRAN program segment to arrange in ascending order a set of N elements, trying to use the least number of passes possible.

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Solution:

One efficient method is known as a bubble sort. With this method, each pair of adjacent elements is compared. If they are in the proper order , they are left alone. If not, the terms are reversed in order before going on to the next pair. The integers are represented in the array ID (N), where N is the total number of elements to be sorted. The variable K permits the termination of the loop: if K is still zero at the beginning of the pass , the outer loop terminates, in-dicating that no switches had to be made , so that the array is sorted, if K equals one, another pass through the list must be made. The program segment is given below: INTERGER I, J, N, ID (N), TEMP. DIMENSION ID (N) K = 1 DO 30 I = N, 2, -1 IF (K.EQ.0) GO TO 50 K = 0 DO 40 J = 1, I - 1 IF (ID (J).LE.ID (J + 1)) GO TO 30 TEMP = ID (J) ID (J) = ID (J + 1) ID (J + 1) = TEMP K = 1 GO TO 30 40CONTINUE 30CONTINUE 50DO 60 m = 1, N WRITE (5, 100) ID (N) 100FORMAT (IX, I4) 60CONTINUE

Question:

What factors limit the number oftrophiclevels in afood chain?

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Solution:

As energy flows through the various food chains,Itis being constantlychanneled into three areas. Some of it goes into production, whichis the creation of new tissues by growth and reproduction. Energy is usedalso for the manufacture of storage products in the form of fats and carbohydrates. Some of the energy is lost from the ecosystem by way of de-composingdead organic materials. The rest of the energy is lost permanentlyto the ecosystem through respiration. The loss of energy due torespiration is very high and only a small fraction of energy is transferred success-fullyfrom onetrophiclevel to the next. Eachtrophiclevel depends onthe preceding level for its energy source. The number of organisms supportableby any giventrophiclevel depends on the efficiency of trans- formingthe energy available in that level to useful energy of the subsequentlevel. Ecological efficiencies may vary widely from one kind of animalto another, even at the sametrophiclevel. It has been shown that theaverage ecological efficiency of any onetrophiclevel is about 10%. The rate at which energy can be transformed by plants is controlled bythe carbon dioxide concentration of the atmosphere, by the amount of photosyntheticsurfaces available to light and by various limiting factors of theenvironment such as water and nutrients. Animals have their energy supply already in the usable form of high-energycompounds. The efficiency of an animaltrophiclevel is primarilya function of the food-getting and digesting process of that trophiclevel. The actual energy flow into primary consumers or herbivores is a smallfraction of the total energy converted by plants because a substantialpart of what they eat cannot be digested and must be returned backto the environment. Moreover, much of the energy converted by vegetationis used by the vegetation itself for respiration. Carnivores face similar but even more rigorous restrictions on their energysupplies, because their energy depends on how many of the herbivoresthey can manage to catch, eat and digest. For secondary carnivoresthe restrictions are even more rigorous and increase through thesuccessivetrophiclevels. Since energy is so important to animals, and is in restricted supply, thenumber of animals that can be supported is determined by the efficiencywith which they utilize energy. From this can be calculated the frequencyof a species and the total number that a given area can support. The number oftrophiclevels or steps in a food chain is normally limited to perhapsfour or five because of the great decrease (90%) in available energyat each level; i.e., only 10% of the energy is transferable to the nexttrophiclevel.

Question:

Find the \lambda's (wavelengths) for the first four hydrogen spectral lines in the Paschen series. Find the wavelength of the series limit.

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Solution:

The Paschen series illustrates the spectrum hydrogen exhibits in the infrared region. An example of a hydrogen line spectrum is shown in the accompanying figure. The series of lines of the Paschen series fit the equation ѵ = 3.29 × 10^15 [(1/9) - (1/b^2)], where v = frequency and b = 4, 5, 6, 7, 8, . . .. Thus, if you want the first four hydrogen spectral lines, you use b = 4, 5, 6, and 7. Your answer would be in terms of frequency, however, you want the wavelengths. To obtain this, employ the relation between frequency and wavelength, c = \lambdaѵ, where c = speed of light, \lambda = wavelength and ѵ = frequency, (c = 2.998 × 10^10 cm/sec.) Once you know ѵ, you can find \lambda. Therefore, 1st line:b = 4, ѵ = 3.290 × 10^15 [(1/9) - (1/16)] = 1.599 × 10^14 sec^-1. 2nd line:b = 5, ѵ = 3.290 × 10^15 [(1/9) - (1/25)] = 2.340 × 10^14 sec^-1. 3rd line:b = 6, ѵ = 3.290 × 10^15 [(1/9) - (1/36)] = 2.742 × 10^14 sec^-1. 4th line:b = 7, ѵ = 3.290 × 10^15 [(1/9) - (1/49)] = 2.984 × 10^14 sec^-1. The series limit is b = infinity. If B \rightarrow \infty, then (1/B^2) \rightarrow 0. Series limit: ѵ = 3.290 × 10^15 [(1/9) - 0] = 3.656 × 10^14 The wavelengths (\lambda) = [{speed of light (c)} / {frequency (ѵ)}] 1st line:\lambda = 1.87 × 10^-4 cm2nd line:\lambda = 1.28 × 10^-4 cm 3rd line:\lambda = 1.09 × 10^-4 cm4th line:\lambda = 1.004 × 10^-4 cm Series limit:.820 × 10^-4 cm.

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Question:

Express [(6,400,000 ) / (400)] in scientific notation

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Solution:

In order to solve this problem, we express the numerator and denominator as the product of a number between 1 and 10 and a power of 10. This is known as scientific notation. Thus 6,400,000= 6.4 × 1,000,000 = 6.4 × 10^6 400= 4 × 100 = 4 × 10^2 Then,[(6,400,000 ) / (400)] = (6.4 × 10^6) / (4 × 10^2) Since (ab/cd) = (a/c) \bullet (b/c): = (6.4/.40) × (10^6 × 10^2) Since (a^x/a^y) = a^x\rule{1em}{1pt}y : = 1.6 × 10^4

Question:

Describe the path of a molecule of sugar from the time it entersthe mouth as part of a molecule of starch, until it reachesthe cytoplasm of the cells lining the small intestine.

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Solution:

Upon entering the mouth, the starch is chewed and mixed with saliva. The saliva dissolves small molecules and coats the larger clumps ofstarch with a solution of the enzyme amylase. As the mass of food is swallowed, the enzyme amylase from the saliva hydrolyzes the starch, degradingit into maltose units. Maltose is a disaccharide composed of two glucosemolecules. Eventually, the acidity of the stomach inhibits the actionsof the salivary amylase, leaving most of the starch undigested. No furtherdigestion of the starch occurs until it reaches the small intestine. There, pancreatic amylase finishes the cleavage of starch into maltose. Intestinal maltase then hydrolyzes the maltose moleculesinto two glucose moleculeseach. Other intestinal enzymes cleave other disaccharides, suchas sucrose, intomonosaccharides. When these free sugar molecules come into contact with the cells liningthe small intestine, they are held by binding enzymes on the cell surface. Glucose is transported through the membrane into the cellular cytoplasm.

Question:

A square coil of 50 cm^2 area consisting of 20 loops is rotated about a transverse axis at the rate of 2 turns per second in a uniform magnetic field of flux density 500 gauss. The coil has a resistance of 20 ohms, (a) What is the average electromotive force induced in the coil during one quarter cycle of a revolution? (b) What is the average current in a complete cycle? (c) How much charge is passed through the coil in this interval?

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Solution:

(a) Note that in one quarter of a revolu-tion, the flux threading the coil is completely changed once. Therefore \varphi = NAB represents the number of lines cut in 1/8 second, since 1/8 second is one quarter of the period of rotation, and since N is the number of loops intercepting the changing flux at all times. Therefore, by Faraday's law, E = [(∆\varphi)/(∆t)] = [{(NAB)/(1/8)}] = [{20 (50 cm^2 ) (500 gauss)}/{(1/8) sec}] = 2 × 5 × 5 × 8 × 10^-4ab-volts = 400 × 10^4ab-volts = 400 × 10^4ab-volts × (1 volt/10^8ab-volt) = 0.04 volt (b) By Ohm's LawI = (E/R) = (.04/20) amperes = .002 amperes (c) But by definition of current I = (Q/t) where Q is the quantity of charge passing through a point in the conductor in a time t. Then Q = It = .002 (1/8) coulomb = .00025 coulomb

Question:

Compare the number of H_2 and N_2 molecules in two containers described as follows: (1) A 2-liter container of Hydrogen filled at 127\textdegreeC and 5 atm. (2) A 5-liter container of nitrogen filled at 27\textdegreeC and 3 atm.

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Solution:

Avogadro's Law states that equal numbers of molecules are contained in equal volumes of different gases if the pressure and temperature are the same. Therefore, if the conditions were the same, then there would be equal numbers of molecules of H_2 and N_2. For reasons of clarity, it is useful to set up a table like that shown below. gas V T P H_2 2 liters 400\textdegreeK 5 atm H_2 5 liters 300\textdegreeK 3 atm By using the general gas equation PV = nRT where n = number of moles, and R = 0.082 atm-liter/\textdegreeK-mole, one can find the number of moles of H_2 and N_2. Thus, for H_2 n = (PV/RT) = [{(5 atm) (2l)} / {(0.082 atm-l/mole-\textdegreeK) (400\textdegreeK)}] = 0.305 moles for N_2 n = (PV/RT) = [{(3 atm) (5 l)}/{(0.082 atm-l/mole-\textdegreeK) (300\textdegreeK)}] = 0.610 moles. There are twice as many moles of N_2 as there are of H_2. Since the number of molecules is directly proportional to the number of moles, then there are twice as many molecules of N_2 as molecules of H_2.

Question:

At what velocity is the mass of a particle twice its rest mass?

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Solution:

When the mass of an object that is travelling at a velocity approaching the speed of light, c, is measured, it is found to be larger them the mass measured when the object is at rest. The mass associated with an object travelling at any velocity v^\ding{217} is called the parti-cle's relativistic mass, and is given by the formula m(v) = {m_o / \surd(1 - v2/ c2)}(1) where m_0 is the rest mass of the particle. We are asked to find the velocity at which the mass of a particle (meaning its relativistic mass) is equal to twice the particle's rest mass. Writing this as an equation, m (v) = 2mo Using (1), this may be written as m_o / \surd{1 -(v2+ c^2)} = 2mo 1 / \surd{1 -(v2+ c^2)} = 2 Multiplying both side by [1/2 \surd{1 - (v2/ c^2)}] , we obtain \surd{1 - (v2/ c^2)} = 1/2 {1 - (v2/ c^2)} =1/4 v^2 / c2= 3/4 v / c= (\surd3) / 2 v / c = 0.866 v = 0.866 × 3 × 1010cm/sec v = 2.664 × 1010cm/sec

Question:

What are spores? How are they formed in mosses andin ferns ?

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/Users/wenhuchen/Documents/Crawler/Biology/F07-0185.htm

Solution:

A spore is a haploid, usually unicellular structure formed by reduction division in the sporangium of asporophyte. When released, it is capable of ger-minating and developing into an entire gametophyte individual , without fusion with another cell. This is in direct contrast to gametes , which, though also haploid, require fusion with another gamete in order to form a new individual, in this case, thesporophyte. The spores of the moss are formed in the sporangium, or capsule, of the mosssporophyte. The capsule is surrounded by an epidermal layer composed of cells similar to those found in the epidermis of higher plants. The inner portion of the sporangium consists of a layer of sterile, supporting tissue covering a core of fertilesporogenouscells. Spore mother cells, each containing the diploid number of chromosomes, develop from thesporogenouscells. Meiosis follows, and each diploid spore mother cell gives rise to four haploid cells, known as the spores. They are the first cells of the game-tophyte generation. When the capsule matures , the upper end forms a lid, which drops off, permitting the libe-ration of spores which are then likely to be dispersed by wind. If a spore falls on a good spot it will, under favorable conditions, germinate and develop into a new gametophyte individual. The fern is a vascular plant, and, unlike the moss, which has a dominant gametophyte generation, the fern has a dominantsporophyte and an inconspicuous gametophyte generation. The relatively large, leafy green plant we call the fern is thesporophytegeneration. Each leaf, or frond , of the fern is subdivided into a large number of leaflets. On the undersurfaces of certain leaflets develop clusters of small, brown sporangia . Depending on the genera and species, sporangia may be grouped insorior grow on definite margins or edges of the leaflet. Within the sporangia, haploid spores are produced from spore mother cells by meiosis . The spores are released at the proper time, fall to the ground, and directly develop into small, flat, green, photosynthetic, heart-shaped gametophytes .

Question:

Design a combinational circuit that accepts a three-bit number and generates an output binary number equal to the square of the input number.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G04-0070.htm

Solution:

The largest number to be squared is decimal seven or 111_2. With decimal seven as an input the output will be 49_10 or 110001_2 hence there will be three inputs, one for each bit, and six outputs. A truth table is made with inputs I_0, I_1, I_2 and outputs A_0, A_1, A_2, A_3, A_4, A_5. I_0 and A_0 are the most significant bits. mintermsInputs I_0I_1I_2 Outputs A_0A_1A_2A_3A_4A_5 0 000 000000 1 001 000001 2 010 000100 3 011 001001 4 100 010000 5 101 011001 6 110 100100 7 111 110001 From the truth table it is seen that A_0 (I_0 , I_1 , I_2) = \sum 6, 7 A_1 (I_0 , I_1 , I_2) = \sum 4, 5, 7 A_2 (I_0 , I_1 , I_2) = \sum 3, 5 A_3 (I_0 , I_1 , I_2) = \sum 2, 6 A_4 (I_0 , I_1 , I_2) = \sum none - connect to Logic 0 A_5 (I_0 , I_1 , I_2) = \sum 1, 3, 5, 7 Use 3-variable K-maps to simplify the expressions A_0 = I_0I_1 A_1 = I_0I_1 + I_0 I_2 A_2 = I_0I_1 I_2 +I_0 I_1 I_2 A_3 = I_1I_2 A_4 = 0 A_5 = I_2 A logical diagram is drawn from the minimized equations. Figure 1 shows the combinational circuit.

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Question:

What is the pH of a 1.0 M solution of the strong electrolyte sodium acetate? The dissociation constant of acetic acid is K_a = 1.8 × 10^-5 mole/liter.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E10-0363.htm

Solution:

The first step is the determination of the hydrolysis constant for sodium acetate. From this we obtain the concentration of hydroxyl contributed by the hydrolysis of sodium acetate. The concentration of hydroniumion, and consequently the pH, is determined by using the water constant. The dissociation of acetic acid (HAc) intohydroniumions and acetate ions (Ac-) may be represented by the equation HAc+ H_2O \rightleftarrows H_3O^+ + Ac^-. The dissociation constant for this reaction is K_a = {[H_3O^+][Ac^-]} / [HAc] This constant and the water constant, K_W = [H_3O^+][OH^-] = 10^-14 mole^2/liter^2 will be used to determine the hydrolysis constant for acetate. Hydrolysis of acetate proceeds according to the following equation: Ac^- + H_2O \rightleftarrowsHAc+ OH^-. The hydrolysis constant isK_h= [HAc][OH^-] / [Ac^-]. This may be rewritten in terms of K_a and K_W as follows: K_h= {[HAc][OH^-]} / [Ac^-] = ({[HAc][OH^-]} / [Ac^-]) × ([H_3O^+] / [H_3O^+]) = ([HAc] / {[Ac^-][H_3O^+]}) × ([H_3O^+][OH^-]) = (1/K_a) × K_W = (K_W/K_a) .Hence, K_h= {[HAc][OH^-]} / [Ac^-] = (K_W/K_a) = [(10^-14 mole^2/liter^2)/(1.8 × 10^-5 mole/liter)] = 5.6 × 10^-10 mole/liter . Let the equilibrium concentration ofHAcformed by the hydrolysis of acetate be x. Since one mole of OH^- is formed per mole ofHAcformed, the equilibrium concentration of OH^- is also x. Furthermore, if we assume that sodium acetate dissociates completely, then the initial con-centration of Ac^- is equal to the concentration of sodium acetate (1.0 M) and the equilibrium concentration of acetate is 1.0 - x. Note that we have neglected the contribution to [OH^-] from the hydrolysis of water. Substituting these concentrations into the expression forK_h, we obtain K_h= 5.6 × 10^-10 mole/liter = {[HAc][OH^-]} / [Ac^-] = [(x \textbullet x)/(1.0 - x)] = [(x^2)/(1.0 - x)] To avoid use of the quadratic formula, we will assume that x is much smaller than 1,0 so that 1.0 - x \cong 1.0. (This assumption will be justified later on in the solution). Hence, we obtain 5.6 × 10^-10 mole/liter = [(x^2)/(1.0 - x)] \cong (x^2/1.0) orx = (1.0 × 5.6 × 10^-10)^1/2 = 2.4 × 10^-5 mole/liter. Since [OH^-] = x, [OH^-] =2.4 × 10^-5 mole/liter. Hence, x is much smaller than 1.0, justifying our earlier assumption We will find [H_3O^+] by use of the water constant, K_W = [H_3O^+][OH^-], or, [H_3O^+] = KW/ [OH^-] = [(10^-14 mole^2/liter^2)/(2.4 × 10^-5 mole/liter)] = 4.2 × 10^-9 mole/liter . The pH is then pH = - log [H_3O^+] = - log (4.2 × 10^-9) = - (- 9.4) = 9.4 .

Question:

Define a hormone. How would you go about proving that a particular gland is responsible for a specific function?

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/Users/wenhuchen/Documents/Crawler/Biology/F21-0530.htm

Solution:

The endocrine system constitutes the second great communicating system of the body, with the first being the nervous system. The endocrine system consists of ductless glands which secrete hormones. A hormone is a chemical substance synthesized by a specific organ or tissue and secreted directly into the blood. The hormone is carried via the circulation to other sites of the body where its actions are exerted. Hormones are typically carried in the blood from the site of production to the site(s) of action, but certain hormones produced by neuro-secretory cells in the hypothalamus act directly on their target areas without passing through the blood. The distance travelled by hormones before reaching their target area varies considerably.In terms of chemical structure, hormones generally fall into three categories:The steriod hormones include the sex hormones and the hormones of the adrenal cortex; the amino acid derivatives(of tyrosine) include the thyroid hormones and hormones of the adrenal medulla; proteins and polypeptides make up the majority of the hormones. The chemical structure determines the mecha-nism of action of the hormone. Hormones serve to control and integrate many body functions such as reproduction, organic metabolism and energy balance, and mineral metabolism. Hormones also regulate a variety of behaviors, including sexual behaviors. To determine whether a gland is responsible for a particular function or behavior, an investigator usually begins by surgically removing the gland and observing the effect upon the animal. The investigator would then replace the gland with one transplanted from a closely related animal, and determine whether the changes induced by removing the gland can be reversed by replacing it. When replacing the gland, the experimenter must ensure that the new gland becomes connected with the vascular system of the recipient so that secretions from the transplanted gland can enter the blood of the recipient. The experimenter may then try feeding dried glands to an animal from which the gland was previously removed. This is done to see if the hormone can be replaced in the body in this manner. The sub-stance in the glands will enter the blood stream via the digestive system and be carried to the target organ by the circulatory system. Finally, the experimenter may make an extract of the gland and purify it to determine its chemical structure. Very often the chemical structure of a substance is very much related to its function. Studying the chemical structure may enable the investigator to deduce a mechanism by which the gland-extract functions on a molecular level. The investigator may also inject the purified gland-extract into an experimental animal devoid of such a gland, and see whether the injection effected replacement of the missing function or behavior. Some hormonal chemicals have additive effects. The investigator may inject a dosage of the purified gland-extract to an intact animal to observe if there was any augmentation of the particular function or behavior under study.

Question:

Assuming the same declarations as in Problem 14, write PROCEDURE INSERT (var S1:string; S2:string, p:integer); which inserts string S2 into string S1 after position P.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G16-0409.htm

Solution:

This is a more generalized version of the CONCAT Procedure. First, as usual, we check to see if the array of size 80 can accommodate both strings. If not, we print out the error message. Otherwise, we move last (length 1- p) elements forward by the length of the second string. This prepares space for the insertion of the second string. This operation is accomplished by the first FOR loop. Then, by the second FOR loop, we actually move S2 into S1, starting with position p + 1. Again, note that I - p indicates the cur-rent position of the character in S2 to be copied into S1. The only thing remaining then is to adjust the length of the string S1. PROCEDURE INSERT (var S1: string; S2 : string); VAR I: integer; BEGIN IF (S1.length + S2.length) > strsize then write1n ('insertion impossible-strings too long') ELSE BEGIN FOR I: = S1.length down to p + 1 DO S1.WORD [I+S2.length]:= S1.WORD[I] ; FOR I:= p+1 to p+S2.length DO S1.WORD [I]: =S2.WORD [I-p]; S1.length:=S1.length + S2.length END END;

Question:

If a person walks 1 km north, 5 km west, 3 km south, and 7 km east, find the resultant displacement vector.

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/Users/wenhuchen/Documents/Crawler/Physics/D01-0007.htm

Solution:

The vector diagram is shown in figure (a). The resultant displacement vector is labelled R\ding{217}. The magnitude of this vector is 2.8 km. The direction, as measured with a protractor, is 45\textdegree south of east, or the tangent may be used to find the direction, since a right triangle is formed. We shall also compute the solution analytically. In figure (b) a closeup of the resultant vector R\ding{217} is shown. We can see from the graph that side A and side B each equal 2 km. Thus, by the Pythagorean theorem: R^2 = A^2 + B^2 = (2 km)^2 + (2 km)^2 = 8 km^2 R = 2 \surd2 km = 2(1.4)km = 2.8 km tan \texttheta = (2 km/2 km) = 1, \texttheta = 45\textdegree R\ding{217} = 2.8 km, 45\textdegree south of east.

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Question:

Calculate the pH at 25\textdegreeC of a solution containing 0.10 M sodium acetate and 0.03 M acetic acid. The apparentpKfor acetic acid at this ionic strength is 4.57.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E12-0433.htm

Solution:

This problem involves the calculation of [H^+] (pH = - log [H^+]) in a mixture of amonoproticacid (acetic acid) and its completely dissociated salt (sodium acetate). Because the addition of a salt represses the ionization of the acid, the concentration ofundissociatedacid is approximately equal to the molar concentration of added weak acid (C_a). The addition of the acid represses the hydrolysis of the salt so that the concentration of anions of the weak acid is approximately equal to the molar concentration of the salt C_s. Using several approximations, thepKfor a weak acid is the pH of a solution containingequimolarquan-tities of salt and acid. However, if the solution does not containequimolarquantities of salt and acid, then the following approximation holds: pH =pK+ log [C_s / C_a] , where C_s and C_a are the concentrations of the salt and the acid, respectively. Thus, in this problem, C_s = 0.10 MC_a = 0.03 M pK = 4.57 andpH = 4.57 + log [0.10 / 0.03] = 4.57 + 0.52 = 5.09.

Question:

A long, horizontal, rigidly supported wire carries a current of 50 A. Directly above it and parallel to it is a fine wire, the weight of which is 0.075 N per meter, which carries a current of 25 A. How far above the first wire should the second wire be strung in order for it to be supported by magnetic repulsion?

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/Users/wenhuchen/Documents/Crawler/Physics/D21-0726.htm

Solution:

If the upper wire is to be supported by magnetic repulsion, the magnetic force per unit length (F/l) must just equal the weight of a unit of length of the wire (mg/l). Further, the currents in the two wires must be in opposite directions in order for the force between the wires to be one of repulsion. Hence (mg/l) = (F/l) but(F/l) = (\mu_0 /2\pi) [(I I')/r](see the figure) \therefore(mg/l) = (\mu_0 /2\pi) [(I I')/r] r = [(\mu_0 l II')/(2\pimg)] = [(2 × 10^-7 N \bullet A^-2 × 50A × 25 A) / (0.075 N \bullet m^-1 )] = 0.33 × 10^-2 m = 0.33 cm. The wires must therefore be very thin in order to allow their centers to be so close together.

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Question:

The earthworm has a central nerve cord running along the entire length of the body. When an earthworm is cut into several parts, severing its nerve cord in the process, each part will go on crawling around for some time. How can each fragment carry out crawling movement even after the nerve cord is severed?

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/Users/wenhuchen/Documents/Crawler/Biology/F12-0300.htm

Solution:

The arrangement of the nervous system in lower invertebrates is quite different from that in higher vertebrates like man. In higher vertebrates, there is an expanded and highly developed anterior end of the spinal cord, forming the brain. The brain coordinates and regu-lates the activities of the entire body. When the brain is separated from the rest of the body, the animal cannot perform any complicated functions. However, in the earthworm, a higher invertebrate, coordinated activities such as crawling can still be ob-served when the body is cut into several transverse sections. This movement is possible because the earth-worm has more than one neural center controlling and coordinating its activities. The nervous system of the earthworm consists of a large, two-lobed aggregation of nerve cells, called the brain, located just above the pharynx in the third segment, and asubpharyngealganglion just below thepharnyxin the fourth segment. A nerve cord connects the brain to the subpharyngeal ganglion and extends from the anterior to the posterior end of the body. In each segment of the body there is a swelling of the nerve cord, called a segmental ganglion. Sensory and motor nerves arise from each segmental ganglion to supply the muscles and organs of that segment. The segmental ganglia coordinate the contraction of the longitudinal and circular muscles of the body wall, so that the worm is able to crawl. When the earthworm is cut into several pieces, thus sever-ing the connection to the brain (that is, the nerve cord), the resulting fragments still contain segmental ganglia which can fire impulses to the muscles of the body wall, resulting in crawling movement.

Question:

A long straight wire in a house carries an alternating current which has a maximum value of 20 amp. Calculate the maximum magnetic induction a distance lm away from the wire.

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/Users/wenhuchen/Documents/Crawler/Physics/D22-0745.htm

Solution:

The magnetic induction at a distance R from a wire contain-ing a current i can be found by using Ampere's Law. Given a closed path c, and a current i through the closed curve, the following re-lation holds: \oint_c B^\ding{217} \textbullet dl^\ding{217} = \mu_0 i where the circle on the integral sign Indicates that c is closed, and the permeability constant \mu_0 = 4\pi × 10^-7 weber/amp-meter . Let the closed curve c of the above integral be in the shape of a circle of radius R, concentric with the current carrying wire as shown in the figure. Ampere's Law then becomes: \int_c B cos \texttheta dl = \int_c B dl Since B^\ding{217} and dl^\ding{217} are collinear and \texttheta = 0 o . B^\ding{217} is constant along the path c, therefore B \int dl = \mu_0 i B \bullet 2\piR = \mu_0 i orB = [(\mu_0 i)/(2\piR)] from which we have B_max = (\mu_0 /2\pi) (i_max /R) = 2 × 10^-7 (\omega/a-m) × (20a/1m) = 4 × 10^-6 (\omega/m^2) This can be compared with the earth's magnetic field, which has an Intensity of the order of 5 × 10^-5 (W/m^2) .

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Question:

Write a BASIC program to solve Laplace's equation (\partial2V/\partialx2) + (\partial2V/\partialy2) = 0. Consider a rectangular 8 by 8 grid with boundary conditions such that 3 sides are assumed to be at a potential of 100. The remaining side is at zero potential as shown.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G21-0533.htm

Solution:

We use the replacement of partial derivatives by central dif-ferences (\partial2V/\partialx2) = (V_i-1,_j - 2V_i,_j + V_i+1,_j ) / (∆x2) and (\partial2V/\partialy2) = (V_i,_j-1 - 2V_i,_j + V_i, _j+1) / (∆y2) Thus [(V_i-1,_j - 2V_i,_j + V_i+1,_j ) / (∆x2)] + [(V_i,_j-1 - 2V_i,_j + V_i, _j+1) / (∆y2)] = 0. But, in the given rectangular grid of little squares, ∆x = ∆y and hence we obtain simply V_ij = (1/4) [(V_i-1,_j + V_i+1,_j + V_i,_j-1 + V_i, _j+1]. Thus when Laplace's equation is satisfied, the value of the function at each grid point is the average of the values of the function at the four closest neighbors. When exterior boundary conditions are specified, the solution, by repeated applications of this algorithm flows out from the boundary and stabilizes when the above condition for V_i,jis met. Initial values of the potential are given in the data statements within the program. Each line, represents one row of data in the grid. The problem specified the boundary values of the potential. Interior values of the potential are guesses. Convergence is studied by considering d = V_ij new -V_ij = = (1/4) [(V_i+1,_j + V_i, _j+1 + V_i-1,_j + V_i, _j-1 -4V_ij]. S is used to accumulate the sum of all changes as we sweep over the grid. In addition, the counter K is used to accumulate the number of passes over the grid. The process is repeated until the total change in any sweep (the variable S) is less than some arbitrary amount. 1\O\O REM ELLIPTIC PARTIAL DIFFERENTIAL EQUATION 1\O1 REM LAPLACE'S EQUATION 11\O FOR I = 1 TO 8 111 FOR J = 1 TO 8 112 READ V(I,J) 113 NEXT J 114 NEXT I 12\O LET K = \O 13\O LET S = \O 14\O FOR I = 2 TO 7 15\O FOR J = 2 TO 7 16\O LET D = (V(I+1,J) + V(I,J+1) + V(I-1,J) + V(I,J-1) -4\textasteriskcenteredV(I,J))/4 161 LET S = S + ABS(D) 162 LET V(I,J) = V(I,J) + D 163 NEXT J 164 NEXT I 17\O LET K = K + 1 18\O IF S > 2 THEN 13\O 2\O\O PRINT 2\O2 FOR I = 1 TO 8 2\O3 FOR J = 1 TO 8 2\O4 PRINT INT(V(I,J)); 2\O5 NEXT J 2\O6 PRINT 2\O8 NEXT I 2\O9 PRINT 212 PRINT "NUMBER OF PASSES FOR CONVERGENCE =\textquotedblright; K 8\O1 DATA 1\O\O, 1\O\O, 1\O\O, 1\O\O, 1\O\O, 1\O\O, 1\O\O, 1\O\O 8\O2 DATA 1\O\O,9\O,9\O,9\O,9\O,9\O,9\O, 1\O\O 8\O3 DATA 1\O\O,8\O,8\O,75,75,8\O,8\O, 1\O\O 8\O4 DATA 1\O\O,75,6\O,5\O,5\O,6\O,75, 1\O\O 8\O5 DATA. 1\O\O,6\O,5\O,3\O,3\O,5\O,6\O, 1\O\O 8\O6 DATA. 1\O\O,5\O,3\O,15,15,3\O,5\O, 1\O\O 8\O7 DATA. 1\O\O,4\O,2\O,5,5,2\O,4\O, 1\O\O 8\O8 DATA 1\O\O,\O,\O,\O,\O,\O,\O, 1\O\O 999 END Sample output is given 1\O\O 1\O\O 1\O\O 1\O\O 1\O\O 1\O\O 1\O\O 1\O\O 1\O\O 97 95 94 94 95 97 1\O\O 1\O\O 94 9\O 88 88 9\O 94 1\O\O 1\O\O 9\O 83 8\O 8\O 83 9\O 1\O\O 1\O\O 84 73 68 68 74 84 1\O\O 1\O\O 73 58 52 52 58 73 1\O\O 1\O\O 52 34 29 29 35 52 1\O\O 1\O\O \O \O \O \O \O \O 1\O\O NUMBER OF PASSES FOR CONVERGENCE = 22

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Question:

A technician is working with dry ice (solid CO_2) in a closed 6000 liter fume hood maintained at a constant pressure of 1.00atmand a constant temperature of 27\textdegreeC. If he has not been conditioned to tolerate CO_2, he will succumb to CO_2 poisoning when the concen-tration rises to 10%, by volume, of the atmosphere. If the ventilation stops, what minimum weight of dry ice must sublimate to constitute a hazard? Assume that there was no CO_2 present initially.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E03-0093.htm

Solution:

To solve this problem we will convert the concentration, 10% by volume, to a partial pressure and then determine what mass of CO_2 gives rise to this pressure. At the point where CO_2 is harmful, it occupies 10% by volume of the total atmosphere. Its partial pressure p is therefore 10% of the total pressure, or p = 10% × 1atm= 0.10 atm. The ideal gas law reads pV=nRT= [m/MW] RT where V is the volume, n the number of moles, R the gas constant, T the absolute temperature, m the mass, and MW the molecular weight, and we have used n = m/MW. Solving for m m = [pV(MW)] / RT Now p = 0.10atm, V = 6000 liters, MW = 44 g/mole of CO_2, R = 0.082 liter-atm/mole-degree, and T = 27\textdegreeC = 300\textdegreeK. Hence, m = [pV(MW)] / RT = (0.10atm× 6000 liters × 44 g/mole) / (0.082 liter-atm/mole-deg × 300\textdegreeK) \cong 1100 g. Thus, 1100 g, or 1.10 kg, must sublimate before there is sufficient CO_2 in the atmosphere to be harmful.

Question:

A convenient unit of energy in atomic physics and in nuclear physics is the electron volt (ev). One electron volt is defined as the potential energy difference of a charge e between two points having a potential difference of one volt. What is the equivalent of electron volts in ergs? What is the kinetic energy of an alpha particle (He^4 nucleus or doubly ionized helium atom) accelerated from rest through a potential difference of 1000 volts?

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/Users/wenhuchen/Documents/Crawler/Physics/D19-0619.htm

Solution:

Potential energy is equal to e ∆\textphi. This is so since the change in energy is \intF \textbulletdsandF =qE so that energy is q \intE \textbulletdsorq ∆\textphi,and here q = e. Therefore, 1 electron volt = e ∆\textphi \cong (4.80x10^-10esu) [(1/300)statvolt] [300 volts = 1statvolt] \cong 1.60 x 10-^-12 erg The kinetic energy of the alpha particle is: 2e(1000 volts) = 2000 electron volts where 2000ev= (2 x 10^3ev)[1/300] (4.80 x 10^-10esu) = 3.2 x 10^-9 erg

Question:

At a certain temperature,Keqfor the reaction 3C_2H_2 \rightleftarrows C_6H_6 is 4. If the equilibrium concentration of C_2H_2 is 0.5 mole/liter, what is the concentration of C6H_6 ?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E09-0306.htm

Solution:

The equilibrium constant (Keq) for this reaction is stated: Keq= {[C6H_6 ] / [C_2H_2 ]^3} where [ ] indicate concentration. The [C_2H_2 ] is bought to the third power because three moles of it react. Equilibrium is defined as the point where no more product is formed and no more reactant is dissipated; thus their concentrations remain constant. Here, one is givenKeqand [C_2H_2 ] and asked to find [C_6H_6 ] . This can be done by substituting the given into the equation for the equilibrium constant. Keq= {[C6H_6 ] / [C_2H_2 ]^3}Keq= 4 [C_2H_2] = 0.5 moles/liter 4 = {[C6H_6 ] / [0.5 ]^3} [C_6 H_6 ] = (0.5)^3 × 4 = 0.5 moles/liter.

Question:

(a) From what height above the bottom of the loop must the car in the figure start in order to just make it around the loop? (b) What is the velocity of the car at point A and at point B?

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/Users/wenhuchen/Documents/Crawler/Physics/D05-0245.htm

Solution:

(a) For the car to just make it around the loop, its speed at the highest point of the loop must be such that the force of gravity on it is sufficient to provide the centripetal force needed to keep it in a circular path. For this to be the case, F_centripetal = (mv^2_c) / R = mg or the velocity must be given by v_c =\surdRg(1) at point C. Neglecting friction, we use conservation of energy and note that the velocity of the car at point D must be the same as at point c, since both correspond to the same net change in potential energy relative to the starting point. The change in potential energy equals the change in kinetic energy and is proportional to the square of the velocity. The change in potential energy from the starting point to point D equated to the corresponding change in kinetic energy yields. mgh = (1/2)m(vD- v_0)^2 = (1/2)mv^2_D v_D = \surd2gh(2) where h is the vertical height of the starting point above points C and D (as shown in figure), and V_0 is the initial velocity of the car, which is zero at the starting point. Equating equations (1) and (2), \surdRg = \surd2gh we haveh = R/2. This indicates that the starting point must be 2R + R/2 = (5/2)R above the bottom of the loop in order for the car to have just enough energy to make it around the loop. (b) Equating the change in potential energy to the kinetic energy of the car at the point in question, as we did in part (a), we find for point A mg(h + 2R) = (1/2)mv^2_A (5/2) Rg = (1/2) v^2_A v_A = \surd(5 gR) for point B,mg(h + R) = (1/2) mv^2_B (3/2) gR = (1/2)v^2_B v_B = \surd(3gR)

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Question:

The concentration of sodium ions (Na+) inside most cells is lowerthan the concentration outside the cells. Why can't this phenomenonbe explained by simple diffusion across the membraneand what process is responsible for this concentrationdifference?

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/Users/wenhuchen/Documents/Crawler/Biology/F02-0061.htm

Solution:

Since the cell membrane is somewhat permeable to sodium ions, simplediffusion would result in a net movement of sodium ions into the cell, until the concen-trations on the two sides of the membrane became equal. Sodium actually does diffuse into the cell rather freely, but as fast asit does so, the cell actively pumps it out again, against the concentrationdifference. The mechanism by which the cell pumps the sodium ions out is calledactive transport. Active transport requires the expenditure of energy forthe work done by the cell in moving molecules against a concentration gradient. Active transport enables a cell to maintain a lower concentration ofsodium inside the cell, and also enables a cell to accumulate certain nutrientsinside the cell at concentrations much higher than the extracellularconcentrations. The exact mechanism of active transport is not known. It has been proposedthat a carrier molecule is involved, which reacts chemically with themolecule that is to be ac-tively transported. This forms a compound whichis soluble in the lipid portion of the membrane and the carrier com- poundthen moves through the membrane against the concen-tration gradientto the other side. The transported molecule is then released, and thecarrier molecule diffuses back to the other side of the membrane whereit picks up another molecule. This process requires energy, since workmust be done in transporting the molecule against a diffusion gradient. The energy is supplied in the form of ATP. The carrier molecules are thought to be integral proteins; proteins whichspan the plasma membrane. These proteins are specific for the moleculesthey transport.

Question:

Explain how the changes in the phytochrome system are related to blooming.

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/Users/wenhuchen/Documents/Crawler/Biology/F09-0238.htm

Solution:

Phytochrome is a pigment found in certain plants in very small amounts. The pigment is thought to be responsible for detecting the light environment of the plant. Phytochrome exists in two different forms - P_660 the inactive form, and P_730, the active form. P660 absorbs red light and is converted to P_730; P_730 absorbs far-red light and is converted back to P_660. In nature, P_660 is converted to P_730 in the daytime (when red wavelengths predominate over far-red). The reverse occurs in the night time. Hence, P_730 predominates in the plant during the day and P_660 predominates during the night. This is summarized in the figure below. This differ-ential response of phytochrome to light and dark is thought to be involved in the phenomenon of photoperiodism in plants. It has been found that red light inhibits flowering in short-night plants but promotes flowering in long-night plants, under conditions during which flowering normally takes place. This experimental observation led many investigators to hypothesize that the P_730 - P_660 inter-conversion might be the plant's time-regulator of flowering. According to this hypothesis, P_730, converted from P_660 by the absorption of red light, would inhibit flowering in short-day plants but promote flowering in long-day plants. Because P_730 accumulates in the day and diminishes at night, short-day plants could flower only if the nights were long enough, during which a great amount of P_730 would be inactivated. Long-day plants, on the contrary, would require short nights, during which the P_730 would not be completely inactivated, so that enough P_730 would remain at the end of the night to promote flowering. This hypothesis was highly regarded in the beginning. However, it was dealt a fatal blow by a later study showing that the reversion of all the P_730 to P_660 requires only three hours; this has been found to hold true in all plants, both long and short-day, that have been studied. What happens during the rest of the dark period? No sat-isfactory answer is as yet available. However, extended days provide P_730 for longer periods. It is generally agreed at present that the time-measuring phenomenon of flowering is not solely controlled by the interconversion of P_660 and P_730 . There seems to be two variables involved. Experimentally it has been found that one is the presence or absence of light, the other is the length of the dark or light period. Phytochromes seem to be responsible for the detection of either light or darkness, but the actual measuring of the time between the moment the plant senses the onset of darkness and the moment it senses the next exposure to light depends on some mechanism, not yet known. Further investigation is needed before we can fully understand the "biological clock\textquotedblright in flowering.

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Question:

The mass of a bullet is 2 grams and its velocity is 30,000 centimeters per second (approximately true for a .22 caliber bullet). What is its kinetic energy?

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/Users/wenhuchen/Documents/Crawler/Physics/D05-0225.htm

Solution:

K.E.= 1/2 Mv^2 = [2 gm × (30,000 cm/sec)^2] / 2 = (30,000)^2 ergs.

Question:

Design a computer program to playHexapawn. What is unusual about this program?

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G23-0561.htm

Solution:

Hexapawnis played on a 3 × 3 chess board. There are six pieces - three white pawns and three black pawns. The movements of the pawns are the same as in chess. The initial position is B B B W W W The object of the game is to get a passed pawn i.e. a pawn on the enemy's last rank. The first one to obtain a passed pawn (i.e. to place his pawn to any starting position of his opponent's pawns) wins. Alternatively, if no more moves are possible, (the position is totally blocked), then the person who made the last move is the winner. The squares are num-bered as follows: 123 456 789 The program forHexapawnis an application of cybernetics. The computer records every unfamiliar position in its memory including all the moves that follow. Now assume that the computer loses a game. It erases the move that led to de-feat. As an example: B Computer You W B W Here, with the computer to move, it could win by playing 6 to 9 (or even 1 to 5). However, if it plays 1 to 4 then you can win by moving 5 to 2. If this same position occurs in another game, the computer (assuming it played the losing move in the first game) will not even consider 1 to 4 but will play 6 to 9 or 1 to 5 instead. The computer has learn-ed from the first game. If the computer gets a position from which all moves have been deleted (they all led to defeat) it erases the move that got it there and resigns. In this way it saves only winning sequences of moves. If it plays long enough and encounters a great variety of unfamiliar positions it will become unbeatable. Note that this method of "learning" cannot be used to create an unbeatable program in chess. The number of different positions in chess is of astronomical order and the chances of a computer encountering an exactly similar position twice are vanishingly small (apart from the openingand elementary end-game positions). Thus the program would have to be modified withstrategicalconcepts and evaluation of would have to be modified withstrategicalconcepts and evaluation of pieces. pieces. 10PRINT "HEXAPAWN" 20DIM B (19,9) M (19,4), S (9), P$ (3) 30W = O: L = 0 40DEF FNR(X) = - 3\textasteriskcentered(X = 1) - (X = 3) - 4\textasteriskcentered(X = 6) - 6\textasteriskcentered(X = 4) - 7\textasteriskcentered(X= 9) - 9\textasteriskcentered(X = 7) + FNS(X) 60DEF FNS(X) = - X\textasteriskcentered(X = 2 OR X = 5 OR X = 8) 70DEF FNM(Y) = Y - INT(Y/10)\textasteriskcentered10 80P$ = "X\bullet0" 90FOR I = 1 TO 19: FOR J = 1 TO 9: READ B(I,J): NEXT J:NEXT I 100FOR I = 1 TO 19: FOR J = 1 TO 4: READ M(I,J): NEXT J: NEXT I 110X = 0: Y = 0 120S(4) = 0: S(5) = 0: S(6) = 0 130S(1) = - 1: S(2) = - 1: S(3) = - 1 140S(7) = 1: S(8) = 1; S(9) = 1 150GOSUB 1200 160PRINT "YOUR MOVE"; 170INPUT M1,M2 180IF M1 = INT (Ml) AND M2 = INT (M2) AND M1 > 0 AND M1 < 10 AND M2 > 0 AND M2 < 10 THEN 220 200PRINT "ILLEGAL COORDINATES\textquotedblright. 210GOTO 160 220IF S (M1) = 1 THEN 240 230PRINT "ILLEGAL MOVE.": GOTO 160 240IF S(M2) = 1 THEN 230 250IF M2 - M1 <> - 3 AND S(M2) <> - 1 THEN 230 260IF M2 > M1 THEN 230 270IF M2 - M1 = - 3 AND (S(M2) <> 0) THEN 230 280IF M2 - M1 < - 4 THEN 230 290IF M1 = 7 AND M2 a 3 THEN 230 300S (M1) = 0 310S (M2) = 1 320GOSUB 1200 330IF S (1) = 1 OR S (2) = 1 OR S (3) = 1 THEN 960 340FOR I = 1 TO 9 342IF S(I) = - 1 THEN 348 344NEXT I 346GOTO 960 348FOR I = 1 TO 9 350IF S (I) <> - 1 THEN 420 360IF S (I + 3) = 0 THEN 440 370IF FNR (I) = I THEN 410 380IF I > 3 THEN 396 390IF S (5) = 1 THEN 440 393GOTO 420 396IF S (8) = 1 THEN 440 400GOTO 420 410IF S (I + 2) = 1 OR S (I + 4) = 1 THEN 440 420NEXT I 430GOTO 960 440FOR I = 1 TO 19 450FOR J = 1 TO 3 460FOR K = 3 TO 1 STEP - 1 470T ((J - 1)\textasteriskcentered3 + K) = B (I,( J - 1)\textasteriskcentered3 + 4 - K) 480NEXT K 490NEXT J 500FOR J = 1 TO 9 510IF S (J) <> B (I,J) THEN 542 520NEXT J 530R = 0 540GOTO 590 542FOR J = 1 TO 9 544IF S (J) <> T (J) THEN 550 546NEXT J 548R = 1 549GOTO 590 550NEXT I 560REM THE TERMINATION OF THIS LOOP IS IMPOSSIBLE 570PRINT "ILLEGAL BOARD PATTERN." 580STOP 590X = I 600FOR I = 1 TO 4 610IF M (X, I) <> 0 THEN 650 620NEXT I 630PRINT "I RESIGN" 640GOTO 960 650Y = INT(RND (1)\textasteriskcentered4 + 1) 660IF M (X,Y) = 0 THEN 650 670IF R <> 0 THEN 730 680PRINT "I MOVE FROM"; STR$ (INT(M(X,Y)/10)); "TO"; STR$ (FNM (M(X,Y))) 700S (INT (M(X,Y)/10) ) = 0 710S (FNM (M(X,Y))) = - 1 720GOTO 770 730PRINT "I MOVE FROM"; STR$ (FNR (INT (M(X, Y) /10))) ; "TO" 740PRINT STR$ (FNR (FNM (M(X,Y)))) 750S (FNR (INT (M(X,Y)/10))) = 0 760S (FNR (FNM (M(X,Y)))) = - 1 770GOSUB 1200 780IF S (7) = - 1 OR S (8) = - 1 OR S (9) = - 1 THEN 1020 790FOR I = 1 TO 9 800IF S(I) = 1 THEN 830 810NEXT I 820GOTO 1020 830FOR I = 1 TO 9 840IF S (I) <> 1 THEN 930 850IF S (I - 3) = 0 THEN 160 860IF FNR (I) = I THEN 920 870IF I < 7 THEN 900 880IF S (5) = - 1 THEN 160 890GOTO 930 900IF S (2) = - 1 THEN 160 910GOTO 930 920IF S (I - 2) = - 1 OR S (I - 4) = - 1 THEN 160 930NEXT I 940PRINT "YOU CAN'T MOVE, SO"; 950GOTO 1020 960PRINT "YOU WIN" 970M (X,Y) = 0 980L = L + 1 990PRINT "I HAVE WON"; W;" AND YOU"; L; 1000"OUT OF"; L + W; "GAMES." 1010PRINT: GOTO 30 1020PRINT "I WIN" 1030W = W + 1 1040GOTO 990 1050DATA- 1, - 1, - 1,1,0,0,0,1,1, - 1, - 1, - 1,0,1,0,1,0,1 1060DATA- 1,0, - 1, - 1,1,0,0,0,1,0, - 1, - 1,1, - 1,0,0,0,1 1070DATA- 1,0, - 1,1,1,0,0,1,0, - 1, - 1,0,1,0,1,0,0,1 1080DATA0, -1, -1,0, -1,1,1,0,0,0, -1, -1, -1,1,1,1,0,0 1090DATA- 1,0, - 1, - 1,0,1,0,1,0,0, - 1, - 1,0,1,0,0,0,1 1100DATA0, - 1, - 1,0,1,0,1,0,0, - 1,0, - 1,1,0,0,0,0,1 1110DATA0,0, - 1, - 1, - 1,1,0,0,0, - 1,0,0,1,1,1,0,0,0 1120DATA0, - 1,0, - 1,1,1,0,0,0, - 1,0,0, - 1, - 1,1,0,0,0 1130DATA0,0, - 1, - 1,1,0,0,0,0,0, - 1,0,1, - 1,0,0,0,0 1140DATA- 1,0,0, - 1,1,0,0,0,0 1150DATA24,25,36,0,14,15,36,0,15,35,36,47,36,58,59,0 1160DATA15,35,36,0,24,25,26,0,26,57,58,0 1170DATA26,35,0,0,47,48,0,0,35,36,0,0,35,36,0,0 1180DATA36,0,0,0,47,58,0,0,15,0,0,0 1190DATA26,47,0,0,47,58,0,0,35,36,47,0,28,58,0,0,15,47,0,0 1200PRINT: FOR I = 1 TO 3 1210FOR J = 1 TO 3 1220PRINT TAB (10); MID$ (P$,S((I - 1)\textasteriskcentered3 + J) + 2,1); 1230NEXT J 1240PRINT: NEXT I 1250PRINT 1260RETURN 1270END

Question:

One end of a cylindrical glass rod is ground to a hemispherical surface of radius R = 2 cm. An object of height h_0 = 1 mm. is placed on the axis of the rod, p = 8 cm to the left of the vertex. Find the image distance q and the image height h_I (a) when the rod is in air, (b) the rod is in water. The indices of refraction of glass and water are 1.50 and 1.33 respectively.

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/Users/wenhuchen/Documents/Crawler/Physics/D27-0866.htm

Solution:

(a) The optical equation for the spherical end of the rod (which is a spherical refracting surface), is given by (n/p) + (n'/q) = {(n' - n)/R} where the object is in the region of refraction index n, the rays are going from the region of refractive index n to the region of refractive index n', and R is the curvature of the refractive surface. The radius R for the glass hemisphere has a positive sign because the center of curvature of the glass surface lies inside the glass, which is the real image region (see Fig. 1). R is negative when the curvature center lies in the virtual image region. As shown in Fig. 1 above, p = +8 cm, R = +2 cm. Therefore, the image distance q is obtained as (1.5/q) + (1/8)cm = {(1.5 - 1)/2 cm} or q = + 12 cm q is positive, i.e., the image is formed at the right of the vertex of the glass surface and is real. The magnification m for this problem is given by m = - (n/n),(q/p) = - (1/1.5) × (12 cm/8 cm) = - 1 that is, the image has the same height as the object, but is inverted, h_I = -h_0 = - 0.1 cm. (b) When the rod is in water, the optical equation yields for q (Fig. 2) (1.5/q) + (1.33/8 cm) = {(1.50 - 1.33)/2 cm} orq = - 18 cm. Since q is negative, the image is formed in the water and is imaginary. For magnification, we have m = - (1.33/1.5) × (-18 cm/8 cm) = 1.99, and the image is erect and has a height h_I = 1.99 h_0 = 0.99 cm.

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Question:

Define a process, and describe its states during its life cycle.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G01-0021.htm

Solution:

A process is internally defined by the operating system. Normally, it is a piece of sequential machine code (a program segment) that is kept track of. This is done by creating a data structure in memory, called Process Control Block (PCB), to keep track of this runnable code segment. The PCB contains the required information about the process: Name or identifica-tion, memory location, size, priority, address of page table (if paged memory management is used), etc. The PCB may additionally contain the "state" of the process, and the context (or the address of the context) of process. The context of a process is the content of the CPU registers at the time of its suspension, so that it can later be resumed to run again. The states of a process, kept track of by the low-level CPU scheduler that is part of the operating system nucleus (kernel) are as shown below: SUBMITTED : The job for which the process needs to be created has been submitted by the user. This job may be waiting in a job spooling queue on the disk. Handled by the high-level job scheduler of the operating system, the job may cause one or more processes to be created. For example, if the job has a READ statement for input, a system program that does this read will have the corresponding process created for it. READY : The PCB for the process has been created and stored in a system table called "ready list." The process has all the re-sources assigned (e.g. memory, devices, etc.) except the CPU. RUNNING : The process has been assigned the CPU resource as well, and is running (executing) . At this point, the entry asso-ciated with the state of the process in its PCB may be marked running, or simply nothing is done. WAITING : The process has been suspended and its PCB has been entered into a queue (the device list, or blocked list) associ-ated with the device for which the I/O is requested. After the I/O is complete, the process is awakened. This means that its PCB is transferred from the device list to the ready list. This does not mean it starts running again. It may later be assigned the CPU to run, depending how the ready list (or queue) of PCBs are manipulated. SWAPPED : The job for which the process has been created has been rescheduled by the high-level job scheduler. This is typi-cally done when the low-level scheduler encounters a problem with the process (e.g. infinite loop, requested additional memory is simply not available, executed too long, more impor-tant processes with higher priorities caused it preemption, etc.). The low-level scheduler calls upon the swapper (or inter-mediate level scheduler) to copy the process code from main memory to disk and release all other resources held by that process. Afterwards, the job for which the process is swapped (sometimes referred to as HOLD state), is rescheduled by the high-level scheduler. Except for the partial output which may or may not be retained (e.g. how do you retain half a page printed?), unfortunately this is as if the job is submitted again. COMPLETED : The process has completed its computations and input- o utput and released the resources held. Its output may be placed on disk and associated with the job for which it is originally created. The solid arrows correspond to the state changes that are important, and normally handled by the low-level scheduler. The dashed arrows indicate the swapping that is not always imple-mented (that is, when a problem is encountered, the process may simply be terminated). The state change from RUNNING to READY state, shown with broken arrow, takes place in time-shared sys-tems where the PCB is placed at the end of the ready queue (list) after the time quantum expires.

Question:

Evaluate the inertial coefficients for a thin uniform spherical shell of mass density a per unit area and thickness d, for an axis through the center of the sphere.

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/Users/wenhuchen/Documents/Crawler/Physics/D04-0177.htm

Solution:

The angular momentum L^\ding{217} of a rotating rigid body of mass M is related to its angular velocity \omega^\ding{217} by: L_xI_xx I_xy I_xz\omega_x L_yI_yx I_yy I_yz\omega_y L_zI_zx I_zy I_zz\omega_z where ' I's are the inertial coefficients given by. I_xy = \int dm xy I_xx = \int dm (y^2 + z^2), etc. The off diagonal elements are zero for the spherical shell as a result of its symmetry. This follows, be-cause for every contribution xy to the integral for I_xy, there is an equal but opposite contribution (- x)y on the sphere. Hence, the integral of xy over a sphere is zero. The diagonal elements are equal since the three axes are equivalent as far as the geometry of the shell is concerned. The ring shown in the figure has a mass dm = \sigma (2\pi r sin \texttheta) (r d\texttheta) hence I_zz = \int dm(y^2 + x^2) = 2\pi \sigma r^2_0\int^\pi d\texttheta sin \texttheta (r^2 sin^2 \texttheta) = 2\pi \sigma r^4 _-1\int^1 d(cos \texttheta)[1 - cos^2 \texttheta] = 2\pi \sigma r^4 (4/3) = (8/3)\pi\sigmar^4 = (4\pi r^2 \sigma)(2/3)r^2 = (2/3)Mr^2 Therefore, I_xx = I_yy = I_zz = (2/3)Mr^2.

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Question:

Consider the changes in momentum produced by the follow- ing forces: (a) A body moving on the x-axis is acted on for 2 sec by a constant force of 10 n toward the right. (b) The body is acted on for 2 sec by a constant force of 10 n toward the right and then for 2 sec by a constant force of 20 n toward the left, (c) The body is acted on for 2 sec by a constant force of 10 n toward the right and then for 1 sec by a constant force of 20 n toward the left.

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/Users/wenhuchen/Documents/Crawler/Physics/D06-0319.htm

Solution:

If the mass of a system (i.e. a group of particles) is constant, we may write F_ext^\ding{217}_ = dP ^\ding{217}/dt(1) where F_ext^\ding{217} is the next external force on the system and P^\ding{217} is the total momentum of the system. (Note that if F_ext^\ding{217} = 0, we obtain the law of conversation of momentum). From (1),we find dP^\ding{217} = F_ext^\ding{217} dt or ^p\ding{217}f\int_ (p)0^\ding{217} dP^\ding{217} =^(t=t)f\int_(t=t)0 F_ext^\ding{217} dt ∆P^\ding{217} = P^\ding{217}_f - P^\ding{217}_0 = ^(t=t)f\int_(t=t)0 F_ext^\ding{217} dt(2) The left side of equation (2) is the change of momentum of the system due to F_ext^\ding{217}, and the right side is called the impulse. If F_ext^\ding{217}_ is time independent, we may write \DeltaP^\ding{217} = F_ext^\ding{217}\Deltat(3) the momentum of any body on which the force acts increases by 20 kg \textbullet m/sec. This change is the same whatever the mass of the body and whatever the magnitude and direction of its initial velocity. Suppose the mass of the body is 2 kg and that it is initially at rest. Its final momentum then equals its change in momentum and its final velocity is 10 m/sec toward the right. Had the body been initially moving toward the right at 5 m/sec, its initial momentum would have been 10 kg \textbullet m/sec, its final momentum 30 kg \textbullet m/sec, and its final velocity 15 m/sec toward the right. Had the body been moving initially toward the left at 5 m/sec, its initial momentum would have been - 10 kg \textbullet m/sec, its final momentum + 10kg \textbullet m/sec, SinceP_f^\ding{217} - P_0^\ding{217} = 20 kg \textbullet m/s P_f^\ding{217} = - 10 kg \textbullet m/s + 20 kg \textbullet m/s = 10 kg \textbullet m/s Hence its final velocity is 5 m/sec toward the right. That is, the constant force of 10 n toward the right would first have brought the body to rest and then given it a velocity in the direction opposite to its initial velocity. (b) The impulse of this force is (+10 n × 2 sec - 20 n × 2 sec) =-20 n \textbulletsec. The momentum of any body on which it acts is decreased by 20kg\textbullet m/sec. (c) The impulse of this force is (+ 10 n × 2 sec - 20 n × 1 sec) = 0. Hence the momentum of any body on which it acts is not changed. Of course, the momentum of the body is increased during the first two seconds but it is decreased by an equal amount in the next second.

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Question:

Discuss the operation of (1) a voltmeter across a 12v. battery with an internal resistance r = 2\Omega, (2) an ammeter in series with a 4\Omega resistor connected to the terminals of the same battery.

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/Users/wenhuchen/Documents/Crawler/Physics/D20-0683.htm

Solution:

(1)Consider a source whose emf \epsilon is constant and equal to 12 volts, and whose internal resistance r is 2 ohms. (The internal resistance of a commercial 12-volt lead storage battery is only a few thousandths of an ohm.) Figure 1 represents the source with a volt-meter V connected between its terminals a and b . A voltmeter reads the potential difference between its terminals. If it is of the conventional type, the voltmeter provides a conducting path between the terminals and so there is current in the source (and through the voltmeter). We shall assume, however, that the resistance of the volt-meter is so large (essentially infinite) that it draws no appreciable current. The source is then an open circuit, corresponding to a source with open terminals and the voltmeter reading V_ab equals the emf \epsilon of the source, or 12 volts. (2)In Fig. 2, an ammeter A and a resistor of resistance R = 4\Omega have been connected to the terminals of the source to form a closed circuit. The total resistance of the circuit is the sum of the re-sistance R, the internal resistance r, and the resistance of the ammeter. The ammeter resistance, however, can be made very small, and we shall assume it so small (essentially zero) that it can be neglected. The ammeter (whatever its resistance) reads the current I through it. The circuit corresponds to a source with a 4\Omega resistance across its terminals. The wires connecting the resistor to the source and the ammeter, shown by straight lines, ideally have zero resistance and hence there is no potential difference between their ends. Thus points a and a' are at the same potential and are electrically equivalent, as are points b and b'. The potential differences V_ab , and V_a'b' . are therefore equal. The current I in the resistor (and hence at all points of the circuit) could be found from the relation I = V_ab /R, if the potential difference V_ab were known. However, V_ab is the terminal voltage of the source, equal to \epsilon - Ir, and since this depends on I it is unknown at the start. We can, however, calculate the current from the circuit equa-tion: I = (\epsilon/R + r) = [(12 volts)/(4\Omega + 2\Omega)] = 2 amp. The potential difference V_ab can now be found by considering a and b either as the terminals of the resistor or as those of the source. If we consider them as the terminals of the resistor, V_a'b' = IR = 2 amp × 4\Omega = 8 volts. If we consider them as the terminals of the source, V_ab = \epsilon - Ir = 12 volts - 2 amp × 2\Omega = 8 volts, The voltmeter therefore reads 8 volts and the ammeter reads 2 amp.

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Question:

Write a program that usesforloop to output the square of numbers 1 through 100.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G17-0430.htm

Solution:

main ( ) { int number; /\textasteriskcenteredinteger whose square is calculate\textasteriskcentered/ int square _ of _ number; /\textasteriskcenteredstorage for square of number\textasteriskcentered/ for (number = 1; number < = 100; number++) /\textasteriskcenteredfor number 1 till 100 do the following\textasteriskcentered/. /\textasteriskcenteredincrement number only after executing following state-ments \textasteriskcentered/ square _ of _ number = number \textasteriskcentered number; printf ("The square of %d is %d\textbackslashn", number, square _ of _ number); }

Question:

The flashlight battery is a zinc and manganese dioxide (MnO_2) dry cell. The anode consists of a zinc can and the cathode of a carbon rod surrounded by a moist mixture of MnO_2, carbon, ammonium chloride (NH_4Cl_2), and zinc chloride (ZnCl_2). The cathode reaction is written ZMnO_2(s) + Zn^2+ + 2e^- \rightarrow ZnMn_2O_4(s) . If the cathode in a typical zinc and manganese dioxide dry cell contains 4.35g of MnO_2, how long can it deliver a steady current of 2.0 milliamperes (mA) before all its chemicals are exhausted?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E16-0573.htm

Solution:

The problem is solved by calculating the amount of charge required to exhaust the supply of MnO_2. and, from this, determining the lifetime of the battery using the relationship charge (coulombs) = current (A) × time (sec), or, time (sec) = [charge (coulombs)] / [current (A)] The cathode reaction indicates that two moles of MnO_2, 2F MnO_2, are consumed for every two moles of electrons present (2F e^-). The number of moles of MnO_2 present is [mass, MnO_2] / [molecular weight, MnO_2] = (4.35) / (87g / mole) = 0.05 mole MnO_2 . Hence, it requires 0.05 mole of electrons (or 0.05F) to consume the 0.05 mole of MnO_2 in the cathode. Converting Faraday's to coulombs (there are 96,500 coulombs in 1F), 0.05 F is equivalent to 0.05F × 96,500 coulombs/F = 4.8 × 10^3 coulombs. The battery is supposed to deliver 2.0 × 10^-3 amp. Therefore, the lifetime of the battery is time= (charge / current) = (4.8 × 10^3coulombs) / (2.0 × 10^-3 amp) = 2.4 × 10^6 sec. Therefore the battery lasts 2.4 × 10^6 sec (about 30 days).

Question:

Describe the most common process of bacterial reproduction .

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/Users/wenhuchen/Documents/Crawler/Biology/F05-0130.htm

Solution:

The most important process in the growth of bacterial populations is binary (transverse) fission. This is a type of cell division in which two identical daughter cells are produced as a result of the division of the parent cell. New cell wall material begins to form on the inner surface of the wall of the parent cell at a point midway along its length and this new wall material invaginates, dividing the cellular material of the parent cell evenly into two halves. The cell is then, separated in two by the completion of the transverse wall. Each daughter cell possesses a complete set of genetic information. The genes, the units of inheritance, are arranged in sequence along a single, circular chromosome composed of DNA. During repro-duction, the DNA is replicated and the two chromosomes move apart into separate nuclear areas in each half of the parent cell before the completion of the transverse cell wall. Each daughter cell has genetic information identical to the parent cell. This process of giving rise to new individuals by cell division is termed asexual reproduction. Although binary fission is the major method of bacterial reproduction, another form of asexual reproduction budding is observed in some bacteria. Budding involves an outgrowth of the parental cell, which enlarges and separates to form a new cell.

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Question:

Write a program which concatenates two strings "I LOVE" and "YOU.\textquotedblright, checks for presence of the character \textquotedblleft.\textquotedblright, replaces it with \textquotedblleft!\textquotedblright, and prints out the final string.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G15-0386.htm

Solution:

Concatenation is the basic operation for combining two strings to form a third. The general form of this operation is represented by the following statements: variable 1 = string 1 variable 2 = string 2 variable 3 = string 1 string 2 There must be at least one blank space on each side of the "=" sign (as well as any other arithmetic operator, except the unary minus) and between strings 1 and 2 in variable 3. The simple pattern matching statement has the form subject pattern where the two fields are separated by at least one blank. This statement tells the computer to check the string indicated by subject forthe occurrence of a string specified by the pattern. A replacement statement has the form subject pattern = object The left part of this statement operates identically with the pattern matching statement. If a match occurs, the object string replaces the pattern string in the subject string. The program follows: WORD 1 = 'I LOVE' WORD 2 = 'YOU.' WORDS = WORD 1 WORD 2 WORDS ' . ' = ' ! ' OUTPUT = WORDS END Note, that END is a label, indicating the end of program. Labels appear beforethe statements and consist of a letter or a digit followed by any number of other characters up to a blank. Also note, that single quotation marks can be used as well as double quotations.

Question:

Predict the reaction of propene with Br_2/CCI_4, H_2O/H_2SO_4, and HCI. Predict the reaction of toluene with these substances.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E21-0771.htm

Solution:

The procedure for solving this problem is as follows: write the structures of toluene and propene; classify these compounds from their structures; once these compounds are classified, you can predict the reaction that will occur with the given reagents Br_2/CCI_4, H_2O/H_2SO_4, and HCI. The structure of propene is H_2C = CH \textemdash CH_3. Notice, it has a double bond which indicates it is an alkene. Alkenes undergo the characteristic reaction of addition to the double bond. Thus, Note: Adding Br_2 breaks the double bond in a CCl_4 solvent. The structural formula of toluene is shown in figure A. The circle inside the ring indicates resonance. Its true structure is a hybrid (figure B) , which means it is stable to addition reactions. Thus, toluene will not undergo any reaction with these reactants.

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Question:

Write a subroutine which computes the roots of the quadratic equation a1x2 + a2x + a3 = 0 according to the quadratic formula: x1,2 = (-a2/2a1) \pm \surd[(a2/2a1)2 - (a3/a1)] ( = [{a2 \pm \surd(a22 - 4a1a3)} / 2a1])

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G21-0517.htm

Solution:

The only difficult part to this problem is taking into ac-count the fact that the character of the solution changes according to the sign of the discriminant, which we define as [a2/2a1]2 - a3/a1. If the discriminant is positive, then the solutions are real numbers given by the above formula. If the discriminant is negative, then the solution consists of the pair of complex conjugate numbers x1,2 = -(a2/2a1) \pm i │DISC│ where DISC is the discriminant. In this latter case, the subroutine should return the value of the real part and \vertDISC\vert for the absolute value of the imaginary part. The flowchart illustrates the required logic. CSOLUTION OF THE QUADRATIC EQUATION CA(1) × X × X + A (2) × X + A (3) = 0 SUBROUTINE QUAD (A, XR1, XR2, X1) DIMENSION A(3) X1 = -A(2)/(2.\textasteriskcenteredA(1)) DISC = X1\textasteriskcenteredX1 - A(3)/A(1) IF (DISC) 10, 20, 20 10X2 = SQRT (-DISC) XR1 = X1 XR2 = X1 XI = X2 GO TO 30 20X2 = SQRT (DISC) XR1 = X1 + X2 XR2 = X1 - X2 XI = 0.0 30RETURN END

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Question:

The density of a gas is measured as 0.222 g/l at 20.0\textdegreeC and 800 torr. What volume will 10.0 g of this gas occupy under standard conditions?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E02-0066.htm

Solution:

Since we know the density at a given temperature and pressure, we can apply the equation P /(\rhoT) = constant, which holds for ideal gases of density \rho at pressure P and absolute temperature T. We must first determine the constant for our system and then use this value along with standard pressure (760 torr) and temperature (273.15\textdegreeK) to determine the density of the gas under standard conditions. From this density, we obtain the volume that 10.0 g of the gas occupies under standard conditions. The value of the constant is, as previously indicated, P /(\rhoT) = [(800 torr)/(0.222 g/l × 20.0\textdegreeC)] = [(800 torr)/(0.222 g/l × 293.15\textdegreeK)] = 12.28 [(torr - l)/(g -\textdegreeK)] Hence, for our gas, (P/\rhoT) = 12.28 [(torr -l)/(g-\textdegreeK)]or\rho = (P/T) × [(1)/(12.28 torr -l/ g -\textdegreeK)]. Under standard conditions, the density of our gas is then \rho = (P/T) × [(1)/(12.28 torr - l/ g - \textdegreeK)] = [(760 torr)/(273.15\textdegreeK)] × [(1)/12.28 torr - l/g -\textdegreeK = 0.2265 g/l. Density is defined as= mass/volume. Hence, volume = mass/\rho, and the volume occupied by 10.0 g of our gas under standard conditions is volume = (mass/\rho) = [(10.0 g)/(.2265 g/l)] = 44.15 l.

Question:

Explain what is meant by an animal's ecological niche, and definecompetitive exclusion.

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/Users/wenhuchen/Documents/Crawler/Biology/F30-0761.htm

Solution:

Ecologically, niche is defined as the functional role and position of anorganism within its ecosystem. The term niche should not be confused withhabitat, which is the physical area where the organism lives. The characteristicsof the habitat help define the niche but do not specify it completely. Each local population of a particular species has a niche that isdefined with many variables. Each has a temperature range along with otherclimatic factors. There are required nutrients and specific activities thatalso help characterize the niche. The principal of competitive exclusion states that unless the niches oftwo species differ, they cannot coexist in the same habitat. Two species oforganisms that occupy the same or similar ecologic niches in different geographicallocations are termed ecological equivalents. The array of speciespresent in a given type of community in differentbiogeographic regionsmay differ widely. However, similar ecosystems tend to develop whereverthere are similar physical habitats. The equivalent functional nichesare occupied by whatever biological groups happen to be present inthe region. Thus,a savannah-type vegetation tends to develop whereverthe climate permits the development of extensive grass-lands, butthe species of grass and the species of animals feeding on the grass may differ significantly in various parts of the world.

Question:

A ball is thrown upward with an initial velocity of 32 ft/sec from the top of a building. Calculate the velocity and the position as functions of the time.

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/Users/wenhuchen/Documents/Crawler/Physics/D03-0081.htm

Solution:

The only force acting on the ball is the grav-itational force, which is directed downward, throughout its motion, therefore, the ball will be accelerated downward, at a = - g = - 32 ft/sec^2. We have chosen the positive direction to be up. We have two directions of motion to consider: first, the upward motion to the max-imum height and then the downward motion toward the ground. Therefore, we must be careful to use the proper signs in our equations. We choose the origin for distance (y=0) at the point from which the ball is thrown. The initial velocity of the ball is v_0 = +32 ft/sec. The ball grad-ually loses velocity until it reaches its maximum height. Then it falls down towards the ground. The equations for velocity and distance therefore become v = v_0 + at = (32 ft/sec) - (32 ft/sec^2) × t y = v_0t + 1/2 at^2 = (32 ft/sec) × t - (16 ft/sec^2) × t^2 From these equations we find t(sec) v(ft/sec) y(ft) 0 32 0 1 0 16 2 -32 0 3 -64 -48 4 -96 -128 After 1 sec, the velocity of the ball has become zero; that is, the maximum height has been reached (16 ft) and the sub-sequent motion is downward. All velocities for t > 1 sec are therefore negative. At t = 2 sec, the ball has returned to its starting point (y = 0) and for all subsequent times, y is negative. The diagrams below show the velocity and the distance as functions of the time.

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Question:

10 cc of battery acid (density 1.21) is diluted to 100 cc. 30 cc of the dilute solution is neutralized by 45 cc of 0.5 N base. What is the normality of the diluted acid? What is its strength in grams per liter? What is the strength of the undiluted 1.21 density acid in g/l?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E26-0879.htm

Solution:

To solve this problem, use the following relationship for the neutralization of acids and bases N_aV_a = N_bV_b where N_a is the normality of the acid at volume V_a and N_b is the normality of the base at volume V_b. Thus, the normal-ity of the diluted acid is N_a = [(N_bV_b) / (V_a)] = {[(0.5 N) (0.045 l)] / [0.030 l]} = 0.75 N. Note: 1 cc = .001 l. The undiluted acid has a density of 1.21 g/cc; thus 10 cc of acid, weighs 12.1 g. Thus, when 90 cc of H_2O is added to form 100 cc of the diluted acid the weight of the entire solution becomes 90 + 12.1 = 102.1 g, but the weight of the acid is still 12.1 g. In a liter of diluted acid, there are {[(12.1 g) / (100cc)] × [10 / 10]} = [(121.0 g) / (1000 cc)] = 121 g/l of acid. For the undiluted acid, the density is: {[(1.21 g) / (1 cc)] × [1000 / 1000]} = [1210 g / 1000 cc] = 1210 g/l.

Question:

A steel tube, whose coefficient of linear expansion is 18 × 10^-6 per \textdegreeC, contains mercury, whose coefficient of absolute expansion is 180 × 10^-6 per \textdegreeC. The volume of mercury contained in the tube is 10^-5 m^3 at 0\textdegreeC, and it is desired that the length of the mercury column should remain constant at all normal temperatures. This is achieved by inserting into the mercury column a rod of silica, whose thermal expansion is negligible. Calculate the volume of the silica rod. (See figure.)

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/Users/wenhuchen/Documents/Crawler/Physics/D12-0455.htm

Solution:

At 0\textdegreeC, let the volume of the silica rod be V_0 , the volume of mercury be V, and the cross-sectional area and length of the column be A_0 and l_0, respectively. Then at t = 0\textdegreeC l_0A_0= V + V_0(1) At any temperature t, V and A_0 will change to their new values as a result of thermal expansion. These new values, are, respectively V' and A, where V' = V(1 + \betat) andA = A_0 (1 + 2\alphat) Here, \alpha and \beta are the coefficient of linear expansion of steel and the coefficient of absolute expansion of mercury. Note that we have imposed the constraint that the column length, l_0, be constant. Hence, at temperature t, we may write l_0 A = V' + V_0 orl_0 A _0(1+ 2\alphat) = V (1 + \betat) + V_0(2) Using (1) in (2) (V + V_0) (1 + 2\alphat) = V(1 + \betat) + V_0. V_0(1 + 2\alphat - 1) = V(1 + \betat - 1 - 2\alphat) orV_0 = [{V(\beta - 2\alpha)t}/2\alphat] = [{V(\beta - 2\alpha)}/2\alpha]. V_0 = [{10^-5 m^3 (180 × 10^-6 - 36 × 10^-6)deg^-1}/(36 × 10^-6 deg^-1)] = [(10^-5 × 144)/36] m^3 = 4 × 10^-5 m^3.

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Question:

(1) Calculate the work done by gravity when a mass of 100 g moves from the origin to r^\ding{217} = (50\^{\i} + 50\^{\j}) cm. (2) What is the change in potential energy in this displacement? (3) If a particle of mass M is projected from the origin with speed v_0 at angle \texttheta with the horizontal, how high will it rise?

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/Users/wenhuchen/Documents/Crawler/Physics/D05-0272.htm

Solution:

(1)Let (\^{\i}, \^{\j}) be the unit vectors along the horizontal and vertical directions respectively, as shown in the figure. The gravitational force is F_G^\ding{217} = - Mg \^{\j} The work W done by F_G^\ding{217} is W = ^(50,50)cm\int_(0,0)cm F_G^\ding{217} \bullet dr^\ding{217} = - Mg ^50 cm\int_0 dy(1) W = - Mg (50 cm) W = - (100 g)(980 cm/s^2)(50 cm) W = - 4.9 × 10^6 ergs(2) The gravitational force does a negative amount of work. The reason for this is that F_G^\ding{217} opposes the upward motion of M from the origin. (2) The definition of potential difference is v_(50,50) - v(0,0)= - ^(50,50)\int_(0,0) F_G^\ding{217} \bullet dr^\ding{217} From (1) and (2) v_(50,50) - v(0,0)= 4.9 × 10^6 ergs (3) In order to find the maximum height h that the particle attains, we relate the energy at the point of projection (x = 0, y = 0) to the energy at y = h. This may be done using the principle of energy conservation. Hence, E_f = E_i (1/2) Mv_f^2 + v_f = (1/2) mv^2_0 + v_0 We may arbitrarily set v = 0 at y = 0. Hence, v_0 = 0. (1/2) Mv_f^2 + Mgh = (1/2) mv0^2(3) Butv_f^2 = v_xf^2 + v_yf^2(4) v_0^2 = v_x0^2 + v_y0^2(5) Because there is no x-component of acceleration, v_x0 = v_xf. Also, at y = h, v_y = 0, hence v_yf = 0. Sub-stituting this data in (4) and (5) v_f^2 = v_x0^2 v_0^2 = vx^2_0 + v_y0^2 Substituting this in (3) (1/2) Mvx^2_0 + Mgh = (1/2) M(vx^2_0 + vy^2_0) orMgh = (1/2) M vy^2_0 orh = (v_y0^2)/(2g) But v_y0 = v_0 sin \texttheta, and h = (v_0^2 sin^2 \texttheta)/(2g)

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Question:

What is wrong with the following statement? "In many desert plants the roots grow very deep because they are searching out for deep underground water to supply the plant." Explain your answer.

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/Users/wenhuchen/Documents/Crawler/Biology/F10-0245.htm

Solution:

Although the observation of root length in desert plants is correct, the explanation given is oversimplified.lt is a teleological interpretation of the observation, or an interpretation in terms of end results. Such a teleological account implies that the plant has the ability to perceive its needs and to respond in order to satisfy these needs. Yet we know that a plant has nothing like a brain or nerves so it would not be possible for it to make such perceptions nor to exhibit goal-directed behavior. The true reason for desert plants having long roots may be the result of natural selection. We know that plant roots grow in response to gravity. This is an intrinsic response to their growth hormoneauxin. Due to the fact that there is a low supply of water in the desert, those plants with short roots may not survive while plants with long roots may be capable of absorbing sufficient amounts of water from the deep underground, and thus survive. Therefore, natural elements favor plants which possess adaptive characteristics for desert survival and restrict those with unfavorable traits. This forms the basis of natural selection. After millions of years, desert plants would have evolved long roots which are essential for their survival. Thus, today we find desert plants with long roots.

Question:

What general trends are observed in the evolution of the brain?

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/Users/wenhuchen/Documents/Crawler/Biology/F20-0506.htm

Solution:

Generally, the higher an animal is on the ladder of evolution, the more complex and heavy is its brain (see Figure 1). In the earthworm, an invertebrate, the 'brain' is merely a ganglion (a clump of nerve cell bodies) of about the same size as the other ganglia present in each segment of the body. Even among the mammals, the size and complexity of the brain vary to a great extent. The weight of the brain of a full grown man is approximately three pounds whereas a gorilla, about four times heavier than man, has a brain that weighs only one and a half pounds. This discrepancy in brain weight can account part-ly for the difference in intellectual capabilities of the two mammals. Increasing size is not the only feature seen in the evolution of the brain. The brains of higher mammals dis-play increasing foldings of their surface. These foldings, or convolutions, are particularly pronounced in man. By stimulating various regions of the brain with an electrode, and observing the corresponding results in behavior, func-tional mapping of the brains of various mammals has been achieved. The mapping shows that different regions of the brain are associated with different functions. There are the areas for motor functions, somatic sensory functions, vision, hearing, olfaction, and so forth. The mapping also shows that the proportion of the total area of the cerebrum devoted to sensory and motor functions differs greatly among species. It has been found that, in general, the more convoluted the surface of the cerebrum (called the cerebral cortex), the smaller the proportion devoted ex-clusively to sensory and motor activities. Since, as mentioned, there is a trend toward increasing convolutions in the evolution of the brain, we can anticipate that higher animals have a more convoluted cerebral cortex, and that at the same time, a smaller proportion of their brain is involved in sensory and motor functions. This is supported by the finding that in the cat the sensory and motor functions occupy a major portion of the cortex, whereas in man, the area of the cortex devoted to those functions is relatively small (see Figure 2). The rest of the brain not associated with motor and sensory activities contain the so-called associative areas. These areas are made up of neurons that are not directly connected to sense organs or muscles but supply inter-connections between the other areas. They are believed to be responsible for the higher intellectual faculties of memory, reasoning, learning, imagination, and personality. They are, in short, responsible for the behavioral con-sequences that most clearly distinguish man from other animals.

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Question:

Debug the following PL/I program. The program is given as a sourcelisting i.e. a written record of the PL/I program suppliedby the computer. STNT 1PROCEDUREOPTION(MAIN); 2DECLARE(A(10),B(10))FIXED (5), (X,Y,Z) FIXED (4), C CHARACTER(5) ; 3PUTPAGE? PUT SKIP(3)EDIT('C','X','Y','Z' , 'W' , 'V','A','R') (COL(20),A,COL(30),A,COL (40),A, COL(50) ,A,COL(60) ,A,COL(70) , A,COL(85) ,A,COL(95) ,A; 5GET EDIT (X, Y,C,W) (X(2) ,F(4) ,X(3) ,F(4) , X(7) ,A(5) , F(2)); 6GET SKIP; 7IF X = 0 & Y = 0 THEN GOTO NOMORE; 8IF X>Y THEN Z=2\textasteriskcenteredX+Y;ELSE Z=2\textasteriskcentered(X+Y)-18; 9V=SQRT(40)(Y-2)-W\textasteriskcenteredX)/(X-Y); 10L \vert :DO I = \vert TO W; 11A(I)=X+Y+3\textasteriskcenteredI; B(I)=X - Y-3\textasteriskcenteredI; 12IF 1=1 PUT SKIP(2)EDIT; EDIT(C,X,Y,Z, 13W,V,A(I),B(I)) (COL(20),A(5),COL(30),F(4) , COL(40) , F(4),COL(50),F(4),COL(60) ,F(2), COL(70),E(13,6) COL(85),F (6),COL(95),F(6)); 14ELSE PUT SKIP EDIT(A(I),B(I))(COL(85),F(5), COL(95),F (5)); 15FIND LI; 16GO TO IN; 17NOMORE; END BUGS;

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G13-0359.htm

Solution:

This program reads sets of edit-directed data, performs some calculationsand prints the results. The written program contains the following errors: 1) Statement 1: The procedure has no name 2) Statement 1: OPTION should be OPTIONS 3) Statement 3: There should be a blank between PUT and PAGE. 4) Statement 7: There should be a blank between GO and TO. 5) Statement 9: There is no arithmetic operator be-tween the constant 40 andthe term (Y-2). 6) Statement 10: The 1 has beenmispunchedas\vert . 7) Statement 11: The minus sign in X-Y is misplaced. 8) Statement 12: The keyword THEN is omitted from the IF statement. 9) Statement 16: There is no statement in the programlabelledIN. 10) Statement 17: BUGS appears nowhere else.

Question:

Discuss the establishment of a new species of plant or animalby the theory of isolation.

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/Users/wenhuchen/Documents/Crawler/Biology/F27-0723.htm

Solution:

It is currently believed that new species of animals and plants may arisebecause some physical separation discontinues the distribution of a speciesfor long periods of time. Such separation may be caused by geographicchanges, such as an emerging piece of land dividing a marine habitat, or climatic changes, such as a heavy drought causing large lakes todivide into nume-rous smaller lakes or rivers to be reduced to isolated seriesof pools. In some species, behavioral patterns may prevent dispersalacross areas that could easily be traversed if the attempt was made. For example, rivers may serve to isolate bird populations on oppositebanks, or a narrow strip of woods may effectively separate two meadowpopulations of butterflies. Such factors, then, may separate a single species population into twoor more isolated groups. Each group will respond to the selection pressureof its own environment. Since mutation is a random process, it is notto be expected that identical mutations will show up in the different populations. A single evolving unit has thus become two independently evolvingunits. As long as such evolutionary units remain isolated, they continueto respond independently to evolutionary forces. Gradually, the twoisolated groups will find their gene pools diverging, and if isolation and independentevolution continue for a long enough period of time, the geneticcomposition of the two groups may become so different that they becomeunable to breed with each other, even when brought very close together. Since two populations must be able to interbreed in nature if they belongto the same species, genetic isolation, brought about in this case bya long duration of physical isolation, causes two new distinct species to arise.

Question:

Five kg of aluminum (c_v= 0.91 J/gm\textdegreeK.) at 250\textdegreeK. are placed in contact with 15 kg of copper (c_v= 0.39 J/gm\textdegreeK.) at 375\textdegreeK. If any transfer of energy to the surroundings is prevented, what will be the final temperature of the metals?

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/Users/wenhuchen/Documents/Crawler/Physics/D13-0477.htm

Solution:

If the final temperature is T the change in the internal energy of the aluminum will be ∆U_AI =mc_v∆T where m is the mass of AI,c_vis the specific heat of AI, and ∆T is the temperature change the sample experiences. Hence ∆U_AI = 5 × 0.91 (T - 250) Similarly, for Cu ∆U_Cu= 15 × 0.39 (T - 375) According to conservation of energy ∆U_Total= ∆U_AI + ∆U_Cu= 0 Thus5 × 0.91(T - 250) + 15 × 0.39(T - 375) = 0 from which we can calculate T = 321\textdegree K.

Question:

In a Van deGraaffaccelerator, protons are accelerated through a potential difference of 5 × 10^6 V. What is the final energy of the protons in electron volts and joules, and what is their mass and their velocity?

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/Users/wenhuchen/Documents/Crawler/Physics/D32-0954.htm

Solution:

The work needed to move a charge q from position a to position b is W_ab=^b\int_aF^\ding{217}_app\bulletd^\ding{217}r whereF^\ding{217}_appis the applied force moving q, andd^\ding{217}ris an element of the path traversed by the particle in moving from a to b. In an electric field, E, the force exerted by the field on q is F^\ding{217} =qE^\ding{217} But the applied force is equal and opposite to the force exerted by E^\ding{217} on q, or F^\ding{217}_app= -qE^\ding{217} Hence, W-_ab = - q W-_ab = - q +^b\int_aE^\ding{217} \textbulletdr^\ding{217} = q [-^b\int_aE^\ding{217} \textbulletdr^\ding{217}] By definition V--_b--V_a= -^b\int_aE^\ding{217} \textbulletdr^\ding{217} where V_a is the potential at point a, and similarly for Vb. Then W_ab = q (V_b - V_a) = q \DeltaV The energy gained by accelerating a particle through a potential difference \DeltaV is the work done on it, or Wab. A proton carries an elementary positive charge. The energy acquired in traveling through a potential dif-ference of 5 MV is, assuming the proton starts at rest, W_ab =q\DeltaV=e\DeltaV= (1.6 × 10^-19 C) (5 × 10^6 V) W_ab = 8 ×10^-13 J To determine the mass at this energy, use the equation E = mc^2. \therefore m = E / C^2 = (8× 10^-13 J) / {(3 × 10^8) m^2 . s^-2} = 0.89 × 10^-29 kg. This is the mass equivalent of the energy acquired. The total mass the particle now possesses is obtained by adding this mass to the rest mass m_0 . That is, M = m + m_0 = (0.89 × 10^-29 + 1.672 × 10^-27) kg = 1.681 × 10^-27 kg Also, M = m0/ \surd{1 - (v2/ c^2)} 1 - (v2/ c^2) =m^20/ M2 1 - (m^20/ M^2)=v2/ c^2 V / C = \surd{1 - (m0/ M)^2} = \surd{1 - (1.672/ 1.681)^2} = 0.1033 \therefore V = 0.1033 × 3 × 10^8 m \bullet s-1 =3.10×10^7 m \bullet s^-1 .

Question:

Write a FORTRAN program to calculate depreciation by the sum of the years digits method.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G18-0440.htm

Solution:

The Internal Revenue Service allows various methods for calculating the depreciation on a piece of machinery. The simplest is the straight-line method. According to this method, if an article costs $1000, has an estimated life of 10 years and $0 scrap value, the depreciation per year will be $100 for 10 years. The straight-line method assumes that machines wear out at a constant rate. The sum of the years digit method on the other hand, is used to compute the depreciation on articles that have the greatest rate of depreciation during the first few years of use. For example, suppose a drilling machine costing $15,000, has a useful life of five years and can be resold for $5,000. The total depreciation is: 15,000cost -5,000trade-in 10,000depreciation The sum of the years 1 through 5 is 1+2+3+4+5 = 15. Thus, the depreciation for the first year is 5/15, for the second year, 4/15, etc. Year Depreciation Book value at the end of the year 1 2 3 4 5 5/15 of 10,000 = 3333.33 4/15 of 10,000 = 2666.67 3/15 of 10,000 = 2000 2/15 of 10,000 = 1333.33 1/15 of 10,000 = 666.67 15,000 - 3333.33= 11666.67 11666.67 - 2666.67 = 9000.00 9000 - 2000= 7000.00 7000 -1333.33= 5666.67 5666.67 -666.67= 5000 The program follows: DIM V(100) READ IYR, COST, SAL TDEP = COST - SAL KSUM = 0 JTIME = IYR 20KSUM = KSUM + JTIME IF (JTIME.EQ.O) GOTO 50 JTIME = JTIME - 1 GOTO 20 50DO 60 I = 1,IYR J = IYR - I + 1 FRA = J/KSUM V(I) = COST - FRA {_\ast} TDEP 60CONTINUE STOP END Program Comments : The READ statement reads in the number of years, the cost of the item, and the salvage value, respectively. TDEP stores the total value to be depreciated. The loop starting at statement 20 calculates the sum of the years. If the number of years IYR, equals 5, then JTIME = 5,4,3,2,1,0 and KSUM = 5+4+3+2+1 = 15. When JTIME = 0, signifying that the sum of years has been calculated, the program jumps to statement 50. This loop first calculates the FRA, the fractional amount of depreciation for each year, and V(I), the book value for each year.

Question:

Using the tables, of standard electrode potentials, list the following ions in order of decreasing ability as oxidizing agents: Fe^3+ , F_2, Pb^2+ , I_2 , Sn^4+ , O_2. Half-reaction E\textdegree, V Li^+ + e^- \rightleftharpoons Li -3.05 K++ e^- \rightleftharpoons K -2.93 Na^+ + e^- \rightleftharpoons Na -2.71 Mg^2+ + 2e^- \rightleftharpoons Mg -2.37 Al^3+ + 3e^- \rightleftharpoons Al -1.66 Mn^2+ + 2e^- \rightleftharpoons Mn -1.18 Zn^2+ + 2e^- \rightleftharpoons Zn -0.76 Cr^3+ + 3e^- \rightleftharpoons Cr -0.74 Fe^2+ + 2e^- \rightleftharpoons Fe -0.44 Cd^2+ + 2e^- \rightleftharpoons Cd -0.40 Co^2+ + 2e^- \rightleftharpoons Co -0.28 Ni^2++ 2e^- \rightleftharpoons Ni -0.250 Sn^2+ + 2e^- \rightleftharpoons Sn -0.14 Pb^2+ + 2e^- \rightleftharpoons Pb -0.13 Fe^3+ + 3e^- \rightleftharpoons Fe -.04 2 H^+ + 2e^- \rightleftharpoons H_2 0(definition) Sn^4+ + 2e^- \rightleftharpoons Sn^2+ 0.15 Cu^2+ + 2e^- \rightleftharpoons Cu 0.34 Fe(CN)_6^3- + e^- \rightleftharpoons Fe(CN)_6^4- 0.46 I_2+ 2e^- \rightleftharpoons 2I^- 0.54 O_2 + 2H^+ + 2e^- \rightleftharpoons H_2O_2 0.68 Fe^3+ + e^- \rightleftharpoons Fe^2+ 0.77 Hg_2^2+ + 2e^- \rightleftharpoons 2Hg 0.79 Ag^+ + e^- \rightleftharpoons Ag 0.80 2Hg^2+ + 2e^- \rightleftharpoons Hg_2^2+ 0.92 Br_2 + 2e^- \rightleftharpoons 2br^- 1.09 O_2(g) + 4H^+ + 4e^- \rightleftharpoons 2H_2O 1.23 Cr_2O_7^2- + 14H^+ + 6e^- \rightleftharpoons 2Cr^3+ + 7H_2O 1.33 Cl_2 + 2e^- \rightleftharpoons 2Cl^- 1.36 MnO_4- + 8H++ 5e^- \rightleftharpoons Mn^2+ + 4H_2O 1.51 Ce^4+ + e^- \rightleftharpoons Ce^3+ 1.61 MnO_4- + 4H++ 3e^- \rightleftharpoons MnO_2(s) + 2H_2O 1.68 H_2O_2 + 2H^+ + 2e^- \rightleftharpoons 2H_2O 1.77 O_3 + 2H^+ + 2e^- \rightleftharpoons O_2 + H_2O 2.07 F_3 + 2e^- \rightleftharpoons 2F^- 2.87

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Solution:

The best oxidizing agent will be the one with the greatest ability to gain electrons (be reduced) and will therefore have the most positive standard electrode potential, E\textdegree . From the tables, Fe^3+ + e^- \leftrightarrows Fe^2+ E\textdegree = + 0.77v F_2(g) + 2e^- \leftrightarrows 2F^- E\textdegree = + 2.87v Pb^2+ + 2e^- \leftrightarrows Pb(s) E\textdegree = - 0.13v I_2(s) + 2e^- \leftrightarrows 2I^- E\textdegree = + 0.54v Sn^4+ + 2e^- \leftrightarrows Sn^2+ E\textdegree = + 0.15v O_2(g) + 2H^+ + 2e^- \leftrightarrows H_2O_2(l) E\textdegree = + 0.68v Thus the substances, in order of decreasing ability as oxidizing agents, areF_2 > Fe^3+ > O_2 > I_2 > Sn^4+ > Pb^2+.

Question:

With two slits spaced 0.2 ran apart, and a screen at a distance of l = 1m, the third bright fringe is found to be displaced h = 7.5 mm from the central fringe. Find the wavelength \lambda of the light used. See the figure.

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Solution:

When the difference D between the path lengths of the rays 1 and 2 is an integral multiple of the wavelength \lambda, one obtains a max-imum (bright fringe) of the Interference pattern on the screen. From the figure we see that D = d sin \texttheta . If I is much larger than the distance between the two slits, we see that \texttheta' \approx \texttheta, where \texttheta' relates the position of the maximum on the screen to the distance between slits and the screen, tan \texttheta' = h/l. The approximation l >> d also means that the \texttheta is small; for which case we have D = d sin \texttheta = 3\lambda orsin \texttheta \approx sin \texttheta' = 3\lambda/d h/l = 3\lambda/d which gives \lambda as \lambda = dh/3l = [(0.75 cm × 0.02 cm) / (3 × 100 cm)] = 5 × 10^-5 cm = 500 × 10^-9m = 500 nm.

Question:

Consider the following structure. struct students { char name [MAX_NAME]; char id [MAX_ID]; float gpa; struct students \textasteriskcenterednext; }; }; Through a program

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Solution:

#define NULL "\textbackslash 0' #define MAX_NAME 30 #define MAX_ID 10 structstudents { charname [MAX__NAME ]; charid [MAX_ID]; floatgpa; structstudents \textasteriskcenterednext; }; }; main ( ) { inti; structstudents \textasteriskcenteredstart, \textasteriskcenteredptr, \textasteriskcenterednew _ ptr; char\textasteriskcenteredcalloc ( ); /\textasteriskcenteredroutine calloc returns a pointer to the location\textasteriskcentered/ /\textasteriskcenteredCreating first element in the list\textasteriskcentered/ if((ptr = (struct students \textasteriskcentered)calloc (1, sizeof (structstudents))) == NULL); { printf ("No memory available \textbackslashn"); exit (1); } ptr \rightarrow next = NULL; start = ptr; scanf ("%s%s%f\textbackslashn", ptr \rightarrow name, ptr \rightarrow id, ptr \rightarrow gpa); /\textasteriskcenteredcreating 9 more elements \textasteriskcentered/ for(i = 1; i <10; i++) { if((new _ ptr = (struct students \textasteriskcentered)calloc (1, sizeof (struct students))) == NULL) { printf ("No memory available \textbackslashn"); exit (1); } new _ ptr \rightarrow next = ptr \rightarrow next; ptr \rightarrow next = new _ ptr; ptr = new _ ptr; scanf ("%s%s%f\textbackslashn", ptr \rightarrow name, ptr \rightarrow id, ptr \rightarrow gpa); } } Explanation of Related Terms Linked List : A linked list is a list of structures. It util-izes the ability of a member of the structure to be a pointer to a structure of the same type. Structures where member points to a structure of the same type are known as self-referential structures. calloc ( ): This is a library routine that allocates space from the heap. The function calloc ( ) returns a pointer to a contiguous block of memory that is used for array or similar structures. It returns NULL value if insufficient space is available for allocation. sizeof ( ): It is an operator in C which returns the number of bytes required to store an object, eg. number _ of _ bytes = sizeof (structstudents); will give us the number of bytes required to store one struc-ture of student. exit ( ): It provides a method of terminating a program. It is one of the library routines i.e. it is a function in the li-brary. It returns several values but most common, are 0 and 1. Value 0 is for normal termination i.e. program completes execu-tion as required. Value 1 is for abnormal termination i.e. program terminates abnormally and does not execute as required. Explanation of Statements 1)char\textasteriskcenteredcalloc (); that has been allocated. It is mainly a function which returns a pointer of the typechar. 2)structstudents \textasteriskcenteredstart, \textasteriskcenteredptr, \textasteriskcenterednew _ ptr This declaration signifies that \textasteriskcenteredstart, \textasteriskcenteredptr and \textasteriskcenterednew _ ptr are pointers to the structure students. 3) ptr = (structstudents \textasteriskcentered) calloc(1, sizeof (structstu-dents)); To follow the concept in this statement we take individual parts and analyze their meaning. a) sizeof (structstudents); will return the number of bytes required to store a struc-ture of students. b) calloc (1, sizeof (structstudents)); Arguments required by calloc ( ) are the number of storage elements to allocate and size of storage area in bytes. c) (struct students \textasteriskcentered) This statement assigns the value returned from the routine calloc ( ) to a pointer to the structure of type students. The asterisk specifies that the value is to be assigned to a pointer. This assigned pointer is to a structure of type students. d) ptr = (structstudents \textasteriskcentered) calloc (1, sizeof (structstudents)); The ptr is assigned to the pointer on the right side of the assignment. i.e. pointer ptr is assigned the value calcu-lated on the right side of the assignment. Note: Assignment is possible only when both the pointers are to the same structure. Here ptr is assigned to another pointer (calculated) to a structure of type students. 4) ptr \rightarrow next ptr \rightarrow name etc. Reference to the members of the structure is through dot notation, i.e. (\textasteriskcenteredptr).name (\textasteriskcenteredptr).next etc. C provides a unique notation for pointers to structures. The notation \rightarrow is used in place of dot notation with pointers to structures. To reference any member in a structure the \rightarrow is used. eg. ptr \rightarrow next ptr \rightarrow name ptr \rightarrow id etc. LINKED-LIST: Represented through diagrams. Note: In element 2 the next is occupied by NULL i.e. next of element 2 is pointing to NULL. Further expansion of this list according to the program will give the required structure. It is suggested that accord-ing to the program the whole list should be drawn to give a clear understanding of concepts involved.

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Question:

What are some of the important properties and characteris-ticsofenyzmes?

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Solution:

An important property of enzymes is their catalytic ability. Enzymes controlthe speed of many chemical reactions that occur in the cell. To understandthe efficiency of an enzyme, one can measure the rate at whichan enzyme operates - also called the turnover number. The turnovernumber is the number of molecules of substrate which is acted uponby a molecule of enzyme per second. Most enzymes have high turnovernumbers and are thus needed in the cell in relatively small amounts. The maximum turnover number ofcatalase, an enzyme which decomposeshydrogen peroxide, is 10^7 molecules/sec. It would require yearsfor an iron atom to accomplish the same task. A second important property of enzymes is their specificity, that is, thenumber of different substrates they are able to act upon. The surface ofthe enzyme reflects this specificity. Each enzyme has a region called a bindingsite to which only certain substrate molecules can bind efficiently. There are varying degrees of spe-cificity:urease, which decomposes urea toammonia and carbon dioxide, will react with no other substance; howeverlipase will hydrolyze the ester bonds of a wide variety of fats. Another aspect of enzymatic activity is the coupling of a spontaneousreaction with a non-spontaneous reaction.An energy- requiringreaction proceeds with an increase in free energy and is non- spontaneous. To drive this reaction, a spontaneous energy-yielding reactionoccurs at the same time. The enzyme acts by harnessing the energyof the energy-yielding reaction and transferring it to the energy- requiringreaction. Thestructure of different enzymes differsigni-ficantly. Some are composedsolely of protein (for example, pepsin). Others consist of two parts: a protein part (also called anapoenzyme) and a non-protein part, eitheran or-ganic coenzyme or an inorganic cofactor, such as a metal ion. Only when both parts are combined can activity occur. There are other important considerations. Enzymes, as catalysts, donot determine the direction a reaction will go, but only the rate at which thereaction reaches equilib-rium. Enzymes are efficient because they are neeededin very little amounts and can be used repeatedly. As enzymes areproteins, they can be permanently inactivated or de-natured by extremesin temperature and pH, and also have an optimal temperature or pHrange within which they work most efficiently.

Question:

Simplify the following expressions according to the commutative law. a) AB+BA + CDE +CDE + ECD b) AB + AC + BA c) (LMN) (AB) (CDE) (MNL) d) F(K+R) + SV + WX+ VS +XW + (R+K)F

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Solution:

The commutative law states that AB = BA ; A+B = B+A Thus, when logic symbols are ANDed or ORed, the order in which they are written does not affect their value. a) Notice that ABandBA are equal to each other andCDE and ECD are equal to each other since only their order is changed. The idempotent law states that A+A = A; AA = A, Thus, any term ANDed or ORed with itself will be equal to itself. The equation is re-written as b) Similarly AB and BA are combined to form AB. Therefore, we get AB + AC c) Remember that the commutative and indempotent laws are also true for the AND operation. Thus LMN and MNL combine to form LMN (LMN) (AB) (CDE) d) In this case: F(K+R) = (R+K)F, SV = VS, and WX=XW, again by the commutative law. Therefore our result is: F (K+R) + SV + WX

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Question:

The screen at the Bijou theatre is 20 feet high with its lower edge 10 feet above the observer's eyelevel. The angle \texttheta at the observer's eye subtends the entire, screen. This angle varies with the distance x of the observer from the plane of the screen. At what distance x is the angle \texttheta greatest?

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Solution:

The problem is a very practical one, for the value obtained for x turns out to be the distance at which the observer gets the best view of the movie. One method involves inverse trigonometric functions. Notice that we can use the triangles constructed in Figure 1 to say: COT \O = X/10 andCOT(\O + \texttheta) = X/30 or\O = ARCCOT X/10 and\O + \texttheta = ARCC0T X/30 Combining terms, we get \texttheta = ARC00T(X/30) - \O \texttheta = ARCCOT(X/30) - ARC00T X/10 To find x for the largest \texttheta, we take derivatives. Remembering that (d/dz) arccot z = -[1 / (1 + z^2)] for 0 < Z < \pi , we can express d \texttheta/dx in terms of x as: d\texttheta/dx = [-{1/(1 + (x/30)^2)}] (1/30) - {[-{1/(1 + (x/10)^2)}] (1/10)} = -[30/{ 900(1 + x^2/900)}] - {- [10/{100(1 + x^2/100)}]} = {-30/(900 + x^2)} + {10/(100 + x^2)} To find x, we set d\texttheta/dx = 0 and solve for x: 0 = {-30/(900 + x^2)} + {10/{100 + x^2)} = {30/(900 + x^2)} = {10/(100 + x^2)} 30(100 + x^2) = 10(900 + x^2) 3000 + 30x^2 = 9000 + 10x^2 30x^2 = 6000 + 10x^2 [(20x^2 = 6000) / 20] x^2 - 300 x = \surd(300) = 17.35 READ (5,1) A,B 1FORMAT () CALL XFIN (X) PHI = ATAN(B/X) ALPHA = ATAN(A/X) THETA = ALPHA - PHI WRITE (6,2) THETA 2FORMAT () STOP SUBROUTINE XFIN (X1)(I), A,B) I = 1 DO 4 X = 1,30,0.1 DALPHA = A/(A\textasteriskcentered\textasteriskcentered2 + X1(I)\textasteriskcentered\textasteriskcentered2) DPHI = B/(B\textasteriskcentered\textasteriskcentered2 + X1(I)\textasteriskcentered\textasteriskcentered2) ZERO (I) = DALPHA - DPHI IF (ZERO(I) .LT.ZERO(I - 1)) GO TO 5 41 = I + 1 5WRITE (6,6) X1(I) 6FORMAT ( ) RETURN END

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Question:

You have an unknown liquid hydrocarbon that may be hexane, 1-hexene or cyclohexene. What simple chemical tests could you use to distinguish which it is?

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Solution:

Organic compounds can be distinguished by their functional groups. A functional group is a group of atoms that characterize a compound's reactions. Thus, to solve this problem, you want to use reactions that clearly distinguish between functional groups. Proceed as follows: Hexane is a saturated hydrocarbon, which means all of its carbons have single bonds and the molecule is made up of carbon and hydrogen only. 1- hexene and cyclohexene both possess a double bond, which means they are unsaturated. A characteristic feature of unsaturated compounds is the fact that they undergo addition reactions. Thus, if you add CI_2 in CCl_4, you get addition to the double bond in bothand CH_3CH_2CH_2CH_2CH = CH_2, but not in CH_3CH_2CH_2CH_2CH_2CH_3. To distinguish between unsaturated compounds, use an ozonolysis reaction. When you add ozone to a double bond it cleaves so that carbonyl compounds are formed, i.e., compounds of the form where R is H and/or carbon group's, likewise with R . If you perform this reaction on 1-hexene, you obtain CH_3CH_2CH_2CH_2CHO (pentanal) and form-aldehyde (CH_2O), the latter is a gas recognizable by its odor, while the reaction on cyclohexene yields only OCHCH_2CH_2CH_2CH_2CHO (hexanedial).

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Question:

Why are marsupials widespread in Australia andalmost nonexistantelsewhere?

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Solution:

The marsupials are a relatively primitive type of mammal in that theiryoung continue their develop-ment in the abdominal pouch for a periodof time even after birth. For a considerable length of time after parturition, the young are extremely dependent upon the mother for protectionand nutrition. If they are removed from the mother's pouch, they willsoon encounter star-vation, cold, predation, and/or death. For this reasonthe marsupials, compared to placental mammals, are at a selective disadvantage. In most parts of the world, the marsupials cannot compete withthe other mammals for survival. But during the Mesozoic Era, the ancestralmarsupials enjoyed a period of competition-free existence in Australia, which was then isolated from the rest of the world. The ancient marsupialsin the isolated continent never had any intense competition fromthe better-adapted placental mammals. Such competition had succeededin eliminating most of the marsupials outside Australia. But in Australia, the marsupials grew, multiplied, radiated into many niches and evolvedinto a variety of better- adapted forms found exclusively there today.

Question:

Which were the first vertebrates to exhibit a hinged jaw and paired fins? Describe the evolution of the hinged jaws of vertebrates.

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Solution:

Roughly four hundred million years ago, there evolved a group of fishlike organisms with an armor of bony scales; usually, but not always, a cartilagenous endoskeleton; both pectoral and pelvic fins, and most notably, hinged jaws. These organisms were grouped in the class Placodermi. The placoderms mark a notable advance in vertebrate evolution in that they possess hinged jaws. The acquisition of jaws was one of the most important events in the history of vertebrates, because it made possible a revolution in the method of feeding and hence in the entire mode of life of early fishes. Studies have indicated that the hinged jaws of the placoderms developed from a set of gill-support bars. Notice that hinged jaws arose independently in two important animal groups, the arthropods and the vertebrates, but that although they are functionally analogous structures, they arose in entirely different ways: from ancestral legs in the insects and from skeletal elements in the wall of the pharyngeal region in vertebrates. Though the placoderms themselves have been extinct for at least 230 million years, two other classes that arose from them are still important elements of our fauna. These are the Chondrichthyes (cartilaginous fishes) and the Osteichthyes (bony fishes).

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Question:

A barge of mass 15,000 kg, made from metal of density 7500 kg\bulletm^-3, is being loaded in a closed dock of sur-face area 6633 m^2 with ore of density 3 g\bulletcm^-3. When 80,000 kg of ore are aboard, the barge sinks. How does the water level in the dock alter? The area of the barge is assumed negligible in comparison with the area of the dock.

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Solution:

Before sinking the total mass of barge plus load was 95,000 kg. Since the barge floated, the up -thrust of the water must have equaled the weight of 95,000 kg. The mass of displaced water was thus 95,000kg and, since the density of water is 10^3 kg/m^3 the volume of the barge is V = m/l where m is the mass of the barge plus its load. Hence V = (95 × 103kg)/(10^3 kg m^3) = 95 m^3 The volume of the material of the barge is, similarly, 15,000 kg/7500 kg\bulletm^-3 = 2 m^3, and the volume of the ore density 3 g\bulletcm^-3 = 3 x 10^3 kg\bulletm^-3 is (80/3) = 26(2/3) m^3. The total volume occupied by the metal of the barge and the ore in the water after sinking is thus 28 2/3 m^3. This is the amount of displaced water after the barge sinks. The displaced water has therefore decreased by 66 1/3 m^3. The water level in the dock thus falls by an amount h, equal to the decrease in volume divided by the surface area of the dock, the surface area of the barge being negligible. Therefore h = (66 1/3)/6633 = (1/100) m = 1 cm.

Question:

A flat coil consisting of 500 turns, each of area 50 cm^2, rotates about a diameter in a uniform field of intensity 0.14 Wb\textbulletm^\rule{1em}{1pt}2,the axis of rotation being perpendicular to the field and the angular velocity of rotation being 150 rad's^\rule{1em}{1pt}1. The coil has a resistance of 5 \Omega, and the in-ducedemfis connected via slip rings and brushes to an external resistance of 10 \Omega. Calculate the peak current flowing and the average power supplied to the 10-\Omega resistor.

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Solution:

Theemfgenerated by the motion is given by the equation \epsilon =NAB\omegasin\omegat. where \omega is the loop's angular velocity, A is the coil's cross sectional area, and N is the number of turns in the coil. Also, B is the field of magnetic induction at the site of the coil. The current flowing is thus, by Ohm's Law, I = (\epsilon/R) = [(NAB\omega)/R] sin\omegat.(1) where R is the coil resistance. The peak value of the current is found when sin\omegatis a maximum [t = \pi/(2\omega)] . Hence, I_max = [(NAB\omega)/R] = (500 × 50 × 10^\rule{1em}{1pt}4 m^2 × 0.14 Wb\textbulletm^\rule{1em}{1pt}2 × 150s ^\rule{1em}{1pt}1) / (15\Omega) = 3.5 A. The average power supplied to the external resistor is P = I^2_rmsR whereI_rmsis the root mean square value of the current given in (1). Now I_rms\equiv \surd()(2) where < > indicates the time average of I^2 over one cycle of I^2. By definition, = (1/T) ^T\int_0 I^2 (t)dt(3) where T is the period of I^2. The period of I^2(t) is the smallest non-zero value of T for which I^2 (t + T) = I^2 (t) Using (1), we find [(N^2A^2B^2\omega^2)/R^2] sin^2 [\omega(t + T)] = [(N^2A^2B^2\omega^2)/R^2] sin^2\omegat orsin^2 (\omegat+\omegaT) = sin^2\omegat(4) Thus, the values of T for which (4) holds are\omegaT= \pm 0, \pm \pi, \pm 2\pi.... Hence, T = \pm 0, \pm \pi/\omega,\pm 2\pi/\omega) The smallest non-zero value of T is then T = (\pi/\omega)(5) Inserting (5) into (3) and utilizing (1) = (\omega/\pi) ^\pi/\omega\int_0 [(N^2A^2B^2\omega^2)/R^2] sin^2\omegatdt = (\omega/\pi) ^\pi/\omega\int_0 [(N^2A^2B^2\omega^2)/R^2] sin^2\omegatd(\omegat) Sincesin^2\omegat= {(1 -cos2\omegat)/2} = [(\omega^2N^2A^2B^2)/(\pi R^2)] [1/2 ^\pi/\omega\int_0 d(\omegat) \rule{1em}{1pt} 1/2 ^\pi/\omega\int_0cos2\omegat d(\omegat)] = [(\omega^2N^2A^2B^2)/(\pi R^2)] [1/2 ^\pi/\omega\int_0 d(\omegat) \rule{1em}{1pt} 1/4 ^\pi/\omega\int_0cos2\omegat d(2\omegat)] = [(\omega^2N^2A^2B^2)/(\pi R^2)] {1/2(\pi - 0) - 1/4 (sin 2\omega × \pi/\omega - sin 0)} = [(\omega^2N^2A^2B^2)/(2R^2)] Finally, from (2) I^2_rms = = [(\omega^2N^2A^2B^2)/(2R^2)] ButI^2_max = [(\omega^2N^2A^2B^2)/(R^2)] whence I^2_rms= (I^2_max)/2 Hence,P = I^2_rms R = 1/2I^2_max R = 1/2 (3.5 A)^2 10 \Omega P = 61.25 W.

Question:

Assuming complete neutralization, calculate the number of milliliters of 0.025 M H_3PO_4 required to neutralize 25 ml of 0.030 M Ca(OH)_2.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E10-0364.htm

Solution:

This problem can be solved by two methods: mole method or equivalent method. Mole Method This method requires one to write out the balanced equation that illustrates the neutralization reaction. The balanced equation is 3Ca(OH)_2 + 2H_3PO_4 \rightarrow Ca_3(PO_4)_2 + 6H_2O. From this equation, one can see that 2 moles of H_3PO_4 react for every 3molesof Ca(OH)_2 . This means that one must first calculate how many moles of Ca(OH)_2 are in-volved, Themolarity of the Ca(OH)_2 is 0.030. (Molarity= no. of moles/liters.) As given, the Ca(OH)_2 solution is 25 ml or 0.025 liters. Therefore, the number of moles of Ca(OH)_2 is (0.030) (0.025) = 0.00075 moles. As the balanced equation indicates, the number of moles of H_3PO_4 is 2/3 the moles of Ca(OH)_2 or 2/3 (0.00075) = 0.00050 moles of H_3PO_4. From the definition ofmolarityfor H_3PO_4 one has 0.025 M = [(0.00050moles)/(liters)]. Themolarity, 0.025, is given. Solving for liters, one obtains 0.020 liters or 20 ml. The key to solving this problem with the mole method, is to write a balanced equation, which will indicate the relative amounts of moles required for complete neutralization. Equivalent Method This method requires that one consider normality and the definition of an equivalent. An equivalent is defined as the molecular weight or mass of an acid or base that furnished one mole of protons (H^+) or hydroxyl (OH^-) ions. For example, the number of equivalents con-tained in a mole of H_2SO_4, is 98/2 or 49, Since each mole of H_2SO_4 produces two protons, divide the molecular weight by 2. The number of equivalents of an acid must equal that of the base in a neutralization reaction. Normality is defined as equivalents of solute per liter. In this problem, it is given that there are 25 ml of 0,03 Ca(OH)_2. To solve the problem, determine how many equivalents are present. The number of moles of Ca(OH)_2 is (0.025)(0.03) or 0,00075 from the definition ofmolarity. The molecular weight of Ca(OH)_2 is 74,08. Therefore, there are (0.00075)(74.08) or 0.06 grams of Ca(OH)_2. The number of equivalents per gram is 74.08/2 since two OH^- can be produced. The number of equivalents is [{(0.00075)(74.08)g} / {(74.08/2) g/equiv}] = 0.0015 equiv of Ca(OH)_2 . This indicates that 0.0015 equivalents of H_3PO_4 are required. The molarityof H_3PO_4, is 0.025 M, which means its normality is 0.075 N, because there are 3ionizableprotons per mole. Recalling the definition of normality, there are 0.075 N = 0.0015 equiv./liters The reason one knows that there is 0.0015 equiv in the H_3PO_4 present is because one knows that for the neutralization to occur the number of equivalents of acid must equal the number of equivalents of base. In this problem, one has already calculated that there are 0.0015 equiv of base present. Thus, volume = 0.20 l or 20 ml.

Question:

A 200 g ice cube is placed in 500 g of water at 20\textdegreeC. Neglecting the effects of the container, what is the resultant situation?

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Solution:

Note that a cube of ice at 0\textdegreeC will lower the temperature of the water, and that for every 80 calories of heat energy absorbed by the ice, one gram will be melted without any change in temperature. If the heat given off by the 500 g of water cooling to 0\textdegreeC exceeds the amount necessary to melt the 200 g of ice, then the water will not cool to 0\textdegreeC. If, how-ever, it is less than sufficient to melt all 200 g of ice, only a fraction of the ice will be melted, and the resultant temperature will be 0\textdegreeC. The amount of heat that must be withdrawn from the water to lower its temperature to 0\textdegreeC is Q = mc ∆T where m is the mass of water, c is its specific heat, and ∆T is the change in temperature that the water experiences. Q = (500 g)(1 cal/g\textdegreeC)(20\textdegreeC) Q = 10000 cal The amount of ice that 10000 calories will melt at 0\textdegreeC is 10000 / 80 = 125 g This is less than 200 g, the original amount of ice. Therefore a 75 g block of ice finds itself floating in water at 0\textdegreeC.

Question:

Using the information of fig. 1, write a program called PRINT-A-RECORD, which will read an input card file and prepare and print a report containing C-CODE, PAYCODE and PRESENT-DDE. CARD RECORD C-COED PERSONAL DATA CREDIT DATA C-NAME C- ADDR-ESS C- PHONE YR- OPENED MAX- CREDIT PRESENT DUE PAY-CODE 9 digits 21 char. 25 char. 10 char. 2 digits 6 digits 2 rt. of dec. 6 digits 2 rt. of dec. 1 digit PRINT-LINE133 character positions 1 HEADING-LINE FILLER 10 spaces H1 " CUST- OMER CODE" FILLER 10 spaces H2 "PAYM-ENT HISTO- RY" FILLER 10 spaces H3 "CURRE- NT BALAN- CE" FILLER to fill line 1 INFO-LINE FILLER 12spaces C-CODE 9 digits FILLER 19 spaces PAYCODE 1 char. FILLER 21 spaces PRESE- NT-DUE Print 6 digits, use $ and dec. FILLER to fill line fig.1,

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G11-0262.htm

Solution:

All the information needed to write the program is given in the group of diagrams supplied with the problem. This program will read an input card file and prepare and print a report containing some of the data from the cards. ASSIGN clause . requirements vary with the system, the general format for the ASSIGN clause is: ASSIGNsystem-name An FD (FILE DESCRIPTION) entry must be included in the File section for each file. The entry must include FD with the file-name, a LABEL RECORDS clause, and may include a DATA RECORD clause. The characters in a file must all be accounted for in its record description entry. The FILLER item is thus used to refer to unused spaces in the printer file. For the PROCEDURE DIVISION new WRITE and MOVE statements are introduced. The WRITE statement accomplishes a MOVE and WRITE in a single COBOL operation. WRITE record-name FROM data-name. This format accomplishes both moving of data-name to record-name and writing of the record to its associated file. The MOVE CORRESPONDING statement; MOVE CORRESPONDING data-name-1 TO data-name-2 moves elementary data items from the data group of data-name-1 to elementary data items;with the exact same name, of the data group data-name-2. As an example, in this problem the statement: MOVE CORRESPONDING CARD- RECORD TO INFO-LINE, will move only the data items C- CODE, PAYCODE, PRESENT-DUE in CARD-RECORD to the corres-ponding items C-CODE, PAYCODE, PRESENT-DUE in INFO- LINE. In the following program, the new concepts introduced in this problem are demonstrated thoroughly

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Question:

A sound is returned as an echo from a distant cliff in 4 sec. A sound is returned as an echo from a distant cliff in 4 sec. How far away is the cliff, assuming the velocity of sound in air to be 1100 ft/sec?

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/Users/wenhuchen/Documents/Crawler/Physics/D26-0826.htm

Solution:

The sound travels the distance (s) between cliff and observer twice. Hence 2s = v × t where v is the velocity of sound, and t is the time required for the entire trip. Therefore, s = (v × t) / 2 = {[(1100 ft/sec) × (4 sec)] / 2} = 2200 ft.

Question:

Certain nerve gases are known to cause a breakdown of cholinesterase. How would these gases affect the human body and why?

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/Users/wenhuchen/Documents/Crawler/Biology/F20-0525.htm

Solution:

Acetylcholine is a chemical responsible for transmitting impulses across synaptic junctions. If the supply of acetylcholine is continuous, repeated stimulation of the postsynaptic neuron will result. Hence, this neuro-transmitter has to be destroyed after performing its function. The substanceacetylcholinesterasespecifically breaks down acetylcholine by splitting off thecholinemoiety from the molecule, thus inactivating it. Certain nerve gases, when inhaled, act to destroy acetylcholinesterase . This action leads to serious con-sequences. Following the breakdown ofacetylcholinesterase, acetylcholine would not be cleaved but would remain in the synaptic cleft and continue to stimulate the postsynaptic neuron. If this neural pathway leads to a muscle, con-tinual excitation of the neuron would produce sustained contraction of that muscle. The person involved may, as a result, enter a state of tremors and spasms, or may even die depending on the quantity of gas inhaled.

Question:

In the figure, a ladder 20 ft. long leans against a verti-cal frictionless wall and makes an angle of 53\textdegree with the horizontal, which is a rough surface. The ladder is in equilibrium. Its weight is 80 lb. and its center of gravity is in the center of the ladder. Find the magnitudes and directions of the forces F_1\ding{217} and F_2\ding{217}.

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Solution:

If the wall is frictionless, F_1\ding{217} is horizontal. The direction F_2\ding{217} is unknown (except in special cases, its direction does not lie along the ladder). Instead of considering its magnitude and direction as unknowns, it is simpler to resolve the force F_2\ding{217} into x- and y-components and solve for these. The magnitude and direction of F_2\ding{217} may then be computed. The first condition of equilibrium states that the net horizontal component of force on an object is zero. Similarly for the net vertical component. Hence, from the figure, \sumF_x = F_2 cos \texttheta - F_1 = 0 \sumF_x = F_2 cos \texttheta - F_1 = 0 (1st condition) \sumF_y = F_2 sin \texttheta - 80 lb. = 0. The second condition of equilibrium states that the net torque acting on a body is zero. If the body is in trans-lational equilibrium (F_net\ding{217} = 0) we may compute the torques about any axis. (Torques coming out of the plane of the figure will be considered positive) The resulting equation is simplest if one selects a point through which two or more forces pass, since these forces then do not appear in the equation. Let us therefore take moments about an axis through point A. \sum\cyrchar\cyrt_A = F_1 × 16 ft. - 80 lb. × 6 ft. = 0.(2nd condition) From the second equation, F_2 sin \texttheta = 80 lb., and from the third, F_1 = (480 lb \textbullet ft)/(16 ft) = 30 lb. Then from the first equation, F_2 cos \texttheta = 30 lb. Hence, F_2 = \surd[(80 lb)^2 + (30 lb)^2] = 85.5 lb. \texttheta = tan^-1 [(80 lb)/(30 lb) = 69.5\textdegree \texttheta = tan^-1 [(80 lb)/(30 lb) = 69.5\textdegree .

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Question:

Describe the function of the sea star's stomach as the animal feeds upon a clam.

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/Users/wenhuchen/Documents/Crawler/Biology/F12-0322.htm

Solution:

The sea star places its arms on the clam, and by applying suction with the tube feet, is able to pry the two shells slightly apart. The sea star thenevertsits cardiac stomach, as it is projected out of its mouth. The stomach then enters the crack between the bivalve shells. Digestive enzymes from the stomach cause the soft body of the clam to be degraded. The clam's adductor muscles are digested and the valves open. The partly digested food is taken into the pyloric stomach and then into the digestiveglands, where digestion is completed and the products are absorbed. After the body of the clam is eaten, the sea star retracts the cardiac stomach. The digestive tract of the sea star consists of a mouth on the lower surface, a short esophagus, an eversible cardiac stomach, a smaller pyloric stomach, a very small intestine, and an anus. Attached to the pyloric stomach are five pairs of large digestive glands. Each pair of digestive glands lies in thecoelomiccavity of one of the arms. Both stomachs fill most of the interior of the central disc. The cardiac stomach is capable of entering a gap between the shells of a clam as small as 0.1 mm. Certain species of sea stars spread theevertedcardiac stomach over the ocean bottom, and digest all types of organic matter encountered. Theevertedstomach of sea stars is also capable of engulfing prey, enabling the prey to be swallowed whole. Sea stars arecarnivousand feed on many marine in-vertebrates, and even small fish. They also feed on dead matter. Some sea stars have extremely restricted diets, and feed only on certain species. Others have a wide range of prey, but exhibit preferences.

Question:

An impure sample of aluminum sulfate, Al_2 (SO_4)_3, is analyzed by forming a precipitate of insoluble barium sulfate, BaSO_4, by reacting aluminum sulfate with an excess of BaCl_2 (to insure complete precipitation). After washing and drying, 2.000 g of BaSO_4 was obtained. If the original sample weighed 1.000 g, what was the per cent of aluminum sulfate in the sample?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E05-0187.htm

Solution:

The problem is solved by calculating the amount of BaSO_4 that would form if the sample were pure, and comparing this with the amount actually formed. Al_2 (SO_4)_3 reacts with BaC1_2 according to the equation Al_2 (SO_4)_3 + 3BaCl_2\rightleftarrows3BaSO_4 + 2AlCl_3. The number of moles of Al_2 (SO_4)_3, assuming the sample is pure, is equal to the mass of the sample (1.000 g) divided by the molecular weight of Al_2 (SO_4)_3, (342.14 g/mole), [(1.000 g) / (342.14 g/mole)] = 0.00292 mole Al_2- (SO_4 )_3 . Since 3 moles of BaSO_4 are formed for every mole of Al_2 (SO_4)_3 reacted, as indicated by the coefficients in the reaction, 1.000 g of pure Al_2 (SO_4)_3 would produce 3 × .00292 mole = 0.00876 mole BaSO_4. The actual number of moles of BaSO_4 formed is equal to the mass of BaSO_4 obtained after washing and drying (2.000 g) divided by the molecular weight of BaSO_4 (233.40 g/mole), [(2.000 g) / (233.40 g/mole)] = 0.00857 mole BaSO_4 formed. The purity of the sample is then given by the ratio of the number of moles BaSO_4 formed divided by the number of moles of BaSO_4 that would have formed from a pure sample, times 100 % : purity = [(0.00857mole) / (0.00876mole)] × 100 % = 97.8 %. The original sample, therefore, contained 97.8 % Al_2 (SO_4)_3 and 100 % - 97.8% = 2.2% impurities.

Question:

In double-slitFraunhoferdiffraction what is the fringe spacing on a screen 50cm away from the slits if they are illuminated with blue light (\lambda = 480nm = 4800 \AA), if d = 0.10mm, and if the slit width a = 0.02mm? What is the linear distance from the central maximum to the first minimum of the fringe envelope?

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/Users/wenhuchen/Documents/Crawler/Physics/D37-1082.htm

Solution:

The intensity pattern is given byl_\texttheta=I_m(cos\beta)^2(sin\alpha/ \alpha)^2 the fringe spacing being determined by the interference factor cos^2\beta. The position of the first minimum of the fringe envelope is given by ∆y = (\lambdaD/ d), where D is the distance of the screen from the slits. Sub-stituting yields ∆y = [{(480 × 10^-9m) (50 × 10^-2m)} / (0.10 × 10^-3m) ] = 2.4 × 10^-3m = 2.4 × 10^-3m = 2.4 = 2.4 mm. The distance to the first minimum of the envelope is determined by the diffraction factor (sin\alpha/ \alpha)^2 inI_\texttheta=I_m(cos\beta)^2 (sin\alpha/ \alpha)^2. The first minimum in this factor occurs for \alpha = \pi. The angle is given by: \alpha = (\pia/ \lambda) sin \texttheta sin\texttheta= (\alpha\lambda/\pia) = (\lambda / a) = [(480 × 10^-9m) / (0.02 × 10^-3m)] = 0.024. This is so small that it is assumed that \texttheta \congsin\texttheta\congtan\texttheta, or y = Dtan\texttheta\cong Dsin\texttheta= (50cm) (0.024) = 1.2cm. There are about ten fringes in the central peak of the fringe envelope.

Question:

What is a pheromone, and how does it differ from a hormone?

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/Users/wenhuchen/Documents/Crawler/Biology/F21-0532.htm

Solution:

The behavior of animals may be influenced by hormones - organic chemicals that are released into the internal environment by endocrine glands which regulate the activities of other tissues located some distances away. Animal behavior is also controlled by pheromones - substances that are secreted by exocrine glands into the external environment. Pheromones influence the behavior of other members of the same species. Pheromones represent a means of communication and of transferring information by smell or taste. Pheromones evoke specific behavioral, developmental or reproductive responses in the recipient; these responses may be of great significance for the survival of the species. Pheromones act in a specific manner upon the recipi-ent's central nervous system, and produce either a tempo-rary or a long-term effect on its development or behavior. Pheromones are of two classes: releaser pheromones and primer pheromones. Among the releaser pheromones are the sex attractants of moths and the trail pheromones secreted by ants, which may cause an immediate behavioral change in conspecific individuals. Primer pheromones act more slowly and play a role in the organism's growth and differentiation. For example, the growth of locusts and the number of reproductive members and soldiers in termite colonies are all controlled by primer pheromones.

Question:

S(s) + O_2 (g) \rightarrow SO_2 (g)∆H = \rule{1em}{1pt} 71.0 Kcal SO_2 (g) +(1/2) O_2 (g) \rightarrow SO_3 (g)∆H = \rule{1em}{1pt} 23.5 Kcal calculation ∆H for the reaction: S(s) + 1(1/2)O_2 (g) \rightarrow SO_3 (g)

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/Users/wenhuchen/Documents/Crawler/Chemistry/E14-0491.htm

Solution:

The heat of a given chemical change is the same whether the reaction proceeds in one or several steps; in other words, the energy change is independent of the path taken by the reaction. The heat for a given reaction is the algebraic sum of the heats of any sequence of reactions which will yield the reaction in question. For this problem, one can add the 2 reactions together to obtain the third. The ∆H is found for the third reaction by adding the ∆H's of the other two. ∆H = \rule{1em}{1pt} 71.0 Kcal + ∆H = \rule{1em}{1pt} 23.5 Kcal \rule{1em}{1pt} 94.5 kcal S(s) +O_2 (g) \rightarrow SO_2 (g) +SO_2 (g) +(1/2) O_2 (g) \rightarrow SO_3 (g) S (s) + O_2 (g) + SO_2 (g) + (1/2)O_2 (g) \rightarrow SO_2(g) + SO_3 (g) This equation can be simplified by subtracting SO_2(g) from each side and adding O_2 and (1/2) O_2 together. You have, therefore S(s) + [O_2(g) + (1/2)O_2(g)] + SO_2(g) \rule{1em}{1pt} SO_2(g) \rightarrow SO_3(g) + SO_2(g) \rule{1em}{1pt} SO_2(g) S(s) + [O_2(g) + (1/2)O_2(g)] + SO_2(g) \rule{1em}{1pt} SO_2(g) \rightarrow SO_3(g) + SO_2(g) \rule{1em}{1pt} SO_2(g) or S(s) + 1 (1/2) O_2 (g) \rightarrow SO_3 (g) or S(s) + 1 (1/2) O_2 (g) \rightarrow SO_3 (g) The ∆H for this reaction is \rule{1em}{1pt} 94.5 Kcal, as shown above.

Question:

What must be the speed v of a lead bullet if it melts when striking a steel slab? The initial temperature of the bullet is T_0 = 300\textdegree K, its melting point is T_1 = 700\textdegree K, its heat of melting q = 5 cal gr^-1, and its specific heat c = 0.03 cal gr^-1 K\textdegree^-1.

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/Users/wenhuchen/Documents/Crawler/Physics/D13-0472.htm

Solution:

Assume that all the kinetic energy of the bullet is transformed to heat energy upon impact. The heat released first raises the temperature of the bullet to its melting point T_1 , and then supplies the heat of melting the lead. The amount of heat Q_1 required to raise the temperature of the bullet to the melting point is Q_1 = mc(T_1 - T_0) = (m gr gr ) × (0.03 cal gr^-1 K\textdegree^-1) (700 - 300)K\textdegree ) × (0.03 cal gr^-1 K\textdegree^-1) (700 - 300)K\textdegree = 12 m cal. where m is the bullet's mass. Melting requires another amount of heat Q_2, where Q_2 =mq = (mgr) × (5 cal gr^-1) = 5 m cal. Therefore, the total amount of heat used up after the collision, is Q = Q_1 + Q_2 = (12m + 5m) cal = 17 m cal. The conservation of energy principle in this problem can be stated as E_kinetic= Q or(1/2) mv^2 ergs = 17 m cal × 4.19 × 10^7 ergs/cal v^2 = 2 × 17 × 4.19 × 10^7 = 14.2 × 10^8 cm^2/sec^2 v = 3.8 × 10^4 cm/sec = 380 m/sec

Question:

NAD+ and NADP+ play important roles in reactions associated with metabolism. Discuss these roles.

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/Users/wenhuchen/Documents/Crawler/Biology/F03-0086.htm

Solution:

In a dehydrogenation reaction, the electrons removed from a substance cannot exist freely but must be transferred to another compound called an electron acceptor. Both nicotinamide adenine dinucleotide (NAD+) and nicotinamide adenine dinucleotide phosphate (NADP+) are electron aceeptors. The functional part of both NAD+ and NADP+ is the nicotinamide ring, which accepts one hydrogen atom and two electrons to become NADH and NADPH respectively NAD+ and NADP+ serve as electron acceptors in the dehydrogenation of substances with the secondary alcohol arangement. Although NADP+ only differs from NAD+ in having three rather than two phosphate groups on the R group, most dehydrogenases require either NAD+ or NADP+ to be present as a coenzyme and will not operate with the other one. NAD+ is used as an electron acceptor in glycolysis, the TCA cycle, and fatty acid oxidation. NADP+ is used as an electron acceptor in the light reaction of photo-synthesis, and in the pentose phosphate pathway (the source of pentoses, necessary constituents of nucleotides). We thus see that the function of NAD+ or NADP+ is to accept hydrogen atoms and electrons and to store energy released from oxidized substances in the form of NADH or NADH. The reverse is true for NADH or NADPH; that is, their function is to give up hydrogen atoms and electrons and energy to reducible compounds. For example, the hy-drogen atoms of NADH in the electron transport chain are ultimately transferred to oxygen, producing high energy phosphate bonds in the form of ATP along the chain. The oxidation of one molecule of NADH yields 3 phosphate bonds; thus the 8 NADHs produced in the TCA cycle yield 24 phos-phate bonds. NADPH is formed from the reduction of NADP+ in the light reaction of photosynthesis. It is subsequently utilized in the dark reactions of carbohydrate synthesis, in which carbon dioxide is used in the production of hexose molecules. To produce one hexose molecule (a 6-carbon sugar) , 12 NADPH and 18 ATP are required, both of which are the products of the light reactions. Thus, both NAD+, NADH and NADP+, NADPH have major importance in reactions associated with metabolic and catabolic functions.

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Question:

How many fringes are formed per millimeter if light beams of wavelength 632.8 nm intersect at an angle of 5\textdegree?

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/Users/wenhuchen/Documents/Crawler/Physics/D37-1083.htm

Solution:

Figure 1 shows the two monochromatic beams of the same wavelength and inclined to each other at an angle \texttheta. These two beams interfere on the photographic plate HH', producing bright and dark lines. Since the beam (1) is seen to be in constant phase across the surface of the hologram plane, the interference pattern, or fringes, will be separated by an amount ∆y, whenever the path difference between the two beams is one wavelength From the triangle CPP', Sin\texttheta = (CP / PP') = (\lambda / ∆y). Therefore, ∆y = (\lambda / sin\texttheta). Hence the number of fringes per mm is equal to [1 / {∆y(mm)}] = [sin\texttheta / {\lambda(mm)}] = (sin 5\textdegree) / (6.328 × 10^-4) = 138(mm)^-1

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Question:

A record player, has a turntable which is a flat plate of radius 12cm and mass 0.25 kg .Calculate the moment of inertia of the turntable about its axis of symmetry, and the torque required to accelerate the turntable to 33.3 rpm in 2 sec. Because his records are warped, the owner of the record player usually places a brass cylinder (radius 4 cm and mass 3 kg) on the center of the record. What torque would then be required to accelerate the turntable?

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/Users/wenhuchen/Documents/Crawler/Physics/D04-0187.htm

Solution:

Torque and angular acceleration are related by \cyrchar\cyrt = I\alpha where \cyrchar\cyrt, I, and \alpha are the torque, moment of inertia, and angular acceleration respectively. First we must calculate the moment of inertia of a cylinder of uniform density \rho (see diagram): The volume of the mass element dM is given by: dV = 2\pi rw dr and since the mass contained in dM is given by dM = \rhodV: dM = 2\pi \rhorw dr Since:I = \intr^2 dM where r is the radial distance from the axis about which we calculate I, we have for the disk: I = ^R\int_0 r^2(2\pi\rhorwdr) = 2\pi\rhow ^R\int_0 r^3 dr = 2\pi\rhor{(r^4)/4} \mid^R_0 = (1/2) \pi\rhowR^4 = (1/2)(\rho\piR^2w)R^2 = (1/2)MR^2 We note that the volume of the disk is given by its surface area multiplied by its thickness: V = \piR^2w and since M = \rhoV: M = \rho\piR^2w where M is the mass of the disk. Thus for the turntable: I_+ = (1/2)(0.25 kg)(0.12m)^2 = 1.8 × 10^-3 kg - m^2 The moment of inertia of the brass cylinder is: I_c = (1/2)(3 kg)(0.04m)^2 = 2.4 × 10^-3 kg - m^2 To calculate the angular acceleration, we must first convert 33.3 rpm to rad/sec. (3.33 revolutions)/(min.) = {33.3(2\pi rad)}/(60 sec) = 3.49 rad/sec \alpha = (3.49 rad/sec)/(2 sec) = 1.74 rad/sec^2 where \alpha is the angular acceleration. The torque that must be applied to the turntable to produce \alpha is: \cyrchar\cyrt = I_t\alpha = (1.8 × 10^-3 kg - m^2)(1.74 red/sec^2) = 3.13 × 10^-3[(kg - m)/(sec^2)] - m = 3.13 ×10^-3 N - m The torque required to accelerate the turntable plus the cylindrical weight is: \cyrchar\cyrt = (I_t + I_c)\alpha =(1.8 × 10^-3kg - m^2 + 2.4 × 10^-3 kg - m^2)(1.74 rad/sec^2) = (4.2 × 10^-3kg - m^2)(1.74 rad/sec^2) = 7.31 × 10^-3N - m.

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Question:

One gram of an unknown gaseous compound of boron (B, atomic weight = 10.8 g/mole) and hydrogen (H, atomic weight = 1.0 g/mole) occupies 0.820 liter at 1.00 atm at 276 K. What is the molecular formula of this compound?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E02-0068.htm

Solution:

The general formula for the compound is B_a H_b where a and b are to be determined. To find these, we must first find the molecular weight of the compound by using the ideal gas law, PV = nRT, where P = pressure, V = volume, n = number of moles, R = gas constant, and T = absolute temperature, n equals the mass, m, divided by the molecular weight MW,PV = nRT = (m/MW) RT,orMW = (mRT/PV) . MW = (mRT/PV) = [(1 g × 0.082 liter - atm/mole-\textdegreeK × 276\textdegreeK) / (1.66 atm × 0.820 liter)] = 27.6 g/mole. To find a and b, we use the relation MW_Ba _Hb = a × atomic weight of B plus b × atomic weight of H, or g/mole = a × 10.8 g/mole + b × 1.0 g/mole. By trial and error, we find that a = 2 and b = 6 (2 × 10.8 g/mole + 6 × 1.0 g/mole = 27.6 g/mole), so that the formula of our compound is B_2 H_6 (diborane) .

Question:

A suburban housewife, who lives at location A of the town street plan in Figure 1, drives to fill up her automobile at the local gas station, located at B. For some reason, all of the horizontal streets run from left to right, and all the vertical streets run from bottom to top. She is curious as to how many possible routes exist between her and the gas station. Write a FORTRAN program to solve her dilemma.

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Solution:

From the definition of the street plan, the direction of movement from each comer can be either up or to the right. Any parti-cular path can be written as a sequence of u's and r's (ups and rights), in which the total number of u's and r's can be written as n and m respectively. Hence, the total number of possible paths is (m + n) ! / n! m! We can write a program to generalize this problem for a street plan of n and m dimensions. The use of the FACT (factorial) function will make the program shorter and easier to understand. INTEGER N,M READ (5,{_\ast}) N,M CCOMPUTE NUMBER OF POSSIBLE PATHS CFROM CORNER TO CORNER PATHS = FACT (M+N) /FACT (N) {_\ast}FACT (M) WRITE (5,100) N,M, PATHS 100FORMAT (1X, 'NUMBER OF POSSIBLE PATHS 1FROM CORNER TO CORNER IN AN' ,I2, 'BY', 2I2, 'MATRIX =',I4) STOP END

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Question:

In the reaction N_2O_5 \rightarrow N_2O_4 + (1/2)O_2, the N_2O_5 decomposes by a first-order mechanism. At 298\textdegreeK, the half-life is 340 minutes. Find the value of the reaction rate constant. Calculate the number of minutes required for the reaction to proceed 70 percent towards completion.

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Solution:

Half-life may be defined as the time necessary for half the particular reactant present initially, in this case N_2O_5, to decompose. For a first order reaction, the rate constant can be expressed in terms of half-life, t_1/2 . Namely, t1/2= .693 / k, where k is the rate constant. (Caution: This expression is onlytrue for a first order reaction.) Solving this expression for k to obtain k = .693 /t_1/2 = .693 / 340 = 2.04 × 10^-3 min^-1. To find the amount of time required for the reaction to proceed to 70% completion, use the fact that t = (2.303 / k) log (C_0 / C) in a first order reaction, where t = time, k = rate constant, C_0 = initial concentration and C = existing concentration. Suppose the initial concentration is X, then at 70% completion, the existing concentration is .30 X_4 Having calculated k, substitute to find t = minutes. t = (2.303/k) log (C_0 /C) = [2.303 / (2.04 × 10^-3)] log {X / .30X} = (1129)(.523) = 590 min.

Question:

Determine the output of the following program READ (2, 3) I, J 3FORMAT (2I11) IF (I - J) 5,6,6 6WRITE (3,10) I 10FORMAT (1X, I11) GO TO 7 5WRITE (3,10) J 7STOP

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G12-0279.htm

Solution:

To analyze the program, start from the top. The READ state-ment tells the computer to use the "card reader" for reading in the values I and J which are specified by the following FORMAT statement, identified by the number 3. The FORMAT statement states that there are eleven columns used on the card for recording each number input (i.e., 2I11). Every blank space among eleven reserved is equivalent to zero. Naturally, zeros before the first digit of the number have no effect on the number, while each blank space at the end increases the number ten times. "IF", is a CONTROL statement. In this problem it has the form IF(e)a, b, c, where e is an expression to be evaluated (in this case: I-J), and a, b, c are three statement numbers. The meaning of such statement is as follows: executed is the onelabelleda. executed is the onelabelledb. executed is the onelabelledc. Therefore, in the program, if the value of I-J is negative, the next statement to be executed is the one numbered 5, which causes the output of J. Following this the STOP statement is executed. If the value of I - J is zero or positive, the next statement to be executed is the one numbered 6,which causes the output of the value of I. The statement following this is another control statement "GO TO". This statement causes a jump in program execution and program pointer goes directly to the statement which is specified by a number following GO TO statement. Therefore it is clear that the program finds and prints out the largest of two numbers.

Question:

For the reaction 2HI(g) \rightleftarrows H_2 (g) + I_2 (g) , the value of the equilibrium constant at 700K is 0.0183. If 3.0 moles of HI are placed in a 5-liter vessel and allowed to decompose according to the above equation, what percentage of the original HI would remainundissociatedat equilibrium?

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Solution:

This problem is an application of the equilibrium expres-sion K = { [H_2] [I_2] } / [HI]^2 . Since two moles of HI are involved in the production of one mole of H_2 and one mole of I_2 , if x moles of H_2 (and therefore x moles of I_2 ) are present at equilibrium, then 2x moles of HI have been consumed and 3\Elzbar2x moles of HI remain. Therefore, at equilibrium, [H_2 ] = [I_2 ]= x moles/ 5 liters and [HI] = (3 \rule{1em}{1pt} 2x) moles/5 liters. Substituting these into the expression for K gives K = 0.0183 = {[H_2] [I_2 ]} / [HI]^2 = {(x/5) (x/5)} / = {(x/5) (x/5)} / [(3 \rule{1em}{1pt} 2x)/5]^2 = [x^2 / (3 \rule{1em}{1pt} 2x)^2 ] × (5^2 / 5^2 ) = [x^2 / (3 \rule{1em}{1pt} 2x)^2] = [x^2 / (9 \rule{1em}{1pt} 12x + 4x^2 )] , or 9(0.0183)\Elzbar 12(0.0183)x + 4(0.0183) x^2 = x^2 0.9268 x^2 + 0.2196x\Elzbar 0.1647 = 0. Using the quadratic formula, x = {\Elzbar0.2196 \pm\surd[(0.2196)^2\Elzbar4(0.9268) (\Elzbar0.1647)]} / {2(0.9268)} or x =\Elzbar 0.56, x = 0.32 . Since negative moles are a nonphysical entity, the first answer is discarded and the second retained. Thus, at equilibrium, [H_2] = x = 0.32 mole [I_2]= x = 0.32 mole [HI] = 3\Elzbar 2x = 3\Elzbar 2(0.32) = 2.36 moles .

Question:

Hard water is often softened by a commercial water softener. Explain, in general, its operation.

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Solution:

Hard water possesses dissolved calcium or magnesium ions. Softening the water means the removal of these ions. Commercial water softeners perform this task by a process known as ion exchange. In this process, water is passed through a bed of cation-exchange resin or zeolite. A cation-exchange resin is an insoluble polymer of polystyrene or other material that possesses sulfonic acid groups attached to it. The resin is activated by NaOH or NaCO_3 that produces the corresponding sodium salt. Calcium and magnesium ions are held more tightly by the resin than are Na^+ ions. Consequently, when the hard water is passed through this resin, the hard-water ions are held by the resin, and the sodium ions are released. This is the operation of a commercial water softener.

Question:

To 1.0 liter of a 1.0M solution of glycine at the iso-electric pH is added 0.3 mole of HCl. What will be the pH of the resultant solution? What would be the pH if 0.3 mole of NaOH were added instead?

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Solution:

When a species is held at its isoelectric pH, the molecules are all neutral. If there is 1.0 liter of 1.0M glycine present, there are 1.0 liter × 1.0 (moles / liter) = 1.0 moles of glycine present. When 0.3 moles of HCl are added to this solution, 0.3 moles of the glycine become protonated and 0.7 moles are left unprotonated. The k^1_a for this reaction is 2.34. One can solve for the resulting pH by using the Henderson-Hasselbach equation, pH = pK^1_a + log [HA] / [H_2A^+] pH = 2.34 + log [(0.7) / (0.3)] = 2.34 + .37 = 2.71. Similar logic is used in solving for the pH when 0.3 moles of NaOH are added. Here 0.3 moles of glycine will be negatively charged and 0.7 moles will remain neutral. The pK^1 for this reaction is 9.6 and the Henderson-Hasselbach equation becomes pH = pK^1_a + log [A^-] /[HA] pH = 9.6 + log [(0.3) / (0.7)] = 9.23.

Question:

A laser emits a monochromatic light beam, of wavelength \lambda, which falls normally on a mirror moving at velocity V. What is the beat frequency between the incident and reflected light?

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Solution:

When a source of light is in motion relative to an observer or vice versa, the light waves exhibit a change in frequency as seen by that observer. This phenomenon is known as the Doppler effect, and the Doppler shift for light is given as v_observed= v\surd[(c \pm v) / (c \mp v)](1) where v is the frequency of the light in an inertial frame of refer-ence, v is the speed of the source relative to the observer, and c is the speed of light. In this problem v is the speed of the mirror. Let us choose the mirror to be moving away from the laser, so that it absorbs the light as if it were an observer moving away at speed v; it then re-emits the light as if it were a source moving away at speed v. In both cases the source and observer are separating from one an-other, so the frequency decreases. To get v_observed, < v = we need the lower signs in equation (1). Now, when the mirror absorbs the light, we have v_absorved= v\surd[(c - v) / (c + v)](2) Similarly, when it re-emits the light, the frequency shift is given by v_emitted=v_absorved\surd[(c - v)/(c + v)] = v[(c - v)/(c + v)].(3) The beat frequency is the difference in frequencies between the in-cident and reflected light: v_beat= v -v_emitted v_beat= v - v[(c - v) / (c + v)] = v[1 - {(c - v) / (c + v)}] = v[2v / (c+ v)] v_beat= c / \lambda[2v / (c+ v)] when v << c, then c + v \cong c. In this casev_beat\cong 2v / \lambda.

Question:

An automobile changes its speed from 0 to 30 m/sec (about 63 mph) in 12 sec and continues to accelerate at this rate. A tire of the auto has a radius of R = 0.4 m. Calculate the angular acceleration of the tire, assuming no slipping, and plot the angular velocity, tangential velocity at the circumference, and centripetal acceler-ation at the circumference as a function of time.

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Solution:

Linear and angular velocity are related by the equation: v = \omegaR where v is linear velocity, \omega the angular velocity, and R the radius of the wheel. We will assume, in this problem, that the velocity increases uniformly since nothing to the contrary has been stated. By definition: a = (∆v)/(∆t)\alpha = (∆\omega)/(∆t) where a and \alpha are the linear and angular accelerations respectively. However: a = (∆v)/(∆t) = (∆\omega)/(∆t) - R = \alphaR We thus have a relationship between a and \alpha. a = (30 m/sec - 0 m/sec)/( 12 sec) = 2.5 m/sec^2 Therefore: \alpha = a/R = (2.5 m/sec^2)/(0.4 m) = 6.25 rad./sec^2 Note that although by dimensional analysis the units in a should reduce to 1/(sec.^2), we express angular acceleration in terms of rad./sec.^2. This is done because of the physical considerations of the problem. We consider the wheel as rotating through a discrete angle measured in radians, in a certain amount of time. However, our rigorous dimensional analysis has not really been violated since radians are dimensionless. As we know, radians are defined using a central angle of a circle. The magnitude of this angle in radians is calculated by dividing the length of the arc that the central angle subtends, by the length of the circle's radius. (See diagram.) \texttheta (in radians) = s/r Both s and r are measured in units of length which cancel in the ratio. We see, therefore, that radians are dimensionless. To calculate the angular velocity, we must remember the formula: \omega = \omega_0 + \alphat where \omega_0 is the initial angular velocity, and t is time. \omega = 0 + (6.25 rad./sec.^2)(t sec.) = 6.25 t rad./sec. The tangential velocity is: v = \omegaR = (6.25 t rad./sec.) (0.4 m) = 2.5 t m/sec. Note that we may now drop the radians from our units. To calculate the centripetal acceleration at the circumference we must remember the formula: a_c = (v^2)/(R) = (\omega^2R^2)/(R) = \omega^2R = (6.25 t rad./sec.)^2 (0.4 m) = 15.6 t2m/sec.^2 w, v, and \alpha are plotted against time in figures 1, 2 and 3.

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Question:

Write a program to count the number of letters and digits in the input data.

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Solution:

PROGRAM CHARACTERCOUNT (INPUT, OUTPUT); VAR LETTERCOUNT, DIGITCOUNT: INTEGER; CH : CHAR; BEGIN LETTERCOUNT: = 0; DIGITCOUNT: = 0; WHILE NOT EOF (INPUT) DO BEGIN READ (CH); IF('A' <=CH) AND (CH<='Z') OR ('a' <=CH) AND (CH<='z') THEN LETTERCOUNT; = LETTERCOUNT + 1 ELSE IF ('0'<=CH) AND (CH<= '9') THEN DIGITCOUNT: = DIGITCOUNT + 1 END; WRITELN (LETTERCOUNT, ' LETTERS', DIGITCOUNT, 'DIGITS.') END. This program assumed that, in the lexicographic ordering, every character between 'A' and 'Z', both inclusive is an upper-case letter, and every character between 'a' and 'z' both inclusive is a lower-case letter. This assumption is valid in most character sets including ASCII, but it is not strictly valid in the EBCDIC character set where certain "control characters " occur among the letters. These control characters are unlikely to occur in normal data, however, so the assumption is not a bad one. The program also assumes that every character between "0" and "9" inclusive is a digit. This assumption is in fact valid in every character set .

Question:

What features enable the ferns to grow to a much largersize than the mosses?

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Solution:

The ferns (DivisionPterophyta) are placed higher up on the evolutionary ladder than the mosses. One obvious difference between the two lies in their sizes. While the mosses rarely grow beyond 15 cm in height , ferns have been found that reach a height of 16 meters, larger by a factor of 100. Ferns are able to attain these heights because of the presence of both more efficient roots and a vascular system. The root of a fern is much more advanced than the rhizoid of a moss. It is elaborated into tissues and varying zones of maturity. From the bottom up, it is composed of a pro-tective cap, an apical region of rapid cell division, a zone of elongation, and above it, a zone of maturation. To increase the total surface area for absorption, the primary root branches and rebranches to form many smaller roots. This well-developed root system of the ferns, in contrast to the simple rhizoids of mosses which are merely filaments of cells, allows for firmer anchorage and more efficient absorption of materials as demanded by a larger plant. The absorbed water and minerals are then conducted up the stem and leaf petioles of the fern to the leaves by the xylem of the vascular system. In addition to xylem , there is the phloem, which transports organic products synthesized in the leaves down to the stem and roots. The xylem and phloem tissues, present in the ferns but not the mosses, make possible long-distance transport of the essential materials required by a big plant. Since the vascular system serves also as supportive tissue, it adds strength and rigidity to a fern plant, enabling it to grow to a large size. In summary, because of the presence of an elaborate root system and a vascular system, the ferns are able to grow to a much larger size than are the mosses.

Question:

The seal over a circular hole of diameter 1 cm in the side of an aquarium tank ruptures. The water level is 1 m above the hole and the tank is standing on a smooth plastic surface. What force must an attendant apply to the tank to prevent it from being set into motion?

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Solution:

For streamline flow of an incompressible, non viscous fluid, Bernoulli's equation can be applied. It states P_1 + (1/2) \rhov^2_1 + \rhogy_1 = P_2 + (1/2) \rhov^2_2 + \rhogy_2 where the subscripts 1 and 2 refer to quantities per-taining to any two points along the flow. The absolute pressure is p, \rho the density, v the velocity, g the gravitational acceleration, and y the elevation above some arbitrary reference level. Take point 1 to be at the surface of the water in the tank, a height h above the hole which is taken to be point 2. The pressures above the tank and outside the hole are both atmospheric pressure. Applying Bernoulli's theorem to this case, we thus have P_a + 0 +\rhogh= P_a + (1/2)\rhov^2 + 0. Here v is the velocity of efflux from the hole, the reference level for height being taken as the horizontal level through the hole. Since the cross section of the tank is very much larger than the area of the hole, the liquid in the tank is assumed to have zero velocity. Thus v = \surd2gh. Let A be the area of the hole. The mass of fluid ejected in timedtis dm =\rhodv=\rhoAdl=\rhoAdl/dtdt= \rho Avdt, and thus the momentum P acquired in timedtis \rhoAv^2dt. The escaping fluid therefore has a rate of change of momentum ofdP/dt= \rhoAv^2 and thus by Newton's second law F =dP/dtthe force causing this is \rhoAv^2. By Newton's third law, an equal and opposite force acts on the tank. Hence, to prevent the tank from moving backward, the attendant must apply to the tank a force of magnitude F = \rhoAv^2 = \rho(\pid^2/4) × 2gh = 10^3 kg/m^3 × \pi/4 × 10^-4 m^2 × 2 × 9.8 m/s^2 × 1 m = 1.54 N.

Question:

A chemist has a solution which contains Zn^2+ and Cu^2+, each at .02M. If the solution is made 1M in H_3O+ and H_2S is bubbled in until the solution is saturated, will a precipitate form? K_sp of ZnS = 1 × 10-^22, K_sp of CuS = 8 × 10-^37, and K_eq for H_2S, when saturated, = 1 × 10-^22 .

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Solution:

A precipitate will form only when the ion product exceeds the experimentally determined solubility product constant (K_sp ). Therefore, compute the ion product of ZnS and CuS to determine if they exceed the stated K_sp 's, 1 × 10-^22 and 8 × 10-^37 , respectively. The equation for the reaction between H_2S and H_2O is H_2S + 2H_2O \rightleftarrows 2H_3O+ + S- . In a saturated solution, the ion product is equal to [H_3O+]^2 + [S^2-] = 1 × 10-^22 = K . You are told that the H_3O+ concentration is 1M. Substituting this into that expression, and solving for [S^2-], you obtain [S^2-] = 1 × 10-^22 M. [S^2-] = 1 × 10-^22 M. You are given the Zn^2+ and Cu^2+ concentrations, which means that the ion products can be computed. For Zn+^2 , [Zn^2+] [S^2-] = (.02) ( 1 × 10-^22 ) = [Zn^2+] [S^2-] = (.02) ( 1 × 10-^22 ) = 2 × 10-^24 = K_spfor ZnS. [Cu^2+] [S^2-] = (.02) ( 1 × 10-^22 ) = 2 × 10-^24 = K_spfor CuS . Since the ion product of ZnS does not exceed the K_sp of ZnS, 1 × 10-^22 , there will be no precipitation. However, the ion product of CuS does exceed the K_sp of CuS, 8 × 10-^37 , which means you will see precipitation of CuS.

Question:

If an object is in a position 8 cm in front of a lens of focal length 16 cm, where and how large is the image? (See figure.)

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Solution:

we refer here to the relationship 1/0 + 1/i = 1/f Here, the image distance, i, is the unknown. The object distance, 0, is 8cm, while the focal length, f, is 16 cm. Thus, 1/0 + 1/i = 1/f 1/8cm + 1/i = 1/16cm 1/i = 1/16cm - 1/8cm 1/i =-(1/16cm) i= - 16cm This means that the image is 16 cm in front of the lens. To calculate the image size, we first calculate the magnification. Thus, m = -i/0 = +16/8 = 2 This means that the image is erect, therefore virtual, and twice as large as the object, or 16 cm high.

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Question:

Explain what the computer takes as the attributes of the fol-lowingvariables by Default Rules of PL/I, if full explicit declarationsare not provided by the programmer. a) A variable named:PERCENT (1:8), b) A variable named:AMOUNT FIXED, c) A variable named:NUMBER COMPLEX, d) A variable named:TOTAL BINARY, e) A variable named:INCANT

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Solution:

a) First of all note that the name of the vari-able; PERCENT starts witha P. The letter P does notliebe-tween the letters I and N (inclusive). Hence, PERCENT is as-signed the following: PERCENT(1:8)DEC, REAL, FLOAT(6); b) The variable AMOUNT has the scale declared as FIXED. Hence, thecomputer completes the attributes as follows: AMOUNT FIXED (5) DEC, REAL. In the above, the precision is made equal to 5, i.e., (5,0). The Base and Mode are supplied as DEC and REAL. c) The variable NUMBER has the mode specified as COMPLEX. Hence, the following remaining attributes are supplied by the computer: Scale :FIXED Base :DEC Precision: (5), i.e., (5,0) Hence, the variable looks like that given below. NUMBERFIXED(5) DEC COMPLEX. d) The missing attributes are supplied for TOTAL BINARY as follows, according to the Default rules: TOTAL FLOAT (21) BINARY REAL; e) The missing attributes are supplied for INCANT as follows: INCANTFIXED(15) BIN, REAL; It should be noted in which order the attributes following a variable name aredeclared, especially the Scale and Preci-sion. The Precision attribute shouldalways follow the Scale attribute, and never vice versa.

Question:

A coil of wire is connected across one gap of a Wheatstone bridge and a temperature-controlled standard 1-\Omega resistor across the other. If the temperature of the coil is 0\textdegreeC, the other arms of the bridge have re-sistances in the ratio 0.923. If the temperature of the coil is 100\textdegreeC the ratio is 1.338. What is the temperature coefficient of resistance of the wire?

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Solution:

A Wheatstone bridge is shown in the figure. When the bridge is balanced, the current through galvanometer G is zero. In this case, then, the po-tential between points a and b is the same, whether we calculate it by path cab or path dab. Using Kirchoff 's Voltage Law around these 2 loops, we find - I_1 R_1 + I_2 R_4 = 0 andI_1 R_2 - I_2 R_3 = 0 Solving each equation separately for the ratio I_1 /I_2 (I_1 /I_2 ) = (R_4 /R_1 ) (I_1 /I_2 ) = (R_3 /R_2 ) and, therefore,(R_4 /R_1 ) = (R_3 /R_2 ) or(R_1 /R_2 ) = (R_4 /R_3 ) Let R_1 be the coil, and R_2 the temperature-controlled standard. In the first case, [{R_1 (0\textdegreeC)}/(1 \Omega)] = .923 orR_1 (0\textdegreeC) = .923\Omega In the second case R_1 (100\textdegreeC) = 1.338 \Omega But R_1 (100\textdegreeC) = R_1 (0\textdegreeC) (1 + \alpha t), where \alpha is the temperature coefficient of resistance of the wire and t is the difference in temperature experienced by the resistor (100\textdegreeC). Thus \alpha = [{R_1 (100\textdegreeC) - R_1 (0\textdegreeC)}/{t R_1 (0\textdegreeC)}] = [(1.338 \Omega - .923 \Omega)/{(100\textdegreeC) (.923 \Omega)}] \alpha = .0045 per \textdegreeC

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Question:

How are the concepts of local and global variables useful in debugging an APL program?

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Solution:

Since APL is an interactive system, the negative feedback on an incorrect program is immediate. An under-standing of local and global variables can help in the quick debugging of APL programs. Variables that retain their values only during function execution are classified as local. By contrast, global var-iables are retained in the workspace before, during and af-ter the execution of a program. To clarify the distinction, consider the following two programs: R \leftarrow 5 \nablaCIRCLE [1]D \leftarrow 2 × R [2]C \leftarrow 3.14159 × D [3]A \leftarrow 3.14159 × R × R [4]R,D,C,A [5]\nabla Here R is a global variable because it is active before the execution of a function. That is, R has already been store before \nablaCIRCLE \nablaCIRCLE 1 R [1]D \leftarrow 2 × R [2]C \leftarrow 3.14159 × D [3]A \leftarrow 3.14159 × R × R [4]R,D,C,A [5]\nabla Now, R, the radius, is a local variable. Suppose R = 5. Then CIRCLE 1 5 cause the values 5, 10, 31.42, 78.54 to be printed. Keying in an R, after execution is completed produces an error message: R VALUE ERROR R \wedge \wedge Once the function has been executed, R is lost. D,C, and A, however, are global variables because they were not defined in the header. Since local variables save space, it is often desirable to use them in a program. Additional local variables can be defined by adding them to the header, with a semicolon be-tween each pair of variables. For example, \nablaCIRCLE R; D; C; A will make all variables local.

Question:

What is the output for the following FORTRAN program? DIMENSION A (8) , PTR (8) DATA A/'A', 'S', 'E', 'Q', 'N', 'U', 'T', '1'/ DATA PTR/4,7,2,6,6,3,0,5/ I = PTR (1) 20IF (I.EQ.0) GO TO 30 WRITE (5,100) A (I) 100FORMAT (1X,A1) I = PTR (I) GO TO 20 30CONTINUE STOP END

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Solution:

This is an elementary problem using pointers. The object is to introduce pointers in a simple way by allowing you to "walk" through the program to find the output. The first executable statement puts the first value of the array PTR in the variable I. If you look at the DATA statement for PTR, you will notice that PTR (1) is 4. I is not equal to zero, so control passes to the WRITE statement, which indicates that the element in A(4) should be outputted. This, as indicated by the DATA statement for A, is repre-sented by the letter Q. Next, the value contained in PTR (4) replaces the initial value of I. The number 6 is in the position of PTR (4) now. Control then passes back to statement 20, and the loop continues until PTR (I) reaches a value of zero. Notice that termination will occur when PTR (7) stores 0 in I. The final output requires only five passes until termination. Fol-low the program and you will see that your "quest" for the answer will not be difficult at all. I^th element of A(I) PTR(I) A(I) 4th 6th 3rd 2nd 7th 4 6 3 2 7 Q U E S T

Question:

If antibodies can be brought into contact with a virus before it attaches itself to the host cell, the virus will be inactivated. Once infection has occurred, the production of interferon may protect the host from ex-tensive tissue destruction. Explain.

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Solution:

Many human diseases have a viral etiology (cause). Among the more common viral diseases are smallpox, chicken pox, mumps, measles, yellow fever, influenza, rabies, poliomyelitis, viral pneumonia, fever blisters (coldsores), and the common cold. Although most viral infections do not respond to treatment with many of the drugs effective against bacterial infection , many are preventable by means of vaccines. Buildup of an adequate supply of antibodies requires some time. During this period, extensive viral infection may occur. Most recoveries from viral diseases occur before full development of the immune response (productionof antibodies). A factor that is important in the recovery from a viral infection is interferon. Interferon is a protein produced by virus- infected cells, which spreads to uninfected cells and makes them more resistant to viral infection. Thus, extensive viral production and resultant tissue damage are prevented. Upon infection, the host cell is induced to syn-thesize interferon, a small protein which is then released into the extracellular fluid and affects the surrounding cells. Interferon binds to surface receptors of uninfected cells . This triggers these cells to synthesize acyto-plasmicenzyme which prevents viral multiplication. Note that the antiviral protein does notprevent entrance of the viral nucleic acid. Interferon produced by a cell infected with a particular virus can prevent healthy cells from becoming infected by almost all other types of viruses . Inter-feron is therefore not virus-specific. However, inter-feron produced by one host species is not effective if ad-ministered to a different species . It is therefore host species specific. Since interferon produced by birds and other mammals cannot be used in treating human beings, it is difficult to obtain large enough quantities of inter-feron to provide effective chemotherapy for viral diseases. (Human donors cannot provide the needed amount of inter-feron.) Interferon is a more rapid response to viral infection than antibody response . Interferon production is initiated within hours after viral infection, reaching a maximum after two days. Within three or four days, interferon production declines as antibodies are produced. Prevention of viral infection by interferon production must be distinguished from another phenomenon - viral interference. Viral interference is observed when an initial viral infection prevents a secondary infection of the same cell by a different virus. The initial virus somehow prevents reproduction of the second virus or inactivates re- ceptors on the cell membrane (there are specific sites for viral attachment ). It may also stimulate production of an inhibitor of the second virus . Viral interference does not prevent uninfected cells from becoming infected . Vaccinationinvolesadministration of an atten-uated (live, yet weakened ) or inactivated strain of virus (or other microorganism), which cannot produce the disease, but which stimulates antibody production and thus prevents infection by the pathogen.

Question:

When a current of 5.36 A is passed for 30 min. through an electro-lytic cell containing a solution of a zinc salt, it is found that 3.27g of zinc metal is deposited at the cathode. What is the equivalent weight of zinc in the salt?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E16-0575.htm

Solution:

To solve this problem one must recall that 1 coulomb = 1A-sec. This means that the number of coulombs passed through the cell is equal to the product of the current ( in amperes ) and the time (in seconds). In our case, since 5.36 A was passed for 30 min. (30 min. × 60 sec/min = 1800 sec), the number of coulombs is 5.36A × 1800 sec = 9650 coulombs. One Faraday (1F) is equivalent to 96,500 coulombs. Hence, converting 96500 coulombs to Faradays gives (9650 coulombs) / (96,500 coulombs / F ) = 0.10 F . The equivalent weight of zinc is the weight deposited by 1Fof charge. Since 0.10Fdeposits 3.27g of zinc, the equivalent weight of zinc is 10 × 3.27g = 32.7g (since 10 × 0.10 F = 1.0 F ).

Question:

Assuming ∆H\textdegree remains constant, calculate the equilibrium constant, K, at 373\textdegreeK, if it equals 1.6 × 101 2at 298\textdegreeK for the reaction 2NO (g) + O_2 (g) \rightleftarrows 2NO_2 (g). The standard enthalpy change for this reaction is - 113 kJ/mole.

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Solution:

This problem can be solved by employing thevan'tHoff equation. This equation allows a determination of an equilibrium constant, if the ∆H\textdegree is known (and constant) and the value of the equilibrium constant at another temperature is known. It states log (K_2 /K_1 ) = [(∆H\textdegree)/(19.15)] [(1/T_1 ) - (1/T_2 )] , where K_2 and K_1 are equilibrium constants, ∆H\textdegree = the standard enthalpy formation, and T_1 and T_2 are tem-peratures in Kelvin. Let K_1 =1.6 × 101 2at T_1 = 298\textdegreeK; T_2 = 373\textdegree; and ∆H = - 113 J/mole. Substitute these values into the equation and solve for K_2 , which will be the solution to this problem. Thus log [(K_2 )/(1.6 × 10^1 2 )] = [(-113, 000)/(19.15)] [(1/298) - (1/373)] Solving for K_2, one obtains K_2 = 1.7 × 10^8 .

Question:

Explain briefly the following terms; (a) isotropic and anisotropic solids, (b) general and directional properties of crystalline substances, (c) a plane, an axis, and a center of symmetry and (d) polymorphism and allotropism.

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Solution:

These properties all pertain to the solid state. (a) An isotropic solid is one in which there are no directional differences in structure. Isotropic solids usually do not form regular crystals, and examples of an isotropic amorphic solids are Plexiglass and glass. An anisotropic solid is one in which its properties depend, in general, upon the particular direction along which the measurement is made. Crystalline substances, for example, are anisotropic. Examples of anisotropic crystals are cubic and tetrahedral lattices, (b) By general properties of crystalline substances, you mean those such as density, specific heat, melting point and chemical composition. These are called "general" due to the fact that none of them involve any special direction. Directional properties refer to those such as refraction and absorption of light, cohesion, expansion and electrical and thermal conductivity, all of which are measured in a special direction, (c) Plane of symmetry is said to exist in a crystal if it can be divided by an imaginary plane passing through the center, such that the two equal parts formed are an exact mirror image of each other. An axis of symmetry is said to exist in a crystal if you can draw a line through the crystal's center and when you proceed to rotate the crystal about this line through 360\textdegree the same appearance of itself is produced more than once. A crystal will have a center of symmetry, if every crystal face has a twin equidistant on the opposite side of this center. The accompanying figures illustrate each of these, (d) Polymorphism refers to the existence of substance having a definite composition yet occurring in more than one crystalline form. When the substance is an element rather a compound, this phenomenon is termed allotropy.

Question:

If a rock sample is found to contain approximately 1 polonium atom ^214 _84 Po for every 8.7 × 10^20 uranium nuclei ^238 _92 U, what is the half-life of polonium.

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Solution:

Each decaying uranium atom produces one polonium nucleus along the decay chain. Let N_u denote the number of uranium atoms decaying slowly (the half-life of ^238U is T_U = 4.5 × 10^9 years), andN_pthat of its decay product polonium. The rate of change ofN_pconsists of two parts. One of these is the rate of increase of Po-nuclei as a result of the decay of U-atoms; (dN_p/dt)_1 =\Elzbar (dN_u/dt) =\lambda_UN_u . where\lambda_uis the decay rate of uranium per nucleus. The enormousness of the ratio of U-atoms to Po-nuclei shows that the decay rate of polonium is much greater than that of uranium. Since uranium decays very slowly, we may regard the ratedN_u/dtas substantially constant over a long period of time. The number of the Po-nuclei decreases as a result of the decay of Po. This decay rate is given by (dN_p/dt)_2 =\Elzbar\lambda_pN_p where\lambda_pis the decay constant of polonium. The total rate of change of the polonium nuclei is (dN_p/dt) = (dN_p/dt)_1 + (dN_p/dt)_2 =\lambda_UN_u\Elzbar\lambda_pN_p or\cong constant\Elzbar\lambda_pN_p(1) For times sufficiently larger than the life-time of polonium, the rate of change of the polonium population, due to uranium decay, will be negligible, and the rate of change of the polonium population is dominated by the fast decay of polonium. This can also be seen from the solution of (1). Let us write (1) as (dN_p/dt) =\Elzbar\lambda_p[N_p\Elzbar { (\lambda_UN_u ) /\lambda_p} ]. If we define N' =N_p\Elzbar { (\lambda_UN_u ) /\lambda_p} , then dN' /dt= (d /dt) (N_p\Elzbar constant) =dN_p/dt ordN' /dt=\Elzbar\lambda_pN' . This is the same as the decay equation for a radioactive material. Its solution, therefore is N' (t) =N'_t= 0e\Elzbar\lambdap^t orN_p(t)\Elzbar { (\lambda_UN_u ) /\lambda_p} = [N_p\Elzbar { (\lambda_UN_u ) /\lambda_p}]t = 0e\Elzbar\lambdap^t . Because of its very slow decay, N_u is approximately the initial number of uranium atoms. Since there are no polonium atoms initially,N_p= 0 at t = 0. We have N_p(t) \approx { (\lambda_UN_u ) /\lambda_p} = (1\Elzbar e\Elzbar\lambdapt) . When t is large, i.e.,t\lambda_p> > 1, e\Elzbar\lambdaptbecomes negligible; N_p\approx { (\lambda_UN_u ) /\lambda_p} .(2) The half-life is defined as T = (1/\lambda) In 2, therefore the half-life of polonium from equation (2) is T_p= (1/\lambda_p) In 2 \approx (N_p/ N_u )[(In 2) /\lambda_U] = (N_p/ N_u ) T_U = [(1 nucleus ) / (8.7 × 10^20 nuclei) ] × (4.5 × 10^9 years) = 5 × 10\Elzbar12year = (5 × 10\Elzbar12year) × (3.1 × 10^7 sec/year) = 1.6 × 10\Elzbar4sec .

Question:

Does an ideal gas get hotter or colder when it expands according to the law pV^2 = Const?

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Solution:

The ideal-gas equation states that pV =nRT where V is the molar volume, R is the gas constant, T is the temperature and n is the number of moles. Therefore, for pV^2 = Const., we have (pV)V = Const. nRTV = Const. orV = (Const.)/(nRT) We see that as the volume increases, temperature must decrease since V is inversely proportional to T.

Question:

A gaseous sample of neon, maintained at constant tem-perature, occupies 500 ml at 2.00 atm. Calculate the volume when the pressure is changed to each of the following:(a)4.00 atm;(b)760 torr;(c)1.8 × 10^-3 torr.

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Solution:

We need a relationship between volume and pressure. Such a relationship is provided by Boyle's Law, which states that the product of pressure P and volume V of an ideal gas is a constant, k, or PV = k. We must first determine k for the neon sample. We are given a value of P corresponding to a given value of V. Therefore, k = PV = 2.00 atm × 500 ml = 2.00 atm × 0.500 l = 1.00 l-atm. Now that k is determined, we can obtain the value of V corresponding to the given values of P by using the formula V = k/P. (a)P = 4.00 atm:V = (k/P) = [(1.00 l-atm.)/(4.00 atm)] = 0.25 l (b)P = 760 torr = 760 torr × [(1 atm)/(760 torr)] = 1 atm: V = (k/P) = [(1.00 l-atm.)/(1atm)] = 1.00 l (c)P = 1.8 × 10^-3 torr = 1.8 × 10^-3 torr × [(1 atm)/(760 torr)] = 2.3 × 10^-6 atm: V = (k/P) = [(1.00 l-atm.)/(2.3 × 10^-6 atm)] = 4.35 × 10^5 l.

Question:

Most membranes of animal cells contain about 60% protein and 40% phosphoglycerides. (a) Calculate the average density of a membrane, assuming that protein has a density of 1.2 g/cm^3 and phosphoglyceride a density of 0.92 (g/cm^3). (b) If a sample of membrane material were centrifuged in NaCl solution of 1.05 specific gravity, would it sediment or float?

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Solution:

The density of any substance is its mass divided by its volume. It is usually written in terms of g/cm^3. (a) The average density of the membrane is made up of 60% protein and 40% phosphoglycerides. From the data given, one knows that 1 cm^3 of protein weighs 1.2 g and that 1 cm^3 of phosphoglyceride weighs 0.92 g. One cm^3 of membrane is made up of .60 × 1.2 g + .40 × 0.92 g or 1.09 g. The average density of the membrane is 1.09 g/cm^3. (b) If a compound is placed in a solution, the compound will sink if its density is greater than that of the solution. The density of the NaCl solution is 1.05 and that of the membrane is 1.09, therefore the membrane will sediment.

Question:

(a) blood plasma, (b) intracellular fluid of muscle, (c) gastric juice (pH = 1.4), (d) tomato juice, grapefruit juice, (f) sea water. Use the accompanying table. pH of some fluids pH Seawater 7.0 - 7.5 Blood plasma 7.4 Interstitial fluid 7.4 Intracellular fluids Muscle 6.1 Liver 6.9 Gastric juice 1.2 - 3.0 Pancreatic juice 7.8 - 8.0 Saliva 6.35 - 6.85 Cow's milk 6.6 Urine 5 - 8

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Solution:

The pH is defined as - log [H^+]. Therefore, [H^+] can be found by calculating 10^- pH. (a)blood plasma - pH = 7.4 7.4 = - log [H^+]; [H^+] = 10^-7.4 = 3.98 × 10^-8 M (b)intracellular fluid - pH = 6.1 6.1 = - log [H^+]; [H^+] = 10^-6.1 = 7.94 × 10^-7 M (c)gastric juice - pH = 1.4 1.4 = - log [H^+]; [H^+] = 10^-1.4 = 3.98 × 10^-2 M (d)tomato juice - pH = 4.3 4.3 = - log [H^+]; [H^+] = 10^-4.3 = 5.01 × 10^-5 M (e)grapefruit juice - pH = 3.2 3.2 = - log [H^+]; [H^+] = 6.31 × 10^-4 M (f)sea water - pH = 7.0 7.0 = - log [H^+]; [H^+] = 1.0 × 10^-7 M.

Question:

What is the kinetic energy of a 1-pound weight that has fallen 16 feet?

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Solution:

Kinetic energy is the energy of an object due to its motion, and it is \textasciigravegiven by (1/2) mv^2 . The mass of a 1 lb object is F = mg m = F/g m = (1 lb) / (32 ft/s^2) = (1/32) slug We now calculate the velocity of the object after falling 16 feet: (1/2) gt^2 = d andgt= v so v = \surd(2gd), Substituting in this equation we have: v = \surd[2 (32 ft/s^2) (16 ft)] = 32 ft/s The kinetic energy is K.E. = (1/2) mv^2 K.E. =(1/2)(1/32) slug (32 ft/s)^2 = 16 ft-lb.

Question:

The result of cleavage is a blastula, typically a ball of cells surrounding a central cavity, or blastocoel. Cell divisions by mitosis continue after the blastula is formed, but generally speaking, the divisions alternate with intervals of cell growth. Hence no further decrease in size of cells takes place. Instead, mass movements of cells take place by which a gastrula is formed. What is a gastrula? Using echinoderms or amphibians as an example, describe the process of gastrulation.

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Solution:

A gastrula is a two or three-layered ball of cells, these layers being known as the germ layers. In amphibians and echinoderms, the gastrula is formed by the inward folding of one side of the blastula, partially obliterating the original cavity, or blastocoel. A new cavity is formed which is open to the outside, and is called the archenteron, or primitive gut. The blastopore is the opening from the archenteron to the outside. It lies on one side of the vegetal pole and is the site where infolding began. The cells which border the blastopore constitute the blastoporal lips. When gastrulation is finally complete, the outer germ layer is the ectoderm, and its inner layer, the endoderm. A third germ layer is present, the mesoderm. It occupies a position intermediate between the ectoderm and endoderm. Loose cells which are commonly present between the germ layers are termed mesenchyme. They are usually thought of as mesoderm, although they may arise from any of the three germ layers. Gastrulation in amphibians and echinoderms is essentially an inward folding and is accomplished by three easily defined processes: invagination; involution, and expansion. During invagination, cells of the vegetal area pocket inward to form the archenteron. The invaginating layer of cells is concave when viewed from out-side. Involution is the process by which the cells that border the vegetal area move toward the blastopore. They progressively roll around beneath the lips of the blastopore, eventually coming to lie within the mass of cells. Expansion is the process by which the cells of the animal hemisphere extend out and converge toward the blastopore. As a result of the coordination between these three processes, an echinoderm or amphibian egg during gastrulation does not cease to be a sphere. Throughout these three processes, the blastopore is changing its shape and position. It first widens from its crescent-shape, to form a semi-circle, then to the shape of a horseshoe, and finally to a complete circle. As it does so, its lips migrate toward the vegetal pole, an area of endodermal cells which is rich in yolk. Some of the yolk cells protrude through the blastopore lips as the yolk plug. This yolk plug is withrawn into the gastrula following gastrulation.

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Question:

Find the degree of ionization of 0.05 M NH_3 in a solution of pH 11.0. K_b = 1.76 × 10^-5.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E10-0350.htm

Solution:

The degree of ionization is the fraction of the total acid or base present (\alpha) that ionizes. Namely, \alpha = (x/c) , where x is the number of moles of acid or base that dis-sociate and c is the original number of moles present. To solve this problem, let it first be stated that NH_3 is a base and that [OH^-] can be calculated from the pH value. The reaction equation is NH_3 + H_2O \rightleftarrows NH_4^+ + OH^-. After obtaining [OH^-], [NH_4^+] will be the only unknown in the equilibrium constant equation, which is K_b = {[NH_4^+][OH^-]}/[NH_3] = 1.76 × 10^-5 . To find [OH^-], note thatpOH= - log [OH^-] and that pH +pOH= 14. Substituting the given value of pH, pOH= 14 - pH = 14 - 11 = 3. Thus, [OH^-] = - log 3 = 1 × 10^-3. Next, remember that for each x moles of NH_3 that dissociates, x moles of NH_4^+ will form. The problem becomes clearer if one observes what is happening from the table below. NH_3 + H_2O \rightleftarrows NH_4^+ + OH^- Before Reaction: 0.05OOO After Reaction: 0.05-xx1 × 10^-3 Upon substituting these values into the K_b equation, one obtains : K_b = {[NH_4^+][OH^-]}/[NH_3] = 1.76 × 10^-5 = [{x (10-3)}/(0.05 - x)] . Solving for [NH_4^+] , x = [NH_4^+] = 8.65 × 10^-4 . [NH_4^+ ] also represents the unknown value in the degree of ionization equation, since its concentration must be the amount of NH_3 that was ionized. Thus, \alpha = [(8.65 × 10^-4)/(0.5)] = 1.73 × 10^-3 .

Question:

Describe the functional units of a typical computer system hardware.

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Solution:

The computer hardware typically consists of functional units shown in Figure 1-3. Take as an example the lower end of the various types of computer systems (see next problem), such as Apple Mac IIx, IBM PS/2 80, and Next Inc. NeXT systems. In such systems, the input device is typically a monitor (a console terminal usually ca-pable of graphics) and optionally another CRT terminal. Strictly speaking, these devices are used for both input and output, where the keyboard is used for input and the Cathode Ray Tube (CRT) screen is used for output. In the example, typically the keyboard is type PC with a number of functions keys for the monitor used as the main system console. Additional terminals that are attached to the RS-232C serial ports of the personal computer may be of plain ASCII type capable of displaying only characters or graphics type capable of low or high resolution graphics. As noted, the user's screen is the output device. The output device may also be a printer attached to either parallel ports (Centronix type with 36 pins) or serial ports (RS-232C type with 25 pins) . The printer normally prints on a standard 8.5 × 11 paper with number of characters per line de-termined by the letter font used. The printers may be of letter- quality or dot-matrix type. The former normally uses daisy-chain character fonts and ribbons much like the typewriters. The dot--matrix printers define characters on a matrix of dots (normally, 5- by-7 or 9-by-12). These, in turn, may be of ribbon or ink-jet types. The ribbon types imprint the pattern through a ribbon, whereas ink-jet types eject special ink through the pattern. Finally, there are laser printers that utilize toners (ink pow-ders) and laser technology to produce high quality print indis-tinguishable from letter quality. The dot-matrix printers are capable of picture output in graphics mode. The communication device is generally a modem that converts signals back and forth for transmission over public telephone lines. The modem may be built-in (e.g. an electronic card inside the PC) or external such as Hayes smartmodem series. The disk may consist of one or two diskette drivers capable of processing 4.5-inch diskettes with capacities of 340 kilo-bytes or 1.2 megabytes. There may also be an optional internal Hard Disk Unit (HDU) with capacities anywhere from 10 to 300 megabytes. In the case of the more modern NeXT personal com-puter, an optical (rather than magnetic) disk unit with 500 megabyte capacity is used, along with an optional magnetic car-tridge tape unit, is used. Note that, in a way analogous to the purchase of automobiles, the computers come with standard (e.g. bundled with basic price) features and accessories, and also optional features and accessories (at additional cost!). The primary memory is Random Access Memory (RAM) that can be read or written. There may, in addition, be ROM memories that contain read-only code to be executed. Such memories may contain frequently used input-output or communication software that is burned (stored permanently) into the ROM. As we move from personal to the more sophisticated computer systems that can support many users, the picture is more complex in terms of a bewildering variety of input, output, and communi-cation devices. For example, in the case of mainframe computers, there may be I/O channels, multiplexers, concentrators, etc. as well as a hierarchy of memories from fast cache memories to slow mass storage units.

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Question:

How many characters can variable names contain in C? How many letters are significant?

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Solution:

Variable names in C are made of letters and digits. The first character must be a letter. The underscore _counts as a letter. Only the first eight characters of the variable name are significant.Eg. , in case of the two given variables, value_of_1and value_of_2, C will interpret these two variable names as equivalent, i.e. referencing a variable called value _ of, since their length is over eight.

Question:

Compare steady-state bacterial growth with synchronous growth. How can one obtain these growth conditions?

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Solution:

When bacteria are cultured in any closed system, such as a tube, flask, or tank, the pattern of growth includes four phases: lag, log, stationary, and death. However, one can set up an environment in which the bacterial population is maintained in the log (expon-ential) phase. This condition is called steady-state growth. Two devices, the turbidostat and the chemostat, are used to maintain the bacteria in a constant environ- ment and promote steady-state growth. The culture is kept at a constant volume by adding fresh media at the same rate old media is removed. This serves to provide a con-tinual replenishment of nutrients and a steady removal of any inhibitory products, maintaining the bacteria in the log phase of growth. The turbidostat involves the use of a photoelectric eye that monitors the turbidity of the culture. The turbi-dity is proportional to the density of the bacterial cells. The turbidostat maintains the turbidity at a specific con-stant value by regulating the flow rate of media (fresh input and old output). The chemostat regulates the level of growth by maintaining a constant, limiting amount of an essential nutrient and draining waste medium. This amount is just sufficient to keep the bacteria in the log phase. A steady-state growth is usually desired for experiments studying metabolism (since the log phase cells are nearly uniform in metabolic activity). Synchronous growth is irrelevant to the steady-state growth in the continuous culture of bacteria. In a popula-tion, all the bacterial cells do not normally divide at the same time (synchronous growth). If they did divide simul-taneously there would be an instantaneous periodic doubling of the bacterial population instead of the constant in-crease. One can manipulate the culture to produce synchro-nous growth but it lasts only a few generations, since some bacteria reproduce out of phase with the others. One may obtain a synchronized culture by inoculating the medium at freezing temperatures. The bacteria will metabolize slowly and become ready to divide. When the temperature is raised, they should all initiate reproduction simultaneously. An-other common method is to filter out the smallest cells of a log phase culture. Since these should be the ones that have just divided, they are rather well synchronized, and will eventually undergo division at approximately the same time. Graphically, synchronous growth appears as follows:

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Question:

Whatfunctionsare served by sound communication in animals?

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Solution:

Communication by sound can convey a variety of messages necessaryto the survival and reproduction of the participants. Sound communicationis of special interest to us as humans because it serves as thefun-damental basis of our language. Other species use forms of auditorylanguage, although not nearly as complex or sophisticated as humanspeech. Mate attraction, species recognition, defense, and territorialclaim are examples of functions of sound communication in animals. The femaleAedesmosquito creates a buzzing sound with her wings, which serves to attract the male mosquito. The sound waves cause themale's antennae to vibrate, stimulating sensory hairs. This allows the maleto locate the female and copulate with her. An interesting adap-tation forthis communication is that in the sexually immature male mosquito, the antennaehairs lie close to the antenna shaft, causing near deafness. This preventssexually immature males from responding to females. When the malemosquito reaches maturity, the antennae hairs become erect, allowinghim to be receptive to the female's characteristic mating sound. The frequency of the buzzing sound is specific for each species of mosquito; thus, sexually mature males will respond only to females of their ownspecies. The mating calls of frogs are also species-specific. In contrastto theAedesmosquito, however, it is the male frog that attracts thefemale by calling. Bird songs are a good example of sound communication being usedfor territorial defense and species recognition. A territory may be definedas, the area defended by an animal. This area centers around the animal'sbreeding ground, nest site, and sources of food or other needs. The use of sound to defend a territory is exemplified in the following experiment. If silent models of wood thrushes are set up within the territoryof a given thrush, they would be attacked by that thrush. If, however, a loudspeaker is set up next to the models, and the char-acteristicspeciessong of another species played, the thrush will not attack. This is because males attack only other males of their own species, since they are the most threatening to their territories. Singing also plays a role in the establishment of territories. A male birdchooses an area and sings loudly in order to warn away other males. During the spring, when boundaries are being established, the male thrushesthat sing most loudly and forcefully are those who successfully acquire the largest territories.

Question:

Given Zn \rightarrow Zn^+2 + 2e^- with E\textdegree = + .763, calculate E for a Zn electrode in which Zn^+2 = .025M.

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Solution:

This problem calls for the use of the Nernst equation, which relates the effect of concentration of ions in a cell on the voltage. It is stated E = E\textdegree -(.0592 / n)log K at: 25\textdegreec where E = potential under conditions other than standard, E\textdegree = standard electrode potential, n = number of electrons gained or lost in the re-action, and K = ratio of concentration of products to reactants. You are given E\textdegree and n = 2, since Zn loses 2 electrons to become Zn^+2 , For this problem, K =[Zn^2+] / [Zn]. You are given that [Zn^+2 ] = .025M. To find [Zn] , you have to know the meaning of standard electrode potential. Standard conditions are defined as 25\textdegreeC and 1 molar in concentration. Thus, if you assume that Zn is in the standard state, then [Zn] = 1M . To find E, substitute these values into the Nernst equation. E = .763 - (.0592 / 2)log (.025 / 1) = .763 - [(.0296)(-1.602)] = .810 V.

Question:

A ray of light enters the face BA of a right-angled prism of refracting material at grazing incidence. It emerges from the adjacent face AC at an angle \texttheta to the normal. If \textphi_C is the critical angle for the material, show that sin \texttheta = cot \textphiC_C. (See figure.) Will a ray always emerge from AC? If not, explain what happens, and deduce for what values of the re-fractive index of the material the ray actually emerges.

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Solution:

Since the ray strikes the prism at grazing incidence, the refracted ray DE must enter at the critical angle \textphi_C. Now angles DAE, ADF, and AEF are all right angles. This follows because the prism is right-angled, and also since DF and FE are normals to the surfaces BA and AC. Thus DFE must be a right angle also, since the four angles of a quadrilateral (ADFE) add up to 360\textdegree. The angles of a triangle must add up to 180\textdegree, and therefore \angleDEF = 90\textdegree - \textphi_C.(1) Applying Snell's law to the refraction at face AC, we have n_1 sin \angleDEF = n_2 sin \texttheta, where n_1 and n_2 are the refractive indices of the prism and air, re-spectively. Since the medium to the right of face AC is air, n_2 = 1, whence n_1 sin \angle DEF = sin \texttheta.(2) At surface BA, we obtain n_2 sin 90\textdegree = n_1 sin \textphi_C or(1) (1) = n_1 sin \textphi_C Thensin \textphi_C = 1/n.(3) Using (3) and (1) in (2) (1/sin \textphi_C) [sin (90 - \textphi_C)] = sin \texttheta But sin (90 - \textphi_C) = cos \textphi_C, whence (cos \textphi_C/sin \textphi_C) = cot \textphi_C = sin \texttheta The ray will emerge from AC only if \angle DEF is less than the critical angle. If it is greater, total internal reflection occurs and the ray is directed along AC. Thus, for the ray to emerge, \angle DEF < \textphi_C. That is, 90\textdegree -\textphi_C < \textphi_C or \textphi_C > 45\textdegree. Using (3) \thereforesin \textphi_C = (1/n_1) > sin 45\textdegree = (1./\surd2)\therefore n_1 <\surd2.

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Question:

What is meant by the autonomic nervous system? What are its subdivisions and to which of the large divisions of the entire nervous system does it belong?

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Solution:

Nerves from the central nervous system (CNS) that innervate the cardiac muscles, smooth muscles and secretory glands form the autonomic nervous system. Structural and physiological differences within the autonomic nervous system are the basis for its further subdivision into sympathetic and parasympathetic systems. Nerves of these two divisions leave the central nervous system at different regions. The sympathetic nerves emerge from the thoracic and lumbar regions of the spinal cord, and the parasympathetic nerves arise from the brainstem and sacral portions of the spinal cord. Thus the sympathetic system is known as the thoracolumbar outflow (originating in T_1 - T_12 and L_1 - L_12), while the parasympathetic system is referred to as the craniosacral outflow (originat-ing in cranial nerves III, VII, IX, and sacral nerves 2-4). The autonomic nervous system contains only motor nerves and is distinguished from the rest of the nervous system by several characteristics. There is no voluntary control by the cerebrum over these nerves. We cannot control our heart beat or the action of the muscles of the stomach or intestines. Another important charac-teristic of the autonomic nervous system is that most internal organs receive a double set of fibers, one set belonging to the sympathetic system and the other to the parasympathetic system. Impulses from the sympat-hetic and parasympathetic nerves always have antagonistic effects on the organs innervated. Thus if one functions to increase a certain activity, the other functions to decrease it. However, many tissues/cells do not receive parasympathetic innervation (i.e. fat cells, most arteri-oles, spleen). The autonomic nervous system is part of a larger unit called the peripheral nervous system. The peripheral nervous system contains two types of nerve fibers: the afferent fibers which convey information from receptors in the periphery to the CNS, and the efferent fibers which carry information from the CNS to the effectors. Effectors are tissues or organs which bring about appropriate responses to certain stimuli, both internal (such as from the brain) and external (such as from the environment). Some examples of effectors are the skeletal muscles, cardiac and smooth muscles, and secretory glands. The peripheral nervous system can be divided functionally into two parts. That part which innervates the skeletal muscles is known as the somatic nervous system, and that which innervates smooth muscle, cardiac muscle and glands is the autonomic nervous system.

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Question:

What temperature on the Centigrade scale correspondsto the common room temperature of 68\textdegreeF?

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Solution:

Because 68 \rule{1em}{1pt} 32 = 36, note that +68\textdegreeF is 36 Fahrenheit degrees above the freezing point of water. 36\textdegreeF above the freezing point corresponds to 5/9(36) = 20\textdegreeC above the freezing point since 1\textdegreeF = 5/9(1\textdegreeC). But since the freezing point is 0\textdegreeC, this temperature is +20\textdegreeC. Thus +68\textdegreeF = 20\textdegreeC. This result could have been obtained directly as follows. The formula for converting temperature in degrees Fahrenheit to degrees Centigrade is \textdegreeC = (5/9)(\textdegreeF - 32) If\textdegree F = 68\textdegree then \textdegreeC = (5/9)(68 - 32) = 20\textdegree

Question:

In sickle-cell anemia, what causes the erythrocytes to assume a sickle shape? Why is the incidence of sickle -cell anemia high in Africa, where the incidence of malaria is also high?

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Solution:

Sickle-cell anemia is caused by an abnormal type of hemoglobin, termed hemoglobin S. Hemoglobin S differs from normal adult hemoglobin in that it contains a different amino acid in one of its polypeptide chains (valineinstead of glutamate in one of the \beta chains). This alteration greatly reduces the solubility of de-oxygenated hemoglobin S. The amino acid substitution creates a site on the surface of the hemoglobin molecule, sometimes referred to as a "sticky patch." Onlyde-oxyhemoglobinS has a complementary site to this sticky patch. Therefore, thedeoxyhemoglobin molecules bind to each other, forming long aggregates which precipitate and distort the normal shape of the erythrocyte. Besides elongating the cell to the familiar sickle or crescent shape, the precipitated hemoglobin also damages the cell membrane, making the cell fragile. Thesickledcells either become trapped in small blood vessels (because of their distorted shape), causing impaired circulation and tissue damage, or they readily lyse because of their fragility. Sometimes, a vicious cycle is initiated when sickled cells cause vascular blockage. The blockage of the vessels creates a region of tissues where the oxygen con-centration is low. This deficiency causes moreoxyhemo-globinS to change into thedeoxyform, which in turn causes moresicklingand further decreases the oxygen concentration. This cycle progresses quickly and the resul-ting anemia is usually severe, often resulting in shock and death. Sickle-cell anemia is genetically transmitted. Sickle-cell anemic are homozygous for an abnormal gene which codes for synthesis of the abnormal hemoglobin S. Heterozygous individuals, who have one normal and one substitute allele, carry the sickle-cell trait. They do not usually show the symptoms of the disease except under conditions of low oxygen such as occurs in high altitude. The incidence of malaria and the frequency of the sickle-cell gene in Africa are correlated. It has been discovered that individuals carrying the sickle-cell trait (heterozygotes) are protected against a lethal form of malaria. This protection acts as a strong selective pressure to maintain the high incidence of the sickle-cell gene in regions where the incidence of malaria is high. This is an example of balanced polymorphism: the hetero zygote is protected against malaria and does not show the symptoms of sickle-cell disease , whereas the sickle-cell homozygote is eliminated by anemia and the normal homozygote is eliminated by malaria. Even though malaria is virtually eliminated in the U.S., there is still a significant incidence of sickle-cell anemia among American Negroes.

Question:

It has been suggested that the time required for light to travel 1 cm be used as a unit of time called the "jiffy". How many seconds correspond to one jiffy? The speed of light is about 3.0 × 10^8 m/sec.

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Solution:

The relationship between distance, speed, and time is distance = speed × time, or, time = [(distance)/(Speed)] The time it takes light to travel 1 cm is: Time = Distance/speed = [(1 cm)/(3 × 10^8 m/sec)] = [(1 cm)/(3 × 10^8 m/sec × 10^2 cm/sec)] = [(1 cm)/(3 × 10^10 cm/sec)] = 3.3 × 10^-11 sec. Hence, 1 jiffy = 3.3 × 10^-11 sec.

Question:

Draw a diagram of the human eye, labeling all parts. Briefly describe the function of each part.

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Solution:

The human eye consists of three layers or tu-nics. The fibrous tunic is the outermost layer of the eye: it consists of the sclera posteriorly and the cornea anteri-orly. The sclera is a protective coat and is the "white of the eye." The cornea is the transparent covering which func-tions in the refraction (bending) of light. The middle coat is the vascular layer. Posteriorly, it consists of the choroid coat. This pigmented layer absorbs excess light and nourishes the retina through its rich vas-culature. The choroid continues anteriorly as the ciliary body. The ciliary body consists of the ciliary muscle, a smooth muscle which functions in the accommodation reflex, and the ciliary process which secretes aqueous humor (fluid) into the anterior part of the eye. Finally, the ciliary body continues anteriorly as the iris. The iris is the colored part of the eye and also contains smooth muscle which func-tions in controlling the size of the pupil - the small open-ing through which light passes. The innermost layer is the nervous tunic. It consists solely of the retina in the back of the eye. The retina con-sists of photoreceptor cells called rods and cones. In the center of the retina is a small depressed area called the fovea centralis, the region of keenest vision. Medial to this is the optic disc. The optic nerve exits the back of the eye at this point. Because there are no photoreceptors at the disc, there is a blind spot in the peripheral field of vision. There is no anterior continuation of the retina. There are two fluid filled cavities inside the eye. The anterior cavity contains an aqueous humor (a watery fluid), and is located between the cornea and the lens. The poste-rior cavity is filled with vitreous humor (a gel-like fluid), and is located between the lens and the retina. The transparent lens which separates these two cavities is re-sponsible for focusing incoming light rays on the retina.

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Question:

Discuss the motion of a freely falling body of mass m taking into account the variation of the gravitational force on the body with its distance from the earth's center. Neglect air resistance.

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/Users/wenhuchen/Documents/Crawler/Physics/D04-0137.htm

Solution:

The gravitational force on the body at a distance r from the earth's center is -Gmm_E/r^2. From Newton's second law its acceleration is a = F/m = - (Gm_E/r^2),(1) where the positive direction is upward Buta =dv/dt= (dv/dr)(dr/dt) a = v(dv/dr) Then, from (1) v(dv/dr) = -Gm_E/r^2, ^(v)2\int_(v)1 (vdv) = -Gm_E^(r)2\int_(r)1 (dr/r^2) where v_1 and v_2 are the velocities at the radial distances r_1 and r_2. It follows that 1/2(v2_2 - v2_1) = -Gm_E[-(1/r)](r)2_(r)1 1/2(v2_2 - v2_1) =Gm_E(1/r_2 - 1/r_1-) v2_2 - v2_1 = 2Gm_E(1/r_2 - 1/r_1)

Question:

What is the expression that corresponds to the case of positron decay?

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/Users/wenhuchen/Documents/Crawler/Physics/D34-1022.htm

Solution:

For a particular mass number A, only one element with this mass number is stable. If an atom of a different element has this mass number, it will decay into the stable atom. A nucleus that has less protons than the stable nucleus has less charge than the stable one. It therefore undergoes \beta^\Elzbar^ decay. This results in the formation of a proton which increases the charge of the atom and the ratio of Z (proton) to A. n \rightarrow p + e\Elzbar+v_e If the nucleus has an excess of protons, it converts them to neutrons in a process known as \beta^+ or positron decay. p \rightarrow n + e^+ +v_e In the above processes the electron (or positron) and the neutrinov_e created are immediately ejected from the nucleus. The positron decay of a nucleus (Z, A) forms the nucleus (Z\Elzbar1, A). Originally, we have a nucleus plus Z atomic electrons with a total (that is, atomic) mass m (Z,A). In the decay process, a positron is emitted and a nucleus with atomic number Z\Elzbar 1 is formed. Since the nuclear charge has decreased by one unit, one of the original Z atomic electrons is superfluous and is shed. Therefore, the final system consists of a nucleus (Z\Elzbar1, A) plus Z\Elzbar 1 atomic electrons, together with the emitted positron and the excess electron. The total mass of the final system is m(Z\Elzbar1,A) + m(\beta^+) + m(e\Elzbar), or m(Z\Elzbar1,A) + 2m since a neutrino has no mass. Hence, the total available energy for positron decay is found from the amount of mass converted to energy. \epsilon\beta^+={ m(Z, A)\Elzbar [ m(Z\Elzbar 1,A) + 2m_e ]} × c^2 =[m(Z,A)\Elzbar m(Z\Elzbar 1,A)] × c^2\Elzbar 2m_ec^2 =∆m × c^2\Elzbar 2m_e c^2 and the maximum positron kinetic energy is 2m_e c^2 = 1.02MeVless than the mass-energy difference between the parent and daughter atoms.

Question:

How does metamorphosis differ from molting? What animals other than arthropods are known to undergo metamorphosis in their development?

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Solution:

In order to grow, arthropods must periodical-ly shed their rigid exoskeletal shell and in its place grow a larger one. This process is termed molting. Thecrustaceamolt many times during development . Each stage resembles the stage before it, and with each successive molt, the animal becomes larger and its body proportions become closer to the adult body proportions until its adult size has been attained. Many insects pass through successive developmental stages that are quite unlike one another. For example, a fly passes from a worm-like larva to a sedentary pupa to an adult form in its maturation sequence. This striking change from the juvenile to the adult form is termed metamorphosis. Certain amphibians undergo metamorphosis. For instance, the larval stage of the frog is the tadpole. Aquatic animals adapted, for sessile lives also often undergo metamorphosis. The tiny ciliated larvae either swim or are carried by sea currents to new locations, where they undergo metamorphosis. An example of this is the sea star, which changes from a bilaterally symmetrical larva to aradially symmetrical adult.

Question:

Find the number of ways that two identical particles (n_i = 2) can be distributed in five states g_i = 5) according to (a) Bose- Einstein and (b) Fermi-Dirac statistics.

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Solution:

(a) Bose statistics determine the number of distinct ways of placing n_i identical particles in g_i states, where the number of particles in a state is not restricted. Consider the system under consideration to be a linear array of (n_i + g_i - 1) holes into which either particles or partitions that separate the states can be inserted. Note that there are (g_i - 1) such partitions to separate the entire cell into g_i states. The remaining n_i holes which are separated into groups by these partitions, represent the distribution of the particles among the various states (see Fig. a). The number of distinct permutations of n_i particles and (g_i - 1) partitions over (n_i + g_i - 1) holes is the same as the number of distinct arrangements of n_i particles among the g_i states. Now, the number of permutations of (n_i + g_i - 1) distinguishable objects is (n_i + g_i - 1) !. But,n_i particles and (g_i - 1) partitions actually are indistinguishable and we must divide the above number by the number of permutations of n_i particles and by the number of permutations of (g_i - 1) partitions. The result gives the number of distinct arrangements. N_BE = [{(n_i + g_i - 1) !} / {n! (g_i - 1) !}] Substituting the numerical values, N_BE = [6!/(2! 4!)] = 15 Two possible distributions are shown in Fig. a. (b) In Fermi statistics a quantum state can contain only one particle. As before, we first find the number of ways of distributing n_i in distinguishable particles over g_i states. The first particle can be put in any of the g_i states, and for each choice of a state, the second particle can be put in any of the (g_i - 1) remaining states, and so on. The last particle will have (g_i - n_i + 1) possible states to choose from. The total number of configurations is g_i (g_i - 1) .... (g_i - n_i + 1) = [(g_i !)/{(g_i - n_i) !}] However, the actual indistinguishability of the particles requires that we divide out the number of identical configurations from this number. n_i particles give rise to n_i! permutations of the part-icles among themselves. Thus, we have N_FD = [(g_i !)/{n_i (g_i - n_i) !} = [5!/(2! 3!)] = 10 Two possible distributions are shown in Fig. b.

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Question:

Calculate the minimum concentration of Br- ion necessary to bring about precipitation of AgBr from a solution in which the concentra-tion of Ag+ ion is 1 × 10-^5 mole per liter. K_sp for AgBr = 4 × 10-^13 .

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/Users/wenhuchen/Documents/Crawler/Chemistry/E11-0400.htm

Solution:

The product of the concentrations of ions in a saturated solution of a relatively insoluble salt, such as AgBr, at a given temperature is constant. This constant is called the solubility prod-uct constant (K_sp ). For AgBr, K_sp can be stated as K_sp = [Ag+] [Br- ] = 4 × 10-^13 . In this problem, we are given the concentration of Ag+ as 1 × 10-^5 M, therefore [Br-] can be found by using the equation for K_sp . [Ag+] [Br-] = 4 × 10-^13 [1 × 10-^5 ] [Br-] = 4 × 10-^13 [Br-] = [4 × 10-^13] / [1 × 10-^5 ] = 4 × 10-^8 M . If the concentration of Br- ions is raised to 4 × 10-^8 M, AgBr will precipitate out, for the addition of any more Br- ions could not be supported by the solubility of the ions in solution, as the K_sp indicates.

Question:

What is the output for each of the following?

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G17-0423.htm

Solution:

The output is as follows:

Question:

An electrical transmission line is strung across a valley between two utility towers. The distance between the towers is 200 m, and the mass of the line is 40 kg. A lineman on one tower strikes the line and a wave travels to the other tower, is reflected, and is detected by the lineman when it returns. If the elapsed time measured by the lineman is 10 s, what is the tension force in the wire?

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Solution:

The distance d traveled by the wave pulse is d = 2 × 200 m = 400m. The speed of the wave is computed using v = (d/t) = (400 m/10s) =40 m/s For the stretched line, the tension F is given by F = \muv^2 where \mu is the inertia characteristic of the wire and is equal to the mass per unit length m/L. Therefore, F = (m/L) v^2 = (mv^2/L) = [{(40 kg) (40 m/s)^2}/(200 m)] = 3.2 × 10^2 N

Question:

Describe the early development and implantation of the human embryo.

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Solution:

The ovum must be fertilized while in the upper regions of the Fallopian tube or oviduct to develop suc-cessfully. Upon fertilization, it begins the series of cell divisions and differentiation which will eventually lead to development of the embryo (see Figure 1). The fertilized ovum continues its passage through the oviduct to the uterus, which requires three to four days. Once in the uterus, the egg floats free in the intrauterine fluid, from which it receives nutrients for several more days. During this time, it continues dividing and begins to differentiate into the blastocyst. The blastocyst stage (a fluid-filled sphere of cells) is attained by the time or shortly after the egg reaches the uterus. The egg would have slowly disintegrated during its passage to the uterus if fertilization had not occurred. While the egg had been traveling to the uterus, the uterus was being prepared under the combined stimula-tion of estrogen and progesterone to receive the egg. If the fertilized egg were to pass too quickly through the oviduct, it would reach the uterus prematurely, when both it and the uterus had not yet reached a suitable state for implantation. For the egg, the necessary event for implantation is the disintegration of the zona pellucida, a remnant of the Figure 1Schematic diagram of the maturation of an egg in a follicle in the ovary following its release (ovulation). a) fertilization in the upper part of the oviduct b) early cleavage as it desends c) late cleavage continuing desent d) blastocyst development in the uterus before implantation e) implantation of blastocyst in the wall of the uterus. follicle. This disintegration is completed at the blasto-cyst stage. The blastocyst is composed of an outer layer of specialized cells called trophoblasts, which are re-sponsible for implantation, and an inner cell mass destined to become the fetus itself. The trophoblast layer, upon disintegration of the zona pellucida, enlarges and adheres to the uterine wall. These trophoblasts secrete lytic en-zymes which will digest the uterine endometrium, allowing the blastocyst to completely embed itself in the uterine lining. This digestion of tissue continues for a period, with the breakdown products of the endometrium serving as a nutrient source for the developing embryo until both the fetal circulation and the placenta have developed. The placenta will form from an intermeshing of the trophoblast layer with the endometrium, and provides for contact and nutrient exchange between the mother and the fetus. In addition to enzymes, the trophoblast secrets hor-mones, including chorionic gonadotropin. This important hormone is necessary for retention of the corpus luteum, which would have involuted if its ovum had not been fert-ilized. The hormonal secretions of the corpus luteum and later the placenta, maintain the uterus in the proper condition for pregnancy and prevent menstruation. Detection of the presence of chorionic gonadotropin in the urine is the basis for most pregnancy tests. Because of the time required for the ovum to traverse the oviduct and for the uterus to be prepared, the total time that elapses from ovulation to implantation is from seven to eleven days.

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Question:

Two wires are attached to a corner fence post with the wires making an angle of 90\textdegree with each other. If each wire pulls on the post with a force of 50 pounds, what is the resultant force acting on the post? See Figure.

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Solution:

As shown in the figure, we complete the par-allelogram. If we measure R and scale it, we find it is equal to about 71 pounds. The angle of the resultant is 45\textdegree from either of the component vectors. If we use the fact that the component vectors are at right angles to each other, we can write R^2 = 50^2 + 50^2 whence R = 71 pounds approximately at 45\textdegree to each wire.

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Question:

At a 4000-m altitude the atmospheric pressure is about 0.605 atm. What boiling point would you expect for water under these conditions?

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Solution:

The pressure and temperature of a liquid that vaporizes are related in the Clausius-Clapeyron equation. This equation is In [(p_1) / (p_2)] = (∆H / R) [(1 / T_2) - (1 / T_1)] where P_1 is the initial pressure, P_2 is the final pressure, ∆H is the heat of vaporization, R is the gas constant, (8.314 J / mole \textdegreeK), T_2 is the boiling point at P_2 and T_1 is the boiling point at P_1. The heat of vaporization of H_2O is 40,600 (J / mole). One can assume for purposes of calculation that P_1 is 1 atm. The boiling point of H_2O at 1 atm is 373\textdegreeK. Solving for the boiling point at 0.605 atm: P_1 = 1 atmT_1 = 373\textdegreeK P_2 = 0.605 atmT_2 = ? ∆H = 40,600 (J / mole) R = 8.314 (J / mole \textdegreeK) ln (1 /0.605) = (40,600 J / mole) / (8.314 J /mole \textdegreeK) [(1 / T_2) - (1 / 373\textdegreeK)] .5025 = 4.88 × 10^3\textdegreeK [(1 / T_2) - 2.681 × 10^-3 / \textdegreeK] (1.0297 × 10^-4) / \textdegreeK = (1 / T_2) - (2.681 × 10^-3) / \textdegreeK (2.784 x 10^-3) / \textdegreeK = (1 / T_2) 359.20\textdegreeK = T_2 The boiling point in \textdegreeC is 359.20\textdegreeK-273 or 86.2\textdegreeC.

Question:

Differentiate between an isometric and an isotonic contraction.

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Solution:

Contraction refers to the active process of generating a force in a muscle. The force exerted by a contracting muscle on an object is known asthe muscle tension, and the force exerted on a muscle by the weight of anobject is known as the load. When a muscle shortens and lifts a load, themuscle contraction is said to be isotonic, since the load remains constantthroughout the period of shortening. When a load is greater than the muscle tension, shortening is prevented, and muscle length remains constant. Likewise, when a load is supportedin a fixed position by the tension of the muscle, the muscle lengthremains constant. This development of muscle tension at a constantmuscle length is said to be an isometric contraction. The in-ternal physiochemicalevents are the same in both isotonic and isometric contraction. Movement of the limbs involves isotonic contractions, whereasmaintaining one's posture requires isometric contractions.

Question:

A container with a pressure of 3 atmospheres and a temperature of 200\textdegree C. contains 36 gm of nitrogen. What is the average speed of the nitrogen molecules? What is the mean distance between molecules (assuming that the container has a cubic shape), and approximately what distance will a molecule travel before it collides with another?

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Solution:

The number of moles of gas present is the mass of gas present divided by the molecular mass of nitrogen, or n = (36/28) = 1.29 The pressure is p = 3 atm = 3 × 1.013 × 10^5 N/m^2 = 3.04 × 10^5 N/m^2 Thus the volume of the container is by the ideal gas law V = nRT/p = (1.29 moles × 8.313 joules/\textdegreek × 473\textdegree k)/(3.04 × 10^5 N/m^2) = (1.29 moles × 8.313 N-m/\textdegreek × 473\textdegree k)/(3.04 ×10^5 N/m^2) = 16.7 × 10^-3 m^3 That is, the volume is a cube with sides 25.6 cm. The total number of molecules is the number of moles of gas times Avogadro's number, or (1.29 moles)[6.02 × 10^23 (molecules/moles)] = 1.29 × 6.02 × 10^23 molecules so that the volume per molecule is (16.7 × 10^-3 m^3)/(1.29 × 6.02 × 10^23) = 2.15 ×10^-26 m^3 = 2.15 × 10^-20 cm^3 This corresponds to a cube with sides 2.8 × 10^-7 cm, which is the mean distance between molecules. The mean square speed is calculated from (1/2)mv^2= (3/2)kT where m is the mass of one molecule of nitrogen,v^2is the mean square speed of the molecule, k is Boltzmann's constant, and T is the absolute temperature. Furthermore, since k = R/N_0 where R is the gas constant and N_o is Avogadro's number, and mN_0 = M where M is the mass of one mole of gas, we have: v^2 = [(3kT)/m] = (3RT)/(N_0m) = (3RT)/M v^2 = (3 × 8.313 joule/\textdegreek mole × 473\textdegree k)/(2.8 × 10^-3 kg/mole) = (3 × 8.313 kg-m^2/s^2 \textdegreek mole × 473\textdegree k)/(28 × 10^-3 kg/mole) = 4.21 × 10^5 m^2/sec^2 Thus the average speed c of a molecule is c = \surd(v^2) = 6.5 × 10^2 m/sec. When two molecules of radius r collide, their centers are a distance 2r apart. Therefore, a molecule travelling along a straight line will collide with other molecules within a distance 2r of its center (we assume a simplified model in which one molecule moves and the others remain stationary and in which all collisions are elastic). During a time ∆t, the molecule which we are following travels a distance c∆t. Thus, the molecules with which our molecule will collide are all found within a cylinder of base radius 2r and height c∆t. This cylinder is called the collision cylinder. If N is the number of molecules in volume V, then the density of molecules is N/V. If we multiply the volume of the cylinder by the density of molecules we will have the number of molecules in the cylinder (4\pir^2c∆tN/V). Dividing this quantity by ∆t we have the number of collisions per unit time (4\pir^2cN/V) . The average time between collisions is the reciprocal of this, called the mean free time (V/4\pir^2cN). If we multiply this by the average velocity, we have the average distance a molecule travels before colliding with another (the mean free path): L = (cV)/(4\pir^2 cN) = V/(4\pir^2N) Substituting our values, we have (in the present case let r = 1.3 × 10^-10 m): L = [16.7 \pi 10^-3 m^3]/[4\pi(1.3 × 10^-10 m)^2(1.29 moles)(6.023 ×10^23 moles^-1) = 1.01 x 10^-7 m

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Question:

Are the daughter nuclei formed during cleavage exactly equivalent , or is there some sort of parceling out of potentialities during cleavage?

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Solution:

There is evidence that there is no segregation of genetic potentialities during cleavage. Experiments have shown that isolated blastomeres (the cleavage cells) are capable of developing into an entire organism . In a classical experiment, the nucleus is removed from a ripe frog's egg with a micropipette. A cell from an advanced stage of embryonic development is then separated from its neighbors. The plasma membrane of the cell is ruptured in the process and the nucleus is injected into the enucleated frog egg. The egg is found to develop normally. When nuclei obtained from late blastula or early gastrula stages are transplanted into enucleated eggs, the eggs can develop normally. Even nuclei from later stages of development, such as from the neural plate (the precursor of the spinal cord and brain) or from cells lining the gut may lead to the development of normal embryos. In humans, there is no segregation of genetic potentialities during early cleavage. Identical twins are thus possible. More experimentation has to be done to determine just how far into development the "no segregation " rule will apply.

Question:

Two parallel plates of area A = 1 m^2 are given equal and opposite charges of Q = 30 microcoulombs each. A sheet of dielectric of permittivity \epsilon = 15 × 10^\rule{1em}{1pt}12 coul^2 / newton-m^2 occupies the space between the plates (see Fig.)A). Compute (a) the resultant electric in-tensity in the dielectric, (b) the induced charge per unit area on the faces of the dielectric.

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/Users/wenhuchen/Documents/Crawler/Physics/D18-0615.htm

Solution:

(a) The charge per unit area on the plates is \sigma= q/A = 30 x 10 ^-6 coul/m2. The resulting electric field between the plates can be obtained by applying Gauss' Law to surface S in Fig. b. \int_s E^\ding{217} \bullet da^\ding{217} = Q / \epsilon In Fig. B, we see that on surface S_1 E^\ding{217} \textbullet da^\ding{217} = \rule{1em}{1pt}E da and on the rest of the surface E^\ding{217} \textbullet da^\ding{217} = 0 since the field through the sides S_2 of the surface is zero. The electric field is constant between the plates. Therefore \int_s E^\ding{217} \bullet da^\ding{217} = \rule{1em}{1pt}E \int_(s)1 da = \rule{1em}{1pt}EA, givingE = \rule{1em}{1pt}Q / \epsilonA = \rule{1em}{1pt}(\sigma / \epsilon ) │E│ = (30 × 10^\rule{1em}{1pt}6 coul / m^2) / (15 × 10^\rule{1em}{1pt}12 coul^2 / n-m^2) = 2 × 10^6 volts / m. (b) The dielectric coefficient is K = \epsilon / \epsilon_0 = (15 × 10^\rule{1em}{1pt}12) / (8.85 × 10^\rule{1em}{1pt}12) = 1.7. The induced charge on the dielectric of a capacitor is given by Q_ind = Q [(1) \rule{1em}{1pt} (1 / K)]. Then the induced charge density in the dielectric is \sigmaind= Q_ind / A = Q / A [(1) \rule{1em}{1pt} (1 / K)] = \sigma[(1) \rule{1em}{1pt} (1 / K)] or\sigma_ind = 30 × 10^\rule{1em}{1pt}6 (1 \rule{1em}{1pt} 1 / 1.7) coul / m^2 = 12.3 × 10^\rule{1em}{1pt}6 coul / m^2.

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Question:

Using Positive Logic, show that the Diode circuit of fig.1 represents a three-input AND gate.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G04-0064.htm

Solution:

A diode is similar to a valve which lets current flow in only one direction. In fig. 1 it is seen that when all inputs are low (OV) all diodes conduct. This action brings V_out to the same voltage as Vin N, hence V_out is low. When one or two inputs are high (\geq 5V) , any diodes corresponding to the low inputs conduct and any diodes corresponding to the high inputs act as insulators. Hence the diodes that conduct bring V_out to the low voltage. The diodes that are insulators have no effect on the out-put voltage. When all inputs are high, all of the diodes act as insulators and only then will the output voltage be high. From this information, it is seen that the output is high only when all of the inputs are high. Using positive logic, this is seen to be an AND gate.

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Question:

2.3 g of ethanol (C_2H_5OH, molecular weight = 46 g/mole) is added to 500 g of water. Determine the molality of the resulting solution.

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Solution:

This problem is a calculation of the molality of an aqueous solution. Molality is equal to the number of moles of solute per kilogram of solvent. In this case the solvent is water and the solute is ethanol. Since 1 kg of water corresponds to twice the amount of water given in the problem it is desirable to calculate the amount of ethanol that would be added to 1 kg of water and still maintain the same concentration of 2.3 g ethanol per 500 g of water. For 2 × 500 g = 1000 g = 1 kg of water, we require 2 × 2.3 g = 4.6 g of ethanol. We now know that our solution of 2.3 g ethanol in 500 g water corresponds to 4.6 g ethanol per 1 kg water, and all that remains to be done is to calculate the number of moles of ethanol in 4.6 g. We do this by dividing by the molecular weight to obtain 4.6 g/46 g/mole = 0.1 mole ethanol. The concentration of the solution is then 0.1 mole ethanol/1 kg water = 0.1 molal.

Question:

A chemist has a certain amount of gas under a pressure of 33.3atm; it occupies 30 l at 273\textdegreeC, For his research, however, the gas must be at standard conditions. Under standard conditions what will the volume of the gas be?

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Solution:

Standard conditions are defined to be 0\textdegreeC and 1 atm. Hence, the gas is cooled and the pressure on it is decreased. The combined gas law relates pressure P, volume V and absolute temperature T. k = (PV/T) where k is a constant that is characteristic to the system. Hence, (P_I V_I )/T_I= (P_F V_F )/T_F where the subscript I indicates the initial values and the subscript F indicates the final states. In this problem one is asked to solve for V_F. The temperatures in \textdegreeC are converted to \textdegreeK by adding 273. T_I= 273\textdegreeC + 273 = 546\textdegreeK T_F =0\textdegreeC + 273 = 273\textdegreeK F Solving for V_F: [{(33.3atm) (30 l)}/(546\textdegreeK)] = [{(1atm) (V_F )}/(2730K)] V_F = 499.5 l.

Question:

Show that Wien' s law and the Rayleigh-Jeans law are special cases of the Planck radiation formula.

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Solution:

The monochromatic energy density within an isothermal blackbody enclosure is given by Planck's relation \psi = [(8\pich\lambda^-5)/{exp (ch/\lambdakT) - 1}](1) where \lambda is the wavelength of the radiation, h is the Planck's constant, and T is the absolute temperature of the blackbody. The Rayleigh-Jeans law was found to con-form to experimental data for low frequencies (or large wavelengths). Making the approximation in (1) that \lambda is very large, then ch/\lambdakTis small and exp (ch/\lambdakT) can be replaced by the first two terms in the expansion of that function in terms of powers of its exponent. Subsequent terms in the expansion will be so small as to be negligible. Hence exp [ch/\lambdakT] - 1 \approx 1 + (ch/\lambdakT) - 1 = (ch/\lambdakT). \Psi \approx 8\pich\lambda^-5 × (\lambdakT/ch) = 8\pikT\lambda^-4 , which is the Rayleigh-Jeans law. Wien's Law diverges from the experimental data in the realm of large wavelengths. Hence, in the limit of small wavelengths, (1) should approach Wien's Law. For small \lambda,ch/\lambdakTis large, and exp (ch/\lambdakT) > > 1. Hence, exp (ch/\lambdakT) - 1 \approx exp (ch/\lambdakT). Therefore, (1) becomes \Psi = [(8\pich\lambda^-5)/{exp (ch/\lambdakT)] which is of the form of Wien's Law \psi = (c_1/\lambda_5) [1/(_eC_2/\lambdaT)] with c_1 = 8\pich,c_2 =ch/k.

Question:

You are given groups of numbers. Using Basic Assember Language. You are to compute the following: Also: \Sigma(x_i)^2 = sum of the squares of each number and (\Sigmax_i)^2 = the square of the sum of the numbers. The input deck of cards containing each group will be di-vided from each other by means of an E card. The E is to be placed in the first column of each partition card. The number cards have the integers right justified in columns 1-3 of the card. The output must be labelled as follows: #(GROUP NUMBER)NSUMAVGVARMAXMIN This is to be printed on a new page, with two spaces between each output line.

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Solution:

Col 1-8 Col 10-14 16-71 1 VARIANCE START 0 2 SAVE (14,12) SAVE REGISTERS 3 BALR 3,0 STORE BASE ADDRESS 4 USING \textasteriskcentered,3 USING REGISTER 3 AS BASE REGISTER 5 ST 13,SAVE+4 6 LA 13,SAVE STORE ADDRESS OF SAVE AREA 7 PRINTLIN HEADING,132 PRINTING THE LABELS 8 AP WRKGPNBR, = P '1' FIRST GROUP NUMBER 9 RDCD READCARD CARDINPA, CRDEOF READ IN A NUMBER, X; 10 CLI CRDATA, C 'E' CHECK TO SEE IF NUM-BER IS A DIVISOR 11 BE CALQUE IF 'E', THEN END OF A GROUP AP WRKGPCNT, = P '1' COUNT ELEMENTS OF GROUP 12 PACK WRKDATA, CRDATA CONVERT NUMBER TO PACKED DECIMAL 13 AP WRKSUM, WRKDATA FINDING SUM OF GROUP 14 ZAP WKDATASQ, WRKDATA PLACE NUMBER IN AN-OTHER AREA 15 MP WKDATASQ, WRKDATA SQUARE NUMBER 16 AP WRKSUMSQ, WKDATASQ FINDING SUM OF THE SQARES 17 ZAP WKDATASQ, ZERO ZERO AREA USED FOR SQUARING 18 CP WRKDATA, WRKMAX FINDING MAX 19 BH GPHIGH STORE MAX 20 CP WRKDATA, WRKMIN FINDING MIN 21 BL GPLOW STORE MIN 22 B RDCD GET NEXT NUMBER IN THE GROUP 23 GPLOW ZAP WRKMIN, WRKDATA STORE NEW MIN 1 - 8 10 - 14 16 - 71 23A B RDCD 24 CALQUE MVC PRTGPNBR, PATTI PREPARE GROUP 25 ED PRTGPNBR, WRKGPNBR NUMBER FOR THE PRINT LINE 26 MVC PRTGPCNT, PATTI PREPARE N FOR 27 ED PRTGPCNT , WRKGPCNT PRINT LINE 28 MVC PRTSUM, PATT2 PREPARE SUM FOR 29 ED PRTSUM, WRKSUM PRINT LINE 30 ZAP WRKSUMST, WRKSUM GET COPY SUM 31 MP WRKSUMST, WRKSUM FINDING THE SQUARE OF THE SUM 32 DP WRKSUMST, WRKGPCNT DIVIDE THE SQUARE OF THE SUM BY N 33 SP WRKSUMSQ, WRKSUMST(5) CALCULATE THE NUMERATOR OF VARIANCE 34 SP WRKGPCNT, = P'1' FIND DENOMINATOR OF VARIANCE 35 DP WRKSUM, WRKGPCNT FINDING VARIANCE 36 AP WRKGPCNT, = P'1' RESTORE N TO FIND AVERAGE 37 DP WRKSUM, WRKGPCNT FIND AVERAGE 38 MVC PRTMAX, PATT2 PREPARING THE 39 ZAP WRKSUM, WRKSUM (2) AVERAGE FOR THE 40 ED PRTAVER, WRKSUM PRINT LINE 41 MVC PRTVAR, PATT2 PREPARING VARIANCE 42 ZAP TEMPHOLD , WRKSUMS Q +V (4 ) FOR THE 43 ED PFTVAR, TEMPHOLD PRINT LINE 44 MVC PRTMAX, PATT2 PREPARING TO PRINT 45 ED PRTMAX, WRKMAX GROUP'SMAX 46 MVC PRTMIN, PATT2 PREPARING TO PRINT 47 ED PRTMIN, WRKMIN GROUP'S MIN 48 PRIMTLIN PRTDETL,133 PRINT OUTPUT LINE 49 AP WRKGPNBR, = P '1' INCREMENT GROUP NUMBER 50 ZAP WRKGPCNT, ZERO 51 ZAP WRKGPCNT, ZERO 52 ZAP WRKSUMSQ, ZERO 53 ZAP WRKMAX, = P '0000' 54 ZAP WRKMIN, = P '9999' 55 B RDCD 56 GPHIGH ZAP WRKMAX, WRKDATA STORE NEW MAX 57 CP WRKDATA, WRKMIN 58 BL GPLOW 59 B RDCD 60 CRDEOF L 13, SAVE + 4 60A RETURN (14, 12) RESTORE REGISTERS \textasteriskcenteredDATA DEFINITION FOR THE CARD INPUT AREA 62 CRDINPA DS 0CL80 63 CRDDATA DS CL3 64 DS CL77 \textasteriskcenteredDATA DEFINITION FOR THE WORK AREAS 66 SAVE DS 18F USE TO SAVE THE CONTENTS REGISTERS 67 WRKDATA DS PL3 68 ZERO DC PL4 '0' 69 WRKSUM DC PL '3' 70 WRKDATASQ DS PL6 71 WRKSUMSQ DC PL6'0' 72 WRKSUMST DC PL6 73 WRKGPNBR DC P '0' 74 WRKGPCNT DC P '0' 75 WRKMAX DC PL3 '0' 76 WRKMIN DC PL3 '9' 77 TEMPHOLD DS PL4 78 PATT1 DC X '40202020' 79 PATT2 DC X '40202021 2020202020' \textasteriskcenteredDATA DEFINITION FOR THE HEADING 81 HEADING DS \varphiCL133 \varphi 1 - 8 10 - 14 16 - 71 72 82 PRTCC1 DC C '1' 82A DC 19C ' ' 83 DC C'# N SUM AVG VAR X 83A MAX MIN' 84 DC 57C ' ' \textasteriskcentered DATA DEFINITION FOR THE PRINTER WORK AREA 86 PRTDETL DS \varphiCL133 \varphi 87 PRTCC2 DC C'-' SKIP TWO SPACES 88 DC 19C ' ' 89 PRTGPNBR DS CL 2 90 DC 3C ' ' 91 PRTGPCNT DS CL2 92 DC 4C '' 93 PRTSUM DS CL 6 94 DC 4C '' 95 PRTAVER DS CL6 96 DC 4C '' 97 PRTVAR DS CL8 98 DC 6C'' 99 PRTMAX DS CL6 100 DC 6C'' 101 PRTMIN DS CL 6 102 DC 50C'' 103 END VARIANCE Comments: For the beginning assembler language programmer, the calculations in his program will generally be done in either of two forms of arithmetic, packed decimal or fixed point binary. Our VARIANCE program written above is an example of an assembler language program which uses only packed de-cimal arithmetic. It should be understood that both forms of arithmetic can be done simultaneously, and, in fact, this is what is done on most occasions. We must first point out that the numbers, as written in the leftmost edge of the program, are simply placed there for the purposes of reference, and are not in any way re-lated to the program itself. The calculations involved in our program have been done in an order corresponding to the listing of our question, and as such, we do not think it necessary to explain the logic of the program. In our explanation, we will discuss only those instructions we think would pose problems for the student. The first six instructions of our program constitute the housekeeping procedures required. The label in Columns 1-8 of Line 1 is the name of the program. The other in-structions we will consider are: 7. PRINTLIN HEADING, 132 - The label 'HEADING\textasteriskcentered is the name of a most cases, specified as the length of a print line (132 or 133 spaces) on an output page. This print area usually contains carriage controls (Column 1), output information (whether defined as constants or moved into the area), and blanks. It is good assembler programming practice to fill up unused portions of a print area with blanks since the computer has a tendency to print undesired characters in 'unblanked' spaces. The above discription given for the "heading' print area applies, in general, to all print areas defined by a data definition. Variations of these areas lie only in their names (chosen by the programmer) and in the fact that data to be printed may be explicitly defined in the area (DC statement), moved into it (MVC statement) or a combination of both. The coding procedures for the way data is to be read from a card are quite similar to those of printing. The only differences are that the length of the input area is 80 bytes (80 columns on a card), and instead of placing blanks in the unused fields, we just define them (DS state-ment - Line 64). 10. CLI CRDATA,C 'E' - This instruction determines if the card read in has an 'E' in column 1. If so, we are at the beginning of a new group, but we must first process (find the variance of) the old group before commencing this one. Lines 14 - 15 and 30 - 31 illustrate the technique used in multiplying (and squaring) two packed decimal numbers. The multiplicand is placed in a field twice as long as that of the multiplier. When the calculation is done, the result will replace the multiplicand. Thus, in making its area twice as long as that of the multiplier, we quard against overflow. In general, when calculations are done in assembler language, Operand - 1 is replaced by the result, so one should always check to see that this area is large enough to re-ceive the largest (most spacious) possible result. For ex-ample, in the DP instructions of Lines 32 and 35, Operand -1 was made large enough to store the resulting quotients and their remainders. The length of these fields can be checked by inspecting them in their various data definitions. In-cidentally, it is good programming practice to somehow in-dicate in which data definition a particular symbol maybe found. This is usually achieved by allowing the first (or last) two or three letters of the symbol to correspond to the name of the data definition. For example, 'CRDDATA' is to be found in the data definition for the card input area, and 'PRTSUM' is located in the definition for the print area. The only other instructions we think necessary to discuss are those involved in the EDITING (ED) of packed de-cimal results to be printed. It is imperative that we edit packed decimal results prior to printing, since our output will not be what we expect unless this is done. Our prob-lem here is one of translation. When the computer reads in our data, it does so in EBCDIC code. This process involves replacing each character of our data with two EBCDIC digits. For example, it reads 1 (data) as F1(ECDIC), E as C5, and so forth. Now, in printing it does the reverse; the informa-tion to be printed is taken as EBCDIC and translated back to 'readable' code. For example, the packed decimal number 406F, taken as EBCDIC, would be translated as Having pointed out the importance of editing, we will now illustrate the way in which it is done. A packed deci-mal number has the form: This is a representation of the number in memory. Two deci-mal digits are stored per byte, except for the rightmost byte, which has a sign factor in its last four bits. The integers are those of our actual answer. As you can see, the storage space alotted a pack decimal number is usually such that unused high order locations are packed with zeros. By first placing a lexadecimal pattern in a print area, a packed number will be translated to the correct EBCDIC code before printing is done. The first byte of this pattern usually contains a hexadecimal 40, which causes suppression of high order zeros. In general, the length of the pattern field is twice that of the packed decimal field. The afore-mentioned 40 of the pattern field may be said to replace the sign factor of the packed decimal field, but it will be in the first byte instead of the last. Each integer in the packed field will then be replaced by a DIGIT SELECTOR (HEX 20) in the pattern field. For example, let us consider the following instructions: MVCPRTSUM,PATTERN EDPRTSUM,WRKSUM Before MVC and ED instructions After instructions:1. MVC PRTSUM, PATTERN The main idea behind editing is to place a hexadeci-mal pattern in the print field, to ensure correctness of the packed decimal data to be printed. The pattern used can be longer than the print field since it will be truncated on the right. Although this means less coding and simple edit-ing, it should not be done unless the portion to be trun-cated is of no importance. One should also note that since there are two digits per byte for a packed decimal field the print field should be at least twice as long. As long as our intentions are to print only the inte-gers from the packed field, it is okay to make the pattern, and hence the printfield, twice as long. But if we intend to place decimal points, commas, and/or any special char-acters in our output, we must use a pattern and a print-field that are more than twice the length of the packed field. Lines 83 and 83A of our program illustrate how a literal is continued from one card to another in a data definition. The X in column 72 of the card is a continu-ation character. Our variance program is one coded to completion. It lacks only the JCL cards and the actual data. But after studying this program, the student should not find it dif-ficult to write a program which uses the packed decimal in-structions.

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Question:

Does a yeast cell metabolize more efficiently in the presence or in the absence of oxygen? Explain.

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Solution:

Under aerobic conditions (in the presence of oxygen), glucose undergoes glycolysis to form pyruvate, which after conversion to acetyl CoA, enters the TCA cycle. The NADH produced is oxidized in the electron transport system provided that oxygen is available as the ultimate electron acceptor. Let us calculate the maximum amount of ATP produced from the oxidation of one mole of glucose. A net gain of 2 ATP is obtained directly in glycolysis (see figure). In addition, 2 NADH are genera-ted. This occurs in the cytoplasm. To enter the electron transport system, the NADH must be transported into the mitochondrion, a process which requires 1 molecule of ATP for every NADH which crosses the mitochondrial membrane. So intead of three ATP being produced per NADH, each cytoplasmic NADH yields only 2 net ATP; that is, a total of 4 ATP are produced by the 2 NADH generated in glycolysis. Recall that two 3-carbon molecules are obtained in glycolysis. Each of these two 3-carbon compounds can eventually enter the TCA cycle as acetyl CoA, and later, their electrons enter the electron transport system. Thus ATP production from the TCA cycle is twice that of a single turn. Two ATP are produced in the cycle via the formation of 2 molecules of GTP from 2 molecules of succinyl CoA. The conversion of pyruvate to acetyl CoA yields one NADH while the Krebs cycle yields 3 NADH. Thus 8 (4 × 2) NADH are produced per glucose at this point. The 8 (4 per turn) NADH formed in the cycle yield 24 ATP (3 ATP per NADH) by means of the electron transport chain. The 2 FADH_2 (1 per turn) produce 2 ATP each via electron transport, giving a total of 4 ATP. Summation of all the ATP from glycolysis, the citric acid cycle, and electron transport leads to a net production of 36 ATP per mole of glucose fully oxi-dized. Under anaerobic conditions, oxygen is not available as a terminal electron acceptor, and the reactions of the electron transport system cease as the reduced intermediates build up (refer to questions 3-20 and 3-26). This leads to an accumulation of the TCA cycle intermediates and, since it cannot enter the cycle, an accumulation of pyruvate.Under these conditions in yeast cells, pyruvate is then decarboxylated to acetaldehyde, which is in turn reduced to ethyl alcohol (ethanol), regenerating oxidized NAD+ for further use in glycolysis. This process is called alcoholic fermentation conditions, this too regenerates NAD+ in order to continue glycolysis for energy production: Pyruvic acid gets reduced to lactic acid as NADH gets oxidized back to NAD+ . (see next question). Under anaerobic conditions then, only 2 ATP molecules are produced, as opposed to the 36 ATP produced when glycolysis is supplemented with aerobic respiration Thus, a yeast cell metabolizes 18 times more efficiently in the presence of oxygen than in the absence of oxygen. ATP yield from the complete oxidation of glucose Reaction sequence ATP yield per Glucose GLYCOLYSIS: GLUCOSE TO PYRUVATE (in the cytoplasm) Phosphorylation of glucose \Elzbar1 Phosphorylation of fructose 6-phosphate \Elzbar1 Dephosphorylation of 2 molecules of 1, 3-DPC +2 Dephosphorylation of 2 molecules of phosphoenolpyruvate +2 2 NADH are formed in the oxidation of 2 molecules of glyceraldehyde 3-phosphate CONVERSION OF PYRUVATE TO ACETYL CoA (inside mitochondria) 2 NADH are formed CITRIC ACID CYCLE (inside mitochondria) Formation of 2 molecules of guanosine triphosphate from 2 molecules of succinyl CoA +2 6 NADH are formed in three oxidation steps. 2 FADH_2 are formed in 1 oxidation step. OXIDATION PHOSPHORYLATION (inside mitochondria) 2 NADH formed in glycolysis; each yields 2 ATP (not 3 ATP each, because of the cost of the shuttle) +4 2 NADH formed in the oxidative decarboxylation of pyruvate; each yields 3 ATP +6 2 FADH formed in the citric acid cycle; each yields 2 ATP +4 6 NADH formed in the citric acid cycle; each yields 3 ATP +18 36 ATP/glucose

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Question:

A woman opens a can of ham and beans and suspects food poisoning . She boils the contents for half an hour. A few hours after eating the ham and beans, nausea, vomit-ing, and diarrhea set in. Explain.

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Solution:

The woman probably suspected that the can of ham and beans was contaminated withClostridiumbotulinum, the cause of the microbial food poisoning known as botulism. These bacteria sometimes, but not always, produce foul odor.Clostridiumbotulinumare usually found in canned, low-acid products since they are anaerobic, growing without the presence of oxygen. The poisoning is due to an extremely potent toxin produced by the bac-teria. The toxin acts upon nerves and causes the paralysis of the pharynx and diaphragm, thus causing respiratory failure and sometimes death (65% of all cases are fatal). The toxin is heat-labile (unstable) and is destroyed by boiling for 15 minutes. Therefore, the woman denatured thebotulinustoxin and her food poisoning cannot have been botulism. She also did not show the symptoms of botulism which affects the neuromuscular junction. There is a flaccid paralysis which may precede cardiac or respiratory failure. Other symptoms include diffi-culty in swallowing and speech. Double vision is often present as well. The woman was probably poisoned by the toxin produced by Staphylococcusaureus , the most common agent causing food intoxication in the U.S. today. The bacteria may have come from food handlers who harbored the organism in the nose, throat, or sores on the skin (boils). The organism is a facultative anaerobe, able to live in the anaerobic environment of the can. Staphylococcal food poisoning usually involves food such as ham and dairy products.The important point in this type of poisoning is that the toxin is heat-stable and would be unaffected by boiling for 30 minutes. Therefore, the woman's precautionary action was not successful. Fortunately, death is rarely caused by this toxemia.

Question:

A slit of width a is placed in front of a lens of focal length 50cm and is illuminated normally with light of wavelength 5.89 × 10^-5 cm. The first minima on either side of the central maximum of the diffraction pattern observed in the focal plane of the lens are separated by 0.20 cm. What is the value of a?

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Solution:

This is an example of Fraunhofer diffraction (see figure). Parallel rays of light are incident on slit NN' from the left. The rays are diffracted diffracted and encounter the lens of focal length f. The rays are then focused on a screen (SS') lying in the lens' focal plane, and a diffraction pattern is observed. The minima of this diffraction pattern are described by the formula sin \texttheta = m\lambda/a where \texttheta locates the minima fringes, a is the slit width, m is the minima number, and \lambda is the wavelength of the light used. In this problem, the first minima on either side of the central maximum is found at angle \texttheta, such that \texttheta = sin^-1 \lambda/a. The angular separation of the 2 minima is then (see figure) \delta = 20 = 2 sin^-1 \lambda/a(1) But, we may write 2 \texttheta \approx DB /f(2) where DB is the linear separation of the 2 minima. Here, we have approximated the arc DB (shown dotted in the figure) by the linear distanceDB. Using (2) in (1) DB/f \approx 2 sin^-1 \lambda/a orsin (DB/f) \approx \lambda/a anda \approx [\lambda/{sin(DB/2f)}] Hencea \approx [{5.89 × 10-5cm} / {sin(.20 cm/100 cm)}] = [{5.89 ×10^-5 cm} / {sin(.0020)}] Since.0020 rad is a very small angle, we may write sin (.0020) \approx .0020 = 2× 10^-3 whencea \approx (5.89 ×10-5cm)/(2 × 10^-3) = 2.495 × 10^-2 cm.

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Question:

The hydra has a unique nervous system. What is this type of system called? Explain how the system operates.

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Solution:

The nervous system of the hydra is primitive. The nerve cells, arranged in an irregular fashion, are located beneath the epidermis and are particularly con-centrated around the mouth. There is no aggregation or coordination of nerve cells to form a brain or spinal cord as in higher animals. Because of the netlike ar-rangement of the nerve cells, the system is called a nerve net. For some time it was thought that these nerve cells lacked synapses, but at the present time research indicates that synapses are indeed present. (Synapse is the junction between the axon of one nerve cell, or neuron, and the dendrite of the next.) It is known that some synapses are symmetrical, that is, both the axon and thedendriticterminals secrete a transmitter substance and an impulse can be initiated in either direction across the synapse; while some are asymmetrical, permitting transmission only in one direction. Impulse in the nerve net can move in either direction along the fibers. The firing of the nerve net results primarily from the summation of impulses from the sensory cells involved, which pick up the external stimuli and the degree to which the response is local or general depends on the strength of the stimulus. When a sensory cell, or receptor, is stimulated, an impulse is relayed to a nerve cell. This in turn relays the impulse to other nerve cells, called effectors, which sti-mulate muscle fibers and nematocyst discharge. The rate of transmission of nerve impulses in hydra is usually quite slow. In spite of the fact that the nerve net is very primitive in comparison to the verte-brate type of nervous system, it is apparently adequate for hydra.

Question:

lf 20 ml of 0.5 N salt solution is diluted to 1 liter, what is the new concentration?

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Solution:

When considering normality one must always keep in mind that it is a concentration defined as the number of equivalents per liter. Since the number of equivalents does not change during dilution, equivalents before dilution = equivalents after dilution, or in other words, N_1V_1 = N_2V_21 liter = 1000 ml where N_1 is the normality of the initial solution, V_1 the initial volume, N_2 the final normality and V_2 is final volume. Solving for the final normality: (0.5) (0.020l) = N_2 (1l) N_2 = [{(0.5) (0.020l)}/(1 l)] = 0.01 N

Question:

A particle of mass M is attached to a string (see the figure) and constrained to move in a horizontal plane (the plane of the dashed line). The particle rotates with velocity v_0 when the length of the string is r_0. How much work is done in shortening the string to r?

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/Users/wenhuchen/Documents/Crawler/Physics/D05-0278.htm

Solution:

The string is stretched under the action of the radial centripetal force which keeps the mass M on its circular path. When we pull in the string we shorten r_0 by increasing the radial force F_centrip^\ding{217} on M. As we know, a force can only produce a torque about the axis of rotation if it has a component perpendicular to the radius which locates the mass M. A purely radial force like F_centrip^\ding{217} has no such component, therefore the angular momentum must remain constant as the string is shortened. Mv_0r_0 = Mvr(1) The kinetic energy at r_0 is (1/2) Mv_0^2; at r it has been increased to (1/2) Mv^2 = (1/2) Mv_0^2 [r_0/r]^2 because v = v_0r_0/r from above. It follows that the work W done from outside in shortening the string from r_0 to r is W = (1/2) Mv^2 - (1/2) Mv_0^2 = (1/2)Mv_0^2 [(r_0/r)^2 - 1](2) This can also be calculated directly as the work done by F_centrip^\ding{217} along the distance r_0 - r; W = ^r\int_r0 F_centrip^\ding{217} \bullet dr^\ding{217} = - ^r\int_r0 F_centrip dr W = -^r\int_r0 dr(Mv^2/r) = - ^r\int_r0 dr(M/r) [(v_0^2r_0^2)/(r^2)] where we have used (1). Hence W = -Mv_0^2r_0^2^r\int_r0 [(dr)/(r^3)] W = (Mv_0^2r_0^2)/(2r^2) \vert^r_r0 = (Mv_0^2r_0^2)/(2) [(1/r^2) - (1/r_0^2)] W = (1/2) Mv_0^2 [(r_0/r)^2 - 1] which is (2). We see that the angular momentum acts on the radial motion as an effective repulsive force. We have to do extra work on the particle on bringing it from large distances to small distances if we require that the angular momentum be conserved in the process.

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Question:

Describe the function of the lateral-line system in fishes.

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Solution:

Just as the sensory hair cells in the semicircular canals of terrestrial vertebrates function in the detection of sound and acceleration, so does the lateral-line system in fishes. The lateral-line system consists of a series of grooves on the sides of a fish. There are sensory hair cells occurring at intervals along the grooves. These sensory cells are pressure sensitive, and enable the fish to detect localized as well as distant water disturbances. The lateral-line system bears evolutionary significance in that the sensory hair cells of terrestrial vertebrates is believed to have evolved from the sensory cells of the archaic lateral-line system in fishes. The lateral-line system of modern fishes functions primarily as an organ of equilibrium. Whether or not a fish can hear in the way that terrestrial vertebrates do is not known.

Question:

The compound calciumcyanamide, CaCN_2, is prepared in con-siderable amounts for use as a solid fertilizer. The solid mixed with the water in the soil slowly adds ammonia and CaCO_3 to the soil. CaCN_2 + 3H_2O CaCO_3 + 2NH3 What weight of ammonia is produced as 200 kg of CaCN2 reacts?

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Solution:

From thestoichiometryof this equation, 1 mole of CaCN_2 produces 2 moles of ammonia. The number of moles of CaCN_2 (MW = 80 g/mole) in 200 kg is [(200,000 g CaCN_2) / (80 g / mole)] = 2,500 moles CaCN2 Thus, (2,500 × 2 =) 5,000 moles of ammonia is produced and the weight of ammonia is its molecular weight (MW = 17 g / mole) times the number of moles produced: (17 g / mole) (5000moles) = 85,000 g = 85 kg of ammonia produced.

Question:

What is the pH of a 0.001 M aqueous solution of the strong electrolyte KOH? The dissociation constant for water is K_w= 10^-14 mole^2/liter^2.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E12-0415.htm

Solution:

We need to find [H_3O^+] in order to find thepH.[H_3O^+] can be determined by substituting into the expression K_W = [H_3O^+] [0H^-], or, [H_3O^+] = K_W / [OH^-] The hydroxyl concentration, [OH^-], is contributed by the dissociation of KOH. This equation for this reaction is KOH \rightarrow K^+ + OH^-. Since KOH is a strong electrolyte, we assume that it dissociates completely. Hence, the concentrations of OH^- and of K^+ are equal to the initial concentration of KOH, or [OH^-] = 0.001 M = 10^-3 M. Thehydronium ion concentration is then [H_3O^+] = K_W / [OH^-] = (10^-14 mole^2/liter^2) / (10^-3 M) = (10^-14 mole^2/liter^2)/( 10^-3 mole/liter) = 10^-11 mole/liter = 10^-11 M. The pH is defined as pH = - log [H_3O^+]. Hence pH = - log [H_3O^+] = - log (10^-11) = - (- 11) = 11.

Question:

Find the radius of the smallest Bohr orbit for the Hydrogen atom.

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Solution:

The Bohr model of the hydrogen atom consists of a single electron of charge -e revolving in a circular orbit about a single proton of charge +e. The electrostatic force of attraction between electron and proton provides the centripetal force that retains the electron in its orbit (see the figure). Bohr's model assumes that the angular momentum of each orbit is re-stricted to integral multiples of Planck's constant divided by 2\pi, L_n = n(h/2\pi). This can be seen by considering the wave nature of the electron. In order to have a stable electronic orbit, the matter wave of the electron must be stationary around the orbit, as shown in the figure. Therefore, the orbit is an integral multiple of the wavelength (de Broglie wavelength) \lambda = h/mv , 2\pir_n = n\lambda or r_n = n(h/2\pimv) where v is the orbital velocity of the electron. This equation can also be written as r_n = (L_n/mv). The electrostatic force on the electron as it moves in the nth circular orbit, can be written as F = (mv^2/r_n) = k(e^2/r^2_n), ormv^2 r_n = ke^2. This is the centripetal force needed to keep the electron in its cir-cular orbit. Bohr's postulate for the quantization of the orbital angular momentum is L_n = mvr_n = n (h/2\pi). Squaring both sides, we have m^2 v^2 r^2_n = n^2 (h^2/4\pi^2). Now, we form the ratio (m^2 v^2 r^2_n/mv^2 r_n) = n^2 [h^2/(4\pi^2 ke^2)] we getmr_n = n^2 [h^2/(4\pi^2 ke^2)] r_n = [h^2/(4\pi^2 ke^2m)] n^2 The smallest orbit will have a radius r_1 = [h^2/(4\pi^2 ke^2m)] When the numerical values are substituted, we find for the first Bohr Radius r_1 = [{(6.63 × 10^-34 J-sec)^2} / {4\pi^2 × (9 × 10^9m/farad) × (1.6 × 10^-19 coul)^2 × (9.11 × 10^-31kg)}] = 5.29 × 10^-11 m. In Bohr's theory, the electron may revolve only in some one of a number of specified orbits.

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Question:

How much work in joules is done when a mass of 150 kilograms is dragged a distance of 10 meters if the coefficient of sliding friction is 0.30?

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Solution:

Work is given by F \textbullet s when the force is constant and is applied in the direction of travel (F being force and s being distance). To calculate the force needed to move the object at constant velocity against the force of friction, we use F =\mu_kinetic\bullet N where N is the normal force which in this case is the weight of the object: N = mg N = 150 kg 9\bullet80 m/s^2 = 1470nt and the force of friction is: F = 0.30 \textbullet (1470 nt)= 441nt The work done = W = Fs = 441nt× 10 m = 4410 joules.

Question:

Heating of NaNO_3 decomposed it to NaNO_2 and O_2. How much NaNO_3 would you have to decompose to produce 1.50 g of O_2 ?

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Solution:

The equation for this reaction is: 2NaNO_3 \rightarrow 2NaNO_2 + O_2 This means that for every mole of O_2 produced, 2molesof NaNO_3 must be decomposed. One is given that 1.50 g of is formed, thus one should determine the number of moles. The number of moles of NaNO_3 needed will be twice this amount. One can solve for the weight by multiplying the number of moles by the molecular weight of the compound. The number of moles of O_2 can be found by dividing 1.50 g by the molecular weight of O_2 (MW = 32.0). no. of moles = (1.50 g) / (32.0 g/mole) = 4.69 × 10^-2 moles. The number of moles of NaNO_3 needed can be found by multiplying the number of moles of O_2 by 2. no. of moles = 2 × 4.69 × 10^-2 moles = 9.38 × 10^-2 moles. One finds the weight of NaNO_3 needed by multiplying the number of moles by the molecular weight (MW = 85). weight of NaNO_3 = 85 g/mole × 9.38 × 10^-2 moles = 7.97 g.

Question:

A car coasts down a long hill and then up a smaller one onto a level surface, where it has a speed of 32 ft/sec. If the car started 200 ft above the lowest point on the track, how far above this low-est point is the level surface? Ignore friction.

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Solution:

The initial velocity of the car is zero. Since there is no friction, the change in potential energy of the car equals its increase in kinetic energy. On the smaller hill of height h, the change in potential energy with respect to the starting point is PE = mg (200 - h) Its kinetic energy is given as kE = (1/2)mv^2 = (1/2)m(32)^2 Equating the two, mg(200 - h)= (1/2)m(32)^2 g(200 - h)= 32(200 - h) = (1/2)(32)^2 200 - h= (1/2)(32) h= 200 - 16 = 184 ft. Therefore in order for the car to have speed 32 ft/sec, the lower hill must be at a height of 184 ft above the lowest point on the track.

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Question:

Differentiate between the hydrophytes,mesophytesand xerophytes . Give examples of each.

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Solution:

Plants have become adapted to grow in environments ranging from verywet to very arid. Botanists classify plants according to their water needs into three groups called hydrophytes,mesophytes, and xerophytes. Hydrophytes grow in a very wet environment, and are either completely aquatic or rooted in water or mud but with stems and leaves above the water for capturing sunlight. Water lilies, pondweeds and cattails are common hydrophytes.Mesophytesare land plants that prefer a habitat with a moderate amount of moisture. Commonmesophytesare beech, maple , oak, dogwood and birch. Xerophytes are plants such as yuccas and cacti that have become adapted to areas where soil water is scarce. They manage to survive with a limited amount of water by minimizing its loss through a number of ways. For example, xerophytes may have a reduced number of stomata, a heavily cutinized epidermis to retard evaporation , or they may develop thick stems and leaves to store water.

Question:

A chemist has a quantity of steam. He wants to determine whether or not it has been polluted with acidic mine drainage (AMD). How should he go about determining this?

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Solution:

AMD occurs yearly when millions of tons of sulfuric acid seep out from coal and copper mines to pollute waterways. It begins by the oxidation of iron pyrite (FeS_2) to sulfate: FeS_2(O_2/H_2O) \rightarrow Fe^2+ + SO^2-_4. The ferrous ions are then oxidized to ferric ions: Fe^2 +(O_2) / (H_2O) \rightarrow Fe^3+. This process can be sped up by microorganisms. The ferric ions can react with water to give hydrogen ions and a precipitate of hydrated ferric oxide: Fe^3+ H_2O \rightarrow Fe_2O_3.H_2O + H^+. Thus, acidic mine drainage is a diluted sulfuric acid solution carrying iron. If the steam has AMD, it must carry iron. Addition of thiocyanate to ferric (iron) solutions will yield an intensely red-colored complex ion. The reaction is as follows: Fe^3+ + SCN^- \rightarrowFeSCN2+ thio-red-colored cyanatecomplex ion Thus, the chemist can detect for AMD pollution in the steam by adding thiocyanate. If a red-colored material is formed, he knows it is contaminated.

Question:

All of the algae have the green pigment chlorophyll, yet the different phyla of algae show a great variety of colors. Explain.

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Solution:

The pigments found in the different algae phyla are extremely varied, and their concentrations result in different colors. The earliest classifications of algae were based on color. Fortunately, later study of algae showed that algae of similar pigmentation also shared other important characteristics and that the older classifications were still valid. In addition to the green pigment chlorophyll, most algae possess pigments of other colors called accessory pigments. The accessory pigments may play a role in absorbing light of various wavelenghts. The energy of these light wavelengths is then passed on to chorophyll. This absorption widens the range of wavelengths of light that can be used for photosynthesis. The accessory pigments phycocyanin and phycoerythrin serve this function in the red algae, the Rhodophyta. These pigments give the Rhodophyta their characteristic red color, although occasionally, they may be black. The red algae often live at great depths in the ocean. The wavelengths absorbed by chlorophyll a do not penetrate to the depths at which the red algae grow. The wavelengths that do penetrate deep enough are mostly those of the central portion of the color spectrum. These wavelengths are readily absorbed by phycoerythrin and phycocyanin. The energy trapped by these pigments is then pass-ed on to chlorophyll a, which utilizes this energy for photo-synthesis. In the green algae, the Chlorophyta, the chlorophyll pig-ments predominate over the yellow and orange carotene and xanthophyll pigments. The predominance of carotene pigments imparts a yellow color to the golden algae, members of the phylum Chrysophyta. The diatoms, the other important class of Chrysophyta, possess the brown pigment fucoxanthin. The Pyrrophyta (dinoflagellates) are yellow-green or brown, due to the presence of fucoxanthin and carotenes. Some red dinoflagellates are poisonous, containing a powerful nerve toxin. The blooming of these algae are responsible for the "red tides" that kill millions of fish. The brown algae, the Phaeophyta, have a predominance of fucoxanthin. These algae range in color from golden brown to dark brown or black. The procaryotic cyanophyta (blue-green algae) have the blue pigment phycocyanin as well as phycoerythrin xanthophyll and carotene. The Euglenophyta contain chlorophyll a and b and some carotenoids.

Question:

A certain platinum-resistance thermometer has a resistance of 9.20 ohms when immersed in a triple-point cell. When the thermometer is placed in surround-ings where its resistance becomes 12.40 ohms, what temperature will it show?

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Solution:

The triple-point of water occurs when water can co-exist in its three forms : liquid, gas, and solid. This can happen at a temperature of 273.16\textdegree K and a water vapor pressure of 4.58 mm-Hg. Since the resistance of a thermometer is directly proportional to the temper-ature, we can write (T_1/R_1) = (T_2/R_2) or T_2 = T_1 (R_2/R_1) T_2 = 273.16\textdegree K (12.40/9.20) = 368.1\textdegree K

Question:

In corn, a strain homozygous for the recessive genes a (green), d (dwarf) and r (normal leaves) was crossed to a strain homozygous for their dominant alleles A (red), D (tall), and R (ragged leaves). Offspring of this cross were then backcrossed to homozygous recessive plants. Listed below are the phenotypes produced from the backcross. Phenotypes (all are -a-d-r) Number 1 ADR 265 2 adr 275 3 ADr 24 4 adR 16 5 Adr 90 6 aDR 70 7 AdRg 120 8 aDr 140 1000 Propose a linkage map with distances between the three genes.

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Solution:

This problem is an application of the concepts discussed before. The F_1 parents of the cross arising from the mating of homozygous parents must beheterozygous heterozygous . The parental types in the progeny are therefore 1 ADR and 2 adr. The rest are recombitants. Consider a cross-over between A and D. From the F_2 the number of recombinants = 90 5 + 706 +120 7 + 140 8 = 420. RF = [(420)/(1000)] = 0.42 This means that A and D are separated by 42 units. Consider a cross-over between D and R. The number of recombinants = 24 3 + 16 4 + 120 7 + 140 8 = 300 RF = [(300)/(1000)] = 0.30 That is, D and R are 30 map units apart. Consider a cross-over between A and R. The number of recombinants = 24 3 + 16 4 + 90 5 + 706 = 200. RF =[(200)/(1000)] = 0.20 That is, there are 20 units between A and R. A genetic map can be constructed by arbitarily making A and D our reference points. R is 30 units away from D. At the same time, R is also 20 units from A. These conditions allow only one arrangement of the genes; that is, with R between A and D. However, the sum of the distances between A and R and between R and D is greater than the distance between A and D. The reason why these distances on the map do not agree is because of the presence of double cross-overs. Double cross-overs can be identified by their lowest frequencies, namely 24 3 and 16 4, in the F_2. The percentage of double cross-overs = [(24+16)/(1000)] = 4% The map distance between A and D should be longer because we have ignored the double cross-overs in our calculation of it. The corrected distance between A and D should be = 42 + (2 × 4) = 50 units. Using this distance, the new genetic map does not show any discrepancy.

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Question:

Write statements in APL to generate the identity matrix of dimension N. This matrix consists of N ones along the main diagonal and zeros everywhere else. Use the rho (\rho) and catenate (,) functions.

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Solution:

We first enter a value for the dimension, say N = 10. We call the matrix V. The following statements will do the job. N \leftarrow 10 V \leftarrow (N,N) \rho (1,N\rho0) V In the above, the statement N \leftarrow 10 initializes N to a value 10. Now, consider the next statement V \leftarrow (N,N) \rho (1,N\rho0) N\rhoO creates a vector of 10 elements (because N = 10). Each element is equal to 0 since the statement N\rhoO is equivalent to: N \rho 0 0 0 0 0 0 0 0 0 0 Note that there are 10 zeroes in the above. Next, (1,Np0) catenates number 1 to the ten zeroes, creating a line 10000000000. (Catenation is a procedure for link-ing program elements together. In APL, it is possible to catenate i) scalars to form a vector, ii) a scalar and a vector to form a vector, iii) several vectors to form a vector, iv) matrices of the same dimension to form a matrix v) N-dimensional arrays to form an array. The symbol for catenation is a comma. Then, (N,N)\rho (1,N\rhoO) creates a 10 by 10 matrix, using the string 1 0 0 0 0 0 0 0 0 0 0 repeatedly. Since there are 11 elements in the string and only 10 vacant spaces in each row of the matrix, the last zero of the string will start the second row, when the string is used the first time. As a result, 1 will move one space to the right in the second row. The third row will then start with 2 zeros, etc. The 10 by 10 matrix, thus created, consists of the elements as shown below

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Question:

Write the micro-operations for the execute cycles of the following register-reference instructions. Instruction Symbol Hexadecimal Code Definition CLA 7800 Clear Accumulator CLE 7400 Clear E register CMA 7200 Complement Accumulator CME 7100 Complement E register INC 7020 Increment Accumulator HLT 7001 Halt Computer

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Solution:

All register reference instructions are specified by an op-code of 111 and a 0 in the I register; thus the control signal specifying the register- reference instruction is q_7I ̅c_2 and a particular step is specified when q_7I ̅c_2t_j . [MBR]_i is true (logic 1). CLA:This instruction is specified by q_7I ̅c_2 and when the MBR appears as or[MBR]_4 = 1 The instruction may be executed during only one timing signal, t_3; hence, q_7I ̅c_2t_3 [MBR]_4 :AC\leftarrow0 F\leftarrow0 The first micro-instruction clears the AC and the second one sets the fetch cycle control signal. The rest of the micro-instructions are found in a similar manner. CLE: q_7I ̅c_2t_3 [MBR]_5 :E\leftarrow0 F\leftarrow0 CMA: q_7I ̅c_2t_3 [MBR]_6 :AC\leftarrow[AC] F\leftarrow0 CME: q_7I ̅c_2t_3 [MBR]_7 :E\leftarrow[E] F\leftarrow0 INC: q_7I ̅c_2t_3 [MBR]_10 :EAC\leftarrow[AC]+1 F\leftarrow0 HLT:The S register designates whether or not the com-puter is running. If s = 1, the computer is on and if s = 0, the computer is stopped; hence, a 0 is stored in S to halt the computer. q_7I ̅c_2t_3 [MBR]_15 :S\leftarrow0 F\leftarrow0

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Question:

Explain how a diode may be used to convert alternating current (AC) into direct current (DC).

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Solution:

Figure (a) shows the arrangement of an AC generator, the diode, and the resistive load represented by a resistor R through which a direct current flows. The polarity of the generator terminals reverses twice during each cycle. (See figure (b), voltage diagram). When the generator lead connected to the negative terminal of the diode is negatively charged, the diode is forward biased and electrons readily flow through the diode, the resistor, and back to the positive terminal of the generator. One half-cycle later, the polarity of the generator lead connected to the negative terminal is positive, the diode is reverse biased, and the current through the diode is essen-tially zero. The current through the resistor will vary like the voltage does during the half-cycle that conduction occurs in the diode, so the current will change as indicated in figure (b).

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Question:

Determine the volume in milliliters of .20 M KMnO_4 required to oxidize 25.0 ml of .40 M FeSO_4, in acidic solution. Assume the reaction which occurs is the oxidation of Fe^2+ by MnO^-_4 to give Fe^+3 and Mn^2+.

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Solution:

This problem can be solved by two methods: the mole and the equivalent methods. The mole method requires consideration of the balanced equation that illustrates the reaction. From the data provided, this equation becomes 5Fe^2+ + MnO^-_4+ 8H_3O^+ \rightarrow 5Fe^3+ + Mn^2+ + 12H_2O. Now 25.0 ml of .40 M FeSO_4 furnishes (.025 liters)(.40 mol / liter) = .010 moles of Fe^2+, since the definition of molarity is molarity is M = (number of moles of solute number of liters) / (number of liters). The balanced equation indicates that the number of moles of MnO^-_4 will be 1/5 that of Fe^2+. As such, the number of moles of MnO_4^- = (.010)(1/5) = .002 moles. Since the KMnO_4 solution has a concentration of .2 M, then the numberof liters required is [.002 mole of MnO^-_4] / [.2 mol/liter)] = .01 liters, which equals 10 ml [.002 mole of MnO^-_4] / [.2 mol/liter)] = .01 liters, which equals 10 ml . The equivalent method functions differently. An equi-valent is defined as that mass of oxidizing or reducing agent that picks up or releases the Avogadro number of electrons. Normality is defined as the number of equi-valents per liter. Since, in going from Fe^2+ to Fe^3+, you lose 1 electron, .40 M FeSO_4 is equal to .40 N FeSO_4. Recalling the definition of normality, you have (.025 liter) (.40 equiv/liter) = .01 equiv. of Fe^2+, the reducing agent. In an oxidation-reduction reaction, the number of equivalents of oxidizing agent must equal that of the reducing agent. This means you must have .01 equiv. of MnO^-_4. You know that for KMnO_4, there exists 1 equiv/liter. Therefore, the number of liters equals (.01equiv.)/(1.0 equiv./liter) = .01 liters or 10 ml.

Question:

Name each of the organic compounds shown in figure A below.

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Solution:

A system of nomenclature called the IUPAC system has been formulated so that all organic molecules may have their structures defined adequately. In this system for hydrocarbons (i.e. compounds containing only hydrogen and carbon atoms) the longest chain is taken as the basic structure, and the carbon atoms are numbered from the end of the chain closest to a branch chain or other modification of simple alkane structure. The position of substituents in the chain are denoted by the number of carbon atom or atoms to which they are attached. In the problem, you are not given alkanes, you are given alkenes, compounds that contain a double bond between a pair of carbon atoms (i.e. unsaturated), and a ring compound. In unsaturated compounds, the rules are the same, except that the position of the double bond is indicated; the numbering starts at the end of the chain nearest the double bond. For rings, name the ring and any substituent present based on which carbon is located. Thus, you proceed as follows: (a) There is only one chain, and it possesses three carbon atoms. The double bond is located on the first carbon, not the second, since you want to use the lowest possible number. As such, you form the molecule propene. Three carbons suggest propane. It's an alkene, so that you change ane to ene. Therefore, propane becomes propene. You have no need to name the position of the double bond here; the double bond can only be in two po-sitions and both yield the same molecule. That is,CH2 = CH - CH_3 is equivalent to CH_3 - CH = CH_2. = CH - CH_3 is equivalent to CH_3 - CH = CH_2. (b) The longest chain containing the double bond has 7 carbon atoms, so that "hep" prefix is suggested. Because it is a double bond, you add the suffix ene to obtain heptene. The double bond is located on the second carbon, not the third, since you want the lowest number. You have, therefore, the 2-heptene. Using this numbering system, you see that CH_3, a methyl group, is located on the fifth carbon. Thus, the name of the molecule is 5-metyl 2 heptene. (c) Has a six carbon chain, which suggests the pre-fix ''hex\textquotedblright. There is a double bond so that the molecule has the suffix ene. Thus, you obtain hexene. The double bond is located on the second carbon, which means the name of the molecule is 2-hexene. (d) This ring compound has a methyl group (CH_3) positioned on a benzene ring. The compound can be called methyl benzene. It is also given the special name of toluene.

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Question:

Write a program in Basic to play the game ofNimwith a com-puter.

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Solution:

The traditional game ofNimis played in the following manner: There are four rows of stones arranged as shown: 0(1) 000(3) 00000(5) 0000000(7) Let A and B be the two players. The players play according to the 3 rules below: 1)On any given turn only objects from one row may be removed. There is no restriction on which row or on how many objects you remove. 2)You cannot skip a move or remove zero objects or remove more objectsthan there are in a row. 3)Opponents take turns removing objects until there are none left. The win option should be specified in the program. The program below allows the number, of piles to be variable, the pile sizesto be variable and also the win option to be either taking the last stone(s) or being left with the last stone(s). 10PRINT "THE GAME OP NIM" 20DIM A (100), B (100,10), D (2) 30PRINT "ENTER WIN OPTION - 1 TO TAKE LAST, 402 TO AVOID LAST"; 50INPUT W 60IF W = 1 THEN 80 70IF W <> 2 THEN 30 80PRINT "ENTER NUMBER OF PILES"; 90INPUT N 100IF N > 100 THEN 80 110IF N < 1 THEN 80 120IF N <> INT (N) THEN 80 130PRINT "ENTER PILE SIZES" 140FOR I = 1 TO N 150PRINTI; 160INPUT A (I) 170IF A (I) > 2000 THEN 150 180IF A (I) <>INT(A (I)) THEN 150 190NEXT I 200PRINT "DO YOU WANT TO MOVE FIRST?"; 210INPUT A$ 220IF A$ = "YES" GOTO 1000 230IF A$ = "NO" GOTO 250 235PRINT "PLEASE, YES OR NO"; 240GOTO 210 250IF W = 1 THEN 490 260LET C = 0 270FOR I = 1 TO N 280IF A (I) = 0 THEN 320 290LET C = C + 1 300IF C = 3 THEN 390 310LET D (C) = I 320NEXT I 330IF C = 2 THEN 470 340IF A (D(1)) > 1 THEN 370 350PRINT "MACHINE LOSES" 360GOTO 1190 370PRINT "MACHINE WINS" 380GOTO 1190 390LET C = 0 400FOR I = 1 TO N 410IF A (I) > 1 THEN 490 420IF A (I) = 0 THEN 440 430LET C = C + 1 440NEXT I 450IF C/2 <> INT (C/2) THEN 350 460GOTO 490 470IF A (D (1)) = 1 THEN 370 480IF A (D (2)) = 1 THEN 370 490FOR I = 1 TO N 500LET E = A (I) 510FOR J = 0 TO 10 520LET F = E/2 530LET B (I, J) = 2\textasteriskcentered(F -INT(F) ) 540LET E = INT (F) 550NEXT J 560NEXT I 570FOR J = 10 TO 0 STEP - 1 580LET C = 0 590LET H = 0 600FOR I = 1 TO N 610IF B (I, J) = 0 THEN 660 620LET C = C + 1 630IF A (I) < = H THEN 660 640LET H = A (I) 650LET G = I 660NEXT I 670IF C/2 <> INT (C/2) THEN 740 680NEXT J 690LET E = INT (N\textasteriskcenteredRND (1) + 1) 700IFA(E) = 0 THEN 690 710LET F = INT (A (E)\textasteriskcenteredRND (1) + 1) 720LET A (E) = A (E) - F 730GOTO 930 740LET A (G) = 0 750FOR J = 0 TO 10 760LET B (G, J) = 0 770LET C = 0 780FOR I = 1 TO N 790IF B (I, J) = 0 THEN 810 800LET C = C + 1 810NEXT I 820LET A (G) = A (G) + 2\textasteriskcentered(C/2 - INT (C/2) )\textasteriskcentered2 \uparrow J 830NEXT J 840IF W = l THEN 930 850LET C = 0 860FOR I = 1 TO N 870IF A (I) > 1 THEN 930 880IF A (I) = 0 THEN 900 890LET C = C + 1 900NEXT I 910IF C/2 <> INT (C/2) THEN 930 920LETA(G) = 1 - A (G) 930PRINT "PILE SIZE" 940FOR I = 1 TO N 950PRINTI; A (I) 960NEXT I 970IF W = 2 THEN 1000 980GOSUB 1120 990IF Z = 1 THEN 370 1000PRINT "YOUR MOVE - PILE, NUMBER TO BE REMOVED"; 1010INPUT X,Y 1020IF X > N THEN 1000 1030IF X < 1 THEN 1000 1040IF X <> INT (X) THEN 1000 1050IF Y > A (X) THEN 1000 1060IF Y < 1 THEN 1000 1070IF Y <> INT (Y) THEN 1000 1080LET A (X) = A (X) - Y 1090GOSUB 1120 1100IF Z = 1 THEN 350 1110GOTO 250 1120LET Z = 0 1130FOR I = 1 TO N 1140IF A (I) =0 THEN 1160 1150RETURN 1160NEXT I 1170LET Z = 1 1180RETURN 1190PRINT "DO YOU WANT TO PLAY ANOTHER GAME" 1200INPUT A$ 1210IF A$ = "YES" THEN 1240 1220IF A$ = "NO" THEN 1250 1230GOTO 1200 1240GOTO 30 1250END.

Question:

Write a FORTRAN program to determine the greatest integerthat does not exceed a given real number X.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G22-0536.htm

Solution:

The greatest integer function f: R\rightarrow Z has as domain the real numbersand co-domain the integers. It is denoted by [X] where X is a real number. For example, [\pi] = [3.1415...] = 3 [e] = [2.7128...] = 2 [-6.01] = -7 ;[8] = 8 . [-6.01] = -7 ;[8] = 8 . Thus, for X \geq 0, [X] is the integer part of X (it is immaterial whether the decimalpart of X is greater than .5 or not). For X < 0, [X] is the integer part of X minus 1. Thus, to find [X], we proceed as follows: 1) Truncate X to an integer, NX 2) If X is nonnegative, [X] = NX 3) If X is negative [X] = NX - 1. CPROGRAM TO DETERMINE THE LARGEST CINTEGER THAT DOES NOT EXCEED A GIVEN CREAL NUMBER X. READ (5,1) X 1FORMAT (F 10.4) NX = X IF (X) 5,3,3 CIF X IS NEGATIVE, IS IT AN INTEGER? CIF SO, PROCEED DIRECTLY TO WRITE. CIF NOT,SUBTRACT1 FROM NX. 5Y = NX IF (X - Y) 7,3,7 7NX = NX - 1 3WRITE (6,9) X 9FORMAT (1HO, 14H REAL NUMBER= ,F 14.4) WRITE (6,11) NX 11FORMAT (1H ,19H GREATEST INTEGER = ,I6) STOP END

Question:

A gaseous substance exhibits a line spectrum, whereas a liquid or a solid substance exhibits a continuous spectrum. Give a qualitative explanation for these phenomena.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E18-0705.htm

Solution:

The basic difference between gases and the condensed phases of matter, which gives rise to these different phenomena, is that in gases the atoms or molecules are sufficiently far apart to have a negligible interaction, whereas in liquids and solids, the inter-action between the constituent atoms and molecules is appreciable. In the gas phase, the molecules are so far apart that they may be considered to be isolated from one an-other and therefore behave independently of one another. Thus, the spectrum arises only from electronic, vibrational, or rotational transitions of individual atoms or molecules, and not from any collective behavior. But since these transitions are well defined energetically, the spectrum exhibits sharp lines. In a liquid or solid, the atoms or molecules are sufficiently close together to affect each other's energy levels. That is, in addition to the individual energy levels of each atom or molecule, there exist energy levels for the collective excitation of groups of atoms or molecules. In general, the interaction of one atom or molecule with another decreases with internuclear distance or intemolecular distance in a continuous fashion, so that a continuous spectrum is observed.

Question:

What are the various methods by which PL/I debugs or notesdefects in a program while processing it?

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G13-0358.htm

Solution:

After a program has been compiled, it enters the execution stage. Here, many situations exist that can cause errors and prevent execution. PL/Ihasfacilities for moni-toring program executions. 1) The CONVERSION condition Suppose Y is define as a numeric variable, but the input format specificationis non-numeric. In this case the system stops the program at thispoint and CONVERSION message is printed out. 2) Overflow and underflow: These conditions occur when a numerical value turns out to be too large ortoo small to beaccomodatedin the ma-chine. If a fixed point number hasmore than 15 digits, the computer prints FIXEDOVERFLOW. With floatingpoint numbers the OVERFLOW condition is raised. This means thecomputer notes the error and automatically increases the limit on the sizeof that number, and continues the program. Finally if a very small valueis divided by a very large number and the number is less (in absolutevalue) than 10^-75, UNDERFLOW oc-curs and an appropriate messageis printed. 3) Division by zero is a special type of overflow. When it is attempted, the ZERODIVIDE condition is printed and the programstops. For example, if X, Y are variables and X = Y, then the instructionSQRT(40\textasteriskcentered(Y - W)/(X - Y)) will stop the program. 4) The usual method of indicating the end of input data is to insert a trailercard with a value that cannot be in the data set. An instruction to exitwhen this value is read is inserted in the main program. If some indica-tionfor the termination of processing is not included, the system will respondwith an ENDFILE condition. 5) Suppose we expect an execution-halting error but wish the processingto continue after it has occurred. This is done through PL/I's ON statement whose general form is ON condition action(1) In (1), 'condition' is some keyword (UNDERFLOW, ENDFILE, etc.). 'action' issome instruction to be performed when the stated condition occurs. Thus, if we want to make sure that SQRT(Y/X) is executed even when X = 0, we insert the statement, ON ZERODIVIDE X = 9999. This should be placed before computations involving X are executed. In (1) the action specified can be any single statement except DECLARE, IP, PROCEDURE, END, DO or another ON. Ex-cept for the ON statement which is a part of the program, the other conditions discussedabove abort the program. But PL/I also gives a diagnosis as to whythe abortion occurred. The next creative run may well produce a normal program.

Question:

How much energy is released in the fission of 1 kg of U235?

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Solution:

Fission can be induced by adding energy to the nucleus in the form of the binding energy of a captured neutron. A typical fission reaction involving U^235 is U^235 + n \ding{217} Ba139+ Kr^95 + 2n . Though the total number of protons and neutrons remains the same, the mass diminishes. The missing mass, arising from neutrons decaying into protons and electrons, is converted into energy. The amount of mass-energy that is converted to kinetic energy in the fission process is approximately 200MeVper nucleus. (This is only about 0.1 percent of the total mass-energy of a uranium nucleus; the other 99.9 percent remains in the masses of the neutrons and protons and is therefore not available for conversion into kinetic energy.) In 1kg of U^235, there are approximately 2.5 × 10^24 atoms. Therefore, the total energy release is \epsilon = (200MeV) × (2.5 × 10^24 ) = 5.0 × 10^26MeV = (5.0 × 10^26MeV) × (1.6 × 10^-6 ergs/Mev) = 8.0 × 10^20 ergs. We can convert this into another popular unit by noting that the explosion of 1 ton of TNT releases approximately 4.1 × 10^16 ergs. Thus, the fission of 1 kg of U^235 releases an amount of energy \epsilon =(8.0 × 10^20 ergs) / (4.1 × 10^16 ergs / ton TNT) \allequal 20 kilotons TNT. This is approximately the size of the original atomic bomb of 1945.

Question:

At standard conditions, it is found that 28.0 g of carbon monoxide occupies 22.4 l. What is the density of carbon monoxide at 20\textdegreeC and 600 torr?

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Solution:

The density of carbon monoxide under STP or standard conditions (pressure = 1 atm or 760 torr and temperature = 0\textdegreeC or 273.15\textdegreeK) is \rho = mass/volume. 28 g of CO, carbon monoxide, represents one mole and a mole of any gas occupies 22.4 liters of volume under standard conditions. Thus, \rho = 28g / 22.4 l = 1.25 g/l. We can apply the equation P /(\rhoT) = constant, which holds for ideal gases of density \rho at pressure P and absolute temperature T. We will first determine the constant for our system and then use this value to calculate the density at 20\textdegreeC = 293.15\textdegreeK and 600 torr. The value of the constant for our system is Constant = (P/\rhoT) = [(760 torr)/(1.25 g/l × 273.15 \textdegreeK)] = 2.225 (torr - l/g -\textdegreeK). Hence, for our system, (p/\rhoT) = 2.225 [(torr - l) / (g - \textdegreeK)]or \rho = (P/T) × [(1)/(2.225 torr - l/g - \textdegreeK)]. At 20\textdegreeC = 293.15\textdegreeK and 600 torr, the density of CO is \rho = (P/T) × [(1)/(2.225 torr - l/g - \textdegreeK)] = [(600 torr)/(293.15\textdegreeK)] × [(1)/(2.225 torr - l/g - \textdegreeK)] = .9198 g/l.

Question:

A 40-lb stone is pushed, on a 30\textdegree incline, to the top of a building 100 feet tall. By how much does its potential energy increase?

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/Users/wenhuchen/Documents/Crawler/Physics/D05-0223.htm

Solution:

The change in a body's gravitational potential energy is the negative of the work done by gravity on the object in displacing it. By definition, this is W = - \intFg^\ding{217} \bullet dr^\ding{217} where Fg^\ding{217} , the force of gravity, is Fg^\ding{217} = -mg(1) The symbolis a unit vector in the positive y direction (see figure). Now dr^\ding{217} = dx+ dy whereis a unit vector in the positive x direction. Then Fg^\ding{217} \bullet dr^\ding{217} = -mg\bullet (dx+ dy) = - mgdy andW = -\int-mhdy(2) We evaluate (2) over the path of motion of the block. If we take the origin of our coordinate system at the foot of the plane, y varies from 0 to 100 ft. Therefore W = mg^100 ft\int_0 dy = mg(100 ft) W = (40 lb)(100 ft) = 4000 ft. lb.

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Question:

In addition to its function in secreting female sex hormones, the human ovary serves as the organ of egg production. Explain the process of egg formation in the mammalian ovary. What happens to the follicle after ovulation?

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/Users/wenhuchen/Documents/Crawler/Biology/F22-0576.htm

Solution:

The mammalian ovary (see figure) consists of two types of cells: the germ cells and the germinal epithelial cells. The germ cells are the oogonia, which will eventually become the eggs. The germinal epithelial cells line the walls of the ovary and will eventually form part of the follicle. At birth, the human ovaries contain an estimated 400,000 immature eggs or oogonia. Thus, in marked contrast to the male, the newborn female already has all the germ cells she will ever have: only a few, perhaps 400, are destined to reach full maturity during her reproductive life. The process of maturation of an oogonium is called oogeneis. Its beginning is marked by the enlargement of the oogonium to form a primary oocyte. Following the formation of the primary oocyte, the prospective egg undergoes the first meiotic division. Although the nuclear division proceeds normally, there is unequal division of the cytoplasm, resulting in one large secondary oocyte and a smaller cell made up almost entirely of nuclear material, called the first polar body. The secondary oocyte also divides unequally in the second meiotic division, producing a large ootid and a small second polar body. The first polar body may or may not divide; in any event, the polar bodies all soon disintegrate. The ootid becomes the mature egg or ovum. The single haploid ovum that results from the meiotic divisions of the primary oocyte contains most of the original cytoplasm and also some stored food. During oogenesis, the developing egg becomes in-creasingly surrounded by a layer of germinal epithelial cells, and by the secondary oocyte stage, it is well enveloped by these cells. The oocyte and its surrounding cells are together termed the follicle. After secondary oocyte formation is completed, about two weeks after the onset of oogenesis, the oocyte alone is released from the follicle in a process called ovulation. This involves the rupture of part of the ovarian wall and the subsequent expulsion of the oocyte from the ovary. The secondary oocyte then enters the Fallopian tube, and remains at the secondary oocyte stage until after fertilization. It is important to realize that the ovary is not directly connected to the Fallopian tube; during ovulation the secondary oocyte is expelled into the body cavity. However, because of the close proximity of the ovary and Fallopian tube and the presence of finger-like projections from the mouth of the Fallopian tube which help direct the oocyte into it, almost all oocytes that are ovulated do end up in the Fallopian tube. The follicular cells which stay behind in the ovary following ovulation enlarge greatly and become filled with a lipid substance; the entire gland-like yellowish structure is known as the corpus luteum (meaning "yellow body"). Under hormonal influence, the corpus luteum secretes progesterone and estrogen, two female sex hormones. These hormones stimulate the proliferation and vascularization of the uterine lining, preparing it for implantation of the embryo. If the ovum is fertilized, the corpus luteum enlarges to form a corpus luteum of pregnancy, which increases its hormone production to maintain pregnancy. If the ovum is not fertilized, the corpus luteum begins to degenerate about nine days after ovulation and is called a corpus luteum of menstruation.

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Question:

How many moles are present in 100 g quantities of each of the following? (a) CaCO_3, (b) H_2O, (c) HCl, (d) Al_2(SO_4)_3?

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Solution:

One can find the number of moles of a specific compound in a certain number of grams of that compound by dividing the number of grams present by the molecular weight. The molecular weight is defined as the weight of one mole or Avogadro's Number (6.02 × 10^23) of particles. number of moles = (weight in grams of sample) / (molecular weight) One calculates the molecular weight of a compound by adding together the molecular weights of the elements present. When calculating the molecular weight, one must take into account the number of atoms of each element in the compound. This is done by multiplying the molecular weight of the element by the number of atoms present of the particular element. This method will be used in the following examples. Once the molecular weight is deter-mined, the number of moles present in 100 g of the compound can be found by using the equation number of moles in 100 g = 100g / ( molecular weight) (a)CaCO_3 There is one atom of Ca present in CaCO_3. Thus, the molecular weight of Ca, 40, is multiplied by one. There is one atom of C. Thus the molecular weight, 12, of C is multiplied by one. Because there are 3 atoms of O present, the molecular weight of O, 16, is multiplied by 3. 1 atom of Ca1 × 40 =40 1 atom of C1 × 12 =12 3 atoms of O3 × 16 =48 molecular weight of CaCO_3= 100 The number of moles of CaCO_3 in 100 g may now be found. number of moles in 100 g = (100g) / (100g/mole) = 1 mole. (b)H_2O (1) Calculation of molecular weight. The MW of H is 1 and the MW of O is 16. 2 atoms of H2 × 1=2 1 atom of O1 × 16 =16 molecular weight of H_2O= 18 (2) Calculation of number of moles in 100 g of H_2O. number of moles in 100 g = (100g) / (18g/mole) = 5.55 moles. (c)HCl MW of H = 1,MW of Cl = 35.5. 1 atom of H1 × 1=1 1 atom of Cl1 × 35.5 =35.5 molecular weight ofHCl= 36.5 number of moles of HCl in 100 g = (100g) / (36.5 g/mole) = 2.74 moles (d) Al_2(S0_4) _3 MW of Al = 27MW of S = 32MW of O = 16 2 atoms of Al2 × 27=54 3 atoms of S3 × 32=96 12 atoms of O12 × 16=192 molecular weight of Al_2(SO_4)_3= 342 number of moles of Al_2(SO_4)_3in 100 g = (100g) / (342g/mole) = .292 moles.

Question:

2.0 g of molybdenum (Mo) combines with oxygen to form 3.0 g of a molybdenum oxide. Calculate the equivalent weight of Mo in this compound.

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Solution:

The equivalent weight of a compound can be determined by use of the Law of Definite Proportions. This law states that when elements combine to form a given compound, they do so in a fixed and invariable ratio by weight. This means that the following ratio is maintained: (weight of Mo in reaction) / (weight of O in reaction) =(equivalent weight of Mo weight) / (equivalent weight of O) The equivalent weight of oxygen is equal to 8. The weight of oxygen in the reaction can be determined by using the Law of Conservation of Mass. This law states that there is no detectable gain or loss of mass in a chemical change. The problem states that 2.0 g of Mo is added to an unknown amount of O to form 3.0 g of a molybdenum oxide. Using the Law of Conservation of Mass the following relation is true: weight of compound = weight of O + weight of Mo 3.0 g = weight of O + 2.0 g 1.0 g = weight of O One can now use the Law of Definite Proportions to solve for the equivalent weight of Mo. (weight of Mo) / (weight of O) = (equivalent weight of Mo) / (equivalent weight of O) (2.0g) / (1.0g) = (equivalent weight of Mo) / 8 equivalent weight of Mo = 8 × (2.0g / 1.0g)= 16 .

Question:

A hospital receives and puts into storage a batch of 100mCi of radioiodine which has a half-life of 8 days. For how long can the batch be kept in storage and still provide a therapeutic dose of 12mCi\textbullet hr?

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Solution:

The decay of a radioactive sample is governed by the relation N = N_0 e\Elzbar\lambdat where \lambda is the decay constant, N_0 is the initial number of atoms in the sample, and N is the number of atoms remaining after time t. Hence, the number of atoms, N', which have decayed is N' = N0\Elzbar N = N_0(1\Elzbar e\Elzbar\lambdat) The activity I of the sample is defined as the number of decays per unit time, or I = (dN') / (dt) = \lambdaN_0e\Elzbar\lambdat=\lambdaN N_0e\Elzbar\lambdat=\lambdaN Defining the initial activity I_0 as I_0 = \lambdaN_0 we may write I = I_0 e\Elzbar\lambdat A "dose" D of the radioactive substance represents a certain number of decays of the parent element. A dose administered from time t to time t' has the value D = I_0^t'\int_te\Elzbar\lambdatdt. If the radioelement may be left in an implant for a long period, t' can be made to tend to infinity. This will give us the maximum value of D. D_max= Limt'\rightarrow\inftyI_0^t'\int_te\Elzbar\lambdatdt Letu =\Elzbar\lambdatdu =\Elzbar\lambdadt ort =\Elzbaru/\lambdadt=\Elzbardu / \lambda Then, when t = t, u =\Elzbar\lambdat, and when t = t', u =\Elzbar\lambdat'. Therefore D_max= Limt'\rightarrow\inftyI_0\Elzbar\lambdat'\int\Elzbar\lambdateu(\Elzbardu / \lambda) D_max=\Elzbar Limt'\rightarrow\inftyI_0 / \lambda\Elzbar\lambdat'\int\Elzbar\lambdateudu D_max=\Elzbar Limt'\rightarrow\inftyI_0 / \lambda\Elzbar\lambdat'[e^ u ]\Elzbar\lambdat D_max=\Elzbar Limt'\rightarrow\inftyI_0 / \lambda [e\Elzbar\lambdat'\Elzbar e\Elzbar\lambdat] D_max=(I_0 / \lambda) e\Elzbar\lambdat Since\lambda = [(In 2) / \tau] D_max=[(I_0 \tau) /(In 2)] e\Elzbar\lambdat where \tau is the half-life of radioiodine. Thus, ifD_max=12mCi\textbullet hr, 12mCi\textbullet hr = [(100mCi× 8 × 24 hr) / (0.6932) ] e\Elzbar\lambdat. e\Elzbar\lambdat= 4.327 × 10\Elzbar4,orexp [\Elzbar {(t In 2) / \tau }] = 4.327 ×10\Elzbar4. Solving for t \Elzbar {(t In 2) / \tau } = In (4.327 × 10\Elzbar4) ort = [\Elzbar\tau / (In 2)] In (4.327 × 10\Elzbar4) t = [ {\tau log [1 / (4.327 × 10\Elzbar4) ]} / log 2 ] = [ (8 × 3.3638) / (0.3010) ] days = 89.4 days. The batch may thus be kept in storage for 89.4 days.

Question:

A racing car passes one end of the grandstand at a speed of 50 ft/sec. It slows down at a constant acceleration a\ding{217}, such that its speed as it passes the other end of the grandstand is 10 ft/sec. (a) If this pro-cess takes 20 seconds, calculate the acceleration a\ding{217} and (b) the length of the grandstand .

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Solution:

(a) For constant acceleration, we have a = change in velocity/time elapsed = ∆v/∆t Therefore a = (v_f- v_i)/∆t = (10 ft/sec - 50.ft/sec)/(20 seconds) = - 2 ft/sec^2 wherev_f_- and v_i are the final and initial velocities, respectively. (b) The length of the grandstand is equal to the distance d the car travels during the 20 seconds. This distance is equal to its final positionx_fminus its initial position x_i and can be found from the kinematics equation x_f = x_i +v_it+ (1/2) at^2. Hence, d =x_f- x_i =v_it+ (1/2) at^2 = (50 ft/sec) (20 sec) + (1/2) (-2 ft/sec^2)(20 sec)^2 = 1000 ft - 400 ft = 600 ft The length d can also be found using d =vtwherevis the aver-age velocity. For constant acceleration,vis given by v = (v_i +v_f)/2 = (50 ft/sec + 10 ft/sec)/2 = 30 ft/sec. Then, d =vt= (30 ft/sec)(20 sec) = 600 ft. which agrees with the first answer.

Question:

What are the products when the following hydrocarbons are oxidized with acidic permanganate solutions? (a) 2-butene; (b) 3,4-dimethy1-3-hexene; (c) 2-methy1-2-butene.

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Solution:

Alkenes readily react with a number of oxidizing agents. A test for the presence of an olefin is the reaction with an acidic solution of permanganate ion (MnO-_4 ). The purple color of permanganate ion disappears as the olefin is oxidized. The course of such a reaction is that the olefin cleaves into two oxidized fragments. A carbon atom with two alkyl groups attached is converted to the carbonyl group of ketone, while a carbon atom with one attached hydrogen becomes the carboxyl group of an acid. This is summarized by the following To solve this problem, write the structural formula of each hydrocarbon, find which alkyl groups correspond to R, R', and R" in the general formula, and then substitute into the general formula. The product of this reaction is acetic acid. The product of this reaction is methyl ethyl ketone. The products of this reaction are acetic acid and acetone.

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Question:

A radio capacitor consists of a stack of five equally spaced plates each of area 0.01 m^2, the separation between neighbors being 2.0 mm. Calculate the capacit-ance (a) if the top and bottom plates are connected to form one conductor and the center three are connected to form the other, and (b) if the top, center, and bottom plates are connected to form one conductor and the other two plates are connected to form the other conductor. (See figs. (a) and (b)).

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Solution:

(a)Since the middle three plates are at the same potential, no field exists between them. The capacitor essentially consists of two capacitors of equal capacitance in parallel, one formed from the top two plates, the other from the bottom two plates (see figure (a)). The capacitance of a parallel plate capacitor is C = (\epsilon_0 A/d) where \epsilon_0 is the permittivity of free space, d is the plate separation, and A is the area of 1 capacitor plate. Since the equivalent capacitance of 2 capacitors in parallel is equal to the sum of their individual capacitances, we obtain C_1 = 2\epsilon_0 (A/d) = [(2 × 8.85 × 10^-12 C^2 \bullet N^-1 \bullet m^-2 × 0.01 m^2 )/(0.002 m)] = 8.85 × 10^-11 F. (b)Here a field exists between any pair of plates. Each of the two connected plates acts as the positive plate for each of two capacitors formed between itself and the outer and middle plates. There are effectively four equal capacitors all connected in parallel. Hence C_2 = 4\epsilon_0 (A/d) = 17.7 × 10^-11 F, which is twice the capacitance of the previous arrangement.

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Question:

Since PL/I can be treated as a modular language, it is very adapt-able to the tenets of structured programming. One of the tenets is the reduction, sometimes even abolition, of the use of the GO TO statement. With this in mind, redesign the following program segments so that the GO TO statements are not necessary: a)IF X = X + Y\textasteriskcentered\textasteriskcentered2 THEN GO TO FIRST; ELSE GO TO NEXT; FIRST: X = X - 1; T = T + X; PUT DATA (T); GO TO FINAL; NEXT: PUT DATA (X,Y); FINAL: C = C + T; b)GET DATA (T); T1 = 0; S = 0; SUM: T1 = T1 + S; S = S + 2; IF T1 = T THEN GO TO B; GO TO SUM; B:PUT DATA. (T,T1,S);

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G08-0172.htm

Solution:

The way to solve this problem Is just by moving segments of statements around. For example, a) may be rewritten thus: IF X = X + Y \textasteriskcentered\textasteriskcentered 2 THEN DO; X = X - 1; T = T + X; PUT DATA. (T); END; ELSE PUT DATA (X); FINAL: C = C + T; The procedure names NEXT and FIRST are unnecessary, so we simplify the code by rendering a simple IF-THEN-ELSE construct. The next program segment can be modified with the use of the DO WHILE: GET DATA (T); T1 = 0; S = 0; DO WHILE (T > T1); T1 = T1 + S; S = S + 2; END; B:PUT DATA (T,S); As before, the procedure names are not always necessary.

Question:

Explain the method of copulation in the earthworms.

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Solution:

During copulation two worms are united, ventral surface to ventral surface, with the anterior ends in opposite directions and the anterior quarter of the length of the bodies overlapping (see figure in previous question). They are held together in this position in part by mucous secretions from a swollen glandular region called the clitellum. The mucus from the clitellum forms a sleeve around the animals. During copulation, each worm dis-charges sperm that pass from the vas deferens into the seminal receptacles of the other worm, through temporary longitudinal furrows that form in the skin. Following the exchange, the worms separate. A few days after copulation, a cocoon is secreted for the deposition of the eggs and sperm. A mucous tube is secreted around the anterior segments, including the clitellum. The clitellum then secretes a tough chitin-like material that encircles the clitellum like-a cigar band; this material forms the cocoon. When completely formed, the cocoon moves forward over the anterior end of the worm. The eggs are discharged from the female gonopores, and they enter the cocoon before it leaves the clitellum. Sperm are deposited in the cocoon as it passes over the seminal receptacles. As the cocoon slips over the head of the worm, and is freed from the body, the mucous tube quickly disintegrates, and the ends of the cocoon constrict and seal themselves. The cocoons are left in the damp soil, where development takes place. Development is direct, that is, there is no larval stage and the eggs develop into tiny worms within the cocoon.

Question:

Perform the following operations (using 1's and 2's complement where necessary). (i) 21 + 7 ; (ii) 45 - 33 ; (iii) - 15 + 6

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Solution:

i)2110101 + 7+ 00111 2811100 ii)45101101 - 33-100001 12001100 Now to do binary subtraction using 1's complement, take the complement of 33: 100001 = 011110. Next add it to 45 Similarly, for the 2's complement method, take the complement of 33 and add binary one to it. 33 = 100001 binary equivalent 011110\leftarrow 1's complement of it + 1 011111\leftarrow 2's complement Then (+ 45)\equiv101101 +(- 33)011111 121001100= 12 drop last digit iii)- 15 +6 - 9 15 = 1111 binary equivalent 00001's complement + 1 00012's complement 6 = 0110binary equivalent - 1510000 ^++ 6001101's complement - 90110 Now as no carry is generated the answer is negative and is in 1's complement form. Taking back its complement, we get the answer in binary equi-valent form. 0110 = 1001 \equiv 9 the correct answer. - 15- 10001 +6+ 001102's complement - 9- 10111 Again, as no carry is generated at the most significant bit position, the answer is negative and is in 2's complement form. Complement and add binary one to obtain the answer in binary equivalent form. 0111 = 1000 + 1 1001= - 9.

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Question:

Suppose that the speed of an electron is 1 × 10^6 m/s and the measurementprecision involves an uncertainty in speed of 1 percent. How precisely can the position of the electronbe simultaneouslydetermined?

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Solution:

The uncertainty in speed is \Deltav= (.01) (1 × 10^6 m/s) = 1 × 10^4 m/s This means that a measurement of the velocity of theelectron will yield a value 1 × 10^6 \pm 1 × 10^4 m/s. The Uncertainty Principle relates the lack of knowledge as to where a particle is located (\Deltax), to the lack of knowledge as to what the momentum of the particle is (\Deltap), when both the position and momentum of the part-icle are measured atthe same time. The relation is \Deltap\Deltax=m\Deltav\Deltax\cong (h/4\pi) wherem is the mass of the particle, and h is Planck's constant. Since the ' massof the electron is 9.1 × 10^-31 kg, then \Deltax= [h/{(\Deltav)4\pim}] = [(6.6 × 10^-34 J s)/{(1 × 10^4 m/s) (4)(3.14)(9.1 × 10^-31 kg) = 5.8 × 10^-9 m This uncertainty in position is at least 100 million times larger than the size ofthe electron itself. There-fore, the uncertainty principle imposes significantlimitations on measurement in the atom-sized world.

Question:

What is the potential energy of a 1-pound weight that has been raised 16 feet?

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Solution:

Potential energy is given by mgs, where mg is the weight of the object that has been raised. But mg is also the net force acting on this object, hence, via Newton's Second Law, we obtain F = mg Therefore P.E. = mgs = Fs Substituting the given data into this relation, we find P.E. = Fs = 1 × 16 = 16 ft-lb.

Question:

A battery of 50 cells is being charged from a dc supply of 230 V and negligible internal resistance. The emf of each cell on charge is 2.3 V, its internal resistance is 0.1 \Omega and the necessary charging current is 6 A. What extra resistance must be inserted in the circuit?

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Solution:

Let R be the extra resistance needed in the circuit. The 50 cells have a total emf of 50 cells × 2.3 V/cell = 115 V and a total internal resistance of 50 cells × 0.1\Omega/cell = 5 \Omega. Let us then represent the battery by an emf source \epsilon_b in series with a resist-ance r. Since the battery is being charged, its polarity will be in the direction opposite to that of the dc supply source. The net emf \epsilon of the circuit is then the difference between the 230 volt dc supply voltage and \epsilon_b = 115 V. The charging current I = 6 A. We apply Ohm's law \epsilon = I(R + r) to obtain (230 - 115)V = 6 A(R + 5\Omega). \thereforeR = [(115 V)/(6 A)] - 5 \Omega = 14.2 \Omega .

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Question:

In the human, many organs in addition to the kidneys perform excretory functions. What are these organs and what do they excrete?

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Solution:

The organs of excretion include the lungs, liver, skin, and the digestive tract, in addition to the kidneys. Water and carbon dioxide, important metabolic wastes, are excreted by the lungs. Bile pigments, hemoglobin, red blood cells, some proteins, and some drugs are broken down by the liver for excretion. Certain metal ions, such as iron and calcium, are excreted by the colon. The sweat glands of the skin are primarily con-cerned with the regulation of body temperature, but they also serve in the excretion of 5 to 10 percent of the metabolic wastes formed in the body. Sweat and urine have similar composition (water, salts, urea and other organic compounds) but the former is much more dilute than the latter, having only about one eighth as much solute matter. The volume of perspiration varies from about 500 ml. on a cool day to as much as 2 to 3 liters on a hot day. While doing hard work at high temperatures, a man may excrete from 3 to 4 liters of sweat in an hour.

Question:

The freezing point constant of toluene is 3.33\textdegreeC per mole per 1000 g. Calculate the freezing point of a solution prepared by dissolving 0.4 mole of solute in 500 g of toluene. The freezing point of toluene is - 95.0\textdegreeC.

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Solution:

The freezing point constant is defined as the number of degrees the freezing point will be lowered per 1000 g of solvent per mole of solute present. The freezing point depression is related to this constant by the follow-ing equation. freezing pt depression =molalityof solute × freezing ptconstant Themolalityis defined as the number of moles per 1000 g of solvent. Here, one is given that 0.4 moles of solute are added to 500 g of solvent, therefore there will be 0.8 moles in 1000 g. (0.4 moles / 500g) = (0.8 moles / 1000 g) Themolalityof the solute is thus 0.8 m. One can now find the freezing point depression. The freezing point constant for toluene is 3.33\textdegree. freezing point depression =molality× 3.33\textdegree = 0.8 × 3.33\textdegree = 2.66\textdegree The freezing point of toluene is thus lowered by 2.66\textdegree. freezing point of solution = (- 95\textdegreeC) - 2.66\textdegree = - 97.66\textdegreeC.

Question:

An Eskimo is about to push along a horizontal snowfield a sled weighing 57.6 lbs carrying a baby seal weighing 70 lbs which he has killed while hunting. The coefficient of static friction between sled and seal is 0.8 and the coefficient of kinetic friction between sled and snow is 0.1. Show that the maximum horizontal force that the Eskimo can apply to the sled without losing the seal is 114.8 lbs. Calculate the acceleration of the sled when this maximum horizontal force is applied.

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Solution:

The seal will slide off the sled when the acceleration provided by P\ding{217} is greater than the maximum acceleration which f_2\ding{217} can provide to the seal. (See figure (A)). The acceleration of the seal is the same as the acceleration of the seal-sled system. (See figure (B) Using Newton's Second Law to calculate the latter, we obtain P - f_1 = (m + M)a(1) where a is the acceleration of the system, taken as positive in the direction of P. The frictional force law is f_1 = u_1N_1(2) where u_1 is the coefficient of kinetic friction between sled and ice, and N1 is the normal force of the ice on the sled-seal system. Sub-stituting (2) in (1) P - u_1N_1 = (m + M)a Since the system is in vertical equilibrium N_1 = (m + M)g and P - u_1(M + m)g = (M + m)a Finally, a = [{P - u_1(m + M)g} / {m + M}](3) is the acceleration of the sled-seal system. Now, applying the Second Law to the system consisting of seal alone, (see figure (B)) we obtain f_2 = ma(4) where a is the acceleration in (3). But, since we require the seal to remain at rest on the sled, f_2 \leq u_2N_2(5) where u_2 is the coefficient of static friction between sled and seal, and N2 is the normal force of the sled on the seal. Inserting (5) in (4) ma \leq u_2N_2(6) Substituting (3) in (6) m[{P - u_1(m + M)g}/(m + M)] \leq u_2N_2 Since the seal is in equilibrium vertically, N_2 = mg and m[{P - u_1(m + M)g} / {m + M}] \leq u_2mg Solving for P P - u_1(m + M)g \leq u_2(m + M)g P \leq (u_1 + u_2)(m + M)g P \leq (.9)(57.6 + 70) lbs. P \leq 114.8 lbs. The maximum value of P is 114.8 lbs.

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Question:

Following fertilization, the zygote begins to cleave. Describe the different cleavage patterns found in the animal kingdom.

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Solution:

Cleavage cells are known as blastomeres. They vary in size and content, principally by reason of differences in the amount and distribution of the yolk and other cytoplasmic inclusions which they contain. In isolecithal eggs, the cleavage cells are of approxi-mately the same size. In telolecithal eggs, such as those of birds, the yolk is more abundant and is con-centrated toward that area of the egg known as the vegetal pole. The opposite side of the egg, an area where the nuclear material is located, is called the animal pole. In this case, the blastomeres nearer the animal pole tend to be smaller and are called micromeres. The cells near the vegetal pole are usually larger and are termed macromeres. As long as the entire egg divides into cells, the cleavage is said to be complete, or holoblastic. In the case of sharks, reptiles, and birds, however, yolk is abundant and fills the egg, except for a thin disc of cytoplasm at the animal pole and an even thinner layer of cytoplasm around the periphery of the egg. In such eggs, cleavage is incomplete and is confined to a small area which surrounds the animal pole. The rest of the egg remains uncleaved. Such incomplete cleavage is termed meroblastic. In the centrolecithal eggs of arthropods, the yolk is concentrated toward the center. After the nucleus divides several times, the offspring nuclei migrateto the periphery of the egg where meroblastic cleavage of the peripheral cytoplasm takes place. There are three basic patterns of cleavage in the animal kingdom: radial cleavage, spiral cleavage, and superficial cleavage. In radial cleavage, the first mitotic spindle elongates in a direction at right angles to the egg axis. The second spindle also elongates transversely to the egg axis and at right angles to the first spindle. The third cleavage spindle elongates in a direction parallel to the egg axis. Consequently, a ball-like stage of eight cleavage cells is formed. This type of cleavage is found in the echinoderms and chordates. In spiral cleavage, the mitotic spindles are oblique to the polar axis of the embryo, and give rise to a spiral arrangement of newly formed cells. This type of cleavage is found in the annelids, mollusks and in some flat- worms. A third type of cleavage, associated with arth-ropods, is known as superficial cleavage. This involves mitotic division of the nucleus without any cleavage, resulting in an uncleaved egg containing a large number of nuclei within the center. Later these nuclei migrate to the periphery where meroblastic cleavage of the egg takes place. The rate of cleavage in all three cases is rapid at first and then slows a bit, and can be regulated by temperature changes.

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Question:

Describe the process of gamete formation and fertilization in the angiosperms.

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Solution:

Reproduction in the flowering plants begins with the development of the gametes. The female gametes, or megaspores, develop within the ovules of the ovary, each ovule being attached to the ovary by a stalk. Embedded deep within the ovule is one cell, called a megasporangium, which enlarges to become the megaspore mother cell (see Figure 1), from which the gametes will be formed. The megaspore mother cell undergoes meiosis to form four megaspores. In most species, three of these disintegrate, leaving one functioning megaspore. This surviving cell then undergoes three mitotic divisions, producing eight nuclei which migrate so that three position themselves at the far end of the now enlarged megaspore and form the antipodal cells; three move towards the micropyle and form the egg and synergid cells. The synergid and antipodal cells are short-lived; their function is obscure. The remaining two central nuclei are the polar nuclei. The entire mature structure is termed the embryo sac, and becomes enclosed by the integu-ment, layers of cells which develop from the megasporangia surrounding the megaspore mother cell. The development of the sperm or microgametophytes begins within the tissues of the anther. Each anther typically contains four pollen sacs or microsporangia (see Figure 2). Early in its development, accelerated cell division in the microsporangia produces numerous microspore mother cells. Each of these microspore mother cells undergoes meiosis, resulting in a tetrad of four microspores. Each microspore then undergoes a mitotic division to form a tube nucleus and a generative nucleus. At this point the structure is termed a pollen grain. When the pollen is mature, the anthers split open and shed the many pollen grains produced. Pollination may now occur by any one of various methods, resulting in the deposition of the pollen onto in the stigma of either the same flower (self-pollination) or another flower (cross--pollination). Upon successful pollination, the pollen produces a pollen tube which digests its way from the stigma down through the style (see Fig. 3) to the ovule. If it has not done so already, the generative nucleus will divide to form two functioning sperm during the journey down the pollen tube to the embryo sac, where they will be released. One sperm will fertilize the egg; the other will migrate and fuse with the two polar nuclei to form the triploid endosperm nucleus. The number of polar nuclei, however, may vary between species, with a consequent variation in the chromosome number of the endosperm. This process of the fusion of sperm nuclei with both the egg and polar nuclei is known as double fertilization. The fertilized egg will develop into the embryo, and the endosperm nucleus into the endosperm, or nutritive tissue for the embryo. The embryo undergoes its first stages of development within the ovary, but eventually becomes dormant and is shed with its surrounding tissues as a seed.

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Question:

A 100-gram piece of ice at 0\textdegreeC is added to 400 grams of water at 30\textdegreeC. Assuming we have a perfectly insulated calorimeter for this mixture, what will be its final temperature when the ice has all been melted?

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Solution:

The heat gained by the ice must equal the heat lost by the water. In addition energy is lost because the ice changes from the solid state to the liquid state, which requires the addition of the heat of fusion (which is 80 cal/gm for ice). The heat gained by the ice is H_g. H_g = (mass x heat of fusion) + (mass × specific heat of ice × change in temperature) =mL+mc∆t H_g = 100 gm × 80 cal/gm + 100 gm × (1 cal/gm × \textdegreec) × (t - 0\textdegree)C where t is the final temperature. The heat lost by the warm water is H_i H_1 =mc∆t H_1 = 400 gm × (1 cal/gm × \textdegreeC) × (30\textdegree - t\textdegree)C Since heat gained must equal heat lost, 100 × 8,000 cal + 100t\textdegree cal = 12,000 - 400t\textdegree cal whence (100 + 400)t\textdegree cal = 12,000 cal - 8000 cal and t = 8 degrees centigrade,

Question:

Photochemicalsmogsare thought to be partially initiated by Photochemicalsmogsare thought to be partially initiated by the photolytic decomposition of NO 2 to form NO and O. When the subsequent reaction 2NO + O 2 \rightarrow 2N0_2 is studied kinetically, it is found that when the initial concentration of O_2 is doubled and that of NO held constant, the initial reaction rate doubles; and when the initial concentration of NO is doubled and that of O 2 held constant, the initial reaction rate quadruples. Write the rate expression for this reaction.

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Solution:

The rate of reaction is equal to some rate constant k multiplied by the concentrations of NO and O_2 raised to the appropriate powers. That is, rate = k [NO]^m [O_2]^n where the exponents m and n are to be determined. When [NO] is held constant and [O_2] is doubled, the rate is doubled. Hence, the rate is directly proportional to [O_2] and n = 1. Thus, if we were to hold [NO] constant and triple [O_2] the rate is tripled. When [O_2] is held constant and [NO] is doubled, the reaction rate is quadrupled. Hence, the rate is second order in [NO] and m = 2. Thus, if we were to hold [O_2] constant and triple [NO], the rate is multiplied by 32= 9. Substituting m = 2 and n = 1 into the rate expression above gives rate = k [NO]^2 [O_2]^1, or, rate = k [NO]^2 [O_2].

Question:

The starter motor in an automobile draws a current of 200A. from a 12-V car battery. What is the electrical power consumed by the motor? If the motor has an efficiency of 80 per cent, how much mechanical power is done by the starter motor?

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Solution:

Electrical power is the rate at which electrical energy is produced or consumed. It is equal to the product of the current in a device and the fall of potential across its terminals or, for the motor, P = VI = (1.2 × 10^1V)(2 × 10^2A) = 2.4 × 10^3W. Since the motor only has an efficiency of 80%, 20% of the available power goes into losses such as heat. The other 80% is available to the motor to do mechanical work. The mechanical power delivered by the motor, is (2.4 × 10^3W)(8 × ^.10^\rule{1em}{1pt}1) = 1.92 × 10^3W.

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Question:

A sample of bristle cone pine wood of age 7,000 \pm 100 years, known by counting growth rings, possesses an activity of 6.6 d(disintegrations) / minute-g of carbon. Calculate the exact age of the wood sample from radiochemical evidence.

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Solution:

Radioactive decay is related to the rate of disintegration and the half-life by the following equation: q_t = q_0 (0.5) q_t = q_0 (0.5) t/(t)1/2 where q_t is the rate of disintegration of the substance con-taining carbon, q_0 is the rate of disintegration of carbon (14 d / min-g), t is the time elapsed during the reaction, and t_1/2 is the half-life of the element concerned. For carbon, t_1/2 = 5730 yrs. Here one needs to solve for t. q_t = q_0 (0.5)^t/t(1/2)q_t= 6.6 d / min - g q_0= 14 d / min - g t= ? t_1/2= 5730 yrs 6.6 d / min - g = 14 d / min - g × (0.5)^t/5730 ^yrs [(6.6 d / min - g) / (14 d / min - g)]= 0.5^t/5730 ^yrs .471= 0.5^t/5730 ^yrs Log .471= (t / 5730 yrs) × log 0.5 - .327 = (t / 5730 yrs) × (- .301) (- .327 / - .301)= (t/ 5730 yrs) t = [(.327 × 5730 yrs) / .301]= 6225.0 yrs. Therefore, the tree is 6,225 years old.

Question:

City A is 100 miles north and 200 miles west of city B. An airplane flies in a direct line between the cities in a time of one hour. What are the vectors that describe the distance of A from B, and the velocity of the airplane?

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Solution:

We will define first a coordinate system with B at the origin (see the figure below). The x-direction is east and the y-direction is north. The vector BA is specified by its coordinates x = -200 mi y = 100 mi or by its magnitude and direction (BA)^2= x^2 + y^2 = [(-200)^2 + (100)^2]mi 2 BA = 100\surd5 mi sin \texttheta = (CA/BA) = (100/100)\surd5\approx(1/\surd5) \approx \texttheta = 26.5\textdegree \textphi = 180\textdegree - \texttheta = 153.5\textdegree The velocities are given in a similar way. Since they are constant V_x = (x/1 hr) = (-200 mi/1 hr) = - 200 mi/hr V_y = (y/1 hr) = (100 mi/1 hr) = 100 mi/hr V^2 = V^2_x + V^2_y = [(-200)^2 + (100)^2]mi^2/hr^2 V = 100\surd5 mi/hr \textphi = 153.5\textdegree

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Question:

A chemist wishes to determine the molecular weight and molecular formula of fructose (a sugar). He places .946 g of it in 150 g of H_2O (water) and finds that the freezing point of water is depressed to -0.0651\textdegreeC. Determine the molecular weight and formula of fructose, assuming that the simplest formula of fructose is (CH_2)O.

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Solution:

To answer this question, you must know the quantitative relationship for the depression of the freezing point, ∆T_f. This relation can be expressed as ∆T_f=kf× m, where ∆T_f= the actual depression,k_f=molalfreezing point constant (-1.86 deg mol^-1 for water) and m =molalityof the solution. Molalityis defined as moles of solute per 1 kg of solvent. In this problem, water is the solvent and the fructose compound is the solute. As such, you can express the freezing point relationship as grams solute ∆T_f=k_f, × [(gramssolutet) / (Molecular weight)] / [(grams solvent)] ×1000, where the numerator is the number of moles of solute. You possess all of the unknowns, except the molecular weight, which can be calculated. In other words, solve for the molecular weight by substitution. Hence. -0.0651\textdegreeC = (-1.86\textdegreeC / mole) × [(.946g /mol.wt.) / (150 g)] × (1000g / kg) mol wt. = (180 g / mole). To determine the molecular formula, you must figure , out what formula has a molecule weight of 180 g, and yet is still a multiple of (CH_2)O. Thus, one determines the weight of 1 (CH_2)O and divides this into 180 g. (MW of (CH_2)O = 30.) no. of (CH_2)O in fructose = (180g) / [30g(CH_2)O] = 6 (CH_2)O. Therefore, the formula for fructose is 6 × (CH_2)O or C_6H_12O_6.

Question:

Compute the sin(x) function, using the notion of the power seriessin(x) = x - (x3/3!)+ (x5/5!)- (x7/7!)+ ...

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Solution:

Power series may also be evaluated via Horner's method. When it is decided how many terms are to be evaluated, the series may betreated as a polynomial. However, this illustration will be useful to introducethe idea of precision. Since the computer performs its own roundingoperations, we want to be able to control the degree of precision. First, let us look at what can happen if we do not specify precision: CCOMPUTE EIGHT TERMS OF THE POWER SERIESSIN(Pl/4) CAND PRINT OUT SIN(X), EACH TERM, ANDSIN(Pl/4) REAL PI, XSQ, SINE, TERM, REALI INTEGER I PI = 3.141592 X = PI/4.0 XSQ = X\textasteriskcentered\textasteriskcentered2 WRITE (5,100) 100FORMAT (1X, 'SIN(X)', 'TERM') TERM = X SINE = TERM. CDO LOOP IS INDENTED FOR CLARITY DO 20 I = 2, 16, 2 WRITE (5,101) SINE, TERM 101FORMAT (1X, E10.9, 14X, E10.9) REALI = I\textasteriskcentered(I+1) TERM = -1.0\textasteriskcentered(TERM\textasteriskcenteredXSQ)/REALI SINE = SINE + TERM 20CONTINUE WRITE (5,102) X 102FORMAT (1X,'VALUE OFSIN(PI/4) EQUALS ',E10.9) STOP END Sample output: SIN(X)TERM 0.7853980E 000.7853980E 00 0.7046525E 00-0.8074546E-01 0.7071429E 000.2490392E-02 0.7071063E 00-0.3657614E-04 0.7071066E 000.3133609E-06 0.7071066E 00-0.1757242E-08 0.7071066E 000.6948429E-11 0.7071066E 00-0.2041018E-13 VALUE OFSIN(PI/4) EQUALS 0.7071069E00 Notice that after the fifth term, the value of sin(x) does not change. We choose the number of terms to be evaluated, but the computer will roundoff each term, thereby creating an accumulation of errors. The conventionchosen here to avoid this error propagation is the DOUBLE PRECISION statement.Let us declare DOUBLE PRECISION SINE, TERM at the beginning of the program.Now, SINE and TERM will be representedinternally by a bit string which is twice as long as in single preci-sion. The round-off error will accumulate in the least significant bits of thedouble precision number. Finally, when the result is re-turned to single precisionat the end of the computation, only one round-off operation is performed. The other modification needed in double precision is in the FORMAT statement: the letter D must re-place the exponential notation E.

Question:

A molecule has no net charge, but its charge dis-tribution may be equivalent to two charges \pm q separated by a small distance a. Then the molecule is said to be a polar molecule, and to have a dipole moment \mu =qa. In an electric field E such molecules, originally of random orientation, will tend to align themselves either parallel orantiparallelto the field. If parallel the potential energy of the molecule is -\muE, ifantiparallel+\muE. Calculate(a) the ratio of the number of molecules in each energy level to the total number, (b) the average energy per molecule of a system of HC1 molecules, if the dipole moment of HC1, \mu, is 3.44 × 10^-30coul-m, and the gas is put into a uniform electric field E = 1.5 × 10^7 V/m, at T = 350 K\textdegree.

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/Users/wenhuchen/Documents/Crawler/Physics/D33-0999.htm

Solution:

(a) We have a two energy level problem. The number of molecules with energies +\muEand -\muErespectively are (N_1/N) = [1/(1 + e ^2E/KT)] (N_2/N) = [(e ^2E/KT)/(1+ e ^2E/KT)] where K is the Boltzmann constant. (\muE/KT) = [{(3.44 × 10^-30coul.m) (1.5 × 10^7 V/m)} / {(1.38 × 10^-33 J/K\textdegree) (350 K\textdegree)}] = 1.07 × 10^-2 Since\muE/KT < < 1, we use the Taylor Series approxima-tion fore^\muE^/KT e^\muE^/KT \approx 1 + (\muE/KT) Thus (N_1/N) \approx [1/(2 + 2 (\muE/KT)] = [1/(2 + 0.0214)] = 0.495 (N_2/N) \approx [{1 + (2\muE/KT)}/{2 + 2 (\muE/KT)}] = [(1 + 0.0214)/(2 + 0.0214)] = 0.505 (b) If U is the internal energy of the gas, the average energy per unit molecule is given as E= (U/N) = [{-\muENtanh(\muE/KT)}/N] = -\muEtan h (\muE/KT) For (\muE/KT) < < 1, we can use the approximation tan h (\muE/KT) \approx(\muE/KT) ThereforeE\approx -\muE\textbullet (\muE/KT) \approx - (3.44 × 10^-30coul.m) (1.5 × 10^7 V/m) × 1.07 × 10^-2 \approx - 5.5 × 10^-25 J.

Question:

Determine themolarityof a 40.0% solution ofHClwhich has a density of 1.20 g/ml.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E08-0287.htm

Solution:

Themolarityis defined as the number of moles of a compound present in 1 liter of solution. Here one is told that the density of this solution is 1.20 g/ml. This means that one ml of the solution weighs 1.20 g. The solution is 40.0%HCl, thus 40,0% of the 1.20 g is made up by HCland 60% by H_2O. Here one finds themolarityof theHClby: 1) determining the total weight of 1 liter of the solution, 2) calculating the weight ofHClpresent, and 3) finding the number of moles ofHClpresent in 1 liter (molarity). Solving: 1) If 1 ml of this solution weight 1.20 g, 1000 ml (1 liter) is equal to 1000 × 1.20 g. weight of solution = density × 1000 ml = 1.20 g/ml × 1000 ml = 1200 g 2) 40.0% of this weight is taken up byHCl. weight ofHCl= weight of solution × .40 = 1200 g × .40 = 480 g. 3) The molecular weight ofHClis 36.5 g/mole, thus the number of moles ofHClpresent in this 1 liter of solution is equal to the weight of the HClpresent divided by its molecular weight. no. of moles ofHClin 1 liter = [(weight ofHCl/ MW] = [(480 g) / (36.5 g/mole)] = 13,15 moles The solution is therefore 13.15 moles / liter inHClor 13.15 M.

Question:

Compute the atmospheric pressure on a day then the height of the barometer is 76.0 cm.

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/Users/wenhuchen/Documents/Crawler/Physics/D10-0416.htm

Solution:

The height of the mercury column of the barometer depends on density \rho and g as well as on the atmospheric pressure. Hence both the density of mercury and the local acceleration of gravity must be known. The density varies with the temperature, and g with the latitude and elevation above sea level. All accurate barometers are provided with a thermometer and with a table or chart from which cor-rections for temperature and elevation can be found. Let us assume g = 980 cm/sec^2 and p = 13.6 gm/cm^3. The pressure due to the atmos-phere supports the weight of mercury in the column of the barometer (see the figure). If the cross sectional area of the column is A, then the weight of mercury in the column is W = mg where m is the mass of the mercury. Since Density (\rho) = mass/volume thenW = \rho(Ah)g where V = Ah, h being the height of mercury in the column. There-fore W/A = \rhogh is the pressure due to the weight of the mercury acting downward. It must equal p_a for equilibrium to be maintained in the fluid, (see figure). Hence p_a = pgh = 13.6 gm/cm^3× 980 cm/sec^2× 76 cm = 1,013,000 dynes/cm^2 (about a million dynes per square centimeter). In British engineering units, 76 cm = 30 in. = 2.5 ft, \rhog = 850 lb/ft^3 p_a = 2120 lb/ft^2 = 14.7 lb/in^2.

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Question:

(a) With what speed must a ball be thrown directly upward so that it remains in the air for 10 seconds? (b) What will be its speed when it hits the ground? (c) How high does the ball rise?

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/Users/wenhuchen/Documents/Crawler/Physics/D03-0082.htm

Solution:

Near the surface of the earth, all objects fall towards its center with a constant acceleration g = 32 ft/sec^2. Therefore, when the ball is thrown, its speed must decrease by 32 ft/sec each second until it reaches 'its maximum height. Then it starts to fall, gaining speed at the rate of 32 ft/sec^2, and retraces its path, hitting the ground with the same speed at which it started the trip upward. This is so because the acceleration is constant and the distance traveled is the same during the rising and falling portions of the motion of the ball. Thus, the average velocity must have the same magnitude in each case and the time required to reach the maximum height must equal the time required to fall back to the ground. (a) Let the upward direction be positive. Then v_0 is positive and the acceleration a is negative. After 10 seconds, v must equal -v_0. From the kinematics equation v = v_0 + at we have for the instant before it hits the ground, -v_0 = v_0 -gt -2v_0 = -gt andv_0 =gt/2 = [(32 ft/sec^2)(10 sec)]/2 = 160 ft/sec (b) The speed of the ball when it hits the ground is -v_0 or - 160 ft/sec. (c) The height reached by the ball can be obtained by realizing that the rise of the ball must take half the total time the ball is in the air, or 5 seconds. The average velocity for this part of the motion must be v = (v-_i +v_f_-)/2 = {(160 ft/sec) + (0 ft/sec)} /2 = 80 ft/sec. The height the ball rises is then d =vt= (80 ft/sec)(5 sec) = 400 ft. This result can also be obtained using the kinematics equation d = v_0t + (1/2) at^2 Substituting values, d = (160 ft/sec)(5 sec) + (1/2) (-32 ft/sec^2)(5 sec)^2 = 800 ft - 400 ft = 400 ft.

Question:

The SLAC accelerator is capable of accelerating electrons to an energy of 4 × 10^10eV. If the number of electrons in the beam corresponds to a current of 6 x 10^\rule{1em}{1pt}5 A, (a) how many electrons arrive at the target each second and (b) how much energy, in joules, is deposited in the target each second?

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/Users/wenhuchen/Documents/Crawler/Physics/D19-0626.htm

Solution:

(a) Since a current of 1 A is defined as the flow of 1 C of charge each second, the beam current 6 × 10 ^\rule{1em}{1pt}5 A implies that 6 x 10^\rule{1em}{1pt}5 C of charge strikes the target each second. The charge of each electron is 1.6×10\rule{1em}{1pt}18c, therefore, the number of electrons N arriv-ing each second is N = (6 ×10^\rule{1em}{1pt}5 c/s / 1.6 × 10^\rule{1em}{1pt}19 C/electron ) = 3.75 x 10^14 electrons/s (b) The energy of the electrons can be expressed in joules: E = (4 × 1010eV) (1.6 × 10\rule{1em}{1pt}19J/eV) = 6.4 × 10\rule{1em}{1pt}9J Therefore, the energy deposited in the target by N electrons per second is E = (6.4 x 10^9j) (3.75 x 10^14 electrons/s) = 2.4 x 10^6 J/s. This energy consumption is about the same as the amount of electrical energy used by a town of 1,000 people.

Question:

Explain the structural and functional aspects of cilia and flagella.

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/Users/wenhuchen/Documents/Crawler/Biology/F02-0045.htm

Solution:

Some cells of both plants and animals have one or more hair-like structuresprojecting from their surfaces. If there are only one or two of theseappen-dages and they are relatively long in proportion to the size of thecell, they are called flagella. If there are many that are short, they are calledcilia. Actually, the basic structure of flagella and cilia is the same. They resemblecentriolesin having nine sets of microtubules arranged in a cylinder. But unlikecentrioles, each set is a doublet rather than triplet of microtubules, and two centralsingletsare present in the center of the cylinder. At the base of the cylinders of cilia and flagella, within the main portionof the cell, is a basal body. The basal body is essential to the functioningof the cilia and flagella. From the basal body fibers project into thecytoplasm, possibly in order to anchor the basal body to the cell. Both cilia and flagella usually function either in moving the cell, or in movingliquids or small particles across the surface of the cell. Flagella movewith an undulating snake-like motion. Cilia beat in coordinated waves. Both move by the contraction of the tubular proteins contained withinthem.

Question:

Microtubules and microfilaments both appear to be involved inintracellular motion and in intracellular structural support. How do the two organelles differ in structure andfunction andhow are they similar?

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/Users/wenhuchen/Documents/Crawler/Biology/F02-0044.htm

Solution:

Microtubules are thin, hollow cylinders, approximately 200 to 300Angstroms in diameter.Micro-filaments are not hollow, and are 50 to 80 Angstroms in diameter.Both microtubules and microfilaments are com- posedof proteins. The protein of microtubules is generally termedtubulin. Tubule proteins can be made to assemble into microtubules in a test tube, ifthe proper reagents are present. Some of the narrower microfilaments havebeen shown to be composed of proteins similar toactin.Actinis a proteininvolved in muscle cellcontration. The composition of thicker microfilamentshas not been completely determined. Microtubules are often distributed in cells in arrangement that suggesttheir role in maintaining cell shape (a "cytoskeletal" role). For example, microtubules are arranged longitudinally within the elongated processesof nerve cells (axons) and thecytoplasmicextensions (pseudopods) of certain protozoa. Microtubules are ar-ranged in a circular bandin the disc-shaped fish red blood cells. Microfilaments may also play astructural role. They are often associated with some specialized region ofthe plasma membrane, such as the absorptive plasma membrane of the intestinalcells, or the portions of the plasma membrane which serve to anchoradjacent cells together so that they can intercommunicate. Microtubules are components of cilia and flagella, and participate in therapid movements of these struc-tures. Microtubules are involved in directionalmovement within the cell during cell division, and are involved incertain types of oriented rapid intracellular movement. Microfilaments seemto be involved in many different types ofcytoplasmicmovement. They are believed to play a role in amoeboid motion of cells, and in the rapidstreaming of the cytoplasm of plant cells about a central vacuole. Close associations between microfilaments and membrane- bound organellessuggests that microfilaments assist in intracellular transport andexchange of materials.

Question:

Write a program to drill the 3 holes in the workpiece shown, and tap the hole #3 with a 3/4"-10 TAP, using a programmable turret drilling machine. Assume that the spindle speeds available are: S01, S02 and S03, viz., 250, 375 and 440 rpm., and the spindle feed rates available are F01, F02 and F03, viz., 2", 1" and 1/2" per min., resp. The tools are mounted as follows: spindle T01 carries a 1" drill, T02 carries a 41/64" drill, T03 carries a 3/4"- 10 TAP, and T04 carries a 1/4" drill. The rest posi-tion of the spindles has a Z-offset of 2", but assume that X and Y offsets are zero.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G09-0219.htm

Solution:

Zero offsets for X and Y mean that the tools are at the origin 0 at the start. A Z-offset of 2" means that the tools are 2" above the workpiece in their idle positions. From the data given, obviously, hole #1 requires spindle T01 carrying the 1" drill. We select a slower speed for this large hole, say S02, i.e., 375 rpm. The feed rate is taken as F02, i.e., 1" per min. The hole #2 requires spindle T04 carrying a 1/4" drill. The speed can be faster, say, S03, i.e., 440 rpm., and the spindle feed rate can be F01, i.e., 2". The hole #3 requires spindle T02 first, and then spindle T03 for the tap. While tapping, we will also require a coolant fluid feature, which is obtained by specifying M06. The program for the machining of the workpiece can be written as follows: OP.NO. `G' `X' `Y' `R' `Z' Auxiliaries N001 G80 X02000 Y01000 S02 F02 T01 G82 R01750 Z02250 N002 G80 X04000 Y01000 S01 F03 T04 G81 R01750 Z03500 N003 G80 X06000 Y02000 S02 F04 T02 G81 R01750 Z03500 N004 G80 X06000 Y02000 S03 F03 T03 M06 G84 R01750 Z02500 G80 X00000 Y00000 M30 In step No. 001, the G80 clears any stray instructions lying about in the storage. The spindle T01, holding a 1" drill moves to the X and Y coordinates. The G82 command makes the spindle rotate at the specified speed S and the spindle descends at a rapid rate by 1-75", so that it is still 1/4" above the workpiece. The spindle then descends at a programmed feed rate F02, for a distance of 1/2" (Z-R = 2.25 - 1.75 = 0.5"). In operation #N002, we perform a G80, change the spindle to T04, move to the X and Y specified, and lower the spindle as specified by R and Z. We perform a G81 drilling operation which differs from the G82 in that in G81 the spindle is withdrawn immediately rather than dwell-ing at the end of the drilling as in G82. In N003, we again do a drilling operation, but by us-ing the T02 spindle which carries a 41/64" drill. Note that as Z = 3.5", the drill goes completely through the 1" thick workpiece and comes out 0.5" on the other side. In N004 we keep X and Y same, and so, perform a tapp-ing operation using spindle T03. Note that the M06 starts the coolant during the tapping operation. Finally, the G80 in the last step clears all previous commands, and moves the spindle to the edge of the work-piece so as to facilitate removal of the workpiece. The M30 command signifies an end of the tape on which the program will be punched, and is also a command to the tape reader on which the tape will be later mounted, to rewind the tape to the starting position so that it is ready to be re-run for the machining of the next workpiece.

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Question:

(a) encode decimal 11 (b) determine if each of the following received words was transmitted correctly, and then decode the word: (i) 0110011 (ii) 0110001 (iii) 0010011

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G06-0130.htm

Solution:

The Hamming system is a type of the parity check system. For a four-bit datum, for example, the Hamming system, generates three check bits. Let the datum be given by (D_0,D_1,D_2,D_3) and the check bits be given by (C_1,C_2,C_3). Then, if +_2 represents mod 2 addition (exclusive-OR sum), then C_1 = D_0 +_2 D_2 +_2 D3(1) C_2 = D_0 +_2 D_1 +_2 D3(2) C_3 = D_0 +_2 D_1 +_2 D2(3) The check and data bits must be interspersed in the correct manner for the decoding and verification to occur correctly. The check digitC_iis placed in bit position 2^(i-1) Bit Position:1234567 Occupying bit: C_1C_2-C_3--- The remaining bit positions are filled sequentially by filling the lowest unfilled bit position with the data bit of highest index: Bit Position:1234567 Occupying bit: C_1C_2D_3C_3D_2D_1D_0 A received word is checked by generating the associated syndrome. For a 4-bit datum encoded into a seven-bit word, the syndrome will be 3 bits (S_3,S_2,S_1), such that S_1 = C_1 +_2 (D_0 +_2 D_2 +_2 D_3)(4) S_2 = C_2 +_2 (D_0 +_2 D_1 +_2 D_3)(5) S_3 = C_3 +_2 (D_0 +_2 D_1 +_2 D_3)(6) Substituting equations (1) - (3) into (4) - (6), respectively, one sees that: S_1 = (D_0 +_2 D_2 +_2 D_3) +_2 (D_0 +_2 D_2 +_2 D_3) = 0 S_2 = (D_0 +_2 D_1 +_2 D_3) +_2 (D_0 +_2 D_1 +_2 D_3) = 0 S_3 = (D_0 +_2 D_1 +_2 D_2) +_2 (D_0 +_2 D_1 +_2 D_2) = 0 if no error has occurred. (Recall that in mod 2 addition if a = b, then a +_2 b = 0) . If an error has occurred, changing either a data or check bit, then one or more of the syndrome bits would not be 0. Equations (4) - (6) can be written as the matrix equation: S = Y . H(7) where S is the vector [S_3 S_2 S_1 ], Y is the received word vector [C_1 C_2 D_3 C_3 D_2 D_1 D_0], \bullet is matrix multiplication (using mod 2 addition),and \mid 001 \mid \mid 010 \mid \mid 011 \mid H =\mid 100 \mid \mid 101 \mid \mid 110 \mid \mid 111 \mid is the parity check matrix. Note that the rows of the parity check matrix are simply the binary representations of decimals 1-7, written in sequence. (a) The four-bit binary representation of decimal 11 is 1011. Substituting into equations (1) - (3) with (D_0,D_1,D_2, D_3) = (1,1,0,1) one D_1,D_2, D_3) = (1,1,0,1) one gets: C_1 = 1 +_2 0 +_2 1 = 0 C_2 = 1 +_2 1 +_2 1 = 1 C_3 = 1 +_2 1 +_2 0 = 0 Therefore, the Hamming code word for decimal 11 is: Bit Position:1234567 Occupying bit:0110011 or 0110011. (b) (i) Substituting into equation (7), with Y = [0110011 ],one gets: \mid 001 \mid \mid 010 \mid \mid 011 \mid S = [0110011 ]\bullet\mid 100 \mid \mid 101 \mid \mid 110 \mid \mid 111 \mid = [0 00] indicating that the message was transmitted correctly. To decode the message, the information bits from the code word must be extracted. Remembering that bit positions 1, 2 and 4 contain the check bits, we extract the information bits from bit positions 3,5,6 and 7 as 1011 = decimal 11. Checking this with the answer obtained in part (a), one can see that the received word was indeed transmitted correctly. (ii) Substituting into equation (7), with Y = [0 110001], one gets: \mid 001 \mid \mid 010 \mid \mid 011 \mid S = [0110011 ]\bullet\mid 100 \mid \mid 101 \mid \mid 110 \mid \mid 111 \mid = [1 1 0 ] \not = [ 000] indicating that an error has occurred. Since this error has effected both S_3 and S_2 but not S_1,the error bit must be the one that occurs in both the occurs in both the computations of S 3 andS2but not S 1 .lookingat equations (4) - (6) one at equations (4) - (6) one sees that the only bit that appears in (5) and (6) but not (4) is D_1. Hence the error bit must be D_1. Note that D_1 is in bit position 6. If S = [1 1 0] is taken as the binary representation of a decimal integer, then binary 110 = decimal 6. If an error has occurred, the syndrome will always point to the bit position of the error. Complementing the bit in position 6, yields: received word:0 110001 corrected word:0 110011 The corrected word is decoded by extracting bits 3,5,6, and 7 as the information bits and converting from binary to decimal: binary 1 0 11= decimal 11. Comparing this result with that of part (i), one sees that, if this were the transmitted word, an error in bit 6 would indeed yield the word received. (iii)Substituting into equation (7), with Y = [0 0 1 0011], one gets: \mid 001 \mid \mid 010 \mid \mid 011 \mid S = [0010011 ]\bullet\mid 100 \mid \mid 101 \mid \mid 110 \mid \mid 111 \mid = [1 1 0 ] \not = [ 000] indicating that an error has occurred. Since binary 0 1 0 = decimal2, the error is in bit position 2, the bit containing C_2. Thus, even errors in the check digits can be detected and corrected. Complementing the bit in position 2, yields: received word:001 0011 corrected word:0 110011 Extracting bits 3,5,6 and 7 as the information bits and converting from binary to decimal one gets: binary 1 0 11= decimal 111. Comparing this result with that of part (i), one sees that if this was the transmitted word, an error in bit 2 would indeed yield the word received. Again, the encoding, decoding and error-correcting scheme is automatically implemented by the machine hardware.

Question:

Derive the formulas for the period and the frequency of a simple harmonic oscillator.

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/Users/wenhuchen/Documents/Crawler/Physics/D09-0383.htm

Solution:

A simple harmonic oscillator in one dimension is characterized by a mass which is acted upon by a force, F, proportional to and directed opposite to its displacement x: F = -kx(1) From Newton's second law: F = ma =m[(d^2x)/(dt^2)] = -kx, [(d^2x)/(dt^2)] =-(k/m)x(2) From the theory of differential equations, a solution for x is: x(t) = Acos(\omegat+ \delta)(3) where A, \omega, and \delta are constants. The constants A and \delta are arbitrary. However, \omega depends upon the physical characteristics of the oscillator, m and k. We can solve for \omega by inserting the expression for x given in equation (3), into equation (2). We must first calculate the second derivative of x with respect to t: dx/dt= - \omega A sin (\omegat+ \delta), [(d^2x)/(dt^2)] = - \omega^2 Acos(\omegat+ \delta) Then, making the substitution: -\omega^2Acos(\omegat+ \delta) = -(k/m) Acos(\omegat+ \delta) Thus,\omega^2 = k/m,\omega =\surd(k/m) Next, we can prove that the oscillator goes through one complete cycle in time T = 2\pi/\omega, by showing that at time t + 2\pi/\omega the displacement x of the oscillator will be the same as at time t: x[t + (2\pi/\omega)]= Acos[\omega{t + (2\pi/\omega)} + \delta] = Acos(\omegat+ 2\pi + \delta) = Acos[(\omegat+ \delta) + 2\pi] = Acos[(\omegat+ \delta) The last step is justified from the laws of trigonometry which state that: cos(y + 2\pi) =cosy where we let y =\omegat+ \delta. The time T it takes the oscillator to complete one cycle is called the period and is represented by T. Thus: T = 2\pi/\omega = 2\pi\surd(m/k) The frequency of oscillation ѵ, is the number of cycles the oscillator goes through per unit time. This is the reciprocal of the period: cycles/sec= 1/(sec/cycle) ѵ = 1/T = \omega/2\pi = 1/2\pi\surd(k/m)

Question:

A solid has a volume of 2.5 liters when the external pressure is 1 atm. The bulk modulus of the material is 2 × 10^12 dynes/cm^2. What is the change in volume when the body is subjected to a pressure of 16atm? What additional energy per unit volume is now stored in the material?

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Solution:

The bulk modulus is defined as the ratio of the net excess pressures ∆p acting on a body and the volume strain ∆V/V_0 resulting from this stress, where V_0 is the original volume. Hence, B = -[(∆P/V)/V_0] B is > 0 since, if ∆p > 0, ∆V < 0 and the solid is compressed ∆p = (16 - 1) = 15atm= 15 × 1.013 × 10^6 dynes cm^-2. Hence the decrease in the volume of the solid is -∆V = ∆pV_0/B = (15 × 1.013 × 10^6 dynes/cm^2 × 2.5 × 10^3 cm^3)/(2 × 10^12 dynes/cm^2) = 18.99 x 10^-3 cm3. For any small change of pressuredp, there will be a change of volumedV, anddp= -BdV/V. The work done on the system in that change and the energy stored in the material isdW= -pdV= (V/B) pdp. In the change mentioned in the question the total work done is W = - ^V2\int_V1 pdV= ^p2\int_p1 (V/B) pdp. The minus sign results because the pressure on the gas opposes the volume changedV. The volume V changes in the process, and account should be taken of this in the integration. In fact the change ∆V is negligible in comparison with V_0, and V may be treated as a constant throughout. Hence W \approx (V_0/B) ^p2\int_p1 pdp= (1/2) (V_0/B) (p^2_2 - p^2_1) The extra energy stored per unit volume is thus W/V_0 =(1/2B) (p^2_2 - p^2_1) = (1/2) × [1/(2 ×10^12 dynes/cm^2)] (16^2 - 1^2) atm^2 × 1.013^2 × 10^12 dynes^2/cm^4 - atm^2 = 65.4 ergs/cm^3.

Question:

Describe two ways in which taxes can be guided.

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/Users/wenhuchen/Documents/Crawler/Biology/F31-0794.htm

Solution:

A taxis is a directed type of orientation in which an animal moves toward or away from a stimulus. But exactly how does the animal detect which direction to move in relation to the stimulus? One simple way is by testing the intensity of the stimulus at different short intervals of time: The animal simply takes momen-tary readings of the stimulus and directs itself toward the strongest reading. If it accidentally moves away from the source, it changes direction until a higher intensity of the stimulus is sensed. For example, dogs locate the source of an odor by repeatedly sniffing as they move. They periodically change their direction so as to continue smelling a high intensity of the odor. A second, more complex taxis guidance system is the instantaneous reading of the stimulus using two receptor organs. Many animals move in response to a stimulus by positioning themselves so that equal intensities of the stimulus are sensed by the two organs. They are then precisely aligned to move either toward or away from the stimulus, depending on whether they are attracted or repelled by it. For example, an animal can orient toward a light source by moving its body, or simply its head, until an equal intensity of illumination is sensed in both eyes. This quidance system can be demonstrated by covering one eye, for instance, the right eye, of a pos-itively phototactic animal. Since the animal senses all the light in its left eye, it moves toward the left. However, its next instantaneous reading directs the animal to the left again, since a balance of illumination was made impossible. This causes the animal to turn to the left indefinitely, resulting in movement in conterclockwise circles! Another example is the phototaxis of a planarian. If two equally bright lights are located at equal distances from the worm, it will move toward a pt. midway between the two lights (see Figure 1) The orientation of the honeybee toward an odor is also one involuing two simultaneous readings of a stimuli. The honeybee tests the strength of an odor simultaneously with both antennae, and then turns in the direction of the antenna that sensed the stronger odor. If an investigator crosses the antennae and glues them into this position, the bee reads the odor as coming from the opposite direction. Should such a bee be placed in a Y-maze with an odor source in one channel, the bee will detect the source as originating from the other channel and move in that direction. (See Figure 2)

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Question:

How much energy in electron volts must be expended to separate the atoms in the potassium iodide molecule, 39_k 127_I, if the ions are originally separated by a distance R = 3.04 × 10^\rule{1em}{1pt}10 m? The potential energy for coulomb's law of force is, E_P = \rule{1em}{1pt}ke^2 / d where e is the electric charge of each ion and d is the distance between the charges. K = 9 × 10^9 N \textbullet m^2 / c^2

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/Users/wenhuchen/Documents/Crawler/Physics/D18-0603.htm

Solution:

When the ions are bonded together their separation distance is R, so the potential energy is, E_P(R) = \rule{1em}{1pt}ke^2 /R = \rule{1em}{1pt}{[{9 × 109N(m^2/ C^2)}(1.6 × 10^\rule{1em}{1pt}19c) 2 ] / (3.04 × 10^\rule{1em}{1pt}10m)} =\rule{1em}{1pt}7.6 × 10^\rule{1em}{1pt}19 J When the ions are very far apart, the separation distance d becomes very large and may be assumed to be infinite. The potential energy for infinite separation distance is zero: E_P (\infty) = \rule{1em}{1pt}ke^2 / \infty = 0 The binding energy of the molecule is the difference between these two potential energies (the amount of energy needed to breakup the molecule). Binding energy = Ep (\infty) \rule{1em}{1pt}E_P(R) = 7.5 × 10^\rule{1em}{1pt}19 J Since I eV = 1.6 × 10^\rule{1em}{1pt}19 J, the energy to separate the atoms may be written, (7.6 × 10^\rule{1em}{1pt}19J)/ (1.6 × 10^\rule{1em}{1pt}19J/ eV) =4.8 eV

Question:

Explain the terms primary, secondary, and tertiary in regards to covalent bonding in organic compounds.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E20-0724.htm

Solution:

If a carbon atom is bound to only one other carbon atom, then the former carbon atom is called primary. If a carbon atom is bonded to two other carbon atoms, then that carbon atom is called secondary. If a carbon atom is bonded to three other carbon atoms, then that carbon atom is called tertiary. Any group that is attached to a primary, secondary, or tertiary carbon is called a primary, secondary, or tertiary group. For example, Carbons 1, 2, and 5 are primary, carbon 4 is secondary, and carbon 3 is tertiary. The hydrogen atoms attached to carbons 1, 2, and 5 are called primary hydrogens, those attached to carbon 4 are called second-ary hydrogens, and those attached to carbon 3 are called tertiary hydrogens. This same principle applies to alcohols, and depending upon where the hydroxyl group (-OH) is attached (that is, primary, secondary, or tertiary carbon), the alcohol is called primary, secondary, or tertiary, respectively.

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Question:

Are all antibodies structurally and functionally similar? Explain.

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/Users/wenhuchen/Documents/Crawler/Biology/F14-0361.htm

Solution:

All antibodies are proteins and are referred to as immunoglobulins. Five major classes of immunoglobulins can be distinguished based on their character-istic structures and separate functional roles: IgG, IgM, IgA, IgD and IgE. IgG immunoglobulins constitute about 80% of all the antibodies in the blood serum. These are the major anti-bodies involved in the destruction of invading micro-organisms. IgM immunoglobulins appear first in response to an antigen; they are later replaced by IgG. IgA im immunoglobulins are found in secretions such as saliva and tears and are involved in the defense of body cavities such as the mouth and vagina. The function of IgD is still unknown, while IgE is thought to be involved in allergic reactions. There are thus five basic types of immunoglobulins. But we are certainly exposed to more than five different antigens. What distinguishes one immunoglobulin molecule from the next? Let us look at the structure of an IgG molecule. It is composed of four polypeptide chains; two heavy chains (mol. wt. \sim 53,000) and two light chains (mol. wt. \sim 22,500). In a given IgG molecule, both heavy chains are identical and both light chains are identical. The polypeptide chains are held together by disulfide (\rule{1em}{1pt}S\rule{1em}{1pt}S\rule{1em}{1pt}) linkages to form a Y-shaped molecule: The ends of each arm of the IgG molecule contain an active site which can combine with a specific antigen. Thus, two identical binding sites permit each antibody molecule to form a complex with two antigens. Both light and heavy chains contain two regions: a "constant" region which has very similar amino acid sequences in all immuno-globulins, and a "variable" region which has different amino acid sequences for each type of antibody. This variability accounts for the specificity of antibodies- that is, one particular antibody molecule can bind only one particular kind of antigen. The specific sequence of amino acids in the variable region at the end of each arm of the antibody molecule determines the spatial configuration of the active site which will only bind one particular antigen.

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Question:

One of the major atmospheric pollutants emitted by fuel combustion power stations is a sulfur oxide mixture general-ly designated as S0_x. It consists mainly of SO_2 but may contain anywhere from 1 to 10% SO_3 computed as percent by weight of the total mass. What would be the approximate range to assign to x?

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Solution:

One finds x by determining the number of 0 atoms present for each S present. For purposes of calcula-tion assume that one always has 100 g of the mixture of S0_2 and SO_3. Thus when 1% of the mixture is SO_3, there will be 1 g of SO_3 present and 99 g of SO_2 present. One then finds the number of moles present of each element. The ratio of S : 0 is then determined. Solving for x when 1% of the mixture is SO_3: (MW of SO_2 = 64, MW of SO_3 = 80). The number of moles present is equal to the number of grams present divided by the molecular weight. For SO_3 : (1g)/(80 g/mole) = 1.25 × 10^-2 moles For SO_2 : (99 g)/(64 g/mole) = 1.55 moles. Because there is one S present in each compound the total amount of S is 1.55 moles + 1.25 × 10^-2 moles or 1.56 moles. In 1 mole of S0_3 there are 3 moles of 0, therefore in 1.25 × 10^-2 moles of SO_3 .there is three times that amount. no. of moles of O in SO_3 = 3 × 1.25 × 10^-2 = 3.75 × 10-2 There are 2 moles of 0 in 1 mole of SO_2, thus in 1.55 moles of SO_2 there are 3.10 moles of 0. The total number of moles of 0 present is equal to the sum of the moles contributed by the SO_2 and the SO_3. total no. of moles of O = 3.75 ×10^-2 + 3.10 = 3.14 moles The formula for this compound is S_1.56 O_3.14. The simplest formula is SO_2._01. When there is 1% of SO_2 present, x = 2.01. x = 3.14/1.56. When there is 10 % of SO_3 present, a similar method is used to find x.

Question:

A 1000-gram mass slides down an inclined plane 81 centi-meters long in 0.60 seconds, starting from rest. What is the force acting on the 1000 grams?

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/Users/wenhuchen/Documents/Crawler/Physics/D04-0105.htm

Solution:

Given the mass of an object, we must know its acceleration in order to calculate the force acting upon it. For an object starting at rest 1/2 at^2 = d where d is the distance travelled. In our case: 1/2 a(0.60s)^2 = 81 cm a = 450 cm/s^2 Therefore F = ma 1000 gm×450 cm/s^2 = 450,000 dynes.

Question:

Describe the bonding in linear, covalent BeCl_2 and planar, covalent BCI_3. What is the difference in the hybrid orbitals used?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E18-0662.htm

Solution:

The solution to this problem involves the hybridization of orbitals. Once this is clear, the bonding in BeCl_2 and BCI_3 will follow. Quantum theory deals with independent orbitals, such as 2s and 2p. This can be applied to a species, like hydrogen, with only 1 electron. However, with atoms that contain more than one electron, different methods must be used. For example, the presence of a 2s electron perturbs a 2p electron, and vice versa, such that a 2s electron makes a 2p electron take on some s-like characteristics. The result is that the hydrogenlike 2s and 2p orbitals are replaced by new orbitals, that contain the combined characteristics of the original orbitals. These new orbitals are called hybrid orbitals. The number of hybrid orbitals resulting from hybridization equals the number of orbitals being mixed together. For example, if one mixes an s and p orbital, one obtains two sp hybrid orbitals. One s and two p = three sp^2 orbitals. One s with three p = four sp^3 orbitals, sp orbitals are linear. sp^2 orbitals assume a planar shape and sp^3 orbitals assume a tetrahedral shape. Solving:It is given that BeCl_2 is linear and co-valent. Since sp orbitals are linear, Be undergoes sp hybridization. If something is linear, the bond angle is 180\textdegree. By understanding hybridization, one also knows the geometry of the molecule. A diagram of the bonding resembles Fig. A. Given that BCI_3 is planar, and since sp_2 hybridization yields a planar structure, B has sp_2 hybridized bonding with angles of 120\textdegree(Fig. B).

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Question:

A cue ball traveling at a speed of 3 m/s collides with a stationary billiard ball and imparts a speed of 1.8 m/s to the billiard ball. If the billiard ball moves in the same direction as the oncoming cue ball, what is the velocity of the cue ball after the collision? Assume that both balls have the same mass.

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/Users/wenhuchen/Documents/Crawler/Physics/D06-0296.htm

Solution:

Linear momentum must be conserved in this isolated, two particle system. Thus, the initial momentum of the system must equal the system's final momentum. Since the collision is 1-dimensional, we may drop the vector nature of momentum and write P_f = P_i mv + m (1.8 m/sec) = m (3 m/sec) + m (0 m/sec) m (v + 1.8 m/sec) = m (3 m/sec) v + 1.8 m/sec = 3 m/sec v = 1.2 m/sec.

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Question:

Write a PROCEDURE DIVISION for the DESCRIPTION, ENVIRON-MENT and DATA divisions of the previous problem. The program should print the customer coder name, and address of the active customers listed in the CARD-FILE. The customer information is located in the file TAPE-FILE. Both CARD-FILE and TAPE-FILE are sequentially ordered by Customer-code.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G11-0264.htm

Solution:

Procedure division is used in this record- matching problem to produce a report. Notice that a new tape is not being created here. When the card file ends, the program can be ended. If the tape file happens to end first, we display an error message and end the program. In the first paragraph, inputs to TAPE and CARD files, and output to PRINT file are opened, allowing data to be passed to and from these files. In paragraphs named PARA2, and RTAPE the CARD and TAPE files are read and an ERROR message displayed if the tape file runs out. In COMPARE, the Customer number is compared with C-CODE. Using decision statements the computer is instruc-ted to go to R-TAPE if Customer number is greater than C-CODE, this means we haven't found the correct customer and will keep searching the tape file until we do. Note that C-CODE is identified in the DATA DIVISION of the previous problem. If the C-CODE is equal to Customer number the com-puter is instructed to go to the paragraph named WRITE- LINE. Here we have found the correct customer information in tape file and thus this paragraph prints the required information. If the CUST-NO does NOT fit these two con-ditions it is an invalid CUST-NO and an error message is displayed. This whole procedure is now repeated until the CARD-FILE or TAPE-FILE is exhausted, whereupon all files are closed and execution is stopped.

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Question:

Given the following program segment, evaluate which elementsof the array A(I) will be a part of the product. PRODUCT = 1; J = 1; N = 12; DOI = 1 BY J TO N; PRODUCT = PRODUCT\textasteriskcenteredA(I); J = 3I - J; N = N - J; Assume, for purposes of the iteration head, either case a), b),c) or d). a) J and N are only evaluated once on initial entry into the loop, b) J is evaluated only once but N is re-evaluated at each entryinto the loop, c) J is re-evaluated at each entry into the loop, but N is evaluatedonly once, and, d) Both J and N are re-evaluated at each entry into the loop.

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Solution:

The working of this program segment can be best illustrated by the followingtables. a) Values stored in the ITERATION HEAD Values obtained during execution of the program JN I = I+J PRODUCT J = 3I - J N = N - J 112 112 112 112 112 112 112 112 112 112 112 112 1 2 3 4 5 6 7 8 9 10 11 12 1 × A (1) 1×A (1)×A(2) 1×A(1)×A(2)×(3) 1×A(1)×...×A(4) 1×A(1)×...×A(5) 1×A(1)×...×A(6) 1×A(1)×...×A(7) 1×A(1)×...×A(8) 1×A(1)×...×A(9) 1×A(1)×...×A(10) 1×A(1)×...×A(11) 1×A(1)×...×A(12) 3×1 - 1 = 2 3×2 - 1 = 5 3×3 - 1 = 8 3×4 - 1 = 11 3×5 - 1 = 14 3×6 - 1 =17 3×7 -1 = 20 3×8 - 1 = 23 3×9 - 1 = 26 3×10-1= 29 3×11-1= 32 3×12 -1= 35 12 - 2 =10 10 - 5 = 5 5 - 8 = - 3 - 3 - 11 = -14 -14- 14 = -28 - 28 - 17 = - 45 - 45 - 20 = - 65 - 65 - 23 = - 88 - 88 - 28 = - 116 - 116 - 29 = - 145 - 145 - 32 = - 177 - 177 -35 = - 212 Note that a new value of J and N is computed on each pass through the loop, however these computed values are not entered in the Iteration head whichalways keeps the values J = l and N = 12 which were obtained on thefirst initial entry into the loop. Thus the Product = 1 ×A(l) ×... ×A(12) ,i.e., all the 12 ele-ments of thearray are multiplied out. Thus here, J is entered in the Iteration Head only the first time. Therefore J is always equal to 1. But N is re-evaluated each time in the Head.The looping ends when I>N. Therefore the first three elements of the array are multiplied out. b) Values stored in the ITERATION HEAD Values obtained during execution of the program J N I = I + J PRODUCT = PRODUCT_XA(I) J = 3I - J N = N - J 1 1 1 1 12 10 5 - 3 1 2 3 4 1_XA(1) 1_XA(1)_XA(2) 1_XA(1)_XA(2)_XA(3) 3_X1 - 1 = 2 3_X2 - 1 = 5 3_X3 - 1 = 8 12 - 2 =10 10 - 5 = 5 5 - 8 = - 3 c) Values stored in the ITERATION HEAD Values obtained during execution of the program J N I = I + J PRODUCT = PRODUCT_XA(I) J = 3I - J N = N - J 1 2 7 29 12 12 12 12 1 3 10 39 1_XA(1) 1_XA(1)_XA(3) 1_XA(1)_XA(3)_XA(10) 3_X1 - 1 = 2 3_X3 - 1 = 7 3_X10 - 1 = 29 12 - 2 =10 10 - 7 = 3 3 - 29 = - 26 d) Values stored in the ITERATION HEAD Values obtained during execution of the program J N I = I + J PRODUCT = PRODUCT_XA(I) J = 3I - J N = N - J 1 2 5 12 10 5 1 3 8 1_XA(1) 1_XA(1)_XA(3) 3_X1 - 1 = 2 3_X2 - 1 = 5 12 - 2 =10 10 - 5 = 5 Here, J is re-evaluated in the iteration head after each loop. The product= A(1) ×A(3)× A(10) . Both J and N are re-evaluated in the Iteration Head after each loop. The product includes only two terms A(1)×A(3).

Question:

Write a pseudocode and a flowchart describing how you would go about analyzing the distribution of sales in a retail store. This will require several program paths, so use the case structure in your description.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G08-0179.htm

Solution:

Let us divide the types of sales into 5 categories, assuming that no sale exceeds $2000: less than $500Case 1 $500 to $999.99Case 2 $1000 to $1499.99Case 3 $1500 to $1999.99Case 4 Pseudocode is another word describing a structured language. It is a precise account of the steps needed to expedite the solution of a problem. We want to add up the number of sales that have taken place within each case, and obtain a percentage for each. Our pseudocode reads: Initialize counters C1, C2, C3, C4, TOTAL Read initial value of SALE If out of data, go to END If error in data-type, write ERROR, and END Divide SALE by 500.00, add 1, and take the integer of this calculation. Store it in K. Choose case K and evaluate: IfK = 1, then C1 = C1 + 1 IfK = 2, then C2 = C2 + 1 IfK = 3, then C3 = C3 + 1 IfK = 4, then C4 = C4 + 1 Read next value of SALE, and continue until SALE has no more elements. Also, increment TOTAL to TOTAL = TOTAL + 1 for each value of sale. To get percentages: For C1, divide C1 by TOTAL then multiply by 100 and place in P1 For C2, divide C2 by TOTAL then multiply by 100 and place in P2 For C3, divide C3 by TOTAL then multiply by 100 and place in P3 For C4, divide C4 by TOTAL then multiply by 100 and place in P4 Output P1, P2, P3, P4. Now we can write a flowchart which may highlight the pseudocode: The pseudocode and flowcharts presented here are general. Keep in mind that there are other forms of pseudocode with stricter rules. What we have presented as pseudocode is a rough draft of the problem. To get from the pseudocode to the flowchart, we used the concept of refinement. This term refers to the development of an algorithm in stages, beginning at a very high and very crude level and systematically refining parts of the algorithm until, finally, it is sufficiently explicated for computer implementation. An analogous procedure in mathematics is the proof of a theorem in stepwise fashion; these steps are called subtheorems or lemmas. In computer programming, these steps are referred to as subprograms.

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Question:

The figure represents a small body of mass m revolving in a horizontal circle with velocity v of constant magnitude at the end of a cord of length L. As the body swings around its path, the cord sweeps over the surface of a cone. The cord makes an angle \texttheta with the vertical, so the radius of the circle in which the body moves is R = L sin \texttheta and the magnitude of the velocity v, equals v = 2\piR/T = (2\piL sin \texttheta)/T, where T is the period of revolution of the motion, the time for one complete revolution. Find T.

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Solution:

The forces exerted on the body when in the posi-tion shown are its weight mg and the tension P in the cord. Let P be resolved into a horizontal component P sin \texttheta and a vertical component P cos \texttheta. The body has no vertical acceleration, so the forces P cos \texttheta and mg are equal, and the resultant inward, radial, or centripetal force is the component P sin \texttheta. Then P sin \texttheta = m(v^2/R), P cos \texttheta = mg. When the first of these equations is divided by the second, we get tan \texttheta = (v^2/Rg). This equation indicates how the angle depends on the velocity v. As v increases, tan \texttheta increases and \texttheta increases. The angle never becomes 90\textdegree, however, since this requires that v = \infty. Making use of the relations R = L sin \texttheta and v = 2\piL sin \texttheta/T, we can also write tan \texttheta= (sin \texttheta)/(cos \texttheta) = v^2/Rg = (4\pi^2L^2 sin^2\texttheta)/(T^2L sin \textthetag) (tan \texttheta)/(sin \texttheta)= 1/cos = (4\pi^2L)/(gT^2) cos \texttheta= (gT^2)/(4\pi^2L) T= 2\pi\surd(L cos \texttheta/g). This equation is similar in form to the expression for the time of swing of a simple pendulum for which T = 2\pi\surd(L/g). Because of this similarity, the present device is called a conical pendulum.

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Question:

Write equations for the following (a) cracking of a hydro-carbon - use C_14 H_30 , (b) incomplete combustion of acety-lene, (c) addition of Br_2 to a mixture of ethane, ethylene, and ethyne.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E21-0760.htm

Solution:

(a) By cracking, you mean the process where petroleum fractions containing a high number of carbon atoms are subjected to high heat and pressure, under which they break down to gasoline and lower carbon number hydro-carbons. Octane-gasoline has the formula C_8H_18 . Thus, you can write the balanced equation for cracking of C_14H_30 as C_14H_30C_8H_18 + C_6H_12 . (b) By complete combustion, you mean the reaction of an organic compound with O_2 to produce water and carbon dioxide. When combustion is incomplete, as in car engines, some carbon monoxide (CO) or free carbon is formed in addition to H_2O and CO_2. An example of incomplete com-bustion with acetylene could have the equation (c) To write equations for this mixture, you need to consider some structural characteristics of the com-ponents. All are organic compounds that belong to different classes. Ethane (C_2H_6) is an alkane (a saturated hydro-carbon) . This means it contains only single bonds between the carbons, and between carbon and hydrogen. Ethylene (C_2H_4) is an alkene (an unsaturated hydrocarbon). It possesses a double bond between the carbon atoms. Ethyne (C_2H_2) or acetylene is an alkyne (an unsaturated hydro-carbon) . It possesses a triple bond between the carbon atoms. Unsaturated hydrocarbons undergo addition re-actions to the double or triple bond, saturated ones do not. Thus, when Br_2 is added to the mixture, ethene and ethyne will undergo an addition reaction. Ethane will not undergo any reaction. Therefore, C_2H_6 (ethane) + Br_2no reaction

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Question:

Define the termsAnthropoidea,Hominidaeand Hominoidea?

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/Users/wenhuchen/Documents/Crawler/Biology/F29-0749.htm

Solution:

Man belongs to the order of mammals called the primates. This order is commonly divided into two suborders \rule{1em}{1pt}Prosimiiand Anthropoidea . TheProsimiiare the primitive primates (lemurs, tarsiers, etc.) and theAnthropoideaare the higher primates. The suborderAnthropoideais divided into thePlatyrrhinior New World monkeys, andCatarrhinior Old World monkeys. Theinfraorderof Catarrhine monkeys is further divided into twosuperfamilies= Cercopithecoidea (tailed anthropoids) andHominoidea(great apes and man). ThesuperfamilyHominoidea(hominoids) is rep-resented by three families;Hylobatidae(gibbons),Pongidae(gorilla, chimpanzee, orangutan) and theHominidae(fossil men and modern man). Kingdom:Animalia Phylum:Chordata(sub: Vertebrata) Class:Mammalia(sub:Theria) Order:Primates Suborder:Anthropoidea Infraorder :Catyrrhinii Superfamily :Hominoidea Family:Hominidae Genus:Homo Species:Sapiens (sub: Sapiens)

Question:

The spectrum of a particular light source consists of lines and bands stretching from a wavelength of 5.0 × 10^-5 cm to 7.5 × 10^-5 cm. When a diffraction grating is illuminated normally with this light it is found that two adjacent spectra formed just overlap, the junction of the two spectra occurring at an angle of 45\textdegree. How many lines per centimeter are ruled on the grating?

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Solution:

The grating formula is d sin \texttheta = n\lambda where A is the wavelength of light incident upon the grating, d is the grating spacing, n is the order number, and 0 locates the maxima of the diffraction pattern. At the angle of 45\textdegree, we have d sin 45\textdegree = m × 7.5 × 10-5 cm, and also d sin 45\textdegree = (m + 1) ×5.0 ×10^-5 cm. (We can see why the smaller wavelength has the larger order number by examining the grating formula, d sin \texttheta = n\lambda. Since 0 and d are the same for both \lambda's, we obtain n\lambda = const. Hence, at a particular \texttheta, the larger the wavelength, the smaller must be n, and vice-versa). \therefore(m + 1)/m = (7.5)/(5.0) = 3/2.\thereforem = 2. The second-order spectrum thus just overlaps with the third. Also, using the first formula above, d = [(2 × 7.5 × 10^-5cm)]/ [sin 45\textdegree] = 2.12 × 10^-4 cm. This is the separation of the rulings. Hence the number of rulings per centimeter, n, is n = 1/d = (10^4)/(2,12 cm) = 4715 per cm.

Question:

In a transistor amplifier circuit it is found that a micro-phone converts sound energy into an electrical current of 0.01 A at a potential of 0.5 V. Compare the power from the microphone with the power delivered to the loudspeaker if the collector current for the particular transistor used is 50 times as great as the base current and the voltage of the amplified signal at the loudspeaker is 2V.

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Solution:

The power delivered by the microphone is equal to the product of the current the microphone produces and the voltage at which it produces this current. Hence, P = VI = (5 × 10^\rule{1em}{1pt}1v) (1 × 10^\rule{1em}{1pt}2 A) = 5 × 10^\rule{1em}{1pt}3 W The loudspeaker is connected to the collector. Therefore, the collector current delivered to the loudspeaker by the transistor is, (50) (1 × 10^\rule{1em}{1pt}2 A) = 5 x 10^\rule{1em}{1pt}1 A so the loudspeaker power is, P = VI = (2V) × (5 × 10^\rule{1em}{1pt}1 A) = 1 W The additional energy required to "drive" the speaker is supplied by the batteries.

Question:

Name the compound shown in figure A by the IUPAC system.

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Solution:

In naming complex open chain organic compounds, first find the longest continuous chain containing the functional groups. Write down the parent name. The parent for this particular compound is heptene. Number the chain starting from the end that will give the smallest prefix numbers for the functional groups. This is shown in figure B. Add the suffix functionality with the appropriate numbering: 4- hepten-2-ol. Add the prefix functionality, remembering to group together like prefixes. Then, double check to make sure a substituent has not been forgotten or one substituent has not been included twice. By following these steps, one arrives at the name for the structure: 4-n-butyl-3, 5-diethyl-4-hepten-2-ol.

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Question:

Calculate themolarityof a solution containing 10.0 grams of sulfuricacid in 500 ml of solution. (MW of H_2SO_4 = 98.1.)

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/Users/wenhuchen/Documents/Crawler/Chemistry/E08-0279.htm

Solution:

Themolarityof a compound in a solution is defined as the number ofmoles of the compound in one liter of the solution. In this problem, one istold that there are 10.0 grams of H_2SO_4, present. One should first calculatethe number of moles that 10.0 g represents. This can be done by dividing10.0 g by the molecular weight of H_2SO_4 . numberof moles= (amount present in grams/molecular weight) numberof moles of H_2SO_4 = [10.0 g/(98.1 g/mole)] = 0.102 moles Sincemolarityis defined as the number of moles in one liter of solution, and since, one is told that there is 0.102molesin 500 ml (1/2 of a liter), one should multiply the number of moles present by 2. This determinesthe number of moles in H_2SO_4 present in 1000 ml. Number of moles in 1000 ml = 2 × 0.102 = 0.204. Becausemolarityis defined as the number of moles in 1 liter, the molarity(M) here is 0.204 M.

Question:

(a) What force T, at an angle of 30\textdegree above the horizontal is required to drag a block of weight w = 20 lb to the right at constant speed along a level surface if the co-efficient of sliding friction between block and surface is 0.20? (b)Determine the line of action of the normal force N\ding{217} exerted on the block by the surface. The block is 1 ft high, 2 ft long, and its center of gravity is at its center.

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/Users/wenhuchen/Documents/Crawler/Physics/D02-0050.htm

Solution:

(a) There can be no net vertical force because there is no accelerated motion upwards; similarly, there is no net horizontal force because the block moves with constant velocity. If T is the rope tension, N, the normal force, and f_k the force of friction, we have using Newton's Second Law, \sumF_x = T cos 30\textdegree - f_k = 0, \sumF_y = T sin 30\textdegree + N - 20 lb = 0.(1) A body may experience translational equilibrium without experiencing rotational equilibrium. The general condition for rotational equilibrium is that the sum of torques taken about the center of mass of the body be zero. However, if no net external force acts, this con-dition is less stringent. The condition in this case is that the sum of torques taken about any point (i.e. C) be zero. Let x represent the distance from point 0 (see diagram) to the line of action of N\ding{217} , and take moments about an axis through 0. Then, from the condition of rotational equilibrium, the net torque about 0 must be zero, or T sin 30\textdegree × 2 ft - T cos 30\textdegree × 1 ft + N × x - 20 lb × 1 ft = 0.(2) We may now solve for N. From (1) and (2) T sin 30\textdegree = 20 lb - N(3) T cos 30\textdegree = f_k.(4) (2 ft) (T sin 30\textdegree) - (1 ft) (T cos 30\textdegree) + (X) (N) = 20 ft \bullet lb(5) But f_k = \mu_kN, where \mu_k is the coefficient of static fraction, Hence T cos 30\textdegree = \mu_kN = .2 N(5) Dividing (3) by(5) tan 30\textdegree = (20 lb - N)/(.2 N) \surd3/3= (20 lb - n)/(.2 N) [(.2\surd3)/3] N + N =20 lb N = (20 lb)/[{(.2\surd3)/3} + 1] = 17. Lb(6) Substituting (6) into (5), we obtain T(\surd3/2) = .2(17.9 lb) orT =(.4/\surd3) (17.9 lb) = 4.15 lb(7) Using (6) and (7) in (5) (2 ft)(4.15 lb)(1/2) - (1 ft)(4.15 lb)(\surd3/2) + x(17.9 lb) = 20 ft \bullet lb x(17.9 lb) = (20 ft \textbullet lb) - (4.15 ft \textbullet lb) + (3.6 ft \textbullet lb) x = (20 ft \bullet lb - 4.15 ft \bullet lb + 3.6 ft \bullet lb)/(17.9 lb) x = 1.08 ft Therefore, the line of action of N must be .08 ft to the right of the center of mass if the block is to maintain its rotational equilibrium.

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Question:

Calculate the maximum kinetic energy of the \beta^- emitted in the radioactive decay of He^6.

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Solution:

To solve this problem, one must first know that \beta^- is an electron of negative unit charge and zero mass. Thus, the equation for this particular decay process is ^6_2He\rightarrow^6_3Li + \beta^-. The kinetic energy of the \beta^- particle comes from the difference in mass (or energy, E = mc^2), between products and reactants. To compute the mass change during this process, only the whole atomic mass of He6 and Li^6 need be considered. Once the difference in mass is known, a con-version factor is used (931.MeV= 1amu) Thus Mass of He^6= 6.01890amu Mass of Li^6=6.01513amu Loss in mass0.00377amu Energy equivalent = (931MeV/ 1amu) (0.00377amu) = 3.51MeV, Therefore, the maximum kinetic energy of the \beta^- particle is 3.51MeV.

Question:

Two motorcycles are at rest and separated by 24.5 ft. They start at the same time in the same direction, the one in the back having an acceleration of 3 ft/sec^2, the one in the front going slower at an acceleration of 2 ft/sec^2. (a) How long does it take for the faster cycle to overtake the slower, (b)How far does the faster machine go before it catches up? (c) How fast is each cycle going at this time?

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Solution:

(a) Both cycles travel for the same length of time t. At the instant the two machines pass each other, the faster one has traveled exactly 24.5 ft more than the slower one. With the subscripts 1 and 2 representing the faster and slower cycles respectively, we have d_1 = v_01t + (1/2)a_1t^2 = (1/2)a_1t^2 d_2 = v_02t + (1/2)a_2t^2 = (1/2)a_2t^2 since the initial velocities v_01 and v_02 are both zero. Now d_1 = d_2 + 24.5 ft. or(1/2)a_1t^2 = (1/2)a_2t^2 + 24.5 ft. Substituting values, we find the time t at which the two cycles pass each other. (1/2)(3 ft/sec^2)(t^2)= (1/2)(2 ft/see^2)(t^2) + 24.5 ft t^2= 49 sec^2 t= 7 sec (b) The distance d_1 traveled by the faster cycle when it passes the slower one is d_1 = (1/2)a_1t^2 = {(1/2)(3 ft/sec)} (7 sec)^2 = 73.5 ft. (c) The velocities of the two cycles can be found from v = v_0 + at Then, as they pass each other, their velocities are v_1 = a_1t = (3 ft/sec^2)(7 sec) = 21 ft/sec v_2 = a_2t = (2 ft/sec^2)(7 sec) = 14 ft/sec .

Question:

A biology student is presented with two jars. One is labelled "bacterial flagella" and the other is tagged "protozoan flagella". He is told they may be improperly labelled. How can he conclusively differentiate the flagella? Also discuss the role of flagella in bacteria

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Solution:

The extremely thin hairlike appendages that protrude through the cell wall of some bacterial cells are called flagella. The length of a flagellum is usually several times the length of the cell, but the diameter of a flagellum is only a fraction of the diameter (the width) of the cell. Baccilli and spirilla usually have flagella, which are rarely found on cocci. Flagella occur singly, in tufts, or equally distributed around the periphery of the cell. Flagella are responsible for the motility of bacteria. Not all bacteria have flagella, and thus not all are motile. The mechanism by which the flagella move the bacterial cell is not completely understood; however, it is proposed that movement requires rotation of the semi-stiff flagellum. Whatever the mechanism, the flagellum moves the cell at a very high speed and it enables the bacterium to travel many times its length per second. Bacterial flagella are structurally different from the flagella Qfi eucaryotic cells. This difference can be used to distinguish the bacterial (procaryotic) flagella from the protozoan (eucaryotic) flagella. By examining a cross section of the flagellum under an electron microscope, one can conclusively determine its identity. The eucaryotic flagellum contains a cytoplasmic matrix, with ten groups of tubular fibrils embedded in the matrix: nine pairs of fibrils around the periphery of the eucaryotic flagellum surround two central single fibrils. (see below). Bacterial flagella consist of only a single fibril and lack the ("9 + 2") structural organization of the eucaryotic flagellum. Most bacterial flagella have the same diameter as a single fibril from a eucaryotic flagellum and hence they are usually thinner.

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Question:

Show how i ) the compression function ii) the expansion function iii) Take and Drop functions are used in APL.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G14-0367.htm

Solution:

Assume the following matrix has been stored in memory \mid61220144153\mid \mid282284441370\mid M =\mid4961451184169474\mid \mid1729405851210961208\mid i) Compression is a dyadic function used to select rows or columns from a given matrix. Its general form is L/ [K]M. L is a binary vector composed of ones and zeros. M is a matrix, K is an index indicating which dimension of M is to be compressed and / separates the two arguments. Suppose it is desired to select the first and third rows only of (1). The function 1 0 1 0 / [1]M gives 61220144153 4961451184169474 The K index indicates the dimension of compression, i.e., [1] = row compression and [2] = column compression. The command 0 1 0 1/[2]M yields, when applied to (1) : 2441 145169 40585961 (The first column here represents all numbers between 1 and 50,000 that are equal to the sum of the factorials of their digits. For example, 145 = 1! + 4! + 5!. The second column gives two pairs of numbers that are the reverse of one another and whose product is a square.) ii) Expansion is a dyadic function used to insert rows and columns of zeros into numeric matrices and rows and columns of blanks into character matrices. Its general form is L/ [K]M LetG = B S H A H T W A D E Q B A R T This is a matrix of names; the first character is the initi-al and the remainder is the last name. To make M more legible, perform the function: 1 0 1111\textbackslash [2]G, which yields B S H A H T W A D E Q B A R T iii) Take and drop: These are dyadic functions used for selecting a submatrix from a given matrix. Their gen-eral form is L+M and L+M. L is a vector containing two components. The first component represents the number of rows to be taken or dropped while the second component gives the number of columns taken or dropped. Suppose we want to select Ramajunam's number (1729) from M. This is a 1×1 sub-matrix. The command 3 - 4\downarrowM causes the top three rows (3) and the right four columns (-4) to be dropped (\downarrow). The result is 1729.

Question:

Given the setup in the figure, what would be the pressure of the gas (in atm) if P_atm is 745 Torr and P_1iq equivalent of a mercury column 3.0 cm high?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E02-0022.htm

Solution:

A manometer is used to measure the pressure of a trapped sample of gas. If the right hand tube is open to the atmosphere, the pressure which is exerted in the right-hand surface is atmospheric pressure, P_atm. If the liquid level is the same in both arms of the tube, the pressures must be equal; otherwise, there would be a flow of liquid from one arm to the other. At the level indicated by the dashed line in the figure, the pressure in the left arm is equal to the pressure of the trapped gas, P_gas , plus the pressure of the column of liquid above the dashed line, P_1iq . One can write p_gas = p_atm - p_1iq Here, one is given that P_atm . is 745 Torr and that P_1iq is equivalent to a mercury column 3.0 cm high. One wishes to find P_gas in atm, thus, one must convert 745 Torr to atm and find the P_1iq . There are 760 Torr in 1 atm, which means no. of atm in P_atm = [(745 Torr) / (760 Torr/1 atm)] = .98 atm One atmosphere pressure supports 76 cm of mercury, thus 3 cm of mercury supports 3/76 atm. Therefore, P_1iq = 3 cm/76 cm/atm = .039 atm. Solving for P_gas P_gas = .980 atm - .039 atm = .941 atm.

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Question:

The passenger compartment of a jet transport is essentially a cylindrical tube of diameter 3 m and length 20 m. It is lined with 3 cm of insulating material of thermal conductivity 10^-4cal \bullet cm^-1 \bullet C deg^-1 \bullet s^-1, and must be maintained at 20\textdegreeC for passenger comfort, although the average outside temperature is - 30\textdegreeC at its operating height. What rate of heating is required in the compartment neglecting the end effects?

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/Users/wenhuchen/Documents/Crawler/Physics/D16-0545.htm

Solution:

The hull of the aircraft is a good conductor of heat and may be considered to be at the outside temperature. The circular cylinder of insulat-ing material has thus a temperature of 20\textdegreeC inside and - 30\textdegreeC outside (see figure). Consider a cylinder of the material at distance R from the center of the craft and of thickness dR. By the normal equation of conductivity, the flow of heat across this infinitesimal cylinder per second is H = KA (dt/dR) where dt/dR is the variation of temperature with R (the temperature gradient) and A is the area of the surface across which the heat flow occurs. Hence A = 2\pi R L andH = 2\pi R L K dt/dR ThereforedR = (2\pi R L K dt)/H where L is the length of the cylinder. Thus dR/R = (2\piKL/H) dt. The quantity H is a constant since the system is in equilibrium. If the same quantity of heat did not pass over every cross section of the insulating mater-ial, heat would build up somewhere and the temperature would rise. This is contrary to the condition that equilibrium shall have been attained. For the whole cylinder of insulating material, then ^R2\int_R1 dR/R =(2\piKL/H) ^t2\int_t1 dt where the temperature at R_1 is t_1, and the temperature at R_2 = t_2 (see where the temperature at R_1 is t_1, and the temperature at R_2 = t_2 (see figure). Then ln (R_2/R_1) = (2\piKL/H)(t_2 - t_1) Hence H = [(2\piKL/H)(t_2 - t_1)]/[ln (R_2/R_1)] = {2\pi × 10^-4 cal \bullet cm^-1 \bullet Cdeg^-1 \bullet s^-1 × 20 × 10^2 cm[20\rule{1em}{1pt} (-30)]\textdegreeC}/{1n (300/294)} = 3100 cal \bullet s^-1 = 12,980 J\bullets^-1 \approx 13.0 kW. Here we have used the fact that 1 cal/s = 4.186 J/s This is the heat which flows through the walls. In order to keep the temperature of the cabin constant, an equal amount of heat power must be supplied to the cabin by external means.

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Question:

Why do cells containlysosomes?

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/Users/wenhuchen/Documents/Crawler/Biology/F02-0042.htm

Solution:

Lysosomesare membrane-bounded bodies in the cell. All lysosomesfunction in, directly or indirectly, intra cellular digestion. The materialto be digested may be of extracellular or intracellular origin. Lysosomescontain enzymes known collectively as acidhydrolases. These enzymes can quickly dissolve all of the major molecules that comprisethe cell, and presumably would do so if they were not confined in structuressurrounded by membranes. One function oflysosomesis to accomplish the self- destruction of injuredcells or cells that have outlived their usefulness.Lysosomesalso destroycertain organelles that are no longer useful.Lysosomesare, in addition, involved in the digestion of materials taken into the cell in membranousvesicle.Lysosomefusewith the membrane of the vesicle so thattheir hydrolytic enzymes are dis-charged into the vesicle and ultimately digestthe material.Lysosomeplaya part in the breakdown of normal cellularwaste products, and in the turnover of cellular constituents. Peroxisomes(ormicrobodies) are other membrane-bound vesicles containingoxidative enzymes.Peroxisomesplay a role in the decompositionof some compounds.

Question:

Write a program that a) Calculates factorial values. b) Assume that factorial of 5 is to be calculated through a routine factorial ( ). c) This routine should use the concept of recursion.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G17-0431.htm

Solution:

If a function is recursive it can call itself. Re-cursion is the name given to the technique of defining a func-tion or a process in terms of itself. Factorial function 0! = 1 1! = 1 n! = n(n - 1)! The end condition in the recursive function is the factorial of1. If this routine is invoked with a value 5, it will check if the value it received was equal to 1. Then it carries out the required factorial functions. main ( ) { int value; value = factorial (5); printf ("The factorial of 5 is = %d\textbackslashn", value); } factorial (number) intnumber; /\textasteriskcenterednumber whose factorial is to be computed\textasteriskcentered/ { if (number ==1) return (1); /\textasteriskcenteredend condition of this recursive function\textasteriskcentered/ else return (number \textasteriskcentered factorial (number - 1)); /\textasteriskcenteredcalls itself until end condition occurs\textasteriskcentered/ }

Question:

A 60-ohm electric lamp is left connected to a 240-volt line for 3.00 min. How much energy is taken from the line?

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/Users/wenhuchen/Documents/Crawler/Physics/D24-0782.htm

Solution:

The current through the resistance of the electric lamp is, by Ohm's Law, I = (V/R) = (240 volts/60 ohms) = 4.0 amp The power (or energy per unit time) dissipated by the resistance is P = (energy/time) = I^2 Ror E = I^2Rt= (4.0 amp)^2× 60 ohm × {3 min × (60 sec/min)} = 1.72 × 10^4 Joules (The unit of time was converted to seconds to make it compatible with the MKS system being used.)

Question:

The pressure in interplanetary space is estimated to be of the order of 10^-16torr. This corresponds to a density of about 4 molecules per cubic centimeter. What temperature must be assigned to the molecules in interplanetary space for this to be true?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E25-0862.htm

Solution:

To solve this problem, one assumes that interplanetary space is an ideal gas at a low concentration and at a very low pressure. This means that the ideal gas law can be used, i.e., PV = n RT The pressure, P, is known 10^-16torr= 10^--16torrx (1atm/760torr) = 1.32 × 10^-19atmand R, the gas constant = 0.082atm-liter/mole-\textdegreeK, is also known. The volume, V, and the number of moles, n, are unknown, but if n is divided by V, then this value becomes a density (moles/liter) and can be determined from the given density of 4 molecules/cm. To do this, (1) convert the number of molecules to moles using Avogadro's number and (2) convert cm^3 to units of l using the conversion 1 cm^3 = 1ml= 0.001l Thus, the density is [(4 molecules)/(1cm^3)] [{1000 cm^3/l} / {6.02 × 10^23 (molecules/mole)}] = 6.64 × 10^-21 moles/l Rearranging the ideal gas low from PV = n RT to P/R \bullet V/n = T, andsubstituting the know values, yields T = [{1.32 × 10^-19atm} / {(0.082 (atm-liter/\textdegreeK-mole)}] × [1 / 6.64 × 10^-21(moles/l)] = 2.42 × 10^2 \textdegreeK T = 242\textdegreeK = 242\textdegreeK - 273\textdegree = - 31\textdegreeC.

Question:

How are whales adapted for their life in cold waters?

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/Users/wenhuchen/Documents/Crawler/Biology/F13-0341.htm

Solution:

Whales are animals that belong to the order Cetacea of the class Mammalia. Only animals belonging to the classesMammaliaand Aves possess a temperature regulating system within their body. They are homeothermic. These animals maintain a nearly constant internal body temperature that is independent of the environmental temperature. Whales, sea cows and pigeons are animals that have this temperature maintenance ability. All animals not belonging to the classMammaliaor Aves arepoikilothermic- their internal body temperature varies directly with the environmental temperature. Another feature of the whale that helps it survive in cold waters is the tremendous amount of subcutaneous fat serving an insulating function. All mammals have a layer of fat underlying the skin. An extreme case of this is seen in the whale.

Question:

Explain why the stomata of a leaf tend to open up during the day and to close at night. What advantage is there in this?

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/Users/wenhuchen/Documents/Crawler/Biology/F10-0254.htm

Solution:

During the day, the plant undergoes photosynthesis, a process which converts carbon dioxide and water into carbohydrates.This results in a drop of the carbon dioxide content of the cell as carbon dioxide is used up. When this happens, the acidity of the cell decreases, promoting the conversion of the insoluble starch to the soluble glucose. This can be explained by the fact that the cellular enzymes that catalyze the conversion of starch to glucose do not function in acidic surroundings. During the night hours, production of carbon dioxide occurs during respiration . In addition, no carbon dioxide is depleted by photosynthe-sis which occurs only during the day. This results in a high carbon dioxide level in the cells. When its levels are high, carbon dioxide will combine with water to form carbonic acid. This will inhibit the conversion of starch to sugar, due to a decrease in thepH.When the carbon dioxide level in the cell is as low as it is during the daytime when photosynthesis occurs, much of the carbonic acid is reconverted to carbon dioxide and water, increasing the alkalinity of the cell and thus raising thepH.The greater alkalinity of the plant cells during the day brings about a higher cellular glucose concentration with the reverse true at night. During the day, the high cellular concentration of glucose causes water to enter the guard cells by osmosis, rendering them turgid. The turgor of the guard cells cause them to bend, opening the stoma as a result . At night, the acidity of the cell inhibits the break-down of starch. Since starch is insoluble in the cell sap, it does not exert an osmotic pressure. In the ab-sence of osmotic pressure, the guard cells lose theirturgor, causing them to collapse and close the stoma. What then is the purpose of having the stoma open in the day and closed at night? In the daytime, when the leaves are photosynthesizing, the carbon dioxide supply in the cell is constantly being depleted and its supply must be replenished in order to maintain photosynthe-sis. Thus it is highly advantageous to have the stomata open during the day, in order for carbon dioxide mole-cules to enter the leaf from the atmosphere through the stomata. Open stomata also result in continuous transpiration, hence the air spaces of the leaf are al-ways kept moist for carbon dioxide to dissolve in. This is important since carbon dioxide must first dissolve in water before it can enter the cell. At night when photosynthesis ceases, an external supply of carbon dioxide is not required. The stomata close to retard water loss by transpiration which may occur rapidly if the night is dry or the temperature is high.

Question:

How can one estimate the total number of genes for a given organism?

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Solution:

Since genetic information is stored as DNA base sequences, the amount of DNA in a cell provides an initial basis for estimating the genetic information available to the cell. It is a well-established fact that chromosomes carry the genetic information for the ordering of amino acid sequences of the proteins that are made in the cell. Proteins consist of one or more polypeptide chains, which are composed of numerous amino acids. A given polypeptide chain is coded for by a specific gene, one gene carrying the information for only one polypeptide. Genes are arranged in linear order on the chromosomes and are composed of a specific number and sequence of DNA base pairs. There is a direct correlation between the size of the gene (or the number of base pairs it contains) and the size of the protein it codes for (or the number of amino acids in that protein). A protein can consist of a few dozen to a few hundred amino acids, with 300-400 amino acids often used as the size for an "average" protein. Three mRNA bases are required to code for a single amino acid, and since mRNA is transcribed from only one of the two complementary DNA strands, this means that three DNA base pairs are needed to code for an amino acid. Thus, a very rough measure of the number of different proteins for which a cell might carry information is obtained by dividing the number of base pairs in its DNA by three, giving a theore-tical number of amino acids in DNA. Dividing that number by the average number of amino acids per polypeptide gives a value for the possible number of polypeptides. If an estimate is made of the number of polypeptides the chromosomes can code for, then the number of genes found on the chromosomes can be estimated from the simple fact that one gene codes for one polypeptide. In order for us to determine the total number of genes in an organism by this method, we must get a rather good estimate of the total number of DNA base pairs contained in the chromosomes. This is rather difficult to do with diploid organisms. In addition to the difficulty caused by the sheer number of chromosomes, large amounts of associated proteins known ashistones, as well as other proteins, are found associated with the chromosomes. Moreover, the general structure of a diploid chromosome is for the most part not stretched out, but folded and convoluted intosupercoils, further complicating the examination of the DNA. The chromosome structure of bacteria and viruses is simpler and better understood. Their chromosome is a pure DNA molecule, having no associated proteins, and is often circular in structure. Bacteria usually contain one DNA molecule per cell and the chromosome can be seen in rather good detail using an electron microscope. Owing to its simple chromosome structure then, let us estimate the total number of genes in the bacterium,Escherichia coli. In order to proceed, we must first know the number of base pairs in the chromosome of E. coli. This figure can be estimated either from the length of the DNA molecule, which has been found to be approximately 1000\mumor 1 mm, or from the molecular weight, which is about 2 × 10^9 daltons. Each base pair has a molecular weight of approximately 660 daltons, and there are about 3000 base pairs per\mumof DNA double helix molecule. Either way of determination (3000 base pairs/\mum× 1000\mum or 2 × 10^9daltons/660daltons/ base pair) indicates that there are approximately three million base pairs in the DNA molecule of E. coli. Dividing this by three, we can find out how many amino acids there are (recall that three DNA base pairs code for an amino acid). This gives us a value of about 1 million amino acids. Since the average protein contains about 400 amino acids, the chromosome of E. coli could possibly code for a maximum of 2500 proteins. (1,000,000 / 400) This figure of 2500 genes may seem like an extremely large amount for one cell to contain. However, E. coli is a relatively complex organism. It is considered complex because it is relatively self-sufficient, living on a mini-mal medium of a glucose solution, from which it produces all other vitamins and nutrients necessary for life. This requires a great deal of enzyme and protein machinery, and thus a relatively large number of genes. The mammalian cell, being more complex, should con-tain a larger number of genes than E. coli. It can be de-termined how much DNA is present per mammalian cell. The amount is approximately 800 times that in E. coli. This number gives us an upper limit of the number of different genes. Thus, a mammalian cell would be capable of synthe-sizing 800 × 2500, or over two million different proteins. However, measurements of the amount of DNA present can be misleading, and we must thus be very cautious about relating DNA content directly to the number of different proteins that may be synthesized by a given cell and there-fore, the number of genes in that cell. There is a definite lack of correspondence between DNA content and the number of genes. While inprocaryotesand lowereucaryotes, most DNAcodes for amino acid sequences, the vast majority of DNA from multicellularorganisms has no apparent genetic function. In Drosophila only about 5% of the total DNA codes for amino acid gene products, while in humans even less DNA is so employed. There are a number of explanations for the presence of this "excess" DNA. There are multiple re-petitive sequences of DNA, some of which are never transcri-bed, although their function is not clear. In Drosophila, about 25% of the DNA shows this kind of repetitive sequence, and in the higherEucaryotes, 10% to 20% of the DNA is re-petitive. Some DNA might play structural roles, such as involvement in chromosome folding or pairing during meiosis. Other regions may have regulatory functions, acting as binding sites for repressor molecules or polymerases. Most genes in higher plants and animals are present in only one copy per haploid genome. Exceptions, however, do exist and there are often multiple copies of some genes. For example, in sea urchins the various genes are present in some 100 to 200 copies. Such repetition of genetic information enables an organism to produce large amounts of a protein if necessary. The existence of DNA base sequences which do not code for proteins, as well as the presence of multiple copies of genes, would make any estimate of the number of genes in a diploid organism highly imprecise.

Question:

Find the number, n, of cycles that the piston of the air pump in Fig. B must go through in order to pump a vessel of volume V from a pressure P_1 to a pressure P_2, if the change in the volume corresponding to one cycle of the piston is v. Assume that the air in the vessel is in good thermal contact with the surroundings.

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Solution:

The expansion of air during pumping is an isothermal (constant temperature) process since, as a result of thermal contact with the surroundings, the air in the vessel will have the same temperature as that of the air outside. The process is shown in Fig. A, where expansion proceeds from the initial state A to the final state B. It is governed by the ideal gas law PV = nRT_1 = Constant. If the volume is increased by an infinitesimal amount dV; then the pressure change can be obtained from d(PV) = 0 PdV + VdP = 0, ordP/P = - dV/V.(1) The volume changes by an amount v during one cycle. This The volume changes by an amount v during one cycle. This corresponds to a decrease ∆P in the pressure of the vessel. Integrating (1) from initial to final parameters of the system during one cycle, we get ^Pi-∆P\int_ Pi dP/P = -^V+v\intVdV/V or1n P ]_PiPi - ∆P= - 1n V]V^V+v 1n [(P_i - ∆P)/P_i] = - 1n [(V + v)/v](2) When the next cycle starts, the pressure of the vessel goes down to P_i - ∆P, but the volume of the vessel is again V. Therefore at the end of the second cycle the pressure becomes P_i - 2∆P and (2) for this case is 1n[(P_i - 2∆P)/(P_i - ∆P)] = - 1n[(V + v)/V](3) Rearranging (3), we have 1n {[(P_i - 2∆P)/P_i]\bullet[P_i/(P_i - ∆P)]}= - 1n [(V + v)/V] 1n [(P_i - 2∆P)/P_i]} - {1n [(P_i - ∆P)/P_i]} = - 1n [(V + v)/V] 1n [(P_i - 2∆P)/P_i] = {- 1n [(V+v)/V]} + {1n [(P_i - ∆P)/P_i]}.(4) Now, we substitute (2) in (4), 1n [(P_i - 2∆P)/P_i] = {- 1n [(V + v)/V]} - {1n[(V + v)/V]} = - 2 1n[(V + v)/V] If we repeat the cycle n times; 1n [(P_i - n∆P)/P_i] = - n 1n[(V + v)/V]. If P_i is the original pressure P_1 of the vessel, then (P_1 - n∆P) is the final pressure P_2, 1n P_2/P_1 = - n 1n[(V + v)/V]. Thus, the number of pumping cycles required to achieve the pressure P_2 is n = [1n p_1/p_2] / [1n (V + v)/V]

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Question:

In an animal with a haploid number of 10, how many chromosomes are present in a)aspermatogonium? b) In the first polar body? c) In the second polar body? d) In the secondaryoocyte? Assume, that the animal is diploid.

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Solution:

In solving this problem, one must keep in mind how meiosis is coordinated with spermatogenesis and oogenesis. a)Spermatogoniaare the male primordial, germ cells. These are the cells that may undergo spermatogenesis to produce haploid gametes. But until spermatogenesis occurs, a spermatogonium is diploid just like any other body cell. Since the haploid number is 10, the number of chromosomes in the diploid spermatogonium is 2 × 10 or 20 chromosomes. b) It is essential to remember that while the polar body is formed as a result of unequal distribution of cytoplasm in meiosis, the chromosomes are still distributed equally between the polar body and the oocyte . Since the first polar body is a product of the first meiotic division, it contains only one of the chromosomes of each homologous pair, since separation of homologous chromosomes has occurred. But daughter chromatids of each chromosome have not separated, so there are 2 identical members in each chromosome. Therefore there are 10 doubled chromo-somes in the first polar body, or 20 chromatids. c) The second polar body results from the second meiotic division. In this division, the duplicate copies of the haploid number of chromosomes separate, forming true haploid cells. Therefore the chromosome number is 10. d) The secondaryoocyteresults from the first meiotic division, along with the first polar body. Since, as we have said, the chromosomes have segregated equally, the secondary oocyte has the same number of chromosomes as the first polar body, and for the same reasons. Therefore, it contains 10 doubled chromosomes or 20 chromatids.

Question:

Compare the ionic bond strength ofNaClto KF; assuming that the radii of Na^+, K^+, F^-, andCl^- are, respectively, 0.097, 0.133, 0.133, and 0.181 NM.

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Solution:

The ionic bond strength is directly proportional to the energy released when positive and negative ions form chemical bonds. To solve, this problem, therefore, compute how much energy is released when Na+ andCl^- ions formNaCland when K^+ and F^- ions form KF. Employ the expression that relates energy to charge and radii: E = 1.44 [(q_1q_2)/ r] where E = energy in electron volts, r = distance between the nuclei of the ions in the molecule, and q_1 and q_2 are the charges on the ions. Note that the charge on each ion is the same. The distance between the nuclei, r, is the sum of the radii of the ions. With these facts in mind, one can determine the amount of energy released in the formation ofNaCl. E_NaCl= 1.44[(q_1q_2) / 0.278]. For KF, E_KF = 1.44[(q_1q_2)/0.266]. Comparing the amount of energy released in each case [E_NaCl/ E_KF ] = [0.266 / 0.278] [0.266 / 0.278] × 100 = 95.6%. Thus, the bond strength ofNaClis 95.6% as strong as the bond strength of KF.

Question:

A circular cylinder of cross-sectional area 100 ft^2 and height 8 ft is closed at the top and open at the bottom and is used as a diving bell, (a) To what depth must it be lowered into water so that the air inside is compressed to 5/6 of its original volume, if the atmospheric pressure at that time is 30 in. of mercury? (b) Air is pumped from the surface to keep the bell full of air. How many moles of air have passed through the pump when it is at the depth calculated above, if the atmospheric temperature is 10\textdegreeC?

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Solution:

Applying Boyle's law to the first part of the problem, we obtain p_0V_0 = p × V_f = p × (5/6) V_0 where p_0 and V_0 are the initial pressure and volume of the air inside the diving bell, p and V_f are the final pressure and volume of the air in the bell. p = (6/5) p_0 Now, the pressure change that the air experiences can only be due to a change in pressure of the water with depth. The relation between pressure (p) and depth (h) in a fluid is given by p = p_0 + \rhogh(1) where p_0 is the ambient pressure at the surface of the fluid (see figure) and \rho is the fluid density. Now, the change in pressure caused by submerging the bell a distance h in the water is p - p_0 = \rhogh(2) But, the change in pressure experienced by the air in the bell is ∆p = (6/5) p_0 - p_0 = (1/5) p_0(3) and this must be caused by the submerging of the bell a distance h in the fluid. Therefore, using (3) and (2) (1/5) p_0 = \rhogh orh = p_0/5\rhog(4) The ambient pressure is given as 30 in. of mercury. This is to be interpreted as meaning that the ambient pressure is equal to the pressure exerted by a column of mercury 30 in. long. Thus p_0 = (30 in.) \rho' g(5) where \rho' is mercury's density. Thus, inserting (5) in (4) h = [(30 in) \rho' g]/[5 \rhog] = [(30 in) \rho']/[5 \rho] Since the relative density of mercury is defined as R = \rho'/ \rho we obtain h = [(30 in) R]/5 = (6 in.) R h = [(30 in) R]/5 = (6 in.) R But R = 13.6, and h = (6 in)(13.6) = (1/2 ft)(13.6) h = 6.8 ft. The water level in the bell is thus at a depth of 6.80 ft below the surface of the fluid. But 1/6 of the bell's volume, V_0, is water. Noting that V_0 = A × 8 ft where A is the bell's cross-sectional area, we find that (1/6)V_0 = [(8 ft) / 6 (1/6)V_0 = [(8 ft) / 6 ] × A is water. Hence, the height of water in the bell is (8/6) ft. (See figure). The depth of the foot of the bell is thus 6.80 ft + 8/6 ft = 8.13 ft. The bell has thus been lowered 8.13 ft into the water. (b) If air filled the whole jar at this depth, the pressure on it would be that due to atmospheric pressure plus 8.13 ft of water. Using (1) p = p_0 + \rhog h p = 1 atm + (1.95 sl/ft^3) (32 ft/s^2) (8.13 ft) p = 1 atm + (62.4 lb/ft^3)(8.13 ft) p = 1 atm + 507.312 lb/ft^2 Since atmospheres can't be added to lb/ft^2, we note that 1 lb/ft^2 = 4.725 × 10^-4 atm Thenp = 1 atm + (507.312 lb/ft^2) (4.725 × 10^-4 atm/lb/ft^2) p = 1.239 atm. The pressure acting on the air in the bell is thus 1.239 atm. The bell's volume, V_0, is constant at V_0 = 8 ft × A = 8 ft × 100 ft^2 V_0 = 800 ft^3 The number of moles in this volume is obtained from the gas equation p V_0 = n R T where R is the gas constant, T is the gas temperature in Kelvin degrees, and n is the number of moles of gas. Hence, at its submerged position n = pV_0/RT = [(1.239 atm) (800 ft^2)]/[.0821 {(liter\bulletatm)/(mole\bullet\textdegreeK)}(283\textdegreek)] where we used the fact that 10\textdegree C = (273 + 10)\textdegreeK = 283\textdegreeK In order to be consistent, we transform 800 ft^3 to liters by noting that 1 ft^3 = 28.32 liters. Then n = (pV_0)/(RT) = [(1.239 atm × 800 × 28.32 liters)/(0.0821 liter\textbulletatm\bulletmole^-1 \bulletK deg^-1 × 283\textdegreeK)] = 1208 moles. On the surface of the water the number of moles in the diving bell was n_0 = p_0V_0/RT = [(1 atm × 800 × 28.32 liters)/(0.0821 liter\textbulletatm\bulletmole^-1 \bullet k deg^-1 × 283\textdegreeK)] = 975 moles. The number of moles that have passed through the pump is thus 1208 - 975 = 233.

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Question:

A container has the dimensions 2.5 m × 50 cm × 60 mm. If it is filled with water at 32\textdegreeF, how many kilocalories will be necessary to heat the water to the boiling point? (212\textdegreeF)

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Solution:

A kilocalorie is defined as the amount of heat necessary to raise the temperature of 1 kg of water 1\textdegreeC. Thus, to solve for the number of kilocalories necessary, one must first solve for the number of grams present and the number of \textdegreeC that the temperature is raised. One cubic cm of water weighs 1 gram. Thus, if one solves for the volume in terms of cubic centimeters, the weight is then quickly determined. One can solve for the volume in cubic centimeters after converting the lengths of dimensions to centimeters. 1 m = 100 cm, therefore 2.5 m = 250 cm. If 1 cm = 10 mm, then 60 mm = 6 cm. Solving for the volume: volume = 250 cm × 50 cm × 6 cm = 7.50 × 10^4 cc. Therefore, the weight of the water is 7.5 × 10^4 g. 1 kg = 1000 g, thus 7.5 × 10^4 g = 75 kg. To convert from \textdegreeF to \textdegreeC one uses the following formula: \textdegreeC = 5/9 (\textdegreeF - 32) Solving for \textdegreeC when t = 32\textdegreeF: \textdegreeC = 5/9 (32 - 32) = 0\textdegreeC. When t = 212\textdegreeF. \textdegreeC = 5/9(212 - 32) = 5/9(180) = 100\textdegreeC. Therefore, the change in temperature is 100\textdegree - 0\textdegree or 100\textdegreeC. Solving for the number of kilocalories needed: no. ofKcal = [(1 kcal)/(\textdegreeC kg)] × 100\textdegreeC × 75 kg = 7.5 × 10^3 Kcal.

Question:

Distinguish between covalent and ionic bonds.

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Solution:

A covalent bond is a bond in which two atoms are held together by a shared pair of electrons. An ionic bond is a bond in which oppositely charged ions are held together by electrical attraction. In general, the electro negativity difference between two elements influences the character of their bond (see Table 1). Table 1: Eletronegativities of main group of elements I A II A III A IV A V A VI A VII A H 2.1 Li Be B C N O F 1.0 1.5 2.0 2.5 3.0 3.5 4.0 Na Mg Al Si P S Cl 0.9 1.2 1.5 1.8 2.1 2.5 3.0 K Ca Ge As Se Br 0.8 1.0 1.8 2.0 2.4 2.8 Rb Sr Sn Sb Te I 0.8 1.0 1.8 1.9 2.1 2.5 Ca Ba Pb Bi 0.7 0.9 1.7 1.8 Electronegativity measures the relative ability of an atom to attract electrons in a covalent bond. Using Pauling's scale, where fluorine is arbitrarily given the value 4.0 units and other elements are assigned values relative to it, anelectronegativitydifference of 1.7 gives the bond 50 percent ionic character and 50 percent covalent character. Therefore, a bond between two atoms with an electronegativity difference of greater than 1.7 units is mostly ionic in character. If the difference is less than 1.7 the bond is predominately covalent.

Question:

Two rods of the same diameter, one made of brass and of length 25 cm, the other of steel and of length50 cm, are 50 cm, are placed end to end and pinned to two rigid supports. The temperature of the rods rises to 40\textdegreeC. What is the stress in each rod? Young'smodulifor steel and brass are 20 × 10^11 steel and brass are 20 × 10^11 dynes \textbullet cm^\rule{1em}{1pt}2 and 10 × 10^11 dynes \textbullet cm ^-2, respectively, and 10^11 dynes \textbullet cm ^-2, respectively, and their respective coefficients of expansion are 1.2 × 10 ^-5 C coefficients of expansion are 1.2 × 10 ^-5 C deg^-1 and 1.8 × 10 ^-5 C deg ^-1. 1.8 × 10 ^-5 C deg ^-1.

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Solution:

The temperature rises and the rods, if permitted to, would expand. Since they are rigidly held, they cannot do so and therefore suffer a compressive stress. The forces in the two rods must be the same. If they were not, then at the interface between them, the forces would not balance, equi-librium would not exist, and the interface would move until the forces were equal. Young's Modulus Y = (Stress/St rain) ,where the stressis the where the stress is the normal force per unit (cross sectional) area acting at the end of the bar, and the strain is the fractional change of length\Deltal l /lof a bar due to the of a bar due to the stress. If Y_B and Y_S are, respectively, Young'smodulii for brass and for for brass and for steel, then, since the stresses are equal Y_B (\Deltal_B/l_B) = Y_S (\Deltal_S/l_S) \textbulletl_Bandl_S are the length of the brass and the steel, respectively, when no stress is applied. But the total decrease in length (\Deltal_B+\Deltal_S) is the amount the rods have not been allowed to expand when the temperature rose. To compute this sum, we use the formula relating the fractional change in length.\Deltal, of a bar to a change in temperature t,\Deltal=l\alphat. l is the original length of the bar, and \alpha is the co-efficient of expansion of the bar. Hence \Deltal_B+\Deltal_S=l_B\alpha_B× 40\textdegreeC +l_S\alpha_S× 40\textdegreeC. But\Deltal_S= (Y_B/Y_S)(l_S/l_B)\Deltal_B. Then \Deltal_B[1 + (Y_B/Y_S)(l_S/l_B)] = (l_B\alpha_B+l_S\alpha_S) × 40\textdegreeC. \therefore\Deltal_B= [{40\textdegreeC × (25 cm × 1.8 × 10^-5 deg ^-1 + 50 cm × 1.2 × 10 ^-5 deg ^-1)}/ {1 + (10 × 10^11/20 × 10^11) × (50/25)}] = 2.1 × 10 ^-2 cm and\Deltal_S= (Y_B/Y_S)(l_S/l_B)\Deltal_B= 1/2 × (50/25) ×\Deltal_B= 2.1 × 10^-2 cm. The stress in each rod is Y_B (\Deltal_B/l_B) = Y_S (\Deltal_S/l_S) = 10 × 10^11 dynes \textbullet cm ^-2 × {(2.1 × 10 ^-2 cm)/25 cm} = 0.84 × 10^9 dyne . cm^-2.

Question:

In the Bohr model of the hydrogen atom, the electron is considered to move around the nuclear proton in a circular orbit that has a radius of 0.53 × 10^\rule{1em}{1pt}8 cm. In what electric field and in what potential does the electron move?

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Solution:

The electric field ,E^\ding{217}, experienced by the electron is radial and has a magnitude of E = e / r^2 where e is the electronic charge of a proton, and r is the dist-ance between electron and proton. E =(4.8 × 10^\rule{1em}{1pt}10 statC) / (0.53 × 10^\rule{1em}{1pt}8 cm)^2 = 1.7 X 10^7 statV / cm which is a very large field indeed. Sparking usually occurs in air when a field strength of 100 statV/cm is reached. The electric poten-tial of the electron is \varphi_E = e / r = (4.8 × 10\rule{1em}{1pt}10 statC) / (0.53 × 10^\rule{1em}{1pt}8 cm) = 0.09 statV which is rather small. The potential difference between the terminals of a flashlight is 1.5 V or 0.005 statV. The electric field E depends on 1/r^2 , whereas the potential \varphi_E depends on 1/r; since r is extremely small (0.5 X 10^\rule{1em}{1pt}8 cm) in the case of the hydrogen atom, the field strength is large while the potential is small.

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Question:

Which is of greater significance in evolution - the individual orthe population? Explain your answer.

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Solution:

Evolution is not the change in an individual during its life time, but thechange in the characteristics of a population over many generations. In other words, on individual cannot evolve, but a population can. The geneticmakeup of an individual is set from the moment of conception, and mostof the changes during its life-time are simply changes in the expressionof the develop-mental potential inherent in its genes. But in a popu-lation, both the genetic makeup (the gene pool) and the expression of thedevelopmental potential can change.The change in allelic frequencies of a given population over time is evolution.

Question:

What are the major tasks performed by the linking loader? Explain the term relocation.

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Solution:

The function of the linking loader is to load a program that has been compiledinto machine code into memory for direct execution by the CPU. The linking loader's major tasks are: 1.todetermine the operational locations of the program; 2.to process and to remove the external symbol dic-tionaries; 3.totranslate all relative addresses, of data and instructions to their absoluteforms and to remove the internal relocation symbol dictionaries; 4.tomake certain determinations relative to channel, input and output unitassignment; 5.tomake certain determinations relative to the num-ber, location, and assignmentof buffers; and 6.tomake the initial placement of data and instruction in the main memoryso that the program may be properly executed. Functionally, tasks (1), (2) and (3) are concerned with program relocation andmemory utilization; tasks (4) and (5) are concerned with input and outputdevice utiliza-tion and task (6) is loading. In addition to these basic facilities, a linking loader may have some new capabilities. For example, chaining,which enables a large collection of programsto be over-layedin a small memory space, and dynamic loading, which enables the loading of a subroutine during execution of the callingprogram. The linking loader performs the process of translatingall theprogram references(e.g. relative address, ex-ternal symbols, internal relocation symbolsand entries) into their absolute form. Absolute form is the form by whichthe CPU can decode. Such a process is known as relocation.

Question:

Two electrolytic cells were placed in series. One was composed of AgNO_3 and the other of CuSO_4,. Electricity was passed through the cells until 1.273g of Ag had been deposited. How much copper was deposited at the same time?

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Solution:

To find out how much copper was deposited, you need to know how many faradays were contained in the electricity that passed through the solution. One faraday, be definition, deposits one gram-equivalent of material. The equivalent weight of a substance is that amount of the substance which 1 mole of electrons (1 faraday) can oxidize or reduce to a neutral species. You are told that 1.273g of Ag are deposited. From this, you can compute the number of faradays. The oxidation state of Ag is +1. Thus, the reaction that deposited Ag had to be (Ag^+) + (e^-)\rightarrow Ag . Thus,1 mole of electrons yields 1 gram-equivalent weight of Ag . In this case, the gram-equivalent weight equals the atomic weight, since 1 mole of Ag is produced with 1 mole of electrons. Recalling the definition of a faraday, you can say that 1 faraday can deposit 107.87g of Ag (the atomic weight). You are told, however, that only 1.273g has been deposited. Thus, the following proportion can be used to find the number of faradays used: (1.273g Ag) / (X Faradays) = (107.87g Ag) / (1 Faraday) Solving, X = .01180 Faradays. Thus, the amount of faradays passed is .01180F, which means the amount of copper deposited depends on this amount of electricity. To deposit copper, you must perform the reaction Cu^2+ + 2e^- \rightarrow Cu you see that it takes 2 moles of electrons to deposit one mole of copper. Thus, one faraday can only deposit one half of a mole copper. Thus, 1 faraday reduces (1 / 2)(63.54) = 31.77 grams. You have, however, only .01180 faradays. Thus, the amount of Cu deposited can be found using the proportion: (1Faraday) / (31.77) = (.1180 Faraday) / (X). Solving for X, X = .3749g. Therefore, 0.3749g of Cu will be deposited at the same time.

Question:

Compute the average kinetic energy inelectronvoltsof a gas molecule at room temperature.

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Solution:

Absolute kinetic energy does not depend on the mass of the molecule ; therefore, the molecules of all gases have the same absolute kinetic energy at a given temperature. KE = (3/2)kT where k is the Boltzmann constant and the temperature T is expressed in the Kelvin scale. Room temperature is 20\textdegreeC or 293\textdegreeK. Substituting values: KE = (3/2) × (8.62 × 10^-5eV/\textdegreeK) × (293\textdegreeK) = 0.038eV

Question:

Many antibodies react by causing agglutination and precipitation of the antigen-antibody complex. Explain why these antibodies and their respective antigens must each have more than one binding site per molecule.

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Solution:

The structure of the IgG molecule has al-ready been visualized in the electron microscope. It is a Y-shaped molecule with one binding site, or valency, at the end of each arm. Each IgG molecule is thus divalent, having two binding sites capable of combining with two antigens. Each antigen molecule also has more than one binding site; that is, it is divalent or multivalent. The reason that both antibodies and antigens are multivalent is to permit a high degree of agglutination (clumping). This allows precipitation of the antibody- antigen complex to occur. If only one binding site existed on each antibody and antigen molecule, then only one molecule of antibody could bind to one molecule of antigen: Precipitation would not occur if the complex were this small. Even if the antibody were divalent, as we know to be so, complex size would still be limited to 3 molecules if the antigen were univalent: Therefore, the antigen molecule must also be at least divalent for there to be extensive bridging of antigenantibody complexes. This large complex of antigens and antibodies, called a lattice, is now able to settle out of the body fluid. It then tends to be engulfed and destroyed by macrophages.

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Question:

For the oxidation of VOby Fe_2O_3 to form V_2O_5 andFeO, what is the weight of one equivalent of VO and of Fe_2O_3?

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Solution:

The equation for this reaction is 2VO + 3Fe_2O_5 \ding{217} V_2O_5 + 6FeO Here V is a reducing agent and Fe is an oxidizing agent. One equivalent of an oxidizing agent is defined as the mass of the substance that picks up the Avogadro number of electrons. One equivalent of a reducing agent is defined as that mass of the substance that releases the Avogadro number of electrons. The oxidation state of O is always - 2, thus the oxidation state of V in VO is + 2. The oxidation state of V in V_2O_5 is + 5 because O contribute (5 × - 2) or - 10, thus 2V must be + 10 and V is + 5. The half-reaction for V is V^+2 \ding{217} v^+5 + 3e^- This means that there are 3 equivalents per mole of VO. One finds the weight of one equivalent by dividing the molecular weight by 3. (MW of VO = 66.94). weight of 1 equiv of V = (66.94 g / mole) / (3 equiv /mole) = 22.31 g one uses a similar method for Fe. The oxidation state of O in Fe_2O_3 is (3 × - 2) or - 6, this means the 2Fe must be + 6 and Fe must be + 6/2 or + 3. The oxidation state of O inFeOis - 2, thus the oxidation state of Fe is + 2. The half-reaction for the Fe is then Fe^+3 + le^- \ding{217} Fe^+2 In Fe_2O_3 there are 2 moles of Fe, therefore the half- reaction becomes 2Fe^+3 + 2e^-\ding{217} 2Fe^+2 There are thus 2 equiv per mole of Fe_2O_3. The weight of one equivalent is equal to the weight of one mole Fe_2O_3 divided by 2. (MW of Fe_2O_3 = 159.70) wt of 1 equiv =(159.7 g / mole) / (2 equiv / mole)= 79.85 g/equiv.

Question:

Simultaneous removal of all nerves to the heart causes the heart rate to increase to about 100 beats per minute. What does this tell us about the regulation of the heart rate? Hormones regulate many bodily functions. Do they influence heart rate?

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Solution:

The excitatory impulses of thesino-atrialnode occur spontaneously without the presence of nervous stimulation or hormones. But the S-A node is always under the influence of both nerves and hormones. Nervous control of the heart rate is found in the vasomotor center in the medulla oblongata of the brain. A large number of sympathetic nerve fibers run from this center to thesino-atrialnode. Sympathetic nerve stimulation increases the heart rate. The parasympathetic nerve fibers which innervate the heart are contained in thevagusnerve, which also runs from the medulla to the S-A node. Stimulation of thevagusnerve has the opposite effect of decreasing the heart rate, and if strong enough, may stop the heart completely. Cut-ting the sympathetic nerves decreases the heart rate, while cut-ting the parasympatheticvagusnerve increases the heart rate. Cutting or removing all nerves from the S-A node causes the heart rate to increase from its normal 70 beats per minute to 100b.p.m. This tells us two things: (1) that the heart rate is influenced by the nervous system; and (2) that the parasympathetic influence is stronger, since cutting both sympathetic and parasympathetic fibers increases the heart rate. This increased rate of 100b.p.m. is the inherent autonomous discharge rate of the S-A node (i.e., without any external influencing factors). There are sensory receptors in the walls of the right atrium superior and inferior vena cava which are stimulated when these walls are stretched by an increased volume of blood. The stimulated recep-tors send impulses to the vasomotor center in the brain which stimulates the sympathetic nerves; this increases the heart rate and stroke volume to compensate for the increased volume of blood in these areas of the heart. In addition, there are sensory receptors in the walls of the aorta and carotid arteries. These receptors are also stimulated by distension of the vessel walls and send impulses to the vasomotor center. Here, however, thevagusnerve is stimulated and the heart rate decreases. The nervous system adjusts the heart rate quickly to the metabolic needs of the body (such as during exercise). It also provides a feedback control to prevent any excessive cardiac response. Sympathetic stimulation acts via the neurotransmitternorepinephrine, while parasympathetic stimulation acts via the neurotransmitter acetylcholine. Besides nervous control, the heart rate is also under hormonal control. Epinephrine, a hormone secreted by the adrenal medulla, is produced in increased amounts during emergencies or "fight-or- flight" situations. Epinephrine causes vasoconstriction in some vascular beds of the body (smooth muscle of the digestive tract) andvasodilationin both skeletal and cardiac muscle. Overall, it causes an increase in blood pressure and an increase in heart rate.Thyroxinealso increases the heart rate but it requires several hours to cause the acceleration. This hormone from the thyroid gland can only affect long-term responses of the heart, unlike the quick action of epinephrine.

Question:

Explain why a four-chambered heart is more efficient than a three-chambered heart. Why does an animal with a two-chambered heart not experience the problem that an animal with a three-chambered heart experiences?

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Solution:

A four-chambered heart is characteristic of "warm-blooded" animals such as birds and mammals. Since these animals maintain a relatively high constant body temperature, they must have a fairly high metabolic rate. To accomplish this, much oxygen must be continually provided to the body's tissues. A four-chambered heart helps to maximize this oxygen transport by keeping the oxygenated blood completely separate from the deoxygenated blood. The right side of the heart, which carries the deoxygenated blood, is separated by a muscular wall, called the septum, from the left side of the heart, which carries .the oxygen-ated blood. In amphibians, the atria are divided into two separate chambers, but a single ventricle exists. This three-chambered heart permits oxygen-rich blood returning from the pulmonary cir-culation to mix with oxygen-poor blood returning from the systemic circulation. It is less efficient than a four-chambered heart because the blood flowing to the tissues is not as oxygen-rich as it could be. Fortunately, amphibians, being cold-blooded, do not have to maintain a constant body temperature and hence do not need the efficiency of warm-blooded hearts. Reptiles also have three-chambered hearts but partial division of the ventricle in these animals has decreased the amount of mixing. Fish, which possess a two-chambered heart, (one atrium and one ventricle) do not have this problem of mixing oxygenated and deoxygenated blood. The blood of fishes is oxygenated in the capillary beds of the gills. This oxygenated blood does not go back to the heart but goes directly into the body circulation. When the blood returns from the body, it drains into the heart, which simply pumps this deoxygenated blood to the gills. No mixing occurs in the two - chambered heart because only deoxygenated blood is passed through the heart.

Question:

Determine the equilibrium constant for the reaction H_2 +I_2 \rightleftarrows2HI the equilibrium concentrations are: H_2 , 0.9 moles/liter; I_2 , 0.4 mole/liter; HI, 0.6 mole/liter.

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Solution:

The equilibrium constant (Keq) is defined as the product of the concentrations of the products divided by the product of the concentrations of the reactants. These concentrations are brought to the power of thestoichiometriccoefficient of that component. For example for the reaction 2A + B \rightleftarrows 3C, the equilibrium constant can be expressed: Keq Keq = = ([C]^3 /{ [A]^2[B]}) where [ ] indicates concentration. One can now express the equilibrium constant for the reaction H_2 +I_2 \rightleftarrows 2HI as Keq= ([HI]^2 /{[H_2] [I_2]}) One is given the concentrations of the components for the system, thus one can solve for the equilibrium constant. Keq= ([HI]^2 /{[H_2] [I_2] })[H_2] = 0.6 mole/liter [I_2 ]= 0.4 mole/liter [HI] = 0.6 mole/liter Keq= ( [0.6 mole/liter]^2 / { [0.6 mole/liter] [0.4 mole/liter] }) = (0.36)/(0.36) = 1.0

Question:

One method being developed to efficiently use solar energy for the heating of homes is the use of sodium sulfate decahydrate(Na_2SO4\bullet 10H_2O). It is placed in a bin on the roof of a house. During the day, the bin stores heat collected from the sun via the endothermic reaction: Na_2SO_4 \bullet 10H_2O \rightarrow NaSO_4 (s) + 10H_2O (l). At temperatures above 32.4\textdegreeC, the reaction goes completely to the right. At night, when the temperature falls below 32.4\textdegreeC, the reaction goes completely to the left. If the efficiency of the system is 100 %, how much heat could 322 kg of Na_2SO_4 \bullet 10H_2O provide to a home at night? Use the following information: Compound Standard Heat of Formation at 298\textdegreeK Na_2SO_4 \bullet 10H_2O (s) - 1033.5 Na_2SO_4 (s) -330.9 H_2O (l) -68.3

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Solution:

This problem can be solved once the amount of heat one mole of Na_2SO_4 \bullet 10H_2O absorbs during the day is known. An endothermic reaction is one that absorbs heat from the surroundings. The amount of heat absorbed can be calculated from the Standard Heats of formation given. The heat of the reaction (i.e. the amount of heat absorbed or released) is equal to the ∆H\textdegree_productsminus ∆H\textdegree_reactants,where ∆H\textdegree = Standard Heats of Formation. During the day (when the reaction absorbs heat and proceeds to the right), Heat of reaction = ∆H\textdegree_products- ∆H\textdegree_reactants = [∆H\textdegree_Na_(2)SO(4)(S) +10∆H\textdegree_H(2)O(l)] - [∆H\textdegree_Na_(2)SO(4)\bulletH(2)O] = (- 330.9 - 683) - (- 1033.5) = + 19.6 Kcal mole^-1. Thus, 1 mole of Na_2SO_4 \textbullet 10H_2O absorbs 19.6 Kcal mole^-1 of heat from the sun during the day. At night, the reaction is reversed, which means it is exothermic, or released heat. If the reaction absorbs 19.6 Kcal mole^-1 of heat when it proceeds to the right, then it must release 19.6 Kcal/mole of heat to the house at night, when it proceeds to the left. The number of moles of Na_2SO_4 \bullet 10H_2O present in 322 kg can be calculated. In 1 kg there are 1,000 g, thus 322 kg = 322 × 10^3 g. A mole is defined as grams (mass)/molecular weight. The molecular weight of Na_2SO_4 \bullet 10H_2O = 322 g/mole. This means that (322 × 10^3) / (322 g/mole) = 1 × 10^3 moles of Na_2SO_4 \bullet 10H_2O is present. Therefore, this amount of Na_2SO_4 \bullet 10H_2O should give (1 × 10^3 moles) (19.6 Kcal/mole) = 1.96 × 10^4Kcal of heat to the house at night.

Question:

Two electrons A and B have speeds of 0.90c and 0.80c, respectively. Find their relative speeds (a) if they are moving in the same direction and (b) if they are moving in opposite directions.

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Solution:

For speeds close to that of light, when adding velocities, the relativistic law must be used. For the relative velocity V_r between two objects A and B, measured relative to A v_r = (v_A - v_B) / [1 + {( v_A - v_B)/c^2}] When V_A and V_B are small, the term {(v_A v_B) / c^2} is small compared to unity and the above equation reduces to the classical expression for relative velocity, v_r = (v_A - v_B) The relativistic equation for relative velocity must be used for speeds close to that of light, since, according to the theory of relativity, the maximum speed V_r between two objects is c, regardless of the reference frame used. (a) For the relative speed between the two electrons A and B, if they move in the same direction v_ab = (0.90c - 0.80c) / [1 - {(0.90c)(0.80c) / c^2}] (0.90c - 0.80c ) / [1 - 0.72c2/ c^2] = 0.10c / 0.28 = 0.36c (b) When the electrons are moving in opposite directions, v_ab = {0.90c - (- 0.80c)} / [1 - {(0.90c)(-0.80c) / c^2}] (0.90c + 0.80c ) / {1 + (0.72c2/ c^2)} = 1.70c / 1.72 = 0.99c If classical physics had been used to compute the relative velocity in the two cases, the relative speeds would have been found to be 0.10c and 1.70c, re-spectively .

Question:

What type of flip-flop does the transistor circuit of fig. 1 represent?

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Solution:

The circuit of fig. 1 is broken down into two identical "sub-circuits" in fig. 2(a) and (b). In sub-circuit (a), if either or both of the transistor bases are connected to 5V (high), current will flow through R_1 to ground and the output Y will be low (OV). If both bases are connected to low (OV), no current will flow and Y_1 will float up to 5V (high). Thus, each sub-circuit is a NOR gate and the output of each gate is connected to the input of the opposite gate. The logic diagram of fig. 1 is shown in fig. 3. The truth table for fig. 3 is shown in fig. 4. InputsPresent StateNext State X_1X_2 Y_1Y_2 Y_1Y_2 00 01 01 00 10 10 01 01 01 01 10 01 10 01 10 10 10 10 11 01 not used 11 10 not used Fig. 4 This is the same truth table for an S-R flip-flop, hence, X_1 = S, X_2 = R, Y_2 = Q, Y_1 =Q

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Question:

How can tissue enzymes and plasma proteins be used in the determination of evolutionary relationships between different species?

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Solution:

By this time it should be clearly understood that genetic change is the basis of evolution. It then follows that evolutionary relationships between species essentially result from genetic relationships. In general, we know that the more closely two species are related, the more they resemble each other in their genetic makeup. Since the biochemical functions in an organism are genetically controlled, the degree of biochemical simarilarity between two species should reflect the degree of similarity in their genetic com-position, and how closely they are related evolutionarily. Since enzymes are proteins whose synthesis is dictated by genes in the chromosomes, the properties of enzymes, such as their comformation and rate of catalysis, are dependent upon the genetic makeup of the organism in which they occur. Enzymes from various species, there-fore, can be compared in their properties, and the degree of genetic similarity between the species can be inferred. It is found that enzymes from animals deemed to be closely related on the basis of anatomic or other evidence reveal very similar patterns of their rates of reaction. Enzymes from remotely related species show very different patterns, as can be expected. Studies of the plasma proteins of various groups of mammals can also be used to infer the evolutionary relationships between species. The degree of similarity of the various plasma proteins can be tested using the antigen-antibody method. In this technique, an animal, such as a rabbit, is given repeated injections of the plasma from a different animal, such as a human (see Figure). The plasma cells of the rabbit produce antibodies specific for human blood-plasma antigens. The antibodies are then isolated from the blood of the rabbit in the serum. A diluted sample of this serum, when mixed with human blood, results in a visible precipitation caused by the combination of antigens and antibodies. By using a number of rabbits, each injected with the blood of different species of mammal, it is possible to obtain a series of antibodies, each specific for the blood proteins of a particular species. The relationship between any two species can be determined by the strength of recognition when the anti-body specific for one species' protein is reacted with the corresponding protein of the other species. The stronger the recognition, the more closely the two are related. For example, suppose one has obtained, by the method discussed, a rabbit- made antibody to human serum albumin. Bovine (sheep) serum albumin can be reacted with this antibody, and the strength of recognition of the antibody determined by the dilution at which preci-pitation (or recognition) no longer occurs. Such a determination could also be made between human serum albumin and that of any other mammal. The relative strengths of recognition enable one to place the diffe-rent species in order of the closeness of their relation-ship to man. The evolutionary relationships determined by this method parallel those determined in other ways. Evolutionary relationships between species can also be characterized by comparing the amino acid sequences of their respective plasma proteins and enzymes. Cyto-chrome c, a respiratory chain enzyme, has been studied extensively in this regard, and many species have been compared. The fact that certain crucial amino acids are consistently present in the same sites in all species suggests that the different cytochrome sequences arose through mutations from an ancestral molecule. As expected, the further apart any two species are evolutionarily, the greater the number of differences in their amino acid sequences. In fact, estimates of the chronological distance between the emergence of one species and another can be made using knowledge of mutation rates and the assumption that any given amino acid difference resulted from a single mutation. Such clear cut relationships however cannot always be found for all proteins studied. Schematic representation of the antigen-antibody technique, The strength of recognition between human serum albumin antibodies (made by the rabbit) and human, monkey, and- bovine serum is indicated by the dilution at which precipitation no longer occurs. According to this, monkey serum albumin is closer to human albumin than is bovine albumin. Studies are also being done which use a measure of the extent of DNA hybridization as an indication of evolutionary relationships. Hybridization is said to occur when a single strand of a species' DNA reforms a double-stranded helix by pairing with a single strand of another species' DNA. The extent to which the pairing occurs is related to the similarities in the two species' DNA molecules. This is direct evidence of genetic si-milarity or dissimilarity between species.

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Question:

Write a FORTRAN program which uses the modified Euler method to simulate the growth of an Isolated species from time t = 0 to t = t_f ; If the population growth rate (per unit of time) is directly proportional to the additional number of individuals the environ-ment could support. Let the number of individuals at time t be N(t), N(0) = N_0, and the constant of proportionality - k, 0 < k < 1. Compare the modified Euler approximations with the exact value.

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Solution:

Let S equal the total number of individuals the environ-ment could support (saturation population). Then, at any time t, the additional number of individuals the environment could support is S - N(t). Thus, if N is the population growth rate, the block diagram for this system would look as shown on figure #1 . or:Ṅ = k (s - N(t))(1) This population model "fills to capacity" using negative feedback control. Also note that equation (1) is dimensionally correct only if k has units (time)^-1. It can be shown that the exact solution of equation (1) is: N(t) = S - (S - N_0 )_e^-kt(2) Note that t \ding{217} \infty \Rightarrow N(t) \ding{217} S and since N_0 \leq S \Rightarrow S - N_0 \geq 0 and e^-kt > 0, (S - N_0)_e^-kt \geq 0 \Rightarrow N(t) \leq Sfor all t, which represents the idea of population "saturation". The program uses POP for N(t), P0P0 for N(0). DATA. T/0,0/ COMMON S,K READ, N,TFIN,ACCUR,S,K,POPO PRINT,T,POPO REALN = N DT = TFIN/REALN POP = POPO SDIF = S - POPO DO 10 I = 1,N T = T + DT CALL MEULER(T,POP,ACCUR,DT) EXACT = S - SDIF\textasteriskcenteredEXP(-K\textasteriskcenteredT) ERROR = ABS ((EXACT-POP)/EXACT)\textasteriskcentered100 PRINT,T,POP,EXACT,ERROR 10CONTINUE STOP END FUNCTION G(W) COMMON S,K G = K\textasteriskcentered(S-W) RETURN END

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Question:

Use Quine-McCluskey Method to minimize the function f (A,B,C,D). f(A,B,C,D) = \summ(2,4,6,8,9,10,12,13,15)

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Solution:

To begin the Quine-McCluskey minimization technique the minterms are grouped according to the number of 1's in the binary representation of the minterm number. This grouping of items is illustrated in fig. 1. Mineterms ABCD 2 4 8 6 9 10 12 13 15 0010 0100 1000 0110 1001 1010 1100 1101 1111 Fig. 1 Minterms 2,4, and 8 form the first group since they both contain a single 1 in their binary representation form. Minterms 6,9,10 and 12 form the second group hence they contain two 1's in their binary representations. The third group which consists of three 1's is the minterm 13. And the fourth group with four 1's, minterm 15. Once this method is formed, using the table obtained, adjacent minterms are found and combined into minterm lists in a minimizing table. In the following tables this procedure is illustrated; In fig. 2 all 4 groups are given. Note here that two terms can be combined if and only if they differ in a single literal. Hence, in Fig, 2 group 1 terms can only be combined with group 2 terms, to form the list given in fig. 3. Note here that in forming fig. 3 several minterms are combined together. Such as m2of group 1, and m_6 of group 2. In these two minterms only the term B differs, and is therefore dropped. Minterm 2 of group 1, and 10 of group 2 differ in A, Minterms 4 of group 1, and 6 of group 2 differ in C; Minterms 4 of group 1, and 12 of group 2 differ in A; Minterms 8 of group 1, and 9 of group 2 differ in D; Minterms 8 of group 1, and 10 of group 2 differ in C; Minterms 8 of group lf and 12 of group 2 differ in B. When all the combinations between the groups 1 and 2 have been made and they have been entered in the list, a line is drawn under these combinations, and combination of terms in group 2 with those in group 3 starts, following the same logic. Next, using the list of fig. 3, the list in fig. 4 is obtained. As before, two terms in list 2 can be obtained only if they differ in a single literal, only terms which have the same literal missing can possibly be combined. Note that in fig. 3 minterm combinations 8, 12 and 9,13 also 8,9 and 12,13 can be combined to yield terms 8,9,12,13 in fig. 4. These terms are checked off in the table of fig. 3 and all the other terms are labeled as Prime Implicants. To determine the smallest number of Prime Implicants required to realize the function, a Prime Implicant Chart is formed as in fig. 5. The double line through the chart between PI_1 and PI_2 is used to separate prime implicants which contain different number of literals. An examination of the minterm columns indicates that minterms 9 and 15 are covered by only one prime implicant. Therefore prime implicants PI_1 and PI_7 must be chosen, and hence they are essential prime implicants. Note that in choosing these two prime implicants, minterms 8,9,12,13, and 15 are also covered. These minterms are shown checked in the table. To cover the remaining minterms 2,4,6, and 10 a reduced prime im-plicant chart is formed as in fig. 6. Prime implicants PI_5, and PI_6 may be omitted because they are covered by PI_4, and PI_3 respectively. Hence ignoring PI_5 and PI_6, for the moment minterms 2,4,6, and 10 can be most efficiently covered by choosing PI_3 and PI_4 . Therefore a minimal realization of the original function would be; f(A,B,C,D) = PI_1 + PI_3 + PI_4 + PI_7 f(A,B,C,D) = 1- 0 - + - 0 1 0 + 0 1 - 0 +1 1 - 1 and using the variable coding; f(A,B,C,D) = AC+BCD+ABD+ ABD 2 4 6 10 PI_2 X X PI_3 X X PI_4 X X PI5 X PI_6 X Fig. 6

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Question:

Describe the various land biomes that are usually encounteredby a traveler going from the equator to the arcticpolar ice cap.

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Solution:

In the equatorial regions of several continents (Africa, Asia, and South America) are tropical rain forests with enormous trees (See figure, problem28-15). The interlocking canopy of leaves blocks out the sun and allowsonly dim light to penetrate. As a result, the ground is sparsely coveredwith small plants. The canopy also interrupts the direct fall of the plentifulrain, but water drips from it to the forest floor much of the time, evenwhen it is not raining. It also shields the lower levels from wind and hencegreatly reduces the rate of evaporation. The lower levels of theforst areconsequently very humid. Temperatures near the forest floor are nearlyconstant. The pronounced differences in conditions at different levels within sucha forest result in a striking degree of vertical stratification. Many speciesof animals and epiphytic plants (plants growing on large trees) occuronly in the canopy, while others occur only in the middle strata and stillothers occur only on the forest floor. Decomposition of fallen leaves anddead wood is so rapid that humus is even missing in spots on the floor. The diversity of life here is the greatest found anywhere on earth. In asingle square mile of the richest tropical rain forests, hundreds of speciesof trees, hundreds more of birds, butterflies, reptiles and amphibians, and dozens of species of mammals make their habitat. Huge areas in both the temperate and tropical regions of the world arecovered by grassland biomes. These are typically areas with relatively lowtotal annual rainfall or uneven seasonal occurrence of rainfall. This typeof climate is unfavorable for forests but suitable for growth of grasses. Temperate and tropical grasslands are remarkably similar in appearance, althoughthe particular species they contain may be very different. In both cases, there are usually vast numbers of large herbivores, which often includethe ungulates (hoofed animals). Burrowing rodents orrodentlike animalsare also often common. North of the tropics, in the temperate regions of Europe and eastern North America, are deciduous forests. Small creepers and epiphytes such aslichens and mosses may be found, but they are not so conspicuous an inthe rain forests. There is richer vegetation on the ground below, which is oftencovered by a carpet of herbs. Such a forest may be called temperate deciduousforest. In those parts of the temperate zone where rainfall is abundantand the summers are relatively long and warm, the climax communitiesare frequently dominated by broad-leaved trees, whose leaveschange color in autumn then fall off in winter and grow back in the spring. North of the deciduous forest of temperate North America and Eurasia is a wide zone dominated by coniferous (cone-bearing) forests sometimesreferred to as the boreal forests. This is the taiga biome. Instead of being bushy-topped, the trees are mostly of the triangular Christmas-tree shape.The trees of the boreal forest are evergreens, cone-bearingand needle-leaved consisting mostly of spruce, fir, and tamarack. They branch over most of their height and are relatively close-packed. The land is dotted by lakes, ponds and bogs. The winters of thetaiga are very cold and during the warm summers, the subsoil thaws andvegetation flourishes. Moose, black bears, wolves, lynx, wolverines, martens, squirrels, and many smaller rodents are important in the taiga communities. Birds are abundant in summer. The territory of the boreal forest is hundreds of miles wide, but eventuallyit meets the completely treeless vegetation of the tundra. The tundrais the most continuous of the earth's biomes, forming a circular bandaround the North pole, interrupted only narrowly by the North Atlantic andthe Bearing Sea. The vegetation of the tundra resembles grassland butit is actually made up of a mixture of lichens, mosses, grasses, sedges andlow growing willows and other shrubs. There are numerous perennial herbs, which are able to withstand frequent freezing and which grow rapidlyduring the brief cool summers, often carpeting the tundra with brightlycolored flowers. A permanent layer of frozen soil lies a few inches toa few feet beneath the surface. It prevents the roots of trees and other deepgrowing plants from becoming established, and it slows the drainage ofsurface water. As a result, the flat portions of the tundra are dotted with shallowlakes and bogs, and the soil between them is exceptionally wet. Reindeer, lemmings, caribou, arctic wolves, foxes and hares are among theprincipal mammals, while polar bears are common on parts of the tundranear the coast. Vast numbers of birds,particularyshorebirds and waterfowl(ducks, geese, etc.), nest on the tundra in summer, but they are notpermanent inhabitants and migrate south for the winter. Insects are incrediblyabundant. In short, the tundra is far from being a barren lifeless land; but the number of different species is quite limited. The succession of biomes between the equator and the Arctic is roughlymirrored in the Southern Hemisphere, although the absence of largelandmasses makes the pattern less complete. There are also some otherbiomes that are more irregularly scattered about the world, but which havea distinctive form. These are the biomes of the desert regions and thesclerophylousbushlands. Thebushlands, common in the chaparral of California,maquisof theMediterranean and themalleeheath of Australia, are made up of differentspecies of plants, but have much in common. They have gnarled andtwisted, rough and thorny plants. The leaves tend to be dark and are hairy, leathery, or thickly cutinized. All grow in places with hot dry summersand relatively cool moist winters. A climatic pattern such as this israre and is always accompanied with vegetation of this form. In places whererainfall, is very low grasses cannot survive as the dominant vegetationand desert biomes occur.Desertsare subject to the most extremetemperature fluctuations of any biome. During the day they are exposedto intense sunlight, and the temperature of both air and soil may risevery high. But in the absence of the moderating influence of abundant vegetation, heat is rapidly lost at night, and a short while after sunset, searingheat has usually given way to bitter cold. Some deserts, such as partsof the Sahara, are nearly barren of vegetation, but more commonly thereare scattered drought-resistant shrubs and succulent plants that can storemuch water in their tissues. Most desert animals are active primarily atnight or during the brief periods in early morning and late afternoon whenthe heat is not so intense. Most of them show numerous remarkable physiologicaland behavioral adaptations for life in their hostile environment. We have thus seen that one moving North or South on theearth's surfacemay pass through a series of different biomes. The major biome typeshave been mentioned with the exception of the comparatively unproductivebiomes such as the polar ice caps. Ecologists often differ in theirclassifications of the major patches of the biosphere because the biomesare strictly defined by the elements put into them. The borders of thebiomes shift as we add orsubstractspecies. The point to remember is thatthe biomes seldom exist as sharply defined patches. They have broad borders, and the species that comprise them have weakly correlated geographical ranges.

Question:

At what distance from the center of the earth does the acceleration due to gravity have one half of the value that it has on the surface of the earth?

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Solution:

Newton's Second Law implies that W = mg. W is the weight of an object of mass m (that is, the grav-itational force of attraction between the earth and the object), and g is the acceleration due to gravity. Then, by Newton's Law of universal gravitation, W =GM_em/R^2 = mg where R is the distance of the object of mass m from the center of the earth, and M_e is the mass of the earth. Therefore g(R) =GM_e/R^2 At the surface of the earth, g(R_e) =GM_e/R2_e But we want g(R) = 1/2 g(R_e). Therefore, (GM_e/R^2) = 1/2(GM_e/R2_e) R^2 = 2R2_e R= \surd(2) R_e = 1,414 × 6.38 × 10^6m = 9.02 × 10^6m The acceleration due to gravity is reduced to one half of its usual value at a distance of 9.02 × 10^6 meters from the center of the earth. This is equivalent to a height of 2.64 × 10^6 meters or 1640 miles above the surface of the earth.

Question:

Distinguish between active and passive immunity. How are they produced? Are vaccines andtoxoidsused to induce active immunity or passive immunity?

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Solution:

Immunity is a natural or acquired resistance to a specific disease. A host is immune to a certain pathogen as long as the antibodies specific for the pathogen are present in his circulatory system. The crucial difference between active and passive immunity lies in the answer to the host's question, "Are these anti-bodies made by me or were they formed in some other organism?" Active immunity results when antibodies are produced by the cells (lymphocytes) of the host as a result of contact with an antigen. The antigen may be a microorganism or its product. Active immunity usually develops slowly within a couple of weeks (as antibody production reaches a maximum level), yet it may last up to many years for some antigens. Active immunity may be induced naturally, while recovering from an infectious disease such as mumps, measles, or chicken pox; or artificially, by vaccines. Vaccines are inactivated or attenuated microorganisms which can still stimulate antibody production. Attenuated microorganisms are living, yet too weak to be virulent. Vaccines are available for typhoid fever, poliomyelitis, rabies, smallpox and various other diseases. Also used to induce active immunity aretoxoids, made by destroying thepoisonous poisonous parts of the toxins produced by some pathogens. The antigenic part of the The antigenic part of the toxin is not affected and can still induce antibody production for protection against diphtheria, tetanus and other diseases. Passive immunity is conferred when antibodies produced by active immunity in one organism are transferred to another. Since no antigen is introduced, there is no stimulation of antibody production; hence, passive immunity is of short duration. It does provide immediate protection, unlike active immunity which requires time for development. Passive immunity is also conferred by both natural or arti-ficial means: naturally, by antibody transfer to the fetus from an immune mother (that is, by placental transfer), or artificially, by injection of serum containing antibodies from an immune individual to a susceptible one. In the latter case, it is not necessary that the donor and the recipient be of the same species. It is, however, dangerous for people who are allergic to animal serum, and so must be used with caution. If the answer to the host's question is "These are my own antibodies," then active immunity is involved; if the answer is "These are not my antibodies," then passive immunity is involved.

Question:

A chemist has one mole of X atoms. He finds that when half of the X atoms transfer one electron to the other half, 409 kJ must be added. If all of the resulting X^- ions are subsequently converted to X+ ions, an additional 733 kJ must be added. Find the ionization potential, and the electron affinity of X.

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Solution:

Two facts must be known before this problem can even be attempted: (1) The ionization potential is the energy required to pull off an electron from an isolated atom, and (2) electron affinity is the energy released when an electron adds to an isolated neutral atom. In this problem, there are two different reactions. The first is one half of a mole of neutral atoms losing electrons, and one half of a mole of neutral atoms gaining electrons. The amount of energy needed for this process to occur is the amount of energy expended to remove the electrons minus the amount of energy released by the atoms accepting the electrons. Energy required = ionization potential of X atoms - electron affinity of X atoms. Let x = ionization potential of entire mole, and let y = electron affinity of entire mole. It is given that the energy required for this process is 409 kJ and that only half the mole of atoms is involved. Thus, one can write (1/2) x - (1/2) y = 409 kJ. In the second process, one half of a mole of X^- ions becomes X+ ions. To do this, one must remove an electron from each X^- ion to form an X atom. This amount of energy needed for this process to occur is equal the opposite of the electron affinity. An electron from the X atom is then removed to form the X^+ ion. The amount of energy necessary for this reaction to occur is equal to the ionization potential. The sum of the reverse of the electron affinity and the ionization potential is equal to the energy for the overall process required. One can therefore write: (1/2) x + (1/2) y = 733. The two processes in this problem are represented by two equations using two variables. These equations are solved simultaneously by adding them together. (1/2) x - (1/2) y = 409 (1/2) x - (1/2) y = 733 x = 1142 KJ/mole Substituting this value back into one of the equations, one can solve for y. y = 324 kJ/mole. 96.48 kJ/mole = 1eV/ particle. Therefore, x = 11.84eV/ atomy = 3.36eV/ atom.

Question:

If two 1 gram masses with equal and opposite velocities of 10^5 cm/sec collide and stick together, what is the additional rest mass of the joined pair?

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Solution:

In figure 1, both particles are moving towards each other with the same velocity in our frame of reference. In figure 2 they have collided and stuck together. The velocity of the joined body would seem to be zero, but we must prove that this is so. By the law of conservation of momentum, the total momentum before and after collision must be conserved. We are told that the velocities of the two particles are equal but opposite and we assume that they are collinear so that the problem is one dimensional making the velocities v and - v respectively. There-fore, the momentum before the collision is mv + m(- v). Since this is equal to 0, the momentum after the collision must also equal 0, or m_final v_final = 0. Since no energy is produced in the collision, no mass can be lost (it can only be gained) so that m_final isnot 0, meaning that v_final is 0. Further, we note that the two particles had total kinetic energy 2 \textbullet 1/2 mv^2 before the collision, and 0 kinetic energy after the collision. Therefore, all this energy must have been converted to heat. Since E = mc^2 in the rest frame, this heat energy has an equivalent mass given by \DeltaE = \DeltaMc^2. Hence, \Deltam =(mv^2 - 0) / c^2 \Deltam = { 1gm (105cm/sec )^2 } / ( 3 ×105cm/sec )^2 } = 1.1 × 10^-11 gm

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Question:

Write a program in FORTRAN to prepare a payroll for a business employing N workers. The following computations must be performed: 1) regular hours and overtime must be totaled. 2) wages equal to regular rate times regular hours plus overtime rate times overtime hours. 3) Deduc-tions must be made for insurance taxes, and government income tax. 4) Net wages are printed for each worker. Assume that tax schedules can be accessed from secondary memory.

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Solution:

In preparing a payroll register, the first step is collecting and processing the time cards. The items 1) - 3) are entered by reading a card. This infor-mation is then used in calculating gross and net pay. Since the wages values require the accuracy of only two decimal places (cents), while the FORTRAN calculations are much more accurate, the rounding should be made by asking for only two places in the FORMAT statement. If the computer uses truncation as the approximation routine, .005 should be added to each value before truncating. To calculate the taxes, a table look-up routine must be performed. Assume that Table 1 has already been loaded into an array. Each time a card is read it will give the salary and number of exemptions. 10FORMAT (F 6.2, I2) 15DIMENSION A (28, 8) CREAD SALARY AND EXEMPTIONS 20READ (3, 10) S,E CEX INDICATES WHERE THE TAX WILL BE EX = E + 2 CC INDICATES ROW OP THE TAX C = 1 CR IS THE LOWEST TAXABLE SALARY R = 80 CTAX NOT IN TABLE IF (S.LT.78) GO TO 140 CCHECKING FOR PROPER ROW 100IF (S.LT.R) GO TO 110 CADVANCING TO NEXT SALARY STEP R = R + 2 CADVANCING TO NEXT ROW IN TABLE C = C + 1 GO TO 100 CPRINTS SALARY AND TAX 110WRITE (4,120) S, A (C,EX) 120FORMAT (F6.2, 5X, F 5.2) GO TO 20 140WRITE (4,150) 150FORMAT ("TAX NOT IN TABLE") GO TO 20 STOP END Note that FORTRAN is not the best choice of language for this type of problem as evidenced by the numerous uncon-ditional GO TO statements scattered throughout the program. If the program had been written in COBOL, the structure would have been easier to follow. Married personWeekly payroll period The wages are the number of withholding exemptions claimed. At Least But less than 0 1 2 3 4 5 6 The amount of tax to be withheld shall be. $ 78 80 82 84 86 88 90 92 94 96 98 100 105 110 115 120 125 130 135 140 145 150 160 170 180 190 200 $80 82 84 86 88 90 92 94 96 98 100 105 110 115 120 125 130 135 140 145 150 160 170 180 190 200 210 $11.10 11.40 11.70 12.00 12.30 12.70 13.00 13.30 13.70 14.00 14.40 15.00 15.80 16.70 17.50 18.40 19.20 20.10 20.90 21.80 22.60 23.90 25.60 27.50 29.50 31.50 33.50 $9.10 9.40 9.70 10.00 10.30 10.60 10.90 11.20 11.50 11.80 12.10 12.70 13.50 14.40 15.20 16.10 16.90 17.80 18.60 19.50 20.30 21.60 23.30 25.00 26.80 28.80 30.80 $7.00 7.30 7.60 7.90 8.20 8.50 8.80 9.10 9.40 9.70 10.00 10.60 11.30 12.10 12.90 13.80 14.60 15.50 16.30 17.20 18.00 19.30 21.00 22.70 24.40 26.10 28.10 $5.00 5.30 5.60 5.90 6.20 6.50 6.80 7.10 7.40 7.70 8.00 8.50 9.30 10.00 10.80 11.50 12.30 13.20 14.00 14.90 15.70 17.00 18.70 20.40 22.10 23.80 25.50 $3.00 3.40 3.60 3.90 4.20 4.50 4.80 5.10 5.40 5.70 6.00 6.50 7.30 8.00 8.80 9.50 10.30 11.00 11.80 12.60 13.50 14.70 16.40 18.10 19.80 21.50 23.20 $1.10 1.40 1.70 1.90 2.20 2.50 2.80 3.10 3.40 3.70 4.00 4.50 5.30 6.00 6.80 7.50 8.30 9.00 9.80 10.50 11.30 12.40 14.10 15.80 17.50 19.20 20.90 $0.00 0.00 0.00 0.10 0.30 0.60 0.90 1.20 1.50 1.70 2.00 2.50 3.20 4.00 4.70 5.50 6.20 7.00 7.70 8.50 9.20 10.40 11.90 13.60 15.30 17.00 18.70

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Question:

Sheets of brass and steel, each of thickness 1 cm, are placed in contact. The outer surface of the brass is kept at 100\textdegreeC and the outer surface of the steel is kept at 0\textdegreeC. What is the temperature of the common interface? The thermal conductivities of brass and steel are in the ratio of 2 : 1.

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Solution:

Once equilibrium conditions have been attained, the same quantity of heat must pass through all sections of the system in unit time. In other words, the heat current flowing through the system is constant; otherwise alterations in the temperature at various points would take place. This would be contrary to the condition that equilibrium had been established. The heat H flowing in unit time across the brass is H = K_1A [(100\textdegreeC - t)/L](1) where K_1 is the thermal conductivity of brass, A is the cross-sectional area of the brass slab, and t is the temperature of the common interface. Heat flows from the inner surface of the steel to its outer surface with the same rate, H = K_2A [(t - 0\textdegreeC)/L](2) where K_2 is the thermal conductivity of steel. From (1) and (2), we get K_1A [(100\textdegreeC - t)/L] = K_2A [(t - 0\textdegreeC)/L] orK_1/K_2 = t/(100\textdegreeC - t) But we are given that K_1/K_2 = 2, hence t/(100\textdegreeC - t) = 2 200\textdegreeC - 2t = t ort = (200\textdegreeC/3) = 66.7\textdegreeC.

Question:

A ball of volume 500 cm^3 is hung from the end of a wire of cross-sectional area 2 × 10^-3 cm^2. Young's modulus for the material of the wire is 7 × 10^11 dynes\bulletcm^-2. When the ball is immersed in water the length of the wire decreases by 0.05 cm. What is the length of the wire?

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/Users/wenhuchen/Documents/Crawler/Physics/D10-0407.htm

Solution:

When the ball is immersed in water, it suffers (according to Archimedes' Principle) an up -thrust equal to the weight of a similar volume of water. Thus immersion causes a compressive force on the wire of magnitude F = 500 cm^3 ×\rhog where \rho is the density of water. Hence F = 500 cm^3 × 1 g\bulletcm^-3 × 981 cm\textbullet s^-2 = 49 × 10^4 dynes. But Young's modulus for the wire is given by the formula Y = (F/A) (∆l/l_0), where ∆l/l_0 is the fractional change in length of the wire, A is the latter's cross- sectional area, and F is the force normal to the cross- section of the wire. Hence l_0 =AY∆l/F = [2 × 10^-3 cm^2 × 7 × 10^11 dynes\bulletcm^-2 × 5 ×10^-2 cm]/[49 × 10^4 dynes] = 142.9 cm.

Question:

If log_10 3 = .4771 and log_104 = .6021, find log_10 12.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E33-0961.htm

Solution:

Since 12 = 3 × 4, log_10 12 = log_10 (3) (4). Since log_b(xy) = log_bx + log_b y log_10(3 × 4)= log_10 3 + log_10 4 = .4771 + .6021

Question:

Phosgene is formed at 25\textdegreeC according to the equation CO (g) + Cl_2 (g) \rightarrow COCl_2 (g). CO (g) + Cl_2 (g) \rightarrow COCl_2 (g). Calculate the free energy change ∆G\textdegree and the equilibrium constant k for this reaction.

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Solution:

The free energy change for the reaction is the difference between the sum of the free energies of formation of the products and the sum of the free energies of forma-tion of the reactants. The free energies of formation of CO (g), Cl_2 (g) and COCI_2 (g) are ∆G\textdegree_CO= - 32.8 Kcal/mole ∆G\textdegree_Cl2= 0 Kcal/mole ∆G\textdegree_COCl2 = 50.3 Kcal/mole. Hence,∆G\textdegree = ∆G\textdegree_COCl2 [∆G\textdegree_CO + ∆G\textdegree_Cl2] = -50.3 Kcal/mole - (- 32.8 Kcal/mole + 0 Kcal/mole) = - 17.5 Kcal/mole = - 17, 500 cal/mole. The equilibrium constant is calculated from ∆G\textdegree = - 2.303 RT log k, where R is the gas constant and T is the absolute temperature (T = 25\textdegreeC + 273 = 298\textdegreeK).Then, ∆G\textdegree = - 2.303 RT log k - 17, 500 cal/mole = - 2.303 × 1, 987 cal/mole - deg × 298\textdegreeK × log k. Solving for k, log k = [(- 17, 500 cal/mole)/(- 2, 303 × 1.987 cal/mole - deg × 298\textdegreeK)] = 12.8, = 12.8, k = 101 2 \bullet 8= 6 × 10^1 2 k = 101 2 \bullet 8= 6 × 10^1 2 .

Question:

The daily protein intake of a typical American diet consists of meat, dairy products and eggs. Its average net protein utilization (NPU) is 70. If the daily protein requirement of the body is 0.3 g per pound of body weight, what is the actual protein requirement for an average American?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E32-0940.htm

Solution:

The biological value of a food is the ratio of the nitrogen retained in the body for growth and main-tenance to the nitrogen absorbed in digestion. This ex-presses the fraction of the protein in the food which is used by the body. Another factor is how much of the protein of a food is digested. This factor is called the digestibility. The NPU of a food is the product of the biological value and the digestibility. The reason that nitrogen content is used as a measure of the amount of protein taken in by the body is because proteins are long chains of amino acids. On one end of each amino acid there is an amino group which contains one nitrogen atom and two hydrogen atoms. Therefore, the amount of nitrogen present is proportional to the amount of protein present. The actual protein requirement is defined by the following equation. [basic protein requirement, grams per pound ] [100/NPU] = actual protein requirement. Substituting and solving for the actual protein re-quirement of an average Americans: [0.3 g / pound] [100/70] = [0.43 g / pound] Hence, a 150 lb man needs to eat [0.43 g / lb] × (150 lb) = 64.5 g of protein per day to avoid degradation of his own body protein.

Question:

In the first second of its flight, a rocket ejects 1/60 of its mass with velocity of 6800 ft/sec. What is the acceleration of the rocket?

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Solution:

The acceleration of the rocket at any time is dv \ding{217} /dt= [(F^\ding{217}_ext)/M] + (u^\ding{217}/M) (dM/dt)(1) where M is the instantaneous mass of the rocket,F^\ding{217}_extis the net external force on the rocket-fuel system, u^\ding{217} is the exhaust velocity of the fuel, and dM /dtis the time rate of change of the mass of the rocket. In our case,F^\ding{217}_ext= -Mg\^{J}where \^{J} is a unit vector pointing up from the surface of the earth. Also u^\ding{217} = -u\^{J}. Then dv \ding{217} /dt= -(Mg/M) \^{J}- (u/M) \^{J} (dM/dt) = \^{J}[g + (u/M) (dM/dt)] Now, after the first second of flight, M = M_0 - (M_0/ 60) = (59/60) M_0 where M_0 is the mass of the rocket at t = 0. Also, dM/dt= - (1/60 M_0 / 1 sec) = - [M_0/(60 sec)] where the negative sign accounts for the fact that M is decreasing as time increases .Hence, dv^\ding{217}/dt= - \^{J} [(32f/S^2) + {(6800 f/s)/(59/60 M_0)} (- {M_0/(60 sec)}) ] = - \^{J}(32 f/s^2 - 115.2 f/s^2) = 83.2 f/s^2 \^{J}

Question:

Explain the basic functions of a base register, and describe howit affects the location of a program segment in the main memoryof the IBM/360-370 computers.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G09-0191.htm

Solution:

The IBM/360-370 computers have 16 general-purpose registers consisting of 32 bits each. Any one of these general-purpose registers can be used as a base register. When a register is used as a base register , the 24 rightmost bits are used as a base address. Only 24 bits are needed for a base address since the main memory of the computer has approximately 2^24 storage locations, and each of our 24 bits represents a power of two. This base is the location at which the program segment was loaded into main memory. Since the base register stores the address of a program in main memory, the location of an entire program may be changed just by altering the contents of this register. Thus, a base re-gister aids in the relocation of a program. Sometimes it is necessary to use more than one base re-gister in a program . This may happen in the implementation of a subroutine. When different base registers are used to address different program segments, the process is referred to as the use of Multiple Base Registers.

Question:

A yo-yo rests on a level surface. A gentle horizontal pull (see the figure) is exerted on the cord so that the yo-yo rolls without slipping. Which way does it move and why?

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Solution:

The forces acting on the yo-yo are the horizontal pull F and the frictional force f f = \muW where \mu is the coefficient of friction. Instantaneous rotation takes place about an axis through the point of contact P (not about the center of the yo-yo, although it might appear so) since the instan-taneous velocity of the contact point is zero. Therefore, the yo-yo rolls in the direction of the pull and its rota-tion is determined by the torque about P, \cyrchar\cyrt = Fh. It should be observed that the frictional force does not contribute to this torque, since f is acting at P.

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Question:

Using the timing diagram given in fig. 1, obtain a minimum two-level NOR network.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G03-0059.htm

Solution:

Using the timing diagram given, a function in mintexm list form may be derived. First it is observed from the timing diagram that the function needed has four variables w,x,y, and z. The values that each variable takes when f(w,x,y,z) is 1 are also shown in the diagram. Therefore the minterms found when f(w,x,y,z) = 1 are used to create the minterm function f(w,x,y,z) = \summ (0,2,5,7,9,10,11,12,14) Using the Kamough Map for the given function and plotting the minterms into the map gives the table of fig. 2, from the definition of NOR networks, it is known that a NOR gate gives the complement of the sum of inputs, therefore the NOR network gives the "Product of the Sum" form as output. The complement of the function f(w,x,y,z) therefore is needed. The K-map forf(w,x,y,z) (complement) is given in fig. 3. The next step now is to minimize the map obtained using f (complement). Going through the standard Karnough map minimizing techniques on the terms that are contained in K-map of fig. 2, we group the terms as In the Kmap of fig. 4. Thee resulting minimized function is obtained as follows; Fromm_1 =w x yz m_3 =w xy z total (term 1) =w xz fromm13= w xyz m_15 = w x y z total (term 2) = w x z fromm4=wxy z m_6 =wxy z total (term 3) =wxz fromm_8 = wx y z and the functionf(w,x,y,z) is; f(w,x,y,z) = wxyz+ wxz +wxz +wxz this function is the minimized form of the complement of f(w,x,y ,z) given initially. To find f =fin "Product of Sum" formf(w,x,y,z) must be complemented. Using DeMorgan's Law; Once this function is obtained, to get the NOR network involution may be used. and the logic for this function is shown in fig. 5

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Question:

Given a two-dimensional array AMATR which contains 10 rows and 10 columns and a one-dimensional array called DIAG which contains 10 elements, write a program segment in FORTRAN to compute the elements of D1AG from DIAG(I) = AMATR(I, I) for I =1,2,.. .,10. Then compute the trace of AMATR, which is defined as trace(AMATR) = 10\sum_i= 1AMATR(I, I) = 10\sum_i= 1DIAG(I).

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Solution:

DIMENSION AMATR (10,10), DIAG (10) TRACE = 0.0 DO 100 I = 1,10 DIAG (I) = AMATR (I, I) TRACE = TRACE + DIAG (I) 100CONTINUE STOP END

Question:

Show that a NOT gate can be achieved with the transistor inverter of fig. 1.

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Solution:

The transistor in figure 1 operates in the following manner: If a voltage, V_BE,is impressed at the emitter with respect to the base; and is impressed at the emitter with respect to the base; and V_BE is greater than a threshold voltage V_\Upsilon, then the collector-emitter junc-tion of the transistor becomes a conductor, or a closed switch as shown in fig 2a. If V_BE is less than V_\Upsilon, then the collector-emitter junction becomes an insulator, or an open switch as shown in fig 2b. Thus the transistor behaves as a switch which can be opened or closed by vary-ing V_BE. It is seen from the circuit of fig. 1 that V_BE = V_in; hence when V_in < V_\Upsilon : switch open V_in > V_\Upsilon : switch closed In many transistorsV_\Upsilonis approximately 0.7V; thus when Vin< 0.7V the V_\Upsilon circuit of figure 1 is seen as the circuit of figure 3. Assuming a negligible voltage drop across resistor R, the output voltage, V_out , is 5V. When V > 0.7V fig. 1 is seen as the circuit of fig. 4. In this case V_out is seen to be OV. In reality however, there is a small voltage, V_CESAT, between collector and emitter which is approximately 0.2V. This occurs because the transistor is not an ideal switch. From the information gathered so far, a circuit function table is made V_in V_out < 0.7V 5V > 0.7V 0.2V If logic 1 is assigned to any voltage above, say, 3V and logic O is assigned to any voltage below 0.7V, the truth table will be: Input Output 0 1 1 0 The truth table proves that the circuit of figure 1 is a logic inverter or a NOT gate.

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Question:

40 g of ice at 0\textdegreeC is mixed with 100 g of water at 60\textdegreeC, What is the final temperature after equilibrium has been established? Heat of fusion of H_2O = 80 cal/g, specific heat = 1 cal/g-\textdegreeC.

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Solution:

In a determination of this kind, the heat lost by the water in cooling must be balanced by the heat gained by the ice in melting and in warming the resulting water (fron the melted ice) to the final temperature. This means that the heat absorbed by the ice must equal the heat lost by the water at 60\textdegreeC. The heat absorbed by the ice is the combination of the heat absorbed by the ice in melting and the heat absorbed by the cold water to the new temperature. The heat absorbed when the ice melts can be calculated by taking into account the heat of fusion. The heat of fusion is defined as the number of calories absorbed when 1 g of solid melts. For water, 80 calories are absorbed for each gram of ice melted. Here 40 g of ice is melted. Therefore, the amount of heat absorbed by the ice is 40 times the heat of fusion. 40 g × 80 cal/g = 3200 cal. As such, 3200 calories are absorbed when the ice melts. To find the amount of heat absorbed, when this water (from the melted ice) is heated to a new temperature, you must consider the specific heat of water. The specific heat is defined as the number of calories needed to raise 1 g of liquid 1 degree. The specific heat of water is 1, which means that for every gram of water raised 1 degree, 1 calorie is absorbed. In calculating the amount of heat absorbed by the water to the new temperature, let t = new temperature. To find the amount of heat absorbed, multply the specific heat by 40 g, because 40 g of water is present, and by the new temperature, t, because the water will be raised t degrees from zero. Amount of heat absorbed by liquid = 40 g × 1cal/g-degree)× t = 40 t cal. Therefore, 40 t calories will be absorbed by the melted ice. Thus, the amount of heat absorbed by the ice when it melts, and by the water when it is heated, is equal to the total amount of heat absorbed by the system, Total amount of heat absorbed = 3200 cal + 40 t cal = (3200 + 40 t) cal. To find the amount of heat lost by the 100 g of water at 60\textdegreeC, when it is cooled to a new temperature, t, the specific heat of water is also used. Here, it means that for every gram of water that is lowered one degree, 1 calorie of heat is released. Therefore, to find the amount of heat released by this water, the specific heat will be multiplied by 100 g, the amount of water present, and by 60-t\textdegree, which is the number of degrees that the temperature will drop. (60 - t)\textdegree × (1 cal/g-degree) × (100 g) = (6000 - 100t)cal. To calculate t, one must set the amount of heat absorbed by the ice equal to the amount of heat lost by the water at 60\textdegreeC. amount of heat absorbed= amount of heat lost amount of heat absorbed= (3200 + 40 t) cal amount of heat lost= (6000 - 100 t) cal (3200 + 40 t) cal= (6000 - 100 t) cal 140 t= 2800 t = 20\textdegree Therefore, the final temperature is 20\textdegreeC.

Question:

In the following pair of geometric isomers, designate which is cis and which is trans.

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Solution:

Isomers, i.e. compounds of the same molecular formula, that differ only in their geometry are termed geometric isomers. Geometric isomerism is one type of stereoisomerism; that is, isomerism due to the arrangement of atoms in a molecule in three dimensions. Geometric isomers are different compounds with different properties. Since the order of attachment of the atoms is the same, they are not structural isomers. The isomer that has the two groups on the same side of the double bond is called the cis isomer and the isomer that has the two groups on the opposite sides is called the trans isomer. In (1), the hydrogens and, as such, the Cl and CH_2CH_2CI groups are located on the same sides of the double bond. Therefore, this is the cis isomer. In (2), however, the groups are located on the opposite side of the double bond. Consequently, this is the trans isomer.

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Question:

If mating is observed to occur freely between organisms of twodifferent populations in the laboratory, and the cross producesviable offspring, can you assume that the two populationsbelong to the same species?

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Solution:

Successful interbreeding in the laboratory does not offer enough evidencethat two populations must be of the same species. It only indicatesthat certain types of intrinsic isolation does not exist between thepopulations. It does not say whether effective inter-breeding will occur innature, nor does it clarify whether or not other kinds of intrinsic isolation arepresent. By definition, two individuals are from the same species if no intrinsicreproductive barrier exists between them to prevent a genetic exchangein nature. Extrinsic isolation, commonly exemplified by geographicseparation, will not cause speciation if intrinsic isolation is not established. In other words, two groups of or-ganisms separated by, say, a mountainrange, cannot be regarded as two species unless they fail to successfullyinterbreed when the separating factor is removed. Whereas interbreeding of two populations under a laboratory setting demonstratesthat certain intrinsic isolating mechanisms, such as mechanicalandgameticisolations, does not exist between them, it does notpreclude the presence of other intrinsic isolating factors that are very muchdependent on the natural environment. For example, under natural conditions,ecogeographic, habitat isolation, or seasonal isolation may exist, yet be inoperative under laboratory conditions. Observation of breedingin the laboratory cannot ascertain whether fertile mating would occur, should individuals of the populations be released from the laboratoryand returned to their own natural habitats, which we assume hereto be continuous with each other. Under this assumption, even thoughno physical barrier is present, the existence of some type of intrinsicisolation which is nature-dependent would necessarily inhibit gene flowfrom one population to the other. In short, since experimental finding cannotpredict accurately what is to happen in nature, and since speciationis a phenomenon of the natural world rather than the artificial laboratory, we cannot assume that two populations that interbreed in the laboratory necessarily would do so in nature.

Question:

Write the necessary statement function(s) to compute the area of a triangle, given 3 sides. Use Hero's formula.

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Solution:

The area of a triangle of sides a, b, and c is given by Area = \surd[s(s - a)(s - b)(s - c)] where s = 1/2(a + b + c) Hence, using one statement function to define s and a second to define Area, we get: S(A,B,C) = 0.5 {_\ast} (A + B + C) X = S(A,B,C) AREA(A,B,C,X) = SQRT(X{_\ast}(X-A){_\ast}(X-B){_\ast}(X-C))

Question:

In an air conditioning process a room is kept at 290\textdegreeK., while In an air conditioning process a room is kept at 290\textdegreeK., while the temperature outside is 305\textdegreeK. The refrigerat-ing machine has compression cylinders operating at 320\textdegreeK. (located outside) and expansion coils inside the house operating at 280\textdegreeK. If the machine operates reversibly, how much work must be done for each transfer of 5000 joules of heat from the house? What entropy changes occur inside and outside the house for this amount of refrigeration?

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Solution:

An engine works by accepting heat at a high temperature, converting part of it into work, and rejecting the remainder at a lower temperature. A re-frigerating machine works the opposite way. By supply-ing to it an amount of work W, it will accept heat Q_1 at a low temperature and reject heat Q_2 = Q_1 + W at a higher temperature. The efficiency \eta of a reversible heat engine is given by \eta = W/Q_1 = (Q_1 - Q_2)/Q_1 = (T_1 - T_2)/T1 where W is the net work done by the machine in one cycle and Q_1 is the heat absorbed in one cycle. Since a refrigerator and a heat engine operate in opposite ways, and Q_2 > Q_1, its efficiency can be defined as \eta = W/Q_2 = (Q_2 - Q_1)/Q_2 = (T_2 - T_1)/T_2 = 1 - (T_1/T_2) where W is the work done on the machine. The sequence of events is shown in the diagram. The machine operates between reservoirs at 280\textdegreeK. and 320\textdegreeK. If Q_1 = 5000 joules, we have \eta= 1 - (280/320) = 0.125 = (Q_2 - Q_1)/Q_2 = (Q_2 - 5000 J)/Q_2 Q_2= 5715 J The work done can be found from \eta = W/Q_2 orW = \etaQ_2 = 0.125 × 5715 J \approx 715 J Irreversible transfers of heat occur when the entropy of the system increases. This occurs for a transfer of heat from a high temperature reservoir to a low temperature reservoir. In this problem, there are two irreversible transfers of heat, one between the inside of the house and the low temperature reservoir, for which entropy,∆S= Q_1[{1/(Texp coil)} - {1/(T_inside)}] = Q_1[(1/280) - (1/290)] = 0.62 J/\textdegreeK. and one between the high temperature reservoir and the outside: ∆S= Q_2[{1/(T_out)} - {1/(T_cylinder)}] = Q_2[(1/305) - (1/320)] = 0.88 J/\textdegreeK.

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Question:

The lymphatic system in man constitutes an extensive network of thin vessels resembling veins. What are the functions of the lymphatic system?

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Solution:

The lymphatic are not part of the circulatory system per se, but constitute a one-way route from interstitial fluid to the blood. The lymphatic system in man constitutes an extensive network of thin-walled vessels resembling the veins. These vessels ultimately drain into veins in the lower neck. One of the function of the lymphatic system is to transfer excess interstitial fluid back to the blood. There is a net movement of plasma out of the capillaries and into the tissues; the lymphatics restore the blood volume by returning this fluid that has filtered out. In addition, the lymphatic system serves to return proteins to the blood. Since the capillaries have a slight permeability to plasma proteins, and since the concentration of proteins in the blood plasma is greater than that in the interstitial fluid, there is a small but steady loss of protein from the blood into the interstitial fluid. This protein returns to the blood via the lymphatics. Should there be a malfunction in the lymphatic system, the interstitial fluid protein concentration would increase to that of the plasma. This eliminates the protein concentration difference between the plasma and the interstitial fluid, thus eliminating the plasma colloid osmotic pressure. Only blood pressure and interstitial fluid pressure remain, and permit the net movement of fluid out of the capillary into the interstitial space (edema). Another function of the lymphatic system is to provide the pathway by which fat and other substance absorbed from the gut reach the blood. It is also believed that certain high-molecular weight hormones reach the blood via the lymphatics. In addition to its function in transport, the lymphatic system plays a critical role in the body's defense mechanism against disease. The lymph nodes, which are found at that junctions of lymph vessels, act as filters and are sites of formation of certain types of white blood cells. Lymph, the fluid in the lymph capillaries, flows slowly through the nodes where invading bacteria are phagocytosed by the cells of the lymph node. Indigestible particles such as dust and soot, which the phagocytic cells cannot destroy, are stored in the nodes. Since the nodes are particularly active during an infection, they often become swollen and sore, as the lymph nodes at the base of the jaw are apt to become during a throat infection. It should be noted that since the lymphatic system is not connected to the arterial protion of the blood circulatory system, lymph is not moved by the hydrostatic pressure developed by the heart. Lymph flow, like that in the veins, depends primarily upon forces external to the vessels. These forces include the contractile action of the skeletal muscle (through which the lymphatics flow) and the effects of respiration on the pressure in the chest cavity. Since the lymphatic have valves similar to those in veins, external pressure would permit only unidirectional flow.

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Question:

a) log_1010b) log_10100c) log_101d) log_10 0.1e) log_10 0.01

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Solution:

The logarithmic expression N = log_bx is equivalent to b^N = x. Hence, a) Let N_1, = log_10 10. Then the logarithmic expression N_1 = log_1010 is equivalent to 10^(N)1 =10. Since 10^1 = 10,N_1 = 1. Therefore, N_1 = 1 = log_1010. b) Let N_2 = log_10 100. Then the logarithmic expression N_2 = log_10100 is equivalent to 10^(N)2 = 100. Since 10^2 = 100, N_2 = 2. Therefore, N_2 = 2 = log_10 100. c) Let N_3 = log_10 1. Then the logarithmic expres-sion N_3 = log_10 1 is equivalent to 10^(N)3 = 1. Since 10^0 =1, N_3 = 0. Therefore, N_3 = 0 = log_101. d) Let N_4 = log_10 0.1 = log_10 (1/10). Then the loga-rithmic expression N_4 = log_1 0.1 = log_10 (1/10) is equiva-lent to 10^(N)4 = (1/10). Since 10^\rule{1em}{1pt}1 = 1/(10^1) = 1/10, N_4 = \rule{1em}{1pt}1. Therefore, N_4 = \rule{1em}{1pt}1 = log_10 0.1. e) Let N_5 = log_10 0.01 = log_10 (1/100). Then the loga-rithmic expression N_5 = log_10 0.01 = log_10 (1/100) is equiva-lent to 10^(N)5 =(1/100). Since 10^\rule{1em}{1pt}2 =(1/10^2) = 1/100, N_5 = \rule{1em}{1pt} 2. Therefore, N_5 = \rule{1em}{1pt}2 = log_10 0.01.

Question:

A block starting from rest slides a distance of 5 meters down an inclined plane which makes an angle of 37\textdegree with the horizontal. The coefficient of friction between block and plane is 0.2. (a) What is the velocity of the block after sliding 5 meters? (b) What would the velocity be if the coefficient of friction were negligible?

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Solution:

The change in potential energy of the block is ∆PE = mgh where h is the height of the block and equals. 5 sin 37\textdegree . Some of this energy goes into doing work against the frictional force f. The fric-tional force is proportional to the normal force N. The block is in equilibrium in the direction perpendicular to the inclined plane. There-fore, the sum of the forces in that direction must equal zero. N - mg cos 37\textdegree = 0 thusN = mg cos 37\textdegree Then the frictional force is F_f = \muN = 0.2 mg cos 37\textdegree where \mu is the coefficient of friction. The energy expended in com-bating this force equals the work done against it. The work equals the product of the frictional force and the distance over which it acts. W = F_fd = (0.2 mg cos 37\textdegree) (5) = mg cos 37\textdegree From the conservation of energy principle, the change in potential energy of an object equals its change in kinetic energy plus the work it does. ∆PE = W + ∆KE = W + (1/2)m (v^2 - v^2_0) Since the initial velocity v_0 of the block is zero, and its height h is d sin 37\textdegree = 5 sin 37\textdegree, ∆PE= mgh = mg cos 37\textdegree + (1/2) mv^2 (1/2) v2= g(h - cos 37\textdegree) = g(5 sin 37\textdegree - cos 37\textdegree) v= \surd[2g(5 sin 37\textdegree - cos 37\textdegree)] = \surd[2(9.8){(5)(0.602) - (0.799)}] = \surd43.3 \approx 6.57 m/sec where v is the final velocity of the block. (b) If friction can be neglected, then we have ∆PE= ∆KE or mgh= (1/2)mv^2 g(5 sin 37\textdegree)= (1/2)v2 v= \surd[(2)(g)(5 sin 37\textdegree)] = \surd[(2)(9.8)(5)(0.602)] = \surd59 \approx 7.68 m/sec

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Question:

The earliest vertebrate fossils date back to the Ordovi-cian period, 500 million years ago. What distinguishes these early vertebrates from later vertebrates?

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Solution:

These early vertebrates belong to the class Agnatha. The organisms of this class have cylindrical bodies up to a meter long, with smooth scaleless skin. Unlike other vertebrates, these organisms lack jaws ("agnatha" means jawless) and paired fins. In addition, they are also the only parasitic vertebrates, feeding primarily on the fishes. Members of this class include the lampreys and hagfishes, and the extinct ostracoderms (the earliest Agnaths). Lampreys and hagfishes constitute the cyclostomes, and have a circular sucking disc around the mouth, which is located on the ventral side of the anterior end. They attach themselves by this disc to other fish, and using the horny teeth on the disc and tongue, bore through the skin and feed on blood and soft tissues of the host. Some may bore completely through the skin and come to lie within the body of the host. The ostracoderms were the only Agnaths that were not parasitic. Since they were the first vertebrates, there were no fish as yet to exploit as food. It is assumed that the ostracoderms obtained food by filtering mud. In the cyclostomes, the notochord persists in the adult as a functional supporting structure. The gill slits and dorsal hollow nerve cord are present in the larva called an ammocoetes and adult, and the vertebrae are rudimentary, consisting only of a series of cartilaginous arches that protect the nerve cord.

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Question:

Assuming the Bohr radius is 0.5292 × 10^-8 cm, what frequency of a light wave has the same wavelength as the Bohr radius?

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Solution:

The question asks one to determine the frequency of a light wave with a wavelength equal to the Bohr radius, .5292 × 10^-8 cm. In answering this question, it should be remember that there is a relation between frequency and wavelength, c =\lambdaѵ, where c = the speed of light (2.998 × 10^10 cm/sec), \lambda = wavelength, and ѵ = frequency. One can ѵ in this equation by substituting in the values of c and \lambda. Thus ѵ = (c/\lambda) = [(2.998 × 10^10 cm/sec)/(.5292 × 10^-8 cm)] = 5.665 × 10^18 sec^-1.

Question:

Translate the following flowchart into a pseudocoded program and a BASIC program:

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Solution:

You probably don't know what this flowchart means at this point. Actually, if you can understand the flowchart symbols alone, you could render the flowchart into pseudocode without understanding the meaning of the problem. This is not advisable, but sometimes programmers are asked to write programs without being given the theory behind the algorithm. If you are familiar with the three basic control constructs - sequencing, looping, decision-making - you can design programs to execute many kinds of tasks. In this example, the logic is straightforward, but the idea behind it may not be so. This flowchart illustrates a program that determines how many of each part is needed for a specific purpose. The process uses the term explosion matrix to find the expense of materials and labor if a contractor plans to build 6 ranches, 11 split-levels, and 15 Tudors in the next year. Assume the average building costs are as given in the table: ExpensesRanchesSplit-LevelsTudors material175001620012400 labor620055004900 We can write the table above and the desired number of homes as the product of two matrices. Thus: \mid175001620012400\mid\mid6\mid\midmaterials\mid \mid620055004900\mid\mid11\mid=\midlabor\mid \mid15 \mid The program outputs the answers in the following format: Ranches Splits Tudors Materials Labor 1 2 3 \bullet \bullet \bullet 10 1 3 5 \bullet \bullet \bullet 19 5 7 9 \bullet \bullet \bullet 23 TOTAL COSTS First, we present the pseudocoded version. Notice that we have referenced array elements with zero subscripts. This was done to ac-commodate this convention in the BASIC language. real A(1,2), B(2,0), C(1,0) output headings: 'RANCHES SPLITS TUDORS MATERIALS LABOR' input data into A M \leftarrow 0.0 L \leftarrow 0.0 do for R = 0 to 9 B(0,0) \leftarrow R + 1 . B(1,0) \leftarrow 2. \textasteriskcentered R + 1 . B(2,0) \leftarrow 2. \textasteriskcentered R + 5. C \leftarrow A \textasteriskcentered B (A, B, and C refer to entire matrices) output B(0,0), B(1,0), B(2,0), C(0,0), C(1,0) m \leftarrow m + C(0,0) L \leftarrow L + C(1,0) end do for output 'TOTAL COSTS', M,L end program In BASIC, we add DATA statements and MAT instructions to calculate the product. The BASIC program is given below: 10DIM A(1, 2), B(2, 0), C(1, 0) 12PRINT 14PRINT "RANCHES\textquotedblright, "SPLITS", "TUDORS", "MATERIALS", "LABOR" 15PRINT 20MAT READ A 25LET M = 0 26LET L = 0 30FOR R = 0 TO 9 34LET B(0,0) = R + 1. 36LET B(1,0) = 2. \textasteriskcentered R + 1. 38LET B(2,0) = 2. \textasteriskcentered R + 5. 40MAT C = A \textasteriskcentered B 45PRINT 50PRINT B(0,0), B(1,0), B(2,0), C(0,0), C(1,0) 54LET M = M + C(0,0) 58LET L = L + C(1,0) 60NEXT R 62PRINT 70PRINT "TOTAL COST = ", M,L 80STOP 82DATA 17500, 16200, 12400 84DATA 6200, 5500, 4900 99END

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Question:

It is known that the heat of vaporization of water is 5 times as great as the heat of fusion. Explain this fact.

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Solution:

The heat of vaporization is the quantity of heat necessary to vaporize 1^4 g. of a liquid substance at its boiling point at constant temperature. The heat of fusion is the quantity of heat necessary to liquefy 1^4 g. of a solid substance at constant temperature at its melting point. With this in mind, let us consider what goes on in each of these phase changes. The volume-increase of the gas in going from a liquid state to a vapor state is much greater than the volume-increase accompanying a solid to liquid transformation. When you increase volume, work is required to overcome the existing external pressure that hinders the volume expansion. Work is defined as the product of the pressure and the change in volume. Since, the volume in-crease is larger in converting from liquid to vapor, more energy is necessary because more work is done. More importantly, however, is that the molecules of a substance in the gaseous state are so much further apart than those in the liquid state. To bring about this separation of particles requires tremendous energy. This is the main reason that the heat of vaporization exceeds the heat of fusion.

Question:

The same quantity of electricity was passed through two separate electrolytic cells. The first of these contained a solution of copper sulfate (CuSO_4) and exhibited the following cathode reaction (reduction): Cu^2+ + 2e^- \rightarrow Cu(s) . The second of these contained a solution of silver nitrate (AgNO_3) and exhibited the following cathode reaction: Ag^+ + e^- \rightarrow Ag(s). If 3.18g of Cu were deposited in the first cell, how much (Ag) was de-posited in the second cell?

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Solution:

The solution to this problem involves calculating how many Faradays were passed through the first cell (and similarly, the second cell) and, then, using this value to calculate the amount of Ag that will be deposited in the second cell. By definition, one Faraday ( 1F ) will deposit the equivalent weight of any element. Also, 1F is equal to the charge of one mole of electrons. From the reaction Cu^2++ 2e^- \rightarrow Cu(s) we see that it takes two moles of electrons (2 F) to deposit one mole of copper. Hence, 1 F will deposit one-half mole of Cu and the equivalent weight of Cu is, therefore, (1 / 2) the atomic weight, or 1 / 2 × 63.6 g / mole = 31.8 g / mole. From the reaction Ag^+ + e^- \rightarrow Ag(s) we see that one mole of electrons (1 F) will deposit one mole of silver, so that the equivalent weight of Ag is equal to its atomic weight, or 108 g / mole. Since 3.18g of Cu were deposited and this is equal to one-tenth the atomic weight (0.1 × 31.8 = 3.18), one-tenth of a Faraday (0.1 F) was passed through the CuSO_4 cell. Since the same quantity of elec-tricity was passed through both cells, 0.1 F was also passed through the AgNO_3 cell. The amount of Ag deposited is then amount Ag = number of moles of electrons × equivalent weight of Ag = 0.1F × 108 g / mole electrons = 0.1 mole electrons × 108 g /mole electrons = 10.8g.

Question:

What is the partial pressure of each gas in a mixture which contains 40 g. He, 56 g. N_2, and 16 g. O_2, if the total pressure of the mixture is 5 atmospheres.

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Solution:

In a mixture of gases, the partial pressure of each gas is proportional to its mole fraction. Here, to calculate the partial pressures of the various gases con-tained in this system, one must first calculate the mole fraction of each component. This is done by calculating the number of moles present of each component and dividing that by the total number of moles present in the system. To calculate the partial pressure of each component, the mole fraction must then be multiplied by the total press-ure of the system. (a) To calculate the number of moles present of each component, divide the number of grams present by the molecular weight of the element. number of moles = number of grams present / molecular weight Moles of He = 40 g / (4 g/mole)= 10 moles Moles of N_2 = 56 g / (28 g/mole) = 2 moles Moles of O_2 = 16 g / (32 g/mole) = 0.5 moles (b) To calculate the total number of moles present, add the number of moles of all components together total number of moles = moles He + moles N_2 + moles O_2 = 10 + 2 + 0.5 = 12.5 moles (c) To calculate the mole fraction of each component, divide the number of moles present of each component by the total number of moles in the system. mole fraction He = 10 / 12.5 = .80 mole fraction N_2 = 2 / 12.5= .16 mole fraction O_2 = 0.5 / 12.5 = .04 (d) To find the partial pressure of each component, multiply the mole fraction by the total pressure in the system. partial pressure He = (.80) × 5atm= 4atm partial pressure N_2 = .16 × 5atm= 0.8atm partial pressure O_2 = .04 × 5atm= 0.2 atm.

Question:

The reaction C_2H_4O \rightarrowCH_4 + CO is a first-order reaction with a specific rate constant of .0123 min^-1 at 415\textdegreeC. Calculate the percent of original C_2H_4O present that will be decomposed after heating at 415\textdegreeC for 1 hour.

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Solution:

First find the percentage of starting material left (as compared to the initial amount C_0), C/C_0 = (100); then the percent decomposition equals 1 - C/C_0 (100) . To find C/C_0 (100), employ the fact that 2.303 log C_0/C = kt , where C_0 = initial concentration, C = existing concentration k = specific rate constant, and t = time in minutes for a first order reaction. The formula can be rewritten as C/C_0 = antilog (- kt / 2.303) . Substitute the values for k and t to obtain C / C_0 = antilog [{(- .0123)(60)} / (2.303)] = antilog (-.320) = .479 . Thus, C/C_0(100) = 100 × .479 = 47.9%. That is, .479 or 47.9% of the original concentration remains after 60 minutes, so that 1 - 47.9% or 52.1% has decomposed.

Question:

A deficiency in which endocrine gland would result in tetany?

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Solution:

Tetany, a disease characterized by skeletal muscular tremors, cramps, and convulsions is caused by an insufficiency of calcium available to cells. It is of great physiological importance that an adequate quantity of calcium ions be present in body cells in order to maintain a normal membrance excitability. In tetany, due to the deficiency of calcium, irritability of nerve and muscle cell membranes is increased. Thus, muscle contractions become spastic and poorly controlled. It is the primary function of parathormone to increase the availability of calcium to the body cells. Parathormone promotes the transport of calcium from the intestinal lumen into the blood, the release of calcium from the bones, and the reabsorption of calcium by the kidney tubules. Should the parathyroid gland be removed, or hypoactivity of the gland occur, the person would suffer from tetany. This would happen because the calcium levels in his blood would decrease. Severe cases of tetany can cause oxygen exhaustion leading to death by asphyxia. If a solution of a calcium salt is injected into the vein of a person in tetanic convulsions, the convulsions will cease.

Question:

Gastrulation in the bird is accomplished by a form of cell migration that differs from that of the frog. Explain.

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Solution:

In birds, the cells from the epiblast (the upper of two embryonic cell layers) migrate down-ward. Initially, a strip of epiblast extending forward in the midline of the embryo from the posterior edge of the area pellucid becomes thickened as the primitive streak, with a narrow furrow, the primitive groove, in its center. Hensen's node, a thickened knot of cells, is present at the anterior end of the primitive streak. It is by the migration of cells from the lateral portion of the epiblast that the primitive streak is formed. The cells of the epiblast invaginate at the primitive streak, move into the space between the epiblast and hypoblast and reach the latter, forming a mass of moving cells. The cells continue to migrate, moving laterally and anteriorly from the primitive streak area. Though cells are migrating, the shape of the primitive streak remains intact. The primitive streak, where invagination occurs, is considered to be homologous to the amphibian blastopore. The cells which form the notochord arise from Henson's node. The notochord is the rod-shaped structure that forms a skeletal axis in chordates which is re-placed in the vertebrates by vertebrae (in frogs, a special area of mesoderm develops into the notochord). Cells which grow laterally and anteriorly from the primitive streak between the epiblast and the hypoblast become the mesoderm. The original hypoblast cells plus other cells that migrate into the lower layer from the primitive streak form the endoderm, which gives rise to the digestive tract and yolk sac. The epiblast forms the ectoderm.

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Question:

A mass M is connected to a linear spring of stiffness K. The masspring system is initially at rest. At time t = t_0, the system is displaced a distance ofx_0 , and released. It can be shown that the motion of the system is described by the following differential equa-tion: Mẍ + Kx(t) = 0 where x(t) is the displacement of the mass and ẍ is the second derivative of x with respect to time. If v_0 is the initial velo-city, construct a model of this system and devise a digital program which simulates the motion of the system from time t = t_0 to time t = t_f.

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Solution:

The general procedure for developing a model for simulation purposes can be summarized as follows: 1.Definition of the problem and the system that is being studied. 2.Making simplifying assumptions concerning the components of the system. 3.Determining the factors outside the system's environment (exogenous factors) that will affect it. 4.Formulation of a mathematical model (if possible). 5.Compiling the relevant data. 6.Construction of a computer program of the model. Applying these steps to this particular problem: 1.The system consists of the mass and the spring. The problem of following the motion of the system can be viewed through the change in displacement, x. So essentially, the problem is to solve the given differential equation and record samples of the values x assumes from t = t_0 to t = t_f . 2.The differential equation given, results from assuming that M and K are time invariant (constant). Also, the mass of M is considered "lumped" at a single point. 3.The differential equation does not include the effect of air resistance for simplicity. 4.The differential equation is a dynamic mathematical model of the system. 5.The data available is, t_0, t_f, x(t_0) = x_0 ; v(t_0) = v_0, K and M. 6.Although analytical methods exist for solving second-order linear differential equations with constant coefficients such as the one given, digital simulation is applicable even when no analytical method applies. The numerical method used here is called the Euler predictor-corrector method (slow, but straightforward and very general). To use it we must transform the equation to a system of simultaneous first-order equations; using the substitution v = x: ẋ = v(t)(1) v̇ = -(K/M) \textasteriskcentered x(t)(2) The flow-chart for the subroutine is as follows. The program closely follows the flow chart: (I/O statements can be appropriately formatted): REAL K,M,N READ K,M,N,T,TFIN, XCOR, VCOR, ACCVR CN IS THE NUMBER OF SAMPLES TO BE CHOSEN DT = (TFIN - T)/N CDT IS THE INCREMENT SIZE 50X = XCOR V = VCOR A = -(K/M){_\ast}X PRINT T,X T = T + DT IF (T.GT.TFIN) STOP XPRED = X +.V {_\ast} DT VPRED = V + A {_\ast} DT 100XDOT = VPRED CEQUATION (1) VDOT = -(K/M) {_\ast} XPRED CEQUATION (2) XCOR = X + 0.5 {_\ast} (V + XDOT) {_\ast} DT VCOR = V + 0.5 {_\ast} (A + VDOT) {_\ast} DT XDIF = ABS (XCOR - XPRED) VDIF = ABS (VCOR - VPRED) IF = (XDIF. LE. ACCUR. AND. VDIF. LE. ACCUR) GO TO 50 XPRED = XCOR VPRED = VCOR GO TO 100 END

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Question:

Two masses m_1 = 5 kg and m_2 = 10 kg have velocities u^\ding{217}_1= 2m/sec in the +x direction and u^\ding{217}_2= 4 m/sec in the +y direction. They collide and stick together. What is their final velocity after collision?

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Solution:

The total x and y components of linear momentum must be conserved after the collision. The mass of the body resulting after the collision is m = m_1 + m_2 and the velocity v^\ding{217} is inclined at angle \texttheta to the x axis. We know that the total momentum vector is unchanged, and we can write down the x and y components of momentum. INITIAL MOMENTUM FINAL MOMENTUM x component m_1u_1 (m_1 + m_2)\cup cos \texttheta y component m_2u_2 (m_1 + m_2)\cup sin \texttheta m_1u_1 = (m_1 + m_2)\cup cos \texttheta m_2u_2 = (m_1 + m_2)\cup sin \texttheta ortan \texttheta = (m_2u_2)/(m_1u_1) = 4 x 10/ 5 x 2 = 4 \texttheta = 75.97\textdegree From the first momentum equation above: \cup = (m_1u_1)/[(m1+ m_2) cos \texttheta] = (2 × 5)/(15 × 0.2424) = 2.750 m/sec.

Question:

What problems are posed in referring to the many forms of protozoans as belonging to one single animal phylum?

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Solution:

The first problem lies in the fact that the unicellular level of organization is the only characteristic by which the phylum can be described; in all other respects the phylum displays extreme diversity. Protozoans exhibit all types of symmetry, a great range of structural complex-ity, and adaptations for all types of environmental condi-tions. Although all of them have remained at the unicel-lular level, they have evolved along numerous lines through the specialization of the protoplasm. Specialization has occurred through the evolution of an array ofsubcellularorganelles. A second problem involves taxonomic organization. As is true of any taxonomic category, aphlyumshould contain members which are derived from a common ancestral form. In the classification of theprotozoans, virtually all motile unicellular organisms have been grouped into a single phylum, with very little regard to evolutionary relationships. It is among the flagellateprotozoansthat the concept of a single subphylum produces the greatest aberration. Most of the unicellular free-living flagellates are organisms which when assembled constitute a collection of largely unrelated forms. A third problem lies in the possession of certain plant-like features, such asautotrophismand the presence of chloroplasts. In fact, certain green flagellate "protozoans" appear to be rather closely related to uni-cellular green algae which belong to the KingdomPlantaeThus, not only is the concept of a single phylum question-able, there is the additional problem of whether the Protozoa is a true animalphlyum.

Question:

A man has a concave shaving mirror whose focal. length is 20 in. How far should the mirror be held from his face in order to give an image of two-fold magnification?

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Solution:

An erect, virtual, magnified image is desired. With q as the distance between the mirror and image, and p the distance between the mirror and the man's face, the equation M = (p M = (p /p) can be used. M represents the ratio of the size of the image to the size of the actual object. This relation between p and q is without regard to sign. Since the image is virtual, it lies behind the mirror. Distances in front of the mirror are positive and distances behind the mirror are 'negative. Therefore q is negative. To compensate for this, a negative sign is placed in front of q so as to make the overall expres-sion positive. For a two-fold magnification, M = (-q/p) = 2 q = -2p. Substitution in the general mirror equation (1/p) + (1/q) = (1/f) gives(1/p) + (1/-2p) = (1/20 in.) {(2 - 1)/2p} = (1/20 in.) p = 10 in.

Question:

Whileencystmentis not, in most cases, a means of reproduction, still it is a means of continuation of the species. Explain howencystmentis used, using acystforming organism as illustration.

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Solution:

Encystmentis used by a number ofprotozoanswhich live in the water. Should their pond become des-iccated or uninhabitable, these organisms would develop hard, protective coatings and the organisms would go into a state of dormancy. In this encysted state they can withstand extremely unfavorable conditions. When favorablecondirions return (such as clean water), the wall of the cyst softens and the organisms escape. Some pathogenicprotozoansdevelop cysts which play a role in the spread of infection.Entamoebahistolyticais a parasitic protozoan which lives in the lining of the human large intestine, causing the disease amoebic dysentery. This parasite secretes a tissue-dissolving enzyme which causes ulceration of the large intestine. Some of theseprotozoans die when they pass from the body into the feces, but some of these organisms develop cysts and continue to live in dry, unfavorable conditions outside the body for a long time. Should such cysts enter the body of another person, the wall of the cysts would dissolve and the parasites would escape into the new host's body. A new in-fection would thus begin.

Question:

Write the micro-operations for the Fetch Cycle. Assume that the computer has the following registers and control signals: M - memory PC - program counter MAR - memory address register MBR - memory buffer register OPR - op-code register F } - cycle identification registers R } I - indirect register The cycle code is shown in Figure 1. Cycle ID Registers FR Cyclecode Cycledescription 00 C_0 Fetch cycle 01 C_1 Indirect cycle 10 C_2 Execute cycle 11 C_3 Interrupt cycle Fig. 1 Each cycle consists of four steps t_0, t_1, t_2, t_3.

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Solution:

During the fetch cycle, the instruction is read from memory. The memory location which contains the instruction is specified in the PC register. In most computers, the instructions are stored in memory sequentially. That is, the next instruction is at the location of the previous instruction plus one; hence, the PC will be incremented during each fetch cycle. The fetch cycle is designated by the C_0 control signal. A step is specified when the control signal expression to the left of the colon is true (Logical 1). C_0t_0 :MAR\leftarrow [PC](Step 1) Note: [] means "contents of" In the first step of the fetch cycle, the contents of the PC, which contains the location of the next instruction to be read from memory, are transferred to the MAR. C_0t_1 :PC \leftarrow [PC]+1 ; MBR \leftarrow [M](Step 2) Two operations take place in the second step of the fetch cycle: (1) The PC is incremented so that it will specify the location of the next instruction when the next fetch cycle occurs. (2) The contents of memory, whose address is specified by the MAR, is transferred to the MBR; hence, the MBR contains the instruction code. These two operations occur simultaneously with both operations having equal importance. C_0t_2 :I \leftarrow [MBR]_0 ;OPR \leftarrow [MBR]_1-3(Step 3) Again, two operations take place in the third step of the fetch cycle: (1) The contents of bit 0 of the MBR is transferred to I. This bit specifies whether or not the data in the instruction will be indirectly addressed. (2) The contents of bits 1 through 3 of the MBR are transferred to the OPR. Once the op-code is in the OPR, it will be decoded into one of eight possible signals, q_0 - q_7. q_7IC_0t_3 :R \leftarrow 1 (Step 4) or(q_7 + I ̅) c_0t_3 : F \leftarrow 1 Different operations take place in the fourth step of the fetch cycle, depending on the status of several con-trol signals, q_7 is excited when the op-code is 111 (an I/O or register-reference instruction) and I is excited when there is an indirect instruction; hence, the first statement states that when there is an indirect memory reference instruction, go to the indirect cycle. The computer goes to the indirect cycle by letting the R-register become 1 and leaving the F-register 0. When F and R are decoded, the c_1 control line is excited, which designates the indirect cycle. The second statement states that when there is a memory-reference or I/O instruction, go to the execute cycle. The computer goes to the execute cycle by letting the F register become 1 and leaving the R register 0. When F and R are decoded, the c_2 control line, which designates the execute cycle, is excited.

Question:

What are hydrogen bonds? Describe fully the importance of hydrogen bonds in the biological world.

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Solution:

A hydrogen bond is a molecular force in which a hydrogen atom is shared between two atoms. Hydrogen bonds occur as a result of the uneven distri-bution of electrons in a polar bond, such as an O-H bond. Here, the bonding electrons are more attracted to the highly electronegative oxygen atom, resulting in a slight positive charge (\delta+) on the hydrogen and a slight negative charge (\delta-) on the oxygen. A hydrogen bond is formed when the relatively positive hydrogen is attracted to a relatively negative atom of some other polarized bond. For example: The atom to which the hydrogen is more tightly linked or specifically the atom with which it forms the polar bond, is called the hydrogen donor, while the other atom is the hydrogen acceptor. In this sense, the hydrogen bond can be thought of as an intermediate type of acid-base reaction. Note, however, that the bond is an electro-static one - no electrons are shared or exchanged, between the hydrogen and the negative dipole of the other molecule of the bond. Hydrogen bonds are highly directional (note the arrows in the figure), and are strongest when all three atoms are colinear. Bond energies of hydrogen bonds are in the range of about 3 to 7 kcal/mole. This is intermediate between the energy of a covalent bond and a van der Waals bond. How-ever, only when the electronegative atoms are either F, O, or N, is the energy of the bond enough to make it important. Only these three atoms are electronegative enough for the necessary attraction to exist. Hydrogen bonds are responsible for the structure of water and its special properties as a biological solvent. There is extensive hydrogen bonding between water molecules, forming what has been called the water matrix. This structure has profound effects on the freezing and boiling points of water, and its solubility properties. Any molecule capable of forming a hydrogen bond can do so with water, and thus a variety of molecules will dissociate and be soluble in water. Hydrogen bonds are also most responsible for the maintenance of the shape of proteins. Since shape is crucial to their function (as both enzymes and struc-tural components), this bonding is extremely important. For example, hydrogen bonds maintain the helical shape of keratin and collagen molecules and gives them their characteristic strength and flexibility. It is hydrogen bonds which hold together the two helices of DNA. Bonding occurs between the base pairs. The intermediate bond strength of the hydrogen bond is ideal for the function of DNA - it is strong enough to give the molecule stability, yet weak enough to be broken with sufficient ease for replication and RNA synthesis.

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Question:

Explain how you would prepare benzoic acid from benzene. Use any inorganic reagents.

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Solution:

One of the most common ways of preparing benzoic acid from benzene is through an intermediate grignard. To plan this synthesis, first write your starting material and desired product. Then, think of inorganic reagents that lead you step by step through intermediates until you obtain the product. A grignard reagent can be represented as R-MgX, where R is any aryl or alkyl group and X is a halogen. To obtain this grignard intermediate, you must halogenate the benzene ring. This can be accomplished by using a solution of halogen and FeX_3, where X is the halogen. Let the halogen be Br. Thus, you have the reaction in Figure B. This is electrophilic substitution of a hydrogen in the ring by a bromine atom. By electrophilic, you mean electron seeking. By addition of Mg, in dry ether, you obtain the grignard reagent. See figure C. Now that you have the grignard, add carbon dioxide. The reaction proceeds as shown in figure D. With the addition of the acid, HBr, the reaction in figure E proceeds to the desired product benzoic acid.

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Question:

In the Ostwald process for the commercial preparation of nitric acid, ammonia gas is burned in oxygen in the presence of a Pt catalyst. The balanced equation is: 4NH_3 + 5O_2^Pt\rightarrow4NO + 6H_2 O What volume of O_2 and what volume of NO is formed in the combustion of 500 liters of NH_3. All gases are under the same conditions of temperature and pressure.

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Solution:

In order to solve this volume-volume problem set up a ratio between the volumes involved. No molecular weights are needed in volume-volume problems. First, one wants to find out how much O_2 is required to react with 500 liters of NH_3. As can be seen from the stoichiometry of the reaction, the ratio of the mole volumes (i.e. how much NH_3 reacts with O_2) is 4 : 5. Using the equation [(volume NH_3 present)/(mole volume NH_3 )] = [(volume O_2 required)/(mole volume O_2 )] [(500 liters) / (4 liters)] = [(volume O_2) / (5 liters)] volume O_2 = 625 liters. To find the amount of NO produced, set up a similar equation. In this case, 4 mole volumes of NH_3 produce 4 mole volumes of NO as indicated by thestoichiometryof the reaction [(volume NH_3 present) / ( mole volume NH_3)] = [(volume NO produced) / (mole volume NO)] [(500 liters) / (4 liters)] = [(volume NO produced) / (4 liters) volume NO produced = 500 liters. It was mentioned in the problem that a Pt catalyst was used. A catalyst is a substance which speeds up the rate at which a reaction occurs by lowering the amount of energy needed to start a reaction (i.e. lowering the activation energy). Catalysts have no effect on the final concentrations of the compounds involved in the reaction. Therefore, the presence of a catalyst does not change the values calculated in this question.

Question:

Distinguish a metallic bond from an ionic bond and from a covalent bond.

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Solution:

The best way to distinguish between these bonds is to define each and provide an illustrative example of each. When an actual transfer of electrons results in the formation of a bond, it can be said that an ionic bond is present. For example, Potassiumsulfurpotassiumsulfur ionionic bond atomsatomionsdue to the (unlike ions due transferattraction of electron from potassiumof unlike to sulfur)ions When a chemical bond is the result of the sharing of electrons, a covalent bond is present. For example: A pure crystal of elemental metal consists of millions of atoms held together by metallic bonds. Metals possess electrons that can easily ionize, i.e., they can be easily freed from the individual metal atoms. This free state of electrons in metals binds all the atoms together in a crystal. The free electrons extend over all the atoms in the crystal and the bonds formed between the electrons and positive nucleus are electrostatic in nature. The electrons can be pictured as a "cloud" that surrounds and engulfs the metal atoms.

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Question:

Define and discuss the terms hardware, software, and firmware .

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Solution:

The coinage of the terms hardware, software, and firmware have military origin, where for example, tanks are considered conventional military hardware. Hardware, software, and firmware are the three principal components of a computer system that was also called Electronic Data Processing (EDP) system in the early days. Hardware is the electromechanical machinery that performs the data manipulations and computations. Software is the pro-grams that direct the operation of the computer. A program is a sequence of instructions that are carried out (executed) by the hardware. Firmware is not a well established term. Loosely, it refers to themicroprogramthat implements the instruction set (machine language) of a computer. The term is sometimes used for software that has been stored permanently (burned into ) the Read-Only Memory (ROM) .

Question:

Calculate the magnitude of the electrostatic force exerted by the proton on the electron in a hydrogen atom and compare it with the weight of the electron.

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Solution:

In a hydrogen atom the nucleus is a single proton and a single electron moves around it at an average distance away of 0.53 × 10^\rule{1em}{1pt}8cm. The charge on each particle is e and the electro-static force is, by Coulomb's Law, F_e = e^2 / R^2 = (4.8 × 10^\rule{1em}{1pt}10 esu)^2 / (0.53 × 10^\rule{1em}{1pt}8 cm)^2 Since 1 dyne \textbullet cm^2 = 1 esu2 F_e = 8.2 × 10^\rule{1em}{1pt}3 dyne The mass of an electron is m_e = 9.11 × 10^\rule{1em}{1pt}28 gram By Newton's Second Law, the gravitational force on m_e is F_g = m_eg where g is the acceleration of m_e due to gravity. F_g is then the weight of m_e. Hence F_g = (9.11 × 10^\rule{1em}{1pt}28gm)(980 cm/s^2) = 8.9 × 10^\rule{1em}{1pt}25 dyne The ratio of the electrostatic force to the weight is F_e / m_eg =(8.2 × 10^\rule{1em}{1pt}3) / (8.9 × 10^\rule{1em}{1pt}25) The electrostatic force is therefore overwhelmingly larger than the weight. We never have to worry about gravitational forces when we are considering the theory of the hydrogen atom.

Question:

Calculate the change in Gibbs free energy (∆G) for the production of 2 NO_2 (g) at 1 atm from N_2O_4 (g) at 10 atm at 25\textdegreeC. The standard Gibbs free energy for the reaction N_2O_4 (g) = 2NO_2 (g) is + 1161 cal at 25\textdegreeC.

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Solution:

If ∆G\textdegree has a positive value, the reactants in their standard states will not react spontaneously to give products in their standard states. By increasing the press-ures or concentrations of the reactants or decreasing the pressures or concentrations of the products, however, it may be possible to make ∆G negative so that the reaction will proceed spontaneously. To calculate ∆G under nonstandard conditions, the following formula is used: ∆G = ∆G\textdegree + RT In \pi_i pvi_I ,where R = gas constant, T = absolute temperature (Celsius + 273) , and \pi_i pvi_i represents the product of the partial pressures of the component gases, each raised to the power that corresponds with its stoichiometric coefficient in the balanced equation. Thus, ∆G = ∆G\textdegree + RT In [(P^2 _NO2 /P_N2O4)] = + 1161 + (1.987 cal\textdegreeK^-1 mole^-1) (298\textdegreeK) (2.303) log (1 atm)^2 /(10 atm) = 1161 - 1364 = - 203 cal. The production of NO_2 (g) at 1 atm from N_2O_4 (g) at 10 atm is a spontaneous process due to the fact that a negative ∆G is obtained. Thus, a continuous process would be thermodynamically feasible if N_2O_4 was maintained at 10 atm and NO_2 was withdrawn by some method so that its partial pressure was maintained at 1 atm.

Question:

The Milky Way galaxy rotates once in 200 million years, and our sun is located about 30,000 light years from the galactic center. As a result, the earth is moving through space relative to the other galaxies. What is the observed Doppler shift in \AA of the hydrogen line of 6563 \AA for light coming from other galaxies? Consider two cases:a) The line of observation is in the direction of the earth's motion; b) The line of observation is perpendicular to the direction of the earth's motion. Ignore other causes of observed Doppler effects.

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Solution:

The relativistic Doppler shift is given by the expression v՛ = v\surd[(1 - u / c) / (1 + u / c)] Multiplying both the numerator and denominator by the factor \surd(1 - u / c), one obtains the expression v՛ = v[(1 - u / c) / \surd{1 - (u / c)^2 } ] where v' is the observed frequency of light of frequency v in the frame of reference of the source; u is the relative source-observer velocity, and c is the speed of light. For case (a), the light is travelling parallel to the earth's motion, and u is given by u = R\Omega , with R = 30,000 light-years, and \Omega = [2\pi / {200(10^6)}](rad / year). Thus, u = [{30,000(2 \pi)} / {200(10^6)}] light- year(rad / year) = 9.425(10^-4)c (where c represents the speed of light) or (u / c) = 9.425(10^-4). Now since u << c, v' \approx v {1 - (u / c)}, or, since v' = c / \lambda' and v = c / \lambda (\lambda = the wavelength of light), 1 / \lambda' \approx 1 / \lambda {1 - (u / c)}, with \lambda = 6563\AA. Thus, \lambda' = 6569.2 \AA, or ∆\lambda = \lambda' - \lambda = 6.2 \AA. b) Where the line of observation is perpendicular to the direction of the earth's motion, the observed frequency shift, which is purely a relativistic effect, is called the transverse Doppler effect. The frequency shift is given by the expression v = v'[1 - (u / c)^2)1 / 2 The derivation of this effect is Indicated in the figure. The generalized Doppler shift Is given by v = v'[{1 + (u / c)cos \texttheta՛ } / \surd{1 -(u/c)^2} ] where, as indicated, the observer (on the earth) is located at point P at rest in the unprimed coordinate system, and the source of light is at rest in the primed system. For transverse observation, \texttheta = \pi / 2, cos \texttheta՛ = -(u / c) Thus,v = v'[{1 - (u / c)^2} / \surd{1 -(u/c)^2} ] = v' [1 - (u / c)^2]1 / 2 or 1 / \lambda = 1 / \lambda՛[1 -(u / c)^2]1 / 2], with (u / c) = 9.425(10^-4) . Thus, \lambda = 6563.003 \AA ,or∆\lambda = 0.003 \AA.

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Question:

At what value of Z would the innermost orbit be expected to be pulled inside a nucleus of radius 1.0 × 10^-13 cm, assuming simple Bohr theory and Bohr radius = 0.5292 × 10^-8 cm? (Assume N = 1.)

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Solution:

NielsBohr stated, for a hydrogen atom, that only certain permitted values for the radius of the electron path exist. He found that the radius, r, of this electron path was equal to a whole number, N, squared, divided by the atomic number, Z, times the Bohr radius, a_0, which is constant. That is r = (N^2/Z) a_0. The above question/ can be answered by using this equation. N is equal to the quantum number corresponding to the energy level occupied by the electron. It has values of n = 1, 2, 3, 4, ... . One is given r and a_0 and is asked to compute Z. The electron occupying the innermost orbital is in thegrouindstate, thus N = 1. Thus, Z = (N^2/r) a_0 = [(1)^2/ 1 × 10^-13 cm)] × .5292 × 10^-8 cm = 5.292 × 10^4.

Question:

A business law class of 25 students takes a midterm exam that has 100 true- or-false questions. The instructor, who is also a computer fanatic, wants to devise a program to obtain a printout with the student's name, score, and letter grade, according to the following schedule: 50or below 51 - 55 56 - 60 61 - 65 66 - 70 71 - 75 76 - 80 81 - 85 86 - 90 91 - 95 96 - 100 F D C- C C+ B- B B+ A- A A+ In addition, the instructor wants to include the total number of students scoring each of the 11 possible grades. How would you use the FORTRAN language to solve his problem?

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Solution:

We need four arrays in this problem. The array KOUNT stores the number of times a grade is achieved by students; eleven cells are to be reserved, one for each grade. The array NGRADE stores the numerical upper limits of the corresponding letter grades. To initialize NGRADE, we use the first data card, filled out in the following format: 50556065707 580859095100 The third array, LETTER, is a two-dimensional array with 11 rows and 2 columns. It stores the 11 possible letter grades. When the grades are to be outputted, LETTER provides the data. As in NGRADE, we use a data card for initialization: FDC- CC+ B- BB+ A- AA+ We set up a DO loop to read in the remaining 25 cards, one for each student. Another loop is nested within the first one to categorize the scores into the proper letter grade range. We output each student's name (the first 20 characters of it) the numerical score, and the letter grade. Finally, the counts for the number of occurrences of each letter grade are printed. DIMENSION KOUNT (11), NGRADE (11), LETTER (2,11) 1 NAME (20) C INITIALIZE KOUNT AND READ TWO C INITIAL DATA CARDS FOR NGRADE AND LETTER DO 5 I = 1,11 KOUNT (I) = 0 5 CONTINUE READ (5,100) NGRADE 100 FORMAT (1113) READ (5,101) LETTER 101 FORMAT (22A1) C READ MASTER CARD DECK DO 10 J = 1,25 READ (5,102) NAME, ISC0RE 102 FORMAT (20A1, I4) DO 20 K = 1,11 IF (ISCORE.LT,NGRADE (K)) GO TO 15 20 CONTINUE GO TO 98 15 WRITE (6,103) NAME, ISCORE, (LETTER (K, 1 J), K = 1,2) 103 FORMAT (1X,20A1, I4, 1X, 2A1) KOUNT (J) = KOUNT (J) + 1 10 CONTINUE WRITE (6,104) KOUNT 104 FORMAT (11(2X,I3)) GO TO 99 98 WRITE (6,105) 105 FORMAT (1X,\textasteriskcenteredDATA ERROR IN DECK\textasteriskcentered) 99 STOP END

Question:

Compare the events of mitosis with the events of meiosis, consider chromosome duplication, centromere duplication, cytoplasmic division and homologous chromosomes in making the comparisons.

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Solution:

In mitosis, the chromosomes are duplicated once, and the cytoplasm divides once. In this way, two identical daughter cells are formed, each with the same chromosome number as the mother cell. In meiosis, however, the chromosomes are duplicated once, but the cytoplasm divides two times, resulting in four daughter cells having only half the diploid chromosomal complement. This difference arises in the fact that there is no real interphase, and thus no duplication of chromosomal material, between the two meiotic divisions. In mitosis, there is no pairing of homologous chromosomes in prophase as there is in meiosis. Identical chromatids joined by their centromere are separated when the centromere divides. In meiosis, duplicated homologous chromosomes pair, forming tetrads. The daughter chromatids of each homolog are joined by a centromere as in mitosis, but it does not split in the first meiotic division. The centromeres of each duplicated member of the homologous pair are joined in the tetrad, and it is these centromeres which separate from one another in anaphase of meiosis I. Thus the first meiotic division results in two haploid daughter cells, each having chromosomes composed of two identical chromatids. Only in meiosis II, after the reduc-tion division has already occurred, does the centromere joining daughter chromatids split as in mitosis, thus separating identical chromosomes.

Question:

In a given population of 610 individuals, the gene frequencies of the L^M and L^N alleles were found to be .62 and .38, respectively. Calculate the number of individuals with M, MN, and N type blood.

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Solution:

The Hardy-Weinberg principle tells us that the proportions of the genotypes in a population are described by the expansion of the binomial equation: (PA +qa)^2 = p^2AA + 2pqAa+ q^2aa, where p represents the frequency of a given allele A and q represents the frequency of its homologous allele, a. In other words, if we were to consider all thematingsin a given generation, a p number of A-containing eggs and a q number of a-containing eggs are fertilized by a q number of A-containing sperm and a g-number of a-containing sperm; and the total possible allelic combinations produced are (pA+qa) × (pA+qa) or (pA+qa)^2. In this instance, p and q are respectively the frequencies of the L^M and L^N alleles. The phenotypic classes for these alleles are M(_VL^M L^M),MN (_VL^M L^M), and N (_VL^N L^N). genotypegenotypegenotype This allelic system is ideal for studying genotypic ratios since the genes arecodominantand theheterozygotestherefore constitute a distinct phenotypic class, ratherthenbeing obscured by the dominant trait. The frequencies of the phenotypic classes in the population can be determined using the binomial expansion: (pL^M+qL^N)^2= (.62L^M + .38L^N)^2 = (.62)^2L^ML^M + 2(.62) (.38) L^ML^N + (.38)^2L^N = .384 M + .471 MN + .144 N Converting the frequencies to actual numbers of individuals, we have (.384)(610)M= 235 M (.471)(610)MN= 287 MN (.144)(610)N= 88 N (1) (610)= 610 TOTAL In this calculation, we multiply each frequency by the total population to obtain the actual numbers of those having each genotype. Note that the frequencies sum up to 1, and the total number of individuals for the three phenotypes equals the total population.

Question:

In what ways are bacteria useful to the dairy industry?

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Solution:

Bacteria are a major cause of disease, but beneficial bacterial species outnumber harmful ones. There are many useful applications microorganisms in the dairy industry. Fermented milk products are made by encouraging the growth of the normal lactic, acid-producing bacteria that are present in the culture. Specific organisms, or mix-tures of them, are used to produce buttermilk, yogurt, and other fermented milk products. The principal bacteria used are species of Streptococcus,Leuconostoc, and Lactobacillus, which produce only lactic acid. Butter is made by churning pasteurized sweet or sour cream with the latter being fermented by streptococci andleuconostocs. Different types of cheeses are made by providing conditions that favor the development of selected types of bacteria and molds. The quality and characteristics of a cheese are determined by the biochemical activities of selected microorganisms. For example , Swiss cheese is produced through the fermentation of lactic acid by bacteria of the genusPropionibacterium. The products of thefermen- tation arepropionicacid, acetic acid, and carbon dioxide. The acids give the characteristic flavor and aroma to the cheese, and the accumulation of carbon dioxide produces the familiar holes. Bacteria are also useful to the dairy industry by providing the means by which certain milk-giving animals acquire nutrients. The ruminants (cattle, sheep, goats, and camels) are a group of herbivorous mammals whose di-gestive system has a compartmentalized stomach, the first section being termed the rumen. In the rumen, there exist certain bacteria and protozoa, which provide enzymes ne-cessary to break down the cellulose acquired from the inges-tion of plant material. The mammal does not normally have the enzymes needed to degrade cellulose into nutrients which the animal can use (the lack of these enzymes in humans also explains why man cannot break down plant mater-ial) . Bacteria are also important to the food industry in other ways. Pickles, sauerkraut, and some sausages are produced in whole or part by microbial fermentations.

Question:

The frequency of the gene for sickle-cell anemia in American Blacks is less than that found in the people living in their ancestral home in Africa. What factors might account for this difference?

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Solution:

The sickle-cell disease is a homozygous recessive trait that usually results in death before reproductive age. In the heterozygous form, the sickle--cell gene is not harmful enough to cause death, but is instead beneficial in certain environments because it gives the carrier an immunity to malaria. In Africa, malaria is a severe problem and the heterozygous indivi-duals have a survival advantage over their fellow Africans. Therefore, the frequency of the sickle-cell gene has been kept fairly constant in the gene pool in Africa. In America, where the incidence of malaria is in-significant, an individual carrying the sickle-cell gene has no survival advantage, and the sickle-cell allele is slowly being lost and diluted in the population. Those with the homozygous sickle-cell genotype usually die, hence the frequency of the sickle-cell allele declines. In addition, with more interracial marriages in America, the sickle-cell genes from blacks are being diluted by the normal genes from the non-Black population. Thus, in the American Black population, a trend is observed in which the frequency of the sickle-cell gene decreases gradually over generations.

Question:

Calculate ∆E for the proposed basis of an absolutely clean source of nuclear energy ^11_5B + ^1_1H\rightarrow^12_6C\rightarrow3 ^4_2He Atomic masses : ^11B = 11.00931, ^4He = 4.00260, ^1H = 1.00783.

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Solution:

The energy of nuclear reactions, ∆E, is calculated from the difference between the masses of products and reactants in accordance with the Einstein Law. Einstein's Law can be stated ∆E = ∆mc^2 , where ∆m is the difference in the masses of the products and the reactants, and c is the speed of light (3 × 10^10 cm/sec). The total reaction here can be written ^11_5B + ^1_1H\rightarrow3 ^4_2He. Thus, ∆m is equal to the mass of ^11_5Band ^1_1H sub-tracted from the mass of 3 ^4_2He. The mass of 3 ^4_2He is equal to 3 times the mass of ^4_2He. ∆m= (3 × m of ^4_2He) - (m of ^11_5B + m of ^1_1H) = (3 × 4.00260) - (11.00931 + 1.00783) = 12.0078 - 12.01714 = - 9.34 × 10^-3 g/mole. One can now solve for ∆E by using ∆m. ∆E= ∆mc^2 = -9.34 × 10^-3 g/mole × (3.0 × 10^10 cm/sec)^2 = - 9.34 ×10^-3 g/mole × 9.0 × 10^20 cm^2/sec^2 = - 8.406 × 10^18 g cm^2/mole sec^2 = - 8.406 × 10^18 ergs/mole. There are 4.18 × 10^10 ergs in 1 Kcal, thus ergs can be converted to Kcal by dividing the number of ergs by the conversion factor 4.18 × 10^10 ergs/Kcal. no. of kcal= (- 8.406 × 10^18 ergs/mole) / (4.18 × 10^10 ergs/kcal) = - 2.01 × 10^8 kcal/mole. ∆E for this reaction is - 2.01 × 10^8 kcal per mole.

Question:

The solubility of iodine in pure water is 0.0279gper 100g of water. In a solution that is originally 0.10M KI, itis possible to dissolve, at the maximum, 1.14g of iodine per 100g of solution. What is K for the reaction I_2 + I- \rightleftarrows I-_3 ?

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Solution:

The concentration of iodine in the given solution is greater than can be dissolved. However, the wording of the problem indicates that there is no precipitate. Thus, the excess iodine must be used up in the production of I-_3 which is soluble. With this in mind, we can solve for K . The equilibrium constant K can be solved from the following equation, (i)K= [I-_3 ] / {[I_2 ] [I-]} . (i)K= [I-_3 ] / {[I_2 ] [I-]} . First, solve for the concentrations of I-_3 , I_2 and I- . The I^- species First, solve for the concentrations of I-_3 , I_2 and I- . The I^- species comes entirely from the dissociation of KI . comes entirely from the dissociation of KI . Since a salt such as KI totally dissociates in aqueous solu-tion, initial Since a salt such as KI totally dissociates in aqueous solu-tion, initial molarity of I- = molarity of KI = 0.10M . molarity of I- = molarity of KI = 0.10M . The initial amount of I_2 added is 1.14g per 100g of water or the The initial amount of I_2 added is 1.14g per 100g of water or the number of moles added equals number of moles added equals (grams of I_2 ) / (mol. wt. I_2 ) = (1.14g) / (254g/mole) (grams of I_2 ) / (mol. wt. I_2 ) = (1.14g) / (254g/mole) = 4.49 × 10-^3 moles per 100g of = 4.49 × 10-^3 moles per 100g of water. However, the water can only hold 0.0279g of iodine per 100g of water. However, the water can only hold 0.0279g of iodine per 100g of water, or water, or (.0279g) / (254g/mole) = 1.1 × 10-^4 moles per 100g of water. (.0279g) / (254g/mole) = 1.1 × 10-^4 moles per 100g of water. Since there is to be no precipitate, the excess I_2 must be used up Since there is to be no precipitate, the excess I_2 must be used up in the production of I-_3 . Since one mole l_2 yields one mole of I_3 , we have in the production of I-_3 . Since one mole l_2 yields one mole of I_3 , we have for the final concentrations: for the final concentrations: (1) concentration of I_2 = maximum concentration of I_2 that can be (1) concentration of I_2 = maximum concentration of I_2 that can be dissolved = 1.1 × 10-^4 moles per 100g of water. Since 1000g of H_2O = 1l , dissolved = 1.1 × 10-^4 moles per 100g of water. Since 1000g of H_2O = 1l , 1.1 × 10-^4moles/100g H_2O = (1.1 × 10-^3 moles) / (1000g H_2O) 1.1 × 10-^4moles/100g H_2O = (1.1 × 10-^3 moles) / (1000g H_2O) = (1.1 × 10-^3 moles) / (1 liter H_2O) . Thus, [I_2 ] = 1.1 × 10-^3 M. = (1.1 × 10-^3 moles) / (1 liter H_2O) . Thus, [I_2 ] = 1.1 × 10-^3 M. (2) concentration of I-_3 = concentration of I_2 reacted. Since there were (2) concentration of I-_3 = concentration of I_2 reacted. Since there were originally4.49 × 10-^3 moles of I_2 /100g H_2O and the final concentration of originally4.49 × 10-^3 moles of I_2 /100g H_2O and the final concentration of l_2 is 1.1× 10-^4moles/100g H_2O, the concentration of I_2 that reacted is the l_2 is 1.1× 10-^4moles/100g H_2O, the concentration of I_2 that reacted is the difference difference [(4.49 × 10-^3 moles) / (100g H_2O)] - [(1.1× 10-^4 moles) / (100g H_2O)] [(4.49 × 10-^3 moles) / (100g H_2O)] - [(1.1× 10-^4 moles) / (100g H_2O)] = (4.38 × 10-^3 moles) / (100g H_2O) = (4.38 × 10-^3 moles) / (100g H_2O) = (4.38 × 10-^2 moles) / (1 liter H_2O). = (4.38 × 10-^2 moles) / (1 liter H_2O). Thus, [I-_3] = 4.38 × 10-^2 M Thus, [I-_3] = 4.38 × 10-^2 M Thus, [I-_3] = 4.38 × 10-^2 M . . (3) concentration of I- = initial concentration I- minus (concen-tration of I- (3) concentration of I- = initial concentration I- minus (concen-tration of I- needed to reacted with the I_2). From (2), the concentration of reacted I_2 needed to reacted with the I_2). From (2), the concentration of reacted I_2 was found to be 4.38 × 10-^2 M. Since one mole of I- is required to react was found to be 4.38 × 10-^2 M. Since one mole of I- is required to react with each mole of I_2 , the concentration of I- needed to reacted with the l_2 with each mole of I_2 , the concentration of I- needed to reacted with the l_2 is also 4.38 × 10-^2 M . is also 4.38 × 10-^2 M . Thus, [I-] = 0.10M -4.38 × 10-^2 M = .0562M. Thus, [I-] = 0.10M -4.38 × 10-^2 M = .0562M. To find K, substitute these values in equation (i): To find K, substitute these values in equation (i): K = (0.0438) / [(0.0011) (0.0562)] = 710 . K = (0.0438) / [(0.0011) (0.0562)] = 710 . The mole fraction of CO_2 in water is The mole fraction of CO_2 in water is X_CO(2) = (number of moles of CO_2 ) / X_CO(2) = (number of moles of CO_2 ) / (number of moles of CO_2 + number of moles of H_2O) (number of moles of CO_2 + number of moles of H_2O) The number of moles of H_2O is. (1000g) / (18.02g/mole) = 55.49 moles The number of moles of H_2O is. (1000g) / (18.02g/mole) = 55.49 moles H_2O . The number of moles of CO_2 has yet to be determined. Thus H_2O . The number of moles of CO_2 has yet to be determined. Thus substi-tuting into Henry's law substi-tuting into Henry's law X_2 = P_2 / K_2 = (760 torr) / (1.25 × 10^6 ) X_2 = P_2 / K_2 = (760 torr) / (1.25 × 10^6 ) = (moles CO_2 ) / [(moles CO_2 ) + (55.49 moles H_2O)]. = (moles CO_2 ) / [(moles CO_2 ) + (55.49 moles H_2O)]. However, since the number of moles of CO_2 may be considered negli-gible However, since the number of moles of CO_2 may be considered negli-gible in comparison with the number of moles of water, then, in comparison with the number of moles of water, then, (760 torr) / (1.25 × 10^6 ) = (moles CO_2 ) / (55.49 moles H_2 (760 torr) / (1.25 × 10^6 ) = (moles CO_2 ) / (55.49 moles H_2 (760 torr) / (1.25 × 10^6 ) = (moles CO_2 ) / (55.49 moles H_2 O). O). Rewriting and solving Rewriting and solving no. of moles of CO_2 = [(55.49) (760)] / (1.25 × 10^6 ) = 3.37 × 10-^2 no. of moles of CO_2 = [(55.49) (760)] / (1.25 × 10^6 ) = 3.37 × 10-^2 There are 1000g of H_2O, or 1 liter of solvent, and therefore, the There are 1000g of H_2O, or 1 liter of solvent, and therefore, the solubility of solubility of CO_2 = (3.37 × 10-^2 moles) / (1 liter) = 3.37 × 10-^2 M . CO_2 = (3.37 × 10-^2 moles) / (1 liter) = 3.37 × 10-^2 M .

Question:

Compare the modes of action of the nervous and endocrine systems. The major endocrine organs in man.

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Solution:

The activities of the various part of the body of higher animals are integrated by the nervous and endocrine systems. The endocrine system consists of a number of ductless glands which secrete hormones. The swift responses of muscles and glands, measured in milliseconds, are typically under nervous control. Nerve impulses are transmitted along pathways consisting of neurons. The hormones secreted by the endocrine glands are transported by the bloodstream to other cells of the body in order to control and regulate their activities. Nervous stimulation is required by some endocrine glands to release their hormones, particularly the posterior pitui-tary gland. The responses controlled by hormones are in gen-eral somewhat slower (measured in minutes, hours, or even weeks), but of longer duration than those under nervous control. The long term adjustments of metabolism, growth and reproduction are typically under endocrine regulation. We already mentioned that hormones travel in the . blood and are therefore able to reach all tissues. This is very different from the nervous system, which can send messages selectively to specific organs. However, the body's response to hormones is highly specific. Despite the ubiquitous distribution of a particular hormone via the blood, only certain types of cells may respond to that hormone. These cells are known as target-organ cells. The central nervous system, particularly the hypo-thalamus plays a critical role in controlling hormone secretion; conversely hormones may markedly influence neural function and behavior as well.

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Question:

A solution containing 6.15 g of cholesterol per 100 ml of solution was made. When a portion of this solution was placed in a 5-cmpolarimetertube, the observed rotation is -1.2\textdegree. Determine the specific rotation of cholesterol.

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Solution:

Apolarimetermeasures the rotation of plane polarized light, which is light whose vibrations take place in only one plane. An optically active substance has the ability to rotate the plane of polarized light. In essence, then, apolarimeter, measures optical activity. Species that possess optical activity have a specific rotation. It is defined in the following equation: specific rotation = [observed rotation (degrees)] /[length (dm) × g/cc] The length of thepolarimetertube must be converted to decimeters, (1 cm = 0.1 dm) Thus, 5 cm = .5 dm. For water, 100 ml = 100 cc. Substituting, you obtain, specific rotation = (-1.2o) / [.5 × (6.15/100)] = -39.02o.

Question:

A turn of \alpha-helix (3.6 residues) is 5,41 \AA in length measured parallel to the helix axis. If your hair grows 6 in. per year, how many amino acid residues must be added to each \alpha-helix in the keratin fiber per second?

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Solution:

Human hair is an example of a keratin fiber, and is found in the \alpha-helix conformation. One is given that in each5 . 41 \AA of length, there are 3.6 residues. After converting 6 in to \AA, one can determine the number of residues that are added per year to each strand. From this, the number of residues added per second can be calculated. Solving for length: 1 in = 2.54 cm, 1 \AA = 10^-8 cm length in \AA = 6 in × 2.54 cm/in× 1 \AA/10^-8 cm = 1.52 × 10^9 A Calculating the number of residues: number of residues = length in \AA × (3.6 residues / 5.41 \AA) = 1.52×10^9 \AA × (3.6 residues / 5.41 \AA) = 1.01 × 10^9 residues There are 1.01 × 10^9 residues produced per year. Solving for the number of residues produced per second residues produced per second = 1.01 × 10^9 (res / year) × (1 year / 365 days) × (1day / 24 hrs) × (1 hr /60 min) × (1min / 60 sec) = 32 res/sec.

Question:

Write a Basic Assembler Language program to compute and print out the first twenty numbers of the Fibonacci Sequence.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G09-0206.htm

Solution:

The Fibonacci numbers are a sequence of numbers beginning with 0 and 1. To compute the following numbers of the series one adds the previous two to form the next. For example, the third number of the series is 0 + 1 = 1. The next is 1 +1 = 2, the next 1 + 2 = 3, the sixth is 2 + 3 = 5 and so on. This simple number sequence is easily programmed in assembler language. We will start off by loading the first two numbers of the sequence into two registers. Next we fall into a loop which prints out the two registers and then adds each register to the other creating the nexttwo num-bers of the sequence. These next two numbers are printed out when the program goes back to the top of the loop and the procedure is repeated. The program follows below: BALR2, 0ESTABLISHES BASE REGISTER USING\textasteriskcentered, 2ESTABLISHES BASE REGISTER LA1, 0REGISTER 1 CONTAINS NUMBER TO BE INCREMENTED BY LOOP LA4, 1REGISTER 4 CONTAINS INCREMENT LA5, 9REGISTER 5 CONTAINS LIMIT L7, = F'0'LOADS FIRST TW0 NUMBERS OF SEQUENCE L8, = F'1'LOADS FIRST TW0 NUMBERS OF SEQUENCE LOOP PRINTOUT7PRINT NUMBERS OF SEQUENCE PRINTOUT8PRINT NUMBERS OF SEQUENCE AR7, 8CREATES NEXT TWO NUMBERS OF SE-QUENCE AR8, 7CREATES NEXT TWO NUMBERS OF SE-QUENCE AR1.4ADDS INCREMENT TO INDEX CR4, 5COMPARE INDEX TO END BLELOOPRETURN FOR NEXT PART OF SERIES IF NOT PAST END END

Question:

What is a cell?

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/Users/wenhuchen/Documents/Crawler/Biology/F02-0030.htm

Solution:

A cell is the fundamental organizational unit of life. One of the most importantgeneralizations of modern biology is the cell theory. There are twocomponents of the cell theory. It states: (1) that all living things are composedof cells and (2) that all cells arise from other cells. Living things arechemical organizations of cells and capable of reproduc-ing themselves. There are many types of cells, and just as many classifications to gowith them. There are plant cells, animal cells,eucaryoticcells, procaryoticcells and many others. Also within each of these divisions, thereare smaller subdivisions pertaining to the spe-cific properties or functionsof the cells. Cells ex-hibit considerable variation in properties basedon different arrangements of components. Cells also vary in size, although most of them fall in the range of 5 to 20m\mu.

Question:

The dirt floor of theShanidarCave in the northern part of Iraq has been examined. Below the layer of soil that contained arrowheads and bone awls was a layer of soil that yielded flint tools and pieces of charcoal. When the charcoal was examined it was discovered that in 1 kg of carbon, approximately 9.4 × 10^2 carbon\Elzbar14 nuclei decayed each second. It is known that in 1 kg of carbon from living material, 1.5 × 10^4 disintegrations of carbon\Elzbar14 occur each second. Use these data to calculate when people of the stone age culture occupied the cave.

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Solution:

The number of nuclei of a radioactive substance at time t, is given by N(t) = N_0 e\Elzbar\lambdat where N_0 is the number of nuclei at t = 0, and \lambda is the decay rate per nucleus. The half-life, T, is defined as the time required for the exponential factor to equal 1/2 ; T = (1/\lambda)ln2 Therefore, we can also write (1) as N(t) = N_0 e\Elzbar(t/T) \textbullet In 2 = N_0 eln(1/2) t/T = = N_0 (1/2)t/T Initially, there were N_0 = 1.5 × 10^4 radioactive carbon atoms decaying per second in a kg. of carbon, when carbon was part of a living material. The measured number of disintegrations per 1 kg. of carbon is found to be N = 9.4 × 10^2 per second. From (2) 9.4 × 10^2 = 1.5 × 10^4 × (1/2)t/T or(1/2)t/T= (9.4 × 10^2 ) / (1.5 × 10^4 ) \approx 1/16 Since (1/2)^4 = 1/16, we have t = 4 T. The half-life of carbon\Elzbar14 is 5730 years. There-fore the tree was burned t = 4 × (5730 years) = 2.3 × 10^4 years ago.

Question:

Explain the 'END OF FILE Condition' method of ending a Explain the 'END OF FILE Condition' method of ending a programin PL/I.

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Solution:

The end-file condition occurs when an attempt is made to acquire datafrom the input file after the last data from the file has already been read. Hence, this end of the file condition has to be specified for the computerby using the key-word ENDFILE, followed by parentheses containingthe name of the file. The file in question could be a deck of cards being v read in througha card reader, in which case we will use the form: ON ENDFILE(SYSIN)... This way we tell the computer what to do when the end- file conditionis reached. For example, the statement: ONENDFILE(SYSIN) GOTO FINISH; tellsthe computer to go to a statementlabelled'FINISH' when the ENDFILE condition is reached on the SYSIN FILE. The following program illustrates the use of thisstate-ment : SUM: PROCOPTIONS(MAIN); DCL (X,SUM)FIXED(4)INIT(0); ONENDFILE(SYSIN) GOTO FINISH; LOOP: GET LIST(X); SUM=SUM+X; GOTO LOOP; FINISH: PUTLIST(SUM); /\textasteriskcentered WE KEEP REPEATING THE LOOP, UNTIL FINALLY WE FIND THAT WHEN THE GET LIST(X) STATEMENT IS EXECUTED, THERE IS NO MORE DATA X LEFT TO BE FETCHED. NOW THE COMPUTER REMEMBERS THAT BEFORE IT HAD STARTED EXECUTING THE LOOPS, IT HAD ALREADY BEEN TOLD THAT ON ENDFILE CONDITION, IT SHOULD GO TO THE STATEMENT LABELLED 'FINISH'.SO THIS IS WHAT IT DOES \textasteriskcentered/ END SUM; Note: The best place for inserting the ENDFILE condition statement is afterthe DCL statement but beforetheGET LIST statement, so that the computer will read it before it does any GET LIST.

Question:

A wave is represented by the equation y = 0.20 sin 0.40\pi (x \rule{1em}{1pt} 60t), where all distances are measured in centimeters and time in seconds. Find: (a) the amplitude, (b) the wavelength, (c) the speed, and (d) the frequency of the wave, (e) What is the displacement at x = 5.5 cm and t = 0.020 sec?

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Solution:

The displacementy of the medium due to wave motion at a positionxand at a timetis (see the figure) y = A sin(2\pi/\lambda) (x - vt) where A is the amplitude, \lambda is the wavelength, andv is the velo-city with which the wave is traveling along the x-axis. If we compare this equation with the expression given in the question, we see that (a)A = 0.20 cm (b)(2\pi/\lambda) = 0.40\pi \lambda = (2/0.40) cm = 5.0 cm (c)v = 60 cm/sec (d)f = (v/\lambda) = {(60 cm/sec)/5.0 cm} (e)y = (0.20 cm) sin [0.40\pi(5.5 - 60 × 0.020)] = (0.20 cm) sin [0.40\pi(5.5 - 1.2)] = (0.20 cm) sin (0.40 × 4.3\pi) = (0.20 cm) sin 1.72\pi = (0.20 cm) (\rule{1em}{1pt}0.77) = \rule{1em}{1pt}0.15 cm

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Question:

A chemist forms magnesium oxide by burning magnesium in oxygen. The oxide obtained weighed 1.2096 grams. It was formed from .7296 g of magnesium. Determine the mass equivalent of magnesium in this reaction.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E04-0135.htm

Solution:

An equivalent is defined as that mass of oxidizing or reducing agent that picks up or releases the Avogadro number of electrons in a particular reaction. One equivalent of any reducing agent reacts with one equivalent of any oxidizing agent. In this problem, the key is to determine the number of equivalents of oxygen involved. Once this is known, you also know the number of equivalents of magnesium. Since the oxide weighed 1.2096 g and the magnesium weighed .7296 g, the mass of the combined oxygen must be 1.2096 - .7296 = .4800 g. Before the oxygen reacted, its oxidation state was zero. After the reaction, however, it was - 2. As such, each oxygen atom gained 2 electrons. Therefore, the Avogadro number of electrons will be taken up by one half of a mole of O atoms. It follows, therefore, that there are 8.000 g per equivalent for oxygen, since 1 mole of oxygen atoms weighs 16 grams. It was found, however, that there were .4800 g of oxygen. As such (.4800 g) /(8.00g/equiv)= .0600 equiv of oxygen . This means that magnesium also has .06 equiv. 0.7296 g of Mg participated in the reaction. Therefore, the grams per equivalent of Mg = (.7296 gMg) / (.060 equiv) = 12.16 g / equiv .

Question:

A 3000-lb car traveling with a speed of 30 mi/hr strikes an obstruction and is brought to rest in 0.10 sec. What is the average force on the car?

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/Users/wenhuchen/Documents/Crawler/Physics/D06-0320.htm

Solution:

The momentum of a body is defined as p = mv. Since F^\ding{217} = ma^\ding{217} = m (dv^\ding{217}/dt) anddp^\ding{217} = d(mv^\ding{217}) = mdv^\ding{217} we have F^\ding{217} dt = dp^\ding{217} . Consider a collision between two masses m_1 and m_2 as shown in the figure. During the collision, the two objects exert forces on each other where F_1 is the force on m_1 due to m_2, and F_2 is the force on m_2 due to m_1. By Newton's third law, these forces at any instant are equal in magnitude but opposite in direction. The change in momentum of m1 resulting from the collision is \Deltap_1^\ding{217} =^t1\int_t0 F_1^\ding{217} dt =F_1 ∆t = mv_1 - mv_0 whereF_1 is the average value of the force F_1^\ding{217} during the time interval of the collision \Deltat = t_1 - t_0. For the car colliding with an obstruction, we can say F\Deltat = mv_1 - mv0 m = W/g = (3000 Ib) / (32 ft/sec^2) = 94 slugs v0= 30 mi/hr = 44 ft/sec F= [m(v_1 - v_0)] / \Deltat = [94 slugs (0 - 44 ft/sec)] / [0.10 sec] = - 4.1 × 10^4 lb.

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Question:

Write a function which utilizes the Newton-Raphsonmethod tocompute the square root of a number.

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Solution:

By applying the Newton-Raphsonmethod x_n+1 =x_n- [f(x_n) / f'(x_n)] tothe case of f(x) = x2 - (root)2 = 0 (where (root)2 is the given number), onearrives at the algorithm below: x_n+1 =x_n- [{x2_n- (root)2} / 2x_n ] = 1/2 [x_n+ {(root)2 /x_n}]. FUNCTION ROOT (WORK) ARG = ABS (WORK) ROOT = ARG/2.0 DO 6 IDX = 1,100 ROOT = (ROOT + ARG/ROOT) /2.0 IF(ABS((R00T\textasteriskcentered\textasteriskcentered2 -ARG)/ARG) - 0.000001) 7,7,6 6CONTINUE 7RETURN END

Question:

At a certain temperature, an equilibrium mixture of NO_2 + SO_2 \rightleftarrows NO + SO_3 is analyzed and found to contain the following molar concentrations: [NO_2 ] = 0.100, [SO_2 ] = 0.300, [NO] = 2.00, [SO_3 ] = 0.600 . If 0.500 moles of SO_2 are introduced at constant temperature, what will be the new concentrations of reactants and products when equilibrium is re-established?

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Solution:

LeChatelier'sprinciple states: If a stress is placed on a system in equilibrium, whereby the equilibrium is altered, that change will take place which tends to relieve or neutralize the effect of the added stress. Thus, in this reaction, if more SO_2 is added more NO and SO_3 will be formed. If stress is placed on the left side of the equation, the reaction will be forced to the right (and vice versa). One can determine the equilibrium constant (Keq) for this reaction by using the concentrations of the original mixture. The equilibrium constant is defined: Keq= {[NO][SO_3]} / {[NO_2 ][SO_2]} where [ ] indicate concentrations. Solving forKeqwith [NO] = 2.00[NO_2] = 0.100 [SO_3 ] = 0.600[SO_2] = 0.300 , one obtains Keq= {(2.00) (0.600)} / {(0.100)(0.300)} = 40.0 . One can solve for the new concentrations by using theKeq. From Le Chatelier'sprinciple, one knows that when S0_2 is added to this mixture, the amounts of NO and SO_3 will increase. Let x = the number of moles by which NO and SO_3 will increase. For each mole of SO_3 and NO formed, one mole of SO_2 and one mole of NO_2 will react, thus the new concentrations of these components will be equal to the original concentrations less x moles. The new concentrations can be stated. [NO]= 2.00 + x [SO_3]= 0.600 + x [NO_2]= 0.100\Elzbar x [SO_2]= 0.300 + the amount added\Elzbar x = 0.300 + 0.500\Elzbar x = 0.800\Elzbar x . Using the formula for the equilibrium constant, one can solve for x. Keq= {[NO] [SO_3 ]} / {[NO_2] [SO_2 ]} = 40.0 . Substituting, 40.0 = {(2.00 + x) (0.600 + x)} / {(0.100\Elzbar x) (0.800\Elzbar x)} 40.0 = (1.20 + 2.6x + x^2) / (.080\Elzbar 0.90x + x^2) (0.80\Elzbar 0.90x + x^2)40 = 1.20 + 2.6x + x^2 3.20\Elzbar 36.0x + 40x^2 = 1.20 + 2.6x + x^2 2.0\Elzbar 38.6x + 38x^2 = 0 . One can use the quadratic formula to solve for x . ax^2 +bx+ c = 0 x = {\Elzbarb \pm \surd(b^2\Elzbar 4ac)} / {2a} 38x^2\Elzbar 38.6x + 2.0 = 0 x = [38.6 \pm \surd{ (38.6)^2\Elzbar 4 × 2.0 × 38} ] / [2 × 38] x = (38.6 \pm 34.44 ) / 76 x = (38.6 + 34.44) / 76 = 0.96 or x = (38.6\Elzbar 34.44 ) / 76 = 0.055 One cannot use x = 0.96 because [NO_2] and [SO_2] will be negative . Concentrations cannot have negative values, which means x = .055. One can now find the new concentrations [NO]= 2.00 + x = 2.055 moles [SO_3]= 0.600 + x = 0.65 moles [NO_2 ]= 0.100\Elzbar x = 0.045 moles [SO_2] = 0.800\Elzbar x = 0.745 moles.

Question:

How do anesthetics act to reduce or eliminate thesensation of pain?

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Solution:

In human beings, the interpretation of sensations (vision, hearing, taste , smell, pain, etc.) takes place in the brain. For example, the rods and cones (receptor cells) of the retina do not "see." Only the combination of rods , cones, optic nerve, and the visual center of the brain lead to the sensation of vision. Since only those nerve impulses that reach the brain can result in sensations , any blockage of the impulses along the nerve fibers by an anesthetic has the same effect as removing the original stimulus entirely. The sense organs, when stimulated, will continue to initiate the impulses, which can be detected by the proper electrical apparatus, but the anesthetic prevents the impulses from reaching their destination in the brain . The tranquilizing effect of anesthetics results from the di-minished amounts of neurotransmitter in the synapse. As a consequence of this, a nerve impulse generated by the receptor cells will not be transmitted to the next neuron. This is because the amount of neurotransmitter released will be insufficient to stimulate the neuron to propagate the nerve impulse. Since the impulses going to the pain center of the brain are blocked, the brain is unable to receive and interpret the original stimulus. Therefore there is no sensation of pain.

Question:

For the following PL/I program, a) Explain how a condition known as 'SUBSCRIPTRANGE CONDI- TION' arises. b) Show what will be printed out if the program is run. c) Write out a corrected program to avoid errors due to this condition . d) Show what the corrected program will print out. EXAMPLE:PROC OPTIONS (MAIN) ; DCL A (5) FIXED (3),(X,Y,Z) FIXED(2); GETLIST(Z) ; DO I = 1 TOZ ; GET LIST(X,Y); A(I) = X\textasteriskcenteredY; PUTLIST(I,A(I)); END; END PROGRAM; Data: 6, 10, 10, 5, 3, 20, 40, 30, 30, 7, 8, 15, 2.

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Solution:

a) The purpose of this example is to illustrate a type of condition in whichan array subscript is calculat-ed beyond the declared range for that array. Notice from the DCL statement that a storage space for an array A of5 elements is to be reserved. The 'GETLIST(Z) ' statement fetches 6 from the DATA CARD as a valuefor Z. Now, this value of Z becomes the upper limit of the DO loop of thenext statement. In the body of this DO loop, a value for X and a value for Y are read fromthe DATA CARD, and the array element A (I) is computed. The program looks straight forward. However, an exam-ination of theprogram reveals the occurrence of a condition known as the 'SUBSCRIPTRANGE CONDITION'. Note that the value of Z obtained from the card is 6. The DO loop willbe performed 6 times. Thus, 6 elements of the arrayA(I) will be computed. But the DCL statement declared the array A (I) as con-sisting of 5 elementsonly. Hence, the program may use ad-ditional storage declared undersome other name, destroying its previous contents. This is undesireable, because those storage elements may have contained some preciousdata which is now destroyed. b) The printout of the computer will be as follows: 100 15 800 900 56 30 Notice that the computer calculated 6 elements for arrayA(5), even thoughA(5) should have had only 5 elements. c) The error due to this condition can be avoided by spe-cially instructing the computer to regard this as a condi-tion to look out for. The instruction is given to the com-puter by attaching a 'KEYWORD' prefix to the first statement of the program, as shown in the corrected program below. Also, the program body must contain an additional statement telling the computer what to do if the given condition does arise. For example, the computer could be told to go to some 'error routine' if the SUBSCRIPTRANGE CONDITION arises, as is done below. The corrected program is as follows: (SUBSCRIPTRANGE): EXAMPLE:PROCOPTIONS(MAIN); DCLA(5) FIXED(3); ON SUBSCRIPTRANGE GOTO (ERROR ROUTINE); GET LIST (Z) ; DO I = 1 TO Z; GET LIST(X,Y); A (I) = X\textasteriskcenteredY: PUTLIST(A(I))? END; END EXAMPLE; d) The computer will now continuously check if the number of array elementscomputed is within the bounds declared for the array. Consequently, now, only 5 elements will be computed and printed out, as follows: 100 15 800 900 56 The program will end now. But, an error message will also be printed out, informingthe programmer that a SUBSCRIPTRANGE CONDITION had occurred during the program.

Question:

A chemist knows that the ∆H\textdegree = - 485 kJ for the reaction 2H_2 (g) + O_2 (g) \rightarrow 2H_2 O (g) and that ∆H\textdegree = - 537 kJ for H_2 (g) + F_2 (g) \rightarrow 2HF (g) . With this information, he calculated the ∆H\textdegree for 2H_2 O(g) + 2F_2 (g) \rightarrow 4HF(g) + O_2 (g) and predicted whether ∆S\textdegree was positive or negative. How?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E15-0526.htm

Solution:

Hess' Law states that the net heat change resulting from a particular chemical reaction is the same, independent of the steps involved in the transformation. Thus, ∆H\textdegree of the sum of two reactions equals the sum of the ∆H\textdegree 's of each reaction. The chemist knows that (i)2H_2 (g) + O_2 (g) \rightarrow 2H_2 O (g)∆H\textdegree = - 485 kJ (ii)H_2 (g) + F_2 (g) \rightarrow 2HF (g)∆H\textdegree = - 537 kJ By doubling (ii) and adding to (i), the chemist obtains the desired equation 2H_2 O\rightarrow2H_2 + O_2 ∆H\textdegree = +485 kJ 2 (H_2 + F_2\rightarrow2HF)∆H\textdegree = -2 (537) kJ 2H_2O + 2F_2 \rightarrow 4HF+ O_2 The ∆H\textdegree for this reaction, according to Hess' Law, equals ∆H\textdegree(i)+ ∆H\textdegree_(ii) × 2 = 485 + 2(- 537) = - 589 kJ/moles. Because the ∆H\textdegree for the overall reaction is negative, the reaction is spontaneous. The ∆S\textdegree is positive because as the reaction proceeds from reactants to products there is an increase in the number of moles present in the system. There are 4 moles of reactants and 5 moles of products.

Question:

Per day, an average adult may lose liquid water in the following liter amounts: as urine 1.0; as sweat, 0.6; as fecal matter, 0.1; in expired air, 0.4. How many gallons of water pass through a human being, assuming these data hold for 60 years?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E32-0932.htm

Solution:

From this data, one can see that 1.0 + 0.6 + 0.1 + 0.4 = 2.10 liters of water are lost per day. In 60 years, there are 365 × 60 = 21,900 days. In 60 years, 21,900 days × 2.10 l/day = 4.60 × 10^4 l of water pass through one person. 1 gallon = 4.224 liters. Converting liters to gallons, 4.60 × 10^4 l × [(1 gallon) / (4.224 liters)] = 1.089 × 10^4 gallon.

Question:

Wrtiea program to find the greatest factor of an integer. Wrtie a program to find the greatest factor of an integer

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G10-0232.htm

Solution:

Here, a trick is used based on the application of the INT function. If the quotient of two integers is also an Integer, the denominator must be a factor (i.e., if N/D = K = INT(N/D), then both D and K are factors of N). The smallest D which satisfies this condition will give us the desired value of the greatest factor K = N/D. What values of D should be tried? The first guess would be 2 to N - 1. A refinement of this guess would be to observe that is the smallest possible factor, while N/2 is the largest. More-over, since \surdN\surdN = N, it is best to examine all D's = 2,3,..., \surdN as possible factors. 1\O READ N 2\O FOR D = 2 TO SQR(N) 3\O IF N/D = INT(N/D) THEN 70 4\O NEXT D 5\O PRINT N; "IS PRIME" 6\O GO TO 1\O 7\O PRINT N/D; "IS THE GREATEST FACTOR OF"; N 8\O GO TO 1\O 9\O DATA 1946, 1949, 1\O\O9, 1\O\O3 1\O\O DATA 11\O\O1, 24\O, 11\O END

Question:

What is meant by character string variables, characterstring constants? Give and explain the rules governing declaration, valueacquisition (GET) and value disposition (PUT).

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G13-0317.htm

Solution:

Consider the following PL/I statement DECLARE STUDENT_NAME CHARACTER (15); GET LIST (STUDENT_NAME); On the data card there is: 'JIMMYbCARTERb', 'TEDbKENNEDYb', b 'RlCHARDbNIXONb'. The declaration specifies STUDENT_NAME as the name of a characterstring variable. A character string variable has character string dataas its value. The GET LIST statement acquires the value 'JIMMYbCARTERbas the value of the vari-able STUDENT_NAME.This constanthas a length of only 13 characters, but the variable is declared a lengthof 15 char-acters. Therefore 2 blanks characters will be added. The storagewill therefore contain: JIMMYbCARTERbbb A character string variable must always be explicitly declared in a program. There are no default rules to asso-ciate character string variables. The character attribute is the word CHARACTER followed by a decimalinteger enclosed in parentheses. The integer value denotes the lengthof the character string. e.g.CHARACTER(10) CHAR(18) CHAR If no length is specified in the attribute as above, the PL/I machine assumesa length of one. The keyword CHARACTER may also be used in abbreviatedform CHAR as shown above. In addition to the character attribute, the attribute VARYING may be usedto indicate that the length specification shows the maximum possible length. But the variable may actually take any length up to the maximum. An abbreviation Of VARYING is VAR. e.g.DECLARE STUDENT_NAME CHARACTER(15) VARYING; GET LIST (STUDENT_NAME); The above GET LIST statement, with the same DATA card as be-fore acquiresthe constant JIMMYbCARTERb The length is 13. But now the 2 extra blanks are .not added in the storage becauseof the specification of the VARYING attribute. If the length attribute specified in the declaration is less than the lengthof the character string constant, then the right-most excess charactersare truncated. e.g.DECLARE CITY_NAME CHAR(5); GET LIST (CITY_NAME); The information punched on the data card is 'CHICAGO' which is 7 characterslong. But as the length attribute is 5, CITY_NAME acquires the valueCHICA. Note that the GET and PUT statements are the same as for numericvariables but the character string constants on the data card must beenclosed within quotes. Also, when we want to output a character string con-stant which is notdeclared in the declare statement, it must be enclosed within quotes e.g.PUT LIST ('THE VALUE IS',- --); PUT LIST ('CUSTOMERS NAME', 'NEW PURCHASE', 'BALANCE');

Question:

If an atom remains in an excited state for 10^-8 sec, what is the uncertainty in the energy of that state?

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/Users/wenhuchen/Documents/Crawler/Physics/D33-0984.htm

Solution:

To solve this we use the Heisenberg uncertainty principle: \DeltaE\Deltat\geq h, where h is Planck's constant. \DeltaE \geq h/∆t = [(6.6 x 10^-27 erg-sec)/(10^-8 sec)] = 6.6 x 10^-19 ergs This is the limit of accuracy with which the energy of such an excited atom can ever be measured. It is called the energy width of the excited state.

Question:

Explain why benzene is a planar molecule, while cyclohexane, C_6H_14 , contains a puckered ring of six carbon atoms.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E21-0770.htm

Solution:

The best way to approach this problem is to consider the geometry of the carbon-carbon bonding. To do this, consider its hybridization. Benzene can be written as shown in figure A. Notice that each carbon has 4 bonds, and is bonded to three other elements (2 other carbons and 1 hydrogen). This suggests sp^2 hybridization, sp^2 hybridization allows bonding to three other elements, and leaves a \textquotedblleftfree" p orbital. This p orbital can overlap with another p orbital to give a pi bond, which serves to form the double bond. With sp^2 hybridization, the molecule must assume a geometry where the orbitals are 120\textdegree apart. This is the only way that the repulsion which occurs when like charges are brought to-gether can be avoided. See figure B. An angle of 120\textdegree creates a planar geometry. This is true because a planar molecule with bond angles of 120\textdegree is a configuration where the 3 atoms involved are the farthest apart. Consider cyclohexane. It can be written as shown in figure C. Each carbon atom, is bound to four other atoms (2 other carbons and 2 hydrogens). This suggests sp^3 hybrodization. With sp^3 hybridization, a molecule must assume a geometry that permits 109\textdegree28' bond angles. The atoms of the bond are arranged in a tetrahedral formation, this allows for the greatest distance between the atoms. The way cyclohexane can accomplish this is to assume a puckered ring of six carbon atoms, as can be seen in figure D.

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Question:

Although it is often possible to see unicellular organisms if a dropof pond water is placed on a microscope slide, it is impossibleto seecuboidalepithelium if a rat kidney is placedunder the microscope. Describe the steps necessary forthe preparation of tissues for light microscopy.

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/Users/wenhuchen/Documents/Crawler/Biology/F02-0064.htm

Solution:

There are four basic steps in the preparation of tissues for light microscopy. (1) Fixing. The tissue is killed and stabilized with chemical agents, sothat structures are preserved with a life-like appearance. Fixing preventspost-mortem changes of tissue structure caused by autolysis (self--digestion) . (2) Embedding. The fixed tissue is embedded in a hard material thatprovides firm support. This material is often paraffin or wax. The tissueis dehydrated and soaked in molten wax which penetrates the tissuesand then is cooled to form a hard matrix. (3) Sectioning. The support provided by the em-bedding material enablesthe tissue to be cut into very thin slices without being crushed by theknife edge. Sectioning tissue permits study of complex structures by providingan unimpeded view of deep layers. (4) Staining. The cells are stained with dyes that color only certain organelles. This provides contrast between different cellular structures, for example, be-tween nucleus and cytoplasm or between mitochondria and othercytoplasmicstructures. Most cellular structures are colorless. If unstained, it is difficult and often impossible to distinguish different cellular regionsby light microscopy. Cytochemistryrefers to staining procedures which specifically color certaincellular substances. For ex-ample, theFeulgenprocedure specificallydyes DNA. DNA reacts with dilute hydrochloric acid to give aldehydegroups; these then react with the colorless Schiff reagent to changethe dye to red. No other cellular components react in the right way withHClto cause the Schiff re-agent to become colored, so only DNA is stained. Othercytochemicaltechniques permit visualization of RNA, of the acidhydrolasesoflysosomes, and of specific lipids, proteins, and carbohydrates.

Question:

Two circular cylinders have the same mass and dimensions, but one is solid while the other is a thin hollow shell. If they are released together to roll without slipping down a plane inclined at 30\textdegree to the horizontal, how far apart will they be after 10 s?

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/Users/wenhuchen/Documents/Crawler/Physics/D04-0202.htm

Solution:

In either case the forces acting on the cylinder are as shown in the diagram. The weight mg^\ding{217} acting downward is split up into its components parallel to and perpendicular to the plane. The other two forces acting on the cylinder are the normal force N^\ding{217} exerted by the plane and the frictional force attempting to prevent motion, F^\ding{217}, which has magnitude \muN. Since the cylinder does not lift from the plane, N = mg cos 30\textdegree. Further, by Newton's second law, mg sin 30\textdegree - \muN = mg(sin 30 - \mu cos 30) = ma, where m is the mass of the cylinder and a the acceleration produced. Rotation about the center of the cylinder also takes place. Since we are dealing with rotations of dis-tributed masses, a relation involving the moment of iner-tia must be used. That relation is the rigid body analogue of Newton's second law, F = ma, or \cyrchar\CYRG = I\alpha where the torque \cyrchar\CYRG corresponds to F, the moment of inertia I corresponds to m, and the angular acceleration \alpha of the cylinder corresponds to a. If an axis is taken about the center of the sphere, then the only torque acting on the sphere is \muNr due to the frictional force \muN. Then \muNr = \mur mg cos 30\textdegree = l\alpha, Since no slipping takes place, the point A is instantan-eously at rest. Hence a = r\alpha, and mg (sin 30\textdegree - \mu cos 30\textdegree) = ma,(1) \mur mg cos 30\textdegree = Ia/r.(2) or, multiplying (1) by r^2 , and (2) by r mgr^2 sin 30 - \mumgr^2 cos 30 = mar^2(3) and\mur^2mg cos 30 = Ia(4) Substituting (4) in (3) mg r^2 sin \texttheta - Ia = mar^2 or a = (mg r^2 sin \texttheta)/(I + mr^2) For the solid cylinder, I = 1/2mr^2 a_1 = (mg r^2 sin \texttheta)/(1/2mr^2 + mr^2 ) = (2/3)g sin \texttheta, and for the hollow cylinder, I = mr^2 a_2 = (mg r^2 sin \texttheta)/(mr^2 + mr^2) = (1/2)g sin \texttheta. The distances traveled in 10 s from rest are, found by using the kinematic equations for constant acceleration. If the top of the inclined plane is taken as the initial position, then s_0 = 0. Also the initial velocity v_0 = 0. ThenS = s_0 + v_0t + (1/2)at^2 . Hence S_1 = 0 + (1/2)a_1t^2 = (1/2) × (2/3) × 32 ft\bullets^-2 ×(1/2) × 100 s^2 = (1600)/3 ft andS_2 = (1/2)a_2t^2 = (1/2) × (1/2) × 32 ft\bullets^-2 ×(1/2) × 100 s^2 = 400 ft. = 400 ft. \therefore S_1 - S_2 = (533.3 - 400)ft = 133.3 ft.

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Question:

When an inch of rain falls on New York, it results in a rainfall of 19.8 × 10^9 liters. If the temperature drops to 0\textdegreeC after this rainfall and the density of rain (H_2O) is 1.00 g/cm_3, how much heat is released when this quantity of water freezes? You may assume ∆H_fus= 1.4 Kcal mole^-1.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E28-0912.htm

Solution:

∆H_fus, is the heat of fusion. It is defined as the amount of heat necessary to melt 1 g of the solid to a liquid at the same temperature. In this problem, water is freezing to ice. To find out the exact amount of heat released, the number of moles of rain that fell must be known. To determine the number of moles of rain that fell, use the fact that its volume is 19.8 × 10^9 liters and density = 1.00 g/cm^3. Mass= (density) (volume) = (1.00 g/cm^3) (1.98 × 10^13 cm^3) = 1.98 × 10^13 g. (note: 1.98 × 10^9 liters is equivalent to 1.98 × 10^13 cm^3 by the conversion factor 1000 cm^3/l). From the mass of the water, the number of moles of H_2O can be computed. number of moles =[grams (mass)] / [molecular weight (M.W.)] The M.W. of water = 18 g/mole. Therefore, in 1.98 × 10^13 grams of water, there are (1.98 × 10^13 g) / (18 g/mole) = 1.1 × 10^12 moles Recalling that∆H_fus= 1.4 Kcal/mole, 1.1 × 10^12 moles of water loses (1.4) (1.1 × 10^12) = 1.54 Kcal of heat when it freezes.

Question:

An automobile traveling at a speed of 30 mi/hr accelerates uniformly to a speed of 60 mi/hr in 10 sec. How far does the automobile travel during the time of accelera-tion?

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/Users/wenhuchen/Documents/Crawler/Physics/D03-0064.htm

Solution:

Converting to ft-sec units, 30(mi/hr) = {30(mi/hr)} × {(5280 ft)/(1 mi)} × (1hr/3600 sec) = 44 ft/sec 60(mi/hr) = 88 ft/sec. Uniform acceleration can be found from the change in velocity divided by the time elapsed during the change. a=\Deltav/\Deltat = {(88 ft/sec) - (44 ft/sec)} / 10 sec = 4.4 ft/sec^2 x= v_0t + (1/2)at^2 = (44 ft/sec) × (10 sec) + (1/2) ×(4.4 ft/sec^2) ×(10 sec) = 440 ft + 220 ft = 660 ft. Suppose next that the automobile, traveling at 60 mi/hr, slows to 20 mi/hr in a period of 20 sec. What was the acceleration? a= (v_2 - v_1)/\Deltat = {(20 mi/hr) - (60 mi/hr)} / (20 sec) = -2 (mi/hr)/sec. The automobile was slowing down during this period so the acceleration is negative.

Question:

An electron of charge - 1.60 x 10^-19coulis in a cir-cular orbit of radius R = 5.5 x 10^-11 m around a proton of equal and opposite charge. The mass of the electron m_e is 9.11 x 10^-31 kg and the mass of the proton m_p is 1.67 × 10^-27 kg. Compare the electrostatic and gravi-tational forces between the two particles. Neglecting relativistic corrections, calculate the speed of the electron and the period of its orbit.

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/Users/wenhuchen/Documents/Crawler/Physics/D19-0629.htm

Solution:

The electrostatic force is: F_E =( k_E k_E ) q^2 ) q^2 / R^2 and the gravitational force is FG= G(m_em_p_ / R^2 ) The ratio is: FE/ FG= [k_E(q^2 / R^2 )] / [G (m_em_p/ R^2 )] = (k_Eq^2 / Gm_em_p) = [(9.10^9 N - m^2 / coul.^2)^ (1.60 × 10-19coul.)^2] / [(6.67×10 ^-11 N-m^2/kg^2)(9.11×10^-31 kg) (1. 67×10^-27 kg)] = 2.26 × 10^39 The electrostatic force is gigantic compared to the gravitational force. If the speed of the electron in orbit is V, then: F_E =k_E(q^2 / R^2) = mea = m_e v2/ R V = q[k_E/m_eR]1/2 = (1.60 × 10-19coul.)[{(9.10^9 N - m^2 /coul.)2}/ {(9.11x 10^-31kg) (5.5 x10^-11 m)}]^1/2 = 2.1 × 10^6 m/sec. The period T is the time it takes for the electron to complete a single orbit: T = 2\piR / V = {2(3.14) (5.5 × 10^-11 m)} / {2.1 × 10^6 m / sec} = 1.64 × 10-16sec.

Question:

The vapor pressures of pure benzene and toluene at 60\textdegreeC are 385 and 139Torr, respectively. Calculate (a) the partial pressures of benzene and toluene, (b) the total vapor pressure of the solution, and (c) the mole fraction of toluene in the vapor above a solution with 0.60 mole fraction toluene.

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Solution:

The vapor pressure of benzene over solutions of benzene and toluene is directly proportional to the mole fraction of benzene in the solution. The vapor pressure of pure benzene is the proportionality constant. This is analogous to the vapor pressure of toluene. This is known asRaoult'slaw. It may be written as P_1 = X_1 P_1\textdegree p_2 = X_2 P_2\textdegree where 1 and 2 refer to components 1 and 2, P_1 and P_2 represent the partial vapor pressure above the solution, P_1\textdegree and P_2\textdegree are the vapor pressures of pure components, X_1 and X_2 are their mole fractions. Solutions are called ideal if they obeyRaoult'slaw. The mole fraction of a component in the vapor is equal to its pressure fraction in the vapor. The total vapor pressure is the sum of the vapor's component partial pressures. To solve this problem one must 1) calculate the partial pressures of benzene and toluene usingRaoult's law . 2) find the total vapor pressure of the solution by adding the partial pressures 3) find the mole fraction of toluene in the vapor. One knows the mole fraction of toluene in the solution is 0.60 and, thus, one also knows the mole fraction of benzene is (1-0.60) or (0.40). UsingRaoult'slaw: P\textdegree_benzene = 385TorrP\textdegree_Toluene= 139 a)P_benzene= (0.40)(385Torr) = 154.0Torr P_toluene= (0.60)(139Torr) = 83.4Torr P_total= (154.0) + (83.4) = 237.4Torr c) The mole fraction of toluene in the vapor = X_toluene,vap= (P_toluene) / (P_toluene+P_benzene) = (83.4) / (237.4) = 0.351.

Question:

(a) A chemist prepares a 0.01 M solution of NaC_2H_3O_2. Find itspH.K_A = 1.8 × 10^-5. (b) Determine the pH of 0.1 M solution of NH_4Cl. K_b = 1.8 × 10^-5.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E12-0440.htm

Solution:

Both parts of this problem involve the hydrolysis of a salt of a weak acid or base to form an acidic or basic solution. This occurs because water can dissociate to H^+ and OH^- ions, and these can react with ions from weak acids and bases. This necessitates a consideration ofK_hyd, the hydrolysis constant. Because both NaC_2H_3O_2 and NH_4Cl are salts, they exist as ions in aqueous solution. The situation in part (a) is C_2H_3O_2^- + H_2O \rightleftarrows HC_2H_3O_2 + OH^- TheK_hydfor this is {[HC_2H_3O_2] [OH^-]} / [C_2H_3O_2^-]. K_hydis also defined asK_w/ K_a. Therefore, K_hyd=K_w/ K_a = {[HC_2H_3O_2] [OH^-]} / [C_2H_3O_2^-]. From the equation that depicts the reaction one can see that the [OH^-] = [HC_2H_3O_2], since for every molecule of OH^-, a molecule of HC_2H_3O_2 must be generated. One can say K_w/ K_a = {[OH^-] [OH^-]} / [C_2H_3O_2^-]or [OH^-]^2 =K_w/ K_a = [C_2H_3O_2^-]. K_w/K_a is small enough so that [C_2H_3O_2^-] at equilibrium will have a value as if hydrolysis had not occurred. Letting B represent [C_2H_3O_2^-] and taking the square root; [OH^-] = \surd{[K_w/ K_a] B}. Next take the negative log of the equation. (Remember thatpOH= - log [OH^-], and thatK_wis the ion-ization constant of water, which is 1.0 × 10^-14). pOH= (1/2)pK_w- (1/2)pK_a+ (1/2)pB However, (1/2)pK_w= 7 andpOH= 14 -pH.As stated, K_w= 1.0 × 10^-14 = [H^+][OH^-] and [H^+] = [OH^-]. Therefore, [OH^-]^2 (or [H^+]^2) = 1.0 × 10^-14 , [OH^-] = 1.0 × 10^-7, and [H^+] = 1.0 × 10^-7 pOH= - log [1 × 10^-7] = 7pOH+ pH = 14. pH = - log [1.0 × 10^-7] = 7. Therefore,pOH= 14 -pH. One can now substitute to obtain pH = 14 - [7 - (1/2)pK_a+ (1/2)pB]or pH = 7 + (1/2)pK_a- (1/2)pB. This is the equation wanted. It is known that B = [C_2H_3O_2^-]. Only when this union is at equilibrium \allequal B can this procedure be used (i.e. the K_a of the acid must be >> 10^-10). The K_a is given as 1.8 × 10^-5 and meets these requirements. ThepK_a= - log [1.8 × 10^-5] = 4.74. B = 0.01 M (given) andpB= 2.00. Now go back to pH = 7 + (1/2)pK_a- (1/2)pBand substitute these values to obtain: pH = 7 + (1/2)(4.74) - (1/2)(2.00) pH = 8.37 Part (b) is worked out in exactly the same way. One ends up having pH = 7 - (1/2)pK_b+ (1/2)pB = 7 - (1/2)(4.74) + (1/2)(1.00) pH = 5.13 where (1/2)pB= - (1/2) log [0.1] = (1/2)(1.00).

Question:

What is a chemical reaction? What is meant by the rate- limiting step of a chemical reaction?

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Solution:

A necessary requirement of a chemical reaction is that the products be chemically different from the reactants. Therefore a chemical reaction is one in which bonds are broken and/or formed. Some chemical reactions have only a single step, with only one bond being formed or broken. In multistep reactions, a certain sequence of bond formation and bond breakages proceed. Usually the formation of a chemical bond releases energy, and the products formed are more stable than the reactants, in that they have a lower potential energy. The breakage of a bond requires some form of energy which is absorbed by the bond, giving products that, together, will exist at a higher potential energy. Thus, a reaction that requires heat to proceed and usually involves breaking bonds is termed an endothermic reaction. A reaction that proceeds with the release of heat and usually involves forming bonds is termed an exothermic reaction. Aside from bond changes, there is a change in potential energy or enthalpy and there can be a change in entropy, or orderliness that accompanies the reaction. There is a tendency for chemical systems to go toward minimum energy or enthalpy and thus achieve maximum stability. There is also a tendency toward maximum disorder or entropy. These two concepts can be combined in a new function called free energy (G) , which is the energy that is available for doing work. G = ∆H - T∆S , where ∆Hrepresents the change in enthalpy ∆Srepresents the change in entropy Tis the temperature in Kelvin ∆Gis the spontaneous change in free energy In order for a chemical reaction to proceed, ∆G must be negative. A loss in enthalpy will be represented by a negative ∆H and a gain in entropy by a positive ∆S. If ∆G is positive, energy would have to be added to the system to drive the reaction forward. Most chemical reactions occurring in the living cell are not simple, one-step reactions. Many contain a series of consecutive steps coupled, or linked, by common intermediates with the product of the first step being the reactant in the second step, and the product of the second step is the reactant in the third, and so on. The ∆G of such reactions is the sum of the ∆G's of the individual steps. Each step of the reaction has its own activation energy, and the one requiring the highest activation complex is kinetically the most un-favorable one. In other words, it will have the slowest rate and will for this reason determine the rate of the entire reaction. This step is better known as the rate-determining (or limiting) step of the reaction.

Question:

In the two pulley systems shown in Fig. 1a, determine the velocity and acceleration of block 3 when blocks 1 and 2 have the velocities and acceleration shown.

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/Users/wenhuchen/Documents/Crawler/Physics/D37-1062.htm

Solution:

Figure 1b defines the positions relative to the fixed pulley A needed to solve this problem. The lengths of the cord are assumed to be constant and the section of cord over the pulleys to remain constant, thus yielding the following constraint equations y_1 + y_B = K_1,(a) (y_2 \rule{1em}{1pt} y_B) + (y_3 \rule{1em}{1pt} y_B) = k_2, ory_2 + y_3 \rule{1em}{1pt} 2_(y)B = K_2.(b) Differentiating each equation (a) and (b) twice yields ẏ_1 + ẏ_B = 0(c) ẏ ̇_1 + ẏ ̇_B = 0(d) ẏ_2 + ẏ_3 \rule{1em}{1pt} 2ẏ_B = 0(e) ẏ ̇_2 + ẏ ̇_3 \rule{1em}{1pt}2ẏ ̇_B = 0(f) These equations lead directly to the desired solutions upon substitution of the information from Figure 1a. If it is supposed that down is positive, then v_1 = ẏ_1 = \rule{1em}{1pt}2 ft / sec,v_2 = ẏ_2 = 3 ft/sec,(g) a_1 = ẏ ̇_1 = 1 ft / sec^2,a_2 = ẏ ̇_2 = \rule{1em}{1pt}2 ft / sec^2. First solve (c) and (d) to find the motion of pulley B. \rule{1em}{1pt}2 ft / sec + ẏ_B = 0 ẏ_B = 2 ft / sec(h) 1 ft / sec^2 + ẏ ̇_B = 0 ẏ ̇_B = \rule{1em}{1pt}1 ft / sec^2.(i) Now using (g), (h) and (i) in equations (e) and (f), it is possible to obtain the desired results: 3ft / sec + ẏ_3 \rule{1em}{1pt} 2 (2 ft / sec) = 0 ẏ_3 = 1 ft / sec (\rule{1em}{1pt}2 ft / sec^2) + ẏ ̇_3 \rule{1em}{1pt} 2 (\rule{1em}{1pt}1 ft / sec^2) = 0 ẏ ̇_3 = 0 Thus, block 3 is moving down with a constant velocity v_3 = 1 ft / sec.

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Question:

Explain briefly the background and applications of COBOL.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G01-0018.htm

Solution:

COBOL, an acronym forCOmmonBusiness-Oriented Lan-guage, is a high-level, procedural language for business appli-cations. The language was first introduced in 1959 by a commit-tee named Conference on Data System Languages (CODASYL) .This committee, sponsored by the Department of DefenseDoD), issued a report in 1960 specifying the first version of COBOL, and the first compiler appeared shortly after. Subsequent revision is-sued in 1961 became the cornerstone of further developments including standard versions by American National Standards In-stitute (ANSI). The following is a quick chronicle of COBOL's background : 1959-CODASYL sponsored byDoD 1960-First COBOL compiler 1961-COBOL-61(First Revision) 1968-ANSI COBOL-68(First Standard) 1974-ANSI COBOL-74(Widely Used Revised Standard) 1985-ANSI COBOL-85(New Standard) COBOL applications are mostly commercial rather than scien-tific. It is still the most widely used language for business applications (accounting, banking, organizational, management information, etc.) accounting for 60% to 80% of the programs written for this purpose.

Question:

Compute the resistance across the terminals in figure A. Compute the resistance across the terminals in figure A.

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/Users/wenhuchen/Documents/Crawler/Physics/D20-0672.htm

Solution:

First we regroup the resistors to simplify our calculations. We are allowed to do this as long as we do not change the orientation of the resistors with respect to each other and to the terminals. An equivalent arrangement is shown in figure B, In figure C we sum the resistors in series. We are left with two re-sistances in parallel which we find from the formula: (1/R_T ) = (1/R_1 ) + (1/R_2 ) = (1/105) + (1/75) = (12/525) = (4/173) where R_T is the total resistance across the terminals. Thus: R_T = (173/4) = 43.25

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Question:

Monochromatic light of wavelength 400m\mu from a distant point source falls on an opaque plate in which there is a small circular opening. As a screen is moved toward the plate from a large distance away, the Fresnel diffraction pattern on the screen first has a dark center when the distance from plate to screen is 160cm. Find the diameter of the central disk in the diffraction pattern if a lens of focal length 160cm is placed just to the right of the circular opening.

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Solution:

We have a circular aperture illuminated by coherent light from a source infinitely distant from the aperture. This implies that all points on the aperture are in phase with all other points. We now place a screen an infinite distance from the aperture and slowly move it towards the aperture observing the center spot of the screen. At some distance the path length from the light at the circumference of the aperture will be just one wavelength out of phase with light from the center of the aperture, which can be represented algebraically as R_1 = R_0 + \lambda in figure (2). We have learned from our study of diffraction that this is just the condition for destructive interference. Now if we move the screen closer to the aperture, the center of the screen will have a bright spot when the path difference between a circum-ferential ray and a central ray is (3\lambda / 2) and dark when there is a 2\lambda path difference, etc. Each of the path differences defines the edges of Fresnel zones. Looking at figure (2) for our problem when R_O is 160cm, r will just be the outer radius of the second Fresnel zone. From figure (2), R^2 _1 = R2_O + r^2(1) andR_1 = R_O + \lambda(because there is destructive interference at R_O = 160 cm.)(2) Substituting for R_1 in equation (1) yields R^2 _O + 2\lambdaR_O + \lambda^2 = R2_O + r^2 The R2_O terms cancel, and so, r^2 = 2\lambdaR_O + \lambda^2 or r =\surd(\lambda^2 + 2\lambdaR_O) = [(400 × 10^-9m)^2 + 2 × 400 × 10^-9m × 1.6m]1 / 2 = 1.13 × 10^-3m,(3) which is the radius of the aperture. If we now place a 160cm focal length lens on the screen side of the aperture, the lens will convert an infinite distance on the screen side to a point 160cm from the aperture (as in figure (1)). Stating it another way, the screen at 160cm from the aperture with a 160cm focal length lens will produce a Fraunhofer diffraction pattern of the source. Now from any standard optics text Fraunhofer diffraction of a circular aperture is (d / 2) / l =1.22\lambda/ 2r(4) where d is the diameter of the central disk; l is the dis-tance from the aperture; \lambda is the wavelength of the light used and r is the radius of the aperture. Now solving for the diameter of the central disk we have d = (l × 1.22 × \lambda) / r where l = 1.6m, \lambda = 400 × 10^-9m, and r = 1.13 × 10^-3m. Henced = (1.6m × 1.22 × 400 × 10^-9m) / (1.13 × 10^-3m) = 6.91 × 10^-4m.

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Question:

In order to attract males for mating, females of many insect species secrete chemical compounds called pheromones. One compound of formula C_19H_38O (282 g/mole) is among these pheromones, 10-^12 g of it must be present to be effective. How many molecules must be present to be effective?

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Solution:

Avogadro's number (6.02 × 10^23) is the number of particles per mole of any substance. To find the number of molecules to be effective, determine the number of moles present, and multiply this quantity by Avogadro's number. Solving: 10-^12 g of pheromones must be present to attract males. The number of moles is equal to number of grams/molecular weight. number of moles = (10-^12 g ) / (282 g/mole) = 3.55 × 10-^15 moles number of molecules = 3.55 × 10-^15 moles × 6.02 × 10^23/mole = 2.12 × 10^9 molecules.

Question:

Chromic oxide (Cr_2 O_3) may be reduced with hydrogen according to the equation Cr_2 O_3 + 3H_2 \rightarrow 2Cr + 3H_2 O (a) What weight of hydrogen would be required to reduce 7.6 g of Cr_2 O_3? (b) For each mole of metallic chromium prepared, how many moles of hydrogen will be required? (c) What weight of metallic chromium can be prepared from one ton of Cr_2 O_3? 1 lb = 454 g.

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Solution:

(a)From the equation for the reaction, one knows that it takes three moles of H_2 to reduce one mole of Cr_2 O_3. Thus, in solving this problem one should first determine the number of moles of Cr_2 O_3 in 7.6 g, then using the ratio (3/1) = [(number of moles of H_2) / (number of moles of Cr_2 O_3)] One can find the number of moles of H_2 necessary to reduce 7.6 g of Cr_2 O_3. After finding the number of moles of H_2 needed, one can obtain the weight by multiplying the number of moles by the molecular weight of H_2. (1)Solving for the number of moles of Cr_2 O_3 in 7.6 g. This is done by dividing 7.6 g by the molecular weight of Cr_2 O_3 (MW = 152). no. of moles = [(7.6 g) / (152 g/mole)] = .05 moles (2)determining the number of moles of H_2 necessary. The ratio (3/1) = [(number of moles of H_2) / (number of moles of Cr_2 O_3 )] will be used. This ratio was made using thestoichiometriccoefficients of the equation for the reaction (3/1) = [(number of moles of H_2) / (.05moles)] no. of moles of H = [(.05ntoles× 3) / (1)] = .15 moles (3) Solving for the weight of H_2. (MW = 2.) The weight of a compound is found by multiplying the number of moles present by the molecular weight. weight of H_2 = .15 moles × 2 g/mole = .30 g. (b) From the equation for the reaction, it is seen that 3 moles of H_2 are needed to form 2 moles of Cr. This means that for every mole of Cr formed 3/2 this amount of H_2 is needed. Thus, 1.5 mole of H_2 is necessary to form one mole of Cr. (c)Using the equation for the reaction, one is told that for every mole of Cr_2 O_3 reduced 2 moles of Cr are formed. Thus, one must determine the number of moles in 1 ton of Cr_2 O_3; there will be twice as many moles of Cr formed. After one knows the number of moles of Cr formed, one can determine its weight by multiplying the number of moles by the molecular weight of Cr. (1)Determining the number of moles of Cr_2 O_3 in 1 ton of the compound (1 ton = 2000 lbs). The number of moles is found by dividing the weight of the Cr_2 O_3 present by its molecular weight. Because the molecular weight is given in grams per mole, it must be converted to pounds per mole before using it to determine the number of moles present. There are 454 g in one pound, thus grams can be converted to pounds by multiplying the number of grams by the con-version factor 1 lb/454 g. (MW of Cr_2 O_3 = 152.) MW of Cr_2 O_3 in lbs = 152 g/mole × 1 lb/454 g = .33 lbs/mole. (2)determining the number of moles of Cr_2 O_3 in 1 ton. The number of moles present can be found by dividing 1 ton (2000 lbs) by the molecular weight in pounds. number of moles = [(2000 lbs) / (.33 lbs/mole)] = 6060 moles Thus, there are twice this many moles of Cr produced, number of moles of Cr = 2 × 6060 moles = 12120 moles (3) finding the weight of Cr formed. One knows that 12,120 moles of Cr are produced. To find the weight of this quantity, the number of moles is multiplied by the molecular weight. To find the weight in pounds, one must first convert the molecular weight from grams to pounds. This is done by multiplying the molecular weight by the conversion factor lb/454 g. This is used because there are 454 g in 1 lb (MW of Cr = 52). MW of Cr in pounds = 52 g/mole × 1 lb/454 g = 0.11 lb/mole. The weight of the Cr formed is now found by multiplying this molecular weight by the number of moles present. weight = 0.11 lb/mole × 12120 moles = 1333 lbs.

Question:

The volume of a sample of gaseous argon maintained at constant pressure was studied as a function of temperature constant pressure was studied as a function of temperature and the following data were obtained and the following data were obtained : Temperature, T (\textdegreeK) Volume, V (l) 2500.005 3000.006 3500.007 Calculate and confirm the Charles' Law constant for this system. Determine the temperature corresponding to a valume of 22.4 l.

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Solution:

Charles' law states that the volume and absolute temperature of an ideal gas are directly pro-portional, i.e. V =kT, k is a constant. To determine the Charles' law constant k, we can use any one of the three sets of temperature-volume measurement. Choosing the first of these, we obtain k = (V/T) = [(0.005)/(250\textdegreek)] = 2.00 × 10^-5 l-\textdegreeK^-1 . This value is confirmed by showing that the other two sets of data give the same value for k: k = (V/T) = [(0.006 l)/(300\textdegreeK)] = 2.00 × 10^-5 l-\textdegreeK^-1 .and k = (V/T) = [(0.007 l)/(350\textdegreeK)] = 2.00 × 10^-5 l-\textdegreeK^-1 . To determine the temperature corresponding to 22.4 l, we solve Charles' law for T, obtaining T = (V/T) = [(22.4 l)/(2.00 × 10^-5 l-\textdegreeK^-1 )] = 1.12 × 10^6 \textdegreeK or about a million degrees Kelvin.

Question:

-The interaction which may be present between particles themselves give rise to internal torques. Show that the sum of all internal torques is zero.

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Solution:

The total torque is defined as N =^N\sum_n_=1r_n× F_n, r_n \ding{217} is the position vector of the particle and F_n^\ding{217} is the force acting upon it. For internal forces F_i F_i = ^N\sumʹ_j=1F_ij, = ^N\sumʹ_j=1F_ij, the sum of the forces on particleifrom all other particles j. (The prime on the summation sign means that the term j =iis excluded.) Thus the internal torque is N_int = \sum_ir_i×F_i= \sum_i\sumʹ_jr_i×F_ij, Here we simply substitute ^N\sumʹ_j=1F_ij^\ding{217} forF_i^\ding{217}. but by relabeling the dummy indicesiand j, we have \sum_i\sumʹ_jr_i×F_ij\equiv \sum_j \sumʹ_ir_j×F_ji, so that N_int= (1/2) \sum_i\sumʹ_j (r_i×F_ij+r_j×F_ji). We now assume that the forces are Newtonian, which means we assumeF_ji= -F_ji. Substituting we have: N_int= (1/2) \sum_i\sumʹ_j [r_i^\ding{217} ×F_ij^\ding{217} +r_j^\ding{217} ×(-F_ij^\ding{217})] Factoring and rearranging we have: N_int= (1/2) \sum_i\sumʹ_j (r_i-r_j)F_ij. Now,r_i^\ding{217} -r_j^\ding{217} is a vector between particleiand particle j. Since the F_ij \ding{217} are central forces (that is, they act along the line joining particlesiand j) they are parallel tor_i^\ding{217} -r_j^\ding{217}. Therefore, (r_i-r_j) ×F_ij= 0, and thus N_int = 0.

Question:

A silicious rock contains the mineral ZnS. To analyze for Zn, a sample of the rock is pulverized and treated with HCl to dissolve the ZnS (silicious matter is insoluable). Zinc is precipitated from solution by the addition of potassium ferrocyanide K_4 Fe (CN)_6. After filtering, the precipitate is dried and weighed. The reactions which occur are ZnS + 2HCl \rightarrow ZnCl_2 + H_2 S 2ZnCl_2 + K_4 Fe (CN)_6 \rightarrow Zn_2 Fe (CN)_6 + 4 KCl If a 2 gram sample of rock yields 0.969 gram of Zn_2Fe(CN)_6, what is the percentage of Zn in the sample? Atomic weight Zn = 65.4, molecular weight Zn_2 Fe (CN)_6 = 342.6.

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Solution:

In this problem, one wants to find the percent of Zn in the original rock sample. One is told that the rock contains some ZnS but not 100 %. One must find the weight of Zn in Zn_2 Fe(CN)_6 for which we are given the total amount present. Since no Zn is lost during the course of the reaction the final and initial weight of Zn is the same. Then, knowing the weight of Zn in the rock, we can calculate the % of Zn in the rock by dividing the Zn weight by the total weight of the rock. The first step is to find the % of Zn in Zn_2 Fe (CN)_6. Use the formula: % Zn = [(atomic weight Zn) ( number of Zn atoms) / (M.W. of Zn_2 Fe (CN)_6 )] × 100 = [{65.4 (2)} / {342.6}] × 100 = 38.2 % or, as a fraction = 0.382. To find the weight of any part of a compound, multiply the decimal fraction by the total weight of compound. In this case, weight of Zn = (0.382) (0.969 g) = 0.360 g. The weight of Zn is constant; thus % of Zn in rock is calculated by 100 × [(weight of Zn)/(total weight of rock)] = (.360)/(2) × 100 = 18 % Zn.

Question:

Discuss feeding mechanisms in differentmolluscs. Describe the differences in structural organization that have enabled the various classes to adapt to different lifestyles.

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/Users/wenhuchen/Documents/Crawler/Biology/F12-0293.htm

Solution:

Thechitonslead a sluggish, nearly sessile life. A horny-toothed organ, theradula, is contained within the pharynx, and is capable of being extended from the mouth.Chitonscrawl slowly on rocks in shallow water, rasping off fragments of algae for food. Gastropods occur in a wide variety of habitats. While most are marine forms, there are many fresh water and some land species. Gastropods have a well-developed head with simple eyes. Most feed on bits of plant or animal tissue that they grate or brush loose with a well-developedradula. Gastropods generally move about slowly. Bivalves lack aradula, and are filter feeders. Sea water is brought to the gills by the siphon. Cilia on the surface of the gills keep the water in motion and food particles are trapped by mucus secreted by the gills. The cilia push the food particles towards the mouth. Most digestion is intracellular. Certain bivalves, such as oysters, are permanently attached to the sea or river floor; others, such as clams, burrow through sand or mud. Scallops are capable of rapid swimming by clapping their two shells together. Enormous amounts of water are fil-tered by bivalves - an average oyster filters 3 liters of sea water per hour. Bivalves do not have a well-devel-oped head or nervous system. The cephalopods are active, predatorymolluscs. The tentacles of squids and octopuses enable them to capture and hold prey. Cephalopods have aradulaas well as two horny beaks in their mouths. These structures enable the cephalopods to kill the prey and tear it to bits. The nervous system is very well-developed. There is a large and complex brain, and image-forming eyes. The muscular cephalopod mantle is fitted with a funnel. By filling the mantle cavity with water and then ejecting it through the funnel, cephalopods attain rapid speeds in swimming.

Question:

The crew of a spacecraft, which is out in space with the rocket motors switched off, experience no weight and can therefore glide through the air inside the craft. The cabin of such a spaceship is a cube of side 15 ft. An astronaut working in one comer requires a tool which is in a cupboard in the diametrically oppos-ite corner of the cabin. What is the minimum distance which he has to glide and at what angle to the floor must he launch himself? If he decided instead to put on boots with magnetic soles which allow him to remain fixed to the metal of the cabin, and thus enable him to walk along the floor and, in the absence of gravitational effects, up the walls and across the ceiling, what is the minimum distance he needs to get to the cupboard?

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Solution:

Figure (a) shows the cabin. Axes have been set up with the x, y and z directions coinciding with the length, breadth, and height of the room. The astronaut must get from point A to point B. The vector A\ding{217} going from the origin 0 to point A is A\ding{217} = (15 ft, 0, 0) The vector from 0 to point B is: B\ding{217} = (0, 15 ft, 15 ft) The vector going from A to B is then: B\ding{217} - A\ding{217} = (0, 15 ft, 15 ft) - (15 ft, 0, 0) = (- 15 ft, 15 ft, 15 ft) Its length is: \vertB\ding{217} - A\ding{217}\vert = \surd[(- 15ft)^2 + (15 ft^2) + (15 ft)] = 15\surd3 ft = 26 ft This is the distance the astronaut must glide. The angle to the floor at which he launches him-self is \texttheta, where tan \texttheta = BC/AC. Point C has coordinates (0, 15 ft, 0). Thus BC has length 15 ft and AC has length \surd(15^2 + 15^2 ft) = 15 \surd2 ft = 21.2 ft. \therefore tan \texttheta = [15 ft/(15 \surd2 ft)] = (1/\surd2) = 0.707or \texttheta = 35.25\textdegree . Figure (b) shows the cabin minus the roof in an exploded diagram; points A, B, and C are again marked in. For convenience a new set of coordinate axes has been chosen. The astronaut walks the same distance along the walls from A to B by any particular route whether the walls are upright or flat as in the ex-ploded diagram. But in the diagram it is much easier to see that the minimum distance from A to B is the straight-line path between the two points. The vector B\ding{217} - A\ding{217} has components 15 ft in the x-direction and 30 ft in the y-direction. Distance \vert B\ding{217} - A\ding{217} \vert therefore equals\surd(15^2 + 30^2 ft) = 33 ft 6(1/2) in. Note that B' is the same point as B in the exploded diagram; AB' is thus an alternative route. There is a further alternative route AB'' which can be seen most clearly in figure (c) in which the wall marked I in figure

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Question:

In a double-slit experiment, D = 0.1 mm and L = 1 m. If yellow light is used, what will be the spacing between adjacent bright lines?

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Solution:

The wavelength of yellow light is approximately 6 × 10^-5 cm. Let the separation between the (n+1)th and m' th maxima be ∆x, as shown in the figure. Then, sin \texttheta_m = [(m)/(D)] \lambda sin \texttheta_m+1 = [(m+1)/D]\lambda The angles \texttheta_m and \texttheta_m+1 are related to the positions h_m ,and h_m+1 of these maxima on the screen; sin \texttheta_m \approx [(h_m+1)/L] sin \texttheta_m+1 \approx h_m/L , hence , sin \texttheta_m+1 - sin \texttheta_m = \lambda/D [(hm+1-h_m)/L] \approx \lambda/D ∆x/L \cong L/D = [{(6 ×10^-5 cm) × (100 cm)} / {10^-2 cm}] = 0.6 cm Thus , the spacing between lines is about 6 mm or (1/2) of an inch.

Question:

Draw the general structure of an amino acid and discuss why it is both an acid and a base.

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Solution:

The general structure of an amino acid can be drawn as: where R represents a side group which can be changed to create any amino acid. Note that the molecule contains two polar groups(COO^- and NH_3^+) and for this reason, amino acids are called dipolar ions or zwitterions. It is these two polar groups which give amino acids the ability to function as both acids and bases. Depending on the pH, amino acids can exist in one of three ionic forms: The pH at which most of the amino acids exist in the zwitterion (dipolar) form is called its isoelectricion point. When the pH is made more acidic, the dipolar ion is converted to the cation I. In this form, the amino acid will act as an acid with the carboxyl group capable of donating its proton. When the pH is made alkaline, the anion III is formed and predominates in solution. In this form, the amino acid will act as a base, with the amino group as the proton acceptor. At its isoelectric point, the amino acid is capable of acting as both an acid and a base, with both ionic forms existing in exact balance with each other. The relative acidity and basicity of an amino acid is strongly influenced by the structure and properties of its particular R group. In addition, the R group can itself act as an acid or base. Thus, each amino acid will have its own characteristic isoelectric point, and its own particular K_a and K_b for its amino and carboxylic acid groups. At a given pH, some amino acids will be acidic (protonated or cationic), while others will be basic (anionic), and others will be neutral (zwitterionic). At physiological pH, most amino acids exist in the dipolar form, with acidity and basicity being determined by each particular R group. In the dipolar form, however the amino acids can react with each other, via their polar groups, to form a covalent peptide linkage: This peptide bond is the means by which large protein molecules are constructed from amino acids.

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Question:

2NaHCO_3(s) \rightarrow Na_2CO_3(s) \rightarrow CO_2 (g) + H_2O(l) , ∆H = 30.92 Kcal/mole. ∆H\textdegree_Na2 CO3 _(s) = - 270.3. ∆H\textdegreeCO2 (g)= - 94.0 and ∆H\textdegreeH2 O_(l) = - 68.4 Kcal/mole, what is the standard enthalpy of formation for NaHCO3(s)?

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Solution:

The ∆H\textdegree for a reaction is an indication of the amount of heat released (or absorbed) when the reactants are converted to products. In general, ∆H\textdegree for a reaction will be the sum of the heats of formation of products minus the sum of the heats of formation of reactants, each of which is multiplied by its coefficient in the equation. Because you are given this sum and the ∆H\textdegree (heats of formation) of all products, the standard enthalpy of formation of NaHCO_3 can be found. Proceed as follows: ∆H\textdegree = ∆H\textdegreeNa2 CO3 (s)+ ∆H\textdegreeCO2 (g)+ ∆H\textdegreeH2 O (l)- 2∆H\textdegreeNaHCO3 (s) Substituting known values, 30.92 = (-270.3) + (- 94.0) + (- 44.8) - 2∆H\textdegree_NaHCO(3) _(s) Solving for ∆H\textdegree_NaHCO3 _(s) , you obtain ∆H\textdegree_NaHCO3 (s) = (- 270-3 - 94.0 - 44.8 - 30.9)/2 = - 440.0/2 = - 220.0 Kcal/mole, which is its standard enthalpy of formation.

Question:

A cycle and rider together weigh 186 lb. When moving at 10 mi/hr on a level road, the rider ceases to pedal and observes that he comes to a stop after traveling 200 yd. What was the average frictional force resisting his motion?

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Solution:

Once the acceleration of the rider as he is stop- ping is known, the frictional force can be found from F = ma, since it is the only force acting on the rider during the deceleration process. Knowing the initial velocity, distance traveled, and final velocity of zero, the acceleration can be found using the kinematics equation: v2_f = 2as + v^2. Sincev_f= 0,-v^2 = 2as. Substituting and converting units to ft-lb-sec, a= -v^2/2s = [-{(10 mi/hr) × (5280 ft/1 mi) × (1 hr/3600 sec)}^2 / {2 × (200 yd) × (3 ft/1 yd)} = [-(14.7 ft/sec)^2 / (1200 ft)] = -0.18 ft/sec^2. m = W/g = 186 lb/(32 ft/sec^2)= 5.8 slugs F = ma = (5.8 slugs) (-0.18ft /sec^2 )= -1.0 lb. The frictional force is negative since it opposes the direction of motion.

Question:

A 100-gram marble strikes a 25-gram marble lying on a smooth horizontal surface squarely. In the impact, the speed of the larger marble is reduced from 100 cm/sec to 60 cm/sec. What is the speed of the smaller marble immediately after impact?

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Solution:

The law of conservation of momentum is applic-able here, as it is in all collision problems. Therefore, Momentum after impact = Momentum before impact. Momentum before impact= MB1× VB1 = 100 gm × 100 cm/sec Momentum after impact= MA1× VA1+ MA2× V_A2 = 100 gm × 60 cm/sec + 25 gm × V_A2 cm/ sec Then, 10,000 gm-cm/sec = 6000 gm-cm/sec + 25 g × v_A2 whence v_A2 = 160 cm/sec.

Question:

The energy of an electron in a certain atom is approxi-mately 1 × 10^-18 J. How long would it take to measure the energy to a precision of 1 per cent?

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Solution:

The energy uncertainty is (.01) (1 × 10^-18 J) = 1 × 10^-20 J. This means that the measurement of the energy of the system will yield a value 1 × 10^-18 \pm 1 × 10^-20 J. One form of the Uncertainty Principle re-lates the lack of certainty as to what the energy of a system is (\DeltaE) , to the length of time needed to measure the energy with this degree of accuracy (\Deltat). \DeltaE\Deltat= (h/4\pi) where h is Planks constant. We are given \DeltaE = 1 × 10^-20 J (i.e. the measured value of energy in the system is E \pm \DeltaE). Then \Deltat\cong [h/{(\DeltaE) (4\pi)}] = [(6.6 × 10^-34Js) / {(1 × 10^-20J) (4\pi)}] = 5.3 × 10^-15 s

Question:

How do we know that ATP is generated at three sites in the electron transport chain? What is the advantage of having a sequential transfer of electrons rather than one single transfer?

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Solution:

The NADH and FADH2 formed in glycolysis, fatty acid oxidation and the citric acid (TCA) cycle are energy-rich molecules. When each of these molecules transfers its pair of electrons to molecular oxygen, a large amount of energy is released. This released energy can be used to generate ATP in oxidative phosphorylation. As electrons flow through the electron transport chain from NADH to O2, ATP is formed at three sites along the chain. How do we know that three separate sites exist instead of one site where 3 ATP are produced? The first evidence is the ATP yield from the oxidation of different substrates. The oxidation of NADH yields 3 ATP but the oxidation of succinate, which forms FADH2, only yields 2 ATP. The electrons from FADH2 enter the chain at coenzyme Q, which is past the first phosphory-lation site. Only 1 ATP is formed when ascorbate (an artificial substrate also called Vitamin C) is oxidized, since its electrons enter at cytochrome c, which is past the first two phosphorylation sites. The P:O ratio (the ratio of organically bound phosphorous produced per mole of oxygen, which is often used as an index of oxida-tive phosphorylation) is 3:2:1 for NADH : succinate : ascorbate, respectively. Hence, by knowing the P:O ratios and that each substrate enters the process at different steps,we see that one mole of ATP is produced at each site. The steps involved can be determined by the free energy change, ∆G, released during electron transfer. The ∆G for electron transfer from NADH to FMN is -12 kcal/mole, from cytochrome b to c, -10 kcal/mole, and from cytochrome a to O2, -24 kcal/mole. These reactions are sufficiently exergonic to drive the synthesis of ATP, whereas the other electron transfer reactions do not release sufficient energy to allow ATP synthesis to take place. There also exist inhibitors which specifically block certain steps in the chain. Rotenone specifically inhibits electron transfer from NADH to coenzyme Q, pre-venting ATP synthesis at the first site. Antimycin A inhibits electron flow at the second site, while cyanide and carbon monoxide block electron transfer at the third site. Each inhibitor has been observed to block all successive steps of the electron transport system. However it is possible to bypass the action of the inhi-bitor. For instance, the action of rotenone can be avoided through the addition of either succinate or ascorbate. Similarly, the action of antimycin A on the second site can be bypassed through the addition of ascorbate. By inhibiting the various sites and measuring the resulting P:O ratios it can be deduced that each site in the electron transport chain is responsible for the production of one molecule of ATP (see figure). It should be noted that the addition of antimycin A will result in the blockage of site 1 as well as site 2. This is due to a backflow of electrons which results in the reduction of site 1, thus inactivating it. Likewise, the addition of cyanide results in the reduction of sites 1 and 2 as well as 3, thus rendering them all incapable of ATP formation. Hence cyanide has the well deserved reputation as one of the world's strongest toxins. The advantage of the sequential transfer over one single transfer is that the sequential reactions divide up the free energy change of the oxidation of NADH, which is highly exergonic. The free energy of oxidation of NADH + (1/2) O2 + H+ \rightleftharpoons H2O + NAD+ is about - 53 kcal/mole. If all this energy were released at once, much of it would be wasted as heat since only 7 kcal are needed to phosphorylate ADP to form ATP. If all the energy from the oxidation of NADH is released at once to form 1 ATP, the efficiency of this system is 7/53 × 100% = 13%. However, if the same energy is released in a stepwise process to form 3 ATP, the efficiency is 21/53 × 100% = 40%. Thus, stepwise reactions increase the efficiency of oxidative phosphorylation from 13% to 40%. In addition, the large amount of heat liberated would very easily destroy enzyme activity and thus is harmful to the cell if released in one burst. The three reactions involved are "coupled" with the phosphorylation of ADP so that the liberated energy is used immediately to drive this otherwise non-spontane- ous reaction. The exact mechanism of oxidative phosphorylation, that is, how the energy liberated is transferred so as to form the high energy phosphate bond of ATP, is un-known. The exact mechanism of oxidative phosphorylation, that is, how the energy liberated is transferred as so to form the high energy phosphate bond of ATP, is unknown. Mitchell proposed a chemiosmotic hypothesis in 1961. The three sites of "ATP synthesis" along the respiratory chain are actually sites where, at each, a pair of protons (2 H+) are pumped from the matrix to the intermembrane space. This creates a proton gradient. The protons flow back onto the matrix through special "lollypop" enzymes. As a pair of H+ flow back in down the proton gradient, one ATP is synthe-sized. Three pairs of H+ are extruded into the intermem-brane space per NADH, so three pairs of H+ can re-enter and thus produce 3 ATPs. The mechanisms is similar for FADH2, except since FADH2 donates their electrons at the second site, only two pairs reenter and two ATPs are synthesized.

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Question:

If a membrane contains 40% by weight of lipids and 60% protein, calculate the molar ratio of lipids to proteins. Assume that the lipid molecules have an average molecular weight of 800 and the proteins an average molecular weight of 50,000.

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Solution:

Assume that one has 100 g of membrane. From the information given, 40 g of this will be composed of lipids and 60 g will be protein. Determine the number of moles present in 40 g of lipids and 60 g of protein. The molar ratio will be equal to the number of moles of lipids divided by the number of moles of protein number of moles = (number of grams) / (molecular weight) no. of moles of lipids = [40g] / [800(g / mole)] = 0.05 moles no. of moles of protein = [60 g] / [50,000 (g / mole)] = 1.2 × 10^-3 moles molar ratio = (lipids/protein)= (0.05 moles) / (1.2 × 10^-3 moles) = 41.7.

Question:

The root mean square (rms) speed of hydrogen (H_2) at a fixed temperature, T, is 1600 m/sec. What is thermsspeed of oxygen (O_2) at the same temperature?

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Solution:

To solve this problem we make use of the following equation for the rmsspeed,v_rms, v_rms= \surd[3 RT / M] , where R is the gas constant, T the absolute temperature, and M the molecular weight. For oxygen and hydrogen we have v_rms_[(O)2] = \surd[3 RT / M _(O)2]andv_rms_[(H)2] = \surd[3 RT / M_(H)2] Dividing the first of these equations by the second gives v_rms_[(O)2] /v_rms_[(H)2] = \surd[3 RT / M_(O)2] /\surd[3 RT / M_(H)2] = \surd{[3 RT / M_(O)2] / [3 RT / M_(H)2]} or, since R is a constant and T is the same for both gases. v_rms_[(O)2] /v_rms_[(H)2] = \surd{M_H(2) / M_(O)2} Solving forv_rms_[(O)2], v_rms_[(O)2] = \surd{M_H(2) / M_(O)2} ×v_rms_[(H)2] = \surd{(2.0 g/mole)/(32.0 g/mole)} × 1600 m/sec = 400 m/sec.

Question:

CICH_2COOH + H_2O \rightleftarrows H_3O^+ + ClCH_2COO^- exists at 25\textdegreeC. K_a = 1.35 × 10^-3. (a) Determine the H_3O^+ concentration for a 0.1 M solution of monochloroaceticacid (CICH_2COOH) in water (H_2O). (b) Can one make the assumption that the dissociated acid is negligible with respect to theundissociatedacid? (c) Calculate to what degree this solution is more acidic than 1.0 M acetic acid (K_a = 1.8 × 10^-5 at 25\textdegreeC).

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Solution:

The concentration of H_3O^+ can be calculated through the equilibrium constant expression. This is a measure of the ratio between the concentrations of products to reactants, each raised to the power of their respective coefficients in the chemical equation. (a) For this reaction, the equilibrium constant, K_a = {[CICH_2COO^-][H_3O^+]}/[CICH_2COOH] .Because K_a = 1.35 × 10^-3 , 1.35 × 10^-3 = {[CICH_2COO^-][H_3O^+]}/[CICH_2COOH] . Letting x = [H_3O^+] at equilibrium, then CICH_2COO^- = x since the reaction shows that the products are formed inequimolaramounts. The original concentration of the CICH_2COOH = 0.1 M. If x moles of H_3O^+ and x moles of CICH_2COO^- form, then, at equilibrium, there are 0.1 - x moles of CICH_2COO^- left. Substituting these variables into the equilibrium constant expression: [(x \textbullet x)/(0.1 - x)] = 1.35 × 10^-3 . Solving for x,x = 0.0110 M. (b) If the dissociated acid is negligible with respect to the amount leftundissociated, then 0.1 - x (from pre-vious part) is equal to 0.1. This means that [(x \textbullet x)/(0.1)] = 1.35 × 10^-3 . Solving for x;x = 0.0116 M = [H_3O^+]. This means that when you make this assumption, [H_3O^+] = 0.0116 M instead of 0.0110 M, the percent error is over 5 %. The error is too high and, therefore, this assumption cannot be made. To answer (c), compare the [H_3O^+] in acetic acid solution of 1.0 M with that of the [H_3O^+] calculated in part (a) . [H_3O^+] for acetic acid is calculated by using the same procedure as in part (a), except that K_a = 1.8 × 10^-5 and the dissociation equation is HOAC + H_2O \rightleftarrows OAC^- + H_3O^+ . After the calculations, [H_3O^+] = 0,00133 M. For acetic acid [H_3O^+] = 0.00133 M and for CICH_2COOH, [H_3O^+] = 0.0110 M. Thus, the ClCH_2COOH has 8.27 times the H_3O^+ concentration of CH_2COOH (acetic acid), and is therefore more acidic. This answer is found by dividing [H_3O^+] dissociated from CICH_2COOH by [H_3O^+] dissociated from CH_3COOH. [(0.0110 M)/(0.00133 M)] = 8.27 .

Question:

In a drag race, a dragster reaches the quarter-mile (402 m) marker with a speed of 80 m/s. What is his acceleration and how long did the run take?

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Solution:

The initial velocity, v_0 = 0, the final velocity, v = 80 m/s, and the distance traveled, d = 402 m, are given. The acceleration a and the time interval t are the unknown observables. From the kinematics equations, a= (v^2 - v^2_0)/2d = [(80 m/s)^2 - (0)]/[(2)(402 m)] = 7.96 m/s^2 t=(v - v_0)/a = [(80 m/s) - (0)]/[ 7.96 m/s^2] = 10.1 s

Question:

Explain how you would prepare aspirin from phenol and acetic acid. You may use any other inorganic reagents.

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Solution:

The best or the most direct method of preparing aspirin from phenol is by use of the Kolbe reaction, which requires a knowledge of the chemistry of phenols. To plan this synthesis, first write your starting material, phenol, and end product, aspirin, and then think of reactions that direct you in a step by step manner to the desired end. These structures are shown in figure A. As mentioned, you want to employ the Kolbe reaction; it produces the carboxylate salt shown in figure B from phenol, which begins to look like aspirin. The procedure is shown in figure C. The hydrogen of the OH group in phenol is acidic, due to the fact that the ring delocalizes the unpaired electrons of oxygen in resonance structures. Thus, when base is present, you have a neutralization reaction to produce sodium phenoxide, a salt, and water. Now, that you have this material, perform the a salt, and water. Now, that you have this material, perform the Kolbe reaction. You have the reaction shown in figure D. If you acidify (by adding a mineral acid like HCI) , you obtain the reaction in figure E. Salicylic acid is produced. You are told that acetic acid is available. The only differ-ence between salicylic acid and aspirin is an ester linkage. Thus, if you perform an esterification reaction with acetic acid on the alcohol portion of salicylic acid, you obtain aspirin. By esterification you mean the reaction of a carboxylic acid (R-COOH) and an alcohol (R-OH) to produce an ester and water. Thus, you perform the reaction in figure F. The acid (H+) is a catalyst, which speeds up the process.

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Question:

How many pounds of air (which is 23.19% O_2 and 75.46 % N_2 by weight) would be needed to burn a pound of gasoline by a reaction whereby C_8 H1 8reacts with O_2 to form CO_2 and H_2 O?

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Solution:

The equation for the reaction is C_8 H1 8+ 12 (1/2) O_2 \rightarrow 8 CO_2 + 9 H_2 O From the equation, one knows that 12.5 moles of O_2 are needed to burn 1 mole of gasoline. To solve for the number of pounds of air necessary to burn 1 pound of gasoline: (1)find the number of moles in 1 pound of C_8 H1 8 (2)determine the number of moles of O_2 needed (3)solve for the number of moles of O_2 in 1 pound of the air (4)calculate the number of pounds of air needed. Solving: (1)The molecular weight of C_8 H_1 8 is 114.23 g/mole. There are 453.50 g in 1 lb, thus one can convert from grams to pounds by multiplying the number of grams by 1 lb/453.50 g. MW of C_8 H1 8in lbs = 114.23 g/mole × [(1 lb) / (453.50 g)] = 2.52 × 10^-1 lbs/mole. One can find the number of moles in 1 lb by dividing 1 lb by the molecular weight of C_8 H1 8in lbs. no. of moles = [(1 lb) / (2.52 × 10^-1 lbs/mole)] = 3.97 moles (2)One needs 12.5 times as much O_2 as C_8 H1 8 no. of moles of O_2 needed= 12.5 × no. of moles C_8 H1 8 = 12.5 × 3.97 moles = 49.63 moles (3)In 1 lb of the air, there is .2319 lb O_2 and .7546 lb N_2 . Thus, to find the number of moles of O_2 in one pound of the air, one must divide .2319 lb by the molecular weight of O_2 in pounds (MW of O_2 = 32 g/mole). MW in lbs of O_2 = 32 g/mole × [(1 lb) / (453.5 g)] = 7.06 × 10^-2 lb/mole no. of moles = [(.2319 lb) / (7.06 × 10^-2 lb/mole)] = 3.29 moles. (4)There are 3.29 moles of O_2 in one lb and 49.63 moles of O_2 are needed to burn 1 lb of gas. no. of lbs of air needed = [(49.3moles) / (3.29 moles/lb)] = 15.10 lbs.

Question:

Discuss how the quantitative measurements of the dioxy-ribonucleic acid content of cells is evidence that DNA is the genetic material.

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Solution:

In the 1940's, A.E.Mirskyand HansRisworking at the Rockefeller Institute, and AndreBoivinand RogerVendrely, working at the University at Strasbourg, independently showed that the amount of DNA per nucleus is constant in all of the body cells of a given organism. By making cell counts and chemical analyses,MirskyandVendrelyshowed that there is about 6 × 10^-9 milligrams of DNA per nucleus in somatic cells, but only 3 ×10^-9 milli-grams of DNA per nucleus in egg cells or sperm cells. In tissues that arepolyploid, (having more than two sets of chromosomes per nucleus), the amount of DNA was found to be a corresponding multiple of the usual amount. From the amount of DNA per cell, one can estimate the number of nucleotide pairs per cell, and thus the amount of gene-tic information in each kind of cell. Only the amount of DNA and the amount of certain basic positively charged proteins calledhistonesare relatively constant from one cell to the next. The amounts of other types of proteins and RNA vary consider- ably from cell to cell. Thus the fact that the amount of DNA, like the number of genes, is constant in all the cells of the body, and the fact that the amount of DNA in germ cells is only half the amount in somatic cells, is strong evidence that DNA is an essential part of the genetic material.

Question:

What advantages do microorganisms have over largerforms of life as subjects for genetic investigation?

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Solution:

If we consider the methods by which genetic studies are carried out , the advantages of using micro-organisms such as bacteria become obvious . Most genetic studies involve breeding and reproduction, and subsequent frequencies and ratios in the offspring often involve several generations . Here, microorganisms are prime experimental subjects, because of their extremely rapid rate of reproduction. Generations can be obtained in a matter of hours, and so genetic experiments can be done in relatively short periods of time. Large populations can be quickly grown, enabling rare events, such as mutations, to occur in observable quantities. Also, determination of genetic frequencies becomes extremely accurate with large numbers. Microorganisms are easily obtained and stored. They live on inexpensive nutrients. This makes them perfectly suited to laboratory studies . In laboratories, the environment can be carefully controlled, and by selecti-vely varying the culture media, one can easily study specific factors of metabolism. Single mutant cells can be isolated from among millions by simply varying the nutrient media. Most of what we know about genes is learned from observations of their phenotypic expressions, and in this respect microorganisms display their biggest ad-vantage. Many microorganism, particularly fungi, have a conspicuous and relatively long - lived haploid generation. In the haploid stage (n), there is only one set of genetic information; that set is expressed phenotypically, whether it be dominant or recessive. In the diploid stage, dominant alleles may mask recessive ones, and the genotype cannot be accurately determined solely from the phenotype. Lengthy procedures such as back - crosses are required to deduce the genotype , but can be eliminated if the test subjects are haploid individuals. Much of recent genetics has involved both micros-copic chromosomal studies and gene effects at the cellular level. Here, the simplicity of the genetic material and metabolism of microorganisms makes them much better candidates for such studies than the complex multicellular organisms.

Question:

Consider the following differential equation with the initial condition y(0) = 1: (dy/dx) = y2 - x2 Develop a FORTRAN program to get a solution for y = y(x) in the in-terval 0 \leq x \leq .5 applying the increment method. Let dx = ∆x = 0.05, and output a table containing the following quantities: x, y, y2, x2, y2 - x2, dy .

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Solution:

Let us first develop the problem from a mathematical view-point. Before we begin solving the equation we must know that a solu-tion exists. Furthermore, this solution should be unique. The exist-ence and uniqueness conditions for the solution of a given differential equation are contained in the following theorems: 1) Existence: Let y' = \varphi(x, y). 0 \leq x \leq 1, - \infty < y < \infty, y(0) = c If \varphi is continuous and bounded, then y = y(x) is a solution of (dy/dx) = \varphi(x, y). 2) Uniqueness: If there exists a constant A (Lipschitz number) such that \vert\varphi(x,y2) - \varphi(x, y1)\vert \leq A \verty2 - y1\vert (where (x, y1), (x,y2) are in the domain), then y = y(x) is a unique solution to y' = \varphi(x, y). The given differential equation is \varphi(x,y) = (dy/dx) = y2 - x20 \leq x \leq 0.5, y(0) = 1 \varphi(x,y) is a difference of continuous functions; hence it is contin-uous. Clearly\vert y2 - x2 \vert is bounded (i.e., less than some constant K) in [0,0.5]. Thus a solution exists. Let A = .3. Then\vert\varphi(x, y2) - \varphi(x, y1)\vert\leq .3(y2 - y1) for all (x, y1), (x, y2) in the domain. Thus the solution is unique. We can rewrite equation (1) in the differential form dy = (y2 - x2)dx(2) The increment method entails the use of differentials to find approxi-mate values of y. Thus, let ∆x = dx, ∆y = dy. Since the initial condition is y(0) = 1, we can substitute into equation (2) and obtain ∆y = (12 - 02)(0.05) ∆y = 0.05 thus, at x = 0.05, y = y0 + ∆y becomes 1 + 0.05 = 1.05. This means that now y(1.00) becomes 1.05. The next value of ∆y becomes, there-fore ∆y = ((1.05)2 - (0.05)2)(0.05) ∆y = 0.055 As you can see, the method readily lends itself to computer implemen-tation. We can use a FORTRAN main program to print the values in tab-ular form. A subroutine, INCR, can do each calculation and pass the values back to the main program for output. Y = 1.000 DX = 0.05 CWRITE TABLE HEADINGS WRITE (5,100) 100FORMAT (2X, 'X', 5X, 'Y', 6X, 'Y\textasteriskcentered\textasteriskcentered2', 3X, 1'X\textasteriskcentered\textasteriskcentered2', 3X, 'Y\textasteriskcentered\textasteriskcentered2-X\textasteriskcentered\textasteriskcentered2', 3X, 'DY') CDO FOR X FROM 1 TO 10 BY 0.05 DO 10 I = 1,10 X = (FLOAT (I) - 1.0)\textasteriskcentered0.05 CALL INCR (DX, DY, X, Y) XX - X\textasteriskcentered\textasteriskcentered2 YY = Y\textasteriskcentered\textasteriskcentered2 DYDX = YY - XX WRITE (5,101) X, Y, YY, XX, DYDX, DY 101FORMAT (1X, F4.2,3(2X,F5.3), 4X, F5.3, 5X, F5.3) 10CONTINUE CEND DO - FOR STOP END CSUBROUTINE INCR TO PERFORM CALCULATIONS SUBROUTINE INCR (DX, DY, X, Y) Y = Y + DY DY = ((Y\textasteriskcentered\textasteriskcentered2) - (X\textasteriskcentered\textasteriskcentered2))\textasteriskcenteredDX RETURN END

Question:

A child of mass m sits in a swing of negligible mass suspen-- ded by a rope of length l . Assume that the dimensions of the child are negligible compared with l. His father pulls the child back until the rope makes an angle of one radian with the vertical, then pushes with a force F = mg along the arc of the circle, releasing at the vertical. (a) How high up will the swing go? (b) How long did the father push? Compare this with the time needed for the swing to reach the vertical position with no push.

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Solution:

The father does work in pushing the swing. The work done increases the total energy of the swing and child as the swing moves from its initial position, \texttheta_0 = 1 radian, to the vertical position where the father stops pushing. From that point on, total energy of the swing and child re-mains constant. Work done by the father is found from W = \int F^\ding{217} \textbullet dr^\ding{217} = \int F dr cos \texttheta(1) where r is the distance along the path. W =^l(\texttheta)0\int_0 mg cos (0\textdegree) dr = mgl\texttheta_0 .(2) (a) Now we can use conservation of energy to find how high the swing goes. (Let the potential of the swing and child be O at the vertical position where the father lets go.) (PE at \texttheta_0) + W = PE at \texttheta = E_total = constant(3) mg (l \rule{1em}{1pt} l cos \texttheta_0) + mgl\texttheta_0 = mg (l \rule{1em}{1pt} l cos \texttheta) mgl (1 \rule{1em}{1pt} cos \texttheta_0 + \texttheta_0) = mgl (1 \rule{1em}{1pt} cos \texttheta) 1 \rule{1em}{1pt} 0.54 + 1 = 1 \rule{1em}{1pt} cos \texttheta cos \texttheta = 0.46 \texttheta = 63\textdegree above the horizon. (b)We need to find the tangential acceleration of the velocity of the swing at the bottom of its arc, which can be found by a second application of the conservation of energy principle as follows: PEat \texttheta+ W = KE_at _bottom = E_total(4) mgl (1 \rule{1em}{1pt} cos \texttheta) + mgl\texttheta_0 = 1/2 mv^2 mg l (1.46) + mgl = 1/2 mv^2 v^2 =2 gl (2.46) = 4.92 gl(5) Application of Newton's Law allows us to find the tangential acceleration, a_T . \sumF_T = ma_T mg + mg sin \texttheta_0 = ma_T(6) But mg sin \texttheta_0 the component of gravity in the tangen-tial direction, does not stay constant. It gets smaller as the angle decreases. Thus, the tangential acceleration is not constant. We can obtain an approximate solution by integrating the gravity force between O and \texttheta_0, and using that equivalent force. ^(\texttheta)0\int_O mg sin \texttheta d \texttheta = \rule{1em}{1pt}mg cos \texttheta │(\texttheta)0_O = 1 rad = [\rule{1em}{1pt}0.54 + 1] mg = 0.46 mg(7) Substituting this value for mg sin \texttheta_0 in Equation (6) yields mg + 0.46 mg = ma_T a_T = 1.46 g. Now to find how long the father pushed write a_T = dv / dt = 1.46 g ^v\int_O dV = 1.46 g^t\int_O dt t = (v/1.46g). Using the value of v from Equation (5) yields t = \surd[(4.92 gl) / (1.46 g)] t = 1.52 \surd(l/g) . The force decelerating the child will be the gravity force. This force will be of the magnitude given in equa-tion (7). Writing F = ma, 0.46 mg = ma a = 0.46 g. The time will again be given by t = v/a = (v/0.46g).(8) If there is no push, conservation of energy yields a velocity of \surd[2 gl (1 \rule{1em}{1pt} cos \texttheta_0)]. Substituting this in equa-tion (8) yields t = \surd[{2 gl (0.46)} / (0.46 g)] t = 2.09 \surd(l / g).

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Question:

Are there comparable changes, in the male, to female menopause?

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Solution:

The changes that occur in the male due to aging are much less drastic than those which occur in the female. The testosterone and pituitary secretions which are initiated when the male reaches puberty continue throughout his adult life. Should the testosterone level decrease, this would result, through the switching-off of the negative-feedback inhibition on the pituitary, in a rise in the pituitary secretions of FSH and LH. Increased levels of FSH and LH stimulate testosterone production. Despite this, there is a steady decrease in testosterone secretion in later life, but this is probably caused by the slow deterioration of testicular function. Despite this decrease, testosterone secretion remains high enough in most men to maintain sexual vigor throughout life, and fertility has been documented in men in their seventies and eighties. Thus, there is usually no complete cessation of reproductive function in males analogous to female menopause.

Question:

A 2-ft-long string which can just support a weight of 16 lb is fixed at one end to a peg on a smooth horizontal surface. The other end is fixed to a mass of 1/2 slug. With what maximum constant speed can the mass rotate about the peg? (See figure).

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Solution:

If the tension in the string exceeds 16 lb, the string will break. Thus the maximum centripetal force that can be exerted on the mass is 16 lb. But if the mass is circling the peg with a velocity v, the centripetal force necessary to keep it in the circle is mv^2/R, where m is 1/2 slug and R is the length of the string, 2 ft. Thus 16 lb = T_max = (mv^2_max) / R = [(1/2) slug × v^2_max]/(2ft) Thereforev^2_max = 64 ft \bullet lb \bullet slug^-1 = 64 ft^2 \bullet s^-2 orv_max= 8 ft \bullet s^-1 .

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Question:

Design a BASIC program to sort 50 random numbers by a method known as radix sorting. This method sorts numbers by examining individual digits, starting with the units column. For this problem, generate random numbers with no more than 4 digits.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G07-0146.htm

Solution:

To understand radix sorting, consider the follow-ing example: 3192, 4365, 6702, 13, 357, 3999, 4132, 1872, 6345, 582. At the first pass, the computer sorts the numbers according to the right- most digits. If two numbers have the same units digit, their order will be unchanged in the new list. 3192, 6702, 4132, 1872, 582, 13, 4365, 6345, 357, 3999. At the next pass - second right-most digits 6702, 0013, 4132, 6345, 0357, 4365, 1872, 0582, 3192, 3999. Similarly, on the third pass the computer compares third digits from the right in every number: 0013, 0357, 0582, 4132, 3192, 6345, 4365, 6702, 1872, 3999. At the fourth pass, the numbers are finally sorted into ascending order. 0013, 0357, 0582, 1872, 3192, 3999, 4132, 4365, 6345, 6702. For solving this problem, we need two arrays: one to hold the unsorted random numbers, and the other to hold the sorted list. Once the random numbers are generated, you check each column in succession, from units to thousands. To print the array after each pass, we will need four sepa-rate executions of the PRINT statement. The values can be printed out using the MAT PRINT command. Notice the par-ameters of the DIM statement in line 240. This means that there will be 10 rows and 5 columns. The reason for the parameters being (9,4) is that since we started our loop at zero, a place must be saved for thezerothvalue. 15REM RADIX SORT: DIGIT-BY-DIGIT COMPARISON 2\ODIM R(49), S(49) 21LET V = 0 25REM FILL ARRAY R WITH 4-DIGIT RANDOM NOS 3\OFOR J = 0 TO 49 4\OLET R (J) = INT (10000\textasteriskcenteredRND (X)) 5\ONEXT J 6\OLET L = 0 7\OLET M = 0 8\OFOR J = 0 T\O 49 85REM IF 1 IS THE FIRST ELEMENT, LEAVE IT ALONE 9\OIF R (J) = 1 THEN 2\O\O 1\O\OLET H = R (J) 11\OFOR K = 0 T\O V 12\OLET Z = INT (H / 10) 125REM G IS THE RIGHTMOST DIGIT 13\OLET G = H - 10\textasteriskcenteredZ 14\OLET H = Z 15\ONEXT K 16\OIF GOM THEN 2\O\O 165REM STORE R (J) IN SORTED ARRAY S (L) 17\OLET S (L) = R (J) 18\OLET R (J) = 1 19\OLET L = L + 1 2\O\ONEXT J 21\OLET M = M + 1 22\OIF M < 10 THEN 80 23\OPRINT 24\OMAT (S) = DIM (9, 4) 25\OPRINT 255REM PRINT SORTED ARRAY S AFTER EACH PASS 26\OMAT PRINT S 27\OMAT S = DIM (49) 28\OMAT R = S 29\OLET V = V + 1 295REM WHEN 4 DIGITS HAVE BEEN CHECKED, END 3\O\OIF V < 4 THEN 6\O 31\OEND

Question:

Silver bromide, AgBr, used in photography, may be prepared from AgNO_3 and NaBr. Calculate the weight of each required for producing 93.3 lb of AgBr. (1 lb = 454 g.)

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/Users/wenhuchen/Documents/Crawler/Chemistry/E05-0177.htm

Solution:

The reaction for the production of AgBr is written AgNO_3 + NaBr \rightarrow AgBr + NaNO_3 This means that one mole of each AgNO_3 and NaBr are needed to form one mole of AgBr. In this problem, to determine the number of pounds of AgNO_3 and NaBr used to form 93.9 lbs of AgBr, one must first determine the number of moles of AgBr in 93.9 lbs. There will be one mole of each NaBr and AgNO_3 for each mole of AgBr formed. Once the number of required moles of NaBr and AgNO_3 are found, the weights of these compounds can be determined by multiplying the number of moles by the molecular weight. To solve this problem: (1)solve for the number of moles of AgBr in 93.9 lbs (2)determine the weights of NaBr and AgNO_3 used. Solving: (1)Molecular weights are given in grams, therefore, the grams should be converted to pounds for use in this problem. MW of AgBr = 188 g/mole. There are 454 grams in one pound, thus grams can be converted to pounds by using the conversion factor 1 lb/454 grams. MW of AgBr in lbs = 188 × 1 lb/454 g = .41 lbs. The number of moles in 93.9 lbs can be found by dividing the 93.9 lbs by the molecular weight, in pounds 93.9 lbs no. of moles of AgBr = [(93.9 lbs) / (.41 lbs/mole)] = 229 moles Therefore, 229 moles of each NaBr and AgNO_3 are needed to produce 93.9 lbs of AgBr. (2)The weight of NaBr and AgNO_3 used is equal to the number of moles times the molecular weight. In this problem, one wishes to find the weight in pounds, not grams, thus the molecular weights must be converted to pounds before the conversion factor 1 lb/454 g can be used. MW of NaBr = 103, MW of AgNO_3 = 170. MW of NaBr in lbs = 103 g/mole × 1 lb/454 g = .23 lbs/mole MW of AgNO_3 in lbs = 170 g/mole × 1 lb/454 g = .37 lbs/mole. Since it has already been calculated that 229 moles of each of these compounds are needed, one can calculate the weight needed of each by multiplying the number of moles by the molecular weight. weight of NaBr = .23 lbs/mole × 229 moles = 52.7 lbs. weight of AgNO_3 = .37 lbs/mole × 229 moles = 84.7 lbs.

Question:

How might you argue that the transmission of impulses does occur in plants?

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/Users/wenhuchen/Documents/Crawler/Biology/F09-0226.htm

Solution:

The reaction of the sensitive plant,Mimosapudica, to touch serves as an excellent argument for the fact that transmission of impulses does indeed occur in plants. The leaf ofMimosais subdivided into many leaflets. Normally the leaves are horizontal. However if one of the leaflets is lightly touched, that leaflet and all other leaflets of the stimulated leaf droop within two or three seconds leading to folding of the entire leaf. In addition, other neighboring leaves will soon fold and droop. The observation that touching one leaf leads other leaves to fold as well demonstrates that the excitation initiated by touching is transmitted from the site of stimulation to neighboring locations. It is now believed that the folding of the leaves results from a decrease in theturgorpressure of the cells at the bases of the leaves. The excitation is transmitted along the sieve tubes of the leaves and stem in the form of an electrical impulse. This impulse involves a temporary physiological change in permeability of the plasma membrane, causing a change in intracellular and extracellular ionic distributions and consequently a change inturgorpressure. The nature of the impulse is fundamentally the same as that of a nerve impulse, although the latter travels about 2400 times faster.

Question:

Write a PL / I Program to initialize stacks or queues and to make insertions or deletions in stacks or queues.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G07-0139.htm

Solution:

In PL/I, since we don't have data structures for stacks and queues we shall use an array of one dimension, along with pointers to simulate the stack and queue opera-tions. DCL STACK (20) type / \textasteriskcentered type in the type of data items to be stored in stack; this statement creates a one dimen-sional array called STACK of 20 elements \textasteriskcentered /. DCL STACK-PTR FIXED DEC (2); / \textasteriskcenteredPointer is required to know whether stack is empty or full \textasteriskcentered /. Insert an item. If STACK_PTR = 20 THEN (go to error routine) (Stack is full) ELSE STACK_PTR = STACK_PTR + 1 STACK(STACK_PTR) =X / \textasteriskcentered Item \textasteriskcentered / We have thus stored a data item X in the location m the stack denoted by STACK (STACK_PTR) , i.e., if for example, the value of STACK_PTR = 6, we insert X in the position STACK(6) in the stack. Deletion of an item. If STACK_PTR = 0 THEN (go to error routine) (Stack is empty) ELSEX = STACK (STACK_PTR) (Delete) STACK_PTR = STACK_PTR -1 After reading out the value of the position STACK (STACK_PTR) into a memory location X, we decrement the STACK_PTR by 1 to keep track of the number of elements still left in the stack. End; Queue:-- Queue is a data structure of type FIFO (First in First out). That is, data is inserted at one end, say the tail, & removed from the other end (head). For this purpose we shall need two pointers and a one dimensional array. / \textasteriskcentered QUEUE: ONE DIMENSIONAL ARRAY \textasteriskcentered / / \textasteriskcentered QUEUE SIZE IS 2\varphi\textasteriskcentered / \varphi / \textasteriskcentered PNTR1: POINTS TO THE TOP ELEMENTS \textasteriskcentered / / \textasteriskcentered PNTR2: POINTS TO THE NEXT AVAILABLE SPACE \textasteriskcentered / / \textasteriskcentered INITIALIZE \textasteriskcentered / DCL QUEUE (20) CHAR(5); DCL (PNTR1, PNTR2)BIN FIXED (15); PNTR1 = 1; PNTR2 = 1; FLAG =\varphi; / \textasteriskcentered FLAG = \Phi INDICATES EMPTY \textasteriskcentered / \varphi / \textasteriskcentered INSERT AN ELEMENT \textasteriskcentered / IF PNTR1 = PNTR2 & FLAG = 1 THEN [go to error routine] / \textasteriskcentered QUEUE FULL \textasteriskcentered / ELSE Do; FLAG = 1, QUEUE (PNTR2) = new element; IF PNTR2 = 2\varphiTHEN PNTR2 = 1; \varphi / \textasteriskcentered THIS IS THE CASE WHEN THE NEXT AVAILABLE SPACE REACHES THE BOTTOM. THE NEXT AVAILABLE WILL GO BACK TO THE TOP \textasteriskcentered / ELSE PNTR2 = PNTR2 + 1; END; / \textasteriskcentered DELETION OF AN ELEMENT \textasteriskcentered / IF FLAG = \Phi [go to error routine]; / \textasteriskcentered QUEUE EMPTY \textasteriskcentered / Element = QUEUE (PNTR1); / \textasteriskcentered DELETE ELEMENT \textasteriskcentered / PNTR1 = PNTR1 + 1; IF PNTR1 > 20 THEN PNTR1 = 1; / \textasteriskcenteredCIRCLE AROUND \textasteriskcentered / IF PNTR2 = PNTR1 THEN FLAG =\varphi/ \textasteriskcentered QUEUE EMPTY \textasteriskcentered / \varphi END;

Question:

A PL/I program has variables U, V, W, X, Y, Z which are de-claredas follows: DCL UBIT(4),V FIXED (4,1), WCHAR(5),X FLOAT(4), YCHAR(3),Z FIXED(3); The Data Card for the same program contains the following in-formation: U = '1101'B, V = 981\textbullet2; U = '1101'B, V = 123\textbullet4, W = '123\textbullet4', X = \textbullet1052E + 02, Y = 'PQR', Z = - 249; a) Show what will be entered into memory storage in the com-puteras a result of the statements: GET DATA; followed alittle later in the program by the statement: GETDATA(Z,V,W,Y,U,X) ; b) In the same program at a later stage, the contents of the memorylocations corresponding to U, V, W, X, Y and Z werechanged to 0100 769\textbullet2 144\textbullet7 \textbullet0129E - 03 BOG - 47 respectively. Show what will be printed out by the following statements: PUTDATA(U,Y); PUT SKIP DATA(X); PUT SKIPDATA(W,V,Z);

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G13-0323.htm

Solution:

a) The GET DATA; statement reads in the values of all the variablesavailable up to the first .semi-colon on the card. Hence, the memorystorage is filled in with con-tents as follows: Name of memory location Content of the memory location U 1101 V 981.2 W X Y Z The next statement, which is GET DATA (U,V,W,X,Y,Z); will read in the valuesof the variables up to the next semicolon on the data card. If the newvalues of the variables are different, the memory contents will be changedaccordingly. Hence, the latest memory contents are as follows: Name of memory location Content of the memory location U 1101 V 123.4 W 123.4 X \textbullet1052e + 02 Y PQR Z - 249 b) As a result of the PUT DATA statements, the following is printed out: U = '0100'BY = ' BOG'; U = '0100'BY = ' BOG'; (BLANK LINE) X = \textbullet0129E - 03; (BLANK LINE) W = '144\textbullet7'V = 769\textbullet2Z = - 47;

Question:

The following data was collected for the reaction OH^- + CH_3COOCH_2CH_3 \rightarrow CH_3COO^- + CH_3CH_2OH : Time, min [CH_3COOCH_2CH_3] [ OH^-] 0 .02000 .02000 5 .01280 .01280 15 .00766 .00766 25 .00540 .00540 35 .00426 .00426 55 .00289 .00289 120 .00137 .00137 (a)Determine whether the reaction is first or second order. (b) Write the rate law expression.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E13-0455.htm

Solution:

The order of a reaction is the number of particles needed to form the transition state. One can tell whether the reaction is first or second order by an investigation into the constancy of the half-life of this reaction over time. It is a characteristic feature of first order reactions that the half-life, the time ne-cessary for half the particular reactant present initially to decompose, is constant over time. Thus, by determining whether the half-life is constant or not, the order of this reaction can be found. From the data, half the material is decomposed between 5 and 15 minutes. In the period, 5 to 15 minutes, the concentration of both [CH_3COOCH_2CH_3] and [OH^-] goes from .0128 to .00766, which means half has decom-posed. One can approximate that the half-life is about 10 minutes. To see if this time of half-life remains constant, notice the time needed for the concentration to go from .0054 to .00270, which is another 1/2 being decomposed. This occurs between t = 25 minutes to approximately t = 55 minutes, a difference of 30 minutes. Thus, the half- life is not constant. It changed from 10 to 30 minutes. Thus, the reaction is not first order. It must be second order. (b)To solve this part, use the information from (a) . If the reaction is second order, the rate depends upon both the concentration of CH_3COOCH_2CH_3 and OH^-. The rate law expression is, thus, rate = {- d[CH_3COOCH_2CH_3]} / [dt] = {-d[OH^-]} / [dt] =k[OH^-][CH_3COOCH_2CH_3 ], where k = rate constant. This equation is, by definition, the rate law expression for a second order reaction with 2 different molecules.

Question:

How can autoradiography be used to show that cells without nuclei do not synthesize RNA?

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/Users/wenhuchen/Documents/Crawler/Biology/F24-0622.htm

Solution:

A valuable technique for tracing events in cells is autoradiography. This method relies on the fact that radioactive precursors (metabolic forerunners) taken up by the cells are incorporated into macromolecules when these cells are grown in a radioactive medium-. The most widely used radioactive precursor, is tritiated thymine. This is thymine (used by the cell to synthesize DNA) made radioactive by the substitution oftritiuip [^3H], a radioac-tive form of hydrogen, for some of its hydrogen atoms. Typically, radioactive precursors are presented to cells. The cells are then sliced into thin strips called sections. The sections are coated with photographic emulsion similar to ordinary camera film. Exposure of camera film to light followed by photographic development leads to reduction of exposed silver salts in the film and to production of metallic silver grains. These grains form the image in the negative. Similarly, exposure to radioactivity and subsequent development produces grains in theautoradiographicemulsion. By examining the sections with an electron microscope, both the underlying structure and the small grains in the emulsion are seen. Autoradiography can be used to show that cells with-out nuclei do not synthesize RNA. This can be shown by growing enucleated cells in a medium containing a radioactive precursor for RNA. One such precursor is tritiated uracil, because uracil is the pyrimidine base found only in RNA. Tritiated uracil can be made by substituting the hydrogen atoms of uracil with tritium atoms. When we make sections of enucleated cells grown in tritiated uracil and examine them under the electron microscope, we observe no grains in the sections. This is because no RNA is produced and therefore no radioactive precursors are incorporated into the cell. However, cells with nuclei, when treated in a similar manner, exhibit grains in their sections. This shows that normal cells do make RNA, whereas cells without nuclei do not synthesize RNA.

Question:

Six students each of mass 80 kg decide to bounce an automobile of mass 1200 kg. They find that if they sit in the auto , the static deflection of the chassis is 1.5 cm. When they press down the bumpers in unison a maximum dynamic deflection of 10 cm is obtained. The auto also oscillates at a frequency of 0.8 sec^-1. Estimate the Q of the automobile suspension , and the damping factor \rho of the shock absorbers .

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/Users/wenhuchen/Documents/Crawler/Physics/D09-0380.htm

Solution:

First we find the spring constant k. The total mass of the students is 6 × 80 = 480 kg; and their weight 480 × 9.8 N = Mg causes a deflection of 1.5 × 10^-2m. According to Hooke's law, W = -kx where x is the extension of the spring of the auto from its equilibrium position , and W is the force (the weight of an object in this case) on the spring . Then k = W/x = Mg/x = (480 kg × 9.8 m/sec^2)/(1.5 × 10^-2m) = 3.14 ×10^5 N/m Next we estimate the work done by the students in pushing the car down. Suppose that in pushing down they take all their weight off their feet. Then in each push a force of (480 × 9.8) N travels through a distance of 0.10 m (10 cm). The work done in each oscillation is W = 480 × 9.8 × 0.10 = 470 joules. The frequency is 0.8 sec^-1, so that the power expended per oscillation is found as follows: p = (work/oscillation)/(time for one oscillation) = W × f = 470 j × 0.8 sec^-1 = 376 watt Here we made use of the relation f = 1/T. The potential energy stored in the suspension is given by H = 1/2 kA^2 whereA is the maximum deflection. Thus H = (1/2) 3.14 × 10^5 N/m × (0.10 m)^2 = 1.57 × 10^3 joules From this we can estimate Q of the system. The Q of the system is a measure of the amount of energy dissipated in the system, per cycle of operation, relative to the energy present in the system. More precisely, we define Q as Q =2\pi(H/P) where H is the energy stored in the system, and P is the energy dissipated , all in one cycle. In our case we have Q =2\pi[(1.57 × 10^3 j)/(470 j)] \cong 21 1.57 × 10^3 j is the potential energy stored in the automobile springs. Then we assume that f is the natural frequency (a moderately good assumption ) we can calculate \rho using an alternate expression for Q. Q can also be described in terms of the damping factor \rho. The damping factor is a measure of the time taken for the oscillations to die out . This in turn depends on the amount of energy dissipated by the system per oscillation. Q then, in terms of time, rather than in terms of energy , is given as follows: Q = \omega_0/\rho = 2\pif/\rho From this, we can find \rho. Then \rho = 2\pif/Q = 2\pi × 0.8/21 = 0.24 sec^-1

Question:

Using either the Subroutine Subprogram or Function Subprogram of the previous two problems obtain the greatest common divisor of L1 and L2, and add it to NN. Assume the Main Program is written, and integer values are assigned to L1, L2, and NN.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G12-0283.htm

Solution:

This is a good example of some of the advantages of using FUNCTION SUBPROGRAMS. We could use the following statements for the Subroutine: CALL GCD (L1, L2, M) NN = NN + M However, for the FUNCTION SUBPROGRAM we only need one statement: NN = NN + IGCD(L1, L2) Notice that the function subprogram is used in expressions wherever needed, in a way similar to the built-in functions such as SQRT (square root), IABS (Absolute Value of I),etc.

Question:

Consider a kinetic experiment in which the chemical re-action is accompanied by a temperature change. Using the Heisenberg uncertainty principle, calculate the uncertainty in a simultaneous measurement of time and temperature.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E18-0678.htm

Solution:

The Heisenberg uncertainty principle relates the uncertainty in energy, ∆E, and the uncertainty in time, ∆t, arising from a simultaneous measurement of the two, by the inequality ∆E∆t \geq h, where h is Planck's constant. Thus, the product of the two uncertainties can-not be less than Planck's constant. In order to apply the Heisenberg uncertainty principle to our problem, we must relate temperature and energy. The internal energy of a substance is approximately proportional to its absolute temperature, T. In fact, for ideal gases, the two are exactly proportional. Let the constant of proportionality for our particular system be c. Then, E = cT and ∆E - c∆T, where ∆T is the un-certainty in the temperature measurement. Then the un-certainty relation becomes: ∆E∆t = c∆T ∆t \geq hor,∆T ∆t \geq h/c.

Question:

A beam of light in air falls on a glass surface at an angle of incidence of 20.0\textdegree. Its angle of refraction in the glass is 12.4\textdegree. What is the index of refraction of the glass?

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/Users/wenhuchen/Documents/Crawler/Physics/D27-0863.htm

Solution:

Since the beam of light passes obliquely from an optically less' dense medium, air, to an optically denser medium, glass, the angle of incidence is larger than the angle of refraction. The equation for indices of refraction is (sin \texttheta_1/sin \texttheta_2) = (n_2/n_1) Since the index of refraction for air is approximately one, this reduces to n_2 = (sin \texttheta_1/sin \texttheta_2) = (sin 20.0\textdegree/sin 12.4\textdegree) = (0.342/0.215) = 1.58.

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Question:

A certain quantity of a gas has a volume of 1200 cm^3 at 27\textdegreeC. What is its volume at 127\textdegreeC if its pressure remains constant?

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/Users/wenhuchen/Documents/Crawler/Physics/D14-0504.htm

Solution:

Charles' law states that for constant pressure the volume of a gas is directly proportional to its absolute temperature (in degrees Kelvin). Converting the temperatures into Kelvin's temperatures by adding 273 we have, T_1 = 27\textdegree + 273\textdegree = 300\textdegree K T_2 = 127\textdegree + 273\textdegree = 400\textdegree K Using Charles' law, we obtain v_1 = kT_1andv_2 = kT_2 where k is a constant. Taking the ratio of these 2 equations, we find v_2/v_1 = T_2/T_1, [v_2/(1200 cm^3)] = (400\textdegree K)/(300\textdegree K) then v_2 = (1200 cm^3 × 400\textdegree K)/300\textdegree K = 1600 cm^3

Question:

What are the major constituents of a productive soil? Whatis the role of each of these constituents in plant growth? What measures can be taken to prevent the loss of topsoil?

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/Users/wenhuchen/Documents/Crawler/Biology/F10-0263.htm

Solution:

A rich, fertile soil should contain sufficient quantities of the essential mineralnutrients. It should also contain an adequate amount of organic materials which is provided by humus. In addition, it should include innumerable bacteria, fungi, and other microorganisms to bring about decay of organic substances and aeration of the soil. The essential minerals play important roles in the structure and functions of the plant. The broadest role of the minerals is their action in catalyzing enzymatic reactions in the cell. In some cases they act as an essential structural component of the enzyme, in others they act as regulators or activators of certain enzymes. Some minerals, absorbed in the ionic form by active transport, function in providing a hydrostatic pressure , orturgorwithin the cell. In other words, their con-centrations inside the cell or in the surrounding medium regulate the movement of water in and out of the cell. Some minerals regulate the permeability and integrity of cell membranes, while others serve as structural components of cell parts. Other minerals constitute structural parts of the electron acceptors of the electron transport system, such as thecytochromesand ferredoxin . Still others act as buffering agents to prevent drastic changes in pH within the cell. Humus, derived from the decaying remains of plants and animals , provides the soil with organic nutrients, among them mainly carbon - and nitrogen-containing compounds. In addition, humus increases the porosity of soil so that proper drainage and aeration can occur, and enhances the ability of the soil to absorb and hold water. To bring about decomposition of the humus, a good soil should contain a large number of bacteria, fungi, and other small organisms. Such organisms, as the earthworms, benefit the soil further by constantly tilling the soil, aerating it, and mixing in additional organic substances. A major conservation problem all over the world is to decrease the amount of valuable topsoil carried away each year by wind and water. Reforestation of mountain slopes, building check dams in gullies to decrease the spread of the run-off water, contour cultivation, terraces and the planting of windbreaks are some of the successful methods that are currently being used to protect the topsoil against erosion.

Question:

What is meant by a joint? What different types of jointsare there ? Why do older people usually complain of stiffness in their joints?

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/Users/wenhuchen/Documents/Crawler/Biology/F19-0485.htm

Solution:

The point of junction between two bones is called a joint. Some joints , such as those between the bones of the skull, are immovable and extremely strong, owing to an intricate intermeshing of the edges of the bones . The truly movable joints of the skeleton are those that give the skeleton its importance in the totaleffectormechanism of locomotion. Some are ball and socket joints, such as the joint where the femur joins the pelvis, or where thehumerusjoins the pectoral girdle. These joints allow free movement in several directions. Both the pelvis and the pectoral girdle contain rounded, concave depressions toaccomodatethe rounded convex heads of the femur andhumerus, respectively. Hinge joints, such as that of the human knee, permit movement in one direction only. The pivot joints at the wrists and ankles allow free-dom of movement intermediate between that of the hinge and the ball and socket types. (Refer to Fig. 1.) The different bones of a joint are held together by connective tissue strands called ligaments. Skeletal muscles, attached to the bones by means of another type of connective tissue strand known as a tendon, produce their effects by bending the skeleton at the movable joints. The ends of each bone at a movable joint are covered with a layer of smooth cartilage . These bearing surfaces are completely enclosed in a liquid-tight capsule , called the bursa. The joint cavity is filled with a liquid lubricant, called the synovial fluid , which is secreted by the syn-ovial membrane lining the cavity. (Refer to Fig. 2) During youth and early maturity the lubricant is replaced as needed , but in middle and old age the supply is often decreased, resulting in joint stiffness and difficulty of movement. A common disability known as bursitis is due to the inflammation of cells lining the bursa, and also results in restrained movement.

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Question:

If a man is inhaling air composed of 19% O_2 and 0.003% CO by volume, what is the ratio of the concentrations of carboxyhemoglobin(HbCO) andoxyhemoglobin(Hb0_2) ? Explain why [CO] should be kept to a minimum.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E23-0828.htm

Solution:

The concentration of carbon monoxide [CO] must be kept to a minimum due tohemoglobins' (Hb) greater affinity for CO than O_2 (oxygen). This occurs by the reaction HbO_2 + CO \rightleftarrowsHbCO+ O_2. Because of this reaction in the presence of CO, less O_2 is available for body cells. It also results in the reduction ofoxy-hemoglobindissociation into hemoglobin and oxygen so that anoxia (oxygen starvation) occurs. The ratio of [HbCO] and [HbO_2] can be calculated by the Haldane equation: [HbCO] / [HbO_2] = M × [P(CO) / P(O_2)], where p(CO) and p (O_2) are the partial pressures (or volume concentrations) of the CO and O_2 gases and M is a constant that depends on the species. Given p(CO), p(O_2) and M= 250, by substituting one obtains [HbCO]/[HbO_2] = 250 × [0.0003 / 0.19] = 0.03947 / 1 = 0.03947.

Question:

In what ways are eggs and sperm adapted structurally for their respective functions in reproduction?

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/Users/wenhuchen/Documents/Crawler/Biology/F22-0575.htm

Solution:

The human sperm consists of a head, a middle piece and a tail. About two-thirds of the head is com-prised of the nucleus. The apical one-third of the head is the acrosome. The acrosome has a conglomerate of enzymes which aid the sperm in penetrating the egg surface in fertilization. Before a sperm can fertilize an egg, it must undergo a process called capacitation followed by the acrosome reaction. Capacitation is the breakdown or removal of a protective coating of the sperm. Following this, small perforations are produced in the acrosome wall, allowing for the escape of enzymes as part of the acrosome reaction. The middle piece contains many mitochondria that provide energy for the movement of the tail. The mito-chondria oxidize nutrients which diffuse into the middle piece from the seminal fluid and thus generate energy for the sperm's motion. The middle piece also contains a centriole which, after fertilization, aids in the division of the zygote. The tail lashes back and forth and propels the sperm. The ovum is massive compared to the sperm, with an abundance of cytoplasm (recall the unequal cytoplasmic division during oogenesis) containing yolk granules. The yolk provides nutrition for the embryo during the early stages of its development. The human ovum is truly large and is just visible to the unaided eye as a tiny speck. Because of its large size and lack of flagella, the egg is nonmotile in contrast to the sperm. The ovum is sur-rounded by two extracellular layers which can be broken down by the enzymatic action of the sperm's acrosome. The inner layer is called the zona pellucida and the outer layer consists of follicular cells which become radially arranged as the corona radiata. The zona pellucida, upon enzymatic degradation by the acrosome, allows the sperm to pass through the egg cell membrane into the cytoplasm. It is also thought to be responsible for the inhibition of entry of more sperm, so that only one sperm will fertilize one egg. The mechanism by which this occurs is not well understood, but it appears to involve some structural changes of the zona pellucida controlled by granules released from the ovum's cytoplasm.

Question:

A circular hoop of radius 0.3 m receives a charge of 0.2 pC (pico Coulomb). Determine the potential and the field strength at the center of the hoop and at a point on the axis of the hoop 0.4 m from the center. Where on the axis is the field strength a maximum?

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/Users/wenhuchen/Documents/Crawler/Physics/D17-0570.htm

Solution:

Consider a small element dl of the hoop carrying a charge dq (see fig. A) . The potential at a point a distance z along the axis of the hoop due to dq is dV = [(dq)/(4\pi\epsilon_0r)] = [dq] / [4\pi\epsilon_0\surd(y^2 + z^2)] The total potential at the point due to the charge on the whole hoop is thus V = \int dV = \int(dq)/(4\pi\epsilon_0r) = (1)/(4\pi\epsilon_0r) \int dq = [q] / [4\pi\epsilon_0\surd(y^2 + z^2)] r is constant, for every line element dl, is equi-distant from the point of interest, q is the total charge of the hoop. Thus the potential at the center of the hoop occurs for z = 0 m, and V_0 = q/4\pi\epsilon_0y = (0.2 × 10^-12 C × 9 × 10^9 N \textbullet m^2 \textbullet C^-2)/(0.3 m) = 6 × 10^-3 V. Also, when z = 0.4 m, V_0.4 = [0.2 × 10^-12 C × 9 × 10^9 N \textbullet m^2 \textbullet C^-2] / [ \surd(0.3^2 + 0.4^2 m)] = [(1.8 × 10^-3)/(0.5)] V = 3.6 × 10^-3 V. The electric intensity at the general point on the axis is E = \rule{1em}{1pt} (dV/dz) = + [ qz/{4\pi\epsilon_0(y^2 + z^2)3/2], From symmetry considerations, the direction of E^\ding{217} must be along the z-axis. Let the electric field due to the charge dq of each segment dl be resolved into a vertical and a horizontal component (see fig. B). From the symmetry of the hoop about its center, the horizontal components of the electric field due to the charges on any two segments dl on opposite ends of the center will cancel one another. The vertical components are in the same direction and add algebraically. The direction, then, of the total E^\ding{217} due to all pairs of segments dl at opposite ends of the center, will be vertically up-ward, or along the positive z-axis. Thus at the center of the hoop where there is no vertical component of the electric field due to the charge dq on a segment dl, E is zero and at the position where z = 0.4 m, using (1) E_0.4 = [9 × 10^9 N \bullet m^2 \bullet C^-2 × 0.2 × 10^-12 C × 0.4 m] / [\surd{(0.3 m)^2 + (0.4 m)^2}]^3 = [0.2 × 10^-12 C × 0.4 m × 9 × 10^9 N \bullet m^2 \bullet c^-2] / [(0.5)^3 m^3] = 5.76 × 10^-3 V \bullet m^-1. Note that when z is positive, E is positive, and when z is negative, E is negative. The direction of E^\ding{217} is thus always along the axis away from the hoop. The electric intensity E is zero when z = 0 and also when z = \infty; E must therefore pass through a maximum value somewhere between these two points. For a maximum dE/dz = 0, or using the product rule for differentiation, [q / {4\pi\epsilon_0(y^2 + z^2)^3/2}] \rule{1em}{1pt} [{3qz^2} / {4\pi\epsilon_0(y^2 + z^2)^5/2}] = 0. \thereforey^2 + z^2 - 3z^2 = 0orz = \pm (y/\surd2). Thus, at a distance along the axis of (0.3/\surd2) m = 0.21 m, E has its maximum value of E_max = [0.2 × 10^-12 C × 0.21 m × 9 × 10^9 N \textbullet m^2 \bullet C^-2] / [0.3^2 + 0.21^2)^3/2 m^3] = 7.70 × 10^-3 V \textbullet m^-1. = 7.70 × 10^-3 V \textbullet m^-1.

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Question:

In the figure, a long straight conductor perpendicular to the plane of the paper carries a current i going into the paper. A bar magnet having point poles of strength m at its ends lies in the plane of the paper. What is the magnitude and direction of the magnetic intensity H at point P?

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Solution:

The assumption that the ends of the magnet can be taken to be point sources of magnetic flux is not a realistic one although it greatly simplifies the calculation. The vectors H_i , H_N , and H_S , as shown in the figure, represent the components of H due respectively to the current, and to the N and S poles of the magnet. Con-sider first H_i . The flux density B at point P, due to the current i in a long straight conductor at a distance "a" from the conductor is known to be B = (\mu_0 /2\pi)(i/a) . In free space, the magnetic field strength H is related to B by H = B/\mu_0 HenceH_i = (1/2\pi) (i/a) . Analogous to the electric field, the magnetic field due to a magnetic pole of strength m at a distance r from the magnetic pole is H = (1/4\pi\mu_0 ) (m/r^2 ) , therefore, the components H_N and H_S are respectively H_N = [1/(4\pi\mu_0)] (m/b^2) H_S = [1/(4\pi\mu_0 )] (m/c^2 ) The resultant of these three vectors is the magnetic intensity H at the point P.

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Question:

In the figure, a block has been placed on an inclined plane and the slope angle \texttheta of the plane has been adjusted until the block slides down the plane at constant speed, once it has been set in motion. Find the angle \texttheta.

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Solution:

The forces on the block are its weight w and the normal and frictional components of the force exerted by the plane. The angle \texttheta of the inclined plane is adjusted until the block slides down the plane. Since motion exist, the friction friction force is f_k = \mu_kN. Take axes perpendicular and parallel to the surface of the plane. Then, applying Newton's Second Law to the x and y components of the block's motion, we obtain (see figure) \sumF_x = \mu_kN - w sin \texttheta = 0 \sumF_y = N - w cos \texttheta = 0 . where \sumF_x and \sumF_y are the x and y components of the net force on the block. Both of these equations are equal to zero because the block accelerates neither parallel nor perpendicular to the plane. Hence \mu_kN = w sin \texttheta , N = w cos \texttheta . Dividing the former by the latter, we get \mu_k = tan \texttheta . It follows that a block, regardless of its weight, slides down an in-clined plane with constant speed if the tangent of the. slope angle of the plane equals the coefficient of kinetic friction. Measurement of this angle then provides a simple experimental method of determining the coefficient of kinetic friction.

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Question:

Explain why the feeding of thyroid glands may cure myxedema, and why the feeding of pancreas does not cure diabetes.

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/Users/wenhuchen/Documents/Crawler/Biology/F21-0540.htm

Solution:

Myxedema is a disease caused by a deficiency of thyroxine, a hormone secreted by the thyroid gland. Thyroxine has the following structural formula: The molecule basically consists of two iodinated amino acid molecules (tyrosine) linked together by an ether bond. One of the tyrosine molecules loses its backbone and part of its sidechain during the synthesis of thyroxine. Although thyroxine is made up of two amino acids, they are not linked together by a peptide bond, which is characteristic of amino acid linkages. It is this fact which allows thyroxine to remain intact when subjected to proteolytic enzymes in the stomach following ingestion. The thyroxine ingested can be used to treat myxedema because it is absorbed by the gut in an unaltered form. Thus the feeding of thyroid glands is one possible way to cure myxedema. The hormone produced by the Islets of Langerhans in the pancreas is insulin. Insulin is one of the smallest proteins known, and like other proteins it consists of a sequence of amino acids linked by peptide bonds. Amino acids join together to form peptide bonds in the following way: Should insulin be ingested and enter the stomach, proteases there would hydrolyze the hormone into individual amino acids, thus destroying the hormone. This is why diabetes cannot be cured by oral administration of insulin. Instead insulin is usually directly introduced into the blood stream.

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Question:

KMnO_4 is added to a 350 ml solution of an approximately 3% H_2O_2 solution. It is found that 3.94 cm^3 of O_2 gas is re-leased. It is collected over H_2O at a pressure (barometric) of 0.941 atm and 28\textdegreeC. Assuming that the pressure due to water vapor is 0.0373 atm, what is the actual concentration of the H_2 O_2 solution?

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Solution:

The concentration of a solution is given in terms of molarity, i.e., moles/liter. This means that the number of moles of H_2 O_2 must be determined, since you already know the volume. To determine the number of moles of H_2 O_2, you know that, when oxidized, 1 mole of H_2 O_2 yields 1 mole of O_2. Thus, if you know the number of moles of O_2 produced, you also know the moles of H_2 O_2 originally present. To find the number of moles of O_2, you must use the equation of state. This tells us PV = nRT, where P = pressure, V = volume, n = number of moles, R = universal gas constant, (.082057 1-atm mol^-1 deg^-1)/ and T = temperature in degrees Kelvin (degree Celsius plus 273\textdegree). Hence, n_O(2) = (PV/RT). All the values needed to find n_O(2) are known or can be calculated. The pressure of O_2 will not be .941 atm, since this figure also includes vapor pressure from water. Therefore, the actual pressure of O_2 is .941 - .0373. That is, the difference between barometric and vapor pressure. Substituting, you obtain n_O(2) = [{(.941 - .0373) (3.94 × 10^-3)} / {(.082) (301)}] = 1.44 × 10^-4 moles of O_2. This means there were originally 1.44 × 10^-4 moles. Now, recalling the definition of molarity, which represents the concentration of a solution, you have [(1.44 × 10^-4 moles)/(.350 liters)] = 4.12 × 10^-4 M which is the concentration of the H_2 O_2 solution.

Question:

The volume of hydrogen evolved during the course of a reaction is measured by the displacement of water as shown in the diagram below. Hydrogen is evolved in flask A and displaces water from flask B into beaker C. If, during a particular run of this experiment in which atmospheric pressure is 765 torr and the water temperature is 293.15\textdegreeK, 65.0 ml of water is displaced, how much water would be displaced at 760 torr and 298.15\textdegreeK? (The equi-librium vapor pressure of water at 293.15\textdegreeK is 17.5 torr and at 298.15\textdegreeK is 23,8 torr.)

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Solution:

We can convert from the volume under non- standard conditions in the problem to the volume under standard conditions by means of the ideal gas equation, PV = nRT (where P = pressure, V = volume, n = number of moles, R = gas constant, and T = absolute temperature) . For our nonstandard conditions we may writeP_nV_n = nRT_n and for final conditions (760 torr and 298.15\textdegreeK) we may write P_sV_s = nRT_s Note that n and R are the same in both cases. Dividing the second equation by the first we obtain P_sV_s / P_nV_n = nRT_s / nRT_n = T_s / T_norV_s = [P_nV_n / V_s] × [T_s / T_n] . The pressure experienced by the hydrogen under non-standard conditions is equal to the pressure exerted by the atmosphere (765 torr) plus that from water vapor pressure at 20\textdegreeC (17,5 torr), or P_n = 765 torr + 17.5 torr = 782.5 torr. Under the final conditions, the pressure experienced by hydrogen is equal to the sum of the atmospheric pressure (760 torr) and the vapor press-ure of water at 298.15\textdegreeK (23.8 torr), or P_s = 760 torr + 23.8 torr = 783.8 torr. We can now calculate the volume of hydrogen under the final conditions. Thus, V_s = [P_nV_n / P_s] × [T_s / T_n] = [(765 torr × 65.0 ml) / 783.8 torr] × [298.15\textdegreeK / 293.15\textdegreeK] = 64.5 ml.

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Question:

The relative rates of diffusion of NH_3(g) and HX (g) can be determined experimentally by simultaneously in-jecting NH_3(g) and HX (g), respectively, into the oppo-site ends of a glass tube and noting where a deposit of NH_4X(s) is formed. Given an 1m tube, how far from the NH_3 injection end would you expect the NH_4X(s) to be formed when HX is a) HF, b) HCl?

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Solution:

The rate of diffusion of a gas is observed to be inversely proportional to the square root of its molecular weight. This is called Graham's law. Rate = constant / \surdMW For HF:According to Graham's Law: the speed of NH_3 is proportional to 1 / \surd17 = .243 (MW of NH_3 = 17) the speed of HF is proportional to 1 / \surd20 = .224 (MW of HF = 20) Thus, for an arbitrary amount of time, let us say 1 min., NH_3 moves .243 cm and HF moves .224 cm. Because the gases are ejected from either end of the tube, after 1 minute the gases move (.243 + .224) cm or .467 cm/min. closer together. When this distance is equal to 100 cm the solid forms. One can determine the time needed by dividing 100 cm by .467 cm/min time after which solid forms = 100 cm / .467 cm/min = 214.13 min. Thus, after 214.13 min. the HF meets the NH_3. The distance the NH_3 would travel is found by multiplying the distance it travels in 1 minute, .243 cm, by the time elapsed till the NH_4F deposits. distance NH_3 travels = .243 cm/min × 214.13 min =52.03 cm. 52.03 cm. b) One solves for HCl in a similar manner MW of HCl =36.5 the speed of HCl is proportional to 1 / \surd36.5 = .165 As previously found, the speed of NH_3 is proportional to 1/\surd17 = .243. Speed is a measure of distance covered over a unit time. Let us designate the speeds here as cm./min. Thus, HCl moves .165 cm/min and NH_3 moves .243 cm/min. The solid forms after the sum of the distances travelled by the two gases is 100 cm. In 1 min. the gases travel a distance of (.165 + .243)cm or .408 cm. The time it takes for the gases to travel 100 cm. is found by dividing 100 cm. by .408 cm/min. time after which the solid will deposit = = 100 cm / .408 cm/min = 245.10 min. The distance the NH_3 travels is then found by multiply-ing 245.10 min by the distance the gas travels in 1 minute. distance travelled by NH_3 = .243 cm/min × 245.10 min = 59.56 cm.

Question:

Discuss the major morphological features of bacteria. Explain why bacteria can produce changes in their environment so quickly.

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Solution:

Among the major characteristics of bacterial cells are their shape, arrangement, and size. These characteristics constitute the morphology of the bacterial cell. Although there are thousands of different species of bacteria, the cells of most bacteria have one of three fundamental shapes: (1) spherical or ellipsoidal, (2) cylindrical or rodlike, and (3) spiral or helically coiled. Spherical bacterial cells are called cocci (singular, coccus). Many of these bacteria form patterns of arrangement which can be used for identification. These patterns can be explained by peculiarities in the multiplication processes of the different bacteria. For example, diplococcal cells divide to form pairs. Streptococcal cells remain attached after dividing and form chains. Staphylococci divide three dimensionally to form irregular clusters of cocci, resembling bunches of grapes. Each of the cells in a diplococcal, streptococcal, or staphylococcal aggregate is an independent organism. Cylindrical or rod like bacterial cells are called bacilli (singular, bacillus). These do not form as wide a variety of arrangements as do cocci, but occasionally they are found in pairs or chains. These patterns do not arise from the multiplication process, but only from the particular stage of growth or growth conditions present, and hence bacilli usually appear as single, unattached cells. There are many variations in the thickness and length of these rod like bacteria. Spiral-shaped bacterial cells are called spirilla (singular, spirillum). Like the bacilli, they usually occur as unattached, individual cells. The spirilla exhibit considerable differences in length and in the frequency and amplitude of the spirals. The average bacterial cell has dimensions of approximately 0.5 to 1.0 \mum by 2.0 to 5.0 \mum. (\mum is the abbreviation for micrometer, which is 1/1000 of a millimeter or 10^-6 meter.) An important consequence of the very small size of a bacterial cell is that the ratio of surface area to volume is extremely high. This ratio allows a very large portion of the bacterial cell to be in contact with its environment.The result is that bacteria are able to rapidly ingest nutrients and growth factors and excrete wastes; their metabolic rate is correspondingly high. This high metabolic rate enables bacteria both to adjust to and to introduce changes in their environment in very short periods of time. These changes may be beneficial to the bacteria producing them, or to other species of bacteria. For instance, the release of carbon dioxide by certain bacteria increases the acidity of the growth medium and favors the growth of bacteria requiring a low pH environment. The most important factor involved in the ability of bacteria to alter their environment is their ability to multiply rapidly. Within the multitudes of bacteria produced by a newly settled bacterium there will be a handful of mutants. Natural selection may allow one of these mutants to be most fit for survival in the changing environment and numerous offspring will be derived from it. This cycle may be repeated, demonstrating the capability of the bacterial species to survive under changing conditions.0

Question:

0.20 g of cuttings from a silver coin (containing Cu) was completely dissolved in HNO_3. Then aNaClsolution was added to precipitateAgCl, which, when collected and dried, weighed 0.2152 g. What is the indicated percent of Ag in the coin?

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Solution:

To solve this problem, we must (1) set up the reactions (2) determine the number of molesAgClpre-cipitated (3) determine the number of moles of Ag used from thestoichiometryof the equations (4) calculate the weight of Ag used (5) calculate the percent of Ag in the coin. The equations are (i)Ag + Cu + HNO_3\ding{217} AgNO_3 + Cu(NO_3)_2 \downarrow (ii)AgNO_3 +NaCl\ding{217}AgCl\downarrow + NaNO_3. Thestoichiometryof these equations says that 1 mole ofAgClis produced from 1 mole of Ag metal. From the data given, theAgClweighed 0.2152 g. Therefore, the number of moles ofAgClprecipitated can be calculated, since the molecular weight ofAgClis 143.4 g/mole. Therefore, [.2152 g / (143.4 g / mole)] = 1.50 × 10^-3 moles ofAgClprecipitated out. Thus, 1.50 × 10^-3moles.ofAg metal are consumed. The weight of Ag metal is the number of moles of Ag times its atomic weight (At. wt. = 107.868 g / mole), which is equal to (1.50 × 10^-3 moles) (107.868 g / mole) = 0.162 g. The percent of Ag metal in the coin is [(0.162 g Ag) / 0.20 g] × 100 = 81 %.

Question:

When 4.90 g of KCLO_3 was heated, it showed a weight loss of 0.384 g. Find the percent of the original KCLO_3 that had decomposed.

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Solution:

To solve this problem, you need to determine how many moles of KCLO_3 decomposed. Once this is determined, calculate the initial number of moles of KCLO_3 and divide to obtain a percentage. To do this, a balanced equation that illustrates this reaction must be written. Such an equation is 2KCLO_3 (s) \rightarrow 2KCL (s) + 3O_2 (g) The weight loss of .384 g must be the amount of O_2 liberated. The number of moles of O_2 is this weight divided by its molecular weight, or [(.384 g) / (32 g/mole)]= .0120 of O_2 . Going back to the original equation, you find that the number of moles of KCLO_3 is 2/3 that of O_2. Therefore, the number of moles of KCLO_3 that reacted is (.0120 mol of O_2) [(2 mol KCLO_3) / (3 mol O_2)] = .0080 moles of KCLO_3. The number of moles of KCLO_3 that you started with, how-ever, is [(4.90 g) / (122.6 g/mole)] =.0400 moles, where 122.6 g/mole is the molecular weight of KCLO_3. Therefore, the percentage decomposition is (.0080)/(.0400) × 100 = 20% .

Question:

What is paleontology? List the various kinds ofpaleontologic evidence. Why is it that fossils are frequently difficult to find?

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Solution:

Paleontology is the science of the finding, cataloguing and interpretationof the abundant and diverse evidence of life in ancient times. Fossils are the most important tools of the paleontologist in studying the past. Fossils can be bones, shells, teeth, and other hard parts of an animalor plant body which have been preserved, and indeed, include any impressionor trace left by some previous organism. As such they constitutethepaleontologicevidence of evolution. Most of the vertebrate fossils found are skeletal parts. From these partspaleontologists can infer the contours of the body, the posture, and thestyle of walking of the ancient animal. Using this information, recon-structionscan be made showing how the animal may have looked in life. Footprints or trails are another type of fossil. These are made by the animalin soft mud, which subsequently hardened. They enable one to inferthe size, body proportion, structures, and style of walking of the animalthat made them. Tools made by early man are also important artifacts. They aid in the determination of how and whenbipedalism evolved(hands had to be free in order to use tools). The discovery of tools alsoserves as invaluable information, providing clues of early man's culture, and can even be used to infer the origin of language. Petrified fossils result from the replacement by minerals of the originalanimal or plant material. The minerals that replace the once living tissuesmay be iron pyrites: silica, calcium carbonate or a variety of other substances. Petrifaction has been shown to be an effective type of preservationfor tissues as old as 300,000,000 years. Molds are still anothertype of fossil, and are formed by the hardening of the material surroundingthe buried organism, followed by the decay and removal of theorganism by ground water. Sometimes the molds became accidentally filledwith minerals, which then hardened to form casts or replicas of the originalorganism. In the far north, frozen fossils of whole animal bodies have been foundin the ground or ice. Old forms of plants, insects and spiders have beenfound preserved in amber, a fossil resin from pine trees. The resin wasat first a soft sap engulfing the fragile insect; then it slowly hardened, preservingthe insect intact. Usually, then, the formation and preservation of a fossil follows the buryingof some structure or entire organism. This may occur at the bottomof a deep body of water, or on land hidden deep in the earth's crust. Generally, fossils are preserved in areas that are well hidden or hardto reach. This is because fossils under shallow water or near the earth'ssurface are usually destroyed due to constant disturbances from naturalforces, such as wind and rain, and from living organisms. Therefore, fossils are generally well concealed and difficult to find.

Question:

Determine the heat needed to raise 60 g ofPbfrom 20\textdegreeC to 360\textdegreeC, given its specific heats are 0.0306 (solid) and 0.0375 (liquid). Its heat of fusion is 5.86 cal/g; its melting point is 327\textdegreeC.

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Solution:

There are three heats involved in finding the amount of heat absorbed in raising the temperature ofPbfrom 20\textdegreeC to 360\textdegreeC. First, the amount of heat absorbed in raising the temperature of the solid from 20\textdegreeC to 327\textdegreeC (its melting point); then, the amount of heat absorbed in melting the compound; last, the amount of heat absorbed in raising the temperature of the liquid from 327\textdegreeC to 360\textdegreeC. The heat absorbed in these three processes are added to-gether to find the amount of heat needed to bringPbfrom 20\textdegreeC to 360\textdegreeC. 1) Raising the temperature of the solid from 20\textdegreeC to 327\textdegreeC. The specific heat of solidPbis used to calculate the amount of heat absorbed in this process. The specific heat is defined by the number of calories absorbed when one gram of mass is raised one degree. The specific heat ofPbas a solid is 0.0306. Here, 60 g ofPbis heated 307\textdegreeC (327 - 20\textdegreeC). Thus, the specific heat is multiplied by 60 and 307. heat absorbed by solid = 60 × 307 × .0306 = 564 cal. 2) Heat needed to melt the solid. To calculate the amount of heat needed to melt the solid, one needs to use the heat of fusion. The heat of fusion is the number of calories necessary to melt one gram of solid. The heat of fusion ofPbis 5.86 cal/g. There are 60 g ofPb, here. Thus, the heat of fusion must be multiplied by 60 g to find the amount of heat needed to melt thePb. heat needed to meltPb= 60 g × 5.86 cal/g = 352 cal. 3) Heat absorbed in raising the temperature of the melted liquid from 327\textdegreeC to 360\textdegreeC. Here the specific heat is used again, but because thePbis now liquid, the specific heat for liquidPbmust be used. This is the amount of heat necessary to raise one gram of liquid one degree. 60 gPbis heated 33\textdegreeC (360\textdegree- 327\textdegree); therefore, the specific heat must be multiplied by 60 and 33 to find the heat absorbed. The specific heat for liquidPbis 0.0375. heat absorbed by liquidPb= 60 × 33 × 0.0375 = 74 cal. The heat absorbed by the total process is the combination of the heats absorbed in these three processes. Heat absorbed by solid564 cal Heat absorbed to melt solid352 cal Heat absorbed by liquid74 cal Total amount of heat needed990 cal.

Question:

Write a skeletal format of C programs.

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Solution:

The following figure provides a skeletal format of C programs. The terms appearing within the square brackets, [], are optional. They can be omitted from the programs if they are not required. [constant declarations] [externals] main ( ) { [program statements] } [additional functions]

Question:

Write a FORTRAN program which reads 100 data cards, each consisting of four letters, and stores them in a vector WORD. Then sort these words into alphabetical order.

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Solution:

Characters are represented internally by a numeric code whose ordering preserves alphabetic precedence. Thus if the integer variables X and Y each contain character data, the expression X.LT.Y will have the value .TRUE, if and only if the character string stored in X precedes in alphabetical order the character string stored in Y. The basic method of the program is to compare successive pairs of four-letter words. Two successive words are inter-changed in the vector if they are not in alphabetical order, i.e., the first is not "less than" the second. After the en-tire vector has been processed, the last word in alphabetical order has been pushed to the end of the vector. The vector is then processed again, and as a result the "second-largest" word appears in the second to last position. After (at most) passes over the data in the vector, the words are arranged in alphabetic order. Note that the logical variable SWT is used to measure whether or not any words are interchanged. If after a given pass an interchange occurs, the sort is not finished and SWT is true. Otherwise, SWT is False, no inter-change occurs, and the sort is complete. INTEGER WORD (1\O\O) LOGICAL SWT DO 1 I = 1, 1\O\O 1 READ2, WORD (I) 2 FORMAT (A4) N = 99 DO 3 I = 1,99 SWT = .FALSE. DO 4 J = 1, N IF (WORD (J).LE.WORD (J + 1)) GO TO 4 ITEMP = WORD (J) WORD (J) = WORD (J + l) WORD (J + l) = ITEMP SWT = .TRUE. 4 CONTINUE IF (.NOT. SWT) GO TO 5 3 N = N - 1 5 PRINT7, WORD 7 FORMAT ('\cyrchar\CYRF' ,1\cyrchar\CYRFA6) STOP END

Question:

What is the impedance of a 1-henry inductor at angular frequencies of 100, 1000 and 10,000rad/sec?

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Solution:

The circuit element is an inductor, therefore the impedance is purely reactive. At an angular frequency of 1000rad/sec, the reactance of a 1-henry inductor is X_L =wL= 10^3 (rad/sec)× 1h = 1000 ohms. At a frequency of 10,000rad/sec the reactance of the same inductor is 10,000 ohms, while at a frequency of 100rad/sec it is only 100 ohms.

Question:

The intestine, especially the small intestine, is a vital organ for absorption of nutrients required by the body. In what ways is it suitable for such a function?

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Solution:

The small intestine is that region of the digestive tract between the stomach and the cecum. Its long, convoluted structure is an adaptation for absorption of nutrients. Structural modifications of the internal surfaces of the small intestine act to increase its surface area for absorption. First, the mucosa lining the intestine is thrown into numerous folds and ridges. Second, small fingerlike projections called villi cover the entire surface of the mucosa (see figure). These villi are richly supplied with blood capillaries and lacteals (for absorption of fats) in order to facilitate absorption of nutrients. Third, individual epithelial cells lining the folds and villi have a "brush-border" on the surface facing the lumen, consisting of countless, closely-packed, cylindrical processes known as microvilli. These microvilli add an enormous amount of surface area to that already present. The total internal surface area of the small intestine is thus incredibly large; this is advantageous for the purpose of absorption. The large intestine also has villi to increase its surface area. However, the number of villi present in the small intestine far exceeds that found in the large intestine. The main function of the large intestine is to absorb water from the undigested food substances and reduce the remains to a semi-solid state before it is expelled through the anus.

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Question:

How does the wing shape of a bird correlate with its distinctive flight?

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/Users/wenhuchen/Documents/Crawler/Biology/F19-0491.htm

Solution:

The shape of a wing is correlated with both the power and type of flight for which it is used. Long, slender, pointed wings, sometimes reaching beyond the tail, are seen in birds having great flying powers and soaring habits, such as gulls, eagles, hawks, and vultures. Birds which do not soar, but which fly by continuous wing strokes, have shorter, more rounded wings. Very short, broad wings occur in the fowls, pheasants, grouses, and quails. These are habitual ground dwellers with feet adapted for running. They occasionally make short powerful flights by rapid wing strokes. Degeneration of the wings to a flightless condition has occurred in a number of birds, such as the penguins and the ostriches. In the ostrich group, many other changes accompany the loss of flight, such as the disap-pearance of the keel of the breast bone, which supports the flight muscles, and the development of strong running legs and feet.

Question:

What is the total binding energy and the binding energy per nucleon of the nucleus of _24Cr^52 which has a mass of 51.95699 amu?

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/Users/wenhuchen/Documents/Crawler/Physics/D36-1043.htm

Solution:

The nucleus of _24Cr^52 is made up of 24 protons and 52 - 24 = 28 neutrons. The combined mass of the particles making up the Cr nucleus is thus \summ =24 × 1.00819 + 28 × 1.00893 = 52.4460 amu,where 24 × 1.00819 + 28 × 1.00893 = 52.4460 amu,where we used the facts that the mass of a proton is 1.00819 amu, and that the neutron mass is 1.00893 amu. The mass defect or the difference in the mass of the constituent particles of _24Cr^52 and an atom of _24Cr^52 is 52.44660 - 51.95699 = 0.48961 amu. The binding energy is, from Einstein's mass - energy relation, E = (mass defect) c2 E = (.48961 amu) (9 × 10^16m2/ s^2) Since1 amu = 1.67 × 10^-27 kg E = (.48961 amu) (1.67 × 10-27amu/kg)(9 × 10^16 m^2/s^2) E = 7.36 × 10^-11 Joules But1 eV = 1.602 × 10^-19 Joules andE= (7.36 × 10^-11 Joules ) / (1.602 × 10^-19 Joules/eV) = 4.6 × 10^8 eV The binding energy per nucleon is thus (4.6 × 10^8 eV) / 52 = 8.9 × 106eV.

Question:

Perform the following conversions from decimal to binary, octal and hexadecimal systems. a) 17_10 b) 1321_10 c) 360_10 d) .75_10 e) .390625_10 f) .9_10

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Solution:

a) (i) The conversion for integer part consists of dividing by the desired base (e.g. 2) repeatedly and saving the remainder: Writing the remainders from right to left we have 10001 17_10 = 10001_2 Another format is the top-down division method as shown below: (ii) Proceeding in the same way for octal conversion As the binary, octaland hexadecimal systems can easily be converted from one form to another, it is only necessary to convert in any one of them and use the triad or quartet methods to obtain the results for remaining bases. Thus, 1321_10 = 2451_8 = 010 100 101 001_2 1321_10 = 010 100 101 001_2 = 0101 0010 1001_2 = 529_16 Thus 360_10 = 550_8 \textbullet Now 550_8 = 101 101 000_2 and 101 101 000_2 = 0001 0110 1000_2 = 168_16 d) (i) The fraction conversion procedure is exactly the opposite of the integer procedure .75_10 to octal .75 ×8 6.00 Take the digit 6, and save it; it's the first digit of the octal num-ber. Now take the decimal part .00 and again multiply it by 8. We get all 0's and hence 0.75_10 = 0.60_8 = 0.110_2 (ii) The binary conversion can be obtained directly or from converted octal numbers as shown above. (iii).75_10 = .110_2 and.110_2 = .1100_2 = .c_16 e) (i).390625_10 = X_8 Thus.390625_10= .31_8 .31_8 = .011 001_2 (ii).390625_10 = .011001_2 (iii).390625_10 = ?_16 Observe that the arrow connecting the 7.2, in the third line and the 3.2 in the eleventh line points out that after multiplying by 8 we shall have a repetition pattern of 1463. Hence, to convert .9_10 to X_8 , the precision must be known, say, up to the 9^th digit after the deci-mal point. Hence, .9_10 = (.7 1463 1463 ...)_8 . ii) The same will be the case for binary conversion. We convert to binary from the obtained octal number .9_10 = (.71463...)8 = (.111 001 100 110 011 ...)_2 For direct conversion iii) Now, to convert 0.9_10 into hexadecimal base, use the binary form already obtained. 0.9_10 = 0.1 1100 1100 1100 11... _2 = 0.1110 0110 0110 0110 ..._2 (grouping by four bits) = 0.E666 ...._16 = 0.E666_16 (up to 4_10 places after the point)

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Question:

A pitcher can throw a baseball weighing 5 ounces so that it will have a velocity of 96 feet per second. What is its kinetic energy?

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/Users/wenhuchen/Documents/Crawler/Physics/D05-0227.htm

Solution:

5 oz = 5/16 lb Therefore, since K. E. = (1/2) mv^2 = (1/2) (W/g)v^2 where W is the weight of the ball. This is equal to [5 lb × (96 ft/sec)^2] / [2 × 16 × (32 ft/sec^2)] = 45 ft-lb.

Question:

What is the difference between multiprogramming and time-- sharing?

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G01-0020.htm

Solution:

Multiprogramming involves more efficient utilization of CPU and I/O devices. This is accomplished by allowing mul-tiple programs that are ready for execution to reside in primary memory. While one program is doing something unrelated to CPU, such as input-output, another one may be running on CPU. This, a multi-programmed operating system, keeps track of programs (ac-tually internally defined program segments called processes or tasks), by assigning and releasing the CPU as a resource. When a program needs to do I/O it is suspended, and another program is assigned the CPU. When the suspended program finishes input- output, it is readied for another CPU assignment. This way, both the I/O devices and CPU are kept busy. Therefore, the efficiency and utilization of resources is higher. A side benefit of multi-programming, as opposed to say loading one program at a time and then running it, is better utilization of the primary memory since it tends to be occupied more. The aim of time-sharing is not efficiency or better re-source usage, but rather, it is being able to support more than one user at a time. Thus, a multi-user operating system uses time-sharing to support as many users as possible (without se-vere performance degradation). First, note that time-sharing requires multiprogramming and not the other way around. With time-sharing, in addition to the multiprogramming where a pro-cess is suspended when, say, it needs to do input-output; each process is suspended regularly upon completion of its time quota (or slice) . The time quota is small, typically in milliseconds, but not so small as to cause too much overhead in suspending and readying processes. It is then possible for multiple users (and their programs) to share the CPU by taking turns at quick time intervals in round robin fashion. Although the CPU operates sequentially by doing only one thing at a time (e.g. run this or that program at a given instant), the time slice is so small as to make the users feel as if they are simultaneously using the CPU. The idea of time-sharing then leads to the idea of "virtual machine." If, on the average, there are ten users on a fast CPU, each user can be regarded as running his/her programs on a com-puter (machine) ten times slower. Note that in a time- sharing system, how fast a program completes depends upon how many users are using the computer at the time. This may not be acceptable in situations where a program must complete within a specified period of real time. An example is the capture of data sent from a remote sensor device.

Question:

A simple pendulum consists of a mass m hung on the end of a string of length L. Find the natural frequency for small oscillations.

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/Users/wenhuchen/Documents/Crawler/Physics/D09-0363.htm

Solution:

We start by drawing a diagram of the forces acting on the mass m. The restoring force in the direction of motion is -mg sin \texttheta. Thus the equation of motion is ma = -mg sin \texttheta. Now we can suppose that \texttheta is small so that we can make the approximation sin \texttheta \cong \texttheta. This is accurate to better than 1 per cent for \texttheta = 15\textdegree and is better than 5 per cent for \texttheta = 30\textdegree. The displacement of the mass is given by the arc s. s = L\texttheta Thus v = ∆s/∆t = L∆\texttheta/∆t = L\omega where\omega = ∆\texttheta/∆t = angular velocity. And the acceleration a is given by a =∆v/∆t = L∆\omega /∆t = L\alpha where \alpha = ∆\omega/∆t = angular acceleration. Then finally the equation of motion reduces to mL\alpha = -mg\texttheta or\alpha + (g/L)\texttheta = 0 The angular acceleration \alpha is proportional to the negative of the angular displacement \texttheta, so that the motion is simple harmonic with natural angular frequency \omega_0 given by \alpha + \omega2_0\texttheta= \upsilon where\omega2_0= g/L

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Question:

A boy sits in a bus holding a balloon by a string. The bus accelerates forward. In which direction will the balloon move?

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/Users/wenhuchen/Documents/Crawler/Physics/D10-0397.htm

Solution:

Our first guess is that the balloon moves backward. However, the balloon actually moves forward. This occurs due to the pressure gradient created by the motion of the bus. When the bus is at rest, the air molecules undergo random motions. On the average the molecules remain in one position. As the bus accelerates, the back of the bus "collects" the air molecules. The front of the bus leaves the air molecules behind. The net result is an increase in air density at the back of the bus, and a decrease in air density at the front. Just as a balloon rises due to the greater pressure at the lower end of the balloon than at the top of the balloon, similarly the greater pressure at the back of the balloon will cause it to move forward.

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Question:

Describe the changes that take place in a larval amphibian which result in adaptation to life on land.

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/Users/wenhuchen/Documents/Crawler/Biology/F13-0336.htm

Solution:

In this problem, the changes that occur in the frog larva which result in the adult frog will be considered. The frog larva is a tadpole which undergoes metamorphosis to become an adult frog. The metamorphosis is a single-step process as opposed to the many molts that an arthropod larva undergoes in reaching the adult form. As in arthropod metamorphosis, amphibian metamorphic processes are hormonally controlled. Thyroxin, secreted by the thyroid gland, is the regulating hormone in amphibian metamorphosis. The tadpole is a completely aquatic organism and breathes via gills. It feeds on microscopic plants in the water. The tadpole also has a tail. During metamorphosis, the gills and gill slits are lost, the lungs develop, forelegs grow out of skin folds, the tail is retracted, the tongue, the eyelids and the tympanic membrane are formed, and the shape of the lens changes. The adult form of the organism, the frog, is a semi-aquatic animal which feeds on insects. The frog breathes by means of its lungs as well as through its thin, moist skin. The skin of amphipians functions as a respiratory organ. This is facilitated by the physical nature of the skin, which must remain moist in order to function as a res-piratory organ. If terrestrial conditions become very dry, the frog's thin, moist skin will be in danger of dessication, and hence, the frog will retreat to the aquatic environment.

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Question:

How much heat is produced in 5 minutes by an electric iron which draws 5 amperes from a 120-volt line?

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Solution:

Work is given in joules by W =Elt To convert this to units of heat (calories) we use the conversion factor of 0.239 [(calorie)/(joule)].This gives us H= [0.239 {(calorie)/(joule)}] EIt EIt E= 120 volts,I = 5 amperes t= 5 minutes = 300 seconds H = [0.239 {(calorie)/(joule)}] × 120 volts × 5 amperes × 300 seconds = 43,000 calories approximately.

Question:

Describe the zoned and packed formats of the binary coded decimal(BCD) representation of the characters and representthe decimal numbers 493 and - 493 in both formatsusing EBCDIC.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G06-0127.htm

Solution:

Each of the representations of the digits in the eight-bit character codesconsists of two fields: a four-bit zone field followed by a four-bit digit field. For digits in EBCDIC the zone field is represented by 1111. The digits4, 9, and 3 in decimal are 0100, 1001, and 0011 respectively. Thus, thenumber 493 in BCD represen-tation looks as follows: 0100 1001 0011, andin zoned BCD representa-tion: 1111 0100 1111 1001 1111 0011 . The equivalentsfor plus and minus signs in EBCDIC are 1100 and 1101 respectively. To indicate the sign of a number in the zoned format, the signequivalent is placed in the zone field of the rightmost digit. Thus, 111101001111100111000011 and 111101001111100111010011 represent+ 493 and -493 respectively in the zoned format (many systems alsoconsider a 1111 in the sign field to be plus, thereby making unsigned numberspositive). While the zone fields are necessary to represent the characters, theyare not necessary when numbers are being manipulated internally. The packed format leads to more efficient memory utilization. To obtain the packed representation from the zoned, two changes haveto be made: 1)allzone fields have to be removed 2)thesign field has to be transposed to the rightmost byte. Thus, 0100100100111100 and 0100100100111101 represent+ 493 and -493 respectively in the packed format.

Question:

Suppose pure line lima bean plants having green pods were crossed with pure line plants having yellow pods. If all the F_1 plants had green pods and were allowed to interbreed, 580 F_2 plants, 435 with green pods and 145 with yellow pods were obtained. Which characteristic is dominant and which is recessive? Of the F_2 plants, how many are homozygous recessive, homozygous dominant and heterozygous? Using G to represent the dominant gene and g to represent the recessive gene, write out a plan showing the segregation of genes from the parents to the F_2 plants.

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Solution:

This example is used to illustrate the basic concepts of genetics and the methods of solving genetic problems. First one must understand the common nomencla-ture used in the study of inheritance: nucleus which contain the hereditary units, the genes. b) gene - the part of a chromosome which codes for a certain hereditary trait. c) genotype - the genetic makeup of an organism, or the set of genes which it possesses. d) phenotype - the outward, visible expression of the hereditary constitution of an organism. e) homologous chromosomes - chromosomes bearing genes for the same characters. f) homozygote - an organism possessing an identical pair of alleles on homologous chromosomes for a given character or for all given characters. g) heterozygote - an organism possessing different alleles on homologous chromosomes for a given character or for all given characters. In solving genetic problems, one uses letters to represent the genotype of the organism. For example, "a" represents the gene for blue color and "A\textquotedblright represents the gene for red color. A capital letter is used for a dominant gene; that is, the phenotype of that gene will be expressed in a heterozygous state. For example, if the genotype Aa is expressed as red, then A is the dominant gene. Small letter a represents the recessive gene; that is, one whose phenotype will be expressed only in the homozygous state. Therefore, aa would be expressed as blue. To solve a genetic problem, one writes down the genotypes of the two parents in the cross. Mendel's First Law tells us what to do. This law, also known as the Law of Independent Segregation, states that genes, the units of heredity, exist in individuals in pairs. In the formation of sex cells or gametes, the two genes separate and pass independent of one another into different gametes, so that each gamete has one, and only one, member of each pair. During fertilization, the gamete of one parent fuses with that of the other parent. Fusion brings the genes from each parent together, giving rise to offspring with paired genes. Now we will illustrate the Law of Segregation as it applies to the problem given. Let G represent the gene for green pod color and g rep-resent the gene for yellow pod color. Since the parents come from pure lines, meaning that the two members of each gene pair are identical, we write the genotype of the parent plant with green pods as GG, and that of the parent plant with yellow pods as gg. Each gamete from the first parent will have one G and each gamete from the second will have one g. (Recall, the Law of Independent Segregati-on). Writing out a schematic cross, we have: The genotype of the F_1 generation is written as Gg because it results from the fusion of the two gametes. We are told that all the F_1 generation from the above cross are green, and we observe that they are all heterozygous. Therefore G, which stands for the gene for green pods, must be dominant (by definition), and g, which stands for the gene for yellow pods, must be reces-sive. To determine the possible genotypes of the F_1 plants, or the second generation offspring, we mate two F_2 plants. The possible gametes derived from the parents are again obtained using Mendel's First Law. But now we obtain two types of gametes, G and g, from each parent, because either gene can come from the parental genotype of Gg. Schematically, then: It is easier to determine the F_2 generation using the Punnet square. The square is constructed as follows: It gives all possible combinations of the parental gametes, so in F_2 we have genotypically: 1 GG:2 Gg:1 gg, Phenotypically: GG is homozygous dominant and green because G is dominant; Gg is hetorozygous and green because G is dominant; and gg is homozygous recessive and yellow because g is recessive. It is important to note that we cannot observe the genotype itself because it lies in the constitution of the gene. Our observations come only from what we actually see, that is, the phenotypic differences. We know from the Punnet square that in the F_2 gene-ration, the ratio of homozygous dominant to the entire progeny is (1/4); the ratio of heterozygous is 2/4 or (1/2) and the ratio of homozygous recessive is Therefore, the number of homozygous dominant (GG) plants is 1/4 × 580or145, the number of heterozygote (Gg) plants is 1/2 × 580or290 and the number of homozygous recessive (gg) plants is 1/4 × 580or145. When added together, they total 580, the number of F_2 plants.

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Question:

Explain how a solar cell is constructed and why it can generate electricity.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E30-0925.htm

Solution:

When certain materials are exposed to light, a current is induced in the material. The effect is called the photo-electric effect. These certain materials are called semiconductors. Using the photo-electric effect, solar cells convert sunlight directly into electricity. A solar cell consists of a solid electrode, attached to the back of a semiconducting material. A semitransparent metal-film electrode is attached to the front of the semi-conductor. In general, cadmium sulfide crystals serve as a semi-conductor in solar cells. A silver electrode is attached to one side of the cadmium sulfide and a thin layer of indium is placed on the opposite side. (See Figure 1.) When light strikes the cadmium sulfide through the indium electrode, electrons are released from cadmium sulfide. The released electrons flow through the external circuit and back to the silver electrode, which creates an electrical current. As long as the light hits the cadmium sulfide, electricity will flow.

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Question:

Find the indicated roots. (a) ^5\surd32 (b) \pm ^4\surd615 (c) ^4\surd\rule{1em}{1pt}125 (d) ^4\surd\rule{1em}{1pt}16.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E33-0954.htm

Solution:

The following two laws of exponents can be used to solve these problems: 1)(^n\surda)^n = (a^1/n)^n = a^1 = a, and 2) (^n\surda)^n = 2) (^n\surda)^n = ^n\surda^n. (a) ^5\surd32 = ^5\surd2^5 = (^5\surd2)^5 = 2. This result is true be-cause (2)^5 = 32, that is, 2 \textbullet 2 \textbullet 2 \textbullet 2 \textbullet 2 = 32. (b) ^4\surd625 = \rule{1em}{1pt} ^4\surd5^4 = (^4\surd5)^4 = 5. This result is true because (5^4) = 625, that is, 5 \textbullet 5 \textbullet 5 \textbullet 5 = 625. \rule{1em}{1pt}^4\surd625 = \rule{1em}{1pt}(^4\surd5^4) = \rule{1em}{1pt} [^4\surd5^4] = \rule{1em}{1pt} [5] = \rule{1em}{1pt} 5. This result is true because (\rule{1em}{1pt}5)^4 = 625, that is, (\rule{1em}{1pt}5) \textbullet (\rule{1em}{1pt}5) \textbullet (\rule{1em}{1pt}5) \textbullet (\rule{1em}{1pt}5) = \rule{1em}{1pt} 625. (c) ^3\surd\rule{1em}{1pt}125 = ^3\surd(\rule{1em}{1pt}5)^3 = (^3\surd\rule{1em}{1pt}5)^3 = \rule{1em}{1pt}5 . This result is true because (\rule{1em}{1pt}5)^3 = \rule{1em}{1pt}125, that is, (\rule{1em}{1pt}5) \textbullet (\rule{1em}{1pt}5) \textbullet (\rule{1em}{1pt}5) = \rule{1em}{1pt}125. (d) There is no solution to ^4\surd\rule{1em}{1pt}16 because any number raised to the fourth power is a positive number, that is, N^4 = (N) \textbullet (N) \textbullet (N) \textbullet (N) = a positive number \not = a negative number, \rule{1em}{1pt}16.

Question:

A walnut-combed rooster is mated to three hens. Hen A, which is walnut-combed, has offspring in ratio of 3 walnut : 1 rose. Hen B, which is pea-combed, has offspring in the ratio of 3 walnut : 3 pea : 1 rose : 1 single. Hen C, which is walnut-combed, has only walnut-combed off-spring. What are the genotypes of the rooster and the three hens?

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Solution:

This problem is an illustration of crosses in which there is gene product interaction. Interaction is usually indicated when the ratios observed in the offspring cannot be explained by simple dominant- recessive or codominant relationships, or linkage. The usual method of interaction is one in which one gene or gene pair masks the expression of another non-allelic gene or gene pair. This is called epistasis. Epistasis can occur in conjunction with dominant-recessive and/or codominant relationships, as it does in this problem. This can best be illustrated by looking at the different genotypes possible. Let R be the dominant allele for rose comb, and r its recessive allele, P be the dominant allele for pea comb and p its recessive allele. These four alleles interact in the following manner. The alleles R and P act codominantly to produce walnut comb. (They are called coepistatic alleles). Each is dominant over both recessive alleles. When neither dominant allele is present, the single comb trait is expressed by the recessive alleles. In summary: R-pp(RRpp; Rrpp)= rose comb rrP-(rrPP? rrPp)= pea comb R-P-(RRPP; RRPp; RrPP; RrPp)= walnut comb rrpp= single comb Let us look at each cross and see what they tell us about the possible genotypes of the rooster and hens involved. Then we can coordinate what we have learned from each cross to determine the exact phenotypes. In the cross between the rooster and Hen A (each walnut and R-P-): Since rose-combed offspring (R-pp) were produced, each parent must have had one recessive P gene to donate. This tells us that the rooster is R-Pp and the Hen R-Pp. In the cross between the rooster (now known to be R-Pp) and Hen B (pea-combed and therefore rrP-): Given F_13 R-P-:3 rrP-:1 R-pp:1 rrpp Walnutpearosesingle The fact that single-combed progeny resulted means that each parent must have both an r and a p to donate. Therefore the R-Pp rooster has to be RrPp and the rrP- hen has to be rrPp. In the cross between the rooster (RrPp) and hen C (walnut-combed and R-P-): Here, since only walnut-combed progeny resulted, hen c can carry no reccessive alleles. Otherwise pea (R-pp) or rose (rrP-) would have been produced Therefore then c is RRPP. We can now go back and solve the first cross. If the rooster is RrPp: If Hen A had an r gene at (-), then the progeny of the rrP- or pea-combed type would have been produced; but none were. So Hen A must be RRPp. Note that we also know that no single-combed progeny were produced. Therefore all 4 recessive alleles cannot be present in the total genotypes of both parents. Since we already have 3 recessive known alleles (-) must be R, and again, Hen A is RRPp. Summarizing: rooster is RrPp, hen Ais RRPp, hen Bis rrPp,and hen Cis RRPP. Doing the crosses verifies the results with the ratios obtained. 1 RRPP 2 RRPp6 walnut 1 RrPP 2 RrPp 1 RRpp2 rose 1 Rrpp Since 6/2 = 3/1, the ratio is 3 walnuts 1 rose. 1 RrPP 2 RrPp3 walnut 1 Rrpp1 rose 1 rrPp3 pea 2 rrPp 1 rrpp1 single The ratio is 3 walnut: 3 pea: 1 rose: 1 single. All the offspring are walnut.

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Question:

Protons of mass 1.67 × 10^-27 kg and moving with a velocity of 2 × 10^7 m \textbullet s^-1 strike a target of mass1 g and specific heat capacity 0.334 cal \textbullet g^-1 \textbullet C deg^-1. The proton stream corresponds to a current of 4.8\muA. At what rate does the temperatureof the target initially rise if one-third of the energy of the protons is converted into heat?

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Solution:

Each proton carries a charge of 1.60 × 10^-19 C. If the current I flowing is 4.8\muA, the number of protons striking the target in 1 s must be n, where I =nQ=(n × 1.60 × 10^-19 C) / (1 s) = 4.8 × 10^-6 A. n = 3.00 × 10^13 protons. In one second the total kinetic energy lost by the protons is KE = n × (1/2) m_pv^2 where (1/2)m_pv^2 is the kinetic energy of each proton. We are given that one-third of this energy is converted into heat in the target. If in one second the temperature rise of the target is t, the heat gained by the target is Q =mct, where c is the specific heat of the target (of mass m) and Q = 1/3 KE. Therefore,mct= (1/3) × (1/2) nm_pv^2 or t = (nm_pv^2) / (6 mc) = (3.00 × 10^13 × 1.67 × 10^-27 kg × 4 × 10^14 m^2 \bullet s^-2) /(6 ×lg × 4.18 J \bullet cal^-1 × 0.334 cal \bullet g^-1 \bullet C deg^-1) = 2.39\textdegreeC.

Question:

A small block of mass m slides along the frictionless loop- the-loop track shown in the figure, (a) If it starts from rest at P, what is the resultant force acting on it at Q? (b) At what height above the bottom of the loop should the block be released so that the force it exerts against the track at the top of the loop equals its weight?

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Solution:

(a) Point Q is at a height R above the ground. Thus, the difference in height between points P and Q is 4R, and the difference in gravitational potential energy of the block between these two points is: mgh_2 - mgh_1= mg(h_2 - h_1) = mg(4R) = 4 mgR Since the block starts from rest at P, its kinetic energy at Q is equal to its change in potential energy, 4 mgR; by the principle of conservation of energy (1/2)mv^2 = 4 mgR v^2 = 8 gR At Q, the only forces acting on the block are its weight, mg, acting downward, and the force N of the track on the block, acting in the radial direction. Since the block is moving in a circular path N = mv^2/R = (8 mgR)/R = 8 mg The loop must exert a force on the block equal to eight times the block's weight. (b) For the block to exert a force equal to its weight against the track at the top of the loop: (mv'^2)/R = 2 mg,v'^2 = 2gR This is the case because gravity exerts a downward force mg on the block. Thus, in order to keep the block moving in a circular path, the rest of the force (= mg) must be exerted by the loop-the-loop. Therefore: mgh = (1/2) mv'^2 h = (v'^2)/(2g) = (2gR)/(2g) = R h = (v'^2)/(2g) = (2gR)/(2g) = R The block must be released at a height R, above the top of the loop or 3R The block must be released at a height R, above the top of the loop or 3R above the bottom of the loop. above the bottom of the loop.

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Question:

Calculate the density of a block of wood which weighs 750 kgand has the dimensions 25 cm × 0.10 m × 50.0 m.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E01-0015.htm

Solution:

The density is a measure of weight per unit volume and is usually expressedin g/cc. Therefore, one must find the weight of this block in gramsand the volume in cubic centimeters. The density is then found by dividingthe weight by the volume. 1 kg = 1,000 g; therefore, 750 kg = 750 × 1,000 g = 7.5 × 10^5 g.To find the volume in cubic centimeters, allof the dimensions must be converted to centimeters first. 1 m = 100 cm; thus, .10 m = 10 cm and 50.0 m = 5,000 cm. Volume = 25 cm × 10 cm × 5,000 cm = 1.25 × 10^6 cc. Solving for the density: density= (weight / volume) = (7.5 × 10^5 g) / (1.25 ×10^6 ) = .60 g/cc.

Question:

Find the transpose of \mid12\mid M =\mid35\mid \mid711\mid Then find M^TM and MM^T using the APL inner product.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G14-0368.htm

Solution:

Matrix transposition is accomplished in APL by means of the monadic operator \textphi. If M is a matrix, \textphi M is its transpose. Assuming that the values of M have already been entered, its transpose is found by: N \leftarrow\textphiM N \mid137\mid \mid2511\mid When a matrix is transposed, the order of its rows and col-umns is reversed. To check this, apply the shape operator (\rho) to M and N: \rhoM 32 \rhoN 23 The inner product function is used in APL to find the pro-duct of two matrices. Its general form is A+. × B. A and B are matrices and + and × are primitive arithmetic func-tions, while is the period character. Two matrices A,B are conformable for multiplication if and only if the number of columns of A equals the number of rows of B. In the given problem, \rhoM= 3 × 2 and\rhoM^T= 2 × 3. Now MM^T means multiplying a (3 × 2) matrix by a (2 × 3) matrix. Since the inner numbers are equal, MM^T is conformable for multiplication. Similarly, M^TM gives (2 × 3) (3 × 2) and is also conformable for multipli-cation. The command N+. × m gives M^TM as 5994 94150 The command M+. × n gives MM^T 51329 133476 2976170.

Question:

On being heated in air, a mixture ofFeOand Fe_3O_4 picks up oxygen so as to be converted completely to Fe_3O_4. If the observed weight gain is 5.00 percent of the initial weight, what must have been the composition of the initial mixture?

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Solution:

One should first determine the weight percent increase when the mixture is 100 %FeOor 100 % Fe_3O_4. 5.0 % increase will be some mixture in between. 5 % will be equal to the sum of the products of the percent weight gained by each compound and the fractions of the mixture that each compound contributes. The reaction ofFeOand O to form Fe_2O_3 is 2FeO + 1/2 O_2 \ding{217} Fe_2O_3 . The percent weight increase for the reaction is found by dividing the weight of Fe_2O_3 by the weight of 2FeO. Weight of 2FeO is 144, weight of Fe_2O_3 is 160. (weight of Fe_3O_4) / (weight of 2FeO) = 160 / 144 = 1.1111 The percent weight increase is equal to (1.1111 - 1.0) × 100 = 11.11 %. This means that for each Fe_2O_3 formed byFeOthere is an 11.11 % weight increase. One can solve for the weight increase for Fe_3O_4 by a similar method. The reaction is 2 Fe_3O_4 + 1/2 O_2 + 3 Fe_2O3 The weight of 2 Fe_3O_4, is 464, the weight of 3Fe_3O_4 is 480. (weight of 3 Fe_3O_4) / (weight of 2 Fe_3O_4) = 480 / 464 = 1.0345 percent weight increase = (1.0345 - 1.0) × 100 = 3.45 %. Thus when Fe_2O_3 is formed from Fe_3O_4, there is a 3.45 % weight increase. The final mixture must have a 5 % weight increase. Let x = fraction of mixture that is FeO. Mixture weight gain = (fractionFeO) × (wt gain percent forFeO) + (fraction Fe_3O_4) × (wt. gain percent for Fe_3O_4) 5 = (x) (11.11) + (1 - x) (3.45) 5 = 11.11x+ 3.45 - 3.45 x 1.55 = 7.66 x 1.55 / 7.66= x .2024 = x .7976 = 1 - x The initial mixture is, therefore, 20.24 %FeOand 79.76 % Fe_3O_4.

Question:

Explain how initialization is done for array variables. Can the statementDCL A(8) FIXED(3) INIT(0) initialize the ele-ments ofthe array A?

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Solution:

In the case of an array, each element of the array must be explicitly initialized. For example, DCLA(8) FIXED(3) INIT((8)0); In the above statement ((8)0) means that all 8 elements are initialized to zero. The given statement DCLA(8) FIXED(3) INIT(O); cannot initialize allthe elements of the array A. It means that only the first element is initializedto 0. The remaining seven elements have no specified initial value. DCL A (2:6, 3:5)FLOAT(3) INIT((9),2); meansthat the first 9 elements are initialized to a value of 2 in the floating pointform. That is, the initial values of the first nine elements are stored in thecomputer memory as 0\textbullet200E+01. Now, the arrayA(2:6, 3:5)has a total numberof elements = (6 - (2 - 1)) × (5 - (3 - 1)) = (6 - 1) × (5 - 2) = 5 × 3 = 15 elements. This means that the remaining (15 - 9) = 6 elements have no ini-tial values specified. Moreover, if a ROW MAJOR ORDER is followed in numbering the arrayelements, as is usual, then, ElementsA(2,3),A(2,4),A(2,5), A(3,3),A(3,4),A(3,5), A(4,3),A(4,4),A(4,5), eachhave an initial value of 0\textbullet200E + 01. And, ElementsA(5,3),A(5,4),A(5,5), A(6,3),A(6,4),A(6,5), haveno initial values specified for them. Finally, consider the following array declaration: DCLB(2,3,2) FIXED(2) INIT((5)0,(4)1,(3)2); In the above, array B is a 2×3×2 = 12 element array. The ele-ments have a FIXED(2) attribute. The initial values of the first 5 elements are stored as 00; of the next 4 elements as 01; and of the last 3 elements as 02. That is, B(1,1,1),B(1,1,2),B(1,2,1),B(1,2,2) and B(1,3,1) have an initialvalue of 0. And, B(1,3,2),B(2,1,1), B.(2,1,2) and B (2,2,1) have an initial value of 1, while B(2,2,2),B(2,3,1) and B(2,3,2) have an initial value of 2.

Question:

Trace the path of the sperm from the testes to its union with the egg in the human.

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Solution:

In order to trace this path, we must first outline some aspects of the human male and female repro-ductive systems, illustrated in Figures 1 and 2, respec-tively. In normal males, the testes lie in the scrotum, a sac attached to the lower anterior wall of the abdomen. The testes reside in the body cavity during early embryonic development, but before birth they descend into the cavities of the scrotum. The inguinal canals are connections be-tween the scrotum and body cavity; these canals are blocked off by connective tissue after the testes descend. The testes are located in the scrotal sacs because the sperm within them require cooler temperatures than the internal body temperature in order to survive and mature. Each testis consists of roughly 1,000 coiled tubules called seminiferous tubules. It is in these tubules that the germ cells produce the sperm. Sertoli cells are also present and nourish the developing sperm. Connected to the seminiferous tubules via fine tubes called vasa efferentia is the epididymis, which is a single, complexly coiled tube in which sperm are stored and mature. The epididymis, which is derived from the embryonic kidney, empties into the vas deferens. This duct passes from the scrotum through the inguinal canal into the abdominal cav-ity, over the urainary bladder to a point where it opens into the ejaculatory duct, which empties into the urethra, the duct that leads from the urinary bladder to the exterior. Thus in the male, the urethra serves as a common passageway for both reproductive and excretory functions. The urethra passes through the penis and is flanked by three columns of erectile tissue. This tissue becomes enlarged with blood during peroids of sexual excitement, causing the penis to become erect. The engorgement of the penis is caused by arterial dilation and increased blood flow at unchanged arterial pressure. Prior to ejaculation, sperm from the testes pass through the vasa efferentia into the epididymis. During ejaculation, sperm from the epididymis are moved through the vas deferens by peristaltic contractions of its walls. Fluids are added to the sperm at the time of ejaculation. These fluids come from three pairs of glands; the seminal vesicles, the prostate glands (which in the human are fused into a single gland), and Cowper's bulbour-ethral glands, The mixture of secretions from these three sets of glands is termed seminal fluid. The semen consists of the sperm and seminal fluid. The seminal fluid consists of mucus (from seminal and bulbourethral secretions) and nutrients (seminal secretions), in a milky alkaline fluid (prostatic secretions). During copulation, the male's penis is inserted into the female vagina and the semen is released there. Sur-rounding the vagina are the labia majora, two folds of fatty tissue covered by skin richly endowed with hair and sebaceous glands. These folds extend dorsally and down, enclosing the openings of the urethra and the vagina and merging behind them. The labia minora - thin folds of tissue devoid of hair - lie within the folds of the labia majora and are usually concealed by them. At the ventral junction of these two is the clitoris, a sensitive, erectile organ, which is the major site of female sexual excitement. The external female sex organs are collectively known as the vulva. The vagina is a single, muscular tube which extends from the exterior to the uterus. From the vagina, the sperm swim, by motion of their flagella, through the cervix, the muscular ring of the uterus which projects into the vagina and pass into the uterus. From there, they enter the Fallopian tubes (also known as the oviducts) where one may fertilize the secondary oocyte, if it is present. If fertilization does occur, the oocyte completes the second meiotic division and the zygote (fertilized egg) is formed. Cleavage of the zygote begins in the Fallopian tube and will have proceeded to a multicellular state by the time the egg enters into the uterus and is implanted.

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Question:

A flexible wire 80 cm long has a mass of 0.40 gm. It is stretched across stops that are 50 cm apart by a force of 500 nt. Find the frequencies with which the wire may vi-brate.

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Solution:

The speed of a wave through a medium is determined by its elasticity and inertia. The elasticity is what causes a restoring force to act on any part of the medium displaced from its equilibrium position. The reaction of the displaced portion of the medium to the re-storing force depends on its inertia. for a stretched wire, the tension T is a measure of its elasticity; the greater the tension, the greater is the elastic restoring force on the displaced portion of wire. The inertia is measured by m, the mass per unit length of the wire. The speed of the wave has been found both analytically and experimentally to be, v = \surd(T/\mu) = \surd[(500 nt)/(0.40 × 10^\rule{1em}{1pt}3 kg/0.80 m)] = \surd[(10^6 m^2)/(sec^2)] = 1000m/sec The wire can only vibrate in a standing wave pattern, since it is stopped at two points (see figure). The incident wave is reflected at points A and B and rein-forces the later incident waves, producing the pattern shown. Hence, \lambda is related to L by, L = (n\lambda)/2(1) Hence, \lambda = (2L)/n Since the velocity of a sound wave is related to its fre-quency f and wavelength by v = f we obtain, (v/f) = (2L)/n or, f = (nv/2L) = [n(10^3 m/s)]/[(2) (50 × 10^\rule{1em}{1pt}2 m)] f = n × 10^3 s^\rule{1em}{1pt}1 Putting in values of n, we find, f_1 = 10^3 s^-1 f_2 = 2 × 10^3 s^\rule{1em}{1pt}1 f_3 = 3 × 10^3 s^-1 The other possible frequencies are the integral multiples of 1000 vibrations/sec, that is, 2000, 3000, 4000,...., vibrations/sec.

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Question:

Fats are not digested in the stomach even though there exists a lipase in the gastric fluid. Explain why.

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Solution:

The general formula of a fat or lipid may be written (RCO_2)_3C_3H_5, where the R's can be the same or different. It may be defined as an ester of glycerol and three long-chaincarboyxlicacids. These ester linkages may be hydrolyzed (degraded) by digestive enzymes called lipases. Such enzymes can function only under certain specific conditions. This is the reason why the lipase in gastric juice cannot digest lipids in the stomach. This lipase has an optimum pH of 7, a neutral solution. In the stomach, however, due to the presence ofHClfor other digestive purposes, there is an acidic environment. Thus, the lipase is virtually inactive under these conditions. As such, the fat will not be digested. But note, even if the lipase were active, lipid would still be undigested. To understand why, consider the following: Fats are insoluble in water, and, as such, tend to form globules in the body. Digestive enzymes cannot penetrate such globules. Only those fat molecules at the surface can be acted upon. The surface area can be increased, however, by breaking the globule into smaller droplets - a process called emulsification. This process occurs in the small intestine, not the stomach, when the fat is mixed with bile, anakalinefluid secreted by the liver and stored in the gall bladder.

Question:

Paper pulp (almost 100 % cellulose) is prepared commercially by digesting wood chips in a hot aqueous solution of calciumbisulfite, Ca(HSO_3)_2. The latter dissolves lignin and resins in the wood, leaving nearly pure cellulose. The sulfite solution is prepared by the following reactions: S + O_2\ding{217} SO_2 SO_2 + H_2O\ding{217} H_2SO_3 CaCO_3 + 2H_2SO_3\ding{217}Ca(HSO_3)_2 + CO_2 + H_2O (a) For every 250 kg of limestone (CaCO_3) used in the process , what weight of sulfur would be required? (b) What weight of Ca(HSO_3)_2 is produced from the limestone and sulfur in part (a)?

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Solution:

From thestoichiometryof the above equations, 1 mole of sulfur yields 1 mole of H_2SO_3 and 1 mole CaCO_3 uses 2 moles of H_2SO_3. Thus, 1 mole of CaCO_3 uses 2 moles of sulfur. The molecular weights of CaCO_3 and S are 100 g / mole and 32 g / mole, respectively. If 250 kg = 250,000 g of CaCO_3 are used then, the number of moles of CaCO_3 is [250,000 g CaCO_3 / (100 g / mole)] = 2500 moles For every 2500 moles CaCO_3 used in the process, twice that amount , or 5000 moles, of sulfur is needed. The weight of sulfur required is (5000moles) (32 g / mole) = 160,000 g = 160 kg. For every mole of CaCO_3 consumed, one mole ofCa(HSO_3)_2 is produced . Thus, if 2500 moles of CaCO_3 is consumed then, 2500molesof Ca( HSO_3)_2 are produced. To obtain the weight ofCa(HSO_3)_2 produced multiply its molecular weight (MW of Ca(HSO_3)_2 = 202 g / mole) by the number of moles produced. The weight ofCa(HSO_3)_2 produced is (2500moles) (202 g / mole) = 505,000 g = 505 kg.

Question:

You are given H, N, O, Net, Ca, Al, and Zn. Determine which of these atoms (in their ground state) are likely to be paramagnetic. Arrange these elements in the order of in-creasing paramagnetism.

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Solution:

Paramagnetic substances possess permanent magnetic moments. An electric current flowing through a wire produces a magnetic field around the wire. Magnetic fields are thus produced by the motion of charged particles. Then, a single spinning electron, in motion around the nucleus, she Id behave like a current flowing in a closed circuit of zero resistance and therefore should act as if it were a small bar magnet with a characteristic permanent magnetic moment. The magnetism of an isolated atom results from two kinds of motion: the orbital motion of the electron around the nucleus, and the spin of the electron around its axis. Two spin orientations are permitted for electrons, (+1/2) and (-1/2). Two electrons occupy each filled orbital and their opposing spins cancel out the magnetic moments, thus, for an atom to be paramagnetic it must contain unpaired electrons. H (Z = 1) : 1s^1 \rightarrow The subshell, s, has only 1 electron as indicated by the superscript number.In the s subshell, you have only 1 orbital. Each orbital can hold 2 electrons. Therefore, this electron is unpaired and H is paramagnetic. N (Z = 7) : 1s^2 2s^2 2p^3 \rightarrow The p subshell has 3 orbitals that contain a total of 3 electrons. Because electrons have the same charge, they try to avoid each other, if possible. Thus, each electron is in a different orbital. Therefore, they are unpaired and N is paramagnetic. O (Z = 8) : 1s^2 2s^2 2P^4 \rightarrow The p subshell has 3 orbitals with 4 electrons. Recalling the information given above, this means that 1 orbital has two electrons. The other 2 orbitals possess one electron each. Thus, there are two unpaired electrons. O is paramagnetic. Ne (Z = 10) : 1s^2 2s^2 2p^6 Here, all orbitals contain two electrons each. No electron is unpaired. Thus, Ne is not paramagnetic. Ca (Z = 20) : 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2. Again, all orbitals have two electrons. Therefore, calcium is not paramagnetic. Al (Z = 13) : ls^2 2s^2 2p^6 3s^2 3p^1 The 3p subshell has only 1 electron for three orbitals, It must be unpaired, as such. It is paramagnetic. Zn (Z = 30): 1s^2 2s^2 2p^6 3s^2 3p^6 3d^10 4s^2.Each orbital has two electrons. Thus, no paramagnetism exists. The order of paramagnetism (increasing) is proportional to the number of unpaired electrons. Thus, H, Al, O and N is the order of increasing paramagnetism.

Question:

What is meant by cellular metabolism? How does metabolismdiffer from anabolism and catabolism?

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Solution:

Cellular metabolism includes the following processes that transform substancesextracted from the environment: degradation, energy productionand biosyn-thesis. All heterotrophic organisms break down materialstaken from their environment, and utilize the product to synthesizenew macromolecules. When materials are broken down, energyis released or stored in the cell; when the products are used in syntheses, energy is expended. There are two general types of metabolism. That part of metabolismby which new macromolecules are synthesized with the consumptionof energy is termed anabolism (Greek,ana= up, as in build up). The degra-dation reactions, which decompose ingested material and releaseenergy, are collectively termed catabolism (Greek,cata= down, asin break down). The degradation of a glucose molecule to carbon dioxideand water during aerobic respiration is an example of catabolism. In the process, 38 molecules of ATP are formed for later use if needed. The degradation of fats is also an example of catabolism. The biosynthesisof proteins (from amino acids) and of carbohydrates like starchor glycogen, (from simple sugars) are two important anabolic processes.

Question:

Contrast the meanings of the terms "gene pool" and "genotype\textquotedblright.

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Solution:

A gene pool is the total genetic information possessed by all the reproductive members of a population of sexually reproducing organisms. As such, it comprises every gene that any organism in that population could possibly carry. The genotype is the genetic constitution of a given individual in a population. It includes only those alleles which that individual actually carries. In a normal diploid organism, there is a maximum of two alleles for any one given locus. In the gene pool, however, there can be any number of alleles for a given locus. For example, human blood type is determined by three alleles, I^A,l^B, and i . The gene pool contains copies of all three alleles, since all these are found throughout the entire population. Any given individual in the population, however, can have at the most two of the three alleles, the combination of which will determine his blood type.

Question:

An old-fashioned water clock consists of a circular cylinder 10 cm in diameter and 25 cm high with a vertical capillary tube 40 cm in length and 0.5 mm in diameter attached to the bottom. The viscosity of water is 0.01 poise. What is the distance between hour divisions at the top of the vessel and at the bottom of the vessel?

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Solution:

The total volume of liquid Q which flows across the entire cross section of a cylindrical tube in time t, is given by Poiseuille's law, Q = (\pi/8) [(R^4)/(\etaL)]∆p t where R is the radius, \eta is the viscosity of the liquid, and ∆p is the pressure difference between the two cross- sectional surfaces separated by a distance L. The water in the capillary flows as a result of the pressures due to the water in the cylinder and its own weight, as shown in the figure. Under its own weight, it would have a rate of flow given by, Q_1/t = (\pi/8) [(R^4)/(\etaL_2)] (\rhogL_2) where \rho is the density of water and g is the gravitational acceleration. The weight of the water exerts a pressure \rho gL, on the upper cross-section of the capillary and gives rise to another pressure difference between the two ends of the capillary. The rate of flow due to this pres-sure difference is, Q_2/t = (\pi/8) [(R^4)/(\etaL_2)] (\rhogL_1) The total rate of flow is therefore Q/t = Q_1/t + Q_2/t = (\pi/8) [(R^4)/(\etaL_2)] \rhog(L_1 + L_2) The quantity of water, Q, flowing from the capillary in time t causes a drop in the level of the cylinder, h. The area of the cylinder is A, and thus Q/t = Ah/t. \thereforeh/t = (\pi/8A) (R^4/\eta) [\rhog(L_1 + L_2)]/L_2 When the cylinder is full, L_1 = 25 cm, and h/t = [\pi × (0.25 × 10^-3m)^4 × 10^3 kg/m^3 × 9.8m/s^2 × 0.65m]/[8 × \pi/4 × 10^-2m^2 × 10^-3Ns/m^2 × 0.40m] = 3.11 x 10^-6 m/s = 1.12cm/hr When the cylinder is empty,L_1 = 0 cm, and, h'/t = [(\piR^4)/(8A\etaL_2)] \rhogL_2 = [(\piR^4)/(8A\etaL_2)] (L_1 + L_2) \bullet (L_2)/(L_1 + L_2) = (h/t) (L_2)/(L_1 + L_2) = (h/t) (0.40/0.65) = 1.12 cm/hr \bullet(0.40/0.65) = 0.69 cm/hr Thus hour divisions are separated by 1.12 cm at the top and 0.69 cm at the bottom. Note that L_1 varies slightly during the hour and, to be quite exact, an integration ought to be performed. The error involved is, however, slight, since the variation in L_1 is very small in compari-son with L_1 + L_2.

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Question:

(a) (z = 11) 1s^2 2s^2 2p^6, (b) (z = 25) 1s^2 2s^2 2p^6 3s^2 3p^6 3d^5, and (c) (z = 29) 1s^2 2s^2 2p^6 3s^2 3p^6 3d^10; provide the proper ionic symbol.

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Solution:

The ionic symbol of an atom is equal to the charge on the atom. This charge is determined by comparing the atomic number Z and the number of electrons as shown in the electronic configuration. The atomic number corre-sponds to the net positive charge on the nucleus and the number of electrons indicates magnitude of the negative charge of the electron cloud. One writes the electronic configuration of hydrogen as 1s^1 1s, indicates the atomic orbital and the superscript 1 indicate that there is one electron in the orbital. Thus, one can determine the number of electrons present by taking the sum of the superscripts. The net charge on an atom is found by adding the net negative charge (the sum of the electrons) and the net positive charge (the atomic number, which is equal to the number of protons) For hydrogen, (Z = 1) 1s^1, the net negative charge is - 1 and the net positive charge is + 1, thus, the atom is neutral and no ionic symbol is used. One uses this method to determine the ionic symbols for the atoms de-scribed in the problem. (a)(Z = 11) 1s^2 2s^2 2p^6. The net negative charge is equal to (- 2) + (- 2) + (- 6) or - 10. The net positive charge is 11. The net charge on the atom is 11 - 10 = + 1. From the periodic table one sees that this atom is Na^+, because the atomic number of sodium is 11. (b)(Z = 25) 1s^2 2s^2 2p^6 3s^2 3p^6 3d^5. Total number of electrons in configuration is2 + 2 + 6 + 2 + 6 + 5 = 23. The net charge on the atom is, then 25 - 23 or + 2. From the periodic table one can determine that this atom is Mn^+2. (c)(Z = 29) 1s^2 2s^2 2p^6 3s^2 3p^6 3d^10. Total number of electrons in configuration = 2 + 2 + 6 + 2 + 6 + 10 = 28. Z = 29 = number of electrons in a neutral atom. Differ-ence:29 - 28 = + 1. The atom is Cu^+1.

Question:

The volume of plasma and other extracellular fluids are regulated by automatic feedback controls. What are these controls?

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Solution:

The mechanisms responsible for the re-gulation-of body fluid volume depend on the fact that an increase in blood volume causes an increase in blood pressure. One such mechanism operates in the capillaries of all tissues. Here, increased blood volume, because it increases blood pressure, tends to increase the pressure within the capillaries. When this occurs, the pressure that tends to move fluid out of the capillaries into the interstitial space becomes greater than the colloid osmotic pressure, the force which tends to move fluid from theinterstitium back into the capillaries (see previous solution). What results is a net movement of fluid out of the capillaries. Eventually, this movement causes a reduction of blood volume, which in turn reduces the blood pressure. Capillary pressure decreases, and there is a net movement of fluid from the extracellular space back into the capillaries. This is due both to the decreased capillary pressure and to a rise in the colloid osmotic pressure in the blood caused by the drop in plasma volume. Fluid thus enters the capil-laries, restoring normal blood volume and normal blood pressure. A second process stabilizing blood volume occurs at the glomerulus-capsule junction. As blood volume increases, the pressure within theglomeruliof the kidneys likewise increases. Filtration pressure, being proportional to theglomerularcapillary pressure, rises also and more urine is subsequently formed. This reduces the fluid volume of the blood, thus decreasing the blood pressure. A third fluid volume regulating process is base on the presence of baroreceptorsin the walls of the arteries in the chest and neck. Increased blood volume exerts a stronger pressure on the arteries, stretching them. Thebaroreceptorsrespond to this stretching of the arterial walls and send impulses to the vasomotor center in the brain. From here, impulses are sent to the smooth muscle in the walls of the afferent renal arterioles, causing them to relax. This relaxation allows the arterioles to dilate and increase the flow of blood into theglomeruli; more blood will be filtered and more water eliminated. By this autonomic feedback mechanism the blood volume can then be restored to normal.

Question:

Persons with kidney trouble usually also have high blood pressure. Explain why. What condition can result if the blood pressure is not restored to normal?

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Solution:

Any kidney disorder which interferes with normal filtration invariably affects blood pressure. Blood pressure is directly related to blood volume; an increase in blood volume will raise blood pressure, and a decrease will lower blood pressure. If for some reason filtration were drastically reduced, the amount of water excreted from the body would drop. This would result in more water retention by the body, increasing both blood volume and blood pressure. Conversely, should filtration be in-creased drastically, water loss from the body would in-crease, blood volume would decrease, and blood pressure would fall. A condition where filtration is increased is less common then a condition where it is decreased. For example, a common disorder in the vasomotor center causes constriction of the renal arterioles, leading to reduced filtration. An increase in blood pressure causes an increase in capillary pressure, favoring movement of material out of the capillary into the surrounding tissue. If the blood pressure remains high, the lymph vessels which normally drain the tissues of excess fluid, become in-effective and fluid begins to accumulate in the surround-ing tissue. This condition is known as edema. Edema may also result from a dietary deficiency of protein. Under such a condition, the colloid osmotic pressure due to protein decreases, and the force holding fluid in the capillaries is reduced. Fluid from the blood diffuses out of the capillaries into the extracellular spaces in tissues, and swelling of the tissues occurs.

Question:

100 ml. of gas are enclosed in a cylinder under a pressure of 760 Torr. What would the volume of the same gas be at a pressure of 1520 Torr?

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Solution:

Since this problem deals with the pressure and volume of a gas at a constant temperature, Boyle's law can be used. Boyle's law states that the volume, V, of a given mass of gas, at constant temperature, varies inversely with the pressure, P. It can be stated as V = k × 1/p, where k is a constant. Hence,k = PV For a particular system, at constant temperature, k is constant. Therefore, if either the pressure or the volume is changed, the other must adjust accordingly. Here,P = 760 TorrandV = 100 m1so k = 760 × 100 = 76000Torr-m1. If P is doubled to 1520 Torr, then k = 76000 Torr-ml = 1520 Torr × V v = [(76000 Torr-ml)/(1520 Torr)] = 50 m Since k is a constant for a given system, another form of Boyle's Law can be expressed as P_1 V_1 = P_2 V_2 This says that the pressure of the original system times the volume of the original system is equal to the new pressure times the new volume. Here, 760 torr × 100 m1 = 1520 torr × V_2 50 m1 = V_2.

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Question:

4x^2 - 7 = 0.

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Solution:

This quadratic equation can be solved for x using the quadratic formula, which applies to equations in the form ax^2 + bx + c = 0 (in our equation b = 0). There is, however, an easier method that we can use: Adding 7 to both members,4x^2 = 7 dividing both sides by 4,x^2 = (7/4) taking the square root of both sides, x = \pm \surd(7/4) = \pm \surd(7/4). The double sign \pm (read "plus or minus") indicates that the two roots of the equation are + (\surd7/4) and \rule{1em}{1pt} (\surd7/4).

Question:

Lithium oxide (Li_2 O, molecular weight = 30 g/mole) re-acts with water (H_2 O, molecular weight = 18 g/mole, density =1.0 g/cm^3) to produce lithium hydroxide (LiOH) according to the following reaction: Li_2 O + H_2 O + 2LiOH. What mass of Li_2 O is required to completely react with 24 liters of H_2 O?

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Solution:

One mole of Li_2 O reacts with exactly one mole of H_2 O, as indicated by the coefficients in the reaction. Hence, if we determine the number of moles of H_2 O in 24 liters, we then know the required number of moles of Li_2 O, and, from this, can calculate the mass of Li_2 O that we need. The number of moles of water is moles = mass/molecular weight. But mass = density × volume, so that the number of moles of water is moles = [(mass) / (molecular weight) = [(density × volume) / (molecular weight)] = [(1.0 g/cm^3 × 24 liters) / (18 g/mole)] = [(1.0 g/cm^3 × 24000 cm^3 ) / (18 g/mole)] = 1333.33 moles Thus, we require 1333.33 moles of Li_2 O to completely react with 24 liters of H_2 O (= 1333.33 moles). Multiplying this by the molecular weight of Li_2 O, we determine that 1333.33 moles × 30 g/mole = 40,000 g = 40 kg of Li_2 O is needed.

Question:

When a metal cylinder of height 14 cm. which is floating upright in mercury is set into vertical oscillation, the period of the motion is found to be .56 s. What is the density of the metal? The density of mercury is 13,600 kg \textbullet m^-3 and g is \pi^2 m \textbullet s^-2.

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Solution:

Let the cylinder have a cross-sectional area A, length 1, and density \rho , and let it float in the mercury of density \rho' immersed to a height y. In this position, shown by the solid line in the diagram, the cylinder is in equilibrium. Hence, the net force on the cylinder, composed of its weight (F_G) and the buoyant force (F_B) is zero. Then, taking the positive direction downward, F_G - F_B = 0 orF_G = F_B(1) ButF_G = m_c g where m_c is the cylinder mass. Since \rho = m_c / [Volume of cylinder (V)] m_c = \rhoV = \rhoAl andF_G = \rhoAlg The buoyant force is equal to the weight of fluid displaced by the object. Hence F_B = m_mg where m_m is the mass of mercury displaced. But m_m = \rho' V' where V' is the fluid volume displaced. Hence, m_m = \rho' A(l - y) andF_B = \rho' A(l - y) g(3) Using (3) and (2) in (1) \rhoAlg = \rho' Ayg If the cylinder is pushed in a further distance x, the up thrust is greater than the weight, and there is a restoring force attempting to return the cylinder to its original position. If a is the downward acceleration, we find, from Newton's Second Law F_net = m_ca Again,F_net = F_G - F_B = m_ca or\rhoAlg - \rho'A(y + x)g = \rhoAla Since\rhoAlg = \rho'Ayg -\rho' Axg = \rhoAla anda = - (\rho'g)/(\rhol) x Comparing this with the equation of motion of a simple harmonic oscillator, we realize that \omega^2 = (\rho'g)/(\rhol) since T = 2\pi/\omega, where T and \omega are the period and angular frequency of the motion, we note that \omega = 2\pi/T and(4\pi^2)/(T^2) = (\rho'g)/(\rhol) \rho = (\rho'gT^2)/(4\pi^2l) = [(13,600 kg \bullet m^-3) (\pi^2 m \bullet s^-2)(.56)^2 s^2]/[(4\pi^2) (.14 m) \rho = 7616 kg \textbullet m^-3.

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Question:

Calculate the uncertainty in the location of a 100 g turtle crossing a road at a velocity of 1 mm/sec, the uncertainty of the latter being 10^-4 mm/sec. (h = 6.626 × 10^-27 erg-sec and 10 mm = 1 cm.)

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Solution:

This problem demands the use of the Heisenberg Uncertainty Principle. This principle states that one can-not simultaneously locate both the position of a particle and its momentum with absolute precision. Whenever one quantity, either position or momentum, is known precisely, the other is known less precisely. Heisenberg showed that the uncertainty in the determination of momentum of a particle, ∆p, and the simultaneous determination of its location, ∆x is related by the equation ∆p∆x\geq h, where h is Planck's constant. Momentum p is equal tomv, where m is mass and v the velocity of the mass, thus ∆p =m∆v. One substitutes this into the Heisenberg inequality attainingm∆v∆x\geq h. One can now solve for ∆x. 1 erg = g cm^2/sec^2, 1 mm = .1 cm. ∆x = h/(m∆v) = [(6.626 × 10^-27 erg-sec) / {(100 g) (10^-4 mm/sec)}] \bullet [(1mm)/(.1 cm)] = 6.626 × 10^-24 cm.

Question:

Water flows into a tank at a constant rate Q. Q is determined by the position of a valve connected to the inflow pipe. Write a digital computer program to simulate this system from time t = 0 to t = t_f if V(O) = V_O . (Consider the walls to be of infinite height, so that overflow is not possible.)

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Solution:

Following the steps outlined in the previous problems: 1.Changes in this system's state can be viewed through changes in the volume V of water in the tank. 2 & 3. Once the valve position is set, Q is determined and will remain unchanged. Thus, it can be considered external to the system consisting of water in the tank and water flowing into the tank. Q is changed only when a new simulation run is being made with a new parameter Q. 4. Using the notation of the preceding problems, the behavior of the system can be diagrammed as shown in Fig. 2 [Q] \rightarrow [V] Fig. 2 V is the accumulation of the rate of change Q (when Q has units of Volume/time). To conform more closely with notation previously developed, let V̇ = the constant Q, and since accumulation of rates of change is mathematically expressed as integration, the complete diagram is shown in Fig. 3 From this, the equation can be written: V̇ = Q or dV/dt = Q(1) 5. Since the solution of (1) can easily be seen to be: V(t) = Q {_\ast} T + V_0(2) ȧ simple program would just substitute values of t into (2) and output V. A more general program would use the Euler predictor-corrector algorithm to solve the first-order differential equation (1). But since the rate of change is constant, the corrected value would always equal the predicted value. The method used here directly simulates the behavior of the system according to observed changes in V for small, discrete changes in t: V_new= V_old + V_added(3) whereVadded= Q {_\ast} change in time = Q {_\ast} (t_new - t_old) = Q {_\ast} ∆t This is the original Euler's method (generally, it is less accurate than This is the original Euler's method (generally, it is less accurate than predictor-corrector method).Thus V_Nis calculated from V_N-1. If V_0 is predictor-corrector method). is calculated from V_N-1. If V_0 is always used for Voldand t_0 = 0 is always used for t_old equation (3) would be reduced to equation (2). The program is as follows: DATA T/0.0/ READ N, TFIN,Q,VOL PRINT T, VOL REALN = N CUSING REALN AVOIDS MIXED-MODE ARITHMETIC DT = (TFIN - T)/REALN DO 50 I = 1,N T = T + DT VOL = VOL + Q {_\ast} DT PRINT T, VOL 50CONTINUE STOP END

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Question:

What mutually beneficial adaptations have arisen in the flowering plants and certain insects?

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Solution:

In the terrestrial environment, insects are of great ecological significance. Insects are crucial for the fertilization of many different species of angiosperms, the flowering plants. Approximately two-thirds of all angiosperms depend upon insects for pollination. Bees, wasps, butterflies, moths and flies are the principal insect pollinators. The three orders represented by these insects have an evolutionary history closely tied to that of flowering plants: both groups underwent an explosive period of development in the Cretaceous period, and both have evolved mechanisms or structures which promote insect pollination. Many insects rely on the nectar produced by plants for food. As the insect sucks up the nectar, the sticky pollen grains adhere to its feet and body. When the insect flies to another plant, the pollen grains stick to the new flower and fertilize the egg. A tremendous amount of adaptation between the flower-ing plants and the insects has occurred. The color, odor, and nectar of flowers all serve to attract insects. As an example, bees are attracted by bright blue arid yellow colors, and by sweet, aromatic, or minty odors. The flowers pollinated by bees have these characteristics, and often have a protruding lip on which the bee can land. Flowers pollinated by the wind, on the other hand, usually lack conspicuous colors, odors, and nectar, and the pollen is light, not sticky. The mouthparts of different insects also show advanced specialization. For example, butter-flies and moths suck liquid nectar, and have mouthparts modified to form a long tube. Bees, which gather both nectar and pollen for food, have mouthparts which are adapted for both sucking and chewing. A truly interesting adaptation for insect pollination is shown in some flowers with floral parts which evolved to resemble female insects of a species. The male insects are attracted to these flowers. The pollen grains are transferred to the bodies of the insects as they attempt to copulate with the insect- resembling flowers. Thus insects help to insure cross-pollination of species of flowers. Some insects eat the fruits of angiosperms and eliminate theundigestible seeds in their waste. This is of significant advantage to the flowering plants in that their seeds can be dispersed over long distances by insects and new environments can be exploited.

Question:

Discuss the problem of the 'coding' of information in the gene. The Genetic code First position (5' end) Second position Third position (3' end) U C A G U U Phe Phe Leu leu C Ser Ser Ser Ser A Tyr Tyr Terminator Terminator G Cys Cys Terminator trp C U Leu Leu Leu Leu C Pro Pro Pro Pro A His His Glu NH_2 Glu NH_2 G Arg Arg Arg Arg A U IIeu IIeu IIeu Met C Thr Thr Thr Thr A Asp NH_2 Asp NH_2 Lys Lys G Ser Ser Arg Arg G U Val Val Val Val C Ala Ala Ala Ala A Asp Asp Asp Asp G Gly Gly Gly Gly

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Solution:

Although the knowledge that genes consist of nucleotide sequences was a major breakthrough in the study of genetics, some new problems arose out of this knowledge. One of the problems was to find out how many nucleotides are needed to code for one amino acid. We know that there are four different types of nucleotides because there are four different nitrogenous bases. We also know that there are 20 kinds of amino acids. Four nucleotides taken two at a time provide only 16 different combinations (4^2 = 16), which are insufficient to code for the 20 different amino acids. Four nucleotides taken three at a time provide 64 different combinations (4^3 = 64). At first glance, this would seem to provide many more terms than are needed, since there are only 20 different amino acids. It was believed at one time that the excess combinations were not used to specify any amino acids. However, there is now strong evidence that all but three of the 64 combinations do, in fact, code for amino acids, and that as many as six different nucleotide triplets may specify the same amino acid. The term degeneracy is used to describe the fact that a given amino acid may be specified by more than onecodon. The fundamental characteristics of the genetic code are now well established: it is a triplet code, with three adjacent nucleotides, termed acodon, specifying each amino acid. A second problem with the genetic code is whether there is overlapping or not. For example, is the sequence CAGAUAGAC read only as CAG, AUA, GAC or can it also be read CAG, AGA, GAU, AUA, UAG, AGA, GAC? Is each nucleotide part of onecodonor three? The amino acid sequences of several mutant forms of the hemoglobin molecule have been analyzed. In each, only a single amino acid in the molecule is sub-stituted. In contrast, if the code were overlapping and a given nucleotide were part of three adjacentcodons, we would expect three adjacent amino acids to be changed. For example, a single substitution (circled) in the sequence CAG A UA GAC, resulting in CAGCUAGAC, would affect only the amino acid specified by AUA if no overlapping occurred. If there were overlapping, such a change would affect 3 amino acids, namely those coded for by AGA, GAU, and AUA. Experi-ments with syntheticpolynucleotideshaving known base sequences have shown conclusively that the code is not over-lapping. One can synthesize a messenger RNA strand con-taining onlyuracil, UUUUU... (polyuridylic acid). For every 3 bases added to the mRNA, only one additional phenylalanine will be incorporated into its peptide chain (the UUUcodon codes for phenylalanine). Finally, the code iscommaless, i.e., no punctuation is necessary, since the code is read beginning at a fixed point, three nucleotides at a time, until the reading mechanism comes to a specific 'termination'codon, which signals the end of the message. The fact that there is punctuation betweencodonsbecomes important in deletion or insertion mutations. Adding or subtracting a nucleotide within a sequence ofcodonsthrows the entire reading frame and can change out of line everycodonpast the point of insertion or deletion. For example, the gene sequence AGAUCUUGG would normally be read as AGA, UCU, UGG and would code for the amino acidsarginine, serine, and tryptophan. The insertion of an extra nucleotide (circled), would result in the sequence AGAU G CUUGG, which would read AGA, UGC, UUG and code forarginine,cysteine, andleucine. The deletion of a nucleotide would have a similar effect. If C were eliminated, the code would read AGA, UUU, and would code forarginine and phenylalanine.

Question:

Calculate the pH of a 0.25 M solution of the salt ethyl-amine hydrochloride, C_2H_5NH_3CI. The dissociation constant for the base, ethylamine (C_2H_5NH_2) is K_b = 5.6 × 10^-4.

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Solution:

The pH will be determined by three processes occurring simultaneously; dissociation of ethylamine, ionization of ethylamine hydrochloride, and hydrolysis of theethylammoniumcation, C_2H_5NH_3^+. All three species are present in an aqueous solution of the salt. Ethylamine hydrochloride ionizes according to the equation C_2H_5NH_3Cl \rightleftarrows C_2H_5NH_3^+ +Cl^-, giving rise to theethylammoniumcation. Theethylammoniumcation hydrolysis according to the equation C_2H_5NH_3^+ + H_2O \rightleftarrows H_3O^+ + C_2H_5NH_2, forming the ethylamine base. Ethylamine base dissociates according to C_2H_5NH_2 + H_2O \rightleftarrows C_2H_5NH_3^+ + OH^-. LetK_wdenote the water constant,K_w= [H_3O^+] [OH^-] = 10^-14, and K_b the dissociation constant of ethylamine, K_b = {[C_2H_5NH_3^+] [OH^-]} / [C_2H_5NH_2] = 5.6 × 10^-4. Then, K _w / K_b = {[H_3O^+] [OH^-]} / {[(C_2H_5NH_3^+) (OH^-)] / [C_2H_5NH_2]} = {[H_3O^+] [C_2H_5NH_2]} / {[C_2H_5NH_3^+]} (10^-14) / (5.6 × 10^-4) = 1.8 × 10^-11 = {[H_3O^+] [C_2H_5NH_2]} / {[C_2H_5NH_3^+]} Note that this is theequlibriumconstant for hydrolysis of the ethylarnmoniumcation. Let x = [C_2H_5NH_2] be the number of moles per liter of ethylamine formed by hydrolysis of theethylarnmoniumcation. Then [H_3O^+] = x, since the balanced equation for hydrolysis states that for every mole of ethylamine formed, one mole of H_3O^+ is formed and one mole of ethylammoniumcationis consumed. If we assume that ethylamine hydrochloride dis-sociates completely, then the number of moles per liter of ethylarnmoniumchloride not hydrolyzed is 0.25 M (from the salt) minus the number of moles per liter which is hydrolyzed (x), so that [C_2H_5NH_3^+] = 0.25 - x. Hence, 1.8 × 10^-11 = {[H_3O^+] [C_2H_5NH_2]} / {[C_2H_5NH_3^+]} = (x \bullet x) / (0.25 - x). Assuming that the number of moles hydrolyzed is much less than 0.25 M, Assuming that the number of moles hydrolyzed is much less than 0.25 M, then 0.25 - x \allequal 0.25 and the algebra simplifies to 1.8 × 10^-11 = (x \bullet x) / (0.25) = x^2 / 0.25 , or x = (0.25 × 1.8 × 10^-11)^1/2 = 2.1 × 10^-6 (which is much smaller than 0.25, justifying our assumption). But x = [H_3O^+] = 2.1 × 10^-6, hence pH = - log [H_3O^+] = - log 2.1 × 10^-6 = 5.7. The pH of 0.25 M ethylamine hydrochloride is thus 5.7.

Question:

Chlorine may be prepared by the action of KClO_3 on HCl, and the reaction may be represented by the equation: KClO_3 + 6HCl \rightarrow KCl + 3Cl_2 + 3H_2 O Calculate the weight of KClO_3 which would be required to produce 1.0 liter of Cl_2 gas at STP. R = .082 liter-atm/mole -\textdegreeK.

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Solution:

From the equation one can see that 1 mole of KClO_3 reacts to form 3 moles of Cl_2. If one can find the number of moles of KClO_3, which will react to form 1 liter of C1_2, then one can find its weight by multiplying the number of moles by the molecular weight of KClO_3. STP (Standard Temperature and Pressure) is defined as 0\textdegreeC and 1 atm. One can find the number of moles of Cl_2 in 1 liter by using the Ideal Gas Law n = (PV/RT) where P is the pressure, V is the volume, R is the gas constant (0.082 liter-atm/mole-\textdegreeK), T is the absolute temperature and n is the number of moles. Here, T is given in \textdegreeC? it can be converted to \textdegreeK by adding 273 to it. You have, then, T = 0 + 273 = 273\textdegreeK. One can now solve for the number of moles of C1_2 produced. n = (PV/RT)P = 1 atm pv V = 1 liter R = 0.082 [(liter-atm) / (mole-\textdegreeK)] T = 273 n = [{1 atm × 1 liter} / {0.082 (liter-atm) / (mole-\textdegreeK) × 273\textdegreeK}] = .045 moles. From the equation for the reaction, the following ratio is determined. (3/1) = [(number of moles of Cl_2 ) / (number of moles of KClO_3 )] One knows that 1 liter contains .045 moles Cl_2 and if you substitute this value into the above ratio, one can find the number of moles of KClO_3 reacted. (3/1) = [(.045 moles) / (number of moles of KClO_3 )] number of moles of KClO_3 = [.045 moles × 1) / (3)] = .015 moles The weight of .015 moles of KClO_3 can be found by multiplying .015 moles by the molecular weight of KClO_3. (MW of KClO_3 = 122.5.) weight of KClO_3 = 122.5 g/mole × .015 moles = 1.84 g.

Question:

Calculate the standard E\textdegree, electrode potential, of the hydrogen- oxygen fuel cell. The standard Gibbs free energy of formation (∆G^0_f)of H_2O(l) = -56,690 cal/mole.

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Solution:

The standard Gibbs free energy of formation (∆G^0_f) is a measure of the energy available to do work when compounds are formed from their elements in the standard state. The electrode potential, i.e., the voltage across the two electrodes in a galvanic cell, such as the hydrogen-oxygen cell, can be computed from ∆G^0_f , by using the equation ∆G^0= -nF E\textdegree , where n = number of electrons transferred and F = a faraday of electricity, which is 23,060 cal v^-1 mole^-1. In a hydrogen-oxygen fuel cell, water is produced from the transfer of 2 electrons. Thus, n = 2. Solving for E\textdegree: -56,690 (cal / mole) = -2(23,060 cal v^-1 mole^-1) E\textdegree . E\textdegree = 1.229 v for this fuel cell.

Question:

What is the maximum speed at which a car can safely round a circular curve of radius 160 ft on a horizontal road if the coefficient of static friction between tires and road is 0.8? The width between the wheels is 5 ft and the center of gravity of the car is 2 ft above the road. Will the car overturn or skid if it just exceeds this speed?

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Solution:

The magnitude of the maximum frictional force that can be brought into play between tires and road is F = \muN, where N^\ding{217} is the normal force exerted by the road on the car and \mu is the coefficient of static friction. But, since there is no upward movement of the car, N^\ding{217} just balances the third force acting on the car, the weight mg^\ding{217}. Hence F = \mumg. This must provide the centripetal force necessary to keep the car in the curve of radius r when it is moving with the maximum permissible speed v. Hence, \mumg = mv^2/r v = \surd(\murg) = \surd(0.8 × 160 ft × 32 ft \bullet s^-2) v = 65 ft \bullet s^-1. The frictional force F^\ding{217} acts in the plane of the road surface and not through the center of mass of the car. In addition to providing the centripetal force necessary to keep the car in the curve, the frictional force must therefore produce a rotational motion about the center of mass. The only point of contact between car and road will then be at O in the diagram. Therefore, N^\ding{217} must act through this point; but the weight of the car of magnitude mg = N still acts through the center of gravity. These two parallel but displaced forces form a couple of positive moment, tending to restore all car wheels to the road and to prevent the overturning. The moment of the frictional force is -\muN = - mv^2/r multiplied by the height of the center of mass above the road. Thus M_1 = -\muN × 2 ft = -\mumg × 2 ft. Assuming that the center of gravity of the car is centrally located, the moment of normal force is N = mg, multiplied by half the width between the wheels. Thus M_2 = mg × 2.5 ft. \therefore M_2 + M_1 = mg × 2.5 ft - 0.8 × mg × 2 ft = mg × 0.9 ft. Since this is positive, \vertM_2\vert >\vertM_1\vert. The restoring moment is therefore greater than the overturning moment at the maximum speed. If this speed is just exceeded, the car does not overturn. It skids, since the centripetal force is not now great enough to provide the acceleration necessary to keep it going round the curve, and the overturning moment is less than the restoring moment.

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Question:

Calculate (a) the number of collisions per second per molecule, (b) the number of collisions per cubic meter per second, and (c) the number of moles of collisions per liter per second of oxygen at a temperature of 25\textdegreeC and pressure of 1 atm. The molecular diameter of oxygen is 3.61×10^-10m.

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Solution:

Molecules of real gas attract one another at large distances and repel one another at very short distances. Assuming that molecules are rigid, noninter-acting spheres with diameter, \sigma, one can drive an approximate equation for the number of collisions per second. It is also assumed that all the molecules travel with the same speed, the arithmetic mean velocityv. If two identical molecules just touch each other, then the distance separating their centers is the molecular diameter \sigma. Thus, a moving molecule collides with other molecules whose centers come within a distance of \sigma. (Hence, the effective collision radius is equal to \sigma, see figure a.) The quantity \pi\sigma^2, is called the collision cross section for the rigid spherical molecule. A molecule moving with a constant velocity ofvwill sweep out (\pi\sigma^ 2v) (meters)^3/sec and strike \pi\sigma2vn molecules per second, where n is the concentration of molecules. In these calculations, we have assumed several things some of which are not completely accurate. Actually the number of collisions per second is off by a factor of \surd2. Thus, the number of collisions per molecule per second is: z = \surd2 \pi\sigma^2vn Since there are n molecules per unit volume, there will be \surd2\pi \sigma^ 2vn^2 collisions per unit time. The number of collisions per unit time per unit volume is given by Z = \surd2/2 \pi\sigma^2vn^2 The factor of one-half comes from the fact that the number of collisions is one-half the number of total collisions. This is true because the number of total collisions is equal to the sum of all of the collisions of all of the particles, but in each collision two molecules are involved which means that in the total number each collision is really counted twice. To solve the problem we must: 1) find the arithmetic mean speed, 2) find the concentration of molecules, n, using the ideal gas law, 3) substitute the values in the equations for z = \surd2 \pi\sigma^2vn and Z = \surd2/2 \pi\sigma^2vn^2. The in-formation known: \sigma = 3.61 × 10^-10mP = 1 atm T = 25\textdegreeC + 273\textdegree = 298\textdegreeK ;M = 32 × 10^-3 Kg/mole = molecular weight of oxygen R = 0.082 [(liter - atm) / (0K - mole)] = 8.312 J/0K-mole. 1)v= [8RT/\piM]^1/2 = [(8) (8.314 J/0K-mole) (298)/\pi(32 × 10^-3 Kg/mol)]^1/2 = 444 m/s. 2)n = N/V = number of molecules / m^3 = concentration According to the ideal gas law, PV =NRT, whereN= moles orN/V = P/RT . Avogardo's number, Na = 6.02 × 10^23 (molecules/mole) so that (NNa)/V = PNa / RT. ButNNa = N so that N/V = n = PNa / RT = {[(1atm)[6.022 ×10^23(molecules/mole)] [10^3(leter/m^3 = {[(1atm)[6.022 ×10^23(molecules/mole)] [10^3(leter/m^3 )]}/ {[0.082 (liter atm / 0K-mole)] (29 8 \textdegreeK)} = 2.46 × 10^25 molecules/m^3 3)z = \surd2 \pi\sigma^2vn = (1.414)(3.14)(3.61×10^-10 m)^2(444 m/s) [2.46×10^23 (molecules/m^3)] = 6.32 × 10 (molecules / s) This is the number of collisions per molecule per second. Z = \surd2/2 \pi\sigma^2vn^2 = (z n)/2 = {[6.32 × 10^9 (molecules/s)] [2.46 × 10^25 (molecules/m^3)]} / 2 = 7.7 × 10^34 (collisions/ m^3 s) To answer part c of the problem, we use the conversion factor [10^-3 (m3/liter)] / [6.02 × 10^23 (molecules/mole)].Thus Z = [(7.7 × 10^34 collisions/m^3-s) (10^-3 m^3/l)] / [6.022 × 10^23 (molecules/mole)] = 1.29 × 10^8 moles/liter-s, which is the number of moles of collisions per liter per second.

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Question:

Suppose that the 2 atoms of a stable molecule oscillate along the line connecting their centers. Treating the system as an harmonic oscillator, calculate the vibration frequency. (Neglect any rotation of the system).

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/Users/wenhuchen/Documents/Crawler/Physics/D09-0391.htm

Solution:

If we treat the given system as an harmonic oscillator, we may assume that the 2 atoms are coupled by a tiny spring connecting their centers. Then, using Newton's Second Law to relate the force on each atom to its acceleration, we obtain F_12^\ding{217} = m_2r̈_2^\ding{217}(t) F_21^\ding{217} = m_1r̈_1^\ding{217}(t)(1) where F_12^\ding{217} is the force acting on atom 2 due to atom 1, and F^\ding{217}_21 is the force on atom 1 due to atom 2. Also, r_2^\ding{217}(t) and r_1^\ding{217}(t) are the initial positions of particles 2 and 1, respectively. (See figure (A)). Now, we may define the relative distance between the 2 atoms as r^\ding{217} \cong r_1^\ding{217}(t) - r_2^\ding{217}_ (t)(2) Looking at figure (B), we observe that, at t = 0, r^\ding{217} - r_0^\ding{217}= r_1^\ding{217}_ (0) - r_2^\ding{217}(0). This is the initial relative separation of the 2 atoms. By Hooke's Law, we may write F_12^\ding{217} = C(r^\ding{217} - r_0^\ding{217}) F_21^\ding{217} = -C(r^\ding{217} - r_0^\ding{217})(3) where r^\ding{217} - r_0^\ding{217}is the extension of the "spring" and C is the spring constant. Note that particle 2 attracts particle 1 and vice versa. In-serting (3) in (1) m_2r̈_2^\ding{217} = C(r^\ding{217} - r_0^\ding{217})(4) m_1r̈_1^\ding{217} = -C(r^\ding{217} - r_0^\ding{217})(5) Multiply (4) by m_1 and (5) by m_2 to obtain m_1m_2r̈_2^\ding{217} = m_1C(r^\ding{217} - r_0^\ding{217})(6) m_1m_2r̈_1^\ding{217} = -m_2C(r^\ding{217} - r_0^\ding{217})(7) Subtracting (6) from (7) m_1m_2r̈_1^\ding{217} - m_1m_2r̈_2^\ding{217}= -m_2C(r^\ding{217} - r_0^\ding{217}) - m_1C(r^\ding{217} -r_0^\ding{217}) orm_1m_2(r̈_1^\ding{217}_ - r̈_2) = -C(r^\ding{217} - r_0^\ding{217})(m_1 + m_2) From (2)r̈_1^\ding{217} - r̈_2^\ding{217} = r̈^\ding{217} and[(m_1m_2)/(m_1 + m_2)] r̈^\ding{217} = -C(r^\ding{217} - r_0^\ding{217}) Now, m_1m_2/m_1+m_2 is defined as the reduced mass u and u r̈^\ding{217} = -C(r\ding{217}- r_0^\ding{217}) Finally r̈^\ding{217} + (C/u)(r\ding{217}- r_0^\ding{217}) = 0(8) Redefining r\ding{217}- r_0^\ding{217} as s^\ding{217}, we note that s̈^\ding{217} = r̈\ding{217}- r̈_0^\ding{217} = r̈^\ding{217} since r_0^\ding{217} = constant. Equation (8) now becomes s̈^\ding{217} + (C/u)s^\ding{217} = 0 which is the equation of motion of a simple harmonic oscillator of an-gular frequency \omega_0 = \surd(C/u)(9) It is known from spectroscopic measurements that the fundamental vibra-tional frequencies of the molecules HF and HCL are \omega_0(HF)= 7.55 × 10^14 rad/sec; \omega_0(HCL)= 5.47 × 10^14 rad/sec; Let us use these data to compare the force constantsC_HF and C_HCL. The reduced mass of HF is, in atomic mass units, 1/u_HF = 1/m_H + 1/m_F = 1/1amu + 1/19amu = (20/19)amu U_HF \approx .950 amu 1/u_HCL = 1/m_H + 1/m_C1 = 1/1amu + 1/35amu = (36/35)amu (Here we have used the atomic mass of the most abundant isotope of chlorine, CL^35.) Notice that the reduced masses are quite close to each other in value. This is because the hydrogen, being lightest, does most of the oscillating. Now from (9) we have for the ratio of the force constants: C_HF/C_HCL = [{(\mu\omega2_0)HF}/{(\mu\omega2_0)HC1}] \cong {54.0 × 10^28 amu/s^2}/{29.0 × 10^28 amu/s^2} \cong 1.86, while for an individual force constant C_HF = {54 × 10^28 amu/s^2}{1.66 × 10^-24g/amu} \approx 9 × 10^5 dynes/cm Here we have inserted the factor which converts the mass from atomic mass units to grams. Is this value of C reasonable? Suppose we stretch the molecule (which is about 1A or 1 × 10^-8 cm in length) by 0.5 A. The work needed to do this would probably be nearly enough to break up the molecule into separate atoms of H and F. By using the formula for the potential energy of a compressed (or extended) spring, the work needed to stretch the molecule 0.5 A should be of the order of magnitude (1/2)C{r - r_0}^2 \approx (1/2){9 × 10^5}{0.5 × 10^-8}^2 \approx 1 × 10^-11 erg This is not unreasonable for an energy of decomposition into separate atoms. Therefore, the calculated value of C is reasonable.

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Question:

What is the advantage of using DMA? When is DMA used?

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Solution:

Computers with the DMA (direct memory access) feature have a direct link between the memory section and the I/O section. Thus, data can be transmitted to and from peripherals without intervention by the CPU. The advantage of using DMA is that the time to transfer data is limited only by the memory access time (usually less than one microsecond). Data transfer to the CPU of a computer without DMA requires several instruction cycles and takes ten to twenty times longer. Figure 1 shows the connections required for direct memory access. The DMA controller located in the I/O section handles data transfers just like the CPU. The "DMA request" control signal ensures that the CPU retains control of memory usage. The Address bus and data bus shown can be the same busses that connect the CPU to memory. The DMA is used with high-speed peripherals such as mag-netic disk, high-speed communication lines, and CRT displays.

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Question:

Explain why the bacterial cell wall is the basis for gram staining .

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/Users/wenhuchen/Documents/Crawler/Biology/F05-0121.htm

Solution:

Gram staining is one of the most important differential staining techniques used to determine differences between bacterial cells. A bacterial cell is fixed to a microscope slide, and the slide is covered with the following solutions in the order listed: crystal violet, iodine solution, alcohol , andsafranin. Bacteria stained by this method fall into two groups: gram-positive , which retain the crystal violet-iodine complex and are dyed a deep violet color, and gram-negative, which lose the crystal violet- iodine complex when treated with alcohol, and are stained by thesafranin, giving them a red color. The cell walls of bacteria are composed ofpeptidoglycan. This layer is thick in gram-positive cells, but thin in gram-negative cells, However, the latter havelipopolysaccharidelayer surrounding the thin peptidoglycan layer. The difference in staining is due to the high lipid content (about 20%) of the cell walls of gram-negative bacteria. (Lipids include fats, oils, steroids and certain other large organic molecules.) During the staining procedure of gram-negative cells, the alcohol treatment extracts the lipid from the cell wall, resulting in increased permeability of the wall. The crystal violet-iodine complex is thus leached from the cell in the alcohol wash . The decolorized cell then takes up the redsafranin. The cell walls of gram-positive bacteria have a lower lipid content, and thus become dehydrated during alcohol treatment. Dehydration causes decreased permeability of the wall, so that the crystal violet-iodine complex cannot leave the cell, and the cell is thus violet-colored. In addition thick peptidoglycan layer of the gram + cell walls preventsdecolorizationby alcohol . Most bacteria fall into one of these two staining groups and gram\rule{1em}{1pt}staining in an important means of bacterial classification.

Question:

When metal M is heated in halogen X_2, a compound MX_n is formed. In a given experiment,1.00 g of titanium reacts with chlorine to give 3.22g of compound. What is the corresponding value of n?

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Solution:

In the compound MX_n , there is one mole of M and n moles of X. In the compound here, one is given that 1.00 g of it is Ti (titanium) and that the compound weighs 3.22 g. This means that 3.22 g - 1.00 g or 2.22 g of the compound is Cl. To find the simplest formula for the compound, one must first determine the number of moles of each component present. This is done by dividing the weight present by the molecular weight no. of moles = (no. of grams) / (MW) For Ti : (MW = 47.9) no. of moles = (1.00g) / (47.9g/mole) = 2.09 × 10^-2 moles. For Cl : (MW = 35.5) no. of moles = (2.22g) / (35.5g/mole) = 6.25 × 10^-2 moles. Since the simplest formula for the compound isTiCl_n, n can be found using the following ratio. (1 Ti) / (n) = (2.09 × 10^-2 moles Ti) / (6.25 × 10^-2 moles Cl) n = (1 Ti × 6.25 × 10^-2 moles Cl) / (2.09 × 10^-2 moles Ti) = 2.99 Cl The formula for the compound is therefore TiCl_3.2.99 Cl is rounded off to the nearest whole number, 3.

Question:

How is the Golgi apparatus related to the endoplasmic reticulum in function ?

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Solution:

The Golgi apparatus is composed of several membrane-bounded, flattened sacs or cisternae arranged in parallel array about 300 \AA apart (see Figure 1) . The sacs are disc-like and often slightly curved. Note the concavity on the trans face near the plasma membrane and the convexity of the cis face are thinner (more like reticulum membrane than like plasma membrane). The function of the Golgi apparatus is best understood in cells involved in protein synthesis and secretion. The protein to be secreted is synthesised on the rough endoplasmic reticulum. Vesicles containing small quan-tities of the synthesized protein bud off from the endplasmic reticulum. These vesicles carry the protein to the convex face of the Goldi complex. In the Goldi appar-atus, the protein is concentrated by the removal of water. In addition, chemical modifications of the protein, such as glycosylation (addition of sugar) occur. The modified protein is released from the concave surface in the form of secretory granules. The secretory granules containing the protein are from the cytoplasm by a membrane that can fuse with the plasma membrane, its content (protein in this case) is expelled from the cell, a process known as excytosis. This process is outlined in Figure 2. Most of the cell organelles are found in a specific arrangement within the cell to complement their function. For example, the Golgi apparatus is usually found near the cell membrane and is associated with the endoplasmic reticulum. Since they are relatively close to each other, transport of materials between them is considerably efficient.

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Question:

What are gallstones and how are they formed?

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/Users/wenhuchen/Documents/Crawler/Biology/F17-0437.htm

Solution:

Gallstones contain about 80% cholesterol. Cholesterol is a steroid molecule that is absorbed from certain digested fats, such as butter, and is also produced in the liver. It is a precursor of many of the body's important steroid hormones (see fig.) Even though cholesterol is vital for proper body functioning, it is often not handled well by our bodies. This is because it is almost completely insoluble in body fluids. Excess cholesterol in the blood may be deposited along the arterial walls and in the heart. This can lead to arteriosclerosis and ultimately, to heart disease. The body's major mechanism for the removal of cholesterol from the blood takes place in the liver, which releases cholesterol into the intestine as part of the bile. Bile contains bile salts and lecithin that emulsify lipids such as cholesterol. In the gallbladder, bile is stored and slowly concentrated by the removal of water. As this happens, the concentration of cholesterol increases to a level near the saturation point. If there is an excess of cholesterol, or insufficient amounts of bile salts or lecithin to keep it soluble, the cholesterol will precipitate, forming a hard solid mass called a gallstone. If a gallstone is small, it will pass through the bile duct and out into the intestine, causing no complications. A larger stone, however, may lodge in the neck of the gall-bladder. This leads to pain and contractile spasms of the gallbladder wall. Since the gallbladder is off to the side of the bile duct, bile can still flow from the liver to the intestine for digestion. If the spasms of the gall-bladder are powerful enough, the stone may be forced out of the gallbladder and become lodged in the bile duct, obstructing it. Complete blockage may lead to a condition known as jaundice, in which the skin yellows due to an accumulation of bile pigments in the body. Since no bile reaches the intestine, there is also decreased emulsification and absorp-tion of fats. Gallstones may sometimes lodge at the point where the bile duct and pancreatic duct join before they enter the duodenum. When this happens, both bile and pancreatic juice are blocked in their passages, resulting in severe impairment of both digestion and absoprtion in the intestinal tract. Gallstones may be removed surgically if they are not dissolved by the body. If necessary, the gallbladder may be removed without any great effect on digestion.

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Question:

A newly discovered planet has twice the density of the earth, but the acceleration due to gravity on its surface is exactly the same as on the surface of the earth. What is its radius?

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/Users/wenhuchen/Documents/Crawler/Physics/D04-0140.htm

Solution:

This problem must be approached carefully. We must express the acceleration due to gravity in terms of the density and the radius of the planet. If the radius is R and the mass of the planet M, then the acceleration due to gravity on its surface is found from Newton's Second Law, F = ma. Consider an object of mass m on the surface of the planet. Then the only force on m is the gravitational force F, and F =GMm/R^2 But a is the acceleration of m due to the planet's gravitational field, org_p. Then g_v= GM/R^2 Assuming the planet is spherical, its volume is the volume of a sphere of radius R: V = (4/3)\piR^3 Since Mass = Volume × Density. M = (4\piR^3\rho/3) where \rho (the Greek letter rho) is the density of the planet. Therefore g_v= {G(4/3)\piR^3\rho}/R^2 = (4\pi/3)GR\rho Similarly, the acceleration due to gravity on the surface of the earth is g= (4/3)\piGR_e_(\rho)e where\rho_eis the density of the earth, and R_e is its radius. If g_v= g Then (4/3)\piGR\rho= (4/3)\piGR_e_(\rho)e Canceling (4/3)\piGon both sides R\rho= R_e(\rho)e If the density of the planet is twice that of the earth, \rho = 2\rho_e So R2\rho_e = R_e(\rho)e Whence R = (1/2)R_e = (1/2) × 6.38 × 10^6 m = 3.19 × 10^6 m The radius of the planet is one half of the radius of the earth, or 3.19 × 10^6 meters.

Question:

What is the radius of the orbit of a 1-MeV proton in a 10^4 - gauss field?

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/Users/wenhuchen/Documents/Crawler/Physics/D21-0709.htm

Solution:

An electron which is moving in a magnetic field B^\ding{217} with velocity v^\ding{217}, experiences a force F^\ding{217} = - (1/c) ev^\ding{217} × B^\ding{217} . As a result of this force, the electron moves on a circular path in the plane of v^\ding{217}. The radius of the orbit is obtained from the expression for the centripetal force F_center = F = [(mv^2 ) / R] or(1/c) evB = [(mv^2 ) / R] R = [(mvc) / eB] . The velocity of a 1- MeV proton is obtained as follows KE = total energy - rest energy = [(m_0 c^2)/{\surd(1 - \beta^2 )}] - m_0 c^2 = [1/{\surd(1 - \beta^2)}] m_0c^2 = 1 MeV = 1.6 × 10^-6ergs The rest energy of a proton is m_0c^2 = 1.67 × 10^-24 gr × (3 × 10^10 cm/ sec)^2 = 1.5 × 10^-3ergs Since KE << m_0 c^2 , we have a non - relativistic motion. In this case, \beta = (v/c) << 1, hence we can use the approximation [1/{\surd(1 - \beta^2 )}] \approx 1 + (1/2) \beta^2 and (1/2) \beta^2 \approx (KE)/(m_0 c^2 ) =[(1.6 × 10^-6 ergs)/(1.5 × 10^-3ergs)] = 1.07 × 10^-3 \beta = \surd(2.14 × 10^-3 ) = 0.046 v = \betac = (0.046 × 3 × 10^10 cm/sec = 1.38 (10^9 cm/sec) In the expression for R we can use the nonrelativistic (rest mass) m^0 for m.Therefore R \approx (m_0 vc)/(eB) Since m_0 = 1.67 × 10^-24ge = 4.8 × 10^-8stat c or R = [(1.67 × 10^-24 g) × (1.38 × 10^9 cm/sec) × (3 × 10^10 cm/sec)] / [(4.8 × 10^-10 statC) × 10^4 gauss] = 14.4 cm

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Question:

At 100\textdegreeC and 720 mm Hg or 720 torr, what is the density of carbon dioxide, CO_2 ?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E02-0064.htm

Solution:

According to the kinetic theory of gases, the parameters of pressure, volume, and temperature of a gas are mutually dependent - that is, a change in one influences the others. In this problem, you want the density of CO_2. Density is defined as mass/volume. Thus, density = (mass of 1 mole CO_2)/(volume of 1 mole CO_2). The mass of one mole of carbon dioxide is 44 g. (This number represents the sum of the atomic weights of all atoms in the gas.) Consequently, this problem involves only the calculation of CO_2 's volume at 100\textdegreeC and 720 torr. At standard temperature and pressure, one mole of CO_2 or any gas occupies 22.4 liters. (Recall that standard temperature and pressure is 760 mm Hg and 0\textdegreeC.) Because the temperature and pressure has been altered, the volume will no longer be 22.4 liters for one mole of CO_2. According to the combined gas law, PV/T = constant. Thus at standard temperature and pressure, [{(760 mm Hg) (22.4 liters)}/(273\textdegree)] = a constant. (You add 273\textdegree to the Celsius temperature, since T must be in Kelvin.) At 100\textdegreeC and 720 mm Hg, you have [{(720 mm Hg) V}/(373\textdegree)] = a constant. According to this law, both expressions are equal, i.e. [{(760 mm Hg) (22.4 liters)}/(273\textdegree)] = {(720 mm Hg) (V)}/(373\textdegree)] Solving for V in this equation, you obtain V = 32.3 liters. The density is, therefore, 44/32.3 = 1.32 g/liter.

Question:

In what ways are human beings not favorable for studies of inheritance?

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/Users/wenhuchen/Documents/Crawler/Biology/F26-0683.htm

Solution:

In order to study inheritance, it is necessary that the phenotypic manifestations of the genes be clear cut and distinct. In this way, correlation of phenotypes and genotypes is direct. This is difficult to do with humans. Many human traits, such as height, are a result of the cooperative action of a number of a genes (polygenetic traits) which are difficult to identify or localize individually. Many broadly defined traits, such as here-ditary deafness, can result from the manifestation of any one of a number of genes. Also, any one gene can often have a variety of phenotypic manifestations (pleiotropism), as can be seen in the widespread bodily effects that may result from the lack of a specific enzyme. Ideally, genetic studies are conducted in controlled environmental situations. This is virtually impossible to do with human beings. Man's environment is far more complex and diversified than that of any other organism and in addition, he himself largely controls it. Unlike most other organisms, not only physiological and natural factors, but psychological and social factors, must be taken into account in any valid investigation. Man also has a much longer life span and generation time than is feasible for genetic studies. His family sizes are far too small for the establishment of good genetic ratios and reliable statistics. Other organisms, such as bacteria and fruit flies, produce large numbers of offspring in relatively short periods of time, making them much more desirable subjects for genetic studies. Furthermore, ethical standards understandably prevent the use of humans for the establishment of standard genetic stocks and the controlled breeding that is a large part of the study of genetics. These obstacles nonetheless do not totally prevent valid and useful human genetic studies. Modern statistical analysis has increased the reliability of human statistics. Pedigree analysis is still a valuable tool for geneticists. In addition, studies using identical twins are employed widely to control for the effect of the environment on the expression of genes. While organisms such as bacteria and fruit flies are by far the best subjects for genetic investigation, studies using human beings will always be essential for the understanding of inheritance in man.

Question:

Define a genetic mutation. After a mutation has occurred in a population, what events must occur if the mutant trait is to becomeestablished in the population?

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Solution:

A genetic mutation occurs as a change in a specific point (allele) or segmentof a chromosome. This gives rise to an altered genotype, which oftenleads to the expression of an altered phenotype. Genetic muta-tions occurconstantly, bringing about a variety of phenotypes in the population, uponwhich natural selection can act to choose the most fit. Because geneticmutations occur in the chromosomes and are therefore inherited, theyare also referred to as the ultimate raw materials of evolution. When a mutation first appears, only one or very few organisms in thepopulation will bear the mutant gene. The mutation will establish itself inthe population only if the mutants survive and breed with other membersof the population. In other words, the mutants must be able to reproduce. Not only must they be able to reproduce, their zygotes must be viableand grow to become fertile adults of the next generation. By breedingof the mutants or mutation carriers either with each other or with normalindividuals, the mutant gene can be transmitted to succes-sive generations. Given the above conditions for establish-ment, over many generations, the mutant gene will appear with greater and greater frequencyand eventually become an establishedconstituantof the population'sgene pool.

Question:

Calculate the volume of 0.3 N base necessary to neutralize 3 liters of 0.01 N nitric acid.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E08-0301.htm

Solution:

For neutralization to occur, there must be the same number of hydrogen ions as there are hydroxide ions. This is shown by the following equation H^+ + OH^- \leftrightarrows H_2O The H^+ ions come from the acid, the OH^- ions from the base. The number of H^+ and OH^- ions are equal to the number of equivalents. Normality is defined as the number of equivalents of acid or base per liter of solution. Normality = (equivalents /liter) An equivalent is the number of grams of the acid or base multiplied by the number of replaceablehydrogensor hydroxides divided by the molecular weight of the acid or base. equivalent = [(grams of solute × no. of replaceable H or OH) / (MW of solute)] The number of replaceablehydrogensorhydrodixesis defined as the number which ionize when the compound is placed in solution. For nitric acid, there is one replaceable hydrogen HNO_3 \leftrightarrows H^+ + NO^-_3 In brief, for neutralization there must be the same equi-valent amount of base as there is acid. Because normality is defined as Normality = (equivalents/liters) equivalents are equal to the normality times the volume. Thus, the normality of the base times its volume equals the normality of the acid times its volume. N_baseV_base=N_acidV_acid_' where N is the normality and V is the volume. Here, 0.3 equiv/liter ×V_base= 0.01 equiv/liter × 3 liters V_base= [(0.01 × 3 liters)/0.3] = 0.1 liter or 100 M.

Question:

A resistor with resistance R is connected in series with the circuit of the previous problem. As before, q_0 is the initial charge on the capacitor and the switch is closed at time t = t_0. It can be shown that the behavior of the circuit is governed by the following differ-ential equation: Lq̇ ̇+ Rq̇ + (1/C)q(t) = 0 where q̇ = i(t) = current at time t. If 1_0 is the initial current, model and simulate the behavior of the circuit from time t = t_0 to t = t_f, using a digital computer program.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G12-0306.htm

Solution:

As in the previous problem, a similarity can be noticed between the given differential equation and that of a differential equation describing harmonic motion (damped). Using the same equi-valences with the addition of . R\rightarrow D, the equation of the damped system can be constructed from the equation of this problem. The programmer needs only to substitute the following as input data in the previous program: USE THE VALUER as D USE THE VALUE(1/C) as K USE THE VALUE(q_0) as XCOR USE THE VALUE (i_0) as VCOR

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Question:

Algae may be microscopic single-celled organisms or enormousmulticellularseaweeds hundreds of feet long. The body of one of these organisms is termed athallus. How does athallusdiffer from a plant body?

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Solution:

A plant body shows differentiation of parts, with its roots, stems, and leaves all varying greatly in structure and in function.Multicellular algae show very little, if any, tissue differentiation. For this reason, the body is termed athallus(from the Greekthallos: a young, undifferentiated shoot or sprout). There is no anatomical basis for distinguishing the leaves or roots ofmulticellularalga, though in some species there may be a functional basis. The plant kingdom has traditionally been divided into two groups: theThallophyta(algae) and theEmbryophyta(land plants). The Embryophyta , in addition to showing tis-sue differentiation, have a life cycle different from that of theThallophyta. The reproductive structures of theThallophytesare usually unicellular, and the reproductive cells lack a protective surrounding wall or jacket of non-reproductive cells.Thallophytezygotes do not develop into embryos until after they are released from the parentthallus. In Embryophyta , the reproductive structures aremulti-cellularand are surrounded by non-reproductive cells. The early stages of embryonic development occur while the zygote is still contained within the parent plant. The body of a fungus is also termed athallus. Fungi, like algae, show little differentiation of parts, and lackmulticellularsex organs with jacket cells. The embryo of a fungus develops outside the parent fungus.

Question:

Show that the principle of conservation of momentum results Show that the principle of conservation of momentum results from the principle of conservation of kinetic energy and the Galilean trans-formation

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/Users/wenhuchen/Documents/Crawler/Physics/D06-0336.htm

Solution:

Suppose we observe a collision from two separate frames of Suppose we observe a collision from two separate frames of reference, S and S', moving with a relative velocity V^\ding{217}see figure (B). see figure (B). The position of a particle relative to S is related to its position relative to S' by see figure (A) . r^\ding{217} = r'^\ding{217} + R^\ding{217}(1) Differentiating (1) with respect to time, v^\ding{217} = v'^\ding{217} + V^\ding{217}(2) where v'^\ding{217} and v^\ding{217} are the velocity of the particle relative to S' and S respectively. Equations (1) and (2) constitute the Galilean trans-formation. We apply the principle of conservation of kinetic energy in frame S see figure (B),or (1/2)M_aV^\ding{217}_aiV^\ding{217}_ai + (1/2)M_bV^\ding{217}_bi \bulletV^\ding{217}_bi = (1/2)M_aV^\ding{217}_afV^\ding{217}_af + (1/2)M_bV^\ding{217}_bf.V^\ding{217}_bf(3) But, from (2) v^\ding{217}_ai = v'^\ding{217}ai+V^\ding{217}v^\ding{217}_af = v ׳ \ding{217} af +V^\ding{217} v^\ding{217}_bi = v ׳ \ding{217} bi +V^\ding{217}v^\ding{217}bf= v'^\ding{217}bf+V^\ding{217}(4) Substituting (4) in (3) (1/2)M_a(v'^\ding{217}_ai + V^\ding{217})\bullet (v'^\ding{217}_ai + V^\ding{217}) + (1/2)M_b(v'^\ding{217}_bi -+ V^\ding{217})\bullet (v'^\ding{217}_bi + V^\ding{217}) = (1/2)M_a(v'^\ding{217}_af + V^\ding{217})\bullet (v'^\ding{217}_af + V^\ding{217}) + (1/2)M_b(v'^\ding{217}_bf + V^\ding{217})\bullet (v'^\ding{217}_bf + V^\ding{217}) or(1/2)M_av'^\ding{217}_ai \bulletv'^\ding{217}_ai + (1/2)M_bv'^\ding{217}_bi \bulletv'^\ding{217}_bi + (1/2)(M_a + M_b)(V^\ding{217}\bulletV^\ding{217}) + M_av'^\ding{217}_ai\bulletV^\ding{217} + M_bv'^\ding{217}_bi \bulletV^\ding{217} = (1/2)M_av'^\ding{217}_af \bulletv'^\ding{217}_af + (1/2)M_bv'^\ding{217}_bf\bulletv'^\ding{217}_bf + (1/2)(M_a+M_b)(V^\ding{217}\bulletV^\ding{217}) + M_av'^\ding{217}_af \bulletV^\ding{217} + M_bv'^\ding{217}_bf \bulletV^\ding{217} Rewriting this last equation: (1/2)M_av'^\ding{217}_ai \bulletv'^\ding{217}_ai + (1/2)M_bv'^\ding{217}_bi \bulletv'^\ding{217}_bi + (M_av'^\ding{217}_ai + M_bv'^\ding{217}_bi)\bulletV^\ding{217} = (1/2)M_av'^\ding{217}_af \bulletv'^\ding{217}_af + (1/2)M_bv'^\ding{217}_bf \bulletv'^\ding{217}_bf + (M_av'^\ding{217}_af + M_bv'^\ding{217}_bf)\bulletV^\ding{217}(5) Now, if conservation of kinetic energy holds (1/2)M_av'^\ding{217}_ai \bulletv'^\ding{217}_ai + (1/2)M_bv'^\ding{217}_bi \bulletv'^\ding{217}_bi = (1/2)M_av'^\ding{217}af\bulletv'^\ding{217}_af + (1/2)M_bv'^\ding{217}_bf\bulletv'^\ding{217}bf and (5) becomes (M_av'^\ding{217}_ai + M_bv'^\ding{217}_bi) \bulletV^\ding{217}_ = (M_av'^\ding{217}_af + M_bv'^\ding{217}_bf) \bulletV^\ding{217}(6) But, sinceV^\ding{217}is an arbitrary vector (6) holds for allV^\ding{217}and we obtain the V^\ding{217} V^\ding{217} principle of conservation of momentum in frame S' M_av'^\ding{217}_ai + M_bv'^\ding{217}bi= M_av'^\ding{217}_af + M_bv'^\ding{217}_bf(7) M_av'^\ding{217}_ai + M_bv'^\ding{217}bi= M_av'^\ding{217}_af + M_bv'^\ding{217}_bf(7) Note also that S' is an arbitrary frame, and, therefore, (7) holds in all frames.k

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Question:

Develop a BASIC program to solve two simultaneous linear equation, AX + BY = CandDX + EY = F. Apply the result to the case A = B = I, C = 3, D = 1, E = F = -1.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G20-0503.htm

Solution:

Consider the pair of equations AX + BY = C(1) DX + EY = F(2) Solving (1) for X yields X = (C - BY) / A, and substituting this results into (2) gives: D(C - BY) / A + EY = F or Y {E - (BD / A)} = F -CD / A or Y {AE -BD} = AF -CD or Y = [(AF -CD) / (AE - BD)] = [(CD - AF) / (BD - AE). Similarly X = [(BF - CE) / (BD - AE)], Provided BD - AE \not = 0. Based on the above, the program looks as follows: 100REM SIMULTANESOUS EQUATION 110READ A, B, C, D, E, F 120LETZ = (B\textasteriskcenteredD - A\textasteriskcenteredE) 130IF Z = 0 THEN 160 140LET X = (B\textasteriskcenteredF - C\textasteriskcenteredE) / Z 141LET Y = (C\textasteriskcenteredD - A\textasteriskcenteredF) / Z 150PRINT "ROOTS ARE", X, Y 151GO TO 999 160PRINT "NO UNIQUE SOLUTION OR INCONSISTENT EQUATION" 180DATA 1, 1, 3, 1, -1, -1 999END

Question:

What weight of Ca(OH)_2, calcium hydroxide is needed to neutralize 28.0 g ofHCl. What weight of phosphoric acid would this weight of Ca(OH)_2 neutralize?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E08-0302.htm

Solution:

When we have neutralization problems, i.e., those where an acid reacts with a base, we must consider the meaning of equivalents. Equivalent weight may be defined as the molecular weight of a substance (grams/mole) divided by the number of protons (H^+) or hydroxyl (OH^-) ions available for reaction. In other words, Equivalent weight = [(molecular weight)/(number of H^+ or OH^-)]. In a neutralization reaction, the number of equi-valents of acid equals the number of equivalents of base. The number of equivalents of acid and base are equal, thus, if we can calculate the number of equivalents of 1 reactant, we automatically know the number of equi-valents present of the other reactant. To find this quantity use the equation no. of equivalents = [(grams)/(grams/equivalent)] The equivalent weight ofHClis [(M.W.)/{no. of H^+ (OH^-)}] = (36.5/1) = 36.5 and Ca(OH)_2 is (74.1/2) = 37 g/equiv. no. of equivalentsHClpresent = [(28 g)/(36.5 g/equiv)] = .767 Therefore, there are also .767 equivalents of Ca(OH)_2. To find the grams of Ca(OH)_2 needed use no. of equiv = [{grams Ca (OH)_2}/(grams/equiv)] .767 = (grams/37) grams Ca(OH)_2 = 28,4 need to neutralize 28 g ofHCl, The second half of the problem is answered in exactly the same way. We know then that there must also be .767 equiv of H_2PO_4 (phosphoric acid) present due to the fact .767 equiv of base Ca(OH)_2 is present. Recall, equivalents of an acid must equal the equivalents of a base for neutralization to occur. The M.W. of H_3PO_4 = 98.0 g, which yields a value of (98/3) = 32.7 grams per equiv. Therefore, No. of equiv = [(grams H_3PO_4)/(grams/equiv)] .767 = (grams/32.7) grams H_3PO_4 = 25.1 g needed to neutralize 28.4 g of Ca(OH)_2.

Question:

A meter reads that a battery is putting out .450 amp in the external circuit of a cell. The cell is involved in the electrolysis of a copper sulfate solution. During the 30.0 min that the current was allowed to flow, a total of .30g of copper metal was deposited at the cell's cathode. Was the meter an accurate measurement of current? (F = 96,500 coul.)

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/Users/wenhuchen/Documents/Crawler/Chemistry/E16-0579.htm

Solution:

The key to solving this problem is the determination of the actual current given the amount of copper deposited in 30.0 minutes. This can be done by employing Faraday's Laws of Electrolysis. Electro-lysis is the phenomenon that occurs when electricity is passed through a solution, such that ions are generated and moved toward an anode or cathode. The laws that govern this are as follows: Masses of substances involved are proportional to the quantity of electricity that flows through the cell. Masses of different substances produced during the process are proportional to their equivalent weight. The electrical equivalent is defined as a Faraday (F). It is capable of reducing one equivalent of positive charge (Avogadro's number of individual unit electric charges). The problem can now be solved. You know how much material is deposited in a given time. Thus, a certain amount of electricity based on current had to be used. The reaction that occurred at the cathode was Cu^2+ + 2e^- \rightarrow Cu. Thus, each mole of copper ion requires 2 moles of electrons or, in other words, 2 faradays (1 Faraday = 96,500 coul). Thus, a total of 193,000 coul. is required per mole of copper. Current = charge/time. If you had one mole of copper, the charge would be 193,000 coul. You have only [0.30g(wt. of Cu)] / [63.55(mw. of Cu)] =4.72 × 10^-3 moles. Thus, charge = 4.72 × 10^-3 (193,000) = 911.075 coul. To determine the current, time must be considered. The time must be in seconds, since current (amps) = [charge (couls)] / [time(secs)]. You are given the time as 30.0 min, which equals 1800 secs. Thus, Current = [(911.075) / (1800)] = .506 amp, The meter said .450, thus, it was inaccurate by [{(.506) - (.450)} / (.506)] × 100 = 11.1%.

Question:

Show by diagrams how genes located on different pairs of chromosomes segregate independently in meiosis in an organism with a diploid number of four.

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/Users/wenhuchen/Documents/Crawler/Biology/F25-0645.htm

Solution:

Because there are two possible ways in which the chromosomes can line up and segregate in meiosis I, there are two possible sets of gametes that could be formed by meiosis. Each set has 2 pairs of identical gametes. There are thus 4 types of gametes possible. We know this from looking at the genotypes of the parental cell (Aa Bb) . Four combinations of these genes are possible:AB, Ab, aB, and ab. All these gametes have been produced in the two possible meiotic divisions. It is important to note, however, that each separate meiosis produced only two types of gametes - AB and ab or Ab and aB - and only in these combinations . Gametic genotype combinations such as AB and Ab or aB and ab could not be produced in the same meiotic division. Although there are eight chromosomes in total, the separation of homologs occurs as a function of duplicated chromosomes. Thus, if in meiosis I, AA is paired with BB, aa must be paired with bb. If AA is paired with bb, then aa must be paired with BB. The genotypes of the gametes produced in a meiotic division are determined in meiosis I, and meiosis II is merely a mitotic type division resulting in the separation of identical daughter chromosomes. In meiotic gamete formation, the randomness of the way the chromosomes line up is one method that ensures genetic variation. Diverse gamates, differing in their combinations of chromosomes, are produced by the independent assortment of chromosomes when they first segregate in meiosis. In a cell with many chromosomes, such as a human cell with 23 pairs, the number of possible gametes that could be produced is very large indeed.

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Question:

Which of the following compounds are saturated? Which are unsaturated? (a) CH_3CH_2CH_2CH_3, (b) CH_2 = CHCH_2CH_3, (c) cyclohexane, (d) cyclohexene, (e) benzene.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E20-0742.htm

Solution:

The most fundamental class of organic compounds is the saturated hydrocarbons. A hydrocarbon is a compound that contains only carbon and hydrogen. Saturated is used to describe the absence of double bonds and / or triple bonds. The term comes from the fact that these compounds do not react with hydrogen because they are saturated with hydrogen. When a double and/or triple bond is present, the compound is termed unsaturated. Thus, compounds a and c are saturated and compounds b, d, and e are unsaturated.

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Question:

There is an appreciative sigh throughout the audience in the "mad\textquotedblright scene of "LuciadiLammermoor", when the soprano hits high C. On the musical scale, high C has a frequency of 1,024 sec-^1 . Assuming the velocity of sound in air is 3.317 × 10^4 cmsec-^1 , compute the wavelength of the high C sound wave.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E30-0919.htm

Solution:

The relationship between the wavelength (\lambda) , frequency (ѵ), and velocity (v) of a sound wave is: \lambda = (v/ѵ) . Given both v and ѵ, substitute these values into this expression and evaluate. The wavelength of high C = = \lambda = (3.317 × 10^4 cmsec-^1) / (1024 sec-^1) = 32.39 cm.

Question:

Explain why simple diffusion is not sufficient to account for the passage of all of the dissolved minerals into the cells of water plants.

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/Users/wenhuchen/Documents/Crawler/Biology/F10-0244.htm

Solution:

Simple diffusion cannot account for the passage of all of the dissolved minerals into the cells of water plants. Plant cells frequently contain minerals in higher concentration than those available in the soil or water in which they are growing. Simple diffusion of materials into the cells cannot occur against a con-centration gradient. Sometimes this gradient can be very steep. Studies of the vacuolar sap ofNitellaclavata, a pond plant, for instance, have shown that the concentration of mineral anions in the sap exceeds that in the pond by 10^4 times! For this to happen, the plant must be able to transport minerals and other materials against this concentration gradient. Such a process is known as active transport. In contrast to simple diffusion which requires no energy expenditure, active transport can occur only with the expenditure of energy. In the living cell, the energy for active transport comes from the energy-rich phosphate bond of the ATP molecule.

Question:

The ^+NH_3CH_2COOH ion (glycine^+) is acationof the amino acidglycine(NH_2CH_2COOH).Glycine^+ undergoes the following two successive dissociations: ^+NH_3CH_2COOHk_1\rightleftarrows H^+ + ^+NH_3CH_2COO^- ; pk_1 = 2.35 ^+NH_3CH_2COO^-k_2\rightleftarrows H^+ + NH_2CH_2COO^- ; pk_2 = 9.78. What is theisoelectricpoint of this molecule (the pH at which the number of positive charges on the molecule equals the number of negative charges).

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/Users/wenhuchen/Documents/Crawler/Chemistry/E10-0373.htm

Solution:

We will approach this problem by considering the equilibrium constants k_1 and k_2. For the first reaction, k_1 is equal to k_1 = {[H^+][NH_3CH_2COO^-]} / [^+NH_3CH_2COOH] and for the secondreation, k_2 is equal to k_2 = {[H^+][NH_2CH_2COO^-]} / [^+NH_3CH_2COO^-] If we take the product of these two constants, the concentration of isoelectricspecies ^+NH_3CH_2COO-cancels out, giving k_1k_2 = ({[H^+][NH_3CH_2COO^-]} / [^+NH_3CH_2COOH]) × ({[H^+][NH_2CH_2COO^-]} / [^+NH_3CH_2COO^-]) = {[H^+]^2[NH_2CH_2COO^-]} / [^+NH_3CH_2COOH] At theisoelectricpoint, the concentrations of all charged species are equal, that is [NH_2CH_2COO^-] = [^+NH_3CH_2COOH]. Hence,[NH_2CH_2COO^-] / [^+NH_3CH_2COOH] = 1 and k_1k_2 = {[H^+]^2[NH_2CH_2COO^-]} / [^+NH_3CH_2COOH] = (H^+)^2 . Taking the logarithm of both sides, (H^+)^2 = k_1k_2 2 log (H^+) = log k_1k_2 = log k_1 + log k_2 . Multiplying this equation by negative one and using the definitions pH = - log (H^+), pk_1 = - log k_1, and pk_2 = - log k_2, we obtain - 2 log [H^+] = - log k_1 + (- log k_2) 2 pH = pk_1 + pk_2,or,pH = (1/2) (pk_1 + pk_2). Thus, theisoelectricpoint is pH = (1/2) (pk_1 + pk_2) = (1/2) (2.35 + 9.78) = 6.07.

Question:

Design a program to play a simple game of solitaire. The game should begin with a deal of nine cards, face up. If any two have the same face value, they are covered with two new cards, also face up. This step is repeated until the deck has been .exhausted, except for one card. If this occurs, the dealer wins. If there are no more pairs showing, the dealer loses. The program should determine an approximation to the probabil-ity of winning.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G23-0557.htm

Solution:

To start, we must decide upon how the individual cards are to be represented. We will use an array of 52 com-ponents, the numbers 1, 2, 3, ... representing the Ace, two, three, ... of spades, and the numbers 14, 15, 16, ... repre-senting the Ace, two, three, ... of hearts, and so on. To determine the suit of a particular card, the expression [(CARD(I) - 1)/13] can be used; a quotient of 0 means spades, a 1 means hearts, and so on. We may find the face value of a particular card via the expression [CARD(I) - 13 × IS], where ISisthe suit. If this expression equals, say, 11, the card is a Jack. (Brackets signify the integer part of the expres-sions, ) To shuffle the deck, we can use "pseudo-pointers." Start by filling the array CARD with the integers 1 through 52. Then, fill another array, RANPTR, with random digits be-tween 1 and 52, including duplicates. Then the shuffling consists of making interchanges, such as CARD (1) with CARD (RANPTR(1)), CARD (2) with CARD(RANPTR(2)), etc. You can generate a new set of random digits for each game to re-shuffle the deck. The program is presented below. Comments explain some of the major control sections. The probabilities outputted at the end are meaningful only if a sufficient number of games are played. Play 25 games and see what outcomes you get. INTEGER RANPTR, CARD, PILE, CHOICE DIMENSION PILE(52), CARD(52), RANPTR(52) DO 10 N = 1,52 CARD (N) = N 10CONTINUE CGENERATE RANDOM NUMBERS TO SHUFFLE DECK 12DO 15 J = 1,52 RANPTR(J) = INT (52\textasteriskcenteredRANDOM(X) + 1) ITEMP = CARD(J) CARD(J) = CARD(RANPTR(J)) CARD(RANPTR(J)) = ITEMP 15CONTINUE CTYPE PLAY IF YOU WISH TO CONTINUE 20READ (5,100) CHOICE 100FORMAT(A5) IF(CHOICE.NE.'PLAY') GO TO 95 I = 1 CREPEAT UNTIL I IS GREATER THAN 9 21CONTINUE PILE(I) = CARD(I) I = I + 1 IF(I.LE.9) GO TO 21 25K = 1 26L = K + 1 28IF(PILE(K).NE.PILE(L)) GO TO 30 CTHEN DEAL TWO CARDS TO COVER PAIR PILE(K) = CARD(I) PILE(L) = CARD(I + 1) I = I + 2 IF(I.LT.51) GO TO 25 NGAME = NGAME + 1 NWIN = NWIN + 1 WRITE(5,101) 101FORMAT(1X, 'DEALER WINS!') GO TO 12 CCHECK ALL VALUES OF K AND L L = L + 1 IF(L.LE.9) GO TO 28 K = K + 1 IF(K.LE.8) GO TO 26 NGAME = NGAME + 1 NLOSS = NLOSS + 1 WRITE(5,102) 102FORMAT(1X, 'SORRY, DEALER LOSES.') GO TO 12 95PWIN = FLOAT (NWIN)/FLOAT (NGAME) 95PWIN = FLOAT (NWIN)/FLOAT (NGAME) PLOSS = FLOAT (NLOSS)/FLOAT (NGAME) PLOSS = FLOAT (NLOSS)/FLOAT (NGAME) WRITE(5,103) NGAME, PWIN 103FORMAT(1X, 'PROBABILITY OF WINNING AFTER', 1I5, 'GAMES IS', F5.4) WRITE(5,104) PLOSS 104FORMAT(1X, 'PROBABILITY OF LOSING IS', F5.4) STOP END

Question:

Why does the number of red blood cells in the human body increase at high altitudes?

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/Users/wenhuchen/Documents/Crawler/Biology/F14-0356.htm

Solution:

It has been observed that the loss of red cells by hemorrhage decreases the ability of the blood to deliver sufficient oxygen, and increased erythrocyte production results. Also, partial destruction of the bone marrow (by x-ray, for example) destroys the sites of erythrocyte production and diminishes the availability of oxygen to the tissues. Increased red cell production by the remaining healthy marrow follows. It seems that the initiator or stimulus for increased erythrocyte production could be a lack of oxygen. This is indeed found to be true. At high altitudes, the partial pressure of oxygen is decreased, and thus less oxygen is delivered to the tissues. Any condition that causes a decrease in the amount of oxygen transported to the tissues causes an increased rate of erythrocyte production. Actually a decreased oxygen concentration in the bone marrow does not directly (by itself) expedite erythrocyte production. In response to decreased oxygen levels, the kidney, liver, and other tissues secrete erythropoietin, also callederythropoieticstimulating factor. This glycoprotein is transported via the blood to the bone marrow, where it stimulates erythrocyte production. When the number of red blood cells has increased to the point where the oxygen level in the tissues is again normal, erythropoietin secretion decreases and a normal amount of erythrocytes are again produced. This negative feedback mechanism prevents an overproduction of red cells, pre-venting the viscosity of the blood from increasing to a dangerous level. It is interesting to note that erythrocyte production and hemoglobin synthesis are not necessarily correlated. An iron deficiency decreases hemoglobin synthesis (since thehemepart of the hemoglobin molecule contains an iron atom). This low hemoglobin level leads to a reduc-tion of oxygen brought to the tissues. As was cited before, a reduced oxygen level causes increased red blood cell production. In this case, the red blood cells have a lower hemoglobin content than normal (they are called hypochromic erythrocytes) and thus are less efficient oxygen carriers.

Question:

(1) aniline from benzene, (2) ethyl amine from ethane.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E21-0781.htm

Solution:

To solve this problem you need to write out the structures of the initial and final compounds. In any synthesis, think of reactions that lead you step by step to the desired result. If you had nitrobenzene, you could go directly to aniline because nitrobenzene can be reduced with metal and acid to aniline. To obtain nitrobenzene, perform a nitration reaction, as shown below in figure B. (2) Ethyl amine \equiv C_2H_5NH_2, Ethane \equiv C_2H_6. Now, if you had ethyl chloride, C_2H_5CI, you could obtain ethyl amine directly, since upon the addition of ammonia (NH_3) ethyl chloride is converted to the corresponding amine. To obtain ethyl chloride, halogenate ethane with Cl_2 and light. C_2H_6 + Cl_2C_2H_5CI + HCI C_2H_5CI + NH_3 \rightarrow C_2H_5NH_2 + HCI .

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Question:

A workman sitting on top of the roof of a house drops his hammer. The roof is smooth and slopes at an angle of 30\textdegree to the horizontal. It is 32 ft long and its lowest point is 32 ft from the ground. How far from the house wall is the hammer when it hits the ground?

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/Users/wenhuchen/Documents/Crawler/Physics/D03-0093.htm

Solution:

Figure A illustrates the first part of the motion. Two forces are acting on the hammer as it slides down the roof; the weight mg\ding{217} acting downward, one component of which, mg cos \texttheta, balances the second force, the normal force exerted by the roof. At the same time, the component parallel to the roof, mg sin \texttheta, is unbalanced and produces the acceleration on the hammer. Apply Newton's second law to the unbalanced force to obtain mg sin \texttheta = ma. Thus the hammer accelerates down the roof with acceleration a = g sin \texttheta . In this case sin \texttheta = sin 30\textdegree = (1/2). The kinematic relation for constant acceleration which does not involve time is used to find the velocity with which the hammer leaves the roof. It is v^2 = v^2_0 + 2as, where v_0,the initial velocity; is 0 and s is the distance the hammer moves on the roof (= 25 ft). Hence, v is obtained from v^2 = 2 × (32/2) ft/sec^2 × 32 ft; that is, v = 32 ft/sec. In the second stage of the fall, the hammer undergoes projectile motion. It drops 32 ft in time t while traveling a distance x horizontally. Let the positive direction of y be taken as downward, and resolve v into its vertical and its horizontal com-ponents: v sin \texttheta and v cos \texttheta, respectively (see fig. B). \texttheta is the same as the angle of the slope of the roof. Since there is no horizontal component of force acting on the hammer when it leaves the roof, there is then no horizontal acceleration. The kinematic equation for constant velocity is then x = (v cos \texttheta)t. The vertical acceleration is the constant acceleration of gravity g. Therefore y = (v sin \texttheta)t + (1/2) gt^2 where t = x/v cos \texttheta. y = v sin \texttheta [x/(v cos \texttheta)] + (g/2) × [(x^2)/(v^2 cos^2 \texttheta)]. [{(x^2 × 32 ft)/sec^2}/{2 × (32 ft/sec)^2 × (3/4)}] + (x/\surd3) - 32 ft = 0 orx^2 + 16 \surd3 × ft - 1536 ft^2= 0 (x + 32 \surd3 ft) (x - 16 \surd3 ft) = 0 x = - 32 \surd3 ftor+16 \surd3 ft. The negative answer is clearly inadmissible. It is the answer that would result if the direction of projection were reversed. Hence the correct answer is x = 16 \surd3 ft = 27.7 ft from the house.

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Question:

Mammalian hemoglobin contains about 0.33 % iron Mammalian hemoglobin contains about 0.33 % iron (Fe, atomic weight = 56 g/mole) by weight. If the (Fe, atomic weight = 56 g/mole) by weight. If the molecular weight of hemoglobin is 68,000 g/mole, how many iron atoms are there in each molecule of hemoglobin?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E04-0159.htm

Solution:

To solve this problem, we consider a one mole sample of hemoglobin. This assumption introduces no error, since the size of the hemoglobin sample does not affect the number of Fe atoms per molecule of hemoglobin. We then calculate the mass of Fe atoms in one mole of hemoglobin and divide by the atomic weight of Fe to obtain the number of moles of Fe per mole of hemoglobin. Since the molecular weight of hemoglobin is 68,000 g/mole, the weight of one mole of hemoglobin is 68,000 g/mole× 1 mole = 68,000 g. The weight of Fe in one mole of hemoglobin is then weight Fe = % Fe by weight × weight of 1 mole ofhemoglobin = 0.33% × 68,000 g = 224.4 g. The number of moles of Fe corresponding to this weight is found by dividing this weight of Fe by the atomic weight of Fe to obtain moles Fe =weight Fe / atomic weight Fe = 224.4g / 56g/mole = 4 moles. Hence, there are 4 moles of Fe per mole of hemoglobin. To convert from number of moles to number of atoms or molecules, we multiply by Avogadro's number: hence, 4 moles of Fe correspond to 4 moles × 6 × 10^23 atoms/mole = 24 × 10^23 atoms of Fe and 1 mole of hemoglobin corresponds to 1 mole × 6 × 10^23 molecules/mole = 6 × 10^23 molecules of hemoglobin. Thus, the ratio of number of atoms of Fe to molecules of hemoglobin is (24 × 10^23 atoms of Fe) / (6 × 10^23 molecules of hemoglobin) =4 atoms Fe / molecule hemoglobin There are 4 atoms of Fe in every molecule of hemoglobin .

Question:

What are the distinguishing features of the different classes of arthropods?

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/Users/wenhuchen/Documents/Crawler/Biology/F12-0302.htm

Solution:

The arthropods are generally divided into the subphyla Chelicerates ,Mandibulates(orCrustacea),Uniramiaand the extinct Trilobita . Thecheliceratebody is divided into two regions, acephalothorax and an abdomen.Chelicerateshave no antennae. The first pair of appendages are mouthparts termed chelicerae. The chelicerae are pincerlike or fanglike. The second pair serves various functions including capturing prey or serving as a sensory device, while the last four pairs of appendages on thecephalothoraxfunction mainly as walking legs. Class Merostomata contains the horseshoe crabs; other members of this class are extinct. The horseshoe crabs have five or six pairs of abdominal appendages thatfunctionasgills. The last abdominal segment ortelsonis long andspinelike. The exoskeleton contains a large amount of calcium salts. The horseshoe crabs are scavengers. Arachnids comprise the other living class ofChelicerates, including spiders, scorpions, mites and ticks. Arachnids are terrestrial. The first pair of legs is modi-fied to function as feeding devices and are calledpedipalps. In most arachnids, the pedipalps seize and tear apart prey. Most arachnids are carnivorous. In spiders, thepedipalpsare poisonous fangs. Abdominal appendages are lost, or else function as book lungs, as in spiders. Arachnids usually have eight simple eyes, each eye having a lens, optic rods, and a retina. The eyes are chiefly for the perception of moving objects, and vision is poor. Themandibulateshave antennae and have mandibles as their first pair of mouthparts. Mandibles usually function in biting and chewing, but are never claw-like as are chelicerae. Maxillae are additional mouthparts found in mostMandibulates. The classCrustaceacontains an enor-mous number of diverse aquatic animals. Crabs, shrimp, lobsters and crayfish are well known crustaceans. There are many small, lesser known crustacean species, such as water fleas, brine shrimp, barnacles, sowbugs ,sandhoppers, and fairy shrimps. Crustaceans have two pairs of antennae, a pair of mandibles, and two pairs of maxillae. Compound eyes are often present in adult crustaceans. The larger crustaceans have a calcareous cuticle. Larger crustaceans have gills, but many smaller crustaceans rely on gas exchange across the body surface. There is enormous diversity among crustaceans. A great range of diet and feeding mechanisms are used. Appendages are modified in many different ways, and the body plan varies greatly. Uniramiaencompasses the next three classes. ClassChilopoda contains the centipedes. Their body consists of a head and trunk. There is a pair of mandibles, two pairs of maxillae, and the first pair of trunk appendages are poisonous claws that enable the centipede to capture prey. All other trunk segments bear a pair of walking legs. Centipedes are terrestrial. They have tracheae andMalpighiantubules. ClassDiplopodacontains the millipedes. The body is divided into head and trunk. The trunk segments bear two pairs of legs. There are mandibles and a single fused pair of maxillae, and no poisonous claws. Millipedes are not carnivorous. Respiration and excretion are similar to that of centipedes and insects. ClassInsectais an enormous group. The insect body has 3 parts: a head, with completely fused segments; a thorax of three segments, each segment bearing a pair of legs; and an abdomen. Two pairs of wings, if present, are attached to the thorax. There is one pair of antennae, and usually a pair of compound eyes. The mouthparts of insects are mandibles, a pair of maxillae, and a lower lip which is formed from fused second maxillae. Respiration is by tracheae and excretion by Malpighian tubules. Insects are the only invertebrates which fly and are mainly terrestrial.

Question:

In the commercial preparation of hydrogen chloride gas, what weight of HCL in grams may be obtained by heating 234 g. of NaCL with excess H_2 SO_4 ? The balanced equation for the reaction is 2NaCL + H_2 SO_4 \rightarrow Na_2 SO_4 + 2HCL Molecular weights:NaCL = 58.5, HCL = 36.5.

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Solution:

This problem can be solved by using either the mole method or the proportion method. Mole method :According to the equation, 2 moles of NaCL produce 2 moles of HCL, thus 1 mole of HCL is obtained for every mole of NaCL used. To use this method, one must first find the number of moles of NaCL reacted because the number of moles of NaCL reacted equals the number of moles of HCL formed. The number of moles of NaCL reacted can be found by dividing the number of grams used by the molecular weight. number of moles = [(number of grams present) /(molecular weight of NaCL)] number of moles = [(234 g) / (58.5 g/mole)] = 4 moles NaCL Since 4 moles of NaCL are reacted, one knows that moles of HCL will be formed. Thus, the weight of the HCL formed is equal to 4 times the molecular weight of HCL. weight of HCL formed = number of moles × molecular weight weight of HC1 formed = 4 moles × 36.5 g/mol = 146 g HCL Proportion method :An alternate method of solution to this problem is also possible. In this method, the molecular weights (multiplied by the proper coefficients) are placed below the formula in the equation and the amounts of substances (given and unknown) are placed above. 234 gX 2 NaCL + H_2 SO_4 \rightarrowNa_2 SO_4 + 2 HCL 2 × 58.5 g2 × 36.5 gr X is the weight of HCL produced. Solving for X [(234 g) / (117 gr)] = (X/73 g) X = [(23.4 g) × (73 g)] / [117 g] = 146 g.

Question:

An electron is separated from a proton by a dis-tance of 10^\rule{1em}{1pt}10 m. What is the gravitational and electro-static potential energy due to the proton at the position of the electron? How much work is required against each field to move the electron to a position of separation 5 × 10^\rule{1em}{1pt}10 m?

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Solution:

The mass of the proton is m_p = 1.67 × 10^\rule{1em}{1pt}27kg and the mass of the electron is m_e = 9.11 × 10^\rule{1em}{1pt}31kg. The magnitude of the charge on each is e = 1.60 x 10^\rule{1em}{1pt}19 coul. The gravitational potential energy is given by U_G = \rule{1em}{1pt}G(m_Pm_e / R) = [(-6.67 × 10^\rule{1em}{1pt}11) × (1.67 × 10^\rule{1em}{1pt}27) × (9.11 × 10^\rule{1em}{1pt}31)] / (R) = [(\rule{1em}{1pt}1.01 × 10^\rule{1em}{1pt}67) / (R)] joules U_E = \rule{1em}{1pt}kE(e^2 / R) = [\rule{1em}{1pt}(8.99 × 10^\rule{1em}{1pt}9) × (1.60 x 10^\rule{1em}{1pt}19)^2 ] / R = [(\rule{1em}{1pt}2.30 × 10^\rule{1em}{1pt}28) / R] joules Thus at R = 10^\rule{1em}{1pt}10 m, U_G = \rule{1em}{1pt}1.01 × 10^\rule{1em}{1pt}57 J and U_E = \rule{1em}{1pt}2.30 × 10^\rule{1em}{1pt}18 J. At R = 5 ×10^\rule{1em}{1pt}10 m, U_G = \rule{1em}{1pt}2.03 × 10^\rule{1em}{1pt}58 J and U_E = \rule{1em}{1pt}4.60 × 10^\rule{1em}{1pt}19 J. The work required to move from one point to the other is W_G = ∆U_G = 8.1 × 10^\rule{1em}{1pt}58 J W_E = ∆U_E = 1.84 × 10^\rule{1em}{1pt}18 J. In both cases work had to be done on the electron, thus W is positive. Notice that W_E is greater than W_G by the enormous factor of about 10^40, which indicates the weakness of the gravitational interaction.

Question:

The population flow between two regions, A and B, is assumed to be proportional to the difference in population density between the areas. Let N_1(t), N_2(t) be the populations in regions A and B respect-ively, with N_1(0) = N_10 and N_2(0) = N_20 . The natural (i.e., in absence of immigration) rates of growth of the regions (per unit of time) are given by the following formulas : Case (1):Ṅ_1 = rN_1(1) Ṅ_2 = (b - mN_2/2N_20)N_2(2) Case (2):Ṅ1= k(S - N_1)(3) Ṅ_2 = r_2N_2(4) Case (3):Ṅ_1 =k(S - N_1)(5) Ṅ_2 = (b - mN_2/2N_20)N2(6) where r,r_2,b,m,k are positive constants, 0 < k < 1 . Write a FORTRAN program which uses the modified Euler method to simulate this system from t = 0 to t = t_f .

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Solution:

A first representation of the system might be the flow graph on figure #1. The rates of flow out of the sources and into the sinks are determined by natural "valves". The rates of flow between N_1 and N2 are determined by valves which are controlled by density levels. Using this information, a detailed diagram of the system is shown on figure #2. The equations for the natural valves are equations (1)-(6). The equation for the inter-region population flow is Q = C (N_1/A - N_2/B)(7) where C is a proportionality constant. Density at A > Density at B \Rightarrow Q > 0. Therefore, the arbitrarily assumed direction of Q is toward B, away from A, and the complete state equations are : Ṅ_1^\textasteriskcentered = Ṅ_1 - Q(8) N_2^\textasteriskcentered= N_2 + Q(9) Case (1):Substitution of equations (1) and (2) into equations (8) and (9), respectively, yields: Ṅ_1^\textasteriskcentered= rN_1 - Q Ṅ_2^\textasteriskcentered = (b - mN_2/2N_20)N_2 + Q Substituting the formula for Q given by equation (7): Ṅ_1\textasteriskcentered = rN_1 - C(N_1/A - N_2/B) = (r - C/A)N_1 + (C/B)N_2 = C_11N_1 + C_12N_2(10) where C_11 = r - C/A and C_12 = C/B can be evaluated outside of the simulation loop, reducing the number of operations per iteration by three: Ṅ_2^\textasteriskcentered = (b - mN_2/2N_20)N_2 + C(N_1/A - N_2/B) = (C/A)N_1 + (b - C/B)N_2 - (m/2N_20)N^2_2 = C_21N_1 + C_22N_2 - C_3N^2_2(11) where C21= C/A, C_22 = b - C/B, C_3 = m/2N_20 . For this coupled, non-linear first-order system, the modified method is first applied to equation (10) and then to equation (11). A logical flag, TWO, is used to determine which equation is being integrated. The variable BIRTH is used for b: REAL T/0.0/,TFIN,DT,REALN,ACCUR,N1,N2, N10,N20,A,B,BIRTH,C,C11,C12,C21 C22,C3,M,R INTEGER I,N LOGICAL TWO/.FALSE./ COMMON N1,N2,TW0,Cll,C12,C21,C22,C3 READ, N, TFIN, ACCUR, N10,N20, A, B, BIRTH, C,M,R PRINT,T,N10,N20 REALN = N DT = TFIN/REALN C11 = R - C/A C12 = C/B C21 = C/A C22 = BIRTH - C/B C3 = M/(2.\textasteriskcenteredN20) N1 = N10 N2 = N20 DO 10 I = 1,N T = T + DT CALL MEULER(T,N1,ACCUR,DT) TWO = .TRUE. CALL MEULER(T,N2,ACCUR,DT) TWO = .FALSE. PRINT,T,N1,N2 10CONTINUE STOP END FUNCTION G(W) REAL G,W,N1,N2,Cll,C12,C21,C22,C3 LOGICAL TWO COMMON N1,N2,TWO,Cll,C12,C21,C22,C3 G = C11\textasteriskcenteredW + C12\textasteriskcenteredN2 IF (TWO) G = C21\textasteriskcenteredN1 + C22\textasteriskcenteredW - C3\textasteriskcenteredW\textasteriskcentered\textasteriskcentered2 RETURN END Case (2): Substituting equations (3) and (4) into equations (8) and (9), respectively, yield: Ṅ_1^\textasteriskcentered = K(S - N_1) - Q Ṅ_2^\textasteriskcentered = r_2N_2 + Q Substituting the formula for Q gives: Ṅ_1^\textasteriskcentered = K(S - N_1) - C(N_1/A - N_2/B) = -(K + C/A)N_1 + (C/B)N_2 + KS = C_11N_1 + C_12N_2 + C_3 where C_11 = -(K + C/A), C_12 = C/B , C_3 = KS Ṅ_2^\textasteriskcentered = r_2N_2 + C(N_1/A - N_2/B) = (C/A)N_1 + (r - C/B)N_2 = C_21N_1 + C_22N_2 where C_21 = C/A, C22= r - C/B. Using the same procedure as used in Case(1): REAL T/0.0/,TFIN,DT,REALN,ACCUR,N1,N2, 1N10,N20,A,B,C,K, R, S,C11,C12,C21 1C22,C3 INTEGER I,N LOGICAL TWO/.FALSE./ COMMON Nl,N2,TWO,Cll,C12,C21,C22,C3 READ,N,TFIN,ACCUR,N10,N20,A,B,C,K,R,S PRINT,T,N10,N20 REALN = N DT = TFIN/REALN C11 = -(K + C/A) C12 = C/B C3 = K\textasteriskcenteredS C21 = C/A C22 = R - C/B N1 = N10 N2 = N20 DO 10 I = 1,N T = T + DT CALL MEULER (T,N1,ACCUR, DT) TWO = .TRUE. CALL MEULER (T,N2,ACCUR,DT) TWO = .FALSE. PRINT, T,N1,N2 10CONTINUE STOP END FUNCTION G(W) REAL G,W,N1,N2,C11,C12,C21,C22,C3 LOGICAL TWO COMMONN1,N2,TWO,C11,C12,C21,C22,C3 G = C11\textasteriskcenteredW + C12\textasteriskcenteredN2 + C3 IF (TWO)G = C21\textasteriskcenteredN1 + C22\textasteriskcenteredW RETURN END Case (3):Substituting equations (5) and (6) into equations (8) and (9), respectively , yield: Ṅ_1^\textasteriskcentered = K(S - N_1 ) - Q = -(K + C/A)N_1 + (C/B)N_2 + KS = C_11N_1 + C_12N_2 + C_3 where C_11 = -(K + C/A), C_12 = C/B, C_3 = KS Ṅ_1^\textasteriskcentered = (b - mN_2/2N_20)N_2 + Q = (C/A)N_1 + (b - C/B)N_2 - (m/2N_20)N^2_2 = C_21N_1 + C_22N_2 - C_4N^2_2 where C_21 = C/A, C_22 =b - C/B, C_4 = m/2N_20 . Using the same procedure as used in Case (1), and using the variable BIRTH for b : REAL T/0.0/,TFIN,DT,REALN,ACCUR,N1,N2,N10, 1N20,A,B,BIRTH,C,C11,C12,C21,C22,C3,C4, 1K,M, S INTERGER I,N LOGICAL TWO/.FALSE./ COMMON N1, N2, TWO, C11, C12, C21, C22, C3, C4 READ, N, TFIN, ACCUR, N10, N20, A, B, BIRTH, C, K, M, S PRINT,T,N10,N20 REALN = N DT = TFIN/REALN C11 = -(K + C/A) C12 = C/B C3= K\textasteriskcenteredS C21 = C/A C22 = BIRTH - C/B C4 = M/(2\textasteriskcenteredN20) N1 = N10 N2 = N20 DO 10 I = 1,N T = T + DT CALL MEULER (T,N1,ACCUR, DT) TWO = .TRUE. CALL MEULER (T,N2,ACCUR,DT) TWO = .FALSE. PRINT,T,N1,N2 10CONTINUE STOP END FUNCTION G(W) REAL G,W,N1,N2,C11,C12,C21,C22,C3,C4 LOGICAL TWO COMMON Nl,N2,TWO,C11,C12,C21,C22,C3,C4 G = C11\textasteriskcenteredW + C12\textasteriskcenteredN2 + C3 IF (TWO) G = C21\textasteriskcenteredN1 + C22\textasteriskcenteredW - C4\textasteriskcenteredW\textasteriskcentered\textasteriskcentered2 RETURN END

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Question:

In corn, the gene R for red color is dominant over the gene r for green color. The gene N for normal seed is dominant over the gene n for tassel seed. A fully heterozygous red plant with normal seeds was crossed with a green plant with tassel seeds, and the following ratios were obtained in the offspring: 124 red, normal: 77 red, tassel: 126 green, tassel: 72 green, normal. Does this indicate linkage? If so, what is the map distance between the two loci?

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Solution:

Let us look at the cross, F RN RN rN rn rn RrNn Rrnn rrNn rrnn 124 red, normal (RrNn) - parental 77 red, tassel (Rrnn) - recombinant 73 green, normal (rrNn) - recombinant 126 green, tassel (rrnn) - parental The phenotypic ratio is indeed indicative of linkage. The two parental types (RrNn and rrnn) are found in much greater numbers than the recombinant types (Rrnn and rrNn). The recombination frequency (RF) is: RF = [(total recombitants) / (total offspring)] = (73 + 77) / (124 + 77 + 126 + 73) = (150) / (400) = .375 An RF of .375 would give a map distance of 0.375 × 100 or 37.5 map units. The two loci are thus located on the same chromosome and are 37.5 map units apart.

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Question:

Discuss the characteristics of the echinoderms arid briefly describe several different classes.

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Solution:

The echinoderms havepentamerousradial sym-metry, that is, the body can be divided into five similar parts along a central axis. Echinoderms have anendo-skeleton composed of calcareousossicles, these may articulate with each other as in sea stars, or may form interlocking plates, as in sand dollars. The skeleton usually bears projecting spines that give the skin a spiny or warty surface. The echinodermcoelomis well developed. The system ofcoelemiccanals and projections, called the water vascular system, is found only in echinoderms. ClassAsteroideacontains the sea stars (starfish). Sea stars have a central disc and usually five arms, although some have more. The mouth is on the lower surface of the disc, and the anus on the upper surface. Locomotion is by means of the tube feet. The surface is studded with many short spines. They have no special respiratory system, but breathe by means of skin gills or dermalbranchiae. ClassOphiuroideacontains the brittle stars. They resemble sea stars, but their arms are longer, more slender, more flexible, and often branched. Locomotion occurs by rapidly lashing the arms, not through the use of tube feet. There is no anus or intestine. Gas exchange occurs throughinvaginatedpouches at the periphery of the disc. The sea urchins and sand dollars are members of Class Echinoidea . Theendoskeletalplates are fused to form a rigid case. The body is spherical or oval, and is covered with long spines. Five bands of tube feet are present, and function in respiration in some species. There is a long, coiled intestine, and an anus. Sea cucumbers, classHolothuroidea, have a much reduced endoskeleton and a leathery body. Many forms have five rows of tube feet. Unlike other echinoderms, they lie on their side. The mouth is surrounded by tentacles attached to the water vascular system. Gas exchange usually occurs through complexly branched respiratory trees attached to the anal opening. ClassCrinoideacontains the sea lilies. They are attached to the sea floor by a long stalk and are sessile. They havelong,feathery, branched arms around the mouth, which is on the upper surface.

Question:

Chromium exists in four isotopic forms. The atomic masses and percent occurrences of these isotopes are listed in the following table: Isotopic mass(amu) Percent occurrence 50 4.31% 52 83.76% 53 9.55% 54 2.38% Calculate the average atomic mass of chromium.

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Solution:

We will make use of the definition of averages: A = p_1M_1 + p_2M_2 + ..... + P_NMN where A is the average value, M_i is the atomic mass of isotope "i" and pi is the corresponding probability of occurrence. For the four isotopes of chromium, we have: M_1 = 50amup_1 = 4.31%= 0.0431 M_1 = 52amup_2 = 83.76% = 0.8376 M_3 = 53amup_3 = 9.55%= 0.0955 M4= 54amup_4 = 2.38%= 0.0238 Hence, the average atomic mass of chromium is A = p_1M_1 + p_2M_2 + p_3M_3 + p_4M_4 = 0.0431× 50amu+ 0.8376 × 52amu+ 0.0955 × 53amu+ 0.0238 × 54 amu = 2.155amu+ 43.555amu+ 5.062amu+ 1.285amu = 52.057amu.

Question:

Discuss the early development of the angiosperm embryo. What primary structure form and what does become in the seeding?

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Solution:

The first cell division that the zygote undergoes following fertilization produces a basal cell and a terminal cell. The basal cell develops into the suspensor, a filament of cells serving as a point of attachment for the embryo within the seed. The suspensor will eventually disintegrate as development of the embryo proceeds. The embryo will develop from the terminal cell, and as the cell undergoes successive divisions, the characteristic structures of the embryo begin to take shape (see figure 1). The most obvious of these are the cotyledons, or primary leaves. Dicots develop two such leaves, monocots only one. In dicots, the cotyledons can serve to either absorb or to both store and absorb food from the endosperm while within the seed. In the monocots, the single cotyledon serves primarily to absorb, rather than store, the endosperm tissue. The portion of the embryo lying along the central axis below the point of attachment of the cotyledons is called the hypocotyl and the part above is called the epicotyl. At this point in its development, the embryo becomes dormant, and will remain so until conditions are favorable for its germi-nation. Upon germination, the hypocotyl elongates and emerges from the seed coat (see Fig. 2). It gives rise to the primitive root or radicle. Since the radicle is strongly and positively geotropic, it grows directly downward into the soil. The arching of the hypocotyl in the seed pulls the cotyledons and epicotyl out of the seed coat. The epicotyl, being negatively geotropic, grows upward out of the soil. It will develop into the stem and leaves. In most dicots, by the time germination occurs, the cotyledons will have completely absorbed the endosperm. They now serve as reserves of food for the growing seedling until it has developed enough chlorophyll to become independent, at which point they shrivel and fall off. In some dicots, the cotyledons do not store nutrient material, but become photosynthetic foliage leaves upon germination. In monocots, the endosperm usually persists even after germination, and the cotyledon continues to absorb the nutrient material for the seedling until it can synthesize its own nutrients.

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Question:

(a) Determine the field intensity midway between two identical positive charges of 200 stat-coulombs each, in oil of dielectric con-stant 5, if the charges are 6 cm apart. (b) What is the answer if one charge is positive and the other is negative ?

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Solution:

(a) Before attempting to calculate the numerical value of the field intensity E at point P due to each charge, note that if unit positive charge is placed at P, the forces acting on it due to each charge will point in opposite directions and will balance each other to give a resultant field intensity of zero. (See figure (a).) E at P is zero. (b) A unit positive charge placed at P will experience a force due to A: (See figure (b)) E_1 = q / Kr^2 = (200 statC / {[(5){(statC2)/(dyne \rule{1em}{1pt} cm^2)}]^ (9cm)^2}) = 4.4 dyne /stat-coulombto the right. K is the dielectric constant of the media, It accounts for the influence the media has on the electric field of a charge. Also due to B: E_2 = 2 / Kr^2 = 4.4 dyne /stat-coulomb also to the right therefore E = E_1 + E_2 = 8.8 dyne / stat-coulomb from + charge to \rule{1em}{1pt} charge.

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Question:

X-radiation of wavelength 0.8560 \AA is scattered from a carbon target. Calculate the change in wavelength produced for radiation scattered at 90, and the energy and direction of the corresponding recoil electrons. (See figure).

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Solution:

In a Compton collision, an x-ray is incident on an unbound (free) target electron, and a collision takes place. After the col-lision, the electron is scattered, but it is observed that the scattered x-ray has a wavelength, \lambda', which is different than the wavelength, \lambda, of the incident x-ray. (Because of the observed particulate behavior of the x-rays, this experiment gave further support to the photon theory of light). The resulting change in wavelength is given by the formula \lambda' - \lambda = 2(h/mc) sin^2(\texttheta/2) , where \texttheta is the angle of scatter, m is the electron rest mass, c is the speed of light and h is Planck's constant. If \texttheta is 90 , \lambda' - \lambda = h/mc = [(6.625 × 1^-34 J \bullet s) / (9.108 × 10^-31kg × 2.998 × 10^8m \bullet s^-1)] = 0.0242 \AA. The energy given to the electron is therefore equal to the loss of energy of the photon. E = h(c/\lambda) - h(c/\lambda') = 6.625 × 10^34J \bullet s × 2.998 × 10^8 m \bullet s^-1× [(10^10/0.8560m) - (10^10/0.8802m)] = 6.377 × 10^-17 J. In order that momentum may be conserved in the collision (see figure), the electron must have a momentum in the initial direction of h/\lambda and a momentum of h/\lambda' at right angles to this. The tangent of the angle \Phi at which the electron is scattered is the ratio of the latter momen-tum to the former, since these momenta are equal to the components of the electron's momentum in the two mutually perpendicular directions. Hence tan \Phi =(h/\lambda')/(h/\lambda) = \lambda/\lambda' =(0.8560 \AA)/(0.8802 \AA) = 0.9725. \Phi = 44012' .

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Question:

An immersion heater in an insulated vessel of negligible heat capacity bringsm_W= 100 g of water to the boiling point from 160C in 7 min. The water is replaced by m_a = 200 g of alcohol, which is heated from the same initial temperature to the boiling point of 78\textdegreeC in 6 min 12 s. Then 30 g are vaporized in 5 min 6 s. Determine the specific heat and the heat of vaporization of alcohol, and the power of the heater.

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/Users/wenhuchen/Documents/Crawler/Physics/D13-0483.htm

Solution:

In 7 min 100 g of water are raised in temperature by (100 -16) \textdegreeC = 84\textdegreeC. The amount of heat provided by the heater to the water is, Q_W = C_Wm_W∆T where C_W is water's specific heat, and M_W is the mass of water. Q_W = (1 cal/gr.\textdegreeC) (100gr) (100\textdegreeC - 16\textdegreeC) = 8.4 × 10^3 cal = (8.4 × 10^3 cal) × (4.186 J/cal) = 3.52 × 10^4 J The rate of delivery of the heat energy is, P = Q_W / t =Q_W / [(7 min) (60 sec/min)]= [(8.4 × 10^3)/(420)] cal/s = 20 cal/s = [(3.52 × 10^4) / 420] J/s = 83.7 W Therefore the power of the heater is 83.7 W, With 200 gr. of alcohol in the vessel, the temperature rises from 16\textdegreeC to 78\textdegreeC in 6 min 12 s. If C_a is the specific heat of alco-hol and M_a is the alcohol's mass, it absorbs, Q_a=C_aM_a(78\textdegreeC - 16\textdegreeC) = C_a (200gr) (62\textdegreeC) amount of heat during the temperature rise. The power of the heater remains the same while heating the water or the alcohol, hence, in calories per second, we have, P =Q_a/ [(6 min) (60s/min) + 12 s] =Q_a/ [372 s] 20 cal/s = [C_a × (200gr) × (620c)] / [372 s] giving c_a = [(20 cal/s) (372 s)] / [200gr(\textasteriskcentered620c)] = 0.6 cal/C\textdegree.gr 30grof alcohol are vaporized in 5 min 6 s = 306 s, hence the amount of heat, Q', required to vaporize it at 78\textdegreeC is, Q ' = p × (306 s) = (20 cal/s) (306 s) = 6.12 × 10^3 cal. If L is the heat of vaporization of alcohol, Q' is given by, Q ' = L (30gr) Then,L = (Q' / 30gr) = [(6.12 × 10^3 cal) / 30gr] = 204 cal/gr

Question:

Obtain the 7's and 8's complement of the following octal numbers: 770, 1263, 00010 and 0000.

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Solution:

In the octal system, 8 = 10, 8^2 = 100, 8^3 = 1000, etc. Thus, the octal system contains only the numbers 0,1,2,3,4,5,6,7. a) The seven's complement of 770 is 777 -770 007 The eight's complement is then 007 + 1 = 010 This could also have been obtained directly: 1000 - 770 010 b) The seven's complement is: 7777 -1263 6514 The eight's complement is: 6514 + 1 = 6515 according to the relationship Radix complement (X) = Radix complement minus one (X) + 1 (In base eight arithmetic, 8 is the radix and the radix complement of a number equals c) The radix minus one complement of 00010 is: 77777 - 00010 77767 and the radix complement is: 77770 Note that although 00010 = 10, the radix complements are different. In the computer, numbers are stored in words of length 12 - 36 bits. In a 12-bit word, the number 10 would be stored as 000000000010 The radix complement is: 777777777770 d) The radix complement of 0000 is: 10000 - 0000 10000 and the radix minus one complement is: 10000 - 1 = 7777.

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Question:

What is the total binding energy of _6C^12 and the average binding energy per nucleon?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E19-0716.htm

Solution:

The mass of an atom, in general, is not equal to the sum of its component masses. The mass of the com-ponent parts (protons, neutrons, and electrons) is slightly greater than the mass of the atom. This difference in mass has an energy equivalent (E = mc^2), which is called the binding energy of the nucleus. Although binding energy refers to the nucleus, it is more convenient to use the mass of the whole atom in calcu-lations. Then,M_n= M -ZM_e, whereM_n, M, and M_e are the nuclear, atomic, and electron masses, respectively, and Z is the atomic number. Since a carbon 6 atom, C_6^12, is made up of 6 protons and 6 electrons (or 6 H^1 atoms) plus 6 neutrons, then the binding energy, (b.e.) can be represented as follows: b.e. =M_n+ZM_e- M b.e. for _6C^12 = 6 [mass of electron and proton (H_1) + mass of neutron] - atomic mass of _6C^12. In other words, a mass difference equation can be written in terms of whole atom masses. Mass of 6 H^1 atoms = 6 × 1.0078= 6.0468 Mass of 6 neutrons = 6 × 1.0087=6.0522 Total mass of component particles= 12.0990 Atomic mass of _6C^12=12.0000 Loss in mass of formation of _6C^12= 0.0990 Binding energy (931.5MeV/ ∆mass) (.0990 ∆mass) = 92.22MeV Since there are 12 nucleons (protons plus neutrons), the average binding energy per nucleon is [(92.22MeV) / (12 nucleons)] = 7.68MeV.

Question:

What is structure of a typical neuron?

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/Users/wenhuchen/Documents/Crawler/Biology/F20-0518.htm

Solution:

Before we get to the answer, it is important to recognize the fact that the term 'typical neuron' is rather vague. Neurons, which are the basic structural and functional units of the nervous systems of multicellular animals, show a great diversity in types. In humans, many types of neurons are present (Figure 1). Nevertheless, often three parts of a neuron can be distinguished: a cell body, an axon, and a group of processes called dendrites. Dendrites are usually rather short and numerous extensions from the cell body. They frequently branch profusely, and their many short terminals may give them a spiny appearance. When stained, they ordinarily show many dark granules. There is usually only one axon per neuron (very rarely two), and it is frequently longer than the dendrites. It may branch extensively, but unlike dendrites it does not have a spiny appearance and does not show dark granules when stained. The most fundamental distinction between dendrites and axons is that dendrites receive excitation from other cells whereas axons generally do not, and that axons can stimulate other cells whereas dendrites cannot. Thus, den-drites carry information to the cell body while axons carry information away from the cell body. Axons may be several feet long in some neurons. A bundle of many axons wrapped together by a sheath of connective tissue is what we commonly call a nerve. Each vertebrate axon is usually enveloped in a myelin sheath formed by special cells, the Schwann cells, that almost com-pletely encircle the axon. Schwann cells play a role in the nutrition of the axons, and provide a conduit within which damaged axons can grow from the cell body back to their original position. The myelin sheath is interrupted at regular intervals; the interruptions are called nodes of Ranvier. At these nodes, the myelin sheath disappears. The myelin functions in speeding up the transmission of impulses in the axon it envelopes. It is crucial to note that the myelin sheath is not a separate layer by itself. Through electron microscopy, it has been proved that the sheath is not a secretion product of the axon or Schwann cells, as was once believed, but a tightly packed spiral of the cell mem-brane of the Schwann cells. Thus the sheath is composed of the lipid from the membrane's bilayer. The nucleus and cyto-plasm of the Schwann cell are pushed aside to form the neu-rolemma. The nodes in the myelin sheath are simply the points at which one Schwann cell ends and another begins (see Figure 2). The neuronal cell body, or perikaryon, contains the nucleus and cytoplasmic organelles distributed around the nucleus. The perikaryon has well-developed endoplasmic reticulum and Golgi apparatus for manufacturing all the substances needed for the maintenance and functioning of the axon and dendrites.

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Question:

Find the pH of a solution of 0.3 M acetic acid. Assume a K_a value of 1.8 × 10^-5.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E12-0418.htm

Solution:

The dissociation of acetic acid into positive and negative ions proceeds as follows: HC_2H_3O_2\rightleftarrowsH^++C_2H_3O_2^- For this particular dissociation the equilibrium constant, K_a, is written as K_a = {[H^+] [C_2H_3O_2^-]} / [HC_2H_3O_2] = 1.8 × 10^-5, where the brackets represent the concentrations of the substances within them. The pH of a solution is defined as the negative log of the hydrogen ion concentration. Mathematically, it is written as pH = - log [H^+]. Thus, to answer this problem, the [H^+] in this solution must be calculated from the equilibrium constant expression and then substituted into the pH equation. From thestoichiometryof the dissociation reaction, for every x molar concentration of H^+ formed there are x molar concentration of C_2H_3O_2^- also formed. Therefore, let x = [H^+] = [C_2H_3O_2^-]. The x molar concentration of these ions comes from the dissociation of the 0.3 M acetic acid. Thus, the exact amount of acetic acid in solution at equilibrium is 0.3 - x However, x is a very small amount as compared with 0.3 M of the acetic acid and is, therefore, approximated as 0.3 M. Substituting these values into the equilibrium expression 1.8 × 10^-5 =(x \bullet x) / (0.3 - x) \allequalx^2/0.3 Solving for x; x = [H^+] = 2.3 × 10^-3 Substituting this value of [H^+] into the pH equation: pH = - log (2.3 × 10^-3) = 2.63.

Question:

Arthropods have an exoskeleton. How is movement accomplished?

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/Users/wenhuchen/Documents/Crawler/Biology/F19-0484.htm

Solution:

Movement in the arthropods is possible in spite of the hard exoskeletonbecause the body is segmented and the segments are joined bya thin layer of flexible chitin. Jointed legs are especially characteristic of the arthropods; they consist of a series of cone-like sections with the small end of one fitting into the large end of the next. Only arthropods and vertebrates have jointed appendages; there are more joints, however, in the arthropod legs because each joint does not have as great a degree of movement as the joint of a vertebrates.

Question:

By the time the blood reaches the veins, the blood pressure is too low to effectively move blood back to the heart. What other mechanisms aid the flow of blood in the veins?

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Solution:

The blood flow in the veins is partially aided by the movements of skeletal muscles near the veins. The middle layer of the venous walls has much less smooth muscle than does the middle layer of the arterial walls. Veins are therefore much (8 X) more distensible and easily collapsible. Most of the veins are surrounded by skeletal muscles, which contract as the body moves. This contraction exerts a pressure on the veins, com-pressing them. The compression of the veins forces the blood to move. As the muscle relaxes, the compressed section of vein fills with the blood, which is again pushed along by the next contraction. This "milking" action of skeletal muscle is shown in the accompanying figure. Why does the blood only move in the direction of the heart? At numerous points along the interior of the vein, there are valves which allow blood to flow in only one direction - towards the heart. Thus when contracting muscle forces the blood to move in the veins, the valves act to ensure unidirectional movement of blood by preventing backflow. When one stands perfectly still for a long period, the milk-ing action of the skeletal muscles is decreased, venous pressure increases in the lower parts of the legs increasing capillary pressure. This forces fluid from the circulatory system into the tissue spaces, causing swelling (edema). The movement of the chest during breathing also aids the flow of blood to the heart. As the diaphragm and chest muscles contract during inspiration, the chest cavity increases in size. The in-creased volume causes a decrease in pressure within the cavity to below atmospheric pressure. This pressure gradient causes air to flow into the lungs. But this decrease in pressure also lowers venous pressure in the chest cavity. During inhalation, blood thus tends to be drawn into the veins in the chest cavity. During exhalation, blood cannot be forced back because of the venous valves.

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Question:

The hump on a camel's back is mostly depot fat. Assuming complete hydrolysis of the fat, and its subsequent oxi-dation to CO_2 and water, calculate the number of kilocalories that would be released from 25 kg of camel fat.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E22-0798.htm

Solution:

Depot fat is usually made up of a mixture of three triglycerides - triolein, tristearin and tripalmitin. These fats are produced by a reaction of three moles of their respective fatty acids (oleic, stearic or palmitic) and one mole of glycerol. The structure of a fatty acid is where R is a long chain saturated or unsaturated alkyl group. The equation for this reaction is It has been found that 9 Kcal of energy is released by each gram of fat hydrolyzed and then oxidized. In this problem, one is asked to find the amount of energy re-leased when 25 kg of fat is oxidized. 25 kg is equal to 25000 g. Therefore, the number of kilocalories produced in this reaction is 25000 g × 9. Kcal/g = 2.25 × 10^5 Kcal.

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Question:

Radioastronomystudies indicate that the clouds of matter between stars consist not only of simple atoms but include molecules that may contain several atoms. From a region of space labeledSgrB by astronomers, photons are observed with a frequency of 1.158 × 10^11 Hz. Determine the rota-tional constant B tor this microwave radiation if the tran-sition is known to be from the j = 1 state to the ground state (j = 0).

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/Users/wenhuchen/Documents/Crawler/Physics/D33-1002.htm

Solution:

The rotational energy of a particle in terms of its angular momentum L is E = [L^2/(2MR^2)] = BL^2 where B is a constant. The quantum mechanical expression for L^2 is j (j + 1)h, whereh= (h/2\pi) and j takes on positive integral values. The letter j also classifies the possible rotational energy states of mass M. The energy of the upper state must therefore be E_U =Bj(j + 1)h= B1(1 + 1)h= 2Bh while the energy of the lower state is E_L = B0 (0 + 1)h= 0. The energy difference between these two states must equal the photon energyhf(conservation of energy): hf= E_U - E_L where f is the frequency of the observed radiation .Hence 2B (h/2\pi) =hf B = \pi × 1.158 × 10^11 Hz. = 3.636 × 10^11 Hz.

Question:

If the radius of the earth were to decrease by 1/2 percent what would be the change in the length of the day? Consider the earth to be a uniform sphere whose moment of inertia is I = 2/5 mR^2.

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/Users/wenhuchen/Documents/Crawler/Physics/D06-0341.htm

Solution:

A dancer's technique of increasing her angular velocity by pulling her arms in to her side conveys the idea of conservation of angular momentum. The dancer is initially rotating at angular velocity \cyrchar\cyromega_0, and she has a moment of inertia I_0. After she pulls her arms in, she has decreased her moment of inertia to I_f. We observe that she has an increased angular velocity, \cyrchar\cyromega_f, and we may therefore write I0\cyrchar\cyromega_0 = I_f \cyrchar\cyromegaf Since I\cyrchar\cyromega is angular momentum, we have motivated the principle of conservation of angular momentum L_0 = Lf where L_0 and L_f are the initial and final angular moment. This general principle holds whenever no external torques act on the system which we are observing. In the present case, we may apply this principle to find the final angular velocity of the earth. We then relate this to the period of the earth's rotation, and, hence, the length of the day. L_0 = Lf I_0 \cyrchar\cyromega_0 = If\cyrchar\cyromega_f [(2/5) mR^2 _0] \cyrchar\cyromega_0 = [(2/5) mR2_f] \cyrchar\cyromega_f R2_0 \cyrchar\cyromega_0 = R2f\cyrchar\cyromega_f \cyrchar\cyromega_f = (\cyrchar\cyromega_0R20)/(R^2_f). But R_f = R_0 - .005R_0 since the final radius of the earth is .5% less than its initial radius. \cyrchar\cyromega_f = (\cyrchar\cyromega_0R_0)/(R_0 - .005R_0)2 \cyrchar\cyromega_f = (\cyrchar\cyromega_0R2_0) / [R2_0(1 - .005)^2] \cyrchar\cyromega_f = \cyrchar\cyromega_0/(.995)^2 = 1.01 \cyrchar\cyromega_0(1) But angular velocity is defined as \cyrchar\cyromega = 2\pi/T Where it takes a body T sec to traverse 2\pi radians of angular distance. Hence \cyrchar\cyromega_f / \cyrchar\cyromega_0 = (2\pi/T_f)(T_0/2\pi) = T_0/T_f(2) Combining (1) and (2) T_0/T_f = 1.01 orT_f = T_0/1.01 But T_0 is the time it took the earth to sweep out 2\pi radians originally, before its radius decreased. This time is 1 day, and T_f = (1 day)/(1.01) = (86400 secs)/(1.01) = 85544.55 secs Hence, the change in the length of a day, is T_f - T_0 = (86400 - 85544.55) secs = 855.45 secs T_f - T_0 = 14.28 minutes.

Question:

What are the advantages of a segmented body plan? Which animal phyla have segmentation?

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Solution:

Segmentation offers the advantage of allowing specialization of different body segments for different functions. More primitive animals have a large number of segments, all very similar to one another. More complex animals have fewer segments, and the specialization of different segments may be so far advanced that the original segmentation of the body plan is obscured. \textasciigraveAnnelida, the phylum containing earthworms, marine worms, and leeches, is the most primitive invertebrate group displaying segmentation. Earthworms and most marine worms, have both internal and external segmentation. The earthworm body is composed of one hundred or more segments. The segments are separated by partitions termed septa. The marine worms often have incomplete septa, and the leeches have lost the septa completely. Except for a few specialized head and tail segments, the appendages and musculature of all segments are exactly the same. Internally, each segment has a pair of excretory organs and its own ventral nerve ganglion. The earthworm has five pairs of muscular tube-like hearts, two pairs of testes, three pairs of seminal vesicles, and one to five pairs of seminal receptacles. Molluscsexhibit embryonic development very similar to that of annelids, and the larval form, called thetrochophore, is very similar in both phyla. This similar-ity suggests a possible evolutionary relationship betweenmolluscsand annelids.Neopilinais the only living species of molluscs showing segmentation; it may, however, only be secondary. Onycophora (Peripatusand relatives) is a very small phylum showing segmentation. Theonycophoranshave a com-bination of annelid and arthropod characteristics, and provide an evolutionary link between these two phyla. Theonycophoranshave many similar segments with 14 to 43 pairs of short,unjointedlegs. Arthropods are a large group of segmented animals that have evolved in many complex ways. More primitive forms have paired, similar appendages oneaahsegment. More advanced forms have appendages modified for a variety of functions such as sensory antennae, feeding apparatus, walking legs, swimming legs, claws, reproductive apparatus, and respiratory structures, depending upon the species. The more advanced arthropods have a greatly reduced number of segments. These arthropods also have a fusion of body segments into distinct regions, such as a head and trunk, or head, thorax and abdomen. Internally, there is a great deal of specialization, and septa between segments have been lost. The first six segments always form a head, usually with a well-developed brain and sense organs. In primitive forms there are ganglia in every segment, but in many groups the ganglia have fused to formganglionic masses. Aquatic arthropods may have one pair or several pairs of excretory organs, orcoxalglands. In most groups of terrestrial arthropods, the excretory organs areMalpighiantubules. Generally, there is only one pair of reproductive organs; centipedes, however may have 24 testes. Chordata is the other phylum characterized by segmen-tation. As with the arthropods, there also has been in-creasing specialization of segments in chordates. In man, the nervous system and the skeletal system aresegmentallyorganized. In primitive fish, the gills and musculature are also segmented.

Question:

10 calories of heat are supplied to 1 g of water. How much does the mass of the water increase?

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Solution:

The increase in total energy of the water is \Delta\epsilon = 10 cal = 10 × (4.19 × 10^7 ergs) = 4.19× 188ergs Then using the mass-energy conversion formula: \Deltam = \Delta\epsilon / c2 = (4.19 × 10^8 ergs) / {3 × 10^10(cm/sec)^2} = 4.7 × 10-13g So that the mass of the water increases from 1 g to 1.00000000000047 g, a negligible increase indeed! But the mass has increased. Where does this additional mass come from? It is just the mass associated with the increase of kinetic energy that has been given to the water molecules by the addition of thermal energy.

Question:

Calculate the weight in grams of sulfuric acid in 2.00 liters of 0.100 molarsolution. (MW of H_2SO_4 = 98.1.)

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/Users/wenhuchen/Documents/Crawler/Chemistry/E08-0280.htm

Solution:

Molarityis defined as the number of moles of a compound in a solution. In this problem, one is trying to find the amount of H_2SO_4 in 2.00 litersof a solution that is 0.100 molar in H_2SO_4 .From the definition of molarity, one can see that in one liter of this solution there is 0.100moles ofH_2SO_4. This means that in 2.00 liters of the same solution, there is twice thatamount. numberof moles of H_2SO_4 in 2 liters = 2 liters × 0.100 moles/liter = 0.200 moles Since one knows that the molecular weight of H_2SO_4 is 98.1 g/mole andthat there are 0.200 moles present in this solution, one can find the numberof grams present by multiplying the MW by the number of moles present. numberof grams = MW × number of moles present numberof grams of H_2SO_4 = 98.1 g/mole × 0.200 moles = 19.6 g.

Question:

What are the functions offlorigenandphytochromein flowering plants?

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Solution:

In the 1930's, M. H.Chailakhyanof Russia was studying flowering in the long-night plantChrysan-themumindicum. Long-night plants will flower only when exposed to darkness for nine or more hours. Chailakhyan showed that if the upper portion of the plant was defoliated and the leaves on the lower part exposed to a long-night induction period the plant would flower. If however, only the upper, defoliated part was kept on long-nights and only the lower, leafy part on short-nights, no flowering occurred, Based on this finding,Chailakhyanproposed the production by the leaves of a naturally-occurring flowering hormone, which he named florigen , the "flower maker." This hormone apparently moves from the leaves to the flower bud, stimu-lating it to mature and flower. Later experiments showed thatflorigenpasses from the leaves to the bud by way of the phloem system, where most organic substances are transported. Unfortunately, nothing is known at present of the chemical composition offlorigennor how it may act to induce flowering. In the late 1950's, a blue, light-sensitive protein pigment was isolated in plants.Borthwickand coworkers gave this pigment the name phytochrome . In the years that followed, it was discovered that phytochrome exists in two forms: P_r (P_660, responds to red light of day) andp_fr(P_730, responds to far-red light of night). P_660, the quiescent form, absorbs red light at a wavelength of 660 nanometers and is converted to P_660. In nature, the P_660 - to - P_730 conversion takes place in daylight, while the P_730 - to - P_660 conversion occurs in the dark. Thus during the day, a plant predominantly containsphyto-chromein the form of P_730, and during the night it con-tains mostly the P_660 form. The prevalence of either form is postulated to provide the plant with a means of detecting whether it is in a light or dark environment. The rate at which P_730 is converted to P_660 provides the plant with a "clock" for measuring the duration of darkness.

Question:

A sulfuric acid solution has a density of 1.8 g/ml and is 90% H_2SO_4 by weight. What weight of H_2SO_4, is present in 1000 ml of the solution? What is themolarityof the solution? the normality?

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Solution:

1) Here one is asked to find the weight of H_2SO_4, in 1000 ml of the solution and told that the solution weighs 1.8 g/ml and that 90% of this weight is made up by H_2SO_4. The total weight of the solution is 1000 ml times the weight of one ml. Hence, weight of solution = 1000 ml × 1.8 g/ml = 1800 g. H_2SO_4 makes up 90% of this weight. weight of H_2SO_4 = .90 × weight of solution = .90 × 1800 g = 1620 g. 2) Themolarityis defined as the number of moles in one liter of solution. One has already found that 1620 g of H_2SO_4 is present in 1 liter of the solution. Thus, to calculatemolarity, one should determine the number of moles present in 1620 g. This is done by dividing 1620 g by the molecular weight of H_2SO_4. (MW of H_2SO_4 = 98.1.) moles of H_2SO_4 present = [(1620 g)/(98.1 g/mole)] = 16.5 moles Since there is 1 liter of solution, themolarityof H_2SO_4 is 16.5 M. 3) The normality is defined as the number of moles ofionizable hydrogensper liter of solution. From the following equation one can see that there are 2ionizablehydrogensfor each molecule of H_2SO_4 . H_2SO_4 \leftrightarrows 2H^+ + SO-_4 Therefore if there are 16.5 moles of H_2SO_4 in one liter of the solution the normality is twice this amount. normality of an acid = no. ofionizableH ×molarity = 2 × 16.5 = 33.0 N.

Question:

Assuming K_W = 1.0 × 10^-14 and K_a = 4.0 × 10^-10, find the pH of a 1.0 MNaCNsolution.

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Solution:

NaCNis a salt that exists in water as Na^+ and CN^-. As such, this problem requires considering the hydrolysis of a salt of a weak acid and strong base. Since water is a weak acid or weak base (which dissociates to H^+ and OH^- ions), it reacts with ions from either weak acids or bases. CN^- + H_2O \rightleftarrows HCN + OH^-. Inherent in this equationigthe ionization of water, that is, H_2O \rightleftarrowsHO-+ H^+ so that it is actually, CN^- + OH^- + H^+ \rightleftarrows HCN + OH^-. The hydrolysis of theNaCNsalt can be expressed byK_hyd, a hydrolysis constant.K_hyd, =K_w/ K_a, where K_a is the ionization of the acid formed andK_wis the ionization constant of water to OH^- + H^+, as previously mentioned. K_hyd=K_w/ K_a = {[HCN][OH^-]} / [CN^-] = (1.0 × 10^-14) / (4.0 × 10^-10). It is given that K_a = 4.0 × 10^-10.K_wis always 1.0 × 10^-14, since a pure water solution must be neutral. As such, the molar concentrations of H^+ and OH^- are both 1.0 × 10^-7 moles/liter. This stems from the fact that [H^+] = [OH^-] with [H^+][OH^-] =K_w= 1.0 × 10^-14. Therefore, [H^+]^2 = 1.0 × 10^-14 or [H^+] = 1.0 × 10^-7. One now has: (1.0 × 10^-14) / (4.0 × 10^-10) = {[HCN] [OH^-]} / [CN^-] The object is to calculate [OH^-]. From [OH^-], one can determine pOH, sincepOH= - log [OH^-]. Once one knowspOH, one can find pH, since 14 = pH +pOH. To find [OH^-], let the concentration of OH^- be x. If this is the case, then [HCN] can also be represented by x. Re-call, that CN^- + H_2O \rightleftarrows HCN + OH^-. For every mole of OH^- generated, a mole of HCN must also be produced. The concentration of [CN^-] can be represented as 1 - x. It is given that the original concentration ofNaCNor CN^-, is 1.0 M. When placed in water, some of CN^- becomes HCN. If HCN is x, then the original concentration of CN^- must be reduced by x. Therefore, [CN^-] has a concentration of 1 - x. It follows that (1.0 × 10^-14) / (4.0 × 10^-10) = (x \bullet x) / (1 - x), if one substitutes the values in the equation (1.0 × 10^-14) / (4.0 × 10^-10) = {[HCN] [OH^-]} / [CN-] The procedure now is to solve for x. x = 5.0 × 10^-3 M = [OH^-]. As previously stated,pOH= - log [OH^-]. pOH= 2.30andpH = 14 -pOH= 11.70.

Question:

Use scientific notation to express each number. (a) 4,375 (b) 186,000 (c) 0.00012 (d) 4,005

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Solution:

A number expressed in scientific notation is written as a product of a number between 1 and 10 and a power of 10. The number between 1 and 10 is obtained by moving the decimal point of the number (actual or implied) the required number of digits. The power of 10, for a number greater than 1, is positive and is one less than the number of digits before the decimal point in the original number. The power of 10, for a number less than 1, is negative and is one more than the number of zeros immediately following the decimal point in the original number. Hence, (a) 4,375 = 4.375 × 10^3(b) 186,000 = 1.86 × 10^5 (c) 0.00012 = 1.2 × 10^-4(d) 4,005 = 4.005 × 10^3.

Question:

A proton is at rest a long way away from the earth. It falls toward the earth under the influence of the earth's gravitational attraction. When it reaches the earth, all its kinetic energy is converted into the energy of a single quantum of light. What is the frequency of this quantum?

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Solution:

When the proton is at an infinite distance from the earth, the mutual gravitational potential energy of the earth-proton system is zero. When the proton reaches the surface of the earth, the mutual potential energy of the system is \varphi_ G = \rule{1em}{1pt}Gm_e^m/R_e where m is the mass of the proton, M_e is the mass of the earth, and R_e is the radius of the earth. The kinetic energy gained by the proton is equal to the potential energy lost, and so, if v is the velocity with which the proton strikes the earth, (1/2)mv^2 =Gm_em/R_e All of this energy is converted into the energy of the quan-tum of light. The energy of a quantum of light is given byhv, where v is the frequency of the light. Therefore hv=GM_e^m/Re v =GM_e^m/hR_e = [(6.67 × 10^\rule{1em}{1pt}11 Nm^2/kg^2)(6 × 10^24 kg)(1.67 × 10^\rule{1em}{1pt}27 kg)] / [(6.63 ×10^\rule{1em}{1pt}34 J.s)(6.37 × 106m)] = 1.6 × 1014sec^\rule{1em}{1pt}1. This frequency is in the infrared.

Question:

The following bases, and their conjugate acids (as the chlorides), are available in the lab: ammonia, NH_3; pyridine, C_5H_5N; ethylamine, CH_3CH_2NH_2. A buffer solution of pH 9 is to be prepared, and the total concentration of buffering reagents is to be 0.5 mole/liter, (a) Choose the best acid- base pair, (b) Give the recipe for preparing one liter of the solution, (c) Calculate the pH of the solution after 0.02 mole ofNaOHhas been added per liter.

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Solution:

This problem deals with the preparation of a buffer solution and then to show the effect the addition of a base has on thepH. If this is a truly buffered system, then, the pH should be only slightly altered when a small quantity of base is added. Since we are using bases it will be more convenient to work initially withpOH. pH is converted topOHby the equation: 14 = pH +pOHorpOH= 14 -pH. Substituting the given value of pH,pOH= 14 - 9 = 5. In part (a) one is asked to choose the strongest base. At a given pOH, the strongest base haspOHabout equal topK_b(the pH at which dissociation occurs). The maximum buffering capacity and minimum sensitivity to pH change occurs when the concentrations of the acid and base are initially equal. From a table ofpK_bvalues, it is found that pK_NH3 = 4.75, pK_C5H5N = 8.81, and pK_CH3CH2NH2 = 3.3 Therefore, NH_3 has apK_b closest to 5, which means it is the best base. To prepare a one liter solution, as requested in part (b) , find the initial concentrations of the components to be added and their ratio. NH_3 is the base and NH_4Cl dissociates to the acid NH_4^+ . The reaction equation is NH_3 + H_2O \rightleftarrows NH_4^+ + OH^- K_b = 1.76 × 10^-5. To find the ratio, set up the equilibrium constant equation: K_b = {[NH_4^+][OH^-]} / [NH_3] = 1.76 × 10^-5. Rewriting to obtain the ratio, one has: [NH_4^+] / [NH_3] = {Kb/ [OH^-] = [(1.76 × 10^-5) / (1.0 × 10^-5)] = 1.76 . ([OH^-] = 1 × 10^-5,sincepOH= 5 andpOH= - log [OH^-].) Therefore, NH_4Cl and NH_3 should be added in the molar ratio of 1.76 : 1. Since it is given that the sum of the concentrations of the buffering reagents equals 0.5 mole/liter, the following equation can be written which says, in effect, that the sum of the parts is equal to the whole [NH_3] + [NH_4^+ ] = 0.5. Let [NH_3] = x and since the ratio is 1.76 : 1, let [NH_4^+] = 1.76 x. This results in x + 1.76 x = 0.5 x = [NH_3] = 0.18 mole/liter 1.76 x = [NH_4^+ ] = 0.32 mole/liter. Therefore, to prepare a 1 liter solution, mix 0.18 moles NH_3 and 0.32 moles NH_4Cl and then add H_2O until a total volume of 1 liter is obtained. Part (c) tells one to add 0.02 MNaOH. Since it is a strong base, complete dissociation occurs giving 0.02 mole Na^+ and 0.02 mole OH^-. The base OH^- then reacts fully with the acid NH_4^+ and converts 0.02 mole NH_4^+ to NH_3. This is more clearly seen below. NH_4^+ + H_2O \rightleftarrows NH_3 + OH^- Before:0.32 mole0.18 mole After:0.32 - 0.02 = 0.30 mole0.18 + 0.02 = 0.20 mole Substituting these new values into the equilibrium constant equation, one arrives at [OH^-] and thenpH.One then has K_b = {[NH_3][OH^-]} / [NH_4^+] . Substituting, [OH^-] = {K_b[NH_4^+]} / [ NH_3] = [{1.76 × 10^-5 (0.20)}/(0.30)] = 1.17 × 10^-5 mole/liter. Thus,pOH= - log [OH^-] = - log [1.17 × 10^-5] = 4.93. Thus, pH = 14 -pOH= 14 - 4.93 = 9.07.

Question:

A volume of 50 liters is filled with helium at 15\textdegree C to a pressure of 100 standard atmospheres. Assuming that the ideal gas equation is still approximately true at this high pressure, calculate approximately the mass of helium required.

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Solution:

From the ideal gas equation, PV =nRT(where P, V, T are the pressure, temperature and volume of the gas, R is a constant, and n is the number of moles of gas in V.) We can find the number of moles required, and from the atomic mass we can calculate the total mass required. Thus, converting all data tocgsunits, we have P = 100 atmospheres × 1.0 × 10^6 (dyne/cm^2 atmosphere) = 1.0 × 10^8 dyne/cm^2 V = 50 liters × 10^3 cm^3/liter = 5.0 × 10^4 cm^3 T = 15\textdegree + 273\textdegree = 288\textdegree K Substituting into the ideal gas equation, 1.0 × 10^8 dyne/cm^2 5.0 × 10^4 cm^3 = n [(8.3 × 10^7 erg)/(\textdegree K \bullet mole)] 288\textdegree K n = (5 × 10^12 dyne \bullet cm)/[(8.3 × 10^7 erg/\textdegreeK \bullet mole)(288\textdegree K)] = 2.09 × 10^2 moles The atomic mass of helium is approximately 4 gm/mole, and since helium is amonoatomicgas, one mole of helium contains 4 gm of matter. Therefore, 210molesof gas contains 210 moles × 4 gm/mole = 840 gm. The helium has a mass of 840 gm.

Question:

Write a FORTRAN program which determines whether or not the given square matrix (up to 20 × 20) is symmetrical.

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Solution:

By definition, a matrix is symmetric if A(i, j) = A(j,i) for alliand j. Hence our approach is to use two nested DO loops to check this condition. In statement number 8, an implied DO loops is used to read in the elements of the matrix. The program looks as follows: INTEGERX (20, 20) 6READ(2, 31) N, ID CN IS MATRIX SIZE, ID IS THE IDENTIFICATION OF THE MATRIX IF (N) 99, 99, 8 8READ (2, 32)[{X (I, J), I = 1, N}, J = 1, N] DO 10 I = 1, N DO 10 J = 1, N IF [X (I, J) - X (J, I)]9, 10, 9 10CONTINUE WRITE (3, 33) ID GO TO 6 9WRITE (3, 34) ID GO TO 6 31FORMAT (I2, 3X, I3) 32FORMAT (I4) 33FORMAT(8HIMATRIX I3, 13H IS SYMMETRIC) 34FORMAT(8HIMATRIX I3, 17H IS NOT SYMMETRIC) 99STOP END

Question:

Describe the kingdoms within which algae are classified.

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Solution:

Algae fall into the Plant,Protistan, andMonerankingdoms. The three phyla divisions of theThallophytes(a major plant division) includes theChlorophyta(green algae),Rhodophyta(red algae), andPhaeophyta (brown algae). TheProtistanalgae include theEuglenophyta (photosynthetic flagellates),Chrysophyta(golden algae and diatoms) , and thePyrrophyta(dinoflagellates) . TheMonerankingdom in-cludes the Cyanophyta (blue-green algae).

Question:

For 1000 seconds, a current of 0.0965 amp is passed through a 50 ml. of 0.1 M NaCl. You have only the reduction of H_2O to H_2 at the cathode and oxidation of Cl^- to Cl_2 at the anode. Determine the average concentration of OH^- in the final solution.

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Solution:

To find out how much OH^- is in the final solution, you must determine the number of faradays that were produced. Then calcu-late the number of moles of OH^- generated per faraday. This problem involves electrolysis, since electrical energy was used to produce a chemical change. Let us compute how many faradays were produced by this current, for according to Faraday's law, this will indicate the number of moles of OH^- that will be produced. There are 96,500 coulombs per faraday. If the current is 0.0965 amp, then you have 0.0965 (coulombs / sec). You used the current for 1000 secs, which means the total number of coulombs is 0.0965 ×1000. Therefore, you have [(0.0965 coulomb/sec) (1000 sec)] / [96,500 coulomb/Faraday] = 0.001 faraday. Now, to generate OH^- from H_2O, the reaction to be followed is 2e^- + 2H_2O \rightarrow H_2(g) + 2OH^- . A faraday is defined as Avogard's number of electrons. This means, therefore, that 2 faradays liberate two moles of OH-, since two moles of electrons are involved. It followes , therefore, that 0.001 faraday would liberate 0.001 mole of OH^- . Assuming the volume of the solution remains 50.00 ml., the final con-centration is (0.001 mol) / (0.05 liter) = 0.02M.

Question:

What factors determine the characteristics ofparticular organismsthat exist in an ecosystem.

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Solution:

Temperature and precipitation are among the most crucial physical factorsin determining what type of vegetation will exist in a certain area. It isoften possible to predict the type of biome at a given locality from the characteristicsof the climate. However, other physical factors are also important. The structure and chemistry of the soil can be equally vital. There are, for example, certain trace elements known to be essential for thefull development of plants. These include boron, chlorine, cobalt, copper, iron, manganese, molybdenum, sodium, vanadium, and zinc. When one or more of these substances is not present in sufficient con-centration, land which should carry a forest according to the temperature- precipitation features, will instead be covered with shrubby vegetation, or be grassland bearing a few scattered trees. Where grasslandwas indicated, there may existsparcevegetation typical of a desert. Any single factor such as the amount of sun-light, temperature, humidity, or the concentration of trace elements is capable of determining thepresence or absence of species and thus the characteristics of the entirebiome. The Law of Minimum is a generalization that is sometimes usedto explain this phenomenon. It states that the factor that is most deficientis the one that determines the presence and absence of species. It does not matter, for example, how favorable temperature and sunlight arein a given locality, or how rich the nutrients and trace elements of the soilare if the precipitation is very low, because the result will still be a desert. Only a consideration of all the climatic and soil characteristics of a regioncan reveal the reasons for the existence of certain plant formations. In turn, animals are indirectly influenced in their distributions by soil types becauseof their dependence upon plants as the source of high-energy organic nutrients.

Question:

Explain the main differences between insect societies and primatesocieties.

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Solution:

The major characteristic differentiating primate societies from insectcolonies is personal recogni-tion of its members. Within primate societiesand most complex vertebrate societies, every member knows everyother member as an individual. Often each member bears some par-ticularrelationship to every other member. Parent-offspring relationshipsare tightly binding and persist a long time. The young may alsobe partly cared for by non-parental members who act as "aunts" and "uncles." Eventually the young primate leaves these personal relationships andestablishes new bonds with its peers. A weak division of labor exists in primate societies. Some members specializein defense, others in foraging. However, each member is physiologicallysimilar to every other member of the same sex. Therefore, thisspecializa-tion of duties is not based upon physiologically different castes. Insect societies, such as those of the ants, wasps, and bees, exist withoutpersonal relationships. No indi-vidual bonds are established during theshort lifespan of the members. The colonies themselves are very large. Only very limited dominance hierarchies exist, whereas in primates, statusis extremely important. Insects show little or no socialization and do notplay. Young primates undergo long periods of socialization and often play. The queen of the insect colony is not recognized personally by the othermembers, but by means of pheromones. If one were to place these pheromoneson a small piece of wood, the other members would react to itas if it were the queen. In contrast to primate societies, strong divisions oflabor exist within insect societies. These divisions are based upon physiologicallydifferent castes. Some members are specialized to perform inreproduction (the queens),in labor (the workers), or in colony defense (thesoldiers). This highly efficient means of organization is sacrificed in primatesocieties for the sake of individual freedom and reproduction.

Question:

If the standard free energy of formation of HI from H_2 and I_2 at 490\textdegreeC is - 12.1 kJ/mole of HI, what is the equilibrium constant for this reaction? Assume R = 8.31 J mol^-1 deg^-1

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Solution:

You are given the chemical reaction H_2 + I_2 \rightleftarrows 2HI and the standard free energy formation ∆G\textdegree = - 12.1 kJ/mole of HI and are asked to find the equilibrium constant K. The equilibrium constant and the standard free energy are related by the formula: (i)∆G\textdegree_490 = - RT In K where R = universal gas constant and T = temperature in Kelvin. Solving for K, you obtain: (ii)lnK = [(∆G\textdegree_490)/(- RT)] In this problem, R = 8.31 J mol^-1 deg^-1, and T = 490\textdegreeC = (490\textdegree + 273\textdegree) K = 763\textdegreeK. ∆G\textdegree_490 is the amount of free energy needed for the reaction H_2 + I_2 \rightleftarrows2HI, i.e., the amount of free energy needed for the production of2moles of HI at 490\textdegreeC. Given that - 12.1 kJ will produce 1 mole of HI at 490\textdegreeC, it follows that ∆G\textdegree_490 = 2 ( - 12.1 kJ/mole) = - 24.2 kJ/mole = - 24, 200 J/mole. Substituting these values in equation (ii), you obtain (iii)In K = - [(∆G\textdegree)/(RT)] = [{- (- 24, 200 J/mole)} /{(8.31 J mol^-1 deg^-1) (763 deg)}] (iv)In K = 3.82 (v)K = 46.

Question:

A 4.0-gm bullet is fired from a 5.0-kg gun with a speed of 600 m/sec. What is the speed of recoil of the gun?

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Solution:

Originally, the momentum of the system consisting of the gun and the bullet is zero. Even if external forces act on the system, the principle of momentum conservation can be applied if the time interval of "collision" is small enough. Therefore we can say that after the bullet has been fired from the gun, the total momentum of the system remains zero. Letting m_1 be the mass of the gun and m_2 the mass of the bullet, with v_1 and v_2 their respectivefinal velocities, we have m_1v_1 + m_2v_2 = 0 v1= - [(m_2)/(m_1)]v2 v1= - [(0.0040 kg)/(5.0 kg)] (600 m/s) = - 0.48 m/sec. where the minus sign indicates that the gun moves in a direction opposite to that of the bullet.

Question:

A solution containingradiophosphorus, P^32 , which is a \beta- emitter with a half-life of 14 days, surrounds a Geiger counter which records 10^3 counts per minute. If the same experiment is performed 28 days later, what counting rate will be obtained?

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Solution:

The number of radioactive atoms still present after time t is given by the radioactive decay formula N = N_0 e\Elzbar\lambdat(1) The number of atoms present after one half-life (\tau) of P^32is, by definition, N_0 /2, whence N (\tau) = N_0 e\Elzbar\lambdat= N_0 /2 ore^\lambdat= 2 Taking the logarithm of both sides of this equation \lambda\tau= l n 2 or\lambda = (1/\tau) l n 2(2) Using (2) in (1) N = N_0 e\Elzbar[(t/\tau)ln2] Hence, the decay rate is dN/dt= [\Elzbar(1/\tau)ln2] N_0 e\Elzbar[(t/\tau)ln2](3) Att = 0,(dN/dt)t = 0= [\Elzbar(1/\tau)ln2] N_0(4) Rewriting (3) using (4) dN/dt= (dN/dt)t = 0e\Elzbar[(t/\tau)ln2](5) At the beginning of the 28 days, the decay rate is (dN/dt)t = 0= 10^3 counts/min. At t = 28 days, dN/dt= (10^3 counts/min) [e\Elzbar(28days/14days)ln2] dN/dt= (10^3 counts/min)( e\Elzbar2ln2) dN/dt= (103counts/min)( e(ln2)(\Elzbar2)) But elnx= x, whence dN/dt= (103counts/min)(2\Elzbar2) dN/dt= 250counts/min. Note that the time involved is two half-lives. Thus the final count-ing rate will be (1/2)^2 = 1/4 of the initial counting rate.

Question:

Show how two clocks, one at the origin and the other a dis-tance d from the origin, may be synchronized.

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Solution:

The velocity of light is the same for all ob-servers regardless of the relative motion between the light source and the observer. If the distance d between the clocks is measured, then the time t required for light to travel from one clock to the other is t = d / c Suppose that a flash of light is produced at the origin when the clock at the origin is recording exactly 12:00 noon. This flash of light will arrive at the other clock when the origin clock reads 12:00 plus d/c. Hence the other clock should be pre-set so that it reads 12:00 o'clock plus d/c. When the flash of light arrives from the origin, the second clock should then be allowed to begin recording time. This process is said to synchronize the two clocks.

Question:

Describe the development of seed.

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Solution:

After fertilization, the zygote undergoes a number of divisions and develops into amulticellularembryo. The triploid endosperm nucleus also undergoes a number of divisions and forms a mass of endosperm cellscarryingahigh content of nutrients. Thisendospermalmass fills the space around the embryo and provides it with nourishment. The sepals, petals, stamens, stigma and style usually wither and fall off after fertilization. The ovule with its contained embryo and endosperm becomes the seed; its wall, or integument, thickens to form the tough outer covering of the seed. The seed has an adaptive importance in dispersing the species to new locations and in enabling it to survive periods of unfavorable environmental conditions. This insures that germination will occur only when favorable growth is possible.

Question:

What is the force between two positive charges of 100\mu coulombs and 200\mu coulombs, respectively, when they are separated a distance of 3 centimeters?

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Solution:

Using Coulomb's Law, F = k[(Q_1Q_2) / r^2 ] we have Q_1 = 100\mu coulombs = 10^\rule{1em}{1pt}4 coulombs Q_2 = 2 x 10^\rule{1em}{1pt}4 coulombs k = electrostatic constant = 9 x 109[(nt\rule{1em}{1pt} m^2) / coulombs^2 ] andr = 3 cm = 3 x 10^\rule{1em}{1pt}2 m ThenF = 9 × 10^9[(nt\rule{1em}{1pt} m^2) / coulombs^2] × [(10^\rule{1em}{1pt}4 coulombs × 2 × 10^\rule{1em}{1pt}4 coulombs ) /(3 x 10^\rule{1em}{1pt}2)^2 m^2 ] = 2 x 10^5nt

Question:

Discuss some of the elements found in advanced animal societies.

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Solution:

A society is a group composed of animals of the same species and organizedin a cooperative manner. The simplest animal society could thereforebe combinations of parents and their offspring. The com-municationbetween adults and young serves to hold this elementary societytogether. Another form of a simple society is a flock of birds, which isan example of a motion group. The members communicate to keep the grouptogether as it travels in search of food. They cooperate in the mutualprotection of members against predators. An element of a more advanced society is leadership. A leader usuallyregulates the activities of the other members and directs the group asit moves. Wolves, elephants, and primates usually have a dominant individualwho watches over the group's activities and leads them from placeto place. Another important element of advanced societies is the territory, an areaoccupied by the group and defended by aggression or warning messages. However, most animals move in a larger area in search of food. This area, called the home range, is often shared with thegroup's neighbors. Territorial behavior involves either the aggressive exclusion of outsidersor the continual vocal signalingthat warns outsiders away. The songsof birds, crickets, and frogs serve as such warning messages. Territory may also be advertised by chemical signals. Rabbits are known toplace characteristically scented fecal pellets at their terri-torial boundaries. The distinctive odor warns intruders that the territory is occupied. The attempts of domesti-cated dogs to urinate on variousland marksand smell the urine and feces of other dogs are just manifestations ofthis territorial behavior. A third characteristic of many advanced societies isa dominance hierarchy. The peck order, which is a set of dominance-subordination relationships, is an example of this. Dominance hierarchies exist in societiesof wasps, bumblebees, hermit crabs, birds and mammals. The conceptof a dominance hierarchy is different from that of leader-ship. In a dominancehierarchy, a subordinate animal may be dominant to other members. In a society characterized by leadership, the leader is the only memberwho exhibits dominance. All these elements of advanced societiesprovide better organization and control for the group, allowing it toexist in its environment more successfully.

Question:

Can I^+ (the iodinecation) be called a Lewis base? Explain your answer.

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Solution:

A Lewis base may be defined as an electron pair donor. Writing out its electronic structure is the best way to answer this question, because it will show the existence of any available electron pairs. The electronic structure of I^+ may be written as [: \"{\i} :]^+. There are three available electron pairs. This might lead one to suspect that it is indeed a Lewis base. But note, I^+ does not have a complete octet of electrons, it does not obey the octet rule. According to this rule, atoms react to obtain an octet (8) of electrons. This con-fers stability. Therefore, I^+ would certainly rather gain two more electrons than lose six. In reality, then, I^+ is an electron pair acceptor. Such substances are called Lewis acids.

Question:

The forces shown in the force-time diagram are applied to a body of mass 5 kg. What is the impulse after 6 sec, and after 12 sec? What are the velocities at these times?

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Solution:

We know that when a force F^\ding{217} acts on a system, for a time interval dt, the impulse resulting is given as \int F^\ding{217} dt. One can interpret impulse as the area under the force-time curve, as shown in Fig. 1. In the case of a constant force, the impulse is simply the product of the force acting on the system and the time interval over which it acts. Thus at t = 6 sec, the area is + 10 x 4 = 40 newton-sec. After t = 7 sec the impulse becomes negative. The total area at t = 12 sec is + 10 × 4 - 5 × 5 = +15 newton-sec. From the force-time curve we can drawn an acceleration- time curve by dividing each value of F(t) by the mass of the body 5 kg. This follows since, by Newton's Second Law, a = F/m. The area under the acceleration-time curve gives the change in velocity between the indicated times, since \Deltav = a\Deltat, by definition for a constant force. Thus at t = 6 sec, v = 2 × 4 = 8 m/sec. At t = 12 sec, v = 2 × 4 - 1 × 5 = 3 m/sec.

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Question:

A gas occupies a volume of 1.0 liter at a temperature of 27\textdegreeC and 500Torrpressure. Calculate the volume of the gas if the temperature is changed to 60\textdegreeC and the press-ure to 700Torr.

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Solution:

For a given mass of gas, the volume is in-versely proportional to the pressure and directly pro-portional to the absolute temperature. This combined law can also be written: (P_1 V_1 )/T_1= (P_2 V_2 )/T_2 where P_2 is the original pressure, V_1 is the original volume, T_1 is the original absolute temperature, P_2 is the new temperature, V_2 is the new volume, and T_2 is the new absolute temperature. In this problem, one is given the original press-ure, temperature, and volume and the new temperature and pressure. One is asked to calculate the new volume. The temperatures are given in \textdegreeC; they must be converted to the absolute scale before using the combined law. This can be done by adding 273 to the temperature in \textdegreeC. Converting the temperature: T_1 = 27\textdegree + 273 = 300\textdegreeK T_2 = 60\textdegree + 273 = 333\textdegreeK Using the combined law: (P_1 V_1 )/T_1= (P_2 V_2 )/T_2 P_1 = 500TorrP_2 = 700Torr V_1 = 1.0 literV_2 = ? T_1 = 300\textdegreeKT_2 = 333\textdegreeK [{(500Torr) (1.0 liter)}/(300\textdegreeK)] = [{(700Torr) (V_2)}/(333\textdegreeK)] V_2 = [{(500Torr) (1.0 liter) (333\textdegreeK)}/{(300\textdegreeK) (700Torr)}] V_2 = 0.79 liter.

Question:

What is meant by a cloaca and what is its relation to the mammals?

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Solution:

Chordates, except for mammals and teleosts, have a cloaca. A cloaca is a common duct through which passes the reproductive, digestive and excretory matters. In the frogs, for example, sperm cells pass from the testes into the kidneys and down the excretory ducts to the cloaca. The reproductive system and the excretory system are very closely related. The evolutionary trend in the vertebrates, with regard to the cloaca, has been one toward increasing separation between the reproductive and excretory systems. Mammals have, instead of a cloaca, ureters which release their exaretory material (i.e., urine) into a urinary bladder. The urethra carries the urine from the bladder to the outside. In the mammals, however, sperms also pass through the urethra. Thus, although there is far more separation in mammalian males than in fish or frogs, the mammalian reproductive and excretory systems still share the urethra and hence do not have separate openings to the outside. Only in mammalian females has complete separation arisen.

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Question:

The following are physical properties of methyl alcohol, CH_3OH: freezing point - 98\textdegreeC; boiling point 65\textdegreeC; specific heat of liquid 0.570 cal/g - degree; heat of fusion 22.0 cal/g; and heat of vaporization 263 cal/g. Calculate the number of kilocalories required to convert one mole of methyl alcohol solid at - 98\textdegreeC to vapor at 65\textdegreeC.

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Solution:

There are three steps in this process: (3) vaporizing the liquid at 65\textdegreeC. calculate the heat absorbed in each process and then take the total. (1) Heat absorbed in melting the solid. To find the amount of heat absorbed when a compound is melted, one uses the heat of fusion. The heat of fusion for methyl alcohol is 22.0 cal/g. This means that for each gram of solid melted 22.0 cal of heat are absorbed. Here, one is melting one mole of CH_3OH. The molecular weight of CH_3OH is 32, thus one will calculate the heat absorbed when 32 g of CH_3OH is melted. no. of cal absorbed = heat of fusion × no. of grams no. of cal absorbed = 22.0 cal/g × 32 g = 704 cal. (2) Heat absorbed in heating the liquid. In determining the amount of heat absorbed in heating the liquid, one uses the specific heat of the liquid. For CH_3OH the specific heat is 0.570 cal/g-degree. This means that for each gram of CH_3OH raised 1 degree, 0.570 calories is absorbed. Here, the liquid is raised 65\textdegree - (- 98\textdegree) or 163\textdegreeC. There are 32 g of CH_3OH heated. no. of cal absorbed= no. of degrees × weight × specific heat no. of cal absorbed= 163\textdegree× 32 g × 0.570 cal/g-\textdegree = 2973 cal. (3) Heat absorbed when the liquid is vaporized. To calculate the amount of heat absorbed when the liquid is vaporized, one uses the heat of vaporization. The heat of vaporization for CH_3OH is 263 cal/g. This means that for each gram of CH_3OH vaporized, 263 calories are absorbed. no. of cal absorbed = heat of vaporization × weight no. of cal absorbed = 263 cal/g × 32 g = 8416 cal. (4) To find the heat absorbed by the whole process, the heat absorbed in these three steps must be added to-gether . total heat absorbed= heat absorbed in melting + heat absorbed in heating the liquid + heat absorbed in vaporization total heat absorbed= 704 cal + 2973 cal + 8416 cal = 12093 cal Calories can be converted to kilocalories by multiplying the number of calories by 1 Kcal/1000 cal, no. of Kcal = no. of cal × 1 Kcal/1000 cal no. of Kcal = 12093 cal × 1 Kcal/1000 cal = 12.09 Kcal.

Question:

At a carnival, the people who go on a certain ride sit in chairs around the rim of a horizontal circular plat-form which is oscillating rapidly with angular simple harmonic motion about a vertical axis through its center. The period of the motion is 2s and the ampli-tude .2 rad. One of the chairs becomes unbolted and just starts to slip when the angular displacement is a maximum. Calculate the coefficient of static friction between chair and platform. (The rim is 12 ft. from the center).

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Solution:

In order to find the coefficient of static friction, we must use Newton's Second Law to relate the acceleration of the seat to the net force acting on it. Now, the platform and seat are oscillating about an axis through C (see figure (A)) with a maximum angular displacement \texttheta_0. Since the motion is simple harmonic, we may write \texttheta = \texttheta_0 cos (\omega_0t + \rho)(1) which gives us the angular displacement of a point on the disc (and, hence, the seat) as a function of time. Note that \rho is an arbitrary constant, which is determined by the way the angular motion begins, and \omega_0 is the angular frequency of the motion. Differentiating (1) twice \omega = d\texttheta/dt = - \texttheta_0 \omega_0 sin (\omega_0t + \rho)(2) \alpha = d\omega/dt = - \texttheta_0 \omega0^2 cos (\omega_0t + \rho)(3) Here, \omega and \alpha are the angular velocity and acceleration of a point on the disc. Since the seat has an angular acceleration \alpha, it also has a tangential acceleration a_t such that a_t = \alphaR = - \texttheta_0 \omega0^2R cos (\omega_0t + \rho)(4) where R is the platform radius (and the distance of the seat from C, see figure (A)). In addition, the seat also has a radial acceleration, a_r pointing inward towards C. The value of a_r is a_r = \omega^2 R = \texttheta0^2 \omega0^2R sin^2 (\omega_0t + \rho)(5) The net acceleration a is then a^2 = at^2 + ar^2 a^2 = \texttheta_0 ^2 \omega04R^2 cos^2 (\omega_0t + \rho) + \texttheta0^4 \omega0^2 R^2 sin^4 (\omega_0t + \rho) a^2 = \texttheta_0^2 \omega04R^2 [cos^2 (\omega_0t + \rho) + \texttheta_0^2 sin^4 (\omega_0t + \rho)] ora = \texttheta_0- \omega_02R [cos^2 (\omega_0t + \rho) + \texttheta_0^2 sin^4 (\omega_0t + \rho)]^1/2(6) If we assume the chair is unbolted, there are 2 sets of forces acting on the chair. In the vertical direction (perpendicular to the disc) 2 forces act - the normal force N of the disc on the chair, and its weight, mg (m is the chair's mass). Since the chair stays on the platform, there is no acceleration in the vertical direction, and N and mg balance: N = mg(7) In the plane of the disc, the only force acting is the force of static friction, f_s. By Newton's Second Law, this must equal the product of the chair's mass and acceleration. Using (6) f_s = m \texttheta_0 \omega0^2 R [cos^2 (\omega_0t + \rho) + \texttheta0^2 sin^4 (\omega_0t + \rho)^1/2] Butf_s \leq u_s N Hence,f_s \leq u_s mg Combining the first, and last equation for f_s, m \texttheta_0 \omega02R [cos^2 (\omega_0t + \rho) + \texttheta0^2 sin^4 (\omega_0t + \rho)]^1/2 \leq u_s mg Solving for u_s u_s \geq [(\texttheta_0 \omega0^2 R)/g] [ cos^2 (\omega_0t + \rho) + \texttheta0^2 sin^4 (\omega_0t + \rho)]^1/2(8) This relation holds as long as the seat is stationary. Once the seat begins to slip, the maximum force of static friction is encountered, and (8) becomes a strict equality: u_s = (\texttheta_0 \omega0^2 R/g)[ cos^2 (\omega_0t + \rho) + \texttheta_0^2 sin^4 (\omega_0t + \rho)]^1/2(9) Note that slippage occurs at the maximum value of \texttheta, or \texttheta_0.For this value of \texttheta, (1) yields \texttheta_0 = \texttheta_0 cos (\omega_-0t + \alpha) whencecos (\omega_-0t + \alpha)= 1 As a result,sin (\omega_-0t + \alpha)= 0 Using these facts in (9) u_s = [(\texttheta_0 \omega0^2 R)/g](1 + 0)^1/2 = (\texttheta_0 \omega0^2 R)/g(10) Note that we don't know \omega_0, but we do know T, the period of the oscillation. By definition \omega_0 = 2\pif = 2\pi/T where f is the oscillation frequency. Using this in (10) u_s = (4\pi^2 \texttheta_0R)/T^2g Substituting the given data u_s = [(4)(9.89)(.2 rad)(12 ft)]/[(2s)^2(32 ft/s^2) u_s = .74 We could have done an easier (but less general) analysis by noting that, when \texttheta = \texttheta_0, the a_r term is zero, for the chair is instantaneously at rest.

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Question:

A 0.100 l container maintained at constant temperature contains 5.0 × 10^10 molecules of an ideal gas. How many molecules remain if the volume is changed to 0.005 l? What volume is occupied by 10,000 molecules at the initial temperature and pressure?

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Solution:

Since only the volume changed and no molecules were added to or withdrawn from the system, the number of molecules at a volume of 0.005 l is the same as that at a volume of 0.100 l, or 5.0 x 10^10 molecules. In the second part of the problem we require a relationship between volume and number of molecules, taking into account the fact that the number of molecules can change. We will obtain such a relationship by modifying the ideal gas equation, PV = nRT, where P = pressure, V = volume, n = number of moles, R = gas constant, and T = absolute temperature. Since the number of moles is equal to the number of molecules, N, divided by Avogradro's number, A, or n = N/A, we can write the ideal gas equation as PV = nRT = N/A RT, or PVA = NRT. The initial and final pressure and temperature are the same. Also, A and r are constants. Denoting the initial volume and number of molecules by V_i and N_i , respectively, and the final volume and number of molecules by V_f and N_f , respectively, we obtain PV_i A = N_i RTandPV_f A = N_f RT. Dividing the second of these by the first we obtain [(PV_f A)/(PV_i A)] = [(N_f RT)/(N_i RT)]or(V_f /V_i ) = (N_f /N_i ), where we have cancelled all the constants. Solving for the final volume we obtain V_f = V_i × (N_f /N_i ) = 0.100 l × [(10,000)/(5 × 10^10)] = 2 × 10^-8 l.

Question:

How isaldosteronerelated to proper kidney functioning? How is its secretion controlled?

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Solution:

Aldosteroneis a hormone produced by the adrenal cortex. It is thought that this hormone stimulates sodiumreabsorptionin the distal convolu-ted tubules of the kidney by altering the permeability of the tubules to sodium. In the complete absence of this hormone, a patient may excrete close to 30gmsof sodium per day, whereas excretion may be virtually zero whenaldosteroneis present in large quantities. The major clue that some substance secreted by the adrenal cortex controls sodium reabsorptioncame from the observation of patients who had missing or diseased adrenal glands. These patients were seen to excrete large quantities of sodium in the urine, despite a decreasedglomerularfiltration rate (which would tend to reduce sodium loss). This indicated that decreased tubularreabsorptionmust be responsible for the sodium loss. Aldosteronesecretion is controlled by reflexes involving the kidneys themselves. Lining the arterioles in the kidney are specialized cells which synthesize and secrete an enzyme known asrenin(not to be confused with rennin, a digestive enzyme secreted by the stomach) into the blood. Renincatalyzes a reaction in whichangiotensinI, a small polypeptide, is split; off from a large plasma protein calledangiotensinogen. Angiotensinogenis synthesized by the liver and is always present in the blood. It is still not certain howreninsecretion is controlled by the kidney. There seems to be multiple nerve inputs to therenin-secreting cells, and their respective roles have not yet been determined. AngiotensinI is converted toAngiotensinII via a dif-ferent enzyme. AngiotensinII then acts to stimulatealdos-teronesecretion by the adrenal gland, and as such consti-tutes the primary input to the adrenal controlling the pro-duction and release of the hormone.Aldosteronethen acts on the kidney tubules to stimulate sodiumreabsorption.AngiotensinII has other effects on the body including stimulat-ing thirst centers to ultimately increase blood volume by drinking, and causing vasoconstriction which increases blood pressure.

Question:

The following observations have been made on a typical day in Los Angeles: The hydrocarbon content in the atmosphere starts to rise at about 6 A.M. It peaks at 8 A.M. and dies off at 10 A.M. A similar peaked curve occurs at 6 P.M. About 2 hours after each peak, the ozone concentration rises. How can you explain these observations?

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Solution:

The peaks in hydrocarbon content occur during the height of the rush hour - 8 A.M. and 6 P.M. This suggests that busses and automobiles may be responsible. Both machines emit large quantities of hydrocarbons into the atmosphere from their exhaust systems. The hydro-carbon content rises at 6 A.M. since, at this time, the rush is just beginning. To understand why the ozone concentration rises after 2 hours, while the hydrocarbon concentration de-creases after 2 hours (10 A.M.), consider the chemical reactions of hydrocarbons with elements that make up the atmosphere. Before doing this, consider the following: It is found that high- temperature combustion generates NO which, on exposure to oxygen, is converted to NO_2 \bullet NO_2 undergoes the following chain reactions: NO_2 + light \rightarrow NO + O O + O_2 \rightarrow O_3 O_3 + NO \rightarrow O_2 + NO_2 The NO_2 undergoes a reaction with light to produce NO and monoatomic oxygen (O), which is highly reactive. This, then reacts with oxygen (diatomic - O_2) to produce O_3 (ozone), a toxic element in the lower atmosphere. However, the ozone reacts with the NO that was previously produced to again form NO_2, which allows the cycle to repeat. Now, based on the above information, consider what happens at rush hour. The cars and busses release hydro-carbons. Among these fragments of hydrocarbons are free radicals which are extremely reactive. They proceed to react with NO, reducing its concentration. When the NO is depleted, there is little of it available to react with ozone to produce NO_2 and O_2. Thus, O_3 builds up. It is found that this process takes 2 hours, which explains why the O_3 concentration increases after the peak hydrocarbon concentration.

Question:

Write a program using PL / I, to sort an integer array into ascending order using a bubble sort.

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Solution:

BUBBLE-SORT Program INTEGER I, J, K, ID (N), TEMP /\textasteriskcentered THE CONTENTS OF THE ARRAY ID AND ITS DIMENSION N ARE ASSUMED GIVEN \textasteriskcentered / I = N; K = 1;/ \textasteriskcentered K IS FLAG \not = 0 ON THE FIRST PASS AND WHEN SWITCHES ARE MADE ON THE PREVIOUS PASS \textasteriskcentered / DO WHILE (I \geq 2) AND K \not = 0; J = 1; K = 0; DO WHILE (J \leq I - 1); IF ID (J) > ID (J + l) THEN DO, TEMP = ID (J); ID (J) = ID (J + 1); ID (J + 1) = TEMP; / \textasteriskcentered SWITCH ENTRIES ID (J) & ID (J + 1) \textasteriskcentered / K = 1; END If; J = J + 1; END DO; I = I - 1;/ \textasteriskcentered COUTER I DECREMTED FROM N TO 2 ON SUCCESSIVE PASSES \textasteriskcentered / END DO; END BUBBLE.SORT;

Question:

What is the output of the following program? PROGRAM TEST (output); VAR I, J, K, COUNT : integer; BEGIN I: = 1; J: = 2; K: = 3; count: = 1; write1n (I div J, K mod J); while count < = 3 DO BEGIN If I < J then If K > J then If (K > I) or (I = J) then write1n (I) ELSE write1n (J) ELSE write1n (K); CASE COUNT OF 1: J: = 0; 2: BEGIN K: = K - 1; J: = J + K END; 3:I: = 2 \textasteriskcentered I; END; count : = count + 1; END {while count........} END.

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Solution:

The first line of output will be the results of two operations: I div J and K mod J. The first is integer division; giving the number of times I contains J. Substitu-ting values for I and J we get 1 div 2 = 0. K mod J is defined as K - (K div J)\textasteriskcenteredJ, or the remainder we get when dividing K by J. 3 mod 2 = 1 Thus the first line of output is 0 1. Then, the program enters the WHILE Loop with I = 1, J = 2, K = 3 and count = l. The first statement is true, so the THEN part is executed. The second IF statement is also true, so the control passes to its THEN part. For the third IF statement to be true, at least one of the conditions, i.e. either K > I or I = J, or both must be true. This is the case, so write1n (I) is executed, outputting 1. Then control passes to the CASE statement. Since COUNT = 1, the statement corresponding to 1 is executed, and J becomes equal to 0. Count is then incremented and becomes equal to 2, going back to the begin-ning of the loop. Count is less than 3, so we go through the loop again. The first IF statement is false this time, and there is no ELSE part corresponding to it. Note that an ELSE always corresponds to the last 'open1 IF before it. Thus the first ELSE statement corresponds to the third IF and the second ELSE to the second IF, and there is no ELSE corresponding to the first IF. Therefore control now passes to the CASE statement, without printing out anything. This time count = 2, so the group of statements corresponding to 2 is executed, setting K to 2, and J to 2 (it had value 0 be-fore). Count is then incremented to 3, which is <=3, so we go through the loop again, with I = 1, J = 2, K = 2, count = 3.The first IF statement is true, the second is not, so the second ELSE is executed, printing out the value of k - 2. Control then passes to the case statement, and the ex-pression corresponding to 3 (the current value of count) is executed setting I to 2. Count then becomes 4, and the loop terminates. The total output of the program is: 1 2

Question:

The function of programcopydeckis to copy a deck of cards, reading from the fileindeckand writing to the fileoutdeck. Blank cards inindeckare not copied tooutdeck. Write the program copydeckto copy a deck of cards.

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Solution:

PROGRAMcopydeck(indeck,outdeck, output); CONST maxcol= 80; blank = ''; TYPE colindex=1 . .maxcol; card= PACKED ARRAY [colindex] of char; cardfile=FILE of card; VAR indeck,outdeck:cardfile; buffer,blankcard:card; column :colindex; BEGIN For column: = 1TomaxcolDO blankcard [ column]:= blank; reset(indeck); rewrite(outdeck); WHILE NOTeof(indeck) DO BEGIN read(indeck, buffer); IF buffer <>blankcard THEN write (outdeck, buffer) END {while} END. {copydeck}

Question:

How is the bonding capacity of an atom related to its electron configuration? Illustrate with an example.

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Solution:

The electrons of an atom are distributed so that they occupy concentric atomic orbitals with each orbital containing two electrons. With increasing distance from the nucleus, there is a corresponding increase in the energy level of the orbital. Each atomic orbital has a characteristic energy level, hence the orbitals are said to be quantized. The quantum theory describes the characteristics of electrons in an atom by assigning four quantum numbers to each electron. The principal quantum number, n, denotes the main energy level or shell in which the electron resides. The main energy levels of an atom are assigned integral values, starting with 1 as the shell closest to the nucleus, and continually increasing as the main energy levels become more distant from the nucleus. Hence, n = 1, 2, 3, ... . The angular momentum quantum number, l, determines the shape of the volume in which the electron is most likely to be found. The shape of this volume can be illustrated with electron cloud diagrams, and depends upon the magnetic and electrostatic inter-actions of the electron with the nucleus and other electrons in neighboring orbitals. These interactions will be dif-ferent for varying distances of the orbital to the nucleus. Consequently, the angular momentum of the electron depends upon the main energy level in which it resides. It has been determined that the quantum numbers l and n are related as follows: l = n - 1. Therefore, l = 0, 1, 2, 3, ..., depending on n. A moving charge generates a magnetic field. Since an electron is an elementary particle with a negative charge, and it moves within some volume described by l, it will propagate a magnetic field. This is accounted for in quantum theory by the magnetic quantum number m, which depends upon l in the following way: m = -l, 0, l, I.e., when l = 0, m = 0; when l = 1, m = -1, 0,1; when l = 2, m = - 2, - 1, 0, 1, 2. One final characteristic of an electron is its rotational motion. An electron spins on its own axis. There are only two possible directions (clockwise and counterclockwise) that an electron can spin about its axis. The spin quantum number, s, accounts for this phenomenon, and the arbitrary values + (1/2) and -(1/2) are used to describe the spin of an electron. It is important to realize that no two paired electrons in the same atomic orbital can have the same spin quantum numbers. This would be a violation of the Pauli Exclusion Principle, which states that no two paired electrons in an atom may have identical sets of quantum numbers. Electrons in the same orbital have the same principal, angular momentum, and magnetic quantum numbers. Therefore their spin quantum number must differ to be in accordance with the Pauli Exclusion Principle. The various atomic orbitals are labeled s, p, d and f. When l = 0, the orbitals are of the s variety; when l = 1 they are p orbitals; when l = 2 they are d orbitals; and when l = 3 they are f orbitals. The number of values of m will determine how many orbitals of each kind will be contained in each main energy level. For l = 0, m = 0 and this means that there is only one s orbital contained in each main energy level. For l, = 1, m = - 1, 0, 1 and hence there are three p orbitals contained in the main energy levels starting with n = 2. It is similarly found that there are five d orbitals and seven f orbitals in the main energy orbitals starting with n = 3 and n = 4, respectively. With this basic knowledge of quantum theory, we can now assign an electron configuration to an atom. Electron configurations show how the electrons within an atom are distributed. From the electron configuration we can deduce the valence of the atom, and hence determine its bonding capacity. An important note should be made here: in determining the electron configuration of an atom, one uses the Aufbau process of filling atomic orbitals. By this process, the electrons fill the orbitals in order of increasing energy, and when there is more than one orbital in an energy sublevel (e.g. p orbitals - there are three of them in a sublevel), the electrons will fill the orbitals so that each one will have one electron before any orbital receives a pair of electrons. The relative energies of the orbitals in a main shell is given as s < p < d < f. We will now assign an electron configuration for oxygen (atomic number = 8) and deduce its bonding capacity from its valence. There are eight electrons in the oxygen atom. In the first main energy level, n = 1, l = 0 and m = 0. There is just one s orbital in this level. Since an orbital can accommodate two electrons, the first main energy level will hold two electrons, leaving us with six electrons to account for. In the second main energy level, n = 2, l = 0, 1 and m = - 1, 0, 1. There is one s orbital (l = 0) and three p orbitals (l = 1) in this level. Since s orbitals are lower in energy than p orbitals, they will be filled first, by two electrons. There are four electrons remaining to be placed in orbitals. Because there are three p orbitals in this level, the electrons will fill them so that, each orbital has one electron before any orbital has a complete pair. Hence the four electrons will occupy the three p orbitals so that one orbital will have a pair of electrons and the other two orbitals will each have one electron. The electron configuration of oxygen can be represented as: Above: The numbers represent the principal quantum number n. Electrons with opposite spins are depicted as arrows (\upharpoonleftand \downharpoonright) . The three p orbitals are labeled p_x, p_y, and p_z for differentiation. The valence of an atom is the number of electrons that can be accommodated via bonding by the atom. The Octet Rule states that all atoms that can obtain an octet of electrons in the valence shell (outermost shell), will try to do so. By looking at the electron configuration of oxygen, we can see that its valence (the number of electrons needed to fill the valence shell)is two. Oxygen complies with the Octet Rule in that it seeks to obtain two electrons to fill up its second main energy level. Hence the bonding capacity (or the unpaired electrons available for bonding) of oxygen is two. Let's now look at the electron configuration of carbon (atomic number = 6) which is one of the most biologically important elements. There are six electrons in a carbon atom. As in oxygen, two electrons will occupy the lone s orbital in the first main energy level. This leaves us with four electrons to account for it. If these remaining electrons filled the orbitals in the second main energy level as was true for oxygen, we would get the electron configuration for carbon: In this case the valence of carbon is four, and the bonding capacity is two. This is because there are two unpaired electrons in the 2p orbitals. It should be noted, however, that this representation of carbon's electron configuration does not comply with the Octet Rule. In this case, a fully bonded carbon atom would contain only six electrons in its valence shell, and has one empty orbital (2p_z). This is an energetically unstable condition. If it is at all possible for carbon to complete its valence shell, it will do so. It so happens that carbon does comply with the Octet Rule, and has all its orbitals in the first two main energy levels filled to capacity when it bonds with other atoms. This is facilitated by the hybridization of atomic orbitals. The electrons in the second main energy level occupy an orbital that is intermediate in energy between the 2s and 2p orbitals. This orbital is called an sp^3 hybrid orbital because it is an orbital formed by one s and three p orbitals. There are four sp^3 orbitals, each containing an unpaired electron. The electron configuration of carbon is shown as: Now it can be seen that carbon has a valence of four, and a bonding capacity of four. In this case the Octet Rule is satisfied.

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Question:

Describe the differences between an element and a compound.

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Solution:

All substances are composed of matter in that they have mass and occupy space. Elements and compounds constitute two general classes of matter. Elements are substances that consist of identical atoms (i.e., atoms with identical atomic numbers). This definition of an element includes all isotopes of that element. Hence 0^18 and 0^16 are both considered to be elemental oxygen. A compound is a substance that is composed of two or more different kinds of atoms (two or more different elements) combined in a definite weight ratio. This fixed composition of various elements, according to law of definite proportions, differentiates a compound from a mixture. Elements are the substituents of compounds. For example, water is a compound composed of the two elements hydrogen and oxygen in the ratio 2:1, respectively. This compound may be written as H_2O, which is the molecular formula of water. The subscript "2" that appears after the hydrogen (H) indicates that in every molecule of water there are two hydrogen atoms. There is no subscript after the oxygen (O) in the mo-lecular formula of water, which indicates that there is only one oxygen atom per molecule of water. Hence water is a compound whose molecules are each made up of two hydrogen atoms and one oxygen atom.

Question:

Calculate ∆S for the conversion of one mole of liquid water to vapor at 100\textdegreeC. Heat of vaporization = 540 cal/g.

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Solution:

When a process occurs at constant temperature, the change in entropy (∆S) is equal to the heat absorbed divided by the absolute temperature at which the change occurs. ∆S = (∆H/T) In this problem, one is given ∆H and one can find T. The absolute temperature is calculated by adding 273 to the temperature in \textdegreeC. T = 273 + 100 = 373\textdegreeK Because one mole of water is reacting here, and the heat of vaporization is given in cal/g, one must multiply the ∆H given by the molecular weight of water to find the ∆H in cal/mole. The molecular weight of water is 18. ∆H = 540 cal/g × 18 g/mole = 9720 cal/mole One can now calculate AS. ∆S = (∆H/T)∆H = 9720 cal/mole T = 373\textdegreeK ∆S = [(9720 cal/mole)/(373\textdegreeK)] = 26.1 cal/mole - \textdegreeK The change in entropy, when one mole of water is vaporized at 100\textdegreeC, is 26.1 cal/mole - \textdegreeK.

Question:

Represent the equivalence function circuit realizations, using: a)NAND gates only b)NOR gates only

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Solution:

a) Using NAND gates: The equivalence function can be written as: A \equiv B= A\bulletB + A'\bulletB'[Using the Sum of Products form] = [(A\bulletB + A'\bulletB') ' ]'[Complementing twice] = [(A\bulletB)' \textbullet (A'\bulletB')']'[Applying DeMorgan's Rules] = (A\uparrowB)\uparrow(A'\uparrowB')[Using the Shaeffer stroke notation] \textasciigraveThe circuit realization is shown in fig. 1. b)Using NOR gates: We can write A \equiv B= (A+B')\bullet (A'+B)[using the Product of Sums form] = [ ( (A+B') \textbullet (A'+B) )' ]'[Complementing twice ] = [(A+B')'; + (A'+B)']'[Applying DeMorgan's rule to the inner primed bracket] = [(A\downarrowB')\downarrow (A' \downarrowB)] . The circuit realization is shown in figure 2.

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Question:

Define the following terms as they relate to Polish string notation: a) infix b) prefix c) postfix Then, convert the following tree into the three types of expressions just defined.

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Solution:

Infix notation refers to the way we normally write expressions, namely, with the operators between the two operands. This is also known as algebraic notation, which is different from Polish notation. Prefix notation refers to the fact that the operator precedes the expression in the string. So, if we have the string \div XY, it means that the division operator acts to divide X by Y. One inconvenience with prefix notation is that to understand the string, you must read from end to beginning. Postfix notation avoids this difficulty. For the string above, postfix notation would be XY \div. The meaning is the same as above, but the operator comes after the two operands. The infix translation of the tree diagram is then (P + Q) × (R - S). This is obvious if you follow the tree up from the terminal nodes to the root node. Prefix translation yields x + PQ - RS. By simply looking at the numbering of the individual nodes, you can see that they correspond with the order of the string. In other words, you can write the following to highlight the correspondence: X + PQ - RS 1234567 Postfix notation looks somewhat different. For this tree, we get PQ + RS - x. This may be read from be-ginning to end as "add P and Q, then subtract S from R, then multiply the sum and the difference together."

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Question:

For sublimation of iodine crystals, I_2 (s) \rightleftarrows I_2 (g), at 25\textdegreeC and atmospheric pressure, it is found that the change in enthalpy, ∆H = 9.41 Kcal/mole and the change in entropy, ∆S = 20.6 cal/deg - mole. At what temperature will solid iodine be in equilibrium with gaseous iodine?

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Solution:

Use the fact that the system is in a state of equilibrium. The change in Gibb's free energy is related to ∆H and ∆S by the equation ∆G = ∆H - T∆S, where T is the absolute temperature of the system. At equilibrium, ∆G = O and T is the equilibrium temperature,T_equil. Hence ∆G = ∆H - T∆S, O = ∆H -T_equil∆S, or,T_equil= [(∆H)/(∆S)] Therefore, T_equil = [(∆H)/(∆S)] = [(9.41 Kcal/mole)/(20.6 cal/deg - mole)] = [(9410 cal/mole)/(20.6 cal/deg - mole = 457 K, or,T_equil= 457 - 273 = 184\textdegreeC.

Question:

We know that radiation and chemicals can cause cancer. Can viruses be a cause of cancer?

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Solution:

Cancer involves an abnormality in the control of cell division and cell function. Cancer cells usually produce cell masses called tumors. These cells lack the normal control systems that shut off unwanted cell division . The factors which give cancer cells their characteristic property of unregulated growth are passed on from parent to progeny cancer cells. For this reason, it has been suggested that genetic changes occur within chromosomal DNA. These changes lead to a heritable cancer phenotype. Somatic mutations (mutations in cells that are not destined to become gametes ) may cause a cancer, if the mutation upsets a normal device controlling cell regulation. Some indication that somatic mutations can induce cancer comes from the fact that manycarcinpgens(agents that cause cancer) are also strong mutagens (agents that cause mutations.) Ultraviolet and ionizing forms of radiation are powerful carcinogens and are also highly mutagenic. Exposure of skin to ultraviolet light frequently results in skin cancer, and x-rays applied to the thyroid are associated with increased incidence of thyroidcancer.Manychemicals are also highly carcinogenic and mutagenic . For example, nitrites and nitrates by themselves cause no genetic or carcinogenic changes, but in cells they may become converted to powerful carcinogens- the nitrosamines. Nitrosamines are also known to be mutagens. Nitrates and nitrites are found in cured meats (frankfurters, bacon, ham) and are used to inhibit growth of anaerobic bacteria . In contrast to the somatic mutation proposal is the hypothesis that most cancers are induced by viruses. Viruses which produce tumors in animals are calledonco-genicviruses. While somatic mutations are suspected to cause a loss of functional genetic material, viruses are vehicles into a cell: they introduce new genetic material that may transform the cell into a cancerous type. In addition to multiplying and lysing their host cell, certain viruses can insert their chromosomes into the host chromosome. In some animal species, this process can transform the cell into a morphologically distinguishable cancer cell. At this point, the virus is in theprophagestate. Therefore, absence of detectable viruses does not provide ample evidence for or against a viral cause of a cancer. In 1911, an RNA virus, called the Rous sarcoma virus, was shown to be the causal agent of a sarcoma (a tumor of connective tissue) in chickens . Other RNA viruses cause sarcomas andleukemias (uncontrolledproliferation of leucocytes) in both birds and mammals. One group of DNA viruses causes warts on the skin of mammals. Other DNA tumor viruses is a mouse virus calledpolyomaand a monkey virus called SV 40. AHerpesvirus, EB (Epstein-Barr) causes infectious mononucleosis and is probably involvedinvolvedinBurkitt'slymphoma, a cancer prevalent in humid tropical regions. Research is being conducted to find viral causes of cancer in humans.

Question:

Two identical wires are stretched by the same tension of 100 N, and each emits a note of frequency 200 cycles\textbullets^\rule{1em}{1pt}1. The tension in one wire is increased by 1 N. Calculate the number of beats heard per second when the wires are plucked.

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Solution:

The frequency of the fundamental note emitted by each wire before the tension change occurs is ѵ = (1/2L) \surd(T/\mu)(1) If T changes, ѵ will also change. We can find the relation between these 2 changes by taking the derivative of (1) with respect to T (dѵ/dT) = [1/(2L)] [1/2 (T/\mu)^\rule{1em}{1pt}1/2 (1/\mu)] (dѵ/dT) = [1/(4L)] \surd(\mu/T) (1/\mu) (dѵ/dT) = [1/(4LT) \surd(T^2\mu/T\mu^2) = [1/(4LT)] \surd(T/\mu) From (1) (dѵ/dT) = ѵ/(2T) Hence,\Deltaѵ\approx (ѵ/2) (\DeltaT/T) where\Deltaѵis the frequency difference induced in the string as a result of a change in tension \DeltaT. In other words,\Deltaѵis the number of beats observed if the string's tension is changed by an amount \DeltaT. Using the given data \Deltaѵ= [(200/2) cycles\textbullets^\rule{1em}{1pt}1] (1N/100N) \Deltaѵ= 1 cycle\textbullets^\rule{1em}{1pt}1

Question:

0.01 liter of 0.3M Na_2SO_4 is mixed with 0.02 liters of a solution that is initially 0.1M ca++ and 0.1MSr++ . Given that K_sp= 2.4 × 10-^5 for CaSO_4 andK_sp= 7.6 × 10-^7 for SrSO_4 , what is the final concentration of Ca++ ,Sr++, and SO--_4 in this solution?

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Solution:

Solve this problem in four steps: (A) Find the relationships involving the final concentrations of Ca++,SO--_4 , andSr++. (B) Find the initial concentrations of the three species. (C) Relate the final concentrations to the initial concentrations. (D) Solve for [Ca++], [SO--_4 ], and [Sr++] . (A)K_sp, the solubility product constant, equals the product of the concentrations of the ions in a saturated solution. Thus, K_sp= CaSO_4 = [Ca++] [SO--_4 ]andK_sp= SrSO_4 = [Sr++] [SO--_4 ] . Combine this with the given : (i) [Ca++] [SO--_4 ] = 2.4 × 10-^5 (ii) [Sr++] [SO--_4 ] = 7.6 × 10-^7 . Equation (i) relates [Ca++] and [SO--_4 ] ; equation (ii) relates [Sr++] and [SO--_4 ] . To find an equation relating [Ca++] and [Sr++], solve (i) for [SO--_4 ]: [SO--_4 ] = (2.4 × 10-^5 ) / [Ca++] . Substitute this into (ii); and solve for [Sr++]. (iii)[Sr++]= [(7.6 × 10-^7 ) / (2.4 × 10-^5 )] [Ca++] = 0.032 [Ca++] . (B) 0.01 liters of 0.3M Na_2SO_4 was initially mixed. The number of moles of Na_2SO_4 present is, therefore, (molarity) (volume) = (0.3) (0.01) = 0.003 moles of Na_2SO_4 . Since the salt dissociates completely, there are 0.003 moles of SO--_4 in the solution. The new volume is (0.02 + 0.01) liters = 0.03 l, after adding the Ca++ andSr++ solu-tion. Thus, the initial molarityof SO--_4 is : (0.003moles)/ (0.03 liter) = 0.1 M. By similar analysis, it follows that the initial concentration ofSr++ = initial concentration of Ca++ = 0.0667M . (C) Let the final concentration of Ca++ equal x. Then the final concentration ofSr++ , according to equation (iii), must be 0.032 [Ca++] = .032s. The change in Ca++ concentration is 0.0667\Elzbar x; the change in Sr++ is 0.0667\Elzbar 0.032 x. To find the final [SO--_4 ], note that one SO--_4 , is lost for each it Ca++ that reacts and one SO--_4 , is lost for eachSr++ that reacts. Thus the change in [SO--_4 ] is the sum of the changes in [Ca++] and [Sr++]. Thus, change in [SO--_4 ] = (0.0667\Elzbar x) + (0.0667\Elzbar .032x) = (0.1334\Elzbar 1.032x). Since the final [SO--_4 ] = initial [SO--_4 ] minus change in [SO--_4 ], the final [SO--_4 ] equal 0.10\Elzbar (0.1334\Elzbar 1.032x) or (iv) [SO--_4 ] = 1.032x\Elzbar 0.0334 (D) Substitute the final values for [Ca++] and [SO--_4 ] in equation (i) and solve for x: (v)[Ca++] [SO--_4 ] = 2.9 × 10-^5 (x) (1.032x\Elzbar 0.0334) = 2.4 × 10-^5 (vi) 1.032x^2\Elzbar 0.0334x\Elzbar 2.4 × 10-^5 = 0 . Use the quadraticformuls (vii)x = [ 0.0334 \pm \surd{(0.0334)^2\Elzbar 4(\Elzbar2.4 × 10-^5 ) (1.032)}] / [2(1.032)] Thus, x = 3.30 × 10-^2 M = [Ca++]. Substitute this value in equation (iii) to obtain (viii) [Sr++] = .032 [Ca++] = .032(3.30 × 10-^2 ) = 1.045 × 10-^3 M . = 1.045 × 10-^3 M . Use this value in equation (ii) (ix)[SO--_4 ] = (7.6 × 10-^7 ) / [Sr++] = (7.6 × 10-^7 ) / (1.045 × 10-^3 ) = 7.15 × 10-^4 M.

Question:

Does the electron probability distribution for 1s and 2s support or refute the Bohr picture of the shell?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E18-0682.htm

Solution:

This question involves the comparison of the Bohr concepts and the electron probability distribution for the 1s and 2s orbitals. Bohr believed that the shells have spherical symmetry with the average distance from the nucleus of the 2s electron being greater than that for the 1s. The electron probability distribution (orbitals) for 1s and 2s shells are, spherical in symmetry with 2s having a greater radius than the 1s. The Bohr picture does agree very well with the electron distribution for 1s. However, the Bohr picture does not describe as accurately the spread-out nature of the true distribution or the node separating the two regions which the 2s electron occupies. This is true because the Bohr picture describes a continuous density of the electron clouds which is not present in the 2s shell. This is seen in the illustration above.

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Question:

A turntable of mass M and radius R^\ding{217} is rotating with angul-ar velocity \omega_a^\ding{217} on frictionless bearings. A spider of mass m^\ding{217} falls vertically on to the rim of the turntable. What is the new angular velocity \omega_b^\ding{217} ? The spider then slowly walks in toward the center of the turntable. What is the angular velocity \omega_c^\ding{217} when the spider is at a distance r^\ding{217} from the center? Assume that, apart from a negligibly small inward velocity along the radius, the spider has no velocity relative to the turntable.

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Solution:

Consider the system which includes the turntable and the spider. Since the bearing is frictionless and the resistance of the air is to be ignored, no external couple acts on this system and its angular momen-tum must always remain the same. Just before the spider lands on the rim (figure a) the spider has no angular motion about the axis AA'^\ding{217} and the angular momentum is contained entirely in the turntable. The turntable is a disc whose moment of inertia about its axis of symmetry is: I_t= (1/2)MR^2 The angular momentum is therefore L= I_t\omega_a = (1/2)MR^2\omega_a When the spider is standing on the rim (figure b) he takes up the motion of the turntable and both have an angular velocity \omega_b^\ding{217}. The moment of inertia of the spider is I_sb= mR^2 since all of the spider's mass is at a distance R^\ding{217} from the center of the turntable. The total moment of inertia of the system is I_b= I_t + I_sb = (1/2)MR^2 + mR^2 = (1/2)(M + 2m) R^2 The angular momentum is L= I_b \omega_b = (1/2)(M + 2m) R^2\omega_b Applying the law of conservation of angular momentum and equating the angular moment a before and after the spider lands, (1/2)(M + 2m) R^2\omega_b= (1/2)MR^2\omega_a \omega_b= [M/(M + 2m)]\omega_a When the spider is at a distance r^\ding{217} from the center (figure c), the angular velocity of both the spider and the turntable is \omega_c^\ding{217}. The moment of inertia of the spider is then I_sc= mr^2 The total moment of inertia is I_c= I_t + I_sc = (1/2)MR^2 + mr^2 The angular momentum is L= I_c\omega_c = [(1/2)MR^2 + mr^2]\omega_c Applying the law of conservation of angular momentum [ (1/2) MR^2 + mr^2] \omegac=(1/2)MR^2 \omegaa \omega_c = [ { (1/2)MR^2 } / {(1/2)MR^2 + mr^2 } ] \omega_a = \omega_a/[1 + (2mr^2)/(MR^2] Check that this agrees with the equation for \omega_b^\ding{217} when r^\ding{217} = R^\ding{217}. As the spider walks inward and r^\ding{217} decreases, the angular velocity increases since angular momentum must remain constant. When the spider reaches the center and r^\ding{217} = 0 \omega_c = \omega_a when r^\ding{217} = 0.

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Question:

Simplify F together with its don't care condition d in (a) sum-of-products form and (b) product-of-sums form. F(A, B, C, D) = \sum (0,1,2,8,9,12,13) d(A, B, C, D) = \sum (10,11,14,15)

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G03-0058.htm

Solution:

F is simplified with a four-variable K map. Each don't care minterm can be treated as a 0 or a 1, whichever can help minimize F the most. (a) The K-map is drawn in fig. 1 with X's representing don't cares. F minterms of 8,9,12 and 13 and don't cares treated as 1's at 10,11,14 and 15 combine to form A. F minterms at 0,1,8, and 9 combine to form BC . F minterms at 0,2, and 8 and a don't care treated as a 1 at 10 combine to formBD. Thus using sum-of- products: F = A +BC+BD (b) Use the K-map of fig. 2, which is the same as fig. 1, but now minimize F with the aid of the don't cares and then use DeMorgan's Law to change F to the product-of-sums form.Fminterms which are 0's for the product-of-sums case, at 4,5,6, and 7 combine to formAB.Fminterms at 3 and 7 and don't cares treated as 0's at 11 and 15 combine to form CD. Thus F=AB + CD Applying DeMorgan's Law gives the product-of-sums form F = (A +B) \textbullet (C+D)

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Question:

What characteristics distinguish humans from the great apes?

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Solution:

The living great apes fall into four groups: gibbons, orangutans, gorillas, and chimpanzees. All are fairly large animals that have no tail, a relatively large skull and brain, and very long arms. The earliest members of the family of man (Hominidae) probably shared a common ancestor with the great apes, in particular the chimp and gorilla. Both fossil evidence and biochemical data indi-cate that gorillas, chimpanzees, and man are more closely related in terms of common ancestry, than any one of them is to orangutans or gibbons. There are distinctive anatomical features that separate man into a class of his own. These anatomical features are directly related to the differences in appearance of men and apes, as well as the differences in behavior, intelligence and physical capabilities. Some of the major differences may be found in the skull and jaws. Though the skulls vary greatly in size in the great apes, they are generally more bony and thicker than man's. The ridges on the skulls of apes are more pronounced in order to support heavier musculature. For example, the brow ridges or supraorbital ridges in apes protrude to a great extent, while in humans this ridge is hardly noticeable. Man has a high vertical forehead while apes do not have this feature. Further, man has a pronounced nose with a distinct bridge and tip. The facial features of apes are more outward, such as the upper and lower jaws, while man's face is relatively flat with the jaws placed further back into the head. The jaws of apes are more massive than man's, showing many ridges for muscle attachment. This indicates a rougher diet and a greater dependence on the chewing mechanism. In comparing the teeth, it can be seen that man's are somewhat smaller, but they are of the same general structure. Apes' canines project more than man's beyond the line of the other teeth. One last point to make concerning the facial differences between man and the apes is that the former has a jutting chin whereas the latter do not. The brain of man is two to three times larger than that of most apes and it is more convoluted, which allows for greater surface area. The human brain, also has a more highly developed cerebral cortex region, where the processes of thought, memory, reasoning and speech are coordinated. The attachment of the skull to the vertebral column in man is located at the center of the base of the skull, while the skull of apes is attached toward the rear. Man balances his head so that he can hold it upright with relative ease while apes require massive neck muscles for this purpose. Keeping the head upright is one of the major requirements for upright posture and walking on two limbs, known asbipedalism. Bipedalism enabled man to free his hands for tool making and other tasks. This ability permitted the biological and cultural evolution of man to take its present course. Orangutans and gibbons live mostly in the trees and accomplish locomotion by swinging from branch to branch. This is called brachiation . They occasionally walk on the ground using all four limbs. Gorillas and chimpanzees spend time on the ground and also inhabit the trees. They run and walk on all four limbs. They are classified as quadrupeds and are sometimes re-ferred to as knuckle walkers because they place the knuckles of their fingers on the ground when they walk. Although gorillas and chimpanzees do use bipedal locomotion for short distances, their anatomy does not allow for very efficientbipedalism, and their posture is only semi-erect when they stand. Man's big toe is moved back in line with his other toes making it non-opposable, unlike his thumb. Apes' feet have very long toes and an opposable big toe. The digits of apes are relatively long compared to their thumbs and big toes, while the digits of man are closer in size and shorter in overall length. The feet of man are flat and have short toes. His feet have developed two arches, with one lengthwise and the other crosswise to better support the weight of his body forbipedalism. The basic skeleton of man, gorilla and chimpanzee are generally the same. They differ mainly in the proportion of certain bones to others, such as that of the four limbs. In apes, the arms are longer than the legs, while in man the arms are much shorter than the legs. The shape of the pelvis, and the general position of the attachment of the legs to it, constitute another reason as to why man is bipedal. Finally, man and apes differ in their diet. The diet of the apes is basically herbivorous with occasional supp-lements of insects, birds, eggs and other small animals. Man, on the other hand, is omnivorous.

Question:

A body is released from rest and falls freely. Compute its position and velocity after 1, 2, 3, and 4 seconds. Take the origin 0 at the elevation of the starting point, the y-axis vertical, and the upward direction as positive.

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Solution:

The initial coordinate y_0 and the initial velo-city v_0 are both zero (see figure). The acceleration is downward, in the negative y-direction, so a = -g = - 32 ft/sec^2. Since the acceleration is constant, we may use the kinematical equations for constant acceleration, or y= v_0t + (1/2)at^2= 0 - (1/2)gt^2 = -16(ft/sec^2) × t^2, v= v_0 + at= 0 - gt = - 32(ft/sec^2) × t. when t = 1 sec, y_1= (-16)(ft/sec^2) × 1 sec^2 = -16 ft, v_1= (-32)(ft/sec^2) × 1 sec = -32 ft/sec. The body is therefore 16 ft below the origin (y is negative) and has a downward velocity (v is negative) of magnitude 32 ft/sec. The position and velocity at 2, 3, and 4 sec are found in the same way. y_2 = (-16)(ft/sec^2) × (2 sec)^2 = (-16)(ft/sec^2) × 4 sec^2 = -64 ft v_2 = (-32)(ft/sec^2) × (2 sec) = -64 ft/sec y_3 = (-16)(ft/sec^2) × (3 sec)^2 = (-16)(ft/sec^2) × 9 sec^2 = -144 ft v_3 = (-32)(ft/sec^2) × (3 sec) = -96 ft/sec y_4 = (-16)(ft/sec^2) × (4 sec)^2 = (-16)(ft/sec^2) × 16 sec^2 = -256 ft v_4 = (-32)(ft/sec^2) × (4 sec) = -128 ft/sec The results are illustrated in the diagram.

Question:

What is a fruit? What roles do fruits play in the dispersion of the seeds?

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Solution:

The ovary of an angiosperm is the basal part of the pistil. The ovary contains the ovules which will become the seeds after fertilization. Concomitant with the development of the zygote into an embryo, the ovary enlarges to form the fruit. A fruit therefore can be defined as a matured ovary, containing the matured ovules. A true fruit is one developed solely from the ovary. An accessory fruit is one developed from sepals, petals, or the receptacle as well as the ovary. The apple, for example, is mostly an enlarged receptacle; only the core is derived from the ovary. All angiosperms have fruits, either true or accessory. This characteristic makes them unique among living things. Fruits represent an adaptation for the dispersal of the seeds by various means, and they may be classified according to this criterion as wind-borne fruits, water-borne fruits, or animal-borne fruits. Wind-borne fruits are light and dry so that they can easily be carried by wind. In the tumbleweeds, the whole plant, or fruiting structure, is blown by the wind and scatters seeds as it goes. Other wind-borne fruits, such as the maple, have evolved wing - like structures. Still others, such as the dandelion, develop aplumelike pappuswhich keeps the light fruits aloft. Water-borne fruits are adapted for floating, either because air is trapped in some part of the fruit, or because the fruit contains corky tissue. The coconut fruit has an outer coat especially adapted for carriage by ocean currents. Rain is another means of fruit dispersal by water, and is particularly important for plants living on hillsides or mountain slopes. Animal-borne fruits are mostly fleshy. This makes them appetizing to vertebrates. When fleshy fruits ripen, they undergo a series of characteristic changes, mediated by the hormone ethylene. Among these are a rise in sugar content, a general softening of the fruit through the breakdown ofpecticsubstances, and often a change in color to conspicuous bright red, yellow, blue, or black. When such fruits are eaten by birds or animals, they spread the seeds that lie within-them either passing them unharmed through their digestive tracts or carrying them as adherent passengers on their fur or feathers. Some fruits are further equipped with prickles, hooks, hairs, or sticky coverings, and so can be transported for long distances by animals. The modifications of seeds for dispersal by animals illustrate an evolutionary adaptation to the coexistence of plant and animal forms.

Question:

The compass of an aircraft indicates that it is headed due north, and its airspeed indicator shows that It is moving through the air at 120 mi/hr. If there is a wind of 50 mi/hr from west to east, what is the velocity of the aircraft relative to the earth?

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Solution:

Let subscript A refer to the aircraft, and subscript F to the moving air. Subscript E refers to the earth. We have given v_AF = 120 mi/hr, due north v\ding{217}_FE = 50 mi/hr, due east, and we wish to find the magnitude and direction of v\ding{217}_AE . By the law of addition of velocities v\ding{217}_AE = v\ding{217}_AF + v\ding{217}_FE . The three relative velocities are shown in the figure. It follows from this diagram that \mid v\ding{217}_AE \mid = 130 mi/hr. Furthermore, tan \varphi = [(50 mi/hr)/(120 mi/hr)] tan \varphi = .4167 and \varphi = 22.5\textdegree The airplane travels at a speed of 130 mi/hr at an angle of 22.5\textdegree east of due north.

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Question:

A mixture of nitrogen and oxygen gas is collected by displacement of water at 30\textdegreeC and 700torrHg press-ure. if the partial pressure of nitrogen is 550torrHg, what is the partial pressure of oxygen? (Vapor pressure of H_2O at 30\textdegreeC = 32torrHg.)

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Solution:

Here, one uses Dalton's Law of partial pressures. This law can be stated: Each of the gases in a gaseous mixture behaves independently of the other gases and exerts its own pressure; the total pressure of the mixture being the sum of the partial pressures exerted by each gas present. Stated algebraically P_total= P_1 + P_2 + ...P_n whereP_totalis the total pressure and p_1, p_2 ...P_nare the partial pressures of the gases present. In this problem, one is told that oxygen and nitrogen are collected over H_2O, which means that there will also be water vapor present in the gaseous mixture, in this case, the equation for the total pressure can be written P_total= P_(O)2 + P_(N)2 + P_(H)2O One is givenP_total, P_(N)2, and P_(H)2O and asked to find P_(O)2. This can be done by using the law of partial pressures P_total= P_(O)2 + P_(N)2 + P_(H)2OP_total= 700torr P_(O)2 = ? 700torr= P_(O)2 + 550torr+ 32torrP_(N)2 = 550torr P_(O)2 = (700 - 550 - 32)torrP_(H)2O = 32torr = 118torr P_(O)2 = 118torr.

Question:

What is meant by an "inborn error of metabolism"? What are some examples of inborn errors of metabolism in man?

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Solution:

The expression of a given structural or functional trait is often the result of a number, perhaps large, of chemical reactions in series, with the product of each serving as the substrate for the next: A \ding{217} B \ding{217} C \ding{217} D. For example, the color of most mammalian skin is due to the pigment melanin (D). Melanin is produced from dihyd- roxyphenylalanine (C), which is derived from tyrosine (B), which is in turn made from phenylalanine (A) , (see figure 1). Each reaction is controlled by an enzyme. The conversion of tyrosine to dihydroxyphenylalanine is mediated by the enzyme tyrosinase. When this enzyme is lacking in an in- dividual, melanin production is blocked, and albanism re-sults. We know that the synthesis of proteins is specified by DNA. It follows that part of a DNA molecule, a gene, codes for the production of an enzyme, which is a protein molecule. In albinos, however, this process is inhibited for the protein product tyrosinase. While the normal gene produces the enzyme tyrosinase, its recessive allele does not. When the recessive allele is in the homozygous state, no tyrosinase is produced in the individual. The absence of a normal functioning enzyme, due to either the absence of or a mutation in the gene for that enzyme, is known as an inborn error of metabolism. Alcaptonuria is another example of an inborn error of metabolism. This disorder is an inherited condition which is characterized by a darkening of the cartilage often associated with a type of arthritis. In addition, the urine of affected individuals turns black upon exposure to air. This is due to the presence of an unusual compo-nent, homogentisic acid, in the urine. Homogentisic acid is a normal intermediate in the metabolism of tyrosine and phenylalanine. In normal individuals, this intermediate is oxidized via an oxidase enzyme, to carbondioxide and water (see figure 2). Alcaptonurics lack this enzyme and conse-quently the acid builds up in the tissues and blood of these individuals and is excreted in large amounts in the urine.

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Question:

What is the diameter of the small piston of a hydraulic press when a force of 20 pounds on it produces a force of 4 tons on the large piston whose diameter is 20 inches, assuming that friction can be neglected? What is the mechanical advantage?

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Solution:

The force exerted on each piston is proportion-al to its area. This means: F_1/F_2 = A_1/A_2 where F_1 is the force on the small cylinder, F_2 the force on the large cylinder, and A_1 and A_2 their respect-ive areas. Therefore, (20 lbs)/(4 tons × 2000 lbs/ton) = [\pir^2] / [\pi(10 in)^2] r_1 = (1/2) in^2 If friction is neglected, then AMA = IMA. The IMA of a hydraulic press is the ratio of the areas of its pistons. IMA = AMA = A_2/A_1 = [\pi(10 in)^2]/[ \pi(1/2 in)^2] = F_2/F_1 = 400.

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Question:

Write a FORTRAN program to print all pairs of amicable numbers up to 10,000. Amicable numbers are defined as pairs of integers, the sum of the factors of one of them being equal to the other.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G22-0539.htm

Solution:

An example of one such pair is 220 and 284. The factors of 220 summed up equal 1 + 2 + 4 + 5 + 10 + 11 + 20 + 22 +44 + 55 + 110, or 284, while the factors of 284 summed up equal 1+2 + 4+71 + 142, or 220. Note that 1 is included as a factor, but the number itself is not! To find all amicable pairs up to 10,000, use the following algorithm: for every N from 1 to 10,000 1) Factor N and find the sum, S, of its factors, including 1, but excluding N. 2) If S is greater than N, find the sum of the factors of S. 3) If this second sum equals N, print the pair N,S, for they are amicables. The flowchart will illustrate the actions taken by the program. The program looks as follows: INTEGER FLAG,Y N = 1 12FLAG = 0 G = FLOAT (N) 16S = 1.0 EVEN = 2.0 20IF ((EVEN\textasteriskcentered\textasteriskcentered2).GT.G) GO TO 55 W = G/EVEN Y = IFI × (W) IF (Y.NE.W) GO TO 45 S = S + EVEN IF (W.EQ. EVEN) GO TO 45 S = S + W 45EVEN = EVEN +1.0 GO TO 20 55IF (FLAG.NE.O) GO TO 72 IF (S.LE.N) GO TO 76 G = S FLAG = 1 GO TO 16 72IF (N.EQ.S) GO TO 90 76N = N + 1 IF (N.LT.10000) GO TO 12 GO TO 99 90WRITE (6, 100) N,G 100FORMAT (1X,'AMICABLE PAIR IS', I4, 'AND', I4) GO TO 76 99STOP END

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Question:

Calculate the solubility of carbon dioxide in water at 25\textdegreeC, where the partial pressure of CO_2 over the solution is 760 torr., using Henry's law constant, K_CO(2) at 250C = 1.25 × 10^6. Assume that a liter of solution contains 1000 grams of water.

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Solution:

Raoult'slaw states that the partial pressure, P_2 , of the component present at lower concentration is directly proportional to its mole fraction, X_2 , for dilute solutions. Those that don't usually obey Henry's law, P_2 = X_2 K_2 . The subscript 2 indicates that the solute (the component at lower concentration)is being considered. The constant K_2 is referred to as Henry's law constant. Fornonidealsolutions, Henry's law holds for the solute in the same range whereRaoult'slaw holds for the solvent. For ideal solutions K_2 = P\textdegree_2 and Henry's law becomes identical withRaoult's law. To solve this problem one must know three things: 1) the value of K_CO(2) , equal to 1.25 × 10^6 ; 2) the partial pressure P_2 , equal to 760torr; 3) the mole fraction of CO_2 in water, which has to be determined.

Question:

The vertebrate skeleton maybe divided into two general parts, the axial skeleton and the appendicular skeleton. Whatbones constitute these in man?

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Solution:

The axial skeleton consists of the skull, ver-tebral column, ribs, sternum and hyoid bone. The primary function of the vertebrate skull is the protection of the brain. The part of the skull that serves this function is the cranium. The rest of the skull is made up of the bones of the face. In all, the human skull is composed of twenty-eight bones, six of which are very small and located in the middle ear. At the time of birth, several of the bones of the cranium are not com-pletely formed, leaving five membraneous regions called fontanelles. These regions are somewhat flexible and can undergo changes in shape as necessary for safe passage of the infant through the birth canal. The human vertebral column, or spine, is made up of 33 separate bones known as vertebrae, which differ in size and shape in different regions of the spine. In the neck region there are 7 cervical vertebrae; in the thorax there are 12 thoracic vertebrae; in the lower back region there are 5 lumbar vertebrae, in the sacral or hip region, 5 fused vertebrae form the sacrum to which the pelvic girdle is attached; and at the end of the vertebral column is the coccyx or tailbone, which consists of four, or possibly five, small fused vertebrae. The vertebrae forming the sacrum and coccyx are separate in childhood, with fusion occurring by adulthood. The coccyx is man's vestige of a tail. A typical vertebra consists of a basal portion, the centrum, and a dorsal ring of bone, the neural arch, which surrounds and protects the delicate spinal cord which runs through it. Each vertebra has projections for the attachment of ribs or muscles or both, and for articulating (joining) with neighboring vertebrae. The first vertebra, the atlas, has rounded depressions on its upper surface into which fit two projections from the base of the skull. This articulation allows for up and down movements of the head. The second vertebra, called the axis, has a pointed projection which fits into the atlas. This type of ar-ticulation allows for the rotation of the head. In man there are 12 pairs of ribs, one pair articu-lating with each of the thoracic vertebrae. These ribs support the chest wall and keep it from collapsing as the diaphragm contracts. Of the twelve pairs of ribs, the first seven are attached ventrally to the breastbone, the next three are attached indirectly by cartilage, and the last two, called "floating ribs", have no attach-ments to the breastbone. The sternum or breastbone consists of three bones - the manubrium, body and xiphoid process - which usually fuse by middle-age. The sternum is the site for the anterior attach-ment of most of the ribs. The ribs and sternum together make up the thoracic cage which functions to protect the heart and lungs. The hyoid bone supports the tongue and its muscles. It has no articulation with other bones, but is held in place by muscles and ligaments. The bones of the girdles and their appendages make up the appendicular skeleton. In the shoulder region the pecto-ral girdle, which is generally larger in males than in fe-males, serves for the attachment of the forelimbs. The pec-toral girdle consists of two collarbones, or clavicles, and two shoulder blades, or scapulas. In the hip region, the pelvic girdle serves for the attachment of the hindlimbs. The pelvic girdle, which is wider in females so as to allow room for fetal development, consists of three fused hip-bones, called the ilium, ischum and pubis, which are at-tached to the sacrum. Collectively, the "hip bone" is called the oscoxae or innominate bone. Articulating with the scapula is the single bone of the upper arm, called the humerus. Articulating with the other end of the humerus are the two bones of the forearm called the radius and the ulna. The radius and ulna permit rotation of the forearm. The ulna has on its end next to the humerus a process often referred to as the "funny bone." The wrist is composed of eight small bones called the carpals. The ar-rangement of these bones permits the rotating movements of the wrist. The palm of the hand consists of 5 bones, known as the metacarpals, each of which articulates with a bone of the finger, called a phalanx. Each finger has three phalanges, with the exception of the thumb, which has two. The pattern of bones in the leg and foot is similar to that in the arm and hand. The upper leg bone, called the femur, articulates with the pelvic girdle. The two lower leg bones are the tibia (shinbone) and fibula, cor-responding to the radius and ulna of the arm, respectively. These two bones are responsible for rotation of the lower leg. Ventral to the joint between the upper and lower leg bones is another bone, the patella or knee cap, which serves as a point of muscle attachment for upper and lower leg muscles. This bone has no counter part in the arm. The ankle contains seven irregularly shaped bones, the tarsals, corresponding to the carpals of the wrist. The foot proper contains five metatarsals, corresponding to the metacarpals of the hand, and the bones in the toes are the phalanges, two in the big toe and three in each of the others.

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Question:

A 200-kg satellite is lifted to an orbit of 2.20 × 10^4 mi radius. How much additional potential energy does it acquire relative to the surface of the earth?

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Solution:

As in the diagram, R is the earth's radius, M is the earth's mass, m is the satellite mass, and r is the distance between the earth's center and the satellite. R = 6.37 × 10^6 m r = 2.20 × 10^4 mi = 3.54 × 10^7m M = 5.98 × 10^24 kg m = 200 kg The additional potential energy is equal to the work done against the earth's gravitational field. At a distance R from the earth's center, that is, on the earth's surface, the satellite has a potential energy, U_surface U_surface = - (GMm/R) In orbit of radius r the potential is U_orbit = - (GMm/r) Then the additional potential energy involved in launching the rocket to its orbit, ∆U, is given by ∆U = U_orbit - U_surface = - (GMm/r) - {- (GMm/R)} = GMm (1/R - 1/r) = (6.67 × 10^-11 nt - m^2/kg^2) (5.98 × 10^24 kg)(200kg) × [1/(6.37 × 10^6 m) - 1/(3.54 × 10^7 m)] = 1.0310^10 joules This is about equal to the work needed to lift an object weighing 3800 tons to a height of 1000 ft above the earth. Note that the change in potential energy of the satellite cannot be found by using U = mgh. This formula applies only to objects near the earth's surface, where g is approximately constant.

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Question:

In order to find how much insulated wire he has left on a bobbin, a scientist measures the total resistance of the wire, finding it to be 5.18 \Omega. He then cuts off a 200-cm length and finds the resistance of this to be 0.35 \Omega. What was initially the length of wire on the bobbin?

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Solution:

The resistance of the wire on the bobbin is related to its length by the formula R_0 = \rhol_0 /A. That is, the resistance is directly proportional to the length (l_0) of the resistor and inversely proportional to the cross-sectional area A of the resistor. \rho is a constant of proportionality (the resistivity). The cut-off length has the same resistivity and cross-sectional area. Hence its resistance is R =\rhol/A. \therefore(l_0 /l) = (R_0 /R)orl_0 = 200 cm × [(5.18\Omega)/(0.35\Omega)] = 2960 cm .

Question:

A thin lens of long focal length is supported horizontal-ly short distance above the flat polished end of a steel cylinder. The cylinder is 5 cm high and its lower end is rigidly held. Newton's rings are produced between the lens and the upper end of the cylinder, using normal-ly incident light of wavelength 6000\textdegreeA, and viewed from above by means of a microscope. When the temperature of the cylinder is raised 25 C deg, 50 rings move past the cross-wires of the microscope. What is the coefficient of linear expansion of steel.

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Solution:

Initially, the cylinder has a length d, and a gap of length t_0 exists between the top of the cylinder and the bottom surface of the lens in the portion of the air wedge viewed in the microscope at the same position as the cross-wires. (Technically, t_0 is the distance of the lens from the cylinder as measured along axis AA' in figure (B). However, since the lens is thin, t_0 is the distance of any point on the bottom surface of the lens from the top of the cylinder). Let us now see how interference fringes are produced. Light is incident upon the lens from above. Light is transmitted through both lens surfaces with no phase change. However, the light reflected from the bottom lens surface and the top of the cylinder are 180\textdegree (or \lambda/2) out of phase. This phase change is due to the reflection from the cylinder. In addition, the wave reflected from the cylinder travels a distance 2t_0greater than the ray reflected from the lens in tra-versing the air gap. If this distance is equal to an odd number of half-wavelengths (odd because the waves are already \lambda/2 out of phase due to reflection) con-structive interference will occur. Hence Constructive, Interference2t_0 = (2n + 1)(\lambda/2)(n = 0, 1, 2, ...) or2t_0 = [n + (1/2)]\lambda(n = 0, 1, 2, ...) Hence, for a bright fringe to appear at the cross -wires, 2t_0 = [{n + (1/2)}] \lambda, where n is an unknown integer. After one heats the cylinder through a temperature difference T, the length of the cylinder is d(1 + \alphaT) , where \alpha is the coefficient of linear expansion of steel, and the gap will have been reduced to t, where t_0 - t = d\alphaT. If a bright fringe is again seen at the position of the cross-wires, then, similarly, 2t = [m +(1/2)]\lambda, where m is an integer, (m \not = n). \therefore2(t_0 - t) = 2d\alphaT = (n - m)\lambda. During the heating process, (n - m) bright fringes must have passed over the cross-wires. Thus \alpha = [(n - m)\lambda] / [2dT] = [(50 × 6 × 10^-5 cm) / (2 × 5 cm × 25 C deg)] = 1.2 × 10^-5 per \textdegreeC.

Question:

For a J-K type, clocked, master-slave flip-flop, give a) a block diagram,b) a timing diagram, c) the Karnaugh Map,d) the excitation equation, e) the state table, and,f) the state diagram.

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Solution:

a) The block diagram is as shown below. The slave gets the inverted clock pulse of what the master gets. The J-K flip-flop differs from the SC flip-flop in that the J-K flip-flop does not have the restriction that the S-C type flip-flop has, i.e., both the inputs of the S-C type flip-flop cannot simultaneously be equal to one. In the J-K type flip-flop, it is permissible to have both the J and K inputs 1. b) The timing diagram is given in Figure 1 for 5 clock pulses up to t_5. Note that y does at the trailing edge of the clock pulse exactly what x does at the leading edge of the clock pulse. Also note that x' and y' are just the reciprocals of x and y respectively. c) The Karnaugh Map for the J-K flip-flop is plotted in Figure 2. The entries in the Map represent the next state y\rightarrowfor a given combination \rightarrow of y, i.e., the present state, and the input J and K. d) The excitation equation is obtained from the Karnaugh map of figure 2, as follows, for the next state y\rightarrow \rightarrow y\rightarrow= Jy' + K'y. \rightarrow e) The Karnaugh Map also helps us to prepare the state table as given in figure 3: JKy y\rightarrow \rightarrow REMARKS 000 001 0 1 NO CHANGE 010 011 0 0 CLEARS 100 101 1 1 SETS 110 111 1 0 FLIPS f) The state diagram is drawn, in figure 4.

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Question:

A) Determine the pH of a solution with a hydrogen ion concentration of 3.5 × 10^-4. B) If a solution has a pH of 4.25, what is the hydrogen ion concentration?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E10-0338.htm

Solution:

To determine the acidity orbasicityof an aqueous solution, the hydrogen ion concentration must be measured. The pH of a solution expresses this concen-tration. pH is defined as the negative logarithm of the hydrogen ion concentration. In other words, pH =(- log [H^+]), where the brackets around H^+ signify con-centration. As such, to solve the problem, you must substitute into the equation. For part "A", you have pH = - log [3.5 × 10^-4] now - log [3.5 × 10^-4] = - log 3.5 - log 10^-4 = - .54 - (- 4) . = - .54 + 4 = 3.46 It follows, then, that the pH = 3.46 for a hydrogen ion concentration of 3.5 × 10^-4, Part "B" is similar, but here you are given the pH and asked to find the ion concen-tration. Therefore, you have 4.25 = - log [H^+] or - 4.25 = log [H^+]. Now. logarithm numbers give only positive mantissas. As such, - 4.25 must be in the form of - 5 + .75. If you take the antilogarithm of each, .75 is 5.6 and - 5 is 10^-5, you obtain a hydrogen ion concentration of 5.6 × 10^-5 mole/liter

Question:

Derive a relation describing the interference effects observed Derive a relation describing the interference effects observed when light is reflected from a thin film. (See figure).

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Solution:

In order to understand the interference effects produced by a thin film, we trace the path of an incident ray of light, as shown in figure (a). The incident ray first encounters surface A, where it is partially reflected and partially absorbed. Since the refractive index of the film is greater than the refractive index of air, the reflected ray undergoes a 180\textdegree phase change. The transmitted part of the incident ray now encounters interface B, where it is partially reflected and par-tially absorbed. However, this time the reflected ray undergoes no phase change since it is traveling from a region of high refractive index to a region of low refractive index. Hence, rays 1 and 2 differ in phase by 180\textdegree , or \lambda_0/2where \lambda_0 is the wavelength of the incident light in free space. In addition to the 180 phase change due to reflection, ray 2 tra-vels a distance 2t greater than ray 1. (This holds only if the rays shown are incident at an angle \o which is very small.) Then, if we want to observe destructive interference, the distance 2t must contain an integral number of wavelengths, N^\lambda, where N = 0, 1, 2,... . But, \lambda is not the wavelength of the light in free space, but rather, the wave-length of light in the film (see figure(b)). However, the wavelengths \lambda_0 and \lambda are related. By definition of the refractive index of the film n = c/v where c is the speed of light in free space, and v is its speed in the film. If \lambda_0 and f_0 are the free space wavelength and frequency of light , we may write c = \lambda_0 f_0 Similarly,v = \lambdaf where \lambda and f are the wavelength and frequency of light in the film. But the frequency of light is the same in all media. Then v = \lambdaf_0 Hencen = (\lambda_0f_0)/(\lambdaf_0) = (\lambda_0)/(\lambda) and\lambda_0 = n\lambda Combining this fact with the previous discussion, we obtain 2t = N\lambda = (N\lambda_0)/(n)N = 0, 1, 2,... destructive interference 2t = [N + (1/2)]\lambda = [{N + (1/2)\lambda_0} / n]N = 0, 1, 2,... constructive interference ort = (N\lambda_0)/(2n)N = 0, 1, 2,... destructive interference t = [{N + (1/2)\lambda_0}/{2n}]N = 0, 1, 2,... constructive interference.

Question:

Simplify [(1600 × 10,000) / (2000)]^1/3.

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Solution:

Observe 1600 = 16 × 100 = 16 × 10^2 10,000 = 10^4 2,000 = 2 × 10^3. Thus, [(1600 × 10,000) / (2000)]^1/3 = [{(16 × 10^2) (10^4)} / {2 × 10^3}]^1/3. Thus, [(1600 × 10,000) / (2000)]^1/3 = [{(16 × 10^2) (10^4)} / {2 × 10^3}]^1/3. Using the associative property, = [{16 × (10^2 × 10^4)} / {2 × 10^3}]^1/3. Recall:a^x \bullet a^y = a^x+y,= [(16 × 10^6) / (2 × 10^3)]^1/3 Since(ab/cd) = (a/c) \bullet (b/c), = Since(ab/cd) = (a/c) \bullet (b/c), = [(16 × 10^6) / (2 × 10^3)]^1/3 Recall(a^x/a^y)= a^x\rule{1em}{1pt}y,= (8 × 10^3)^1/3 Since(ab)^x = a^xb^x,= 8^1/3 × 10^3(1/3) = 2 × 10^1 = 20.

Question:

The flame test for barium involves placing a barium-containing compound in a Bunsen burner flame. The barium compound then decomposes to give barium atoms which sub-sequently undergo an electronic transition of energy 3.62 × 10^-12 erg. What color flame would this give rise to?

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Solution:

To solve this problem we use the relationship E =hc/\lambda, where E is the energy of the transition, h is Planck's constant, c is the speed of light, and \lambda is the wavelength of the emitted light. Solving for \lambda, \lambda = (hc/E) = [(6.62 × 10^-27 erg-sec × 3.0 × 10^10 cm/sec)/(3.62 × 10^-12 erg)] = 5.490 × 10^-5 cm. Converting to \AA, \lambda = 5.490 × 10^-5 cm = 5.490 × 10^-5 cm × 10^8 \AA/cm = 5490 \AA, which corresponds to a green flame.

Question:

A cyclotron has an oscillator frequency of 11.4 Mc \bullet s^-1 and a radius of 60 cm. What magnetic induction is required to accelerate protons of mass 1.67 x 10^-27 kg and charge 1.6 × 10^-1 9 C, and what is the final energy that they acquire? What error is made by assuming that the mass of the protons remains constant?

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Solution:

The force on a particle of charge q, travelling with velocity v^\ding{217}, in a field of magnetic induction B^\ding{217}, is F =qv^\ding{217} × B^\ding{217} Assuming that B^\ding{217} and v^\ding{217} are perpendicular, F =qvB(1) Since a particle in this situation will execute a circular orbit, F is a centripetal force and the particle's acceleration is a = (v^2 /R). Using Newton's Second Law, F = ma = (mv^2 /R)(2) where m is the mass of the particle, and R is the radius of its orbit. Inserting (2) in (1) (MV^2)/ R = qvB or,solving for v v = (qBr) / m(1) Now, we must eliminate v, since it is unknown.But, since the path is circular, and v is constant,we may write v = (2\piR)/T where T is the time it takes the particle to traverse its orbit once, or the period of the motion. Since T = (1/f) where f is the frequency of the motion v = (2\piR)/f(2) Substituting (2) in (1) gives us B in terms of known quantities. 2\piRf = (qBR)/m Solving for B B = (2\pifm)/q = [(2\pi × 11.4 × 10^6 s^-1 × 1.67 × 10^-27 kg)/(1.6 × 10^-19 C)] = 0.748Wb\bulletm^-2 . where we have used the fact that in the cyclotron, the orbital frequency of the particle equals the oscillator frequency. The final energy of the protons is, using (1) (1/2) mv^2 = [(q^2 B^2 R^2 )/(2m)] = [(1.6 × 10^-19)^2 C^2 × (0.748)^2 Wb^2 \bullet m^-4 × 0.6^2 m^2) /(2 × 1.67 × 10^-27 kg)] = 0.154 × 10^-11 J Since E = mc^2 , this energy is equivalent to an increase of mass ∆m = [(0.154 × 10^-11 J)/(9 × 101 6m^2 \bullet s^-2 )] = 0.017 × 10^-27 kg . The error is thus (∆m/m) × 100 = [(0.017)/(1.67)] × 100 = 1.02% .

Question:

Euglena is a common green flagellate protozoan found in fresh water ponds. Describe briefly the method of locomotion, nutrition, and asexual reproduction in this organism.

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Solution:

Normally, locomotion in Euglena is produced by undulating movements of the flagellum. These movements of the flagellum draw the organism after it in a charac-teristic spiral path. Actually, Euglena has two flagella, but only one extends from the body and is used for loco-motion. With regard to nutrition, Euglena carries on autotrophic nutrition, as do green plants, synthesizing food from inorganic substances in the presence of light. However, under condition of total darkness for long periods of time, in a medium containing necessary nutrients, Euglena will shift to a heterotrophic mode of nutrition. Under these conditions the chloroplasts, the organelles which carry on photosynthesis, disappear, and the organ-isms live by absorbing the necessary nutrients from the surrounding medium through their cell membranes. When returned to the light environment, the chloroplasts reappear, and the autotrophic mode of living is resumed. This behavior is one of the reasons why Euglena is said to be both plant-like and animal-like. The life cycle of Euglena involves both an active phase, during which the organism moves about, and an en-cysted phase, during which the organism is rounded up with a protective cyst membrane surrounding it. Asexual reproduction can occur in both stages. The division is typically a longitudinal binary fission. The nucleus divides by mitosis, and then the cytoplasm divides, forming two cells each with a nucleus.

Question:

Suppose the plates in the figures have an area of 2000 cm^2 or 0.20 m^2, and are 1 cm or 10^\rule{1em}{1pt}2 m apart. The potential difference between them in vacuum, V\ding{217}_o, is 3000 volts, \ding{217} and it decreases to 1000 volts when a sheet of dielectric 1 cm thick is inserted between the plates. Compute the following: (a) the relative permittivity K of the di-electric, (b) its permittivity \epsilon, (c) its susceptibility X, (d) the electric intensity between the plates in vacuum, (e) the resultant electric intensity in the di-electric, (f) the electric intensity set up by the bound charges, and (g) the ratio of the surface density of bound charge, \sigma_b, to that of free charge, \sigma_f.

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Solution:

The dielectric is polarized when inserted between the plates and the polarization charges (or bound charges) effectively neutralize some of the surface charges on the plates, as shown in figure 2. This then reduces the field strength in the dielectric, thus lower-ing the voltage difference across the plates, since the latter is proportional to the electric field (V = Ed). What remains unchanged in this case is the true charge density on the plates, i.e. the total charge on the plates. The ratio of the field in the dielectric to that in the vacuum gives the relative permittivity of the dielectric (a) K = (E / E_0) = (V / V_0) = (3000 / 1000) = 3 (b) \epsilon = K\epsilon_0 = 3\epsilon_0 (c) X = \epsilon_ \rule{1em}{1pt} \epsilon_0 = 2\epsilon0 (d) E_0 = V_0 / լ= (3 × 10^3 volts) / (10^\rule{1em}{1pt}2 m) = 3 × 10^5 (volts / m) (e) E = V / լ = (1 × 10^3 volts) / (10^\rule{1em}{1pt}2 m) = 10^5 (volts / m) The bound charges of the dielectric set up a new electric field E_b in the slab which opposes the electric field E0due to the plate charges. The new field E is the resultant of these two (f)E_b = E_0 \rule{1em}{1pt} E = 2 × 10^5 (volts/m). The surface charge density \sigmais given by (g)\sigma = \epsilon_0E_0 = 3 × 10^5 \epsilon_0 (coul / m^2) The bound charges on the surface of the slab have a density \sigma_b = \epsilon_0E_b = \epsilon_0(E_0 \rule{1em}{1pt} E) = \epsilon_0E_0(k\rule{1em}{1pt}1) = XE = 2 × 10^5 \epsilon_0 (coul / m^2), and we get [(\sigma_b) / (\sigma_f)] = (2 / 3).

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Question:

When light is incident on a thin slit, a diffraction pattern, shown in figure (a), is produced. Find an expression which give the location of the minima of diffraction pattern in terms of the angle \psi.(see figure (b)).

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Solution:

In this problem, we consider only Fraunhoferdiffraction. (See figure (b).) This type of diffraction is characterized by parallel, incident light rays encountering a slit, and being diffracted, again parallel to one another. In practice, this restriction may be effected in 2 ways. The light source is placed very far away from the slit, and the screen is far from the slit. Another method employs a converging lens to the left of the slit, and a converging lens to the right. Light rays incident upon the first converging lens leave par-allel to one another, encounter the slit, and are diffracted parallel to one another (see figure (c)). They then are focused by the second con-verging lens on screen F, and the diffraction pattern results. Using figure (c), we can derive an equation locating the minima of the diffraction pattern. Let us focus on rays coming from points A and B. Since the incident light is in phase, the only phase differencebetween r_1 and r_2 must occur because r_2 travels a larger distance r_1. (Here, we assume that the rays are effectively parallel). If this distance, BB', is equal to (1/2) \lambda, where \lambda is the wavelength of light, the rays r_1 and r_2 will interfere destructively and p_1 will be a minimum. Since BB' = a/2 sin \psi, minima will be observed when a/2 sin \psi = \lambda/2 Hence, a sin \psi = \lambdaminima. This formula gives the 1st order minima of the diffraction pattern. Now, examine figure (d), where the slit has been divided into 4 equal portions of length a/4. Looking at rays r_1, and r_2, we note that the only phase difference between them occurs because they travel unequal distances. Ray r_2 travels a distance BB' greater than r_1. If BB' = \lambda/2, r1and r_2 will destructively interfere. Furthermore, r_3 and r_4 will also destructively interfere since DH = \lambda/2. Hence, each pair of similar rays will destructively interfere, and p_2 will be a minimum of the diffraction pattern. Therefore, since BB' = a/4 sin \psi the second order minimum is described by Thena/4 sin \psi = \lambda/2minimum a sin \psi = 2\lambdaminimum. In general, if we divide the slit into n equal segments, where n is an even integer, we will find anda/n sin \psi =\lambda/2minima a sin \psi = n\lambda/2miniman even sincen = 2m wherem = 1 , 2 ,... we obtain a sin \psi = m \lambdaminima m = 1 , 2 ,... Note that n must be even because each pair of rays emanating from the slit interfere.

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Question:

When a uranium nucleus ^235 _92 U captures a neutron, fission occurs. If one of the fission fragments formed is the krypton nucleus ^95 _36 Kr, identify what nuclei are formed as the krypton decays to the stable nucleus ^95 _42 Mo by a succession of \beta-decays .

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/Users/wenhuchen/Documents/Crawler/Physics/D34-1031.htm

Solution:

When a nucleus of atomic number Z and atomic weight A decays into another nucleus by emitting an electron, the atomic weight of the new nucleus has the same A but its atomic number is Z + 1, ^A _Z X\rightarrow^A _Z+1 Y + e^\Elzbar +v_e .(1) Nucleus X has charge +Ze, where e is the unit charge. The charge of Y is + (Z + 1)e and that of the electron is - e. The neutrino (ѵ_e) is achargeless particle. Therefore the total charge of the decay products is + (Z + 1)e + (\Elzbare) + 0 = +Ze This is equal to the charge of the initial particle, due to the law of conservation of charge. By using the decay equation (1) and the periodic table, we find the following intermediate decays for the transition ^95 _36 Kr \rightarrow^95 _42 Mo. ^95 _36 Kr \rightarrow^95 _37Rb+e^\Elzbar +ѵ_eRb- rubidium ^95 _37Rb\rightarrow^95 _38Sr+e^\Elzbar +ѵ_eSr- strontium ^95 _38Sr\rightarrow^95 _39 Y+e^\Elzbar +ѵ_eY-yittrium ^95 _39 Y\rightarrow^95 _40Zr+e^\Elzbar +ѵ_eZr- zirconium ^95 _40Zr\rightarrow^95 _41Nb+e^\Elzbar +ѵ_eNb- niobium ^95 _41Nb\rightarrow^95 _42 Mo +e^\Elzbar +ѵ_eMo- Molybdenum

Question:

The photoneutron effect is the absorption of a photon by a heavy nucleus with subsequent emission of a neutron. The threshold energy of the photon is found to be about 10 MeV. Assume that the neutron is removed from the nucleus by applying to it a constant force which has to move it through a distance of 10^\rule{1em}{1pt}13 cm outward. Calculate the order of magni-tude of this "average nuclear force" and compare it with the electrostatic force between two protons separated by 1.9 × 10^\rule{1em}{1pt}13 cm.

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Solution:

The minimum energy needed to remove the neutron from the nucleus (the threshold energy) is 10 MeV. This is the work which must be done by the constant force to "pluck out" the neutron. Work done by force = 10 MeV = 10 × 106eV × [(1.602 × 10^\rule{1em}{1pt}12 erg) / 1eV] = 1.602 × 10^\rule{1em}{1pt}5 erg But,Work = Force x distance because, in this case, the force is constant and parallel to the displacement. Hence, Force × (10^\rule{1em}{1pt}13cm) = 1.602 × 10^\rule{1em}{1pt}5 erg Force = 1.602 × 10^8 dyne. Notice, first, that this force, which is present inside a single nucleus, is equivalent to the weight of about 160 kilograms. It could support the weight of two men. The electrostatic force between the two protons (in e.g.s. units) is Electrostatic force = e^2 / R2 = [(4.8 × 10^\rule{1em}{1pt}10 esu) / (1.9 × 10^6 cm)]^2 = 6.4 × 10^6 dyne. The "average nuclear force" is about 25 times larger than the electrostatic force. Since, in reality, the nuclear force falls off very rapidly as the neutron is pulled out-ward, the force in the initial position is considerably larger than the average.

Question:

a) Carefully examine the given switching network, find its output, and explain each operation. b) Draw an equivalent network which will give the same output function as in part (a), using the minimum amount of gates.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G03-0053.htm

Solution:

The circuit and the output of each gate is shown in fig. 1. Gates 1 and 2 are EXCLUSIVE OR gates, 3 and 4 are NOR gates, 5 is an OR gate, 6, and 7 NAND gates. The output of gate 7 is determined by the use of outputs of gates 5, and 6, and according to the definition of NAND gates (i.e., f(a, b) =ab). Therefore the output function f(a, b, c) is: By the use of Boolean algebra theorems, this function can be reduced as follows: = (a\oplusb)(b\oplusc) + (a+b) \bullet (a + c)(Demorgan's law) = (ab+ab ) (bc+bc) + (a+b) (a + c)(Definatoin of Exclusive OR function) = abbc+ abbc +abc+abbc + (a+b) (a + c) (Distributive property) =bc +ab +ac + ab(Demorgan's law) =bc +ab + ab(Demorgan's law) =bc + a\oplusb (Definatoin of exclusive OR ) This final function can now be translated into a switching network by the use of AND, OR, and EXCLUSIVE OR gates. The first termbc can be obtained by an AND gate, and the second term with an Exclusive OR gate. These functions then must be added by the use of an OR gate. The equivalent switching network and its truth table is given in fig. 2. Note that original and the equivalent circuits both give the same results, only the latter is in reduced form. abc b c a\oplusb f(a, b, c) 000 001 010 011 100 101 110 111 0 1 0 0 0 1 0 0 0 0 1 1 1 1 0 0 0 1 1 1 1 1 0 0

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Question:

What evidence regarding the nature of the genetic material has been obtained from experiments with bac-teriophage?

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/Users/wenhuchen/Documents/Crawler/Biology/F24-0608.htm

Solution:

By using radioactive tracers, A.D. Hershey and M. Chase furnished proof that DNA is the genetic material. This evidence was obtained by use of the bacterium Escherichia coli; and the bacteriophage which attack it. The bacteriophage were grown on a medium containing limited amounts of radioactive sulfur and phosphorous. The radioactive labels used were phospho-rous 32 (^32P) and sulfur 35 (^35S). Phage proteins do not contain appreciable amounts of phosphorus. Therefore, only the DNA was labeled with ^32P. Similarly, only the protein envelope around the phage was labeled with ^35S.By this method, the DNA phage and proteins werediffe-rentially labeled. Following growth of the phage and incorporation of the radioactive labels, the radioactive viruses were exposed to non-radioactive bacteria. After absorption and infection of the bacteria, the phage were separated from the bacterial cells by centrifugation. The location of the ^32P labeled DNA and ^35S labeled proteins of the virus were then determined. The phosphorus label was found to be associated with the bacterial cells and the sulfur label was in the protein coats left in the medium. This indicated that the DNA had penetrated the cells but that the protein coats of the phage was left outside the walls of the bacteria and had been separated from the bacterial cells during centrifugation. Only the labeled DNA was passed on to the bacteria. The significant conclusion of this experiment is that only the DNA of the virus entered the host bacteria. All the protein of the phage remained outside the bac-teria cells wall. No protein was in the bacteria during viral replication. This experiment illustrates that it is DNA and not protein that is the genetic material. (See diagram) After the DNA part of the virus was reproduced within the bacterial cell, new protein was synthesized and became associated with the DNA units. New infective virus particles were thus formed. When the host (bac-terium) lysed, numerous infectious virus particles emerged, ready to enter other bacterial cells and repeat the cycle.

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Question:

What are the importance of desmids, diatoms and dinoflagellates to man?

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/Users/wenhuchen/Documents/Crawler/Biology/F06-0163.htm

Solution:

Desmids, diatoms and dinoflagellates are all unicellular algae. Desmids are nonmotile, fresh-water green algae, commonly found in lakes and rivers. Desmids have symmetrical, curved, spiny or lacy bodies with a constriction in the middle of the cell. Under the microscope they look like snowflakes. Diatoms, members of Chrysophyta, are found in fresh and salt water. They have two shelled, siliceous cell walls and store food as leucosin and oil. The cell walls are ornamented with fine ridges, lines, and pores that are either radically symmetrical or bilaterally symmetrical along the long axis of the cell. Diatoms lack flagella, but are capable of slow, gliding motion. Dinoflagellates, the majority of the Pyrrophyta, are surrounded by a shell consisting of thick, interlocking plates. All are motile and have two flagella. A number of species lack chlorophyll. These are the heterotrophs which feed on particulate organic matter. Dinoflagellates are mainly marine organisms. All of these organisms play extremely important roles in aquatic food webs. Plankton is defined as small aquatic organisms floating or drifting near the surface. Phyto-plankton are photosynthetic autotrophs and the zooplankton are heterotrophs. Desmids are important freshwater phyto-plankton. Diatoms are the most abundant component of marine plankton (a gallon of sea water often contains one or two million diatoms). Diatoms are also found in abundance in many rivers. Dinoflagellates are second only to the diatoms as primary producers of organic matter in the marine en-vironment. Probably three quarters of all the organic matter in the world is synthesized by diatoms and dino-flagellates. In addition to the production of organic material, these algae have the primary responsibility for the continued production of molecular oxygen via photosyn-thesis. Respiration by animals utilizes oxygen. If the supply were not constantly being replenished, the oxygen on earth would be exhausted. Phytoplankton are essential to aquatic food chains; they are eaten by zooplankton, by invertebrates, and by some fish. The organisms which eat phytoplankton are in turn eaten by other organisms. Ultimately all aquatic life depends on the phytoplankton. Since terrestrial animals rely ultimately on the oceans, rivers, and lakes for a large part of their food source, the desmids, diatoms, and dinoflagellates are of crucial importance to land life as well. Diatoms are also important to man by virtue of their glasslike cell walls. When the cells die, their silica-impregnated shells sink to the bottom of the sea, and do not decay. These shells accumulate in large quantities and geologic uplifts bring the diatomaceous earth to the surface, where it is mined and used commercially. Diatomaceous earth is used as a fine abrasive in detergents, toothpastes, and polishes. It is also used as a filtering agent, and as a component in insulating bricks and soundproofing products. Diatoms utilize oils as reserve material, and it is widely believed that petroleum is derived from the oil of diatoms that lived in past geologic ages.

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Question:

A 200 lb man hangs from the middle of a tightly stretched rope so that the angle between the rope and the horizontal direction is 5\textdegree, as shown in Figure A. Calculate the tension in the rope. (Figure B).

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/Users/wenhuchen/Documents/Crawler/Physics/D02-0024.htm

Solution:

Since the two sections of the rope are symmetrical with respect to the man, the tensions in them must have the same magnitude, (Fig. B.) This can be arrived at by summing the forces in the horizon-tal direction and setting them equal to zero since the system is in equilibrium. Then \sumF_x = T_1 cos 5\textdegree - T_2 cos 5\textdegree = 0 andT_1 = T_2 = T Considering the forces in the vertical direction, \sumF_y = T sin 5\textdegree + T sin 5\textdegree - 200 1b = 0 200 lb = 2T sin 5\textdegree = 2T(0.0871) T= (200)/[(2)(0.0871)] = 1150 lbs. Note the significant force that can be exerted on objects at either end of the rope by this arrangement. The tension in the rope is over five times the weight of the man. Had the angle been as small as 1\textdegree, the tension would have been T= 200/(2 sin 1\textdegree) = 200/[(2)(0.0174)] = 5730 lbs. This technique for exerting a large force would only be useful to move something a very small distance, since any motion of one end of the rope would change the small angle considerably and the ten-sion would decrease accordingly.

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Question:

How much electricity will be required to decompose 454g of water by electrolysis? The overall reaction is 2H_2O \rightarrow 2H_2 + O_2.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E16-0572.htm

Solution:

Whenever a problem deals with weights and electricity, the solution involves an application of Faraday's Law: The passage of 1 faraday of electricity (96,500 coulombs) causes 1 equivalent weight of matter to be oxidized (the loss of 1 electrons) at one electrode and the reduction (the gain of 1 electrons) of 1 equivalent weight at the other electrode. Equivalent weight may be defined as molecular weight divided by number of moles of hydrogen transferred. To solve this problem, therefore, calculate the number of equivalents present in 454g of water. Water has a molecular weight of 18g/mole, but since 2 H's are transferred, water has an equivalent weight of 9g. Therefore, the number of equivalents is [total weight / equivalent weight] = [454 / 9] = 50.4 equiv. Recalling that 1 faraday of electricity is used per equivalent, 50.4 equivalents times 1 Faraday/equivalent = 50.4 faradays of electricity required to decompose 454g of water by electrolysis.

Question:

How may the gene pool be altered?

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/Users/wenhuchen/Documents/Crawler/Biology/F26-0697.htm

Solution:

The total genetic material of a given population is termed the gene pool. The Hardy-Weinberg principle tells us that the relative gene frequencies within the gene pool of a population will remain constant from generation to generation unless certain factors alter the equilibrium of gene frequencies in the gene pool. Such factors include mutation, natural selection, migration, random genetic drift, and meiotic drive. Mutations can change one allele to another, such as A to a, and if they are recurrent or ''one-way', the relative frequencies of the two alleles will be changed by an increase in the proportion of a at the expense of A. Owing to the low mutation rate of most genes, such direct change in allelic frequency alone probably does not cause significant change in the gene pool. More important is the fact that mutations are a source of new genes, and thus traits, within a population. Together with sexual reproduction, which creates new combinations of existing genes, mutations provide the variation which is the basis for the operation of natural selection. Selection is the nonrandom differential retaining of favored genotypes. Unlike mutation, which operates directly on a gene to alter its frequency, selection indirectly alters a gene's frequency by acting on its carriers as a function of their ability to reproduce viable offspring. For example, if individuals carrying gene A are more successful in reproduction than indivi-duals carrying gene B, the frequency of the former gene will tend to increase generation after generation at the expense of the latter. Any trait which gives an organism a better chance at survival in a given environment, will increase that organism's ability to grow and reproduce, and increase the proportion of the gene for that trait in the gene pool. By the same token, any gene which confers a disadvantage to its carrier within its environment will decrease that organism's chance of survival to reproductive age, and thus decrease the frequency of that gene in the gene pool. This is the principle underlying natural selection. Migration acts both directly and indirectly to alter gene frequencies. Directly, a population may receive alleles through immigration of individuals from a nearby population. The effectiveness of immigration in changing allelic frequencies is dependent on two factors: the difference in gene frequencies between the two populations and the proportion of migrant genes that are incorporated. Alternatively, there can be emigration of members from a population, which, depending on the size of the emigration and whether or not it is selective, can result in a change in allelic frequencies in the gene pool. Indirectly, migration acts as a source of variation, similar to mutation, upon which the forces of natural selection can operate. Migration can also enhance natural selection within a population by upsetting the equilibrium that may exist among the given genotypes in a population, concerning an advantage by sheer numbers to a given group. It can blur the effects of natural se-lection by replacing genes removed by selection. Allelic frequencies may fluctuate purely by chance about their mean from generation to generation. This is termed random genetic drift. Its effect on the gene pool of a large population is negligible, but in a small effectively interbreeding population, chance alteration inMendelianratios can have a significant effect on gene frequencies, and may lead to the fixation of one allele and loss of another. For example, isolated communities within a given population have been found to have dif-ferent frequencies for blood group alleles than the population as a whole. Chance fluctuations in allelic frequency presumably caused these changes. Another factor that may alter allelic frequencies is meiotic drive. This is the term for preferential segregation of genes that may occur in meiosis. For example, if a particular chromosome is continually segregated to the polar body in femalegametogenesis, its genes would tend to be excluded from the gene pool since the polar bodies are nonfunctional and will disintegrate. There is significant evidence that, due to physical differences between certain homologous chromosomes, preferential selection of one over the other often occurs at other than random proportions.

Question:

S + 3/2 O_2 \rightleftarrows SO_3 The heat formation of SO_3 at 25\textdegreeC is - 94.45 Kcal/mole and the standard molar entropy changes for S, O_2, and SO_3 at 25\textdegreeC are 7.62, 49.0 and 61.24 cal/mole \textdegreeK, re-spectively.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E15-0552.htm

Solution:

One is given the values for ∆H and S which can be related by the equation ∆G\textdegree = ∆H\textdegree - T∆S, where ∆G is the change in free energy, ∆H is the change in the heats of formation, T is the absolute temperature, and ∆S is the change in entropy. After calculating ∆G\textdegree, one can solve for the equilibrium constant (K) by using the equation ∆G\textdegree = - RT In (K) where R = 1.987 cal/mole \textdegreeK. One can find ∆S by remembering that it is equal to the sum of the ∆S values of the reactants subtracted from the ∆S values of the products. In solving for ∆S one multiplies the ∆S values by its coefficient in the equation for the reaction. Solving for ∆S: ∆S - ∆S\textdegree _products - ∆S\textdegree _reactants ∆S = ∆S\textdegree_SO3 - [∆S\textdegree_S + 3/2∆S\textdegree_O2) ∆S = 61.24 cal/mole \textdegreeK ∆S = 61.24 cal/mole \textdegreeK - [7.62 cal/mole \textdegreeK + 3/2(49 cal/mole \textdegreeK)] - [7.62 cal/mole \textdegreeK + 3/2(49 cal/mole \textdegreeK)] = 61.24 - 7.62 - 73.5 = -19.88 cal/mole \textdegreeK = 61.24 - 7.62 - 73.5 = -19.88 cal/mole \textdegreeK Solving for ∆G\textdegree:T = 25\textdegreeC + 273 = 298\textdegreeK ∆G\textdegree = ∆H\textdegree - T∆S\textdegree ∆G\textdegree = - 94, 450 cal/mole - (298\textdegreeK) (- 19.88 cal/mole -\textdegreeK) = - 94, 450 cal/mole + 5924 cal/mole = - 88,525 cal/mole Solving for K: ∆G\textdegree = - RT In K - 88, 525 cal/mole = (- 1.987 cal/mole \textdegreeK) (298\textdegreeK) (In K) In K = (- 88, 525 cal/mole)/(- 592 cal/mole) In K = (- 88, 525 cal/mole)/(- 592 cal/mole) In K = 150 K = 1.39 × 106 5.

Question:

Compare the navigational system of dolphins and bats with that of the fish Gymnarchus (electric eel)

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/Users/wenhuchen/Documents/Crawler/Biology/F31-0803.htm

Solution:

Most navigational systems in organisms consist of detecting stimuli originating from another object. However some animals bounce their own signals off the object and use the echoes of these signals to navigate. Although dolphins use a variety of sounds to communi-cate many kinds of information, they also use sound to navigate and locate food. They possess a type of sonar system in which high-frequency sounds are emitted. The dolphins can then detect the echoes after the sound waves strike objects. The dolphin uses these echoes to determine the object's location by noting the direction of the echoes and how long they take to return. This type of naviga-tion, called echolocation, is also used by bats. Bats are able to fly rapidly through deep caves in near total darkness without bumping into objects. The bat does this by emitting very rapid clicking sounds and detecting the echoes from obstacles. This is very similar to echoloca-tion in dolphins except that the bat's sounds are at a higher frequency. Most bat frequencies are 50 to 100 kilohertz per second; these ultrasounds are out of human range (up to 20 kilohertz per second). Bats, like dol-phins, use echolocation to locate food, mainly insects. However, some moths have evolved ears capable of detecting the clicks of bats and can thus dodge the bat's detection system. Electric eel (Gymnarchus) generates an electric field around its body. Foreign objects, whether good or bad conductors, distort field. Eel detects changes with receptors in its head. The electric eel Gymnarchus also has a sophisticated form of energy-emitting navigation. However, the signals used are electrical, not auditory (See Figure). The eels generate these electrical impulses from special organs in the tail. A dipole field is thus formed between the head (which is positive) and the tail (which is negative). Objects which enter the electric field surrounding the body, distort the field. The eel detects these changes using special sensory organs located in the head. By using this form of sensory perception the eel can navigate in murky waters.

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Question:

A plane wave of monochromatic light of wavelength 5893\textdegree A passes through a slit 0.500 mm wide and forms a diffraction pattern on a screen 1.00 m away from the slit and parallel to it. Compute the separation of the first dark bands on either side of the central bright band, that is, the width of the central bright band.

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Solution:

According to Huygens' principle, each point in the slit may be considered to behave as a single point source of light. Consider a pair of corresponding points in each half of the slit as shown in the figure. For a dark band to occur (corresponding to destructive inter-ference) at P, the two rays must be out of phase by 180\textdegree or \lambda/2. This phase difference results because ray 2 travels a distance ∆l, greater than ray 1 (see figure). Therefore, since P is a point of destructive interference ∆l = \lambda/2 = b/2 sin \texttheta \lambda= b sin \texttheta. If L is much greater than b, we can make the following approximations: The hypotenuse of the large right triangle is equal to L. and in considering any two corresponding points in the two halves of the slit, the angle \texttheta remains unchanged. Then, we can write sin \texttheta = x/L. Butsin \texttheta =\lambda/b and x/L = \lambda/b or x= \lambda/L = [{(0.00005893 cm) (10)} / {0.0500 cm}] = 0.118 cm where x is the distance of the first dark band from the middle of the central bright band. The separation of the two first-order dark bands is 2x = 2 (0.118 cm) = 0.236 cm = 2.36 mm.

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Question:

A vertical spring has an unstretched length L. When a mass m hangs at rest from its lower end, its length increases to L + l. Find the period of small vertical oscillations of m (figure).

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Solution:

When the mass is hanging at rest (figure b), the extension of the spring is l and the force exerted by it on m is, according to Hooke's law, S = -kl. The positive direction is downward, but S is an upward force and therefore negative. Since the mass has no acceleration, this upward force exerted by the spring must be exactly counterbalanced by the weight, mg and by Newton's Second Law, F_net = -kl + mg = 0. Hence, mg = kl .(1) Suppose that, during the oscillation, the mass is a distance x below its equilibrium position, so that the ex-tension of the spring is l + x (figure c). The force exert-ed on m by the spring is then S' = -k(l + x) = -kl -kx. The total force F on m is the sum of S' and the weight mg. F = S' + mg F = -kl -kx + mg. According to equation (1) -kl cancels +mg, so F = -kx(2) where x is the extension of the spring from its equilibrium position (figure c). In order to find the period of small vertical oscillations of m, we must solve the equation of motion (2) for x(t). Noting that F = ma = m[(d^2x)/(dt)^2] we may write, from (2) m[(d^2x)/(dt)^2] = -kx [(d^2x)/(dt)^2] + (k/m)x = 0(3) We define \omega^2 _0 = k/m.(4) using (3) and (4), [(d^2x)/(dt)^2] + \omega2_0x = 0(5) This is a linear, second order differential equation for x in terms of the variable t. A valid method of solu-tion is to make an educated guess for x (t), substitute this guess into (5), and see if an identity results. If so, x(t) is the solution of (5). A good guess for x(t) is x(t) = A cos(at + \Phi).(6) where a, A and \Phi are arbitrary constants. Noting that [dx(t)]/dt = -aA sin(at + \Phi) [d^2x(t)]/dt^2 = -a^2A cos(at + \Phi) and substituting these results into (5), we obtain -a^2A cos(at + \Phi) + \omega2_0 A cos(at + \Phi) = 0 which is an identity if \omega2_0 = a^2. Hence, x(t) is a solution if \omega_0 = a and x(t) = A cos(\omega_0t + \Phi). To find the period of the motion, note that the cosine function goes through one complete cycle of variation when its argument (\omega_0t + \Phi) has gone through 2\pi radians. Here, the change in the argument is (\omega_0t_f + \Phi) -(\omega_0t_0 + \Phi) or\omega_0(tf\Elzbar t_0) and this must equal 2\pi radians, or \omega_0(tf\Elzbar t_0) = 2\pi (tf\Elzbar t_0) = 2\pi/\omega_0.(7) But t_f - t_0 is the time required for cosine to go through one cycle, which is defined as the period of the function (T). Using (7) and (4), T = 2\pi/\omega_0 = 2\pi\surd(m/k).(8) From (1),mg= kl ork= mg/l Inserting this in (8), we obtain T= 2\pi\surd(l/g).

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Question:

Show that, for a given dipole, V and E cannot have the same magnitude in MKS units at distances less than 2 m from the dipole. Suppose that the distance is \surd5 m; determine the directions along which V and E are equal in magnitude.

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Solution:

The expression for the magnitude of the potential due to a dipole is V =P cos \texttheta / 4\pi\epsilon_0r^2 where p is the dipole moment (p = ql) of the dipole, r is the distance from the dipole to the point at which we calculate V, and \texttheta is as shown in the figure. 1 / 4\pi\epsilon_0 is a constant equal to 9 \textbullet 10^9 N \textbullet m^2/c^2. The magnitude of the electric field intensity is E = (p / 4\pi\epsilon_0r^3 ) \surd(4 cos^2 \texttheta + sin^2 \texttheta) If these are to be equal in magnitude, cos \texttheta = \surd(4 cos^2 \texttheta + sin^2 \texttheta) / r orr^2 = (4 cos^2 \texttheta + sin^2 \texttheta) / cos^2 \texttheta = 4 + tan^2 \texttheta The minimum value of r^2 occurs when tan \texttheta = 0. Hence the minimum value of r for V and E to be equal in magnitude occurs for r^2 = 4; that is, r = 2 m, in MKS units. If r = \surd5 m, then V and E are equal in magnitude when (\surd5)^2 = (4 cos^2 \texttheta + sin^2 \texttheta) / cos^2 \texttheta 5 cos^2\texttheta = 4 cos^2 \texttheta + sin^2 \texttheta cos^2 \texttheta = sin^2 \texttheta Thus \texttheta = 45\textdegree, 135\textdegree, 225\textdegree, or 315\textdegree.

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Question:

Show how two metallic balls that are mounted on insulating glass stands may be electro statically charged with equal amounts but opposite sign charges.

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Solution:

The two metal balls are assumed to be initially uncharged and touching each other. (Any charge on them may first be removed by touching them to the earth. This will provide a path for the charge on the spheres to move to the ground). A charged piece of amber is brought near one of the balls (B) as shown in the Figure. The negative charge of the amber will repel the electrons in the metal and cause them to move to the far side of A, leaving B charged positively. If the balls are now separated, A retains a negative charge and B has an equal amount of positive charge. This method of charging is called charging by induction, because it was not necessary to touch the objects being electrified with a charged object (the amber). The charge distribution is induced by the electrical forces associated with the excess electrons present on the surface of the amber.

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Question:

What is your understanding of a substitution reaction mechanism? Illustrate your answer with a specific example.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E21-0752.htm

Solution:

A substitution reaction is one in which one atom or group of atoms replaces another. In this type of reaction, the overall change taking place in the alkane is that one sigma bond is broken and a new sigma bond is formed. At elevated temperatures, or in the presence of ultraviolet light, one or more hydrogen atoms in an alkane molecule may be replaced by atoms of chlorine, bromine, or other similar atoms. The substitution of a chlorine atom for hydrogen in alkane molecules proceeds by a free-radical mechanism and is a chain reaction. In the initiation step of chlorination of methane, for example, high temperature or light furnishes energy for dissociating chlorine molecules into chlorine radicals as shown (unpaired electrons are shown with a dot): (1) Cl_2 \bullet Cl + \textbullet Cl The chlorine atom collides with a methane molecule to form a methyl radical and a molecule of hydrogen chloride. (2) CH_4 + \textbullet Cl \textbullet CH_3 + HCl Subsequent collision of a methyl radical with a chlorine molecule produces the desired compound,CH_3 Cl, and another chlorine atom (3)\textbullet CH_3 + Cl_2 CH_3 Cl + \textbullet Cl, so that the whole chain may be repeated. Equations 2 and 3 are called propagation steps in the chain reaction. Termination steps of this chain re-action are 4 and 5: \textbullet Cl + \textbullet Cl Cl_2 (4)\textbullet Cl + \textbullet CH_3 CH_3 CI (5)\textbullet CH_3 + \bullet CH_3 CH_3 CH_3 These chain-terminating reactions tend to stop the reaction between methane and chlorine by consuming radicals necessary for the propagation steps.

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Question:

In a typical experiment performed to measure the mechanical (electrical) equivalent of heat the following data were obtained: resistance of the coil, 55 ohms; applied voltage, 110 volts; mass of water, 153 gm; mass of calorimeter, 60 gm; specific heat of calorimeter, 0.10 cal/(gmC\textdegree); time of run, 1.25 min; initial temperature of water, 10.0\textdegreeC; final temperature, 35.0\textdegreeC. Find the value of J.

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/Users/wenhuchen/Documents/Crawler/Physics/D13-0484.htm

Solution:

The current through the coil resistance is, by Ohm's Law I = V/R = (110 volts)/(55.0 ohms) = 2.00 amp The power (or energy per unit time) development by the coil is p = (energy dissipated) / time = I^2 R In 1.25 min. the energy dissipated by the coil, to the surrounding water and calorimeter container, is E = I^2 R t = (2.00 amp)^2 (55.0 ohm) [1.25 min×(60 sec/1 min)] = 16.5 × 10^3 Joules By the principle of conservation of energy, this electrical energy generated by the coil is converted entirely to heat energy. This heat energy is absorbed by the surrounding water and calorimeter container. As a consequence, the temperature of the water-calorimeter system increases from the initial temperature, 10.0\textdegreeC to the final temperature, 35.0\textdegreeC. Consequently, the heat energy absorbed by the two bodies in this system is given by Q = (M_\cyrchar\cyromega C_\cyrchar\cyromega + M_c C_c) (T_F - T_i)(1) where C_\cyrchar\cyromega and C_c are the specific heats (that is, the amount of heat required to raise the temperature of one gram of the substance one degree centigrade) of water and of the container, respectively. Substituting the given values in (1), and noting that C_\cyrchar\cyromega is defined as 1 cal/gm\textdegreeC, Q = [(153 gm) (1 cal/gm\textdegreeC) + (60 gm) (.10 cal/gm\textdegreeC) ] [35.0\textdegreeC - 10.0\textdegreeC] = 3975 cal. But from the principle of conservation of energy, electrical energy = heat energy or 16.5 × 10^3 Joules = 3975 cal. Therefore, in order to find how many Joules are equivalent to 1 calorie, we develop the proportion J = (x Joules)/(1 cal) = [(16.5 × 10^3 Joules) / (3975 cal)] orJ = 4.15 Joules/cal. Alternatively, 4.15 Joules = 1 cal.

Question:

Determine which of the atoms in each pair possess a partial positive charge and which a partial negative, (a) the O-F bond, (b) the O-N bond, (c) the O-S bond.Electronegativity values for these elements can be found from a table of electronegativities .

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Solution:

Electronegativityis the tendency of an atom to attract shared electrons in a chemical covalent bond. Since electrons are negatively charged, to find the partial charge on each atom of the pair consult a table for theelectronegativityvalues of the atoms in the molecules. The atom with the higher value will have a greater tendency to attract electrons, and will, thus, have a partial negative charge. Because the over-all bond is neutral, the other atom must have a partial positive charge. For part (a), F has a higherelectronegativitythan O, which means that F will have a partial negative charge (\delta^-) and O will have a partial positive charge (^+\delta). To show this the molecule can be written (O^\delta^+ -F^\delta^-). Similar logic is used in working out parts b and c. (b) O - N bond: Theelectronegativityof O is 3.5 and of N is 3.0, thus O is more negative than N. The molecule is written (O^\delta^- -N^\delta^+). (c) O - S bond: Theelectronegativityof O is 3.5 and of S is 2.5; therefore, O is the more negative of this pair, (O^\delta^- -S^\delta^+).

Question:

A railway gun whose mass is 70,000 kg fires a 500-kg artillery shell at an angle of 45\textdegree and with a muzzle velocity of 200 m/sec. Calculate the recoil velocity of the gun

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Solution:

In this problem momentum must be conserved in both the horizontal and vertical directions. Let us refer to the bullet by using the subscript 1, and the gun by the subscript 2. We can state this conser-vation of momentum in the horizontal direction as follows: P_1x = p2x or m_1v_1x = m_2v2x but V_1x = v1cos 45\textdegree Therefore, we may express p_1x as, P_1x = m_1v_1 cos 45\textdegree = (500 kg) × (200 m/sec) × 0.707 = 7.07 × 10^4 kg-m/sec This must equal (except for the sign) the recoil momen-tum of the gun which moves only horizontally: P_2 = m_2v_2 = -7.07 × 10^4 kg-m/sec Therefore, v^2 = - [ {7.07 × (10^4 kg-m/sec)} / {7 × 104kg} ] \cong - 1m/sec or, approximately 2 mi/rh. What has happened to the vertical component of the recoil momentum, P_1y = 7.07 x 10^4 kg-m/sec ? Since the railway platform is in contact with the Earth, the Earth absorbs the ver-tical momentum. The Earth does recoil, but because of the extremely large value of the Earth's mass compared to that of the railway gun, the recoil velocity cannot be measured.

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Question:

An electron with angular momentum L = 6 × 10^-35 N-sec and total energy H_0 = -4 × 10^-18 J is in orbit around a proton. What are the distance of nearest approach and farthest separation?

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Solution:

The electrostatic potential is given by V_E = -k_E(e^2/R) where the negative sign indicates its attractive nature. The kinetic energy of the electron is k.E. = (1/2)mv^2 = [(m^2u^2R^2)/(2mR^2)] since, by definition of angular momentum, L =mvR. The total energy of the electron in its orbit is H_0 =k.E. + V_E = (L^2/2mR^2) - [k_E(e^2/R)] orH_0 = (L^2/2mR^2) - (k_Ee^2)/R R^2H_0 = (L^2/2m) - k_Ee^2R R^2 = (L^2/2mH_0) - [(k_Ee^2R)/(H_0)] R^2 = [(k_Ee^2R)/H_0] - (L^2/2mH_0) = 0 Solving for R via the quadratic equation R = - [(k_Ee^2/H_0) \pm \surd{(k^2_Ee^4/H^2_0) + (2L^2/mH_0)}] / 2 or2R = -[(k_Ee^2/H_0) \pm {(k^2_Ee^4/H^2_0) + (2L^2/mH_0)}^1/2] (k_Ee^2/H_0) = [{9 × 10^9 (N\bulletm^2/c)}(2.56 × 10^-30 c^2)] / [ -4 × 10^-18 J] (k_Ee^2/H_0) = -5.76 × 10^-11 m 2L^2/mH_0= [2(36 × 10^-70 N^2 \bullet s^2)] / [(9.1 × 10^-31 kg) (.4 × 10^-18 J)] 2L^2/mH_0= -1.98 × 10^-21 m^2 Thus 2R = 5.76 × 10^-11 m \pm (3.32 × 10^-21 m^2 - 1.98 × 10^-21 m^2)^1/2 = 5.76 × 10^-11 m \pm 3.66 × 10^-11 Thus R = 1.05 × 10^-11 m or 4.71 × 10^-11 m Which gives the nearest approach and farthest separation.

Question:

What is an operon? With the aid of diagrams, show how a known operon works.

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Solution:

A controllable unit of transcription is called an operon. An operon consists of a binding site for RNA polymerase (a promoter), a binding site for a specific repressor (an operator), and one or more structural genes (see figure). These structural genes are usually associated in their functions, and their coordinated synthesis is de-sirable. Thus they are linked together in the operon, and when they are transcribed, a single mRNA molecule is produced. The promotor and operator act as controls for this trans-cription, and the entire system is controlled by an associated regulatory gene, which codes for the repressor molecule of the operon. A known example of an operon is the lactose (lac) operon, present in E. coli. This operon contains the structural genes for \beta-galactosidase, an enzyme which cleaves lactose into glucose and galactose; galactoside permease, which allows for the increased rate of entry of lactose into the bacteria; and galactoside acetylase, whose function is not known. When lactose is absent from the medium, these enzymes are not needed, and so their levels are low. But when lactose is added to the medium containing these bacteria, the production of the enzymes increases tremendously, almost a thousand fold. How is the manufacture of this prodigious quantity of enzyme initiated and controlled? The regulatory gene (1) for the lac operon is transcribed constitutively, that is, its product, the repressor, is constantly being synthesized. This repressor binds to the operator(o) and blocks the transcription of the structural genes. That is why the levels of the three enzymes are normally so low. The situation is altered when lactose is introduced into the medium. Lactose molecules move into the nucleus of the bacterium, and bind to any repressor molecules in such a way that the repressors are inactivated. An inactivated rep-ressor cannot bind to the operator. It therefore cannot block the transcription of the structural genes by RNApoly-merase which binds to the promotor. Messenger RNA now transcribed from the genes of the operon is translated on the ribosomes into the three enzymes of lactose metabolism. As long as the inducer (lactose) is present, the synthesis of the enzymes can occur. But when the inducer is depleted by metabolism, repressor proteins (remember that synthesis of the lac repressor isconstitutive) become free to again bind to the operator and effectively turn off the synthesis of the enzymes. This model implies that the usual condition of the lactose operon is to be turned off. This is accomplished through the control of the regulator gene, which constitu-tively produces the repressor. In other words, the system is under negative control. Because the operon can be turned on by the presence of lactose (the inducer), it is said to be inducible. An important implication is that some genes, such as (1), produce proteins whose function is to regulate the expression of other genes, while certain other genes, namely operators and promoters, very probably do not code for any proteins at all.

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Question:

Figure A shows a strut AB, of length l , pivoted at end A, attached to a wall by a cable, and carrying a load w at end B. The weights of the strut and of the cable are negligible. Suppose the weight w, and the angles \texttheta_1 and \texttheta_2 are known. What is the direction of C\ding{217}?

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Solution:

The system shown in figure A is in equilibrium. Hence, the net external force on the system must be zero. Also, the net external torque acting on the system about any point must be zero. Then, using figures (a) and (b) T_1\ding{217} + T_2\ding{217} + C\ding{217} = 0(1) and, taking torques about point A l(T_2 sin \texttheta_1) -l(T_1 sin \texttheta_2) = 0(2) Changing (1) into 2 scalar equations using figure B, C sin \varphi + T_2 sin \texttheta_1 = T_1 sin \texttheta_2(3) C cos \varphi = T_2 cos \texttheta_1 + T_1 cos \texttheta_2 To find the direction of C\ding{217}, we must solve for \varphi. Using (3) C sin \varphi = T_1 sin \texttheta_2 - T_2 sin \texttheta_1 C cos \varphi = T_1 cos \texttheta_2 + T_2 cos \texttheta_1 Dividing these last 2 equations tan \varphi = (T_1 sin \texttheta_2 - T_2 sin \texttheta_1)/(T_1 cos \texttheta_2 + T_2 cos \texttheta_1)(4) Solving (2) for T_1 T_1 = (T_2 sin \texttheta_1)/(sin \texttheta_2) Substituting this equation in (4) tan \varphi = [{(T_2 sin \texttheta_1)/(sin \texttheta_2)} sin \texttheta_2 - T_2 sin \texttheta_1]/ [{(T_2 sin \texttheta_1)/(sin \texttheta_2)} cos \texttheta_2 + T_2 cos \texttheta_1] andtan \varphi = 0 \varphi = 0\textdegree Hence, C\ding{217} is directed along the strut AB.

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Question:

At the National Air Races in Reno, Nevada, the planes fly a course around a series of pylons. In a turn around a pylon, the maximum centripetal acceleration that the pilot can withstand is about 6g. For a plane traveling at 150 m/s, what is the turning radius for an acceleration of 6g?

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Solution:

The speed is v = 1.5 × 10^2 m/s, and the acceleration a = 6g = (6)(9.8 m/s^2) = 5.88 × 10^1 m/s^2. R = v^2/a = (1.5 × 10^2 m/s)^2/(5.88 × 10^1 m/s^2) = 3.83 × 10^2 m. This is the minimum turning radius.

Question:

Two inertial systems S and S' have a common x-axis and parallel y-axis, and S' moves with a speed of 0.6c relative to S in the direction of the x-axis. An observer in S sees a rocket moving with a speed of 0.1c in the positive y-direction. What is the speed and direction of the rocket as seen by an observer in the S'-system?

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Solution:

According to the Lorentz transformation, the coordinates of an event as recorded in S' (x', y', z', t') are related to the coordinates of an event as recorded in s (x, y, z, t) by x' = (x -vt) / \surd{1 - (v^2c^2)}y' = y,z' = z, t' = {t - ( vx / c^2)} / \surd{1 - (v^2c^2)} . Here, v is the relative velocity of the 2 frames of reference. It follows that the velocity of an object with coordinate x' in frame S' is u'_x = dx' / dt' = {(dx/dt') - v(dt/dt') } / \surd{1 - (v^2c^2)} =[ (dx/dt) × {(dt/dt') - v(dt/dt') }] / \surd{1 - (v^2c^2)} But, from the initial equations, we have t = \surd{1 -(v^2c^2)} t' + (vx/c^2) dt / dt' = \surd{1 - (v^2c^2)} + (v/c^2) (dx/dt') dt / dt' = \surd{1 - (v^2c^2)} + (v/c^2) (dx/dt) × dt/dt' Since dx/dt = u_x^, the velocity of the same object relative to S, we have \thereforedt / dt' {1 - vu_x / c^2} = \surd{1 - (v^2c^2)} or dt / dt' = \surd{1 -(v^2c^2)} / [1- (u_xv/ c^2)] u'_x = [{(dx/dt) - v} (dt/dt')] / \surd{1 - (v^2c^2)} = [(u_x - v) / \surd{1 - (v^2c^2)}] × [\surd{1 - (v^2c^2)} / {1 - (u_xv / c^2)} = (u_x - v) / {1 - (u_xv / c^2)} .(1) Similarly , u'_y_ = (dy' /dt) × (dt /dt') = (dy/dt) × (dt / dt') = [u_y \surd{1 - (v^2c^2)} ] / {1 - (u_xv / c^2)} . (2) In this problem, S sees the rocket move in the y direction with speed .1c. Hence,u_x= 0,u_y= 0.1c, and v = 0.6c . Using (1) (2) \thereforeu'_x = - 0.6c, u'y= 0.1c × \surd(1 - 0.6^2) = 0.08c . To an observer in S', therefore, the rocket appears to have components of velocity of -0.6c in the x-direction and of 0.08c in the y-direction. It thus appears to have a velocity of \surd(0.6^2 + 0.08^2c) = 0.605c in a direction making an angle of tan^-1 (0.08/0.60) =tan^-1 0.133, that is an angle of 7\textdegree36' with the negative direction of the x-axis.

Question:

What is the angular deflection of a light beam or photon which passes by the sun at its edge?

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/Users/wenhuchen/Documents/Crawler/Physics/D06-0337.htm

Solution:

This problem involves a photon moving with the velocity of light in a gravitational field. We do not get the correct answer without doing a careful calculation using special relativity, but we can get the order of magnitude of the correct answer by a naive calculation. Suppose that the photon has an effective mass M_L; it will turn out that M_L drops out of the calculation of the deflection and thus we do not have to know what it is. Let the light beam pass the sun at a distance of closest approach, r_0, as measured from the center of the sun. We suppose that the deflection will turn out to be very small, so that r_0 is essentially the same as if the light beam were not deflected. The force F on the photon at the position (r_0, y) is F = -[(G M_S M_L)/(r^2 _0 + y^2)] The transverse component, F_X, is F_X = - [(GM_SM_L)/(r^2 _0 + y^2)] cos \texttheta = - [(GM_SM_L)/(r^2 _0 + y^2)] [r /(r^2 _0 + y^2)]1/2 = - GM_SM_L [r_0/(r^2 _0 + y^2)^3/2] where y is measured from the point P ,as in the figure The final value of the transverse velocity component v_x of the photon has the value given by M_LV_X = \int F_X dt = \int F_X (dy/v_y ) \cong 1/c \int F_X dy so that v_x \cong - [(2GM_Sr_0)/C ]^\infty\int_0[dy/(r^2 _0 + y^2)^3/2 ] \cong - [(2GM_S)/(cr_0)]

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Question:

An automobile tire of volume 5.6 × 10^3 cc is filled with nitrogen to a gauge pressure of 29 psi at room tempera-ture 300\textdegree K. How much gas does the tire contain? If, during a trip, the temperature of the tire rises to 320\textdegree K., what will be the pressure?

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Solution:

First we convert the data to MKS units. Gauge pressure is the pressure above atmospheric pressure (14.7 psi). Thus the total pressure is p = 29 + 14.7 = 43.7 psi Using the conversion factor 1 psi = 6.9 × 10^3 N/m^2, p = 43.7 × 6.9 × 10^3 N/m^2 = 3.02 × 10^5 N/m^2. The volume isv = 5.6 × 10^3 cm3 = 5.6 × 10^-3 m^3. The amount of gas in moles is n = (pV/RT) = (3.02 × 105× 5.6 × 10^-3)/(8.31 × 300) = 0.68. The molecular weight of nitrogen (N_2) is 28 gm. Since there are 0.68 moles of nitrogen in the tire, the mass in grams is m N_2 = 0.68 moles N_2 × 28 gm/mole N_2 = 19.0 gm. As the temperature rises, n and V are constant (for this example); therefore the ideal gas law (p_1V_1)/(n_1T_1) = (p_2V_2)/(n_2T_2) reduces to (p_2/p_1) = (T_2/T_1) Substituting values, p_2/(43.7 psi) = 320/300 p_2 = 46.6 psi The gauge pressure will be 46.6 - 14.7 = 31.9 psi, and this will be the pressure read by the service station attendant.

Question:

Consider an atom containing one electron and having an atomic number Z. It is observed that light of a wave-length of 0.1 nm will excite this atom from the ground state to infinite ionization. Determine the value of Z.

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Solution:

We know the wavelength \lambda of the exciting light and the quantum numbers of the ground state (n_g = 1) and the ionized state (n_i = \infty). We must determine Z, the number of protons in the nucleus of the atom. There is a relationship to find Z using n_g and n_i. It is provided by the Bohr theory of the atom. It is: (c/\lambda) = [(Z^2 e^2) / (2ha_0)] [(1/n_g ^2) - (1/n_i ^2)], where c is the speed of light (c = 2.9969 × 10^10 cm/sec), e is the electronic charge (e = 4.80 × 10^-10 esu), h is Planck's constant (h = 6.626 × 10^-27 erg-sec), and ao is the Bohr radius (a_0 = 0.05292 × 10^-7 cm). If we combine some of these constants, we arrive at the value e^2/2ha_0 = 3.290 × 10^15 sec^-1. Hence, our expression becomes (c/\lambda) = Z^2 [(e^2) / (2ha_0)] [(1/n_g ^2) - (1/n_i ^2)] = Z^2 × 3.290 × 10^15 sec^-1 [(1/n_g ^2) - (1/n_i ^2)]. To solve this problem, we must solve this equation for Z and substitute c, n_g, n_i, and \lambda (\lambda = 0.1 nm = 0.1 × 10^-9 m = 10^-10 m = 10^-8 cm) into the resulting equation. Solving for Z, we obtain Z^2 = (c/\lambda) [3.290 × 10^15 sec^-1 {(1/n_g ^2) - (1/n_i ^2)}]^-1,or Z = [(c) / {\lambda × 3.290 × 10^15 sec^-1 [(1/n_g ^2) - (1/n_i ^2)]}]^1/2 Substituting the values of c, \lambda, n_g, and n_i. into this expression yields Z = [(2.9969 × 10^10 cm/sec) / {(10^-8cm × 3.290 × 10^15 sec^-1 (1/1) - (1/\infty)}]^1/2 \cong 30, where we used the equality 1/\infty = 0. Hence, this atom must have 30 protons.

Question:

What is the radius of gyration of a slender rod of mass m and length L about an axis perpendicular to its length and passing through the center?

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Solution:

The radius of gyration is the distance from the axis of rotation to the radius at which we may consider all the object's mass to be concentrated in a thin hoop, as the diagram shows. For a thin hoop of radius k_0, the moment of inertia, I_0, is mk_0^2, where m is the mass of the hoop. To find k_0, we set I of the object equal to I_0 the moment of inertia of the equivalent hoop I = I_0 or(1/12)m1^2 = mk_0^2 Hence k_0 = [\surd{(1/12)mL^2}] / m = L/(2\surd3) = 0.289 L. The radius of gyration, like the moment of inertia, depends on the location of the axis. Note carefully that, in general, the mass of a body can not be considered as concentrated at its center of gravity for the purpose of computing its moment of inertia. For example, when a rod is pivoted about its center, the distance from the axis to the center of gravity is zero, although the radius of gyration is L/(2\surd3).

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Question:

How can radioactive isotopes be used in biological research such as tracing the pathways of compounds synthesized in the living cell?

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Solution:

Labeling substance with radioactive isotopes is a powerful method for studying specific metabolic pathways, as well as being applicable to many other areas of biological research. Radioactive labeling has been used, for example, in determining the rates of metabolic processes in intact organisms, and also in determining whether a given metabolic pathway is the predominant route, for a given metabolite. It is used in establishing whether a pathway which has been studied in a test tube is the same reaction that occurs in vivo. Labeling is done by taking the compound in question and substituting in a radioactive isotope for one of the naturallyoccuringelements comprising the compound.(see Table). The radioactive isotope, when incorporated into a biological system will undergo spontaneousemmisionof radiation while taking part in the same biological acti-vities as its non-radioactive counterpart. Although this is not observable by eye, there are many known methods by which these emissions can be detected. The methods commonly employed include Geiger counters, photographic methods, fluorescent methods, and cloud chambers. In this way, the course of the reaction can be followed. The value of radioactive tracers in biological research can be seen by considering the photosynthetic equation: 6 CO_2 + 6 H_2O(sunlight/chlorophyll)\rightarrowC_6H_12O_6 + 6 O_2. Although in the equation the photosynthetic process appears to be a simple reaction, it actually follows a very complicated course. This complex reaction, like most others, proceeds through a series of changes involving one or two molecules at a time. Researchers were able to use 14 6 C, which was incorporated into CO_2, to determine the course of this reaction. They monitored the radio-activity through the use of a Geiger counter. Some isotopes useful as radioactive tracers in biological system. naturallyoccuringelement radioactive isotopes half - life 1 1 H 3 1 H 12.1 years 12 6 C 14 6 C 5,700 years 23 11 Na 24 11 Na 15 hours 31 15 P 32 15 P 14.3 days 32 16 S 35 16 S 87.1 days 35 17 Cl 36 17 Cl 3.1 × 10^5 years 39 19 K 42 19 K 12.5hours 40 20 Ca 45 20 Ca 152 days 56 26 Fe 59 26 Fe 45 days 127 53 I 131 53 I 8 days

Question:

Minimize the following minterm function containing 'don't cares' using the Karnaugh-Map. f(A,B,C,D) = \summ(5,6,7,8,9) + d(10,11,12,13,14,15)

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Solution:

f(A,B,C,D) = \summ(5,6,7,8,9) + d(10,11,12,13,14,15) . In the design of digital circuits one often encounters cases in which the switching function is not completely specified. In other words, a function may be required to contain certain minterms, omit certain minterms, with the remaining minterms being optional; that is certain minterms may be included in the logic design if they help simplify the logic circuit. A minterm which is optional is called a don't care minterm. The plotting of the given function f(A,B,C,D) is shown in fig. 1 . In the use of don't cares there is one additional rule which may be used in mapping. Recall that the don't cares by definition can be either 0 or 1. Hence in minimizing terms in Sum of Product form, the don't cares may be chosen to be 1 if in doing so the set of blocks on the map which can be grouped together is larger than would other-wise be possible without including the don't cares. In other words, with regard to don't cares one can take it or leave it, depending on whether they do or do not aid in the simplification of the function. In the map of fig. 1 the following minterms and don't cares may be grouped. m_8 = AB C D m_9 = AB CD d_11 = ABC D d_10 = ABCD d_12 = A BC D d_13 = A BCD d_15 = A B C D d_14 = A B CD and by the use of Boolean theorem A\bulletA= 0, terms B,C, and D are eliminated. The result from this group would be A. The next group is minterms 5 and 7, and don't cares 13 and 15. m_5 =ABCD m_7 =AB C D d_13 = A BCD d_15 = A B C D With the same argument, the eliminated terms are A and C. Leaving the resulting terms BD. The third and the last group consists of minterms 7 and 6, and don't cares 15 and 14. m_6 =AB CD m_7 =AB C D d14= A B CD d_15 = A B C D Giving the result BC. The resulting total function can now be written as the sum reduced terms that are found, namely; f(A, B, C, D) = A + BD + BC Note that this function is much simpler to deal with than it would have been without the inclusion of don't cares.

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Question:

Explain the events which take place during glycolysis.

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Solution:

Glycolysis is the series of matabolic reactions by which glucose is converted to pyruvate, with the concur-rent formation of ATP. Glycolysis occurs in the cytoplasm of the cell, and the presence of oxygen is unnecessary. Glucose is first "activated," or phosphorylated by a high- energy phosphate from ATP (See reaction 1). The product, glucose-6- phosphate, undergoes rearrangement to fructose- 6-phospate, which is subsequently phosphorylated by another ATP to yield fructose-1, 6- diphosphate (reactions 2 and 3). This hexose is then split into two three- carbon sugars, glyceraldehyde-3-phosphate (also called PGAL) and dihydroxyacetone, phosphate (see reaction 4). Only PGAL can be directly degraded in glycolysis; dihydroxyacetone phosphate is reversibly converted into PGAL by enzyme action (reaction 4a). Since two molecules of PGAL are thus produced per molecule of glucose oxidized, the products of the subsequent reactions can be considered "doubled" in amount 1 glucose \textemdash\textemdash\ding{217} 2 PGAL PGAL gets oxidized and phosphorylated. NAD+, the coenzyme in the dehydrogenase enzyme which catalyzed this step, gets reduced to NADH, and 1,3-diphosphoglycerate is formed (reac-tion 5). The energy-rich phosphate at carbon 1 of 1,3 diphosphoglycerate reacts with ADP to form ATP and 3-phospho- glycerate. This undergoes rearrangement to 2- phosphoglycerate, which is subsequently dehydrated, forming an energy- rich phosphate: phosphoenolpyruvate (PEP) (reaction 8). Finally, this phosphate group is transferred to ADP, yield-ing ATP and pyruvate, (reaction 9) Since two molecules of PGAL are formed per molecule of glucose, four ATP molecules are produced during glycolysis. The net yield of ATP is only 2, since 2 ATP were utilized in initiating glycolysis (reactions 1 and 3 ). Pyruvate is then converted to acetyl coenzyme A which enters the Krebs cycle. In addition, two molecules of NADH are pro-duced per molecule of glucose. Hence, the net result of glycolysis is that glucose is degraded to pyruvate with the net formation of of 2 ATP and 2 NADH. The process of glycolysis can be summarized as follows: glucose + 2ADP + 2Pi + 2NAD+ \textemdash\textemdash> 2 pyruvate + 2ATP + 2NADH + 2H+

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Question:

Why is the organism that causes smallpox not used in the vaccine against smallpox?

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/Users/wenhuchen/Documents/Crawler/Biology/F05-0158.htm

Solution:

Smallpox (variola) is an acute infectious disease that is caused by a virus . The disease is spread by droplet infection or by handling articles infected by a smallpox patient. The virus is thought to lodge in the nasopharynx (the part of the pharynx just dorsal to the soft palate), where it proliferates and spreads to the blood (viremia), enabling it to infect the skin and other tissues.Tnedisease is characterized by an initial fever followed by the appearance of pustules (small fluid-filled eruptions) on the skin , which regress and leave the scars character-istic of smallpox. It has long been known that contracting smallpox protects one against a second attack. For centuries, the Chinese inoculated the skin of a healthy person with the material from lesions on a smallpox patient. The variola virus, present in this material, would stimulate antibody production in the inoculated individual to give the person immunity to smallpox. However, the use of the actual virulentvariolavirus may be dangerous if too much is administered. In 1796, it was discovered by Jenner that inoculation with thevacciniavirus from cowpox lesions would confer immunity to smallpox as well as to cowpox. This immunization procedure is safe because smallpox is never produced. This vaccination procedure is practised today, usingvacciniavirus taken from cows, sheep, or chick embryos . Although the United States and many other countries have eliminated smallpox, outbreaks can occur as long as reservoirs of the variola virus still exist (e.g., Africa, India, and Southeast Asia). However, today the smallpox virus is basically confined to research laboratories. People are usually vaccinated very early in life then revaccinated every 3 to 5 years. Prevention is vital since there is no specific treatment for smallpox .

Question:

Name and briefly describe the commonly used input and output devices.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G01-0012.htm

Solution:

Console Monitors: In the early days, teletype consoles that use 132-column wide paper were used as the primary interface device between the operator and the computer. It looked very much like the 132- column wide paper printers that are not used as much these days, but it differed from the hardcopy line printer (LP) since it was capable of both input and output. Both input and output was one line at a time. Consoles were primarily used for booting up and maintenance of the system, and for important operating system messages. Nowadays, the CRT monitors that are directly connected to the computer system are used for the same purpose. The monitors that look like the small-screen TV along with keyboards that attach to them are widely used in PCs. These monitors are ca-pable ofdotmappedgraphics where the screen contains small pixels (dots) with dimensions that usually range from 640-by-400 low resolution to 1280-by-900 high resolution. Otherwise, the screen is capable of 80-by-25 character display in line (non--graphics) mode. In addition, the monitors may have either mono-chrome or colored screens. The number and shades of colors de-pend upon the specific make and model. Terminals : Perhaps the most frequently used input-output device is the CRT terminals with a keyboard and a screen. Strictly speaking, the keyboard is used for input and the video screen is used for output. The keyboards conform to basic standards such as ASCII, ANSI, and PC, but are often specialized for the conve-nient user interface. The screen may be used in character or graphics mode. In character mode, it can display characters such as ASCII set on 24 rows (lines) and 80 columns (characters per line). In some cases, there may be 25th row (line) to display function keys on the screen (so called soft function keys used in word processing systems) . There may also be 132-column mode, where the characters are smaller and hard to read. This is re-ally a remnant of the old days when 132-column wide papers were used to print out the reports. Tape Units : Although not used as frequently, except for backup of large amounts of data, the magnetic tapes are supported by tape units (drives) that have two reels feeding each other. The reels contain tapes with densities ranging from 800 bpi (bits per inch) to 6250 bpi. The access rate (e.g. how fast the tape is read or written) depends upon the manufacture of the drive, except that the access is almost always sequential. This means that it is not possible to access records (80 or more bytes data) directly. Disk Units : Also referred to as secondary storage (as opposed to the primary memory or RAM), disks are used for both input and output. Disks allow direct (or random) access to any block of data. A block of data is generally 512 to 1024 bytes and physi-cally correspond to a disk sector that can directly by homed in by the disk head. The more precise access to data within a block is done by first reading the entire block into the primary memory. The disks have capacities that depend upon the type of computer (e.g. mainframe, mini, PC, etc.) and generally range from 10 megabytes to about half gigabytes. The personal comput-ers have non- removable hard disks referred to as HDU (Hard Disk Unit) . The capacities are, as of the 1990s, anywhere from 10 to 80 megabytes. The optical disks are usually read-only (e.g. allow storage only once), but have capacities ranging from half gigabytes to four gigabytes. Diskettes : As its name implies, a diskette is a small magnetic disk. These are also called floppy disks or simply floppies. The most common sizes are either 4.25-in. by 4.25 in. standard or 3.50-in. by 3.50-in mini diskettes. A high-density standard diskette can store up to 1.2 megabytes and requires an appropri-ate diskette drive (floppy unit) for processing. A double-den-sity standard diskette has 360-kilobyte capacity. The minidiskettes have capacities of 720 kilobytes and 1.44 mega-bytes. Printers : Although the so called line printers (LPs) that print on 132- column wide paper still exist and are used in some cases with mainframe andsuperminisystems, most printers are much more compact and print on standard (8.5-by-ll inch) paper. The number of characters per line is determined by the font size 10 (10 characters per inch), normally 80 characters per line is printed. The printers are attached to the computer through printer ports (e.g. communication interface for transferring bits of information). The ports are of 36-pinCentronixtype for paral-lel interface and of 25-pin RS-232 type for serial interfaces. The printers may be of letter-quality or dot-matrix type. The former normally uses daisywheel character fonts and serial- impact upon ribbons much like the typewriters. There is less flexibility to define characters and fonts, but the quality is letter-quality. The dot-matrix printers define characters on a matrix of size 5-by-7 (coarse) or 9-by-12 (fine). These are less expensive printers based on older technology and imprint the pattern through ribbon (reels or cartridge). The ink-jet print-ers use special inks in liquid or solid forms ejected through a matrix with higher resolution (12-by- 15, 18-by-36, or better). Using newer technology, ink-jet printers produce almost letter quality prints, and can also print in colors. The dot-matrix and ink-jet printers are capable of graphics output, and allow easy definition of fonts by software. Thermal printers utilize two different printing methods. The older, direct thermal types form characters and images by burning a heat-sensitive paper with heated pins. Thermal-transfer printers, on the other hand, employ thin-film or thick-filmprintheadsto melt and deposit on paper a wax-based in contained on amylarribbon. The thermal printers are most appropriate for color printing that is very useful in facsimile market. Finally, at the higher end, there are more expensive and reliable laser printers that produce fine quality prints in a flexible way. These printers utilize VLSI and laser technologies and ink powders referred to as toner. Hewlett-Packard's LaserJet is a typical example of these printers that are presently most popular and much sought after despite high cost. One shortcoming is that these printers are not yet capable of color prints.

Question:

In a Wilson cloud chamber at a temperature of 20\textdegreeC, particle tracks are made visible by causing condensation on ions by an approximately reversible adiabatic expansion of the volume in the ratio (final volume)/(initial volume) = 1.375 to 1. The ratio of the specific heats of the gas is 1.41. Esti-mate the gas temperature after the expansion.

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/Users/wenhuchen/Documents/Crawler/Physics/D14-0522.htm

Solution:

The adiabatic expansion of an ideal gas obeys the law pV^\Upsilon = constant where \Upsilon is the ratio of the specific heats at constant volume and at constant pressure, \Upsilon = C_p/C_v. Also, by the ideal gas law P =nRT/V, we also havepV/T = constant (for the number of moles of gas within the chamber is constant). Then TV^\Upsilon-1 = constant. If T_1 and V_1 are the initial temperature and volume respectively, and T_2, V_2 are the final ones, T_1V_1^\Upsilon-1 = T_2V_2^\Upsilon-1 orT_2/T_1 = (V_1/V_2)\Upsilon-1 Taking the logarithm of both sides, log (T_2/T_1) = (\Upsilon - 1) log (V_1/V_2) = 0.41 log (1/1.375) = - 0.0567 =1.9433. From the logarithmic table, we get T_2/T_1 = 0.878. T_2 = 0.878 T_1 = 0.878 × 293\textdegreeK = 257.2\textdegreeK = -15.8\textdegreeC.

Question:

Using the pertinent information from the previous problem, find the third and fourth standard moments of ungrouped data in an observed distribution.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G19-0482.htm

Solution:

The third and fourth standard moments are dimensionless. Unlike the first and second moments, these incorporate the fraction (X_i - M) / S, which renders these moments independent of units. The third standard moment, W, is a measure of asymmetry or skewnessof the distribution, and the formula is given as: W= {^N\sum_i=1 [(X_i - M)/S]^3} / N = (1/ S^3) [{N\sum_i_=1 (X_i - M)^3} / N ] \sum = {N\sum_i=1 (X_i - M)^3} / S^3N The fourth standard moment, K, is a measure of peakedness of the distribution curve, and is given by: K= {^N\sum_i_=1 [(X_i - M)]^4} / N = (1/ S^4)[ {N\sum_i_=1 (X_i - M)^4} / N ] \sum = {N\sum_i_=1 (X_i - M)^4} / S^4N \sum These equations will now be translated into BASIC, assuming that the mean M and the standard deviation S are known: 1\O REM COMPUTE THIRD AND FOURTH STANDARD 2\O REM MOMENTS 3\O LET Y = \O 4\O LET Z = \O 6\O READ X 7\O IF X < \O THEN 12\O 8\O LET Y = Y + (X - M)\uparrow3 9\O LET Z = Z + (X - M)\uparrow4 11\O GO TO 6\O 12\O LET W = Y / (S\uparrow3 \textasteriskcentered N) 13\O LET K = Z / (S\uparrow4 \textasteriskcentered N) 14\O REM PRINT OUT RESULTS 15\O PRINT "N ="; N 16\O PRINT "3RD STANDARD MOMENT ="; W 17\O PRINT "4TH STANDARD MOMENT ="; K 18\O DATA 75, 22, 14, 83, 16, 12, 17, -1 19\O END Note, that the value of N is not specified in the program. One can either substitute N with a specific value, or add a READ N and DATA... statements.

Question:

Unlike Amoeba, Paramecium has a permanent structure, or organelle, that functions in feeding. Describe this organelle and the method of feeding and digestion in Paramecium.

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/Users/wenhuchen/Documents/Crawler/Biology/F11-0271.htm

Solution:

The feeding organelle in Paramecium begins with an oral groove. This oral groove is a ciliated channel located on one side of the organism. Food particles are swept into the oral groove by water currents pro-duced by the beating cilia, and are carried down the groove to a point called the cytostome, which can be thought of as the mouth. Food is carried through the cytostome into the cytopharynx, which is also lined with cilia. As food accumulates at the lower end of the cytopharynx, a food vacuole forms around it. The vacuole eventually breaks away and begins to move toward the anterior of the cell. Lysosomes fuse with the food vacuole and secrete digestive enzymes into the vacuole. As digestion proceeds, the end products, such as sugars and amino acids, diffuse across the membrane of the vacuole into the surrounding cytoplasm, and the vacuole begins to move toward the posterior end of the cell. When at the posterior end of the cell, the vacuole fuses with a struc-ture there called the anal pore. Undigested material in the vacuole is expelled to the outside through the anal pore. In addition to serving as the digestive organ, the food vacuole also serves, by its movement, to distribute the products of digestion to all parts of the cell.

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Question:

Develop a BASIC program to use Euclid's Algorithm to find the greatest common divisor of the two positive integers A and B.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G22-0545.htm

Solution:

Briefly, Euclid's Algorithm is a method which takes the smaller of two numbers and divides it into the larger number. The remainder from this division then becomes the smaller number, and the process is repeated. When the remainder reaches zero, the last smaller number is the greatest common divisor. A flowchart of the program's logic is given below. The program is given below: 100 REM GREATEST COMMON DIVISOR 110 READ A,B 120 IF A < = B THEN 130 125 GO TO 140 130 IF A > 0 THEN 150 131 GO TO 160 140 LET C = B 141 LET B = A 142 LET A = C 143 GO TO 130 150 LET B = B - A 151 GO TO 120 160 PRINT "THE GREATEST COMMON DIVISOR IS", B 800 DATA 16,36 999 END

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Question:

Write a FORTRAN program to read in an initial cost, salvage value , expected lifetime, number of years to be depreciated, and a code. The code indicates the function to be performed namely 1 -straight-linedepreciation 2 -decliningbalance depreciation 3 - sum-of-the-years-digits depreciation The program will calculate the depreciation in accordance with the code number you choose.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G18-0442.htm

Solution:

To do this problem, we assume the cost is not equal to $999.99. (We use the value of $999.99 to indicate that there are no more DATA cards to be read, and the program should be terminated.) We can set up a DO-WHILE construct here: processing will continue until all the cards are read in. Input values as well as the output, statements have been omitted so that you may come up with your own ideas on that. DIMENSION VALUE (100) 15READ (3,100) COST, SALV, EXLIFE, IYRS, ICODE IF (COST. EQ. 999.99) GO TO 99 IF (ICODE. EQ. 1)GOTO 50 IF (ICODE. EQ. 2)GOTO 70 CBY DEFAULT, THIS SECTION TAKES CARE OF ICODE = 3 DEP = COST - SALV JSUM = 0 KT = EXLIFE 20JSUM = JSUM + KT IF (KT.EQ.0) GO TO 30 KT = KT - 1 GO TO 20 30DO 40 M = 1, IYRS J = IYRS - 1 FRA = J/IYRS VALUE (M) = FRA{_\ast}DEP 40CONTINUE GO TO 15 CTHIS SECTION TAKES CARE OF ICODE = 1 50ANNDEP = DEP/ELIFE DO 60 K = 1, IYRS COST = COST - ANNDEP 60VALUE(K) = COST GO TO 15 CTHIS SECTION TAKES CARE OF ICODE = 2 DO 80 I = 1, IYRS DEP = VALU\textasteriskcentered.1667 VALU = VALU - DEP VALUE(I) = VALU 80CONTINUE GO TO 15 99STOP END

Question:

A bud on the side of a stem of an orange tree will only develop into a small scale if it is left on the stem. If this bud is removed and placed against the cambium layer of the main stem of a seedling and the terminal portion of the seedling is cut off, the bud may grow to form the entire crown of an orange tree. How can this be explained on the basis ofauxinaction?

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/Users/wenhuchen/Documents/Crawler/Biology/F09-0231.htm

Solution:

One of the many effects ofauxinis its action as a lateral bud inhibitor. The hormone, produced in the apicalmeristem, inhibits the growth of lateral buds, resulting in apical dominance or preferential growth at the apex. In the case above, a bud on the side of a stem of an orange tree will produce only a small scale if it is left on the tree. This is due to the inhibitory action ofauxinon the lateral bud, preventing the bud from actively growing and differentiating. If the bud is removed from the parent tree and grafted onto the side of a seedling, very likely the bud will not grow and develop if the dominant apicalmeristemof the seedling is intact. If, however, the apex is removed from the seedling, and the bud is grafted correctly so that theprocambiumlayer of the bud is continuous with that of the seedling, the bud will grow to form the entire crown of an orange tree. This is because removal the apex of the seedling leads to the elimination of the apical dominance, and lateral buds are now freed from the inhibitory influence ofauxin. Continuation of theprocambiumlayers of the bud and of the seedling is important, for the vascular system is derived from the procambium. If theprocambiumlayers are not continuous, the derived vascular systems will not be continuous, and the bud, soon deprived of water and minerals, will eventually die.

Question:

Explain the action of1the following procedure which includes afunction procedure within itself. SALES:PROCEDURE; DCL 1 SALES_FORCE (500) ; 2NAME; 3 FIRST_NAMECHAR(10), 3 LAST_NAMECHAR(20), 2 BASE_SALARY FIXED (6,2) 2 PERCENT_COMMISSION FIXED(2,0), A:IF BASE_SALARY(SALESMAN_INDEX)= BASE_SALARY (SALESMAN_ INDEX) THEN GOTO L; ELSE GOTO A; SALESMAN_INDEX: PROCEDURE; GETLIST(F_NAME, L_NAME); DO I = 1 TO 500 WHILE (F_NAME (I) \lnot = FIRST_NAME (I) \vert L_NAME (I)\lnot=LAST_NAME (I)); END; IF I>500 THEN DO; PUT LIST (F_NAME\vert\vert L_NAME, 'b b NOT FOUND') GOTO ERROR; END; RETURN(I) ; END SALESMAN_INDEX; L:END SALES;

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G13-0356.htm

Solution:

In the above program, 'SALES' is the label of the main procedure. The declaration shows that the variable SALES-FORCE is a 500 elementarray. The first executable statement of the program islabelled 'A'. But, in order to evaluate this statement expression, the computer comesacross the variable 'SALESMAN_INDEX' on the left hand side of theexpression of the statementlabelled'A'. Now, SALESMAN_INDEX is notin the DCL statement. Therefore, the computer checks to see if it is a procedurereference. On looking down, it sees that there is a procedure called'SALESMAN_INDEX'. Hence, the program jumps to that label. As a part of this procedure, the computer reads the data card for valuesbf variables F_NAME and L_NAME. Then, if I is evaluated, it returnsthe value of I, say I_1 as a consequence of the statement RETURN(I). That is, the value of the variable SALESMAN_INDEX in the statementlabelledA is made equal to the value of I. Now, the program comesto the right hand side of the statementlabelled'A'. Once again it seesthe function procedure reference SALESMAN_INDEX. . Hence, the programagain goes to the procedure SALESMAN_INDEX and evaluates anew value of I, say, I_2. This new value of I is returned to the right hand sideof the statementlabelled'A'. Now we have all the information required to evalu-ate the conditionalpart of statement 'A', A:IF BASE_SALARY (I_1) = BASE_SALARY (1_2)THEN ... The program is, given a list of F_NAME and L_NAME, it checks whether theperson having particular F_NAME, L_NAME is in sales force and if so wego to L. Here L islabelledto end but we can transfer control to any othercomputation e.g., calculation of commission, etc. bylabellingthat computa-tionas L. Hence, if the above condition is satisfied then we go to thestatementlabelledL, which is the END SALES statement. But, if the abovecondition is not satisfied, then we go to the ELSE part of the statement, which asks us to go back to A. This means we evaluate all over againthe function procedure SALESMAN_INDEX for new values of F_NAME and L_NAME read from data cards, and continue as before.

Question:

The ability to roll the tongue is conferred by a dom-inant gene while its recessive allele fails to confer this ability. If 16% of the students in a school can-not roll their tongues, what percentage of students are heterozygous.

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/Users/wenhuchen/Documents/Crawler/Biology/F25-0653.htm

Solution:

The 16 percent of the students that cannot roll their tongues must have a homozygous recessive genotype because the trait is determined by a dominant allele. The homozygous dominant or heterozygous condition will give one the ability to roll his tongue. If we represent the dominant allele by R and the recessive allele by r, then the three possible genotypes will be RR,Rr, andrr. We are told that the non-tongue-rolling phenotype and thus the homozygous recessive genotype (rr) occurs with a frequency of 16% or .16. To find the frequency of the recessive allele r we take the square root of .16 \surd(.16) = .40 Thus, .4 is the frequency of the non-roller allele. Knowing that the frequencies of the two alleles must add to 1 (i.e. , : R + r = 1) we can calculatethe frequency of the roller allele (R) to be .60 (i.e., R + .40 = 1; R = 1 - .40; R = .60). Squaring .60 will give us the frequency of the homozygous dominant rollers (RR). It is .36 (.60^2 = .36).The total homozygous proportion of the student population can be obtained by adding the homozygous recessive's frequency (rr) with the homozygous dominant's frequency (RR). 0.36 + 0.16 = 0.52. To obtain the heterozygous proportion of the population we can use the fact that total population = homozygous population + heterozygous population. Thus, 1 = .52 + heterozygous population or, heterozygous population = 1 - .52 = .48. Therefore, 48 per cent of the students areheterozygotesand have the ability to roll their tongues.

Question:

Use the definite integral to find the velocity and coordinate, at any time t of a body moving on the x-axis with constant acceleration. The initial velocity is v_0 and the initial coordinate is zero.

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/Users/wenhuchen/Documents/Crawler/Physics/D03-0074.htm

Solution:

Acceleration is defined as the time rate of change of velo-city. Since we are concerned with motion along the x-axis, we may neglect the vector nature of acceleration (a\ding{217}), velocity (v\ding{217}) and position (r\ding{217}) and write a = dv/dt dv = adt Because a is constant ^v2\int_v1 dv = at2\int_t1 dt As the limits of integration, we that v_1 = v_0 at t_1 = 0, and v_2 = v at t_2 = t. Then ^v\int_v0 dv=at\int_0 dt v - v_0= at v= v_0 + at Butv= dx/dt = v_0 + at where x is the position of the body along the x-axis. Then dx= v_0dt + at dt ^x1\int_x2 dx= v_0\int^t2_t1 dt + a\int^t2_t1 t dt(1) We take x_1 = x_0 at t_1 = 0, and x_2 = x att_2 = t, whence ^x\int_x0 dx= v_0t\int_0 dt + at\int_0 t dt x - x_0= v_0t + (at^2)/2 x= x_0 + v_0t + (1/2)at^2 We can also obtain this result by noting that evaluating the integral in (1) is equivalent to finding the area under the velocity vs. time curve shown in the figure.

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Question:

Design the block diagram of the accumulator and code the micro-operations that go into it.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G05-0123.htm

Solution:

Each micro-operation and its associated control function describes what logic operations need to be car-ried out. For example, the statement: C_0t_2 :OPR\leftarrow[MBR]_1-3 implies that a two-input AND gate and a three bit register (called OPR) are needed. Figure 1 shows the block diagram implementation of this statement. To design the block diagram of the accumulator, a list is made of all the micro-operations whose results go into the AC and their associated control functions. Control function Micro-operation Definition q_0c_2t_2 : AC\leftarrow[AC] \wedge [MBR] AND with MBR q_1c_2t_2 + q_2c_2t_2 : AC\leftarrow[AC] + [MBR] Add q_2c_2t_1 + q_7I ̅c_2t_3 [MBR]_4 : AC\leftarrow0 Clear q_7I ̅c_2t_3 [MBR]_6 : AC\leftarrow[AC] Complement q_7I ̅c_2t_3 [MBR]_8 : AC_n+1\leftarrow[AC]_n, AC_0\leftarrow[E] Shift-right q_7I ̅c_2t_3 [MBR]_9 : AC_15\leftarrow[E]; AC_n-1\leftarrow[AC]_n Shift-left q_7I ̅c_2t_3 [MBR]_10 : AC\leftarrow[AC]+1 Increment q_7Ic_2t_3 [MBR]_4 : AC_8-15 \leftarrow[INPR] Transfer INPR Figures 2 and 3 show the block diagram of the accumulator. Two figures are used to simplify the circuit, however, the total circuit is the combination of Figures 2 and 3. The accumulator register is an integrated circuit package capable of storing, shifting, complementing, incrementing and clearing a sixteen bit number. The add and transfer operations are shown in Figure 3.

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Question:

The length of a microscope tube is 10 centimeters. The focal length of the objective is 0.5 centimeters and the focal length of the eyepiece is 2 centimeters. What is the magnifying power of the microscope?

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/Users/wenhuchen/Documents/Crawler/Physics/D28-0877.htm

Solution:

The magnification of this compound microscope is equal to the product of the magnifications of its component lenses or M = M_1 × M_2 which in this case is (25cm/f_0) × (10cm /f_E). We use 25 cm since in most cases we want the image to be formed at that distance from the ocular. Therefore, the solution is: (25L)/(f_0f_E)_ = (25 × 10) / (0.5 × 2) = 250

Question:

What components of the plasma can affect the blood pressure? Explain how they affect it. What other solutes are present in the plasma?

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/Users/wenhuchen/Documents/Crawler/Biology/F14-0345.htm

Solution:

Although the blood appears to be a red-colored homogeneous fluid as it flows from a wound, it is actually composed of cells contained within a liquid matrix. The yellowish, extracellular liquid matrix of the blood is called plasma. Suspended in it are three major types of formed elements: the red blood cells, or erythrocyte?, the white blood cells, or leukocytes, and the platelets, small disc-shaped cell fragments important in the clotting process. The cellular portion constitutes about 45% of the whole volume of blood, while plasma constitutes the other 55%. Since the cellular part of the blood has a higher specific gravity (the density of a substance relative to that of water) than plasma (1.09 vs. 1.03), the two may be separated by centrifugation. Although the plasma consists of 90% water, there are many substances dissolved in it. The concentrations of two of these substances, inorganic salts and plasma proteins, can affect the blood pressure. The major inorganiccations(positively charged ions) in the plasma are sodium (Na^+), calcium (Ca^2+), potassium (K^+), and magnesium (Mg^2+). The major inorganic anions (negatively charged ions) in the plasma include chloride (Cl^-), bicarbonate (HCO_3^-) , phosphate (PO_4^3-), and sulfate (SO_4^2-). The major salt is sodium chloride. In mammals, inorganic ions and salts make up .9% of the plasma by weight. Usually, the plasma concentration of these substances remain relatively stable due to a variety of regulatory mechanisms, such as the action of the kidneys and other excretory organs. But if the total concentration of dissol-ved substances is appreciably changed, serious disturbances can occur. For example, the concentration of sodium chloride (NaCl) and sodium bicarbonate (NaHCO_3) determines the osmotic pressure of the plasma relative to the extracellular medium bathing the cells of the body. Normally, the osmotic pressure is balanced; however, if the concentrations of these salts are changed, the osmotic pressure of the plasma is changed and an osmotic gradient established. Water will then enter or leave the plasma by diffusion from the extracellular fluid. The resulting water loss or uptake will change the total blood volume. As with any fluid in a closed system, the volume of the blood directly affects the pressure it exerts on the walls of the vessel containing it. If the concentration of sodium chloride in the plasma is increased, water will tend to diffuse into the blood by osmosis, increasing its volume, which in turn increases the blood pressure. The proper balance of the individual ions is also essential to the normal functioning of nerves, muscles, and cell membranes. A deficiency of calcium ions, for instance, may cause severe muscular spasms, a condition called tetany. Certain ions, such as bicarbonate, affect the pH of the plasma, which is normally slightly alkaline at 7.4. Other important substances in the plasma that can affect blood pressure are the plasma proteins. Plasma proteins, which constitute about 9% of the plasma by weight, are of three types: fibrinogen, albumins and globulins. These substances contribute to the osmotic pressure of the plasma since they are too large to readily pass through the blood vessels. Their concentration affects the amount of water which diffuses into the blood, thereby regulating the general water balance in the body. In addition, these proteins determine the viscosity of the plasma. Viscosity is a measure of the resistance to flow caused by the friction between molecules as they pass one another. The heart can maintain normal blood pressure only if the viscosity of the blood is normal. For example, if the viscosity were to decrease, there would be less resistance to flow. There is a physical law which states that an increased flow rate (due to decreased resistance) will exert more pressure on the walls of the vessels transporting the fluid. Hence, this decreased viscosity leads to increased blood pressure; the viscosity of the blood is therefore inversely proportional to the blood pressure. In addition to inorganic ions and plasma proteins, plasma also contains organic nutrients such as glucose, fats, amino acids and phospholipids. It carries nitrogenous waste products, such as urea, to the kidneys. Hormones are also transported in the plasma. Dissolved gases are also transported, including carbon dioxide and oxygen, the gases involved in respiration. Most of these molecules are frequently bound to specific sites within the red blood cells, and not transported freely in the plasma.

Question:

How is the hemoglobin of a human fetus similar to that of a llama that lives high in the Andes Mountains (as com-pared to the average mammal?)

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/Users/wenhuchen/Documents/Crawler/Biology/F14-0353.htm

Solution:

Animals such as the South American llama live at high altitudes where the partial pressure of oxygen is significantly lower than at sea level. For example, at an altitude of 10,000 feet, the partial pressure of oxygen drops to 110 torrs (from 159 torrs at sea level). These animals have evolved a different kind of hemoglobin that has a higher affinity for oxygen than does the hemoglobin of the average mammal living at sea level. Their hemo-globin picks up oxygen more easily so that sufficient oxygen can be obtained at high altitudes. Human fetal hemoglobin is similar to llama hemoglobin in that it is also adapted to pick up oxygen in a medium which has a partial pressure of oxygen lower than normal. Fetal hemoglobin thus has a higher affinity for oxygen than adult hemoglobin. This adaptation is significant since the fetus obtains its oxygen from the mother's blood. For the hemoglobin of the fetus to be able to take oxygen from the hemoglobin of the mother, it must have a higher affinity for the oxygen than does the mother's hemoglobin. Fetal hemoglobin is therefore oxygenated at the expense of adult hemoglobin in the maternal circulation. What accounts for the higher oxygen affinity? The structure of fetal hemoglobin is slightly different from that of adult hemoglobin. This altered structure binds an organic phosphate called 2,3\rule{1em}{1pt} diphosphoglycerate (DPG) less strongly than does adult hemoglobin. DPG, which is present in human erythrocytes, reduces the oxygen affinity of hemoglobin: DPG reduces the oxygen affinity by stabilizing the deoxy-hemoglobin structure; the binding of oxygen is thus less favored. Since fetal hemoglobin binds DPG more weakly than does adult hemoglobin, its deoxygenated form does not undergo this degree of stabilization; its oxygen affin-ity is therefore enhanced. The oxygen dissociation curves of both fetal and maternal are shown in the following diagram: For fetal hemoglobin, the oxygen dissociation curve is shifted to the left, indicating a higher oxygen affinity. When an adult human adapts to high altitude, there is an increase in DPG in the red blood cells which shifts the curve to the right, favoring unloading of oxygen from hemoglobin at any given partial pressure.

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Question:

What is the difference between a basic assemblerlanguage macroinstruction and a machine or imperative instruction. Indicate what, if any, relationship exists between them.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G09-0195.htm

Solution:

In assembler language, as in any other computer language, we oftenfind it necessary to repeat a contiguous group or block of instructions. This block, of course, may consist of any combination of codespermitted by the assem-bler. Having to do repetition, the assembler languagepro-grammer will find a macro instruction facility useful. Macro instructions(or just Macros) are single abbreviations for groups of instructions. So in writing a macro instruc-tion, the programmer in effect definesa single line of code to represent a group of instructions. Each timethis line occurs in the program, the macro facility will sub-stitute the blockof instructions. This process is similar to calling a subroutine in a highlevel language. After the substitution, this block of code is translated toma-chine language. Hence, unlike the assembler instruction,which is translateddirectly to object code, the macro must first be expanded (substitutingthe block of code for the macro is referred to as expanding themacro) before its translation to object code. Although the macro may contain other types of instruc-tions, we can clearlysee that the assembler instruction is usually a subset of the macro.

Question:

Using the speed of sound at sea level as approximately 750 miles per hour, what is the velocity of a jet plane traveling at Mach Number 2.2?

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/Users/wenhuchen/Documents/Crawler/Physics/D26-0827.htm

Solution:

Mach Number=velocity of body / velocity of sound,and velocity of body = Mach Number x velocity of sound, whence Velocity of sound = 750 mph x 2.2 = 1650 mph = 2200 ft/sec approximately.

Question:

Describe the usage of a standard computer terminal in terms of its keyboard for input and screen for output.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G01-0005.htm

Solution:

A typical keyboard of a terminal (or a monitor that is directly connected to the computer) consists of about 100 keys in a way analogous to typewriter keyboard. The keyboards, as of 1990s, come in three different types: a) PC keyboards. In addition to keys for typing letters and numbers, moving the cursor, and rolling the screen, these keyboards contain function keys that can be programmed to input a predetermined sequence of characters, b)ASCII keyboards. These are simpler keyboards and may not have function keys or may have function keys that are preprogrammed . c) ANSI keyboards. These keyboards do much the same thing as the ASCII keyboards, except that they must adhere to the ANSI standards for various control character and escape sequences (sequence of characters entered after an escape key) for changing the behavior of the terminal itself (e.g. reverse screen, blinking characters, etc). The screen typically is 80 columns by 24 lines. There may be a 25th line for displaying terminal status information or current function keys when the function keys are programmable by soft-ware. The screen may be monochrome (actually green or amber) or color. With colored terminals , the screen colors can be set for background (the main 80-by-24 or 80-by-25 area), borders (the area surrounding the main area), and foreground (the characters typed).

Question:

A charged oil drop of mass 2.5 × 10^\rule{1em}{1pt}4 g is in the space between the two plates, each of area 175 cm^2, of a parallel- plate capacitor. When the upper plate has a charge of 4.5 × 10^\rule{1em}{1pt}7 C and the lower plate an equal negative charge, the drop remains stationary. What charge does it carry?

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/Users/wenhuchen/Documents/Crawler/Physics/D18-0610.htm

Solution:

The electric intensity between equal and oppositely charged parallel plates is given by the equation E = \sigma / \epsilon_0, where \sigma is the surface charge density on one of the plates. By definition, \sigma = Q / A and E = Q/A\epsilon_0, where Q and A are the charge on, and area of, the positive plate. The force on the oil drop is, by definition of the electric field intensity, F = qE = qQ / A\epsilon_0, where q is the oil drop's charge. Since this force balances the weight of the drop, mg = qQ/A\epsilon_0, or q = mg A\epsilon_0 / Q = [(2.5 × 10^\rule{1em}{1pt}7 kg)× (9.8 m \bullet s^\rule{1em}{1pt}2) × (175 × 10^\rule{1em}{1pt}12 m^2) × (8.85 × 10^\rule{1em}{1pt}12C^2 \bullet m^\rule{1em}{1pt}2 \bullet N^\rule{1em}{1pt}1)] / (4.5 × 10^\rule{1em}{1pt}7c) = 8.43 × 10\rule{1em}{1pt}13c.

Question:

Three moles of a diatomic perfect gas are allowed to expand at constant pressure. The initial volume is 1.3 m^3 and the initial temperature is 350\textdegreeK. If 10,000 Joules are transferred to the gas as heat, what are the final volume and temperature?

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/Users/wenhuchen/Documents/Crawler/Physics/D15-0534.htm

Solution:

Since the process of heat (Q) addition occurs at constant pressure, we may write Q = \mu C_p(T_2 - T_1) where \mu is the number of moles of gas in the system, C_p is the molar specific heat at constant pressure, and T_2 - T_1 is the temperature difference between the 2 equilibrium states [see figures (a) and (b)]. Now, Q is given (10,000 Joules), as is \mu and T. Hence, we may solve for T_2 as a function of C_p. If we can calcul-ate the value of C_p from kinetic theory, we will have obtained T_2. We now perform the appropriate calculation. Consider a gas moving from one equilibrium state to another, via some thermodynamic process. We assume that the process occurs at constant volume. Using the First Law of Thermodynamics, we obtain ∆U = Q - W(1) Here, ∆U is the change in internal energy of the gas during the process, and Q and W are the heat added to and the work done by the gas, respectively, during the process. Writing (1) in differential form dU = dQ - dW(2) In general, the element of work done by the gas in an expansion is, by definition dW = F^\ding{217} \textbullet ds^\ding{217} where F^\ding{217} is the force the gas exerts on the piston [see figure (a)] and ds^\ding{217} is the element of distance the piston moves during the expansion. Since F^\ding{217} acts perpendi-cular to the face of the piston (that is, F^\ding{217} and ds^\ding{217} are parallel), we obtain dW = F ds(3) But F may be written in terms of the pressure p, that the gas exerts on the piston face of area A F = pA Using this in (3) dW = p ds A = p dV where dV is the differential change in volume of the gas during the expansion. Substituting this in (2), we find dU = dQ - p dV(4) Applying this equation to the above-mentioned isovolumic process, we obtain dU = dQ since dV = 0. By defintion, however dQ = \mu C_v dT where C_v is the molar specific heat of the gas at constant volume, and dT is the differential temperature change the gas experiences due to the addition of heat dQ. Then dU = \mu C_v dT(5) We now assume that the change in internal energy of a gas is only a function of the temperature differ-ence experienced by the gas. Then, no matter what thermodynamic process the gas experiences, (5) holds. Consider next an isobaric thermodynamic process. Again, we apply (4) dU = dQ - p dV(4) Since the process occurs at constant pressure, dQ = \mu C_p dT(6) where C_p is the molar heat capacity at constant pressure. Furthermore, from the ideal gas law pV = \muRT If p is constant, p (dV/dT) = \muR orp dV = \mu R dT(7) Using (6), (7), and (5) in (4) \mu C_v dT = \mu C_p dT - \mu R dT orC_p = C_v + R(8) Equation (8) relates the molar specific heat at constant pressure to the molar specific heat at constant volume. All the derivations up to now have been necessary in order to obtain certain relations involving molar specific heats, namely equations (5) and (8). We now turn to an examination of the internal energy of a diatomic gas. Each method of energy storage of a diatomic molecule is called a degree of freedom. If we view a diatomic molecule as being dumbell-shaped, then it has 5 degrees of freedom [see figures (c) and (d)] . The molecule may move translationally in 3 directions (x, y, z) with 3 kinetic energies [(1/2) mv^2_x, (1/2) mv^2_y, (1/2) mv^2_z). Furthermore, it may rotate about 3 axes (x, y, z), again, with 3 kinetic energies [(1/2)I_x\omega^2_x, (1/2)I_y\omega^2_y, (1/2)I_z\omega^2_z]. However, the rotational kinetic energy about the y axis is negligible [see figure (d)] because I_y < < I_x , I_z . Hence a diatomic molecule has 5 independent methods of energy absorption, or 5 degrees of freedom. Notice one important fact: each of these kinetic energy terms has the same form, mathematical-ly. That is, they are all of the form of a positive constant times the square of a variable which has a domain extending from - \infty to + \infty. The theorem of equipartition of energy tells us that, when Newtonian mechanics holds, and the number of gas particles is large, each term of this form has the same average value per molecule, namely, (1/2) kT. In other words, each degree of freedom of a gas molecule contributes an amount (1/2) kT to the internal energy of the gas. For a diatomic gas, then, the internal energy per molecule is U_i = 5 (1/2)(kT) = (5/2) kT The internal energy for \mu moles of molecules is U = \muN_0U_i = (5/2) \mukN_0T = (5/2) \muRT Using this in (5), we may solve for C_v (1/\mu)[d{(5/2) \muRT}/{dT}] = C_v C_v = (5/2)R Using this fact in (8) C_p = C_v + R = [(5 R)/2] + R = (7/2)R Getting back to the original problem, use this value of C_p in the first equation Q = \mu C_p(T_2 - T_1) = (7/2)\muR(T_2 - T_-1) orT_2 = T_1 + Q/[(7/2)\muR] = 350\textdegreeK + [10,000 Joules] / [(7/2)(3 moles)(8.31Joules/mole \textdegreeK)] T_2 = 350\textdegreeK + 114\textdegreeK = 464\textdegreeK T_2 = 350\textdegreeK + 114\textdegreeK = 464\textdegreeK Using the ideal gas law for the 2 equilibrium states pV_1 = \muRT_1 pV_2 = \muRT_2 orV_2/V_1 = T_2/T_1 whence V_2 = (T_2/T_1)V_1 = (464/350)1.3 m^3 = 1.72 m^3.

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Question:

1.0g of scandium (Sc) combines with oxygen (O) to form 1.5338 g of oxide. Assuming the oxide contains two atoms of scandium for every three atoms of oxygen, calculate the atomic weight of Sc. Oxygen has an atomic weight of 15.9994amu.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E04-0125.htm

Solution:

This problem can be solved once a proportion is set up. The oxide weighs 1.5338 g and the Sc, 1.0 g. There-fore, the weight of the oxygen that reacted must be 1.5338 - 1= .5338 g. The problem stated that 3 oxygen atoms reacted for every 2 Sc atoms. Therefore, the total atomic weight of the oxygen in the compound is 3 × 15.9994. Let y be the atomic weight of Sc, then 2y is the total weight involved in the formation of the oxide. There-fore, there is .5338 g of oxygenwnose atoms (3 of them) have an atomic weight of 3 × 15.9994, reacting with 1 g of Sc whose atoms weigh 2y. This can be represented as (.5338 g ) / { 3(15.9994)} =(1.00g) / (2y) Solving for y, results in 44.96, which is the atomic weight of scandium .

Question:

A proton is released from rest at a distance of 1 A from another proton. What is the kinetic energy when the protons have moved in-finitely far apart? What is the terminal velocity of one of the protons if the other is kept at rest? If both are free to move, what is their velocity?

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/Users/wenhuchen/Documents/Crawler/Physics/D19-0628.htm

Solution:

When protons are moved infinitely far apart, the mutual potential energy of the protons is zero. Thus they have only kinetic energy. By conservation of energy we know that the kinetic energy must equal the original potential energy, which is EP= e^2/r \approx (4.8 × 10^\rule{1em}{1pt}10Statcoulomb)2/ 1/10 ^\rule{1em}{1pt}8 cm \approx 23 × 10^\rule{1em}{1pt}12 erg Here, r is the original separation of the protons, and e is the unit of electronic charge. The terminal velocity of the moving proton is given by (using conservation of energy) E_k= (1/2) mv^2 \approx 23 X 10^\rule{1em}{1pt}12 erg v^2 \approx (2 × 23 × 10^\rule{1em}{1pt}12 erg) / (1.67 ×10^\rule{1em}{1pt}24 g) \approx 27 × 10^12(cm/s)^2 or v \approx 5 × 10^6 cm/s Using MKS units E_p= [9 × 10^9 (N-m^2 / kg^2)][(1.6 × 10^-19coul)^2 / (10^-10m)] \approx 23 ×10-19J v^2_ \approx (2× 23 ×10-19) / (1.67 × 10^-27) \approx 27 × 108m^2 / s2 v \approx 5 × 10^4 m / s If both protons are free to move, each proton will have equal and op-posite momenta as a result of conservation of the initial momentum, which is zero in this case. The particles are identical, therefore their speeds and their kinetic energies will be the same. The total energy of the system in the final state is (1/2) Mv^21+ 1/2 Mv^2 _2 = Mv^2 \approx 23 ×10^-12 erg Hence the final speed of each particle is v \approx (5 × 106cm/s) / \surd2 \approx 3.5 × 10^6 cm / s

Question:

In a cloud-chamber photograph a proton is seen to have under-gone an elastic collision, its track being deviated by 60\textdegree. The struck particle makes a track at an angle of 30\textdegree with the incident proton direction. What mass does this particle possess? (See figure).

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/Users/wenhuchen/Documents/Crawler/Physics/D06-0315.htm

Solution:

Let the incident proton have mass m and velocity u^\ding{217}, the velocity becoming v^\ding{217} after scatter. Let the struck particle of mass M acquire velocity V^\ding{217} after the collision. Then, by the principle of conservation of energy, (1/2)mu^2 = (1/2)mv^2 +(1/2)MV^2 and since momentum is conserved both parallel and perpendicular to the original direction of travel of the proton mu = mv cos 60\textdegree + MV cos 30\textdegree and mv sin 60\textdegree = MV sin 30\textdegree . Thus V = (m/M) v [(sin 60\textdegree)/(sin 30\textdegree)] = \surd3 (m/M) v. Substituting into the other two equations gives (1/2) mu^2 = (1/2) mv^2 + (3/2) (m^2/M)v^2 and mu = (1/2) mv + (3/2) mv u^2 = v^2 + 3 (m/M) v^2 = v^2[1+ (3m/M)](1) u = (1/2) v + (3/2) v = v[(1/2) + (3/2)] or u^2 = v^2[(1/2) + (3/2)]^2(2) Equating (1) and (2) v^2 [(1/2) + (3/2)]^2 = v^2[1+ (3m/M)] or 1/4 + 9/4 + 3/2 = 1 + (3m/M) m/M = 1 and the struck particle must have been a hydrogen nucleus (a proton).

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Question:

Write a FORTRAN subroutine to find approximate solutions to the second-order initial-value problems: i) d^2y/dx^2 = g_1(y); y(x_0) = y_0, y'(x_0) = y'_0(1) ii) d^2y/dx^2 = g_2(y,y'); y(x_0) = y_0, y'(x_0) = y'_0(2) iii) d^2y/dx^2 = g_3(x,y,y'); y(x_0) = y_0, y'(x_0) = y'0(3) using the modified Euler (predictor-corrector)method.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G24-0577.htm

Solution:

The difference equation to approximate y_i+1 using the modified Euler method is: y^(k+1)_i+1,c =y_i+ 1/2 [f(x_1 , y_1) + f(x_i+1,y^(k)_i+1)]\Deltaxwhere y^(0)_i+1^ =y_i+ f(x_i +y_i)\Deltax. 1)Rewrite equation (1) as a system of simultaneous first-order equations: y' =dy/dx= z(4) z' =dz/dx= g_1(y)(5) with initial conditions y(x_0) = x_0, z(x_0) = y'(x_0) = y'_0 . Now use the modified Euler method to find solutions to both equations (4) and (5): SUBROUTINE EULER1 (N,XN,X,YCOR,ZCOR, ACCUR) REAL N DELTAX = (XN - X)/N 50Y - YCOR Z = ZCOR G1OLD - G1(Y) PRINT X,Y,Z X = X + DELTAX IF(X.GT.XN) GO TO 5 YPRED = Y + Z \textasteriskcentered DELTAX ZPRED = Z + G1 (YPRED)\textasteriskcenteredDELTAX 100YCOR \textasteriskcentered Y + 0.5 \textasteriskcentered(Z + ZPRED)\textasteriskcenteredDELTAX ZCOR = Z + 0.5 \textasteriskcentered(G1QLD + G1(YPRED))\textasteriskcenteredDELTAX YDIF = ABS(YCOR - YPRED) ZD IF = ABS (ZCOR - ZPRED) IF(YDIF.LE.ACCUR.AND.ZDIF.LE.ACCUR) GO TO 50 YPRED = YCOR ZPRED = ZCOR GO TO 100 5RETURN END ii)Rewrite equation (2) as a system of simultaneous first-order equations: y' =dy/dx= z, z(x_0) = y'(x_0) = y'_0(6) z' = dz dz /dx= g_2(y,z); y(x_0) = y0(7) /dx= g_2(y,z); y(x_0) = y0(7) Now use the modified Euler method to find approximate solutions to both (6) and (7): SUBROUTINE EULER2 (N,XN,X,YCOR, ZCGR,ACCUR) REAL N DELTAX = (XN - X)/N 50Y = YCOR Z = ZCOR G20LD = G2(Y,Z) PRINT, X,Y,Z X = X + DELTAX IF(X.GT.XN) GO TO 5 YPRED = Y + Z\textasteriskcenteredDELTAX ZPRED = Z + G2(YPRED,Z)\textasteriskcenteredDELTAX 100YCOR = Y + 0.5\textasteriskcentered(Z + ZPRED)\textasteriskcenteredDELTAX ZCOR = Z + 0.5\textasteriskcentered(G20LD + G2 (YPRED, ZPRED)) \textasteriskcenteredDELTAX YD IF = ABS (YCOR - YPRED) ZDIF = ABS(ZCOR - ZPRED) IF(YDIF.LE.ACCUR.AND. ZDIF.LE.ACCUR) GO TO 50 YPRED = YCOR ZPRED = ZCOR GO TO 100 5RETURN END Rewrite equation (3) as a system of simultaneous first-order equations: y' =dy/dx= z; z(x_0) = y'(x_0) = y'_0(8) z' =dz/dx= g_3(x,y,z); y(x_0) = y_0(9) Now use the modified Euler method to find approximate solutions to both (8) and (9): SUBROUTINE EULER3(N,XN,X,YCOR,ZCOR,ACCUR) REAL N DELTAX - (XN - X)/N 50Y = YCOR Z = ZCOR G30LD = G3(X,Y,Z) PRINT,X,Y,Z X = X + DELTAX IF(X.GT.XN) GO TO 5 YPRED = Y + Z\textasteriskcenteredDELTAX ZPRED = Z + G2 (X,YPRED,Z)\textasteriskcenteredDELTAX 100YCOR = Y + 0.5\textasteriskcentered(Z + ZPRED)\textasteriskcenteredDELTAX ZCOR = Z + 0.5\textasteriskcentered(G2OLD +G2(X,YPRED,ZPRED))\textasteriskcenteredDELTAX YDIF = ABS(YCOR - YPRED) ZDIF = ABS (ZCOR - ZPRED) IF(YDIF.LE.ACCUR.AND.ZDIF.LE.ACCUR) GO TO 50 YPRED = YCOR ZPRED = ZCOR GO TO 100 5RETURN END

Question:

What is the freezing point of a solution of 92 g of alcohol (C_2H_5OH) and 500 g of H_20 ?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E07-0238.htm

Solution:

The freezing point is dependent on the number of solute particles. One mole of a substance dissolved in 1000 g of water lowers the freezing point 1.86\textdegree. One uses the following equation to find the freezing point depression: freezing pt. depression =molalityof solute × 1.86\textdegree. Now, one must find themolalityof the alcohol. Themolalityis defined as the number of moles in 1000 g of H_2O. In this solution, there are 92 g of alcohol present. The number of moles is found by dividing 92 g by the molecular weight of the alcohol. MW of C_2H_5OH = 46. no. of moles = (92.0g) / (46g / mole) = 2 moles There are 2 moles in 500 g of water. In 1000 g of H_2O, there would be twice this amount or 4 moles. Therefore, themolalityof the alcohol is 4 m. One can now find the freezing point depression. freezing pt. depression =molality× 1.86\textdegree = 4 × 1.86\textdegree = 7.44\textdegree. The freezing point of H_2O is 0\textdegreeC. freezing pt of solution = (freezing pt of H_2O) - (freezing pt of depression) = 0\textdegree - 7.44\textdegree = - 7.44\textdegree.

Question:

Given thatK_spfor Mg(OH)_2 is 1.2 × 10-^11 , calculate the solubility of this compound in grams per 100ml of solution. The reaction equation is Mg(OH)_2 \rightleftarrows Mg+^2 + 2OH- .

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/Users/wenhuchen/Documents/Crawler/Chemistry/E11-0388.htm

Solution:

K_spis the solubility product constant; it measures the equilibrium established between the ions in the saturated solution and the excess solid phase. Knowing theK_sp, we can calculate the solu-bility of the compound. TheK_spequation for general compound A_x B_y is K_sp= [A]^x[B]^y. From the equation, it can be seen that if x moles per liter of Mg(OH)_2 dissolves, x moles of Mg+^2 and 2x moles of OH form per liter, K_sp= [Mg+^2 ] [OH-]^2 1.2 × 10-^11 = x(2x)^2 = 4x^3 . Solving, x = 1.4 × 10-^4 mole/liter of Mg(OH)_2dissolved. One is asked, however, for grams per 100ml of solution. 1.4 × 10-^4 mole/liter can be converted to grams/100ml by the following method: 100ml = .1 liters since there exists 1000ml = 1 liter. If one has 1.4 × 10-^4 moles in 1 liter, then in .1 liters, there are (.1)( 1.4 × 10-^4 ) = 1.4 × 10-^5 moles. The molecular weight of Mg(OH)_2= 58.312g/mole. Therefore, 1.4 × 10-^5 moles/100ml translates into (1.4 × 10-^5 [moles/100ml]) (58.312g/mole ) = 8.16 × 10-^4 grams/100ml.

Question:

Find the energy equivalent to 1amu.

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/Users/wenhuchen/Documents/Crawler/Physics/D31-0917.htm

Solution:

The atomic mass unit is 1/12 of the mass of a c^12_6 atom and is equal to 1.660 x 10^-24 gm. The amount of energy released by the conversion of this mass to energy is, E = ∆mc^2 = 1.660 × 10^-24 gm × [2.998 × 10^10(cm/sec)]^2 = 14.94 × 10^-4 erg = 14.94 ×10^-11 joule Converting to units of electron volts, since the energies on the molecular level are small, 1Mev= 10^6 volts × 1.602 × 10^-19coul =1.602 × 10^-13 joule 1amu= [(14.94 ×10^-11) / (1.602 × 10^-13)]Mev = 931 Mev

Question:

An item worth $30,000 initially and having a life of 15 years isdepreciated according to the double declining balance method. Calculate the amount of depreciation in each of the firstfive years.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G18-0441.htm

Solution:

The depreciation during the nth year can be expressed as D = (2C/N) [1 - (2/N)] ^n-1 where D is the depreciation for a particular year. C is the original cost. N is the life span of the machine. nis the particular year for which we are calculating the depreciation. Hence wecamwrite a statement function to compute D. values of D for n = 1,2,3,4,5will be stored in an array A. DIMENSION A (5) REAL D, C, N, K DO 100 J = 1,5 A (J) = D (30000.,15., J) 100CONTINUE END FUNCTION D (C, N, K) D = 2{_\ast}C/N{_\ast} (1.-2./N){_\ast}{_\ast}(K-1.) RETURN END

Question:

A man standing symmetrically in front of a plane mirror with beveled edges can see three images of his eyes when he is 3 ft from the mirror [see figure (a)]. The mirror is silvered on the back, is 2 ft 6 in. wide, and is made of glass of refractive index 1.54. What is the angle of bevel of the edges?

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/Users/wenhuchen/Documents/Crawler/Physics/D27-0867.htm

Solution:

The man can only see an image of his eyes if light leaves them, strikes the mirror, and is re-flected back along the same path. The central image is thus formed by light traversing the perpendicular from his eyes to the mirror. The outer images are formed by light striking the beveled edges at the point A [see figure (b)] at an angle of incidence \textphi such that the angle of refraction \textphi' makes the refracted ray strike the silvered surface normally. This must be the case if the ray of light is to leave the beveled edge by the same path with which it arrived. The angle \textphi' lies between the normal to the beveled edge and the normal to the back surface. Since \angleSAX = 90\textdegree [see figure (b)] \textphi' + \angle DAX = 90\textdegree But \angle DAX = 90\textdegree - \texttheta Hence \textphi' = 90\textdegree - 90\textdegree + \texttheta = \texttheta Draw BA, a construction line at A parallel to the back of the mirror. Angle BAC is also equal to \texttheta. But by Snell's Law n_1 sin \textphi = n sin \textphi', where n_1 is the refractive index of air (n_1 = 1) and n is that of glass. Then sin \textphi = n sin \textphi' = n sin \texttheta. Also \alpha = \texttheta +(90 - \textphi). [See figure (b)]. sin[90 - (\alpha - \texttheta)] = n sin \texttheta. But sin (90 - \psi) = cos \psi and cos (\alpha - \texttheta) = n sin \texttheta. By the trigonometric relation for double angles cos \alpha cos \texttheta + sin \alpha sin \texttheta = n sin \texttheta. cos \alpha + sin \alpha tan \texttheta = n tan \texttheta cos \alpha = tan \texttheta [n - sin \alpha] tan \texttheta = {cos \alpha/(n - sin \alpha} Looking at figure (a) cos \alpha = {1(1/4) ft}/{\surd[1 (1/4) ft]^2 + (3 ft)^2} = (5/13) sin \alpha = [(3 ft)/{[1 (1/4) ft]^2 + (3 ft)^2}] = (12/13) whencetan \texttheta = [(5/13)/{1.54 - (12/13)}] = 0.625. \texttheta = 32\textdegree.

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Question:

Write the structural formulas of the hydrogenation products of the following olefins: (a) propylene; (b) 2-methy1-2- pentene; (c) 1-methyl cyclohexene.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E21-0757.htm

Solution:

The most important reactions of alkenes (olefins) involve addition of reagents to the double bonds. An alkene can be converted to an alkane by addition of hydrogen to the double bond. These reactions are usually carried out by using a high pressure of hydrogen gas in the presence of a catalyst such as finely divided platinum, palladium, or nickel. To solve this problem, draw the structural formulas of each alkene. The addition of hydrogen breaks the double bond, and the H_2 molecule attaches itself to the carbon atoms involved. Thus,

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Question:

When two tuning forks are sounded simultaneously, a beat note of 5 cycles per second is heard. If one of the forks has a known frequency of 256 cycles per second, and if a small piece of adhesive tape fastened to this fork reduces the beat note to 3 cycles per second, what is the frequency of the other fork?

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/Users/wenhuchen/Documents/Crawler/Physics/D26-0831.htm

Solution:

This problem involves the phenomenon of beats. When two similar waves are superimposed, the beat frequency represents the numerical difference in their frequencies. Hence, for the case in question, n = (256 \pm 5)cycles/sec where n represents the unknown frequency. It appears that n has two possible values, either 251 or 261. Now, when the standard fork is loaded with the tape, its frequency will decrease. Since the beat frequency is then reduced to 3 cycles per second, the unknown frequency must be less rather than more than 256. Hence, n = 251.

Question:

A chemist mixes .5 moles of acetic acid (HC_2H_3O_2) and .5 moles of HCN with enough water to make a one liter solution. Calculate the final concentrations of H_3O^+ and OH^-. Assume the following constants: KHC2 H3 O2= 1.8 × 10^-5, K_HCN = 4 × 10^-10, K_W = 1.0 × 10^-14.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E12-0411.htm

Solution:

To solve this problem, note the simultaneousequilibria. Once this is recognized, you can set up equilibrium constant expres-sions which measure theseequilibriain terms of the concentrations of the species involved. There are three simultaneousequilibriain the final solution. Two derive from the acids HCN and HC_2H_3O_2 donating their protons to form hydroniumions in water. The third stems from the ionization of water to hydroniumand hydroxyl ions. You have: (1) HC_2H_3O_2 + H_2O \rightleftarrows H_3O^+ + C_2H_3O_2^- (2) HCN + H_2O \rightleftarrows H_3O^+ + CN^- (3) H_2O + H_2O \rightleftarrows H_3O^+ + OH^-. Each reaction contributes H_3O^+ ions. However, only the acetic acid produces an appreciable concentration of H_3O^+. This can be determined by inspection of the dissociation constants. The larger a dissociation constant, the greater the dissociation of the species. The dissociation constant of acetic acid, 1.8 × 10^-5, is the largest, and, therefore, its [H_3O^+] contribution is the greatest. As such, you can neglect the H_3O^+ contribution from HCN and water. Let the H_3O^+ contribution from acetic acid = x. If you started with a concentration of .5M for acetic acid, then, at equilibrium, it becomes .5-x. From the dissociation of acetic acid in water, it also becomes evident that the concentration of C_2H_3O_2-can be represented by x, since the equation states H_3O^+ and C_2H_3O_2^- will be produced inequimolarquantities. You have HC_2H_3O_2 + H_2O \rightleftarrows H_3O^+ + C_2H_3O_2^-, thus, K_HOAc= {[H_3O^+] [C_2H_3O_2^-]} / [HC_2H_3O_2] = (x.x) / (.5-x) = 1.8 × 10^-5. Solving for x, using the quadratic equation, you obtain x = 3.0 × 10^-3M = [H_3O^+] . To find [OH^-], use the fact that K_W = [OH^-] [H_3O^+] = 1 × 10^-14. Since you know [H_3O^+], you can substitute to find [OH^-]. Thus, [OH^-] = 3.3 × 10^-12M.

Question:

One of the functions of the kidney is to regulate the amount of sodium in the body. What functions does sodium perform in the body?

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Solution:

Sodium is essential to the body for proper functioning. Sodium is responsible for changes in mem-brane potential, and is thus required in physiological processes involving excitation of cells. Recall that the plasma membrane is differentially permeable to potassium and sodium ions. The membrane also has a pumping mechan-ism which actively pumps sodium ions out of the cell. The net effect is a higher concentration of sodium outside the cell and a lower concentration inside the cell. Although potassium ions are also distributed unequally on the two sides of the membrane, the difference is not as great as that of the sodium ions. Thus there is a net negative charge inside the cell as compared to the outside. This electrical potential is the resting cell membrane potential. A change in the distribution of sodium ions will disturb the resting potential. In nerve cells, this change generates a nerve impulse. In muscle cells, a change in sodium ion distribution may lead to muscle con-traction. Sodium is therefore important in transmission of nerve impulses and in muscle contraction. (Refer to the appropriate solutions for discussions of these mechanisms.) Sodium is also important for the maintenance of the total solute concentration, orosmolarity, of the plasma and other body fluids. Any change in the sodium content of the plasma or extracellular fluid can have drastic effects on the body cells. For example, should the sodium concentration in the plasma increase, a gradient will be created whereby water would moveosmoticallyout of the red blood cells, resulting in their crenation.

Question:

If two fruit flies, heterozygous for genes of one allelic pair, were bred together and had 200 offspring a) about how many would have the dominant phenotype? b) Of these offspring, some will be homozygous dominant and some heterozygous. How is it possible to establish which is which?

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/Users/wenhuchen/Documents/Crawler/Biology/F25-0655.htm

Solution:

Let A be the dominant gene and a be the recessive gene. Since the flies are heterozygous, each must therefore be Aa. a)In the cross, we have: homozygoushetero-homozygous dominantzygousrecessive The proportion of offspring expressing the dominant phenotype is 1/4 + 2/4 = 3/4. Therefore, the expected number with the dominant phenotype is 3/4 × 200 or 150. b) To establish which of the dominant phenotypes are homozygous and which are heterozygous, we have to perform a test cross; i.e., crossing one fly whose genotype is to be determined with a homozygous recessive fly. If the test fly is homozygous dominant: All the progeny will show the dominant phenotype. This implies that the test fly is homozygous dominant. If the test fly is heterozygous, then in the back cross: So half the progeny will show dominant phenotype and the other half will show recessive phenotype. This is indicative that the test fly was heterozygous. Therefore, by performing a test cross between a fly whose genotype we do not know and a homozygous recessive fly, we can determine if the fly is homozygous or heter-ozygous by looking at the phenotypes of the progeny.

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Question:

The following MN blood types were determined from the entire population of a small isolated mountain village. M N MN Total 53 4 29 86 What is the frequency of the L^M and L^N alleles in the population?

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/Users/wenhuchen/Documents/Crawler/Biology/F26-0700.htm

Solution:

According to the Hardy-Weinberg principle, we know that we can express the distribution of the MN blood type phenotypes in the population as p^2 + 2 pq + q^2 = 1, where p equals the frequency of the L^M allele and a equals the frequency of the L^N allele. (Remember that is phenotypically M, L^ML^N is MN, and L^NL^N, N.) p^2 is the frequency, or proportion, of M individuals in the population, 2 pq the proportion of MN individuals, and q^2 the proportion of N individuals. Therefore, for this case: q^2 = 53/86= .616 q^2 = 4/86= .047 2 pq = 29/86= .337 1.0 1.0 From this we can determine the frequencies of the alleles themselves. frequency L^M = p = \surd.616 = .785 frequency L^N = q = \surd.047= .215 1.0 1.0 As a check: 2 pq = 2(.785) (.215) = .337

Question:

Express the rest mass energy of an electron in ergs and in electron volts.

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/Users/wenhuchen/Documents/Crawler/Physics/D32-0929.htm

Solution:

The rest mass energy of any object is given by the product of the mass of that object as observed by an observer travelling with the object (m_0), and c^2 (c = the speed of light). In ergs, Erest= m0c^2= (9.11 × 10^-28 g)(9 × 10^20 cm^2/s^2) m0c2= 8.2 ×10^-7 erg Since 1 erg = 6.24 × 10^11ev E_rest = m_0c^2 = (8.2 × 10^-7 erg) {6.24 × 10^11 (eV/erg)} m_0 c2= 0.512 × 10^6 eV The rest mass energy of an electron is 0.512 MeV.

Question:

Estimate the radius of the nucleus of mercury. The radius R varies with atomic weight A according to R =1.2 × 10^-13 A^1/3 cm.

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Solution:

For mercury, A = 200, thus R = 1.2 × 10^-13^3\surd(200) = 7 × 10^-13 cm. The variation of R with A quoted above holds for any nucleus. From it we can see that the density of nuclear matter is the same for every nucleus. That is, density = (mass)/(volume) = m/v. The mass of a nucleus of atomic weight A is m = 1.66 × 10^-27 A. If the nucleus is assumed to be spherical with radius R, its velocity will be V = [(4\piR^3)/(3)] = (4\pi/3) × (1.2 × 10^-13)^3 A. Thus the density does not depend on A, and is approximately Density = (3/4\pi) × [(1.66 × 10^-27)/(1.2 × 10^-13)^3] = 2.3 × 10^11 (gm/cm)^3. This is a truly fantastic density; one cubic centimeter of nuclear matter has a mass of about 200,000 tons !

Question:

Explain what is meant by a Hash Table, and a Hashing func-tion. Explain various methods of inserting entries in the table.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G07-0162.htm

Solution:

A Hash Table is a type of data structure in the computer memory. Stocks, lists, arrays, etc., are other examples of data structures. The advantage of a hash table over many other types of data structures is that the operations of insertion, dele-tion, searching, etc., are much more efficient, especially when the number of elements to be inserted in the structure is extremely large or initially the number of elements is not known. In essence, a Hash Table is a sequential block of memory space set aside for the table. Let us assume we reserve 1000 memory words for the Hash Table. Now, in order to determine where we should insert a par-ticular data item, within any one of the 1000 words, we make use of a Hashing function. A Hashing function can be any arbitrary function selected by us, which operates on the given data item, and extracts from it a memory address, within our reserved block of storage. For example, if our data items are names of persons, we could define our Hashing Function arbitrarily as follows: a) Give weights in ascending order from 1 to 26 to each of the letters of the alphabet from A to Z. i.e., A = 1, B = 2, ..., Y = 25, Z = 26. b) Add up the weights of the letters in a given data item (Eg., if our data item is the name NANCY the sum of weights is: N + A + N + C + Y = 14 + 1 + 14 + 3 + 25 = 57). c) Divide the sum by 7. d) Multiply by 10. e) Round off to the nearest integer, to get the value of the Hashing function. Thus, once a value of the Hashing function is obtained, this corresponds to the position, within our reserved block of memory, into which the data item is to be stored. Observe that the important aspect about hashing func-tions is not how it is devised, but whether it distributes the different data items quite uniformly within our reserved memory block. Conversely, if we want to search whether a given data item is in the Hash Table or not, we subject the data item to the same Hashing Function and obtain an address. We can then immediately access the address thus found and check if the data item we are looking for is in that location. Many times it happens that the Hashing function calcu-lates the same memory address for more than one data item. Thus, when we go to the address location, we may find that there already is some other data item there. Or, while search-ing, we may find that there is some other data item at the loca-tion calculated instead of the one which we are searching for. In this case we adopt any one of three methods: a) We multiply the data item by a pre-determined fixed constant and re- valuate the Hashing Function, and see if now we find an empty location for insertion, or the given data item while searching. We repeat the process till success, using a set of pre-determined constants. b) Beginning at the calculated location, start a sequential scan up or down the table till we either succeed or come to the end of the table. c) Bucket Hashing: Associated with each of the memory words in our Hash Table is a pointer. When a Hash function calculates an address and we find that the addressed location is already full, we search for any other vacant memory location, say within the general storage area of the memory, outside the Hash Table block. Data is stored in the first empty location and its address is stored into the pointer position of our calculated hash table address. e.g., NANCY and FRED has the same hash function, say 10. NANCY was previously stored in location 10. Now to store FRED with hash function 10 we search for vacant storage. Store FRED in that location and the address of that loca-tion is placed in the pointer position of NANCY. We now store our data item in this empty location. If a third data item also happens to have the Hash Table address, we search for another empty location in the general storage area, store the data in it, and move the pointer of the second data item to this third data item, forming a long list or bucket. While searching too, we search down the bucket.

Question:

What are acid-base reactions? In what way are they analogous to oxidation-reduction reactions?

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Solution:

An acid-base reaction involves proton transfer between two species: the proton donor and the proton acceptor. The proton donor is better known as a Bronsted acid and the proton acceptor as a Bronsted base. The general equation for an acid-base reaction can be written as: Proton donor \textemdash\textemdash\textemdash\textemdash\rightarrow H^+ + proton acceptor . For example, the following reaction occurs between acetic acid (CH_3 COOH) and water (H_2O): CH_3COOH + H_2O \textemdash\textemdash\textemdash\textemdash\rightarrow CH_3COO^- + H_3O^+ Acetic acid is the Bronsted acid because it donates a proton to become the acetate anion, and water is the Bronsted base since it accepts a proton to form the hydronium ion. The acetate anion now has the capacity to accept a proton, and therefore is a Bronsted base by definition. The acetic acid and the acetate anion together constitute a conjugate acid- base pair. The generalized formula of these types of reactions would be: Bronsted acid_J + Bronsted base_K \textemdash\textemdash\rightarrow conjugate Bronsted base_J + conjugate Bronsted acid_K where the compounds written with the subscript, J are on conjugate acid-base pair and the compounds written with the subscript K are the other conjugate acid-base pair. An oxidation-reduction reaction includes the transfer of electron(s), rather than proton(s), between two species. We can write the following general equation for oxidation- reduction (also known as redox) reactions, analogous to the one we wrote for acid-base reactions: Electron donor \textemdash\textemdash\rightarrow e^- + electron acceptor The electron donor is the species that is undergoing oxidation (loss of electrons), and is also called the reducing agent (i.e., one that reduces another species).The electron acceptor is the species undergoing reduction (gain of electrons), and is also known as the oxidizing agent (i.e., one species that oxidizes another) . The electron donor and acceptor together constitute a conjugate redox pair or couple. In some oxidation- reduction reactions, the transfer of one or more electrons is made via the trans-fer of hydrogen. Dehydrogenation is thus equivalent to oxidation. One example of a redox reaction is the one be-tween NAD^+ and NADH + H^+: (NAD^+, nicotinamide adenine dinucleotide, will be discussed at length in a later chapter.) As different acids and bases differ in their tendency respectively to dissociate and accept protons, different reducing and oxidizing agents differ in their tendency to lose and gain electrons or hydrogen, respectively. The acidity of a species is measured by its pH, while the electron-donating ability of a reducing agent is measured by its standard oxidation-reduction potential.

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Question:

What is meant by evolution? Explain briefly the concepts of inorganicand organic evolutions.

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Solution:

Evolution, in the broadest sense, means a gradual, orderly change fromone condition to another. Evolution may be either inorganic or organic. The planets and stars, the earth's topography, the chemical compoundsof the universe, and even the chemical elements and their subatomicparticles have undergone gradual, orderly changes. This is referredto as inorganic evolution since these changes do not involve living matter. Evolution that involves changes in the substance of living organismsis known as organic (meaning "living") evolution. The principle oforganic evolution states that all the various plants and animals existing atthe present time have descended from other, usually simpler, organismsby gradual modifications accumulated in successive generations. These modifications may be anatomical, physiological, biochemical, reproductive, or any other aspects of a species, that could havebeen instrumental in the emergence, proliferation, or complete extinctionof a species.

Question:

A variable to be used in a program is called BIRTH-DATE. This variable is not an independent variable, that is to say it has subdivisions such as MONTH, which is three char-acters long, DAY, which is two characters long, and YEAR, which is four characters long. Write group and elementary items to describe this variable.

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Solution:

The variable BIRTH-DATE will be described in DATA DIVISION with level number 01, which refers to the highest level of a data item, and coded with margin A. Following this, using margin B and level number 02 (actually any level number from 02-49 could be used) the subdivisions are defined as shown in fig. 1. PIC stands for Picture and describes the length and type of the item. Each X corresponds to one character in the ele-mentary item descriptions, thus YEAR is 4 characters long. The use of the X in the item description means that the item is alphanumeric, it can be any combination of characters, numbers and special characters.

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Question:

Write pseudocode and flowcharts to represent the LOOP-EXIT IF-ENDLOOP construct. What rudimentary control constructs are used to build this larger construct?

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Solution:

In pseudocode, we have the following rendition: do for I = 1 through J tasks A - M if (X) then exit do for tasks N - Z end do for A DO-FOR loop is set up, in which tasks A through M are executed. Then, an IF-THEN construct interrupts the program flow to evaluate some con-dition X. If X is satisfied, no more looping needs to be done, so program control jumps out of the loop. The following flowchart will help you visualize the process:

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Question:

How can one measure the volume of blood in a man's body (without draining it out)?

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/Users/wenhuchen/Documents/Crawler/Biology/F14-0346.htm

Solution:

The volume of any fluid can be measured by placing a known quantity of a test substance in the fluid, allowing it to disperse evenly throughout the fluid and then measuring the extent to which it has become diluted (the final concentration of the substance). All one need know is the total quantity of the test substance (usually a dye) injected and its final concentration in the fluid after dispersion. Dyes that are used for blood volume measurements must be able to stay in the plasma and not diffuse out into the tissues. Dyes that are usually used are those which bind to plasma proteins, these proteins being too large to diffuse out of the plasma. The most universally employed dye is called T-1824 or Evan's blue. A known amount of dye is injected into the blood, where it combines with the plasma proteins and disperses throughout the circulatory system within approximately ten minutes. A sample of blood is then taken and the red blood cells removed by centri-fugation. The volume of the remaining plasma and the concentration of dispersed dye in this plasma sample are measured. (The concentration is determined using a spec-trophotometer, an instrument which measures the absorbance of light by a solution. The higher the absorbance, the greater the concentration of the solution). Knowing the amount of dye in this plasma sample and the initial amount of dye injected, we can calculate the total plasma volume using a simple ratio: (amount of dye in sample) / (volume of plasma sample) = (total amount of dye injected) / (total plasma volume) For example, if 2000 units of dye were injected, and we determined that there were 10 units of dye in a 15 milliliter sample of plasma after injection, then the total volume of plasma would be: 10/15= (2000) / (total plasma volume). total plasma volume= [2000(15)] / [10] = 3000 ml. or 3 liters. Since this is only the plasma volume and not the whole blood volume, we must also determine the volume of the cellular blood component. The hematocrit of blood refers to the percentage of cellular components in the blood. Blood hematocrit is determined by centrifuging blood in a calibrated tube so that one can directly read the percentage of cells from the level of packed cells. The hematocrit of a normal man is about 42, whereas that of a normal woman is about 38. If the percentage of cells is 42%, the per-centage of plasma is 58%. The total blood volume of this normal man can then be calculated: .58 x = 3000 ml. where x = total blood volume (ml). x \approx 5200 ml. \approx 5.2 l. The volume of blood in this man's body would be about 5200millititersor about 5.2 liters. There is a small degree of error (\sim 1%) in this method arising from the loss of dye from the circulatory system during the dilution. Some dye is excreted into the urine and some is lost from the circulatory system by leakage of plasma proteins.

Question:

a) 1 1 0 1 1 1 0 1 1_2 into base 10 b) 4 5 7_8 into base 2 c) 7 B 3_16 into base 8 d) 1 2 4 2 5_10 into base 16

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G02-0023.htm

Solution:

We will develop programs for these conversions later. For this exercise, you should figure out the conversions by hand. 1 1 0 1 1 1 0 1 1_2 may be converted into base 10 by considering this fact: Each digit in a base two number may be thought of as a switch, a zero indicating "off", and a one indicating "on". Also note that each digit corresponds, in base 10, to a power of two. To clarify, look at the procedure: 2^8 2^7 2^6 2^5 2^4 2^3 2^2 2^1 2^0 1 1 0 1 1 1 0 1 1_2 (1×256)+(1×128)+(0×64)+(1×32)+(1×16)+(1×8)+(0×4)+(1×2)+(1×1) = 443. If the switch is "on", then you add the corresponding power of two. If not, you add a zero. Follow the next conversion closely. b) 457_8 may be converted into base two by using the notion of triads. Triads are three-bit groups of zeros and ones which correspond to octal (base 8) and decimal (base 10) numbers. This table can, with some prac-tice, be committed to memory: Binary (2) Octal (8) Decimal (10) Hexadecimal (16) 0 0 0 0 0 0 0 0 1 1 1 1 0 1 0 2 2 2 0 1 1 3 3 3 1 0 0 4 4 4 1 0 1 5 5 5 1 1 0 6 6 6 1 1 1 7 7 7 So, taking each digit of 457_8 separately, we get: 4_8 = 100_2 5_8 = 101_2 7_8 = 111_2 Which becomes100101111_2 = 457_8 Which becomes c) 7B3_16 is a hexadecimal number. Letters are needed to replace number, since the base is 16. The following chart will help: BINARY OCTAL DECIMAL HEXADECIMAL 1 0 0 0 10 8 8 1 0 0 1 11 9 9 1 0 1 0 12 10 A 1 0 1 1 13 11 B 1 1 0 0 14 12 C 1 1 0 1 15 13 D 1 1 1 0 16 14 E 1 1 1 1 17 15 F 1 0 0 0 0 20 16 10 It may be easier to follow if we convert 7B3_16 into decimal first. We get: It may be easier to follow if we convert 7B3_16 into decimal first. We get: 16^216^116^0 7B3_8 (7×256) + (11×16) + (3×1) = 1971_10 Now, convert to octal using this procedure: Divide 8, the base to be used, into 1971. Find the remainder of this division and save it as the first digit of the new octal number. Then divide 8 into the quo-tient of the previous division, and repeat the procedure. The follow-ing is an illustration: d) 1 2 4 2 5_10 may be converted to base 16 by this method: Take the highest power of 16 contained in 12425. Then, subtract that value and try the next lowest power. Continue the process until all the digits are accounted for. 1 2 4 2 5 16^3 × 3 =1 2 2 8 8 1 3 7 16^1 × 8 =1 2 8 9 16^0 × 9 =9 0 Answer: 12425_10 = 3089_16

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Question:

A bell circuit consists of 150 feet of No. 18 wire which has a resistance of 1 ohm, a bell which has a resistance of 5 ohms, and a battery of two dry cells, each having an emf of 1.5 volts. What is the current through the circuit, assuming that the resistance of the battery is negligible?

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/Users/wenhuchen/Documents/Crawler/Physics/D20-0669.htm

Solution:

It may be of assistance to draw a diagram (see diagram). Note that R_1 = resistance of wire (which is represented in the schematic diagram as a resistor in the circuit). R_2 = resistance of bell in the schematic diagram. Since the bell and the wire are in series, the total resistance is the sum of 5 ohms and 1 ohm; that is, 6 ohms. From Ohm's law I = (E/R) = [(3 volts)/6 ohms)] = 0.5 amp.

Question:

Which of the following compounds can exhibit geometric and/or optical isomerism? (a) H_2C = C(Cl)CH_3;(b)ClFC = CHCl; (c) CH_3CH_2CH = CHCH(CH_3)_2; (d) CH_3CHCICOOH;(e) HC \equiv CCH = CHCl; (f) ClCH= CHCHClCH_3; (g) cyclohexene;(h) H_3CN=NCH_3; (i) [NH (CH_3) (C_2H_5)(C_3H_7)]^+Cl^-.

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Solution:

Optical isomerism refers to the ability of a substance to rotate plane polarized light. Such substances exist as enantiomers (species which are interchangeable only by the breaking and the reforming of bonds and not by only the rotation about bonds). When a mirror is placed between enantiomers, it shows them to be mirror images of each other. If the mirror image of a compound is not equivalent to the original compound, the original compound and its mirror image are said to be non-super- imposable or enantiomeric. If a certain compound possesses an enantiomer, it must possess a chiral center. For a molecule to be chiral, it must have a carbon atom attached to four different groups. Thus, you can identify optical isomerism by detecting whether there exists a chiral carbon atom in the molecule. Only (d), (f), and (i) fit into this general category: and similarly with (i). In (i) the nitrogen atom is the chiral center. Geometric isomerism refers to cis-trans isomerism. When a compound possesses a double bond, it is possible for geometric isomerism to exist. A geometric isomer is said to exist when two different groups are bonded to each of the two carbon atoms forming the double bond. To determine whether a compound is the cis or the trans isomer, these groups are first assigned priority in order of atomic number. The atom with the highest atomic number is given highest priority. A plane is then imagined along the length of the double bond. If the atoms of greatest priority are both on one side of the plane the isomer is said to be cis. This can be seen in the diagram below. (The numbers indicate priority ratings.) Only (b), (c), (e), and (h) fit this definition. Notice: Similar results can be shown in the others.

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Question:

A pulley system consisting of two fixed and two movable pulleys (making four supporting strands) is used to lift a weight. If the system has an efficiency of 60%, how much weight can be lifted by an applied effort of 50 pounds?

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Solution:

With four supporting strands the IMA = 4. IAM = imaginary mechanical advantage AMA = actual mechanical advantage E = effort R = resistance (weight of load) Since efficiency = AMA/IMA , 0.60 = AMA/4, whenceAMA = 2.4. Since AMA = R/E , 2.4 = R/(50 lb), whenceR = 120 lb.

Question:

Mendel believed that hereditary factors were always either dominant or recessive. How might he have altered this view had he performed the following cross? When pure line sweet peas with red flowers are crossed with pure line plants having white flowers, all the F_1 plants have pink flowers.

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Solution:

Let R be the gene for red color and W be the gene for white color. In the cross: After observing such a cross, Mendel could not have proposed the concepts of dominance or recessiveness, because there is evidence for neither in the results. It is possible that he might have proposed the idea of "blending"; saying that the heterozygous genotype is the result of a genotypic blending of the two alleles. This would be erroneous however, because the two alleles still act and separate independently. This could be evidenced if a cross between two F_1 plants were done: Two pink flowered plants have produced not only pink offspring, but also offspring having the red and white homozygous traits. Therefore the genes are still sepa-rating independently in the heterozygotes. Mendel could have proposed that the two genes products interracted to form some sort of phenotypically blended product, but this is not exactly what occurs. If one looked closely at the pink flowers of the hetero-zygote, the independent action of each of the alleles is obvious. The pink color is not the result of some sort of blending to produce pink pigment, but results from the independent expressions of the red and the white pigments in the flower. The flower appears pink because of the interspersion of the red and white pigment granules in the petal. The two alleles are said to be codominant. In such a case, neither allele is expressed in preference to the other. Each allele is capable of some degree of expression of its own trait in the heterozygous state.

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Question:

Develop a FORTRAN subroutine that sorts a list of N elements and arranges them in ascending order.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G07-0154.htm

Solution:

This method is called a selection sort. It is not the most efficient method in terms of computer time, but it is simple to understand. It compares the first value of the array Y (N) to each successive value. If the first value is greater than the Nth value, they are interchanged. Then, the second value is compared to the others in succession, and so on. The subroutine is as follows: SUBROUTINE SORT (Y, N) DIMENSION Y (1) CY IS THE ARRAY TO BE SORTED N1 = N - 1 DO 10 I = 1, N1 J = I + 1 DO 20 K = J, N IF (Y (I).LE.Y (K)) GO TO 20 TEMP = Y (I) Y (I) = Y (K) Y (K) = TEMP 20CONTINUE 10CONTINUE RETURN END

Question:

A metal bar of length 1 m falls from rest under gravity While remaining horizontal with its ends pointing toward the magnetic east and west. What is the potential difference between its ends when it has fallen 10 m? The horizontal component of the earth's magnetic induction is 1.7 × 10^-5 Wb \bullet m^-2.

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Solution:

Figure (A) shows the bar, of length l, moving with a velocity v^\ding{217} in a field of magnetic in-duction B^\ding{217}. Each charge Q of the bar will experience a force F^\ding{217} = Qv^\ding{217} × B^\ding{217} where v^\ding{217}' is the velicity of Q. Note that, at t = 0, when we let go of the bar, v^\ding{217}' = v^\ding{217}, the velocity of the bar due to its free-fall motion. The force at t = 0, then, accelerates the charges Q along the bar. Hence, at any time t after t = 0, the charge Q has 2 components of velocity - one is parallel to the bar, and the other component is equal to v^\ding{217}. The first component of v^\ding{217}' will tend to curve the paths of the charges in the bar [see figure (A)]. However, since Q and B^\ding{217} are very small, the curvature of the actual path of the charges will be extremely large, and we may approximate the trajectory of Q by a straight line parallel to the bar. Furthermore, F^\ding{217} will be small, so that v^\ding{217}' will be very close to v^\ding{217}. We may then write v^\ding{217}' \approx v^\ding{217} andF^\ding{217} = Qv^\ding{217} × B^\ding{217} where v^\ding{217} is the bar's velocity. This means that the net effect of F^\ding{217} is not to change the velocity of Q by an appreciable amount, but, rather, to separate the oppos-itely charged particles in the bar. These charges will accumulate at points a and b, and will then set up an e.m.f. in the bar. We now evaluate this e.m.f. By definition, the work done by F^\ding{217} on a charge Q in moving it aIong an arbitrary path from point a to point b is W = ^b\int_a F^\ding{217} \bullet ds^\ding{217} = ^b\int_a (Qv^\ding{217} × B^\ding{217}) \bullet ds^\ding{217} where ds^\ding{217} is a differential element of the path. Hence, (W/Q) =^b\int_a (v^\ding{217} × B^\ding{217}) \bullet ds^\ding{217}(1) Now, an agent that performs work on charge carriers (thereby instituting a current) can be viewed as an e.m.f., e, where \epsilon = (W/Q)(2) and W/Q is the work done by the agent on charge Q. In our problem, the agent is the magnetic field B^\ding{217}, doing work through the force F^\ding{217}. Using (1) in (2), we obtain \epsilon = ^b\int_a (v^\ding{217} × B^\ding{217}) \bullet ds^\ding{217}(3) Now, v^\ding{217} and B^\ding{217} are uniform throughout the bar, and we may rewrite (3) as \epsilon = (v^\ding{217} × B^\ding{217}) \bullet ^b\int_ads^\ding{217} Using the coordinate system shown in figure (a) v^\ding{217} = - v\^{\i} B^\ding{217} = B \cyrchar{\'\CYRK} ds^\ding{217} = ds \^{\j} and\epsilon = (- v\^{\i} × B \cyrchar{\'\CYRK}) \textbullet ^b\int_a ds \^{\j} \epsilon = v B \^{\j} \textbullet ^b\int_a ds \^{\j} = vB (b - a) But b - a = l, and we find \epsilon = v B l Since the motion of the bar is free fall, the velocity of the bar is v^2 = v^2_0 - 2 gs, where v_0 is the initial velocity of zero. When s is 10 m, v^2 = (- 2 × 9.8 m \bullet s^-2 ) × (- 10 m) orv = 14 m \bullet s^-1 . The magnitude of the induced e.m.f. is then \epsilon = vBl = 14 m \bullet s^-1 × 1.7 × 10^-5 wb \bullet m^-2 × 1 m = 2.38 × 10^-4 V. (b) We may also do this problem as indicated in figure (b). We consider the bar to be connected to an imaginary external circuit (AGCD). As the bar moves down, the flux through this imaginary circuit changes. Using Faraday's Law, the induced e.m.f. is \epsilon = - [(d\varphi)/(dt)](4) where \varphi is the magnetic flux through the circuit, or \textasciigrave\varphi = \intB^\ding{217} \textbullet dA^\ding{217} . This integral is evaluated over the area enclosed by the circuit. dA^\ding{217} is a vector element of area [see figure (B)]. \varphi = B^\ding{217} \textbullet \int dA^\ding{217} Again, using the coordinate system shown in figure (A) B^\ding{217} = B \cyrchar{\'\CYRK}dA^\ding{217} = dA \cyrchar{\'\CYRK} and\varphi = B \int dA = BA But the area A is a function of time. Measuring the height of the bar from the surface of the earth by y, we obtain A = y l and\varphi = B y l(5) Using (5) in (4) \epsilon = - (d/dt) (\bulletByl) = - Bl (dy/dt) = - Blv where v is the velocity of the bar. The magnitude of e is then \epsilon = 2.38 × 10^-4 V.

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Question:

It was once thought that living things could form directly out of non-living matter. People believed that discarded food would be transformed into bacteria and molds. An experiment was prepared as illustrated in the figure below. Tube A was left untouched. Tubes B, C and D were prepared as illustrated and then boiled for 15 minutes in a beaker of water. Sample A spoiled in one day, B in two days, C in one week, and D remained fresh until the cork and the "S\textquotedblright tube were removed. How do these results disprove the following statements: change into bacteria Why does the soup in C eventually spoil while that in D remains fresh? How do these results help to show thatbacteria are living creatures which can exist floating in the airaround us?

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/Users/wenhuchen/Documents/Crawler/Biology/F05-0146.htm

Solution:

Bacteria are microorganisms found everywhere around us. They are responsible for turning food sour either by way of fermentation or putrefaction. Fermentation is the decomposition of carbohydrates by bacteria while putrefaction is the decay of proteins by the same agents. Although the result in tube A suggests that meat soup spontaneously generates bacteria, the outcome in tube D shows this to be incorrect because it shows no sign of decay even though it contains boiled meat soup. The process of boiling meat soup does not prevent the soup from spoiling as shown by tube B. The soup in tube B does spoil, disproving statement (b). The result in tube C disproves the claim that boiling the soup alters the air because although the tube is sealed after boiling, the content spoils after a week. It should be noted that the process of boiling kills the bacteria already present in the soup. Tube A spoils in a day because of the activity of the bacteria already present in it. Tube B shows that although the soup is free from bacteria after boiling, it still spoils because it is exposed to the bacteria in the air. Tube C, with a cotton seal across its mouth, restricts the exposure of the soup to the air, thus reducing the chances of the bacteria getting into the soup and causing it to decay. But the cotton seal is not airtight and air does get into the soup gradually. That is why the soup spoils in a week. Tube D has a special "S" tube attached to the mouth of it. Although the tube has direct access to the air, all the bacteria present in the air are trapped in the bend of the "S" tube. In this way, tube D can remain fresh inde-finitely, because bacteria cannot get into the soup. By this series of experiments, we conclude that ex-posed food does not spoil spontaneously. It decays because the bacteria that are present in the atmosphere have settled on and interacted with the food and produced decay.

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Question:

Two bulbs of equal volume joined by a narrow tube of negli-gible volume contain hydrogen at 0\textdegreeC and 1 atm pressure. 1) What is the pressure of the gas when one of the bulbs is immersed in steam at 100\textdegreeC and the other in liquid oxygen at -190\textdegreeC? 2) The volume of each bulb is 10^-3 m3and the density of hydrogen is 0.09 kg\textbullet m^-3 at 0\textdegreeC and 1 atm. What mass of hydrogen passes along the connecting tube?

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Solution:

1) When the two bulbs are at different tempera-tures, one bulb contains n_1 moles at temperature T_1 occupy-ing volume V, and the other n_2 moles at temperature T_2 also occupying volume V. Once equilibrium has been attained, both must be at the same pressure p. Originally, the gas in the bulbs was at T_0 = 0\textdegreeC = 273\textdegreeK with the pressure p_0 = 1 atm. It had (n_1 + n_2) moles in a volume of 2V. The gas equation in this case is pV = nRT where R is the gas constant. Then p_0(2V) = (n_1 + n_2)RT_0 or P_0V = [(n_1 + n_2) / 2]RT_0(1) When the bulbs are immersed in steam and liquid oxygen, the gas equation for each bulb is pV = n_1RT_1, T_1 = 100\textdegreeC = 373\textdegreeK(2) pV = n_2RT_2, T_2 = -190\textdegreeC = 83\textdegreeK(3) Dividing (2) by (1), we get, p/p_0 = [n_1RT_1] / [(1/2)(n_1 + n_2)RT_0] = [2n_1T_1] / [(n_1 + n_2)T_0](4) From (2) and (3) , we see that pV = n_1RT_1 = n_2RT_2 orn_2/n_1 = T_1/T_2(5) Substituting in (4), p/p_0 = {2T_-1} / {[1 + T_1/T_2]T_0} = {2 × 273\textdegreeK} / {[1 + (373\textdegreeK/83\textdegreeK)]273\textdegreeK} = 0.497. Hence p = 0.497p_0 = 0.497 atm. At the initial temperature T_0, one bulb contained (1/2)(n_1 + n_2) moles and at temperature T_1 it had n_1 moles. Since T_1 = 100\textdegreeC > T_0 = 0\textdegreeC, gas in this bulb expands and some of it flows into the other bulb. The number of moles that pass along the connecting tube is (1/2)(n_1 + n_2) - n_1 = (1/2)(n_2 - n_1). The equation (n_2/n_1) = (T_1/T_2) from (5) is identical to (n_2 - n_1) / (n1+ n_2) = (T_1 - T_2) / (T2+ T_1)or [(1/2)(n_2 - n_1)] / [(1/2)(n_1 + n_2)] = (T_1 - T_2) / (T2+ T_1) = (373\textdegreeK - 83\textdegreeK) / (373\textdegreeK + 83\textdegreeK) = 290/456. Thus 290/456 of the mass in the bulb at 0\textdegreeC passed along the tube as the temperature varied from 0\textdegreeC to 100\textdegreeC. But each bulb held 10^-3 m^3 at 0\textdegreeC, corresponding to a mass of 10^-3 m^3 × 0.09 kg \textbullet m^-3 = 9 × 10^-5 kg. The mass passing along the tube is thus (290/456) × 9 × 10^-5 kg = 5.72 × 10^-5 kg = 0.0572 g.

Question:

According to the free electron gas theory the average kinetic energy, E, of the free electrons in a metal has been shown to be (3/5)E_F where E_F is the Fermi energy of the metal. Use this information to estimate the average speed of the valence electrons in copper. (The Fermi energy of copper is 7eV).

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Solution:

At temperature T = 0 (absolute zero), the electrons have zero kinetic energy, therefore they will have the smallest possible energies at that temperature. However, theexlusionprinciple states that no two electrons can be in the same quantum state. Therefore, even at T = 0, the electrons in general must have lowest energies which are different from each other and from zero. The highest of these available "lowest energies" is called the Fermi energy. For a piece of copper metal the average energy is E = (3/5)(7eV) = 4.2eV Since 1eV= 1.6 × 10^-19 J, the energy may be expressed in joules: E = (4.2eV)(1.6 × 10^-19 J/eV) = 6.7 × 10^-19 J The kinetic energy may be writtennonrelativisticallyas E = (1/2)mv^2 The average speed v of a free electron is found by solving this expression for v; v^2 = (2E/m) = [{(2)(6.7 × 10^-19 J)}/(9.11 × 10^-31 kg)] = 1.47 × 10^12 m^2/s^2 orv = 1.21 × 10^6 m/s

Question:

Calculate the percent weight of each element in magnesium chloride, MgCl_2.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E04-0148.htm

Solution:

As can be seen from the formula, for each molecule of Mg there are 2 molecules ofClthat must combine to form the compound. The way to solve a percent composition problem is first to find the total molecular weight. Then divide the weight of each component by the total molecular weight to find the percentages. First set up a table as shown below. Element Number of Atoms or Moles Atomic Weight Total AtomicWeight Mg 1 24.3 24.3 Cl 2 35.45 70.9 M.W. ofMgCl_2 = 24.3 + 70.9 = 95.2 To find percent composition use the formula % composition = [( Total Atomic weight) / (total molecular weight)] × 100 % Mg = (24.3 / 95.2) × 100 = 25.5 % %Cl=(70.9 / 95.2) × 100 = 74.5 %. To double check that the answer is correct, see if the sum of the percentages is 100 %.

Question:

You have two 1-liter containers connected to each other by a valve which is closed. In one container, you have liquid water in equilibrium with water vapor at 25\textdegreeC. The other container contains a vacuum. Suddenly, you open the valve. Discuss the changes that take place, assuming temperature is constant with regard to (a) the vapor pressure, (b) the concentration of the water molecules in the vapor, (c) the number of molecules in the vapor state.

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Solution:

The vapor pressure is the pressure exerted by the gas molecules when they are in equilibrium with the liquid. When the valve is opened some of the gas molecules will move to the empty container. At this point the press-ure will be less than the equilibrium pressure because the concentration of the gas molecules will be lowered. Very quickly, though, the equilibrium will be attained again by the action of more liquid molecules vaporizing. Therefore, the vapor pressure and the concentration of the gaseous molecules of the system remains essentially unchanged. Since the concentration of the gaseous molecules remains unchanged when the volume of the system is doubled, the number of molecules must also be doubled. This is true because concentration is an expression of the number of molecules per unit volume.

Question:

A uniform drawbridge has a weight W of 3600 lb and is 20 ft long. It is hinged at one end and a chain is attached to the center of the other end. The draw-bridge is lowered by letting out the chain over a pulley which is located in the castle wall 34.6 ft above the hinge. When the drawbridge is horizontal but has not yet touched the ground, what is the force F\ding{217} acting on it at the hinge?

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/Users/wenhuchen/Documents/Crawler/Physics/D02-0040.htm

Solution:

In the position stated in the problem, three forces are acting on the drawbridge: the weight acting downward, the tension in the chain, and the reaction at the hinge. The direction of W\ding{217} is as shown in the figure, and must act through the center point of the drawbridge, since the drawbridge is uniform. The tension T\ding{217} acts along the chain which makes an angle \texttheta with the drawbridge, where tan \texttheta is the height of the chain's pulley above the hinge divided by the length of the drawbridge: tan = \texttheta (34.6 ft)/(20 ft) = 1.73 Thus:\texttheta =60\textdegree Since the three forces are in equilibrium, the horizontal component of F\ding{217} must be equal and opposite to that of T\ding{217} (since W\ding{217} has no horizontal component) and the sum of the vertical components of T\ding{217} and F\ding{217} must equal W\ding{217}. The latter bit of information alone doesn't help us to solve for the vertical component of F\ding{217}. However, we know that the moments about any point in the drawbridge must cancel. Taking moments about point C, we note that the vertical components of T\ding{217} and F\ding{217} both act over moment arms of the same length (10 ft). This being the case we know that the vertical components of T\ding{217} and F\ding{217} must be equal. Since both the horizontal and vertical components of F\ding{217} are equal in magnitude to those of T\ding{217}, F\ding{217} must also make an angle \texttheta = 60\textdegree with the drawbridge. Thus, since we know that the vertical component of F\ding{217} equals half the weight of the drawbridge: F sin \texttheta= (1/2) W F= W/(2 sin \texttheta) = (3,600 lb)/(2 sin 60\textdegree) =(3,600 lb)/{2 (\surd3/2)} = 2,079 lb.

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Question:

List and briefly describe the steps of program development.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G01-0002.htm

Solution:

A program is developed in a larger context of a soft-ware system. The program development process and the resulting program must fit into the overall scheme of things (analogous to assembly line operations in manufacturing cars) . Above all, the program must function correctly and reliably, and should have clearly defined interfaces to other programs. In the larger context, development of a program (a "unit" in software develop-ment jargon) requires the following steps: Definition : This is normally done by the systems analyst who talks to the user (or customer) about requirements, and supplies the programmer with input and output data layouts (specs). Design : With specifications on hand, the programmer designs the program using tools such as flowcharts, hierarchy charts, andpseudocode. Coding : The programmer then enters the program into a file called the source program file, according to the syntax rules of the computer language. Compiling : The programmer verifies the syntax to make sure the compiler produces the object program file without serious diag-nostics (error messages). Linking : The object program file is then linked to other object files and references to symbols not defined locally by the pro-gram are resolved. The linker or link editor used for this pur-pose produces an executable load module in the machine's lan-guage. The "load module" is ready to be loaded into the memory and run. Testing : The programmer runs the executable load module using test data that represents application. He checks to make sure the specifications are met with correct outputs. This phase unmasks faults in the design and logic of the program that are referred to as "bugs." The debugging tools are used to identify the source of faults so that they can be fixed (directly or through modification of the source program) . Note that, after testing phase is over, and the program is operational on the field (say, processing cash machine transactions), new faults or bugs may be found. Thus debugging is a continuous process of keeping a program up-to-date and correct in regards to expected results. Documenting : Unless self-documenting, the program must be de-scribed (narrated) so that it can be understood and modified if necessary by other people. However, it is more important to document subsystems that consist of numerous interrelated pro-grams because of greater complexity.

Question:

Calculate E\textdegree for the cell in which the following reaction occurs. 2AI + 3NiCl_2 - 2AICI_3 + 3Ni. First, indicate the direction of the electron flow.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E16-0589.htm

Solution:

This problem is solved by finding the oxidation and re-duction reactions which occur. Knowing this, we can solve both questions. First, we must determine the metals involved. They are AI and Ni, for which the half-reactions are AI \rightarrow Al^+3 + 3r^-E\textdegree_oxid = 1.66 Ni^+2 + 2e^- \rightarrow NiE\textdegree_red = (-.25) Since AI is losing electrons (oxidation), and Ni is gaining electrons (reduction), the electron flow is from AI to Ni E\textdegree, the standard electrode potential is the sum of the E\textdegree's for the half-reactions listed above E\textdegree = E\textdegree_ox + E\textdegree_red = 1.66 + (-.25) = 1.41 volts.

Question:

Define the term "chain reaction". Using a specific example, distinguish between a chain-starting, a chain-propagating, and a chain-terminating step. Discuss the energy absorbed or released in the first and last of these.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E13-0473.htm

Solution:

In chemical kinetics, a chain reaction occurs when an intermediate species that is consumed in one step is regenerated in a later step. As a result, there is a sequence of steps which endlessly repeat themselves, like links in a chain, until the chain is ended or the starting material exhausted. A specific example of a chain reaction, is the formation of methylchloride from methane (CH_4) and chlorine gas (Cl_2). Overall, the reaction is: CH_4 + CI_2 \rightarrow CH_3CL(methyl chloride) + HCl The mechanism of this chain reaction is as follows: (1)CI_2\rightarrow 2Cl \textbullet (2) CH + Cl \textbullet\rightarrow CH_3 \textbullet + HCl (3) CH3\textbullet + Cl_2 \rightarrow CH_3CI + Cl \textbullet A chain-starting step encompasses one or more stable molecules and produces highly reactive species, such as single atoms. In this example, (1) fits this description. A stable Cl_2 molecule forms single chlorine radicals (Cl \textbullet) that immediately react in step (2). Cl_2 does not go to Cl \textbullet spontaneously. Energy must be supplied in the form of heat or light. Chain-starting steps generally require energy. A chain-propagating step involves the reaction between a molecule and one of the highly reactive species, producing at least one highly reactive species. Reactions (2) and (3) fit this description. In (2), the reactive species CH_3 \textbullet , methyl radical, is produced via the reaction between the molecule CH_4 and reactive species Cl \textbullet . In (3), this species, i.e. , CH_3\textbullet reacts with CI_2 to produce an-other reactive species, chlorine radical (Cl \textbullet). Note: this chlorine radical can go back to react with CH_4 to repeat the entire process, i.e., (2) and (3) over and over again, until you have the chain-terminating step. Chain termination occurs when one of the highly reactive species reacts with another highly reactive species, or with the wall and no energetic species result. In our example, such a step would be the reaction of two chlorine radicals to produce chlorine gas, Cl \textbullet+ Cl \textbullet \rightarrow CI_2. Since CI_2 is not a reactive species, the chain is terminated. Notice that this is the reverse of CI_2\rightarrow 2C1\textbullet , whichrequired energy. Thus, a chain-terminating step will generally release energy.

Question:

If a 1-kg object could be converted entirely into energy, how long could a 100-Wlight bulb be illuminated?A 100-W bulb uses 100 J of energy each second.

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Solution:

Einstein's mass-energy relation can be used to calculate the energy derived from this mass: E = mc^2 = (1kg) (3 × 108m/s)2= 9 × 10^16 J . A 100 Watt bulb uses 100 Joules of energy in 1 second, by definition of the watt. Hence, for every 100 Joules of energy supplied, the bulb remains lit for 1 second. The time t the bulb is lighted by 9 × 10^16 Joules is t = (9 ×10^16J)/ (1 × 10^2J/ s)2= 9 × 1014s One year is approximately 3.1 x 10^7 , so the time may be written t = (9 ×10^14s)/ (3.1 × 10^7s/year)= 2.9 × 107years Mass is indeed a very compact form of energy !

Question:

Explain how the absorption of a digested fat differs from the absorption of digested starch.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E23-0836.htm

Solution:

After fat is digested, you have a mixture of glycerol and fatty acids as well as mono- anddiglycerides. These products adhere to bile salts and pass into the epithelial cells that line the intestines. At this point, they are reassembled into triglycerides. They leave and enter the lymphatic fluid as components of fat droplets orchylo-microns. The lymphatic system is connected to the venous circulation, so that thechylomicronsenter the blood, giving it a milky appearance. Starch absorption - The end products of carbohydrates, such as starch after digestion are glucose, fructose, andgalactoseand other six carbon sugars. These are absorbed through thevilliof the small intestine and enter the venous circulation, where an enzyme present in red blood cells converts some of them to glucose. Others are con-verted to glucose by differentenzymes,inthe liver. In summary, the differences are as follows: Starch must be hydrolyzed before it is absorbed, while fats do not have to be. The other major difference in absorption is that digested starch is absorbed into the bloodstream directly, whereas fats are absorbed into the lymphatic system.

Question:

What do the words 'gymnosperm' and 'angiosperm' mean? What characteristics distinguish conifers from flowering plants?

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Solution:

The term gymnosperm means "naked seeds" and the term angiosperm means "enclosed seeds." The gymnosperms and angiosperms are both seed-producing plants which differ in the degree of protective covering pro-vided to the seeds by the structures producing them. The seeds of angiosperms are formed inside a fruit, and the seed covering is developed from the wall of the ovule of the flower. The seeds of gymnosperms are borne in various ways, usually on cones, but they are never really enclosed as are angiosperm seeds. Although embedded in the cone, the seeds lie open to the outside on the cone scales, and are not contained within any modified protective tissue. Conifers are classified as members of the group gymnosperm, while flowering plants are classified as angiosperms. An easy way to distinguish conifers from flowering plants is to compare the gross anatomy of their leaves. Needle-like leaves having a heavily cutinized epidermis are peculiar to the conifers. Due to their small amount of surface area (as compared to the broad, flat leaves of angiosperms) and their waxy coat, these leaves enable the conifers to survive hot summers and cold winters and to withstand the mechanical abrasions of storms. Most angiosperms cannot. Both conifers and angiosperms utilize xylem for the transport of water. The two types of xylem cells through which water is conducted are tracheidsand vessel elements (see 9-16). If we cut open the trunk of a conifer, we will find that its xylem consists almost entirely oftracheidswith bordered pits. On the other hand, the xylem of many flowering plants are composed of bothtracheidsand vessel elements, although the vessel elements predominate. Conifers can also be distinguished from flowering plants by their reproductive structures. Whereas angiosperms produce flowers and fruits, conifers produce cones which are formed from spirally arranged scale-like leaves bearing either seeds or pollen on the inner surfaces. Seed-bearing cones are referred to as female or ovulate cones; pollen-bearing cones are called male or pollen cones. The method of fertilization differs in these two groups as well. The flower of the angiosperm has its pistil constructed in such a way that the germinating pollen tube must grow through both the stigma and the style in order to reach the egg. In the conifers, pollen lands on the surface of the ovule and its tube grows directly into the ovule. Moreover, there is the phenomenon of double fertilization in flowering plants, which gives rise to a diploid zygote and a triploid endosperm. Since fertilization is a single process in the conifers, the endosperm consists of the haploid tissue of the female gametophyte and is thus quite different from the triploid endosperm cells of the angiosperms.

Question:

Some sex-linked traits are expressed more often in girls than in boys, while others are expressed more often in boys than in girls.

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Solution:

The X chromosome carries not only the genes for sex, but also many other genes not related to sex, such as the gene for color blindness or the gene for hemophilia. Such genes are called sex-linked, because they are located on the sex chromosome. The Y chromo-some does not carry any genes that we know of other than those which are related to the expression of the male sex. Girls have two X chromosomes. Boys have one X chromosome and one Y chromosome. Each individual inherits one sex chromosome from each parent, so that a girl gets one X chromosome from her mother and one X chromosome from her father. A boy receives an X chromosome from his mother and a Y chromosome from his father. (His mother cannot give him a Y chromosome, as she has only X chromo-somes to give). Since the genes for sex-linked traits are carried only on the X chromosome , it follows that a girl would have twice the chance of receiving a sex-linked gene than would a boy, because she receives two X chromosomes, while a boy receive only one. Thus a girl has a chance to receive a sex-linked gene if either of her parents carry that gene. A boy, however , could only receive such a gene from his mother, even if his father carried that gene on his X chromosome. Although a girl has a greater chance than a boy of receiving a sex- linked gene, the chance that either of them will express thetraitcoded for by that gene varies according to the dominant or recessive nature of the gene . For example, if the gene for a sex-linked trait is dominant, that trait would be more commonly expressed in girls than in boys. Because it is dominant , only one copy of the gene is necessary for its expression. Since girls have a greater chance of receiving a copy of a sex-linked gene, they have a greater chance of expressing its trait. If a sex-linked trait is recessive, a boy would have a greater chance of expressing the trait. In order for a girl to express that trait, she would have to have two copies of the recessive gene, because its expression would be masked by the presence of a normal X chromosome or one with a dominant allele. Thus both her parents would have to carry the gene. A boy however, need only have one copy of the gene in order to express its trait , because his Y chromosome does not carry any genes that would mask the recessive gene. So only his mother need carry the gene. The chances of this happening are much greater than the chance that two people carrying the gene will mate and have a girl, so the trait is more commonly expressed in boys. It is important to note that girls still have a greater chance of receiving a single copy of a recessive gene, although they may not express it. Such individuals are called carriers, and their frequency in the population is greater than that of individuals expressing the trait, be they male or female.

Question:

A star of mass 2 × 10^30 kg moving with a velocity of 2 × 10^4 m/sec collides with a second star of mass 5 × 10^30 kg moving with a velocity of 3 × 10^4 m/sec in a direction at right angles to the first star. If the two join together, what is their common velocity?

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Solution:

The figure shows the situation before and after the collision. The momentum of the first star is p_1 = m_1v_1 = (2 × 10^30 kg) (2 × 10^4 m/sec) = 4 × 10^34 kg-m/sec and is depicted in a vertical direction. The momentum of the second star is p^2 = m_2v_2 = (5 × 10^30 kg) (3 × 10^4 m/sec) = 15 × 10^34 kg-m/sec and is depicted in a horizontal direction. The vector diagram shows how to add the initial moment of the two stars. The total momentum of the system of the two stars before the collision is represented by the re-sultant in the triangle of vectors. According to the law of conservation of momentum this also represents the momentum of the single star resulting from the coa-lescence. Applying Pythagoras' theorem to the triangle of vectors, the magnitude of the total momentum is [\surd(4^2 + 15^2)] × 10^34 = 15.52 × 10^34 kg-m/sec. Since the combined mass is 7 × 10^30 kg, the velocity after coalescence is (15.52 × 10^34) / (7 x 10^30) = 2.22 × 10^4 m/sec. To obtain the angle \texttheta, notice that tan \texttheta = [(4 × 10^34) / (15 x 10^34)] = 0.2667 Whence\texttheta = 14.93\textdegree. Therefore, the single star resulting from the coa-lescence moves with a velocity of 2.22 × 10^4 m/sec in a direction making an angle of 14.93\textdegree with the direc-tion in which the second star was initially moving.

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Question:

Consider two large parallel plates of area 1m^2 separated by 1mm. (a) What is the capacitance of the system? (b) Suppose that the capacitor is charged so that there is a charge +q on the upper surfaceand \rule{1em}{1pt}q on the lower where q = 10^\rule{1em}{1pt}3 coul. How much work must be done to charge the capacitor? (c) What is the force between the plates?

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Solution:

(a) The capacitance of a parallel plate capacitor is given by: C = (\epsilon_0A) / d = A / (4\pik_Ed) where A is the area of each of the plates, d is the separation of the plates, and k_E = 1 / (4\pi\epsilon_0 ) (c)1m^2 / [4(3.14)(9 ×10^9 nt \rule{1em}{1pt} m^2 coul^2) (10^\rule{1em}{1pt}3 m)] = 8.05 × 10^\rule{1em}{1pt}9 (coul^2 \rule{1em}{1pt}sec^2/kg\rule{1em}{1pt}m^2 ) = 8.05 × 10\rule{1em}{1pt}9farads (b) The work done in charging the capacitor equals the energy stored in it: W = E = (1/2) (q^2/c) = (1/2)[(10^\rule{1em}{1pt}3 coul)^2 /{8.05×10^\rule{1em}{1pt}9(coul^2\rule{1em}{1pt}sec^2/kg\rule{1em}{1pt}m^2)}] = 62.1(kg\rule{1em}{1pt}m^2/sec^2) = 62.1 joules (c)To find the force between the plates, we must first calculate the field between the plates. We construct as a Gaussian surface, a rec-tangular box with one face within the top plate and another between the plates, as in the diagram (a). Since the plates are separated by a distance which is small compared to their length, the field between them is uniform. Therefore: \intE^\ding{217} \bullet ds^\ding{217} = q / \epsilon_0 = \sigmaA / \epsilon_0, E = \sigmaA /A\epsilon0= \sigma / \epsilon_0 where A is the area of the face of the Gaussian surface between the plates. We see that since the vertical sides are parallel to the field, they make no contribution to the flux. The face within the top plate also makes no contribution to the flux since the field within a conductor is zero. Next, we must calculate the force on an infinitesimal charge element dq on the bottom plate. Part of the field between the plates is due to charge elements such as this one. Therefore, in order to obtain the net external field acting on dq (which we use to calculate the force on dq) we must subtract the field due to dq, from the field between the plates. We construct as a Gaussian surface a rectangular box (see figure (b)) with horizontal faces very close to the surface of dq. Close to dq, the field is vertical and uniform. This occurs because, at this distance, dq ap-pears to be a long sheet of charge. Since dq has an almost infinitesimal width, the flux through the vertical faces is negligible. By Gauss's law: \intE^\ding{217} \bullet dS^\ding{217} = 2dAE_1 = dq / \epsilon_0 = \sigmadA /\epsilon_0, E_1 = \sigma / 2\epsilon_0 Thus, the net external field acting on dq is: E' = E \rule{1em}{1pt} E_1 = (\sigma / \epsilon_0) \rule{1em}{1pt} (\sigma / 2\epsilon_0) = (\sigma / 2\epsilon_0) The force on dq then is: dF = E'dq = (\sigma / 2\epsilon_0)dq Therefore the total force on a plate is F = \intdF = \surd(\sigma / 2\epsilon_0)dq = (\sigma / 2\epsilon_0)Q = [(Q/A ) / (2\epsilon_0)]Q = q^2 / (2\epsilon_0A) F = (10^\rule{1em}{1pt}3 coul)^2 / [2(8.85 × 10^\rule{1em}{1pt}12 farads/m)( 1m^2)] = (5.65 × 10^4 [coul^2/{( coul^2 \rule{1em}{1pt}sec^2 / kg\rule{1em}{1pt}m^2)\rule{1em}{1pt}m}] = 5.65 × 10^4 (kg\rule{1em}{1pt}m / sec^2) = 5.65 × 10^4 N

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Question:

It is known that at 4.6 mm Hg, the boiling point of water is lowered to 0\textdegreeC. But at 4.6 mm Hg, the freezing point is still 0\textdegreeC. Explain (a) why the freezing point was not altered substantially by a change in pressure, and (b) the paradox of water freezing and boiling at the same temperature.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E07-0250.htm

Solution:

To solve this problem, you want to consider what would alter the temperature of any phase change and see if the condition exists in this case. The temperature of any phase transition is affected by pressure only because of the accompanying volume change. Since the difference in density between solid and liquid is usually small, as compared to liquid and gas, a pressure change will not alter to any significant degree the freezing point, since the volume change will be small. Thus, the boiling point was altered, but not the freezing point. The paradox of both the boiling point and freezing point being at the same point can be explained by a discussion of the triple point. At this point, some molecules are going from liquid to solid and others from solid to liquid, and similarly for the gas-liquid and liquid-solid transformations. An equilibrium is established when an equal number of molecules are going each way in all three processes. Thus, at the triple point, you have a set of conditions in which all three states may exist in equilibrium with each other. The freezing and boiling points, when at the same tem-perature, suggest such an equilibrium. Thus, the existence of one temperature for both boiling and freezing point can occur at a given pressure.

Question:

A chemist decides to react 2 g of VO (vanadium oxide) with 5.75 g of Fe_2 O_3 to produce V_2 O_5 andFeO. How many grams of V_2 O_5 can be obtained?

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Solution:

It is extremely important to write out a balanced equation that illustrates this chemical re-action. The balanced equation for this reaction is: 2VO + 3Fe_2 O_3 \rightarrow 6FeO + V_2 O_5 The coefficients tell us the relative number of moles of each reactant and the relative number of moles of each product. For example, 2molesof VO will react for every three moles of Fe_2 O_3. To find the amount of V_2 O_5, therefore, you must first determine how many moles of V_2 O_5 will be generated. To do this you must know the number of moles of the reactants. moles of VO= [(weight in grams) / (molecular weight)] = [(2.00 g) / (66.94 g/mole)] = .0299 moles moles of Fe_2 O_3 = [(5.75 g) / (159.69 g/mole)] = .0360 moles According to the equation, the number of moles of VO must be 2/3 the number of moles of Fe_2 O_3 . However, 2/3 of .0360 = .0240, but you only have .0299 moles. This means that VO is in excess. In other words, for the amount of Fe_2 O_3 present, there exists more VO than will react. As such, the products produced will depend on the number of moles of Fe_2 O_3, and not on VO. Fe_2 O_3 is called the limiting reagent. In general, then, if you have two quantities that are known, and want to compute a third, you determine the limiting reagent. The unknown quantity will be dependent upon it only. As stated earlier, the number of moles of Fe_2 O_3 was .0360. According to the reaction, for every 3molesof Fe_2 O_3, one mole of V_2 O_5 is generated. Therefore, the number of moles of V_2 O_5 = 1/3 (.0360) = .0120. Recalling the definition of the mole, you have for V_2 O_5 : .0120 moles = no. of grams/molecular weight. The molecular weight of V_2 O_5 = 181.9. Therefore the number of grams of V_2 O_5 obtained is .0120 moles (181.9 g/mole) = 2.18 g.

Question:

What are the typical periods in the unfolding of abehavioral act?

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Solution:

Most behavior is goal-oriented; that is, the final behavioral act is directedat some specific object or state. There are usually three phases toa behavioral act. The first phase is called appetitive behavior. The animalorients its behavior toward some specific goal which is determined byits physiological needs. A hungry animal searches for food; a sexually matureanimal searches for a mate. It is frequently difficult to identify the natureof the appetitive be-havior until the goal is observed. The exploratorybehavior of a wolf looking for a mate is very similar to its behaviorwhen looking for prey. Once the appetitive behavior has enabled the animal to reach its goal, another type of behavioral act -the second phase- ensues. The sign stimuli, which in the above examples could be the sight of food or the presenceof a mate, evoke the appropriate response called the consummatoryact. Eating is theconsummatoryact of feeding behavior; copulation, theconsummatoryact of sexual behavior. The last phase is called quiescence. After theconsummatoryact is performed, an animal in this phase is not likely to be responsive to the samesign stimuli a second time. It usually slows down or stops its appetitivebehavior; thus, a well-fed animal no longer searches for prey butmay pursue some other goal. An animal in the quiescent phase also willnot perform additionalconsummatoryacts when presented with the samegoal. If, for example, a recently-fed animal is presented with food it usuallywill not eat. Eventually, the quiescence phase is terminated and appetitivebehavior gradually resumes.

Question:

A lunar module usedAerozine50 as fuel and nitrogen tetroxide (N_2 O_4, molecular weight = 92.0 g/mole) as oxidizer.Aerozine50 consists of 50 % by weight of hydrazine (N_2 H_4, molecular weight = 32.0 g/mole) and 50 % by weight of unsymmetricaldimethylhydrazine ((CH_3)_2 N_2 H_2, molecular weight = 60.0 g/mole). The chief exhaust product was water (H_2 O, molecular weight = 18.0 g/mole). Two of the reactions that led to the formation of water are the following: 2N_2 H_4 + N_2 O_4\rightarrow3N_2 + 4H_2 O (CH_3)_2 N_2 H_2 + 2N_2 O_4\rightarrow2CO_2 + 3N_2 + 4H_2 O. If we assume that these reactions were the only ones in which water was formed, how much water was produced by the ascent of the lunar module if 2200 kg ofAerozine50 were consumed in the process?

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Solution:

Aerozine50 consists of N_2 H_4 and (CH_3)_2 N_2 H_2. From the first reaction, we see that 2 moles of H_2 O are produced per mole of N_2 H_4 consumed, and, from the second reaction, we see that 4 moles of H_2 O are produced per mole of (CH_3)_2 N_2 H_2 consumed. Thus, if we determine the number of moles of N_2 H_4 and the number of moles of (CH_3)_2 N_2 H_4 in 2200 kg ofAerozine50, we can calculate the number of moles of water, and, from this, the mass of water produced, since moles = grams (mass) /molecular weight. Since N_2 H_4, and (CH_3)_2 N_2 H_4 each form 50% ofAerozine50 by weight, the mass of each component in 2200 kg ofAerozine50 is mass N_2 H_4 = 50 % × 2200 kg = 1100 kg = 1.1 × 10^6 g. mass (CH_3)_2 N_2 H_4 = 50 % × 2200 kg = 1100 kg = 1.1 × 10^6 g. (there are 1000g per kg.) To convert mass to moles, we divide by the molecular weight. Hence, 2200 kg ofAerozine50 contains moles N_2 H_4 = 1.1 × 10^6 g/32.0 g/mole = 3.4 × 10^4 moles N_2 H_4 moles (CH_3)_2 N_2 H_2 = 1.1 × 10^6 g/60.0 g/mole = 1.8 × 10^4 moles (CH_3)_2 N_2 H_2. 3.4 × 10^4 moles of N_2 H_4 produces 2 × 3.4 × 10^4 = 6.8 × 10^4 moles of H_2 O and 1.8 × 10^4 moles of (CH_3)_2 N_2 H_2 produces 4 × 1.8 × 10^4 = 7.2 × 10^4 moles of H_2 O. The total number of moles of H_2 O produced is 6.8 × 10^4 + 7.2 × 10^4 = 1.4 × 10^5 moles. To convert this to mass, we multiply by the molecular weight of water. Hence, 1.4 × 10^5 moles × 18.0 g/mole = 2.5 × 10^6 g = 2.5 × 10^3 kg of water were produced.

Question:

The single DNA molecule in an E. coli chromosome (MW = 2.80 × 10^9) contains about 4.5 million mononucleotide units which are spaced about 3.4 \AA apart. Calculate the total length of this DNA molecule and compare it with the length of the E. coli cell.

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Solution:

There are 4.5 million mononucleotide units spaced 3.4 \AA apart in a single E. coli DNA molecule. To solve the for length of this molecule, multiply the number of units by the length of the space. length = (4.5 × 10^6 units) (3.4 \AA/unit) = 1.53 × 10^7 \AA. The length of the E. coli cell is 20,000 \AA. Therefore, the DNA molecule is (1.53 × 10^7) / 20,000 or 765 times as long as the cell.

Question:

How is a survivorship curve used in determining the age distribution of a population?

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Solution:

The age distribution of a population is the proportion of individuals belonging to each age group. When a population is allowed to exist in a stable environment for several generations, so that its birth and death rates become balanced, its age distribution becomes stable. The stable age distribution differs greatly from species to species, and is expressed by a survivor-ship curve. The oyster is an example of a species in which large numbers of young are produced, but the majority die in a short time. Only a tiny fraction succeeds in attaching themselves to a rock or to some other support which is the necessary step for continuing the life cycle. The survival rate among oysters after reaching this point is much higher. The hydra exemplifies species with constant mortality rates. In this case, an individual is just as likely to die when it is one year old as when it is one day old. Man and fruit flies, in contrast to the hydra and oyster, are species with a definite period of senescence. If provided with a good environment, most individuals live to a certain age in reasonably sound health. Then diseases and infirmities of old age begin to set in, and death becomes increasingly probable with each passing year. Changes in environmental conditions may radically alter the shape of the survivorship curve for any given population, and the altered mortality rates in turn have profound effects on the age distributions and on its future size. For example, the chief cause for the enormous increase in the population size of human beings has been a great reduction in mortality during the early life stages as a result of improvements in sanitation, nutrition and medical care. These improvements have caused a shift in the human survivorship curve from one intermediate between the curves C and D in primitive societies to one approaching A in most advanced societies.

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Question:

Calculate the binding energy of the hydrogen atom in its ground state.

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Solution:

The potential energy of the electron in the nth Bohr orbit is V_n= k(e^2/r_n) The kinetic energy can be obtained as a function ofr_nby using Newton's 2nd law. The electrostatic force between the electron and the nucleus causes the centripetal acceleration. Therefore F = ma = (mv^2/r_n) = (ke^2/r^2_n) where v is the velocity of the electron. The kinetic energy is T_n= (1/2)mv^2 = (1/2)k (e^2/r_n) For a circular orbit the kinetic energy is therefore equal to half of the magnitude of the potential energy. The total energy (binding energy) becomes E_n =T_n+V_n= - (1/2)V_n= - (1/2)k(e^2/r_n). For the first orbit (n = 1), r_1 = [h^2/(4\pi^2 kme^2)]. If we use the CGS unit system k = 1. The binding energy for the ground state then becomes E_1 = - [(4\pi^2k^2me^4) / 2h^2] = - (me^4/2h^2) = - [{(9.1 × 10^-28 g) × (4.80 × 10^-10statC)^4} / {2 × (1.05 × 10^-27 erg-sec)^2}] = - 2.18 × 10^-11 erg × [1/(1.60 × 10^-12 erg/eV)] = - 13.6eV This (negative) energy is the total energy of the state, and so, the binding energy (that energy that must be supplied to raise the total energy to zero and thereby release the electron) is 13.6eV.

Question:

What is the maximum number of entries that can be stored in a binary tree if the longest path from the root to any node does not exceed N? Find the general equation for the aver-age search time for any entry in a tree with N levels.

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Solution:

If the longest path is N, the tree will have N levels below the root node. At most, one node can be at the root, two nodes at the next level, four nodes at the next, etc. This expands geometrically, until the total number of nodes possible is 1 + 2 + 4 + . . . 2^N-1 = 2^N - 1 If we want to search the nodes for a particular entry, each time we descend to the next level, we must perform another comparison. For example, if we have 3 levels, the following distribution of search times is realized: 1 node takes 1 comparison 2 nodes take 2 comparisons 4 nodes take 3 comparisons The average over the seven nodes is given simply by multi-plying nodes by comparisons, taking the sum, and dividing by the total number of nodes. Hence, (1 × 1 + 2 × 2 + 4 × 3) / 7 = 2.43 We do not include the unit of time here, for it could bemilli-, micro-, or nanoseconds, according to the machine doing the comparisons. The general equation for N levels is given by N - 1 + (N / (2^N - 1))

Question:

There are some 25,000ribosomesin an E. coli cell. If the structucalproteins of theseribosomeswere stretched out end to end as fully extended polypeptide chains, how many times could they encircle the E. coli cell?Assume that theribosomesare 180 \AA in diameter, with a specific gravity of 1.0, and that they contain 40% protein. Assume that the E. coli cell is a sphere 1\mu in diameter.

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Solution:

Ribosomesare complexes of nucleic acids and protein, and are found in the cytoplasm of the cell. This problem is solved by using the following series of steps. (a) the volume of a ribosome (b) the total weight of all theribosomespresent and the weight of the protein (c) the number of amino acid residues in the protein (d) the length of the extended protein chain and the number of times it will encircle an E. coli cell. Solving: (a) A ribosome is assumed to be spherical and, thus, its volume can be found by using the formula 4/3 \pir^3, where r is the radius of the sphere. If the diameter of a ribosome is 180 \AA its radius is 90 \AA or 90 × 10^-8 cm. volume of one ribosome = (4/3) \pi (90 × 10^-8 cm)^3 = 3.06 × 10^-18 cm^3 (b) Because the specific gravity of the ribosome is 1.0, one ribosome weighs 3.06 × 10^-18 g. There are 25,000ribosomespresent. total weight = 3.06 × 10^-18 g/ribosome × 25,000ribosomes = 7.65 × 10-14g The protein comprises 40% of the ribosome. weight of the protein = (.4) (7.65 × 10^-14 g) = 3.06 × 10^-14 g (c) The molecular weight of an average amino acid residue is 120. One can find the number of moles of amino acid residues present by dividing the weight of the protein by 120. The number of residues is found by multiplying the number of moles by 6.02 × 10^23 residues/mole. no. of amino acid residues = (3.06 × 10^-14 g) / (120 g/mole) × 6.02 × 1023(residues/mole) = 1.54 × 108residues (d) When a protein is fully extended, each amino acid residue contributes 3.6 \AA or 3.6 × 10^-4 \mu to the length of the chain. length of the chain = 3.6 × 10^-4 \mu × 1.54 × 108 = 5.54 × 104\mu The diameter of the E. coli is 1\pi, which means its radius (1/2) 1\pi = 0.5 \mu. As such the circumference of the E. coli equals 2\pir, where r is radius, or 2\pi(0.5\mu) =\pi\mu= 3.14\mu. Therefore, the protein, when fully extended, can be wrapped around the E. coli (5.54 × 10^4) / 3.14 = 1. 76 × 104times.

Question:

The ability to roll the tongue into almost a complete circle is conferred by a dominant gene, while its recessive allele fails to confer this ability. A man and his wife can both roll their tongues and are surprised to find that their son cannot. Explain this by showing the genotypes of all three persons.

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Solution:

Let us represent the dominant allele for the trait of tongue-rolling by R, and the recessive allele by r. We know that if a gene is dominant for a trait, it will always be expressed if it is present. Since the son cannot roll his tongue, he cannot possess the dominant gene for this trait, and his genotype must be rr. This means that, each parent must have at least one recessive allele to donate to their son. Also, each parent must have a dominant allele for this trait because they have the ability to roll their tongues. Thus the genotype of parents for this trait must be Rr. We can illustrate this by looking at the mating of two such parents, The offspring will be obtained in the following ratios F1/4 RR,Homozygous dominant;tongue roller. 1/2 Rr,Heterozygous;tongue roller. 1/4rr,Homozygous recessive;non-tongue roller. We see then that there is a one-in-four chance that two parents who are heterozygous for a dominant trait will produce offspring without that trait. It is not unlikely then, that the parents in this problem could have had a son who does not have the ability to roll his tongue.

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Question:

Write a program to create, initialize, insert, delete,search, etc., a Hash Table, using PL / I.

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Solution:

PL / Ihasno Hash Table as a standard data struc-ture type. Hence, we shall use a two-dimensional array to simulate a Hash Table. Creation of Hash Table: DCLHASHTAB(N, 2)CHAR (6) VAR, MARKER FIXED (3); / \textasteriskcenteredWehave created an array named HASHTAB of N, say, 180 data items each of which can be a character variable, ofupto6 characters. This is our Hash Table. \textasteriskcentered / / \textasteriskcenteredThevariable MARKER will be useful later as a pointer \textasteriskcentered / / \textasteriskcenteredForany location J of the Table, the first element HASH-TAB (J, 1) will store a data item and the second element HASHTAB(J, 2) will store a pointer , if necessary, to the next linked element having the same HASH address . \textasteriskcentered / Search and Insert: / \textasteriskcenteredAssumethat the value of the HASHING function has turned out to be J \textasteriskcentered / IF J<1\mid >N THEN GO TO ('recalculate' error routine); / \textasteriskcenteredWecheck if the Hashed address is within bounds of our table \textasteriskcentered / DO WHILE HASHTAB (J, 1)\rceil= 0; / \textasteriskcentered We check if location is empty \textasteriskcentered / MARKER = J J = J + 1; / \textasteriskcentered If not empty, look at the next location \textasteriskcentered / IF J = N THEN J = 1; / \textasteriskcentered If we reach the end then we go round to the top of the array \textasteriskcentered / If J = MARKER THEN GO TO (table full error routine); / \textasteriskcenteredWecome back around to the starting point again, which indicates the Table has no empty spaces left \textasteriskcentered / HASHTAB( MARKER, 2) = J; / \textasteriskcenteredPointer of location calculated by the hashing function points to the new location J, \textasteriskcentered / End; / \textasteriskcentered End of DO WHILE \textasteriskcentered / HASHTAB (J, 1) = DATA; / \textasteriskcentered If we find the new location HASHTAB (J, 1) to be empty we do not enter the DO WHILE block, and jump to this statement, where we insert the data item. \textasteriskcentered / Search and Read Out; / \textasteriskcenteredAssumethat we are looking for a data item denoted by KEY, and whose HASH function value has been calculated as J \textasteriskcentered / DO WHILE HASHTAB (J, 1)\rceil= KEY; MARKER = J / \textasteriskcenteredWestore the original value of J in the variable MARKER \textasteriskcentered / J = HASHTAB (J, 2); / \textasteriskcenteredWechange the value of J to that pointed to by the pointer part of J \textasteriskcentered / If j = 0 THEN GO TO ('not available' error routine); / \textasteriskcenteredThismeans there are no more linkages to search, and the data item is not in the table \textasteriskcentered /. End; / \textasteriskcentered End of DO WHILE \textasteriskcentered / Y = HASHTAB (J, 1); / \textasteriskcentered If the position denoted by J is equal to the Key, we do not enter the DO WHILE, but come to this statement, and read out the value written in the location into some required location Y \textasteriskcentered / HASHTAB (MARKER, 2) = HASHTAB (J, 2); / \textasteriskcenteredWenow complete the pointer linkages of the chain by making the pointer of the previous element , stored in Marker, to point to the next element to the deleted element \textasteriskcentered / End;

Question:

Differentiate between sapwood and heartwood and between spring wood and summer wood. What are the vascular rays?

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Solution:

The sapwood is the youngest, outermost portion of the xylem in a woody plant. It contains tracheids and vessels which function in carrying water and minerals from the root to the leaves. The heartwood is the older, inner portion of the xylem whose tracheids and vessels have lost the ability of conduction. The heartwood serves to increase the strength of the stem, and to support the increasing load of foliage as the tree grows. The spring wood and summer wood together make up the secondary xylem ring (see figure) of each growing season. They account chiefly for the increase in diameter of a woody tree trunk. They are derived from the vascular cambium, which also gives rise to the secondary phloem. At the beginning of each growing season, the cambial cells divide to produce relatively large, thin-walled xylem cells, which make up the spring wood (early wood). Toward the end of each season, the xylem cells produced are smaller and have thicker walls. These smaller cells constitute the summer wood (late wood). Ordinarily, there is only one growth season per year and abrupt change in cell size between the spring wood and summer wood clearly marks off distinct alternate circular zones called annual rings. The number of annual rings found in the crosssection of a woody trunk is now used as an indicator of the age of the tree. In addition, because the width of the annual rings varies according to the climatic conditions prevailing when ring was formed, it is possible to infer what the climate was at a particular time by examining the rings of old trees. The vascular rays are rows of small, parenchymal cells that lie at right angles to the long axis of the stem or root, reaching radially from the secondary xylem to the secondary phloem. They are, like the secondary xylem and phloem, derived from the vascular cambium, and extend in length as the plant increases in girth. The vascular rays conduct water and minerals from the secon-dary xylem to the vascular cambium, the secondary phloem, and the cortical cells. Nutrients form the secondary phloem are channeled to the vascular cambium, the living cells of the secondary xylem, and the cells of the pith (if living).

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Question:

The time required for a volume of gas, X, to effuse through a small hole was 112.2 sec. The time required for the same volume of oxygen was 84.7 sec. Calculate the molecular weight of gas X.

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Solution:

This problem involves the rate of effusion through a small hole and the method to solve it involves a modified statement of Graham's Law. High densitygaeseffuse more slowly than low density gases. The time required for effusion is inversely proportional to the rate of effusion. It is mathematically stated as r_1 / r_2 = t_2 / t_1 =\surd(M.W._2) / \surd(M.W._1) where, r_1, r_2 are the rates of the two gases;t_2, t_1 are the times; and M.W._2 and M.W._1, = molecular weights. Knowing 3 of the 4 values, one can use this equation to find the 4th value. We can write t_(O)2 / t_X t_X = = \surd(M.W._(O)2) / \surd(M.W._X) 84.7 sec / 112.2 sec = \surd32 / \surdX M.W._X = 56.2amu. Molecular Weight is expressed in terms ofa.m.u, (atomic mass units).

Question:

Write a complete COBOL program to update an indexed sequential file that contains records of customers who have charge accounts at a department store. The update infor-mation will include new customer records to be added to the file charges to be added to customer's accounts, and credits to be subtracted from others. a) Write the complete division to identify the program.

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Solution:

IDENTIFICATION DIVISION. In the COBOL program this identifies the program to the user, and the computer. Besides the mandatory section header and program-ID. sentences the programmer may use optional sentences such as DATE-WRITTEN., AUTHOR., DATE-COMPILED., and REMARKS, which may be used to clarify the program's main idea to any future users.

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Question:

What is an action potential? Discuss the physical and electrochemical changes during an action potential.

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/Users/wenhuchen/Documents/Crawler/Biology/F20-0523.htm

Solution:

We know that an unstimulated nerve cell exhibits a resting potential of about 60 millivolts across its membrane. An active pumping mechanism within the mem-brane causes the cell interior to accumulate a high concen tration of K^+ ions and the exterior a high concentration of Na^+ ions. Since the resting membrane is 50 to 75 times more permeable to K^+ than to Na^+, more K^+ moves by simple diffu-sion out of the cell than Na^+ moves into the cell in the resting state, and hence a potential difference exists across the membrane. Diffusion moves ions down their concen-tration gradients, which were established by the active transport pump. When a region of the axonal membrane is stimulated by a neurotransmitter, that region undergoes a set of electrochemical changes that constitutes an action potential. An action potential can be detected as a spike on a recording device, with a rising phase, a peak, and a falling phase, each corresponding to the characteristic flow of ions across the membrane of the axon. The duration of an action potential in a neuron is about 1 msec (.001 sec). During an action potential, the permeability of the membrane to Na^+ and K^+ ions is markedly altered. Initially, the membrane permeability to Na^+ undergoes a thousand--fold increase, whereas that to K^+ remains relatively un-changed. Consequently Na^+ ions rush into the cell. If one inserts an electrode into this region of the axon, one will find that the potential at the inside of the membrane starts to rise, as the influx of positive (Na^+) ions reduces the negativity of the cell interior. Soon, the inside of the membrane registers a net positive charge. The positivity rises until a peak is reached at about + 40 millivolts. This phase of the action poten-tial, when the membrane potential approaches or even rises above zero, is known as the depolarizing (rising) phase. At the peak of the action potential the increased sodium permeability is rapidly turned off, and immediately following this the permeability of the membrane to K+ suddenly increases (see Figure). Sodium entry stops and an efflux of K+ results due to the concentration gradient of K^+. The membrane potential starts to move toward zero, then drops below zero and finally restores the pre-exci-tation, or resting, state (at - 60 mV) . This phase of the action potential, when the membrane potential moves toward its resting level, is called the repolarizing (declining) phase. How depolarization is brought to a stop and repolarization is brought about is explained by a rapid shutting off of the increased sodium permeability (sodium inactivation).

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Question:

The most widely accepted theory of muscle contraction is the sliding filament theory. What is the major point of this theory?

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Solution:

The major premise of the sliding filament theory is that muscle contraction occurs as the result of the sliding of the thick and thin filaments past one another; the lengths of the individual filaments remain unchanged. Thus the width of the A band remains constant, corresponding to the constant length of the thick filaments. The I band narrows as the thin filaments approach the center of the sarcomere. As the thin filaments move past the thick filaments, the width of the H zone between the ends of the thin filaments becomes smaller, and may disappear altogether when the thin filaments meet at the center of the sarcomere. With further shortening, new banding patterns appear as thin filaments from opposite ends of the sarcomere begin to overlap. The shortening of the sarcomeres in a myofibril is the direct cause of the shortening of the whole muscle. The question arises as to which structures actually produce the sliding of the filaments. The answer is the myosin cross bridges. These cross bridges are actually part of the myosin molecules which compose the thick filaments. The bridges swivel in an arc around their fixed positions on the surface of the thick filaments, much like the oars of a boat. When bound to the actin filaments, the movement of the cross bridges causes the sliding of the thick and thin filaments past each other. Since one movement of a cross bridge will produce only a small displacement of the filaments relative to each other, the cross bridges must undergo many repeated cycles of movement during contraction.

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Question:

Isendosporeformation in bacteria a method of reproduction ? Explain.

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Solution:

To answer this question, it is necessary to examine the results of spore formation. Some bacteria have the capacity to transform themselves into highly resistant cells calledendospores. In a process known as sporulation , bacteria form theseintracelluarspores in order to survive adverse conditions, such as extremely dry, hot, or cold environments. Each smallendosporedevelops within a vegetative cell. The vegetative cell is the form in which these bacteria grow and reproduce. Each endospore contains DNA in addition to essential materials derived from the vegetative cell. All bacterial spores containdipicolinicacid, a substance not found in the vegetative form. It is believed that a complex of calcium ion,dipicolinicacid, andpeptidoglycanforms the cortex or outer layer of theendospore. This layer or coat helps the spore to resist the destructive effects of both physical and chemical agents. Thedipicolinic acid/calcium complex may play a role in resuming metabolism during germination. The spore is generally oval or spherical in shape and smaller than the bacterial cell. Once theendosporeis mature, the remainder of the vegetative cell may shrink and disintegrate. When the spores are transferred to an environment favorable for growth, they germinate and break out of the spore wall, and the germinating spore develops into a new vegetative cell. Theendosporeis incapable of growth or multiplication. Endospore formation is neither a kind of reproduction nor a means of multiplication, since only one spore is formed per bacterial cell. During spore formation, a singleendosporeis present within the bacterium. The remaining portion of the vegetative cell dies off, while theendospore remains to later germinate into a new vegetative cell. This new vegetative cell is identical to the old one because it contains the same DNA. Spores only represent a dormant phase during the life of the bacterial cell. This phase is initiated by adverse environmental conditions. Bacteria in the generaBacillusandClostridiumare partially characterized by their ability to formendospores. Spores ofBacillus anthracis , thebacteria causing anthrax (primarily a disease of grazing animals), can germinate 30 years after they were formed.

Question:

Sea divers are aware of a danger known as the "bends". Explain the physiological mechanism of the bends.

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Solution:

In addition to hypoxia (lack of oxygen at the tissue level), decompression sickness may result from a rapid decrease in barometric pressure. In this event bubbles of nitrogen gas form in the blood and other tissue fluids, on the condition that the barometric press-ure drops below the total pressure of all gases dissolved in the body fluids. This might cause dizziness, paralysis, and unconsciousness, and it is this set of symptoms that describes the condition known as the "bends". Deep sea divers are greatly affected by the bends. Divers descend to depths where the pressure may be three times as high as atmospheric pressure. Under high pressure, the solubility of gases (particularly nitrogen) in the tissue fluids increases. As divers rise rapidly to the surface of the water, the accompanying sharp drop in barometric pressure causes nitrogen to diffuse out of the blood as bubbles, resulting in decompression sickness.

Question:

The temperatures of three different liquids are maintained at 15\textdegreeC, 20\textdegreeC, and 25\textdegreeC, respectively. When equal masses of the first two liquids are mixed, the final temperature is 18\textdegreeC, and when equal masses of the last two liquids are mixed, the final temperature is 24\textdegreeC. What tempera-ture will be achieved by mixing equal masses of the first and the last liquid?

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Solution:

Let the mass used in all cases be m, and label the specific heat capacities of the liquids c_1, c_2, and c_3, respectively. The heat Q which must be supplied to a body of mass m and specific heat c to raise its temperature through an increment ∆t is given by Q = m c ∆T In the first mixing, the heat lost by the second liquid must equal the heat gained by the first. Thus mc_2 × (20 - 18)\textdegreeC = mc_1 × (18 - 15)\textdegreeC or 2c_2 = 3c_1. Similarly, for the second mixing, mc_3 × (25 - 24)\textdegreeC = mc_2 × (24 - 20)\textdegreeC orc_3 = 4c_2. It follows that c_3 = 6c_1. If the third mixing produces a final temperature t, then one applies the same argument as before, to obtain mc_3 × (25\textdegreeC - t) = mc_1 × (t-15\textdegreeC). \therefore6c_1(25\textdegreeC - t) = C_1(t - 15\textdegreeC) 150\textdegreeC - 6t = t - 15\textdegreeC. \thereforet = (165/7) \textdegreeC = 23 (4/7) \textdegreeC.

Question:

It is a fact that the second ionization potential of alkali atoms falls off more rapidly with increasing atomic number than does the first ionization potential. Why?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E17-0627.htm

Solution:

The first ionization potential is the energy necessary to remove one electron from a neutral atom in thegasesousstate. After this first electron is removed, the atom becomes an ion with a +1 ionization state. Second ionization potential refers to the energy necessary to remove an electron from the +1 ion in the gaseous state. As the atomic numbers of elements increase the atomic radii of the atoms increase. The atomic radius determines the distance from the nucleus to the outermost electron of the atom. The first and second ionization potentials should fall off as the atomic radius increases. As the radius increases, the outer electron, which is the one that will be removed, is at a greater distance from the nucleus. Thus, the "pull" on it by the nucleus, is less than that of an electron closer to the nucleus. Hence, less energy should be required to remove an electron that is far away from the nucleus than one that is close. In considering the second ionization potential, one has an added factor involved; the element is charged. The concentration of charge will be decreased as the atomic radius increases. This increased diffusion of charge with the increased atomic number is the reason for the added fall off of the second ionization potential as compared to first ionization potential .

Question:

The turntable of a record player is accelerated from rest to a speed of 33.3 rpm in 2 sec. What is the angular acceleration?

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Solution:

The angular kinematics equation for constant acceleration \omega = \omega_0 +\alphat can be used. The initial velocity \omega_0 is zero. The final angular velocity after t = 2 sec is \omega = 2\pif = 2\pi × [(33.3 rev/min )/(60 sec/min) = 2\pi × 0.556 sec^-1 = 3.48 sec^-1 The angular acceleration is then \alpha = (\omega - \omega_0)/(t) = (3.48 sec^-1 - 0 sec^-1)/(2 sec) = 1.74 sec^-2

Question:

At normal body temperature, 37\textdegreeC (98.6\textdegreeF), the ionization constant of water,K_w, is 2.42 × 10^-14 moles^2/liter^2. A physician injects a neutral saline solution into a patient. What will be the pH of this solution when it has come into thermal equilibrium with the patient's body?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E12-0430.htm

Solution:

To solve this problem we must employ the definition ofK_w, i.e.K_w = [H_3O^+] [OH^-]. At 37\textdegreeC, we are given K_w = 2.42 × 10^-14 mole^2/liter^2 = [H_3O^+] [OH^-]. Since, for neutral solution, [H_3O^+] = [OH^-], we can set x = [H_3O^+] = [OH^-]. Then, x^2 = [H_3O^+] [OH^-] = 2.42 × 10^-14 mole^2/liter^2or x = (2.42 × 10^-14 mole^2/liter^2)^1/2 = 1.56 × 10^-7 mole/liter. Hence, [H_3O^+] = x = 1.56 × 10^-7 mole/liter and the pH of the solution is pH = - log [H_3O^+] = - log (1.56 × 10^-7) = - (- 6.807) = 6.807 \allequal 6.8.

Question:

The skin of all vertebrates consists of many layers of cells. It is the largest organ of the human body. Describe the major physical characteristics of skin. What structures are derived from human skin?

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/Users/wenhuchen/Documents/Crawler/Biology/F19-0460.htm

Solution:

Human skin is composed of a comparatively thin, outer layer, the epidermis, which is free of blood vessels, and an inner, thick layer, the dermis, which is packed with blood vessels and nerve endings. The epidermis is a stratified epithelium whose thickness varies in different parts of the body. It is thickest on the soles of the feet and the palms of the hands. The epidermis of the palms and fingers has numerous ridges, forming whorls and loops in very specific patterns. These unique finger-prints and palmprints are determined genetically, and result primarily from the orientation of the underlying fibers in the dermis. The outermost layers of the epidermis are composed of dead cells which are constantly being sloughed off and replaced by cells from beneath. As each cell is pushed outward by active cell division in the deeper layers of the epidermis, it is compressed into a flat (squamous), scalelike epithelial cell. Such cells synthesize large amounts of the fibrous protein,keratin, which serves to toughen the epidermis and make it more durable. Scattered at the juncture between the deeper layers of the epidermis and the dermis are melanocytes, cells that produce the pigment melanin. Melanin serves as a protective device for the body by absorbing ultraviolet rays from the sun. Tanning results from an increase in melanin production as a result of exposure to ultraviolet radiation. All humans have about the same number of melanocytes in their skin. The difference between light and dark races is under genetic control and is due to the fact that mela-nocytes of dark races produce more melanin. The juncture of the dermis with the epidermis is uneven. The dermis throws projections called papillae into the epidermis. The dermis is much thicker than the epider-mis, and is composed largely of connective tissue. The lower level of the dermis, called the subcutaneous layer, is connected with the underlying muscle and is composed of many fat cells and a more loosely woven network of fibers. This part of the dermis is one of the principle sites of body fat deposits, which help preserve body heat. The subcutaneous layer also determines the amount of possible skin movement. The hair and nails are derivatives of skin, and develop from inpocketings of cells from the inner layer of the epidermis. Hair follicles are found throughout the entire dermal layer, except on the palms, soles, and a few other regions. Individual hairs are formed in the hair follicles, which have their roots deep within the dermis. At the bottom of each follicle, a papilla of connective tissue projects into the follicle. The epithelial cells above this papilla constitute the hair root and, by cell division form the shaft of the hair, which ultimately extends beyond the surface of the skin. The hair cells of the shaft secrete keratin, then die and form a compact mass that becomes the hair. Growth occurs at the bottom of the follicle only. Associated with each hair follicle is one or more sebaceous glands, the secretions of which make the surface of the skin and hair more pliable. Like the sweat glands, the sebaceous glands are derived from the embryonic epidermis but are located in the dermis. To each hair follicle is attached smooth muscle called arrector pili, which pulls the hair erect lipon contraction. Nails grow in a manner similar to hair. Both hair follicles and nails develop from inpocketings of cells from the inner layer of the epidermis. The translucent, densely packed, dead cells of the nails allow the underlying capillaries to show through and give the nails their normal pink color. Like the hair and nails found in man, the feathers, hoofs, claws, scales, and horns found in other vertebrates are also derivatives of the skin.

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Question:

A computer with a teletype as an I/O device has the fol-lowing I/O registers INPR:Input Register - 8 bits OUTR:Output Register - 8 bits FGI:Input Flag - 1 bit FGO:Output Flag - 1 bit IEN:Interrupt Enable - 1 bit Describe how each register is used.

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Solution:

The teletype is able to encode and decode an alpha-numeric symbol to an eight bit word. The eight bit word is stored in the Input Register (INPR) if it is going from the teletype to the computer. The eight bit word is stored in the output register (OUTR) if it is going from the computer to the teletype. The input flag (FGI) is set (1) when an eight bit word enters the input register from the teletype. The computer is constantly checking the FGI. When the FGI is set, the computer transfers the eight bit word from the INPR to the AC, and the FGI is cleared (0). The INPR will not accept a word from the teletype until the FGI is clear. The output flag, FGO, is originally set. When the com-puter wants to send data out to the terminal it checks the FGO to see if it is set. If the FGO is set, the computer sends the contents of the AC to the OUTR and clears the FGO. The FGO remains cleared until the 8 bit word is decoded and printed. The teletype clears the FGO after the word is printed. The computer cannot send a word to the OUTR when the FGO is cleared because this condition indicates that the teletype is still in the process of printing the previous character. TheInput/Outputprocess just described is very wasteful because the computer, which operates much faster than the teletype, cannot perform any useful operations while the teletype is operating. Thus, the computer wastes time checking the state of the flags. With the IEN, however, the computer does not have to check the teletype because the teletype can now tell the computer when it is ready to accept or send information. Thus, the computer can run other tasks when previously it had to wait. The IEN can be set or cleared by the pro-grammer's instruction. When IEN is cleared the computer cannot be interrupted by the flags. When IEN is set the flags can interrupt the computer.

Question:

What is the intensity of the electric field 10 cm from a negative point charge of 500 stat-coulombs in air?

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Solution:

The electrostatic force on a positive test charge q' at a distance r from a charge Q is, by Coulomb's law, (in the CGS system of units) F^\ding{217} = k (Qq')/(r^2) The electric field intensity E is defined as the force per unit charge, or E^\ding{217}= F^\ding{217}/q' =kQ/r^2 E^\ding{217} points in the direction the force on the test charge acts. In a vacuum, k = 1 (to a good approximation, k = 1 for air as well), therefore the electric field 10 cm from a point charge of 500 stat-coulomb is E^\ding{217} = (1) [(500 stat-coul)/(10 cm)^2 ] = 5 dyne/stat-coul pointing directly toward the negative charge.

Question:

a) .1_10 = (.19999...)_16 b) .875_10 = (.513)_6 c) .2_10 = (.001100110011...)_2

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Solution:

a) To convert fractions from decimal to hexadecimal, we begin the followingprocedure: Multiply the base you want (in this case 16) by the fractionto be converted. .100 × 16 = 1.600 Now, the first digit to the left of the decimal point becomes the first digit to theright of the hexadecimal point. Thus, we begin constructing our hexadecimalfraction with .1 . Subtract the first digit from the product obtained above. 1.600 - 1.000 .600 Repeat the multiplication process. Then you will have obtained the next digitof the hexadecimal fraction. .600 × 16 = 9.600 The digit 9 becomes the next digit on the right side of the hexadecimal point. Do the subtraction as before and you will see that our fraction is now.19. The process will repeat infinitely because there will always be a remainderleft from the subtraction. Hence, we prove the original equality. b) In general, we can express a fraction f converted to base b as the following: f = a_-1 × b^-1 + a_-2 × b^-2 + a_-3 × b^-3+ ... wherea_-1, a_-2,... are digits from 0 to b - 1. If we do this for .875_10, we get thesethree steps: Multiplying .875 by 6 yields the integer 5 plus the fraction.250. Multiplying .250 by 6 yields the integer 1 plus the fraction .500. Multiplying .500 by 6 yields the Integer part 3 plus the fraction .000. Putting this all together, we prove that .875_10 = (.513)_6. c) Conversions from decimal to binary are simple, but the student often getscaught up in the tedium of writing only 1's and 0's. For this ex-ample, wewill bore you only with the first four steps, since the digits repeat after thatanyway: Multiplying .2 by 2 yields the integer 0 plus the fraction.4. Multiplying .4 by 2 yields the integer 0 plus the fraction.8. Multiplying .8 by 2 yields the integer 1 plus the fraction.6. Multiplying .6 by 2 yields the integer 1 plus the fraction.2. Another notation for repeating digitsis todraw a line over those digitswhich repeat. Hence, .2 = (.00110011)_2-.

Question:

A sample of hydrogen is collected in a bottle over water. By carefully raising and lowering the bottle, the height of the water outside is adjusted so that it is just even with the water level inside (see figure) . When a sample of gas was collected the initial conditions were: volume = 425 ml, pressure = 753 mm and the temperature of the water (and thus, the gas also) = 34\textdegreeC. Calculate the volume of the hydrogen if it were dry and at a pressure of 760 mm and a temperature of 0\textdegreeC (STP).

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Solution:

First, apply Dalton's Law and then make use of the Combined Gas Law to obtain the desired answer. Dalton's Law states that the total pressure of a mixture of gases, P_total is equal to the sum of the individual partial pressures, P_(H)2 and P_(H)2O In this example the individual gases are hydrogen and water vapor. Therefore, the total pressure is the sum of these two gases. At 35\textdegreeC, the partial pressure of H_2O vapor is 40 mm, P_total = P_(H)2+ P(H)2O P_(H)2 =P_total -P_(H)2O = 753 - 40 = 713 mm This means that if no water vapor was present the H_2 gas would fill the container at 713 mm of pressure. Since pressure is indirectly proportional to the volume of H_2 gas, the effect of changing the pressure of 713 mm to STP pressure (760 mm) is to decrease the volume. In addition, however, the temperature of the H_2 gas is directly proportional to the volume of the gas, so that the effect of changing the temperature of (34\textdegreeC + 273\textdegree =) 307\textdegreeK to STP temperature (273\textdegreeK) is also to de-crease the volume of the gas. Therefore, we can predict that the net effect of changing the pressure and tem-perature of the gas to STP conditions is to decrease the volume. Using the combined gas law equation to solve for the dry H_2 gas volume, V_2 , [P_1V_1] / T_1 = [P_2V_2] / T_2 whereP_1 = 713 mmP_2 = 760 mm V_1 = 425 mlV_2 = volume at STP T_1 = 3070KT_2 = 2730K Thus, substituting, V_2 =[P_1V_1T_2] / [P_2T_1] = [(713 mm) (425 ml) (273\textdegreeK)] / [(760 mm) (307\textdegreeK)] = 355 ml. We see that this answer is in total agreement with our prediction that the volume at STP, V_2 = 355 ml, is less than the volume at the initial conditions, V_1 = 425 ml.

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Question:

Write a program to simulate a coin toss experiment of 50 tosses. Repeat the experiment ten times. Count the number of heads in each experiment and print out this value.

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Solution:

We know that if we say LET F = INT(2\textasteriskcenteredRND + 1) then F takes on values of 1 and 2 randomly corresponding to heads and tails. Hence we will loop through this statement to produce the required simulation: 2 FOR Y = 1 TO 1\O 5 LET C = \O 1\O FOR X = 1 TO 5\O 2\O LET F= INT (2\textasteriskcenteredRND + 1) 3\O IF F = 1 THEN 6\O 4\O PRINT "T"; 5\O GO TO 1\O\O 58 REM C COUNTS THE NUMBER OF HEADS 6\O LET C = C + 1 7\O PRINT "H"; 1\O\O NEXT X 11\O PRINT 12\O PRINT C; "HEADS OUT OF 5\O FLIPS" 125 NEXT Y 13\O END

Question:

Parasitic organisms often have complex life cycles invol-ving two or more host species. The nematode (round worm) causing elephantiasis has man as its primary host and the mosquito as its intermediate host. Describe the life cycle of this nematode.

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Solution:

Adultfilariaworms of the genus-speciesWucheriabancrofti, inhabit the lymph glands of man, par-ticularly in the upper leg or hip. Large numbers of these nematodes block the lymph glands and prevent the return of fluid to the blood stream. This causes enormous swel-ling of the legs, a condition known as elephantiasis. Blocking of the lymph glands also predisposes the host to microbial infections. Thesefilaraworms mature in the host's lymph glands where they later lay their eggs. The larval nematodes then migrate to the peripheral blood stream. The migration to the peripheral circulation has been found to correlate temporally with the activity ofmosquitos. When a mosquito of a certain species bites the infected person, the larvalfilariaerespond by entering the host's peripheral blood-stream. When othermosquitos subsequently bite the host and suck his blood, some larvae move into the bodies of these intermediate hosts. Inside a mosquito, the larvae migrate from the new host's digestive tract to its thoracic muscles, where they undergo further development. After a certain period, the larvae move to the proboscis of the mosquito. The proboscis consists of the upper lip or labium, the mandibles, maxillae, and thehypopharynx, which are long and sharp, modified for piercing man's skin and sucking his blood. The immaturefilariaeare then introduced into another human whom the mosquito bites. Inside this primary host, they are carried by the circula-tory system to the lymph glands, where they grow to adults and reach sexual maturity.

Question:

At 90\textdegreeC, benzene has a vapor pressure of 1022 Torr, and toluene has a vapor pressure of 406 Torr. From this information, calculate the composition of the benzene- toluene solution that will boil at 1 atm pressure and 90\textdegreeC, assuming that the solution is ideal.

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Solution:

A solution boils when the sum of the partial pressures of the components becomes equal to the applied pressure (i.e. total pressure). To solve this problem, one must realize that the applied pressure is atmospheric pressure, 760 Torr. One can determine the partial pressure of benzene (P_a\textdegree) and of toluene (P_b\textdegree) by using Raouit's law. Raoult's law states that the partial pressure p of a gas is equal to its vapor pressure P\textdegree times its mole fraction X in the solution. p = p\textdegreeX Dalton's law of partial pressure states that the sum of the partial pressures of a system is equal to the total pressure of the system. For this particular benzene- toluene solution to boil the total pressure of the system must equal atmospheric pressure 760 Torr. If one lets the mole fraction of benzene equal X_a, then the mole fraction toluene is equal to 1 -X_a. One determines the partial pressures by substituting into Raoult's law p_a = p_a\textdegree(x_a ) P_b = P_b\textdegree(1 - x_a) Then, substituting into Dalton's law: 760 Torr = P_a + P_b 760 Torr = 1022 X_a + 406 (1 - X_a) = 1022 X_a + 406 - 406 X_a X_a = 0.574 The mole fraction of benzene in the liquid is 0.574 and the mole fraction of toluene is (1 -0.574 ) = 0.426.

Question:

Two cells, one of emf 1.2 V and internal resistance 0.5 \Omega, the other of emf 2 V and internal resistance 0.1 \Omega, are connected in parallel as shown in the figure and the combination connected in series with an external resistance of 5 \Omega . What current passes through this external resistance?

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Solution:

The diagram is labeled and current values have been inserted in each part of the circuit. Apply-ing Kirchhoff's current law to point A, we have I_1 + I_2 = I_3(1) Note that in the figure, boxes have been drawn representing the 2 batteries in the circuit. The given voltages, \epsilon_1 and \epsilon_2, are the voltages of the batteries assuming zero internal resistance. (The terminal volt-ages are V_BC and V_FE , not \epsilon_1 and \epsilon_2 .) Applying Kirchhoff's voltage law to the closed circuit containing both cells, and then to the closed circuit through the lower cell and the external re-sistance (see figure for the directions in which the loops are traversed) we have + \epsilon_2 - \epsilon_1 + (.5 \Omega) I_1 - (.1 \Omega) (I_2 ) = 0 and+ \epsilon_2 - (5 \Omega) (I_3) - (.1 \Omega) (I_2 ) = 0 Hence,\epsilon_2 - \epsilon_1 = (2 - 1.2)V = (.1 \Omega) (I_2 ) - (.5 \Omega) (I_1 ) \epsilon_2 = 2V = (5 \Omega) (I_3 ) + (.1 \Omega) (I_2 ) or, upon multiplication of both sides of each equation by 10, 8 V = (1 \Omega) (I_2 ) - (5 \Omega) (I_1 )(2) 20 V = (1 \Omega) (I_2 ) + (50 \Omega) (I_3 )(3) Substituting equation (1) in (3) 20 V = (1 \Omega) (I_2 ) + (50 \Omega) (I_1 + I_2) 20 V = (51 \Omega) (I_2 ) + (50 \Omega) I_1(4) Multiplying (2) by 10, 80 V = (10 \Omega) (I_2 ) - (50 \Omega) (I_1 )(5) Adding (4) and (5), we may solve for I_2 100 V = (61 \Omega) I_2 andI_2 = [(100 V)/(61 \Omega)] = 1.64 A(6) since 1 ampere = 1 volt/ohm. Substituting (6) in (3), we obtain I_3, 20 V = (1 \Omega) (1.64 A) + (50 \Omega) (I_3 ) [(18.46 V)/(50 \Omega)] = I_3 orI_3 \approx .37 A .

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Question:

Find the first five energyeigenvaluesfor a hydrogen atom. (An equation in which an operation on a function gives a constant times the function is called aneigenvalueequation. The function is called aneigenfunctionof the operator and the constant is called aneigenvalue. TheSchroedinger equation haseigenfunctionswhich are the possible spatial portions of the wave functions andeigenvalueswhich are the possible values of the energies involved.)

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Solution:

Quantum mechanics extends classical mechanics into microscopic phenomena by taking quantum considerations into account.Schroedinger developed a wave equation des-cribing both the wave and matter aspects of matter. To find the energy values of theeigenstatesof hydrogen, consider theSchroedingerequation expressed in spherical co-ordinates . \psi(r, \texttheta, \textphi) = R (r) y (\texttheta, \textphi). In general, the total energy of the hydrogen electron is due to its angular momentum (rotation with radius r), its linear momentum along the variable radius r, and its potential energy due only to its distance r from the nucleus. TheSchroedingerequation for the radial part R(r) of the wavefunctioncan be obtained at once as follows. The angu-lar momentum equals L =mvr. The rotational energy of the particle at radius r is (1/2)mv^2 = (1/2)m[(L/mr)]^2 = (L^2/2mr^2). Rotational energy, the angular motion of the particle, is related to the y(\texttheta, \textphi) part of thewavefunction. The angular momentum of the electron in a stable state around the nucleus (energyeigenstate) is quantized as L^2 = l (l + 1)h^2l = 0,1,2,... Wherehis Planck's constant divided by 2\pi. As far as the radial motion along r is concerned, the contribution of the rotational motion to the total energy of the particle, [{L(L + 1)h^2}/2mr^2] , depends only on the radial position r. Potential energy can be viewed as any energy which de-pends on a distance r and reduces to zero at infinite dis-tance. This permits us to consider the energy due to rota-tion as a form of potential energy. Hence, the one- dimensional radial motion takes place in an effective potential (important since it is expressed in terms of only one variable r). U (r) = [L(L + 1)h^2)/2m_er^2] + v (r)(1) where V(r) is the electrostatic potential. The radial part of theSchroedingerequation is given by - (h^2/2m_e) (1/r^2) (d/dr)(r^2 R') + U (r) R (r) = ER (r)(2) where E is the total energy (eigenvalue) corresponding to an orbit (eigenstate) of radius r and R' = [{dr(r)}/dr]. For the lowest energy states, let l equal zero. Also, for an electron a distance r from a nucleus of charge Ze, the potential energy is U (r) = [(-KZe^2)/r] = (-Ke^2/r)(3) where K is the Coulomb constant and Z = 1 for hydrogen. We therefore have (-h^2/2m_e) (1/r^2) (d/dr)(r^2R') - (Ke^2/r)R = ER.(4) Let\lambda^2 = (-2m_eE/h^2).(5) We assume \lambda^2 is positive since the energy E must be negative if the electron is bound to the nucleus. Multiplying (4) by \lambda^2/E we have (1/r^2) (d/dr)(r^2R') + [(2m_eKe^2)/h^2] (R/r) = \lambda^2R.(6) A solution of Eq. (6) has to approach zero as r approaches \infty since the energy is zero at infinite distance. Rewriting the equation, R\textquotedblright + [(2R/r) + {(2m_eKe^2)/h^2} (R/r)] = \lambda^2R.(7) For large r, the terms in parentheses are insignificant and we have R" = \lambda^2R(8) which is satisfied by R \approx e-\lambdar(9) which goes to zero when r, the distance of the electron from the nucleus, is infinite. This solution will be true for all r if the terms in parentheses in (7) cancel each other. To find the value of \lambda which makes this true, we set these terms equal to zero. (2R'/r) + [(2m_eKe^2)/h^2] (R/r) = 0. We know that R' = -\lambdae^-\lambdar= -\lambdaR. Therefore we have (-2\lambdaR/r) + [(2m_eKe^2)/h^2] (R/r) = 0(10) and \lambda = [(m_eKe^2)/h^2] = (1/0.529 A)(11) sinceh^2/m_eKe2 \approx 0.529 A is the Bohr radius. Therefore this value of \lambda corresponds to the ground energy level (n = 1). This minimum energy is found from (5) E_1 = (-h^2\lambda^2/2m_e) = - 13.6eV. The energyeigenvaluescorresponding to theeigenfunctionsinvolved in the well-behaved solutions to the radial equa-tion (1) are E_n = (E_1/n^2)(12) where n is the quantum number. Therefore E_1 = - 13.6eV E_2 = [(- 13.6eV)/(2)^2] = -3.4eV E_3 = [(- 13.6eV)/(3)^2] = -1.51eV E_4 = [(- 13.6eV)/(4)^2] = -0.85eV E_5 = [(- 13.6eV)/(5)^2] = -0.54eV.

Question:

A drag racer achieves an acceleration of 32 (mi/hr)/sec. Compare this value with g.

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Solution:

To remain consistent with the units in which g is given, we convert to m/sec^2. a= 32 [mi/(hr-sec)] = 32(mi/hr) × (1/sec) × (5280 ft/1 mi) × (1 hr/3600 sec) = 32 × (5280/3600) (ft/sec^2) = 1.46 × 32 (ft/sec^2) = 1.46 g This acceleration is about the maximum that can be achieved by a vehicle that travels on wheels and depends on the friction between the wheels and the road for its thrust. A vehicle moves according to Newton's third law of motion: Every action has an equal but opposite re-action. The wheels of a car exert a force on the road in the backward direction. The reaction force is the force which acts in the direction opposite to the friction on the wheels, pushing the car forward. There is a maximum reaction force that the road can exert on the wheels, limited by the coefficient of friction which depends on the smoothness of the road, and by the weight of the car. This maximum thrust which the frictional force can exert on the car, limits its maximum accelera-tion. Attempts to surpass this maximum value by using a more powerful engine will result merely in spinning tires. (Rocket-powered cars and sleds can, of course, achieve much greater accelerations.)

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Question:

Find the coefficient \sigma of surface tension of a liquid if it rises to a height h = 32.6 mm. in a capillary of dia-meter D = 1 mm. The density of the liquid is \delta = 1 gr/cm^3. The contact angle of the surface film is zero.

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Solution:

The surface force acting along the circumference of the water surface in the tube (as shown in the figure) supports the weight of the water column. Surface force is given by S = \sigma\piD. Hydrostatic equilibrium requires that S = W, \sigma\piD = g × mass = g\delta × volume = g\deltah\pi D^2/4. where g is the acceleration due to gravity. The coefficient of surface tension is \sigma= (1/4) g \delta h D = (1/4) × (980 cm/S^2) × (1 gr/cm^3) × (3.26 cm) × (10^-1 cm) = 80.4 dyne/cm = 80.4 dyne/cm

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Question:

Find the displacement ratio of a screw jack (Fig.) whose threads have a pitch p and whose handle has a length R. (b) If p = 0.15 in, R = 18 ft, and the jack has an efficiency of 30 percent, find the force needed to lift a load of 3300 lb.

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Solution:

(a) A screw is a cylinder with an inclined plane wrapped around it. The distance between two adjacent threads is called the pitch (p) of the screw, as shown in the figure. As the handle is turned through one complete revolution, the weight moves through distance p. At the same time, the man's hand moves a distance 2\piR. The displacement ratio DR, is the distance the man's hand moves divided by the resultant displacement of the screw: DR = (2\piR)/p = [(2\pi) (1.8 ft) (12 in/ft)] / [0.15 in] = 144 (b) The efficiency is defined as the ratio of work output to work input. The work output for the screwjack is equal to the product of the weight W and the distance it is moved. The work input is the force F the man exerts multiplied by the distance (2\piR) through which he moves the handle. For a displacement p of the weight, the efficiency is e = Wp/[F (2\piR)] Substituting the known values, we find the force the man exerts to be F = Wp/[e(2\piR)] = [(3300 lb)(0.15 in)] / [(0.30) (2\pi) (1.8 ft) (12 in/ft)] = 77 lb.

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Question:

An object is 4 inches from a concave lens whose focal length is -12 inches. Where will the image be?

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Solution:

It may be useful to construct a ray-diagram first (see diagram). We draw two rays - one parallel to the axis and one through the center of the lens. From the diagram it can be seen that the image is virtual and that it is smaller than the object. We now attempt a mathematical solution: D_0 is 4 inches and f is - 12 inches. Substituting in 1/d_0 + 1/D_I = 1/f 1/4 in. + 1/D_I = 1/ -12 in. Solving DI= - 3in. Hence DI= -3 in. which means that since D_I is negative, the image is 3 inches from the lens on the same side on the lens as the object and is virtual.

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Question:

A person has two strains of mice which are genetically different. How can he successfully perform a skin graft on a mouse of one strain using skin from a mouse of the other strain?

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Solution:

The answer to this question will show how immunological capabilities develop. Since the two strains are genetically different, they should have very few histocompatibility genes in common. Therefore any grafting attempt should be unsuccessful, since the recipient mouse's cellular immunity would respond to the antigens on the donor's tissue cells. The solution to the problem is based on an observation made during the 1940's. It was discovered that non-identical twin calves each had the antigens normal to his own red blood cells and also the antigens of his twin's red blood cells. (The antigens on red blood cells determine the blood type.) Each calf, moreover, was not destroying the antigens of its twin. It was shown that the circulatory systems of the twins had been interconnected during their early embryonic develop-ment. Apparently, by being exposed to the other twin's antigens early in the calf's fetal development, it later recognizes these antigens as self. The immunological mechanisms had not developed at the time the antigens were first shared. This lack of immunological responsiveness to antigens present when the antibody-forming system is being developed is called immunological tolerance. We can thus perform successful grafting experiments on the mice by exposing a mouse of one strain to the antigens of a mouse from the other strain. If an early embryo were injected with cells from a mouse of the other strain, the newborn mouse would accept grafts from that same mouse. Because the cells were injected before development of the antibody-forming system (which can last for several weeks after birth), the adult mouse will not form antibodies against the antigen of those cells. According to one view, if developing lymphocytes encounter their corresponding antigen at some critical stage in their development, they fail to mature and die. This not only explains why the mouse does not have anti-bodies against the antigens of the donor's tissue, but also why the mouse does not have antibodies complementary to its own antigens. Any lymphocyte which would recog-nize "self" antigens would be destroyed during fetal development. Lymphocytes would mature normally if their antigen were absent from the system. These mature lymphocytes are then able to recognize nonself, or foreign antigens. Another method of increasing acceptance of tissue grafts is to remove the thymus from the newborn mice. The thymus is the organ where T lymphocytes, responsible for cellular immunity involved in transplants, become differen-tiated. These so-called "nude" mice, which lack thymuses, are incapable of producing T lymphocytes. Although they die within a few months, many of them can accept tissue grafts from different strains of mice, and even rats.

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Question:

The entire power from a 100-hp automobile engine is used to agitate 50 kg of water thermally insulated from its surroundings. How long will it take for the temperature of the water to rise 10 Celsius degrees?

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Solution:

Since 1 hp = 746 watts (W), the power avail-able is 7.46 × 10^4 W. All this power is turned to heat in the agitation of the water. A watt equals a joule/sec. Therefore the rate at which heat is supplied to the water is [(7.46 × 10^4 joules/sec) / (4.186 joules/sec)] = 1.782 × 10^4 cal/sec. In time \cyrchar\cyrt the heat supplied is thus 1.782 × 10^4 cal/s × \cyrchar\cyrt. The temperature rise takes place according to the following equation, the specific heat capacity of water being assumed constant and of value 1 cal/g\bullet C deg over the range of temperature considered: Q = mc(t_2 - t_1), the symbols having their usual significance. Thus 1.782 \cyrchar\cyrt × 10^4 cal/s = 5 × 10^4 g × 1 cal/g\bulletCdeg × 10 C deg or\cyrchar\cyrt = [50 / 1.782] s = 28 s.

Question:

Complete the program defined earlier in the previous three problems by writing the PROCEDURE DIVISION. The trans-action records for this program each contain a TRANSACTION- CODE based on the type of transaction. A code of 1 indi-cates a new customer's record is to be added to the file. A code of 2 indicates a charge to the customer's account, and a 3 indicates a credit, either from payment or return. The program must specify different paths for each trans-action. In coding your solution, write the three routines ADD-RECORD, ADD-CHARGE, and SUBTRACT-CREDIT first. Then write the basic program branching to and from routines as necessary. Any time an INVALID KEY clause is activated while a record is being replaced in the file, display "INVALID KEY", C-NR, then transfer control to READ- UPDATE- INFO. Refer back to the Data Division as needed.

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Solution:

The complete procedure division is shown in fig. 1.In the OPENING paragraph Indexed-file is opened as I/O , allowing us to write data to and from it. The statement is a conditional Go To statement. In this pro-gram paragraph 1 (ADD RECORD) is specified when TRANSACTION- CODE is a 1, paragraph 2 (ADD-CHARGE) is specified when TRANSACTION-CODE is a 2, etc. The statements after INVALID KEY are executed whenever there is an error while writing or reading from an indexed file. The error messages are printed using the ERROR-LISTING PARAGRAPH, which displays the error message and the record which contained the error.

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Question:

Write a function to evaluate f(x) = e ^\rule{1em}{1pt}(x)2 (1 - x^3 /3). Show calculation of f(3).

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Solution:

The simplest type of function is the so-called statement function. Making use of this concept, we can immediately write: F(x) = EXP(\rule{1em}{1pt}X{_\ast}X){_\ast} (1. \rule{1em}{1pt} (X{_\ast}{_\ast}3)/3.) Y = F(3.0)

Question:

The cells of humans and most animals have become adapted to surviving only within a relatively small range of hydrogen ion concentrations. What is the role of the kidneys in regulating hydrogen ion concentration?

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Solution:

Most metabolic reactions are sensitive to the hydrogen ion concentration of the fluid in which they occur, due primarily to the marked influence of protons on enzyme function. Accordingly, the hydrogen concentration of the extracellular fluid is one of the most critically and delicately regulated chemical levels in the entire body. The kidneys regulate the hydrogen ion concentration of the extracellular fluids by excreting either acidic or basic constituents when the levels deviate from normal. This regulation is achieved by exchanging hydrogen ions for sodium ions in the tubular fluid. Sodium ions usually combine with bicarbonate ions in the tubular fluid to form sodium bicarbonate. However, when excess hydrogen ions are secreted into the urine, they also combine with the bi-carbonate there, essentially replacing the sodium ions to form carbonic acid. The carbonic acid formed dissociates to produce CO_2 and water: H^+ + HCO_3^-\rightleftarrowsH_2CO_3\rightleftarrowsCO_2+ H_2O bicarbonatecarbonic acidbicarbonate The CO_2 is reabsorbed into the blood from the tubules and transported to the lungs where it is eliminated during expiration. The decrease in the concentration of bicarbonate ions in the tubular fluid represents a net excretion of hydrogen ions. Usually the amount of hydrogen ions ex-changed for sodium in the distal convoluted tubules is equivalent to the amount of bicarbonate ions in the tubular fluid. If the plasma becomes very acidic, the quantity of bicarbonate ions in the kidney filtrate is insufficient to react with the abnormally high quantity of hydrogen ions. When this happens, the hydrogen ions combine with other buffers, in the tubular fluid, such as phosphate, and are excreted. If the extracellular fluids become too alkaline, the amount of bicarbonate in theglomerularfiltrate becomes greater than the amount of hydrogen ions secreted by the tubules. Theunreactedbicarbonate is simply ex-creted as sodium bicarbonate. The loss of sodium bi-carbonate makes the body fluid more acidic, returning the acid-base balance to normal. In addition to the bicarbonate technique, the kidney has another mechanism for coping with excessive acidity which involves the secretion of ammonia by the kidney tubules. Should the hydrogen ion concentration in the filtrate be too great for the bicarbonate to handle, ammonia would be secreted into the tubules. Here, the ammonia combines with hydrogen ions to produce ammonium ions (NH_4^+), which are then excreted. By combining with hydrogen ions, the ammonia effectively buffers the acidic tubular fluid by removing the hydrogen ions from the fluid. The more hydrogen ions there are in the tubule, the greater is the conversion of ammonia to ammonium ions.

Question:

Calculate the hydrogen ion (H^+) concentrations of the following fluids: (a) blood plasma, (b) intracellular fluid of muscle, (c) gastric juice (pH - 1.4), (d) tomato juice, (e) grapefruit juice, (f) sea water. Use the accompanying table. pH of some fluids pH Seawater 7.0 - 7.5 Blood plasma 7.4 Interstitial fluid 7.4 Intracellular fluid: Muscle 6.1 Liver 6.9 Gastric juice 1.2 - 3.0 Pancreatic juice 7.8 - 8.0 Saliva 6.35 - 6.85 Cow's milk 6.6 Urine 5 - 8 Tomato juice 4.3 Grapefruit juice 3.2 Soft drink (cola) 2.8 Lemon Juice 2.3

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Solution:

pH is defined as the logarithm (log_10) of the reciprocal of the proton concentration, or log (1/[H^+]). This expression can be rewritten a -log[H^+]. Rearranging the equation pH = -log [H^+] gives us the means to solve for [H^+] : [H^+] = 10^-pH (a) To find the [H^+] of blood plasma: The pH of blood plasma is given in the table to be 7.4.After inserting this into the equation, the [H^+] is 10^-7.4 or 3.98 × 10^-8 Molar. (b) [H^+] in intracellular fluid of muscle: The pH of intracellular fluid of muscle = 6.1 \therefore [H^+] = 10^-PH = 10^-6.1 = 7.94 × 10^-7 M. (c) H+ in gastric juice pH of gastric juice = 1.4 \therefore [H^+] = 10^-1.4 = 3.98 × 10^-2 M. (d) [H^+] in tomato juice pH of tomato juice = 4.3 \therefore[H^+] = 10^-4.3 = 5.01 × 10^-5 M. (e) [H^+] in grapefruit juice pH of grapefruit juice = 3.2 \therefore [H^+] = 10^-3.2 = 6.31 × 10^-4 M. (f) [H^+] in seawater pH of seawater = 7.0 [H^+] = 10^-7 = 1.0 × 10^-7 M.

Question:

Using the following table of bond energies, calculate the energy change in the following: (a) 2H_2 (g) + O_2(g) \rightarrow 2H_2O(g) (b) CH_4 (g) + 2O_2(g) \rightarrow CO_2 (g) + 2H_2O (g) (c) CH_4 (g) + Cl_2 (g) \rightarrow CH_3CI (g) + HCl(g) (d) C_2H_6 (g) + Cl_2 (g) \rightarrow C_2H_5CI(g) + HCl(g). Bond Energies (in kcal per mole) H\rule{1em}{1pt}H 104 C\rule{1em}{1pt}O 83 H\rule{1em}{1pt}F 135 C=O 178 H\rule{1em}{1pt}CI 103 C\rule{1em}{1pt}CI 79 H\rule{1em}{1pt}Br 88 C\rule{1em}{1pt}F 105 H\rule{1em}{1pt}I 71 SI\rule{1em}{1pt}O 106 Li \rule{1em}{1pt}H 58 SI\rule{1em}{1pt}F 136 CI\rule{1em}{1pt}CI 58 C\rule{1em}{1pt}C 83 C\rule{1em}{1pt}H 87 C=C 146 O\rule{1em}{1pt}H 111 C\equivC 199 O=O 118 N\equivN 225 P\rule{1em}{1pt}CI 78

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Solution:

When using the bond energies to calculate the net energy change in a reaction, the net energy change is equal to the total bond energy of the bonds formed sub-tracted from the bond energy of bonds broken. This method is illustrated in the following examples. (a) 2H_2(g) + O_2 (g) + 2H_2O(g) There are two H_2 bonds and 1 O_2 bond broken and 4 O-H bonds formed here. 2H\rule{1em}{1pt}H(g) + O = O(g) \rightarrow 2 H\rule{1em}{1pt}O\rule{1em}{1pt}H(g) net bond energy = (2× bond energy of H\rule{1em}{1pt}H) + (1 × bond energy of O=O) \rule{1em}{1pt} (4 x bond energy of O\rule{1em}{1pt}H) The following bond energies can be found in the table. Bond Bond energy (Kcal/mole) H\rule{1em}{1pt}H 104 O=O 118 O\rule{1em}{1pt}H 111 The net energy or heat evolved in the reaction can be found using these values. net bond energy= [(2 × 104) + (1 × 118)] \rule{1em}{1pt} (4 × 111) ∆H= (208 + 118) \textemdash 444 = \textemdash118 Kcal. Thus when this reaction occurs 118 Kcal are released. The following examples will be solved in a similar manner. (b) CH_4 (g) + 2O_2 \rightarrow CO_2 (g) + 2H_2O(g) This reaction can be rewritten (g) + 2 0=0 (g) \rightarrow O=C=O (g) + 2H\rule{1em}{1pt}O\rule{1em}{1pt}H(g) This means that there are 4 C\rule{1em}{1pt}H bonds and 2 O=O bonds broken and 2 O=C bonds and 4 O\rule{1em}{1pt}H bonds formed. net energy = [(4 × bond energy of C\rule{1em}{1pt}H) + (2 × bond energy of O=O)] \rule{1em}{1pt} [(2 × bond energy of C=O) + (4 × bond energy of O\rule{1em}{1pt}H)] The following bond energies are used. Bond Bond energy (Kcal/mole) O\rule{1em}{1pt}H 111 O=C 178 O=O 118 C\rule{1em}{1pt}H 87 One can now determine the heat evolved in this re-action. net energy = [(4 × 87) + (2 × 118)] - [(2 × 178) + 4 × 111]. ∆H = \rule{1em}{1pt} 216 Kcal. c) CH_4 (g) + Cl_2 (g) \rightarrow CH_3CI (g) + HCl (g) This equation can be rewritten H_3C\rule{1em}{1pt}H (g) + Cl\rule{1em}{1pt}Cl (g) \rightarrow H_3C\rule{1em}{1pt}Cl (g) + H-Cl (g) This means that there is 1 C\rule{1em}{1pt}H and 1 Cl\rule{1em}{1pt}Cl bond broken and 1 C\rule{1em}{1pt}Cl and 1 H\rule{1em}{1pt}Cl bond formed. net energy = (1 × bond energy of C\rule{1em}{1pt}H + 1 × bond energy of Cl\rule{1em}{1pt}Cl) \rule{1em}{1pt} (1 × bond energy of C\rule{1em}{1pt}Cl + 1 × bond energy of H\rule{1em}{1pt}Cl) The following bond energies are used. Bond Bond energy (Kcal/mole) C \rule{1em}{1pt} H 87 Cl \rule{1em}{1pt} Cl 58 C \rule{1em}{1pt} Cl 79 H \rule{1em}{1pt} Cl 103 The heat evolved can now be determined. net energy= ∆H = [(1 × 87) + (1 × 58)] \rule{1em}{1pt} [(1 × 79) + (1 × 103)] = (87 + 58) - (79 + 103) = \rule{1em}{1pt} 37 Kcal. d) C_2 H_6 (g) + Cl_2 (g) \rightarrow C_2H_5CI (g) + HCl (g) This can be rewritten as Thus, there is 1 C\rule{1em}{1pt}H and 1 Cl\rule{1em}{1pt}Cl bond broken and 1 C\rule{1em}{1pt}Cl and 1 H\rule{1em}{1pt}Cl bond formed. The net energy for the reaction can be determined by using the following values. Bond Bond energy (Kcal/mole) C \rule{1em}{1pt} H 87 Cl - Cl 58 C - Cl 79 H \rule{1em}{1pt} Cl 103 The net energy can now be found, net energy= ∆H = [(1 × 87) + (1 × 58)] - [(1 × 79) + (1 × 103)] = (87 + 58) - (79 + 103) = \rule{1em}{1pt} 37 Kcal.

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Question:

A factory chimney of height h = 50 m carries off smoke at a temperature T1= 60\textdegree C. Find the static pressure ∆P producing the draught in the chimney. The air temperature is T_0 = 0\textdegree C. The density of air is d_0 = 1.29 × 10^-3 g cm-3at 0\textdegree C.

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Solution:

Let us assume that initially the chimney has no smoke. The pressure A (see the figure) is the sum of the air pressure P_B at B plus the pressure at A due to the air column in the chimney P_A = P_B + gd_0h where g is the gravitational acceleration, and d_0 is the density of air at T_0 = 0\textdegreeC. When hot smoke fills the chimney, the temperature inside becomes T_1. We can take the density of the smoke air mixture as approximately equal to that of the air d1at T_1. The pressure at A in this case, is P'_A = P_B + gd_1h. Right before A, where hot air is being produced, pres-sure is equal to the air pressure P_A. Since P'_A < P_A, hot air is pushed upward by the difference ∆P = P_A - P'_A= gh(d_0 - d_1). We can arrive at this result in a simpler manner if we realize that the warm air in the chimney is less dense than the surrounding cold air and is in touch with the cold air at the two ends of the chimney. Hence, as a result of Archimedes' principle for fluids, hot air will be buoyed up with a force equal to the weight of the cold air displaced by the warm air. The net force ∆F acting on the warm air column therefore equals the difference between this buoyant force and the weight of the warm air ∆F = gd_0V - gd_1V = gAh(d_0 - d_1). The resulting draught pressure ∆P is ∆P = ∆F/A = gh(d_0 - d_1). Let us assume that air can be thought of as an ideal gas. We do not expect the pressure inside the chimney to differ appreciably from the outside pressure since it is mainly the higher temperature that determines the volume of the warm air: V \approx Constant × temperature. Under the assumption that pressure is approximately constant, the volume of a gas at two different temperatures T_1 and T_0 obeys the relation V_0/V_1 = T_0/T_1. The mass of air in both cases is the same Mass = V_0d_0 = V_1d_1. giving V_0/V_1 = d_1/d0 The density of warm air at T_1\textdegree C is related to the density of air at 0\textdegreeC by d_1 = d_0(T_0/T_1). The final expression for ∆P is therefore ∆P = ghd_0[1 - (d_1/d_0)] = ghd_0[1 - (T_0/T_1)]. or= (980 cm/sec^2) × (50 m × 10^2 cm/m) × (1.29 × 10^-3 gm/cm^3) × [1 - 273\textdegreeK/(273\textdegree + 60\textdegree)K] = 1.14 × 10^3 dynes/cm^2.

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Question:

How does a plasma cell make a specific antibody when stimulated by a specific antigen?

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Solution:

We still do not know the exact mechanism of antibody stimulation, although several theories have been proposed. One theory widely accepted some thirty to forty years ago was called the instructive theory of antibody formation. According to this theory, the antigen is needed to supply the information necessary for synthesis of the specific antibody. It was thought that the antigen com-bines with a newly synthesized antibody chain before it folds into a particular three-dimensional form. The developing antibody molecule would fold around the antigen, creating a complementary shape at the antibody's binding sites. The specificity of the antibody was then assumed to be due to the spatial configuration of its chains and not to the amino acid sequence of the antibody. This theory was questioned when it was discovered that dif-ferent antibodies have different amino acid sequences. Moreover, if we simultaneously inject several different antigens into an animal, we find that each plasma cell produces only one specific antibody. According to the instructive theory, if several different antigens enter a particular plasma cell, we would expect that cell to produce different antibodies (since each antigen acts as a template for antibody formation). This, however, is not the case. It has now been determined that the presence of an antigen simply increases the number of plasma cells which produce the corresponding antibody specific for that antigen. The presence of the antigen only stimulates division and differentiation of those lymphocytes which synthesize antibodies specific for that antigen. This theory thus assumes that there pre-exists a very large number of differentiated small lymphocytes already "programmed" genetically to produce one specific type of antibody. The sole function of the antigen is to stimulate the division of those lymphocytes which will produce the corresponding antibody. This generally accepted, theory is called the theory of clonal selection. A clone is a group of identical cells all descended from a common ancestral cell. How does a particular lymphocyte recognize the antigen that its antibodies are specific for? Both B and T lymphocytes have immunoglobulin molecules on the surface of their plasma membranes. These antibodies are positioned in such a way that the binding sites face outward and can thus bind antigens. Each particular lymphocyte contains only one type of surface antibody; when the antibody inter-acts with the corresponding antigen, the lymphocyte is stimulated to proliferate and produce more of this specific antibody. The exact mechanism of how the antigen stimulates division is still unknown. The clonal selection theory of antibody formation can be outlined as follows:

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Question:

A silver wire has a resistance of 1.25 ohms at 0\textdegreeC and a temperature coefficient of resistance of 0.00375/C\textdegree . To what temperature must the wire be raised to double the resistance?

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Solution:

The change in resistanceR_t- R_0 is directly proportional to the change in temperature and the original resistance R_0. It is known from experiment that ∆R = \alphaR_0 ∆t . Substituting values, we have ∆t = [(∆R)/(\alphaR_0 )] = [(R_t- R_0 )/(\alphaR_0 )] t - 0\textdegree C = [{(2.50 - 1.25) ohms}/(0.00375/C\textdegree × 1.25 ohms)] sinceR_t= 2R_0 = 2(1.25) = 2.50 t = 266\textdegreeC .

Question:

What evolutionary advanced features are present in Selaginella but not in the ferns?

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Solution:

The genusSelaginellahas four advanced features which are not found in most of the other lower vascular plants, including the ferns. First, vessels which are missing in all but two genera of ferns are found in the xylem of many species ofSelaginella. Second, theSelaginella gametophyte is greatly dependent on thesporophyte, it is highly reduced in size and complexity, and itdevelopesentirely within the spore cell. The fern gametophyte, though also small and inconspicuous, is, however, a photosynthetic, independent organism. Third,Selaginellahas acquiredheterospory, an evolutionary advanced feature of higher plants. In contrast to the ferns which are homosporous ,Selaginellabears two types of sporangia and gives rise to two types of spores. As their names imply, themegasporanqiumforms larger spores calledmagaspores, and the micro-sporangium forms smaller spores called microspores. Last, theSelaginellaembryo, unlike that of the ferns, is equipped with a suspensor, afootlikestructure that grows into the surrounding gametophytic tissue to absorb food. A similar, analogous structure by the same name is found in the gymnosperm embryo. These four advanced features of theSelaginelladiscussed above suggest thatSelaginellamay have evolved in a closer line with the higher plants than have the ferns.

Question:

One liter of an ideal gas under a pressure of 1 atm is ex-panded isothermally until its volume is doubled. It is then compressed to its original volume at constant pressure and further compressed isothermally to its original pressure. Plot the process on a p-V diagram and calculate the total work done on the gas. If 50 J of heat were removed during the constant-pressure process, what would be the total change in internal energy?

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Solution:

During an isothermal change, T is constant. The work done on the gas in such a change is W = -^Vf\intVip dV. This is negative because the pressure on v_i the gas is in opposition to the volume change. From the ideal gas law, pV = nRT, we get W = -^Vf\int_Vi (nRT/V) dV = -nRT ^Vf\intVidV/V = -nRT ln (V_f/V_i) = -p_iV_i 1n(V_f/V_i) = -p_iV_i 1n(p_i/p_f). where the subscripts i and f refer to initial and final states respectively. Thus the work done on the gas in the first change is [from (1) to (2), as shown in the figure] W_1 = -p_1V11n(V_2/V_1) = -(1 atm ×1.013 × 10^6 dynes/cm^2 \bullet atm) × (1 liter × 10^3 cm^3/lit)×1n 2 = - 1.013 × 10^6 dynes/cm^2 × 10^3 cm^3 × 1n 2 = - 7.022 × 10^8 ergs = - 70.22 J. Further, since the volume is doubled, by the application of Boyle's law p_1V_1 =p_2V_2 we see that the pressure is halved at (2) p_2 = p_1 (V_1/V_2) = (1/2)p_1. The work done on the gas in the second change is [from (2) to (3)] W_2 = -p_2 ^V3\int_V2 dV = -p_2(V_3 - V_2) = (1/2)p_1(V_1 - 2V_1) = (1/2)p_1V_1 = (1.013 × 10^6 dynes/cm^2 × 10^3 cm^3) / (2 x 10^7ergs/J) = 50.65 J. The work done on the gas in the final change from (3) to (4) is W_3 = -p_3V_3 1n (p_3/p_4) = -(1/2)p_1v_1 1n(P_2/P_1) = (1/2)p_1V_1 1n (1/2) = (1/2) p_1V_1 1n 2 = [1.013 × 10^6 dynes/cm^2 × 10^3 cm^3 × in(2)] / [2 x 10^7ergs/J] = 35.11 J The total work done on the gas is thus W1+ W_2 + W_3 = (50.65 + 35.11 - 70.22)J = 15.54 J. In the first and third processes the temperature does not change. In an ideal gas the internal energy depends only on the temperature, so that no change of internal energy takes place in the first and third processes. Any work done on the gas in these changes is equal to the heat transfer taking place. The second process is isobaric. The change in internal energy during the process is given by the first law of thermodynamics as ∆U = Q - W, where Q is the heat energy added to the system and W the work done by the system. Hence ∆U = -50 J - (-50.65 J) = +0.65 J. The internal energy thus increases by 0.65 J during the three processes.

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Question:

Trace the evolution of man using the existing classifications and characteristics of fossil men.

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Solution:

The fossils of the Dryopithecine date back to about 30 million years ago during the oligocene era. It is believed that there once existed several species of Dryopithecus: Dryopithecus africanus which was thought to be ancestral to the chimpanzee, Dryopithecus major - thought to be ancestral to the gorilla, and Dryopithecus sivolensis thought to be ancestral to the present day orangutan. Although not enough information has been collected to tell exactly when and where the human lineage first started, it is believed that it may have had its roots in the Dryopithe-cine line. The apelike Dryopithecus was probably the ancestor of man. By the middle of Miocene (about 20 million years ago), the evolutionary lines leading to the various types of modern anthropoids were distinct. The first set of fossils found to be considered in the hominid line, representing Ramapithecus, dates back to the upper Miocene epoch (about 10 to 14 million years ago). It must be pointed out, however that this is supported only by vague evidence since only the jaws and teeth from Ramapithecus have been found. The canine and incisor teeth are relatively small in man, and this may be an important distinction from the apes. If Ramapithecus did not use his teeth for defense and his diet changed to one that required more chewing, then this change may have been correlated with the increased use of hands. Since no fossil skull of Ramapithecus has been found, little can be learned about his intelligence. If he did reply more on his hands, there is the implication that he might stand at least semi-erect and use bipedalism at times. A gap of about five million years in man's evolutio-nary history is found between Ramapithecus and the next distinct homonid form, called Australapithecus, who appeared about 5.5 to 1.2 million years ago. Relatively large portions and numbers of the Australopithecine skeleton have been found scattered throughout South and East Africa. An important advance toward the modern human condition shown by the Australopithecines is the acquisition of some degree of bipedalism. They might have stood and walked completely erect or perhaps they were fully erect only while running. From fossil evidence it is deduced that Australopithecus was relatively small, an adult weighing about 75 pounds. His dental anatomy was humanlike, especially that of the canines. His cranial capacity was very small and comparable to that of a present day gorilla. It must however be noted here that although the Australopithecine cranial capacity is comparable to a gorilla's, the ratio of his brain size to body weight is much higher that the gorilla's proportion. Australopithecus probably was not very intelligent but was obviously developing human characteristics and some form of culture. He probably lived in a group and had a daily hunting-gathering mode of living. This is supported by the finding of bones of small animals and evidence of gathered fruits, nuts, and shoots at or around excavated sites. Stone tools are found also at these sites and are of a style either known as pebble tools or as the Oldowan industry. Therefore, Australopithecus was a toolmaker as well as a tool user, a feature much less common among animals other than man himself. A later stage in human evolution is represented by a group of fossil men, including Java and Peking man, that can be classified as Homo erectus. This species first appeared about 1.2 million years ago and existed till about years ago. As compared to Australopithecus, Homo erectus had greater body size, increased cranial capacity, reduced facial shape and dental structure, and the beginning of a human type culture. A major feature found in Homo erectus is the elaboration of cultural adaptations that originated in the earlier Australopithecine stage. The cultural associations of Homo erectus, in the making of tools (called Acheulian tools) and in social activities, are more complex than those associated with the Australopithecines. By the Homo erectus stage of human evolution, adap-tations based on culture are seemingly well established. If culture is the basis of human adaptation, then natural selection would favor a more efficient culture. In these fossil men, an enlarged brain is related to an increased capacity for more complex cultural skills and behavior patterns. Reduction in facial and dental structure also may be linked to more efficient use of tools and cooking of food. Homo erectus appear to have led the same kind of life, including the hunting and gathering that characterizes some groups of modern man. Because of their similarities to modern men, these fossil men are included within the genus Homo. The ancestral line, or lines, leading to modern man becomes hazy approximately 300,000 years ago. From around to 150,000 years ago, the fossil record is rather scanty. However, a collection of fossils found in Terra Amata, France, which is about 200,000 to 35,000 years old, gives us a decent picture of man's transition from Homo erectus to Homo sapiens. The structures of these skeletal remains clearly show an increase in brain size, as well as the increasing likeness of facial features to those of modern man. The Homo sapiens neanderthalensis, known as the cavemen, is a hominid group with skulls that were large and massive with a thick, bony ridge over the eyes and a receeding forehead. Their nose was broad and short and there was almost no chin at all. Their cranial capacity was as large as, or larger than that of modern man. In general, these characteristics are intermediate between those of Homo erectus and those of Homo sapiens. The population of Neanderthal men that inhabited regions that are now Europe, North Africa and the Middle East some 75,000 to 35,000 years ago seemed to be both culturally and physically modern. They were a widely spreading species - their remains have been found over a wide area including parts of western Europe, Asia and Africa, indicating that they were able to adapt to various climates. They were a society of small family groups who lived primarily in caves, used fire, made flint weapons of the Mousterian tool tradition, hunted a variety of game ranging in size up to the mammoth and rhinoceros, and buried their dead re-verently with food, flowers and ornaments. This last mentio-ned fact indicates that they were capable of abstract thinking including the concept of a life after death. Reconstruction of one Neanderthal skeleton caused these men to be depicted as standing stooped, with their knees bent forward. However, it has been determined later that this portrayal of Neanderthal man is incorrect. The skeleton used to form this illustration actually come from a man who died at a very old age with rickets, which accounts for the skeleton's deformation. It is now known through examination of many other skeletons that Neanderthal man stood as erect as modern man and was probably just as efficient a bipedal walker. Presently, Neanderthal man is considered to belong to a sub-species of Homo sapiens.

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Question:

In a pound-for-pound comparison, which would you prefer as a rocket fuel, hydrogen (H2) or dimethylhydrazine 2 (CH_3)_2NNH2? H_2(g) + (1/2)O_2 (g) \rightarrow H2O(l), 2 2 O(l), ∆H= - 68.3Kcal/mole (CH_3)2NNH_2 + 4O_2 (g)\rightarrow N_2(g) + H 2 \rightarrow N_2 4H_2O(g) + 2CO_2(g), ∆H = - 404.9 Kcal/mole.

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Solution:

The material that is the better fuel will release more energy per unit weight in its reaction with oxygen. Calculate the energy released per unit weight for H_2 and (CH_3)_2NNH_2. You can do this by using the heats of formation and enthalpy changes given. Calculate the total mass or weight of materials from moles that react and divide the ∆H by it to obtain the amount of energy re-leased per unit weight. For hydrogen: 1 mole of H_2 reacts with (1/2) mole of oxygen (1 mole H_2) (1.008 g H_2/mole H_2) + ((1/2) mole O_2) (16.00 g O_2/mole O_2) = 1.008 g + 8.00 g = 9.01 g. ∆H = - 68.3 Kcal. Thus, (-68.3) / (9.01) = - 7.58 Kcal/g. For dimethylhydrazine: [1 mole (CH_3)_2NNH_2] [60.10 g (CH_3)_2NNH_2/mole (CH_3)_2NNH_2] + (4 moles O_2)(16.00 g. O_2 /mole O_2) = 60.10 g + 64.00 g = 124.10 g. ∆H = - 404.9. The amount of heat released per unit weight with (CH_3)_2NNH_2 + O_2 = (- 404.9 Kcal) / (124.10 g) = - 3.26 kcal/g. Hydrogen yields over twice as much energy per gram of fuel mass, and is thus preferable in this respect.

Question:

The constant b in vanderWaals' equation for helium is 23.4 cm^3 \textbullet mole^-1. Obtain an estimate of the diameter of a helium molecule.

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Solution:

At low densities, real gases obey the ideal gas law PV =\muRT. when the density of a real gas increases, we can no longer neglect the fact that the molecules occupy a fraction of the volume available to the gas and that the range of molecular forces is greater than the diameters of the molecules. Vander Waals developed an equation to take these factors into account. Assume the molecules have a diameter d. Then they can't approach within a distance d/2 of the walls and a distance d from the center of another molecule. Therefore, the volume available to the molecules is less than the volume V of the container that is used in the ideal gas law. The free volume per mole is less than the "geometric" volume per mole, V/\mu, by a volume b. Modi-fying the ideal gas law, p(v - b) = RT where v is the "geometric" volume per mole, V/\mu. To account for the intermolecular forces, we note that they vary as the square of the number of particles per unit volume. In-versely, they then vary as the square of the volume per mole, as (1/v^2). The gas acts as though it experienced a pressure in excess of that applied externally. This pres-sure is equal to a constant a times (1/v^2). The excess pressure (a/v)^2 causes the gas to occupy less volume than if it was ideal. Further modifying the ideal gas law gives vanderWaals equation of state of a gas, [p + (a/v^2)](v - b) = RT. The equation linking the constant b in vanderWaals equation to the molecular diameteris b = (2/3)N_0\pid^3. \therefored^3 = 3b/(2\piN_0) = (3 × 23.4 cm3\bullet mole^-1)/(2\pi × 6.02 × 10^23 mole^-1) or d = ^3\surd[(3 × 23.4)/(2\pi × 6.02 × 10^23)cm] = 2.65 A.

Question:

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Solution:

To begin evaluating trees, we look for terminal nodes, those which contain numerical values or variables at the end of the tree. Then, the rule becomes "left node, right node, operator." Starting at the leftmost node, we begin combining operands with their operators. The tree be-gins to. lose its leaves as the process of combination con-tinues. Let us evaluate the first tree, combining nodes as we go along. Starting with the leftmost node, we see that /./ is to be divided by 4.0. After evaluation, we get the following tree: Note that the left side is now complete. Turning to the right side, we may combine 5.1 and 3.7 under the subtraction operator to obtain 1.4. A new node is created containing 1.4 because of this operation. Moving up the tree from the right, we can combine 2.4 and 1.4 under the addition operator to obtain 3.8. The tree now has the following shape: To complete it, we simply perform the multiplication to obtain the final answer 1.045. This tree introduces the \uparrow operator, which signifies the exponential operation. Starting once again from the left, we have 5.0 raised to the 1^st power, which yields 5.0. Then, we move up the tree to find that the next task to be done is to subtract 2.2 from 5.0, which yields 2.8. At this stage, the tree is We now start with the right side to find that we must add 3.9 to 1.8, giving us the answer 5.7. We take this answer and move up to the next node, which indicates that we are to subtract 5.7 from 2.9. After this operation, we have which evaluates to zero.

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Question:

A red signal flare for use on railroads is made by mixing strontium nitrate, carbon, and sulfur in the proportions as shown by the equation 2Sr (NO_3)_2 + 3C + 2S \ding{217} Complete and balance the equation (no atmospheric oxygen is necessary).

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Solution:

Until the beginning of the twentieth century, large amounts of KNO3 (saltpeter) were used in the pre-paration of black gunpowder. This powder as prepared by mixing powdered saltpeter, carbon, and sulfur. When ignited, sulfur and carbon are very rapidly oxidized by KNO_3, pro-ducing several gaseous products. 4KNO_3 + S + 4C \ding{217} 2K_2O + 2N_2 + SO_2 + 4CO_2 + heat This equation can be applied to solve this problem. The only difference is that a red flare usesSr(NO_3)_2 to give it a distinctive color and is mixed in different proportions with carbon and sulfur.Sr(NO_3)_2 also oxidizes carbon and sulfur to form the same gaseous products as black gunpowder. Thus, the products for the oxidation of carbon and sulfur are CO2 and SO_2 and the product of reduction is N_2 (going from N^+5 in NO^-_3 to N0 in N_2). The complete reaction is: 2Sr (NO_3)_2 + 3C + 2S \ding{217} 2SrO + 2N_2 + 3CO_2 + 2SO_2.

Question:

Find the pH of a 0.2 M solution of formic acid. K_a = 1.76 × 10^-4.

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Solution:

To solve the problem, first note that the acid dissociates according to the general equation HA+H_2O\rightleftarrowsH^++A^-, where HA is the acid and H^+ and A^- are the dissociation products. To find the pH, calculate [H^+]. One can see from the equation that equal con-centrations of H^+ and A^- form when HA dissociates. At equilibrium, x moles of HA dissociate into x moles of H^+ and x moles of A^-. HA+H_2O\rightleftarrowsH^++A^- Before:0.2OO After:0.2-xxx Now, set up the equilibrium constant equation, which is K_a = [products] / [reactants]= {[H+][A^-]} / [HA] = 1.76 × 10^-4 Before substituting in the values, simplify the calculation by assuming that 0.2 - x \allequal 0.2. This is a valid assumption since the acid is very weak and only a very small percentage dissociates. Thus, one has K_a = {(x)(x)} / (0.2) = 1.76 × 10^-4 . Solving: x = [H^+] = 5.93 × 10^-3. H^+ concentration is then converted to pH according to the equation, pH = - log [H^+] = - log 5.93 ×10^-3 = - [- 3 + log 5.93] = 3 - 0.77 = 2.23.

Question:

What are the properties of actin and myosin which produce the cyclic activity of the cross bridges responsible for contraction? What causes rigor mortis?

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Solution:

Myosin, the larger of the two molecules, is shaped like a lollypop (see Figure 1). The myosin molecules are arranged within the thick filaments so that they are oriented tail-to-tail in the two halves of the filament; the globular ends extend to the sides, forming the cross briges which bind to the reactive site on the actin molecule. Actin is a globular-shaped molecule having a reactive site on its surface that is able to combine with myosin. These globular proteins are arranged in two chains which are The globular end of the myosin molecule, in addition to being a binding site for the actin molecule, contains a separate binding site for ATP. This active site has ATPase activity, and the reaction that is catalyzed is the hydrolysis of ATP: H_2O + ATP \textemdash\textemdash> ADP + Pi However, myosin alone has a very low ATPase activity. It appears that an allosteric change occurs in the active site of myosin ATPase when the myosin cross bridge combines with actin in the thin filaments, considerably increasing the ATPase activity. The energy that is released from the splitting of ATP produces cross bridge movement by an as yet unknown mechanism. It is believed that the oscillatory movements of myosin cross bridges produce the relative movement of thick and thin filaments, resulting ultimately in the shortening of a muscle fiber (see Figure 3). Since many cycles of activity are needed to produce the degree of shortening observed during muscle contrac-tion, the myosin bridge must be able to detach from the actin and then rebind again. This is accomplished by the binding of ATP to the myosin in the cross bridge, forming what is known as a low-energy complex. The low-energy complex has only a weak affinity for actin; the actin-myosin bond is broken, allowing the cross bridges to dis-sociate from actin. Shortly after this event, a conformational change occurs in the myosin - ATP complex and a high energy complex is formed. The high energy complex has a very high affinity for actin, and the cross bridges are able to rebind to the actin. In this manner, the cross bridges are able to bind and dissociate from actin in a cycle of coordinated actions. This cycle may be summarized in the following sequence of events: A = actinM= myosin M-ATP\textemdash\textemdash\textemdash\textemdash>M^\textasteriskcentered - ATP (low-energy complex)(high-energy complex capable of binding action) A + M^\textasteriskcentered -ATP\textemdash\textemdash\textemdash\textemdash>A - M^\textasteriskcentered -ATP (with actin bound, myosin is able to split ATP) A - M^\textasteriskcentered -ATP \textemdash\textemdash\textemdash\textemdash>A - M + ADP + Pi (as ATP is split, cross bridge movement occurs) A - M + ATP \textemdash\textemdash\textemdash\textemdash>A + M-ATP (low-energy complex dissociates from actin) Rigor mortis is a phenomenon in which the muscles of the body become very stiff and rigid after death. It results directly from the loss of ATP in the dead muscle cells; the myosin crossbridges are unable to combine with actin and those bonds already formed cannot be broken - thus the rigid condition. At the molecular level, we can identify two specific roles for ATP: 1) to provide energy for movement of cross bridge, and 2) to dissociate actin from the myosin cross bridges during the contraction cycle of the bridges. ATP is also needed to restore Ca^+ in the sarcoplasmic reti-culum following contraction (see next question). Two regulatory proteins, troponin and tropomyosin, are associated with actin. During nervous stimulation of a muscle, there is an increase in free intracellular calcium ions: calcium diffuses in from the terminal cisternae and from the extracellular fluid in the T tubules. Calcium binds to troponin which causes tropomyosin to shift its position along the actin helix. This exposes the binding site on ac-tin for myosin.

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Question:

A chemist mixes equal volumes of .01M Na_2C_2O_4 and .001M BaCl_2together. Assuming theK_spof BaC_2O_4 is 1.2 × 10-^7 , will a precipitate form?

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Solution:

A precipitate will only form when the ion concentration product for the particular substances exceed their givenK_sp. In other words, a precipitate will form if [Ba^2+] [C_2O_4^2-] is greater than theK_sp= 1.2 × 10-^7 . Thus, to obtain this ion product, calculate the concentrations of Ba^2+ and C_2O_4^2- in the mixture. One is given that the initial concentrations of C_2O_4^2- and Ba^2+ in the solution are .01M Na_2C_2O_4 and .001M BaCl_2 . When they are mixed , however , the volume double . Thus, in the mixture, the concentrations of BaCl_2 and Na_2C_2O_4 are halved. (Remember, M = moles/liter , so that if the volume increases, M = concentration decreases). Therefore, after mixing, the concentrations are .005M of Na_2C_2O_4 and .0005M of BaCl_2 . As such, the ion product of [Ba^2+] [C_2O_4^2-] = (.0005) (.005) = 2.5 × 10-^6 . 2.5 × 10-^6 is greater than theK_spof 1.2 × 10-^7 , which indicates that a precipitate will form.

Question:

"Vascular plants are like animals because they both have circulatory systems." Using your knowledge of the vascular system of plants, discuss why the statement is basically incorrect.

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Solution:

The vascular system of a plant and the circulatory system of an animal are similar only in the sense that both systems have a common function of trans-portation of food and inorganic salts. The differences between these two systems are really more important than their similarity. Water moves through the xylem of a plant in a continuous stream and is constantly lost to the atmosphere as water vapor; this loss of water is called transpiration. Blood circulates inside an animal's body but, unless hemorrhage occurs, it is not continuously lost to the outside. It is true that under high temperature the body tends to lose water and salts to the outside through pores in the skin and through the urine, but this is a much slower process than transpiration. Compared to the 100 gallons of water lost by a large tree in a day, the sweat excreted by an animal is negligible. Moreover, the mechanism of flow in the vascular system is not comparable to the one found in animals. In plants, there are no pumps to drive the fluid along the vessels; in animals, a heart or analogous organs are always present to pump blood throughout the body. In addition, while the circulatory system of higher animals is designed to deliver oxygen to all parts of the body, only a very limited amount of oxygen moves through the vascular system of plants. Instead, oxygen enters the plant body by diffusion through the specialized pores or thin surfaces of the plant. Further-more, the physical relationship between the xylem and phloem is fundamentally different from that between arteries and veins. Whereas arteries are connected to the veins by networks of capillaries, so that arterial blood is continuous with venous blood, the xylem and phloem are basically independent of each other, and very few substances pass from one to the other. Finally, while in the circulatory system of animals a large tube branches successively into smaller and smaller tubules, all xylem and phloem tubes are small and do not vary in their radii during their courses. They occur in bundles and it is the number of tubes per bundle which changes: in the lower part of the stem there are many per bundle; in the upper part there are fewer per bundle, some having entered the branches of the stems.

Question:

What is the relation between the total energy and the angular momentum for a 2-body system, each body exe- cuting a circular orbit about the system center of mass?

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Solution:

In order to solve this problem we must transform the given 2-body problem to an equivalent one-body problem. To do this, we find the equation of motion of each mass shown in the figure. Using Newton's Second Law and his Law of Universal Gravitation, we obtain: F1 2^\ding{217} = m_2r_2^\ding{217} = [Gm_1m_2(r_1^\ding{217} - r_2^\ding{217})]/[\vert r_1^\ding{217} - r_2^\ding{217}\vert^3](1) F2 1^\ding{217} = m_1r_1^\ding{217} = - [{Gm_1m_2(r_1^\ding{217} - r_2^\ding{217})} / {\vert r_1^\ding{217} - r_2^\ding{217}\vert^3}](2) where F_12^\ding{217} is the force exerted on 2 by 1, and similar-ly for F_21^\ding{217}. Rewriting r_2^\ding{217} = (1/m_2)[{Gm_1m_2(r_1^\ding{217} - r_2^\ding{217})} / {\vertr_1^\ding{217} - r_2^\ding{217}\vert^3} ](3) r_1^\ding{217} = -(1/m_1)[{Gm_1m_2(r_1^\ding{217} - r_2^\ding{217})} / {\vertr_1^\ding{217} - r_2^\ding{217}\vert^3}](4) Subtracting (3) from (4), and using the figure to realize that r_1^\ding{217} - r_2^\ding{217} = r^\ding{217}, we obtain r_1^\ding{217} - r_2^\ding{217} = r^\ding{217}= -(1/m_1) [(Gm_1m_2r^\ding{217})/(r^3)] - (1/m_2) [(Gm_1m_2r^\ding{217})/(r^3)] or r^\ding{217} = -[(Gm_1m_2r^\ding{217})/(r^3)] [(1/m_1) + (1/m_2)] Defining the reduced mass \mu as 1/\mu = (1/m_1) + (1/m_2)(5) we find \mur^\ding{217} = -[(Gm_1m_2r^\ding{217})/(r^3)](6) This equation is a one body equation describing the motion of a particle of mass \mu under the influence of a gravitational force. Now, to further reduce the problem, assume that each mass shown in the figure rotates in a circular orbit with the given angular velocities about Q, the center of mass. Using Newton's Second Law for each mass, (m_1v ׳ 1 2 )/(r ׳ 1 ) = (Gm_1m_2)/(r^2)(7) (m_1v_2 ׳ 2 )/(r_2 ׳ ) = (Gm_1m_2)/(r^2)(8) where the primed variables are measured with respect to the point Q. By definition of the center of mass m_1r_1 ׳ \ding{217} + m_2r_2 ׳ \ding{217} = 0 orr_1 ׳ \ding{217} = -(m_2r_2 ׳ \ding{217} )/(m_1)(9) Furthermore,r\ding{217}= r_1 ׳ \ding{217} - r_2 ׳ \ding{217} orr_2 ׳ \ding{217} = r_1 ׳ \ding{217} - r^\ding{217}(10) Inserting (10) in (9), and solving for r_1 ׳ \ding{217} r_1 ׳ \ding{217} = - [(m_2)/(m_1)] r_2 ׳ \ding{217} = -[(m_2)/(m_1)]( r_1 ׳ \ding{217} - r^\ding{217}) r_1 ׳ \ding{217} [1 + (m_2)/(m_1)] = [(m_2)/(m_1)] r^\ding{217} r_1 ׳ \ding{217} = [m_2/m_1r^\ding{217}] / [1 + (m_2)/(m_1)] = [m_2r^\ding{217}] / [m_1 + m_2](11) But using (5) \mu = (m_1m_2)/(m_1 + m_2) Hence, (9) becomes r_1 ׳ \ding{217} = (\mu/m_1) r^\ding{217}(12) Similarly,r_2 ׳ \ding{217} = -(\mu/m_2) r^\ding{217}(13) Hencer_1 ׳ \ding{217} = (\mu/m_1) r r_2 ׳ \ding{217} = (\mu/m_2) r(14) Using (14) in (7) and (8) (m^2_1v_1 ׳ 2 )/\mur = (Gm_1m_2)/(r^2) (m^2_2v_2 ׳ 2 )/\mur = (Gm_1m_2)/(r^2) or(m_1v_1 ׳ 2 )/2= (\muGm_2)/(2r) (m_2v_2 ׳ 2 )/2= (\muGm_1)/(2r) Therefore, the net kinetic energy of the system relative to the center of the mass is T = (1/2) m_iv_1 ׳ 2 + (1/2) m_2v_2 ׳ 2 = [(\muG)/(2r)] (m_1+ m_2)(15) T = (Gm_1m_2)/(2r)(16) by definition of \mu. The total energy is E = T + V = (Gm_1m_2)/(2r) - (Gm_1m_2)/(r) E = - [(Gm_1m_2)/(r)](17) To remove the variable r, we replace it with the angular momentum J^\ding{217} as follows. The total system angular momentum is (relative to Q) J^\ding{217} = r_1 ׳ \ding{217} × m_1v_1 ׳ \ding{217} + r_2 ׳ \ding{217} × m_2v_2 ׳ \ding{217} Since r_1'^\ding{217} and v_i'^\ding{217} are perpendicular, and similarly for r_2 ׳ \ding{217} and v_2'^\ding{217}, we obtain J = m_1r_1 ׳ v_1 ׳ + m_2r_2 ׳ v_2 ׳ (18) From (12) and (13) v_1'^\ding{217} = (\mu/m_1) v^\ding{217}r_1'^\ding{217} = (\mu/m_1) r^\ding{217} v_2'^\ding{217} = -(\mu/m_2) v^\ding{217}r_2'^\ding{217} = -(\mu/m_2) r^\ding{217} orv_1' = (\mu/m_1) v^\ding{217}r_1'= (\mu/m_1) r^\ding{217} v_2' = -(\mu/m_2) v^\ding{217}r_2'= -(\mu/m_2) r^\ding{217}(19) using (19) in (18) J = (m_1)[(\mu/m_1) r] [(\mu/m_1) v] + (m_2)[(\mu/m_2) r] [(\mu/m_2) v] J = (\mu^2/m_1) vr + (\mu^2/m_2) vr = \mu^2 vr[(1/(m_1) + (1/m_2)] By definition of \mu J = \muvr(20) We now eliminate v in (20) so that we may sub-stitute (20) in place of r in (17). We know, from (15) that T = (1/2) m_1v_1^'2 + (1/2) m_2v_2^'2 = (\muG/2r) (m_1 + m_2)(15) Substituting forv_1', v_2'from (19) in (15), Substituting for from (19) in (15), T = (1/2) m_1- [(\mu^2/m_1^2)v^2] + (1/2) m_2 [(\mu^2/m_2^2)v^2] = (\muG/2r) (m_1 + m_2) T = [(\mu^2v^2)/(2m_1)] + [(\mu^2v^2)/(2m_2)] = (\muG/2r) (m_1 + m_2) T = [(\mu^2v^2)/(2)][(1/m_1)+(1/m_2)] = (\muG/2r) (m_1 + m_2) By definition of \mu T = [(\muv^2)/(2)] = (\muG/2r) (m_1 + m_2)(21) T = [(\muv^2)/(2)] = (\muG/2r) (m_1 + m_2)(21) Hencev^2 = (G/r)(m_1 + m_2)(22) Using (22) in (20) J = \mur \surd[{G(m_1 + m_2)}/r] orJ^2 = \mu^2 r G(m_1 + m_2) Thereforer = (J^2) / [\mu^2G(m_1 + m_2)](23) Inserting (23) in (17) E = -[(Gm_1m_2)/2] [ {(\mu^2G(m_1 + m_2)} / J^2] E = -[{G^2 m_1m_2(m_1 + m_2) \mu^2} / 2J^2] Finally, m_1m_2(m_1 + m_2)\mu^2 = m_1m_2(m_1 + m_2) [(m_1m_2) / (m_1 + m_2)] \mu orm_1m_2(m_1 + m_2)\mu^2 = m^2_1m^2_2 \mu ThenE = - [(G^2 \mu m^2_1m^2_2)/(2 J^2)]

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Question:

In the trans form of nitrous acid, all the atoms are in the same plane, but the HO is directed away from the other O. The O-H distance is 0.098 nm; the distance from the central N to the hydroxyl O is 0.146 nm; and the distance to the other O is 0.120 nm. If the H-O-N bond angle is 105\textdegree and the O-N-O bond angle is 118\textdegree, how far is the H from the other O?

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Solution:

The formula for nitrous acid is HNO 2 .Thestructure of the trans The structure of the trans form as described in the problem is: One is asked to find the distance from O_2 to H. This can be done by solving for the length of O_2C and CH. The sum of the squares of these two lengths is equal to the square of the length of O_2H. This can be done through the use of trigonometry. To find O_2C, one must find NA and AC. O_2C is equal to the sum of O_2N, NA, and AC. The measure of angle ONA is equal to 180\textdegree - 118\textdegree or 62\textdegree, since they are supplementary. cos 60o= 0.469 = NA / NO_1 = NA / 0.146 nm [adjacent /hypotenuse] 0.469= (NA / 0.146 nm) NA= 0.069 nm O_2A= NA + O_2N = 0.069 nm + 0.120 nm = 0.189 nm AC=O_1B, since they are the opposite sides of a rectangle. \angle BO_1H = 90o - \angle HO_1A(\angle = angle) \angle NO_1A = 90o - \angleO_1NA = 90o - 62o = 28o \angle HO_1A = \angle HO_1A - \angle NO_1A = 105o - 28o = 77o \angle BO_1H = 90o - 77o = 13o from above. cos 13o =OB/OH= 0.974 (O_1B/O_1H) = (O_1B/ 0.098 nm) = 0.974 O_1B= (0.098 nm) × (0.974) = 0.95 nm Thus,AC= 0.095 nm. Solving forO_2C: O_2C=O_2N+NA+AC = 0.120 nm + 0.069 nm + 0.095 nm = 0.284 nm HC=OA-BH sin \angle ONA= (O_1A/NO_1)\angle ONA = 62o sin 62o= (O_1A/NO_1) = 0.833 (O_1A/NO_1)=0.883 OA= (0.883) ×NO_1= (0.883) × (0.146 nm) = 1.129 nm sin \angle BOH= (BH/O_1H)\angle BOH = 13o sin 13o= (BH/ 0.098 nm) = 0.225 BH= (0.098 nm) × (0.225) = 0.22 nm HC= 0.129 nm - 0.022 nm = 0.107 nm (O_2H)^2= (O_2C)^2 + (HC)^2 (O_2H)^2= (0.284 nm)^2 + (0.107 nm)^2 (O_2H)^2= 0.081 nm^2 + 0.011 nm^2 = 0.092 nm^2 (O_2H)^2= 0.303 nm.

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Question:

An explorer, lost in. the desert, stumbled on an abandoned automobile whose gas tank still held 5 liters of fuel. Gasoline has a density of 0.67 g / cc. Assuming gasoline to be composed of C_7H_16 molecules, how many liters of water could he produce by burning the fuel and trapping the products ? The reaction is C_7H_16(l) + O_2(g) \ding{217} CO_2(g) + H_2O(g)

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/Users/wenhuchen/Documents/Crawler/Chemistry/E27-0906.htm

Solution:

There is one mole of H_2O produced by each mole of C_7H_16 burned. To find the amount of water formed by burning the gasoline determine the number of moles of fuel present. Once the number of moles formed is determined , the number of grams of H_2O can be found. There are 5 liters of gasoline present. It is given that 1 cc of C_7H16 weighs 0.67 g. In 1 liter there are 1,000 cc, therefore the gasoline weighs 5000 cc × 0.67 g / cc = 3350 g One finds the number of moles present by dividing the total weight by the molecular weight. (MW of C_7H_16 = 100). moles= [3350 g / (100 g / mole)] = 33.50 moles. Because 33.50 moles of C_7H_16 are present, 33.50molesof H_2O can be formed. The number of grams of H_2O formed is found by multiplying the molecular weight of H_2O by the number of moles present (MW of H_2O = 18). grams = 33.50 moles × 18.0 g / mole = 603.0 g.

Question:

Write the structure for methyl- 6- chloro- 5,8- diethyl- 8- iodo- 3,4,4- trimethyl- 2,6- decadienoate.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E20-0748.htm

Solution:

To draw this structure one should follow certain rules. (1) Identify and draw the parent name of the compound. The parent is decane, (2) Add the suffix functionality. Note that in this compound two different types of functional groups are designated in the suffix endings, double bonds and an ester. Start with the double bonds: -2, 6-decadienoate. Now consider the ester group.The carbonyl (C = O) of the ester must be at carbon number 1. The alkyl portion of the ester precedes the name (and the prefixes) as a separate word. In this case it is methyl (CH_3) so that the structure becomes (3) Add the prefix functionality in the order given, 6-chloro-5,8- diethyl-8 iodo-3,4,4-trimethyl-. Next, fill in the hydrogens, and the structure is complete.

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Question:

The SO_2 content in air may be determined by passing the air through water to produce a solution of sulfurous acid, H_2SO_3. The subsequent reaction of this solution with potassium permanganate, KMnO_4, establishes the amount of SO_2 absorbed from the air. (a) Write the balanced equation for the re-action of SO_2 with water to produce H_2SO_3. (b) Write the balanced equation for the reaction of H_2SO_3 (or SO^2-_3) with KMnO_4 (or MnO^-_4) to form MnSO_4, K+ and H_2SO_4, among other products, (c) If 1000 liters of air are passed through water and the resulting solution requires 2.5 × 10^-5 moles of KMnO^5 for complete reaction (part b) , what is the con-centration (in ppm) of SO2 in the air. Assume STP conditions.

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Solution:

(a) An SO_2 molecule reacts with a water molecule to form an H_2SO_3 molecule according .to the following equation: SO_2 + H_2O + H_2SO3 However, H_2SO_3 is a diprotic acid and dissociates into 2 protons and a sulfate ion HO_2SO_3 \rightarrow 2H^+ + SO_3^= Thus, for every mole of SO_2 consumed 1 mole of SO_3 = are produced. (b) This reaction is a typical redox reaction and the equation must be balanced using the standard rules. From the problem, one knows initially that H_2SO_3 + KMnO_4 \rightarrowMnSO_4 + H_2SO_4 + K^+ This reaction is not balanced in terms of charge and elemental constituents. The first rule is to separate the above reaction into a reduction and an oxidation reaction. red:KMnO_4 \rightarrow MnSO_4 + K^+ Mn is reduced from +7 to +2 state. ox:H_2SO_3 \rightarrow H_2SO4 S is oxidized from +4 to +6 state. Next one balances the oxygen and hydrogen atoms in these reactions using OH^-, H^+, and H_2O molecules on either side of the equation depending upon whether the reaction is acidic or basic. Once the masses are balanced, electrons are used to balance charges and are placed on the appropri-ate side of the equation. Thus, red:KMnO_4 \rightarrow MnSO_4 + K+ The potassium ion and the SO_4 ion do not play significant roles in the balancing of this equation and are therefore unnecessary. MnO^-_4 \rightarrow Mn^+2 This reaction is acidic and therefore will be using H^+ and H_2O. 8H++ MnO^-_4 \rightarrow Mn^+2 + 4H_2O The H2O is placed on the right to balance the in MnO^-_4; the H in H_2O is balanced on the left with H^+. equation will be balanced in terms of elements and when electrons are added. Thus, 8H^+ + 5e-+ MnO^-_4 \rightarrow Mn^++ + 4H_2O ox:H_2SO_3 \rightarrow H_2SO4 the hydrogen ions are unnecessary and play insignificant roles in balancing this equation; therefore, SO_3=\rightarrow SO_4^= Next place water on the left and balance using a sufficient number of hydrogen ions on the right. H_2O + SO_3^= \rightarrow SO_4^= + 2H^+ Now balance off the charge using electrons: H_2O + SO_3^= \rightarrow SO_4^= + 2H++ 2e- Once both oxidation and reduction processes are balanced, they are added together in the following manner. red:[8H^+ + 5e-+ MnO^-_4 \rightarrow Mn^++ + 4H_2O] × 2 ox:[H_2O + SO_3^= \rightarrow SO_4^= + 2H++ 2e^-] × 5 These reactions are solved simultaneously to eliminate the electrons in both reactions. Thus, 16H^+ + 2MnO^-_4 \rightarrow 2Mn^++ + 8H_20 5H_2O + 5S03^=\rightarrow 5SO_4^= + 10H^+ Total:6H^+ + 2MnO^-_4 + 5SO_3^= \rightarrow 2Mn^++ + 5SO_4=+ 3H_2O (c) To solve this part of the problem, one must know the number of moles of SO_2 in air is equal to the number of moles of SO_3^= that are represented in the equation for the reaction with KMnO_4. This amount is determined from the stoichiometry. For every mole of SO_2 consumed, 1 mole of SO_3^= is formed, and for every 1 mole of SO_3^= consumed, 2/5 mole of MnO^-_4 is also consumed. Thus, number moles of SO_2 = 5/2 (2.5 × 10^-5) = 6.25 × 10^-5 moles To obtain a concentration in units of ppm (parts per million), one needs to know the volume of SO_2 consumed and the total volume of air. The total volume of air is 1 × 10^3 liters. The volume of SO_2 consumed is volume of SO_2 = (6.25 × 10^-5 moles)(22.4 liter/mole) = 1.4 × 10^-3 l, since, at STP, one mole of gas will occupy 22.4 liters. Thus, the ratio of SO_2 volume to total volume of air is (1.4 × 10^-3 liter) / (1 × 103liter) = 1.4 × 10^-6. To convert this ratio into ppm, we make the denominator 10^6 liters, (1.4 × 10^-3 liter) / (1 × 103liter) × (10^3/10^3) = (1.4) / (1 × 10^6). The concentration of SO_2 in the air is 1.4 ppm.

Question:

What are the three fundamental logical structures used in computer programming? Give a brief example of each by drawing flowcharts.

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Solution:

The three cornerstones of structured programming are: 1) Simple sequencing 2) Decision-making 3) Repetition or looping By implementing these concepts in the form of distinct modules, you will find that your program can be understood readily and, if problems occur, you will be able to debug your program with greater ease. The looping construct is also known as a DO-WHILE structure. The looping continues while some condition remains satisfied. When the condition is violated, the loop terminates.

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Question:

Write in FORTRAN a complex function to find the complex numberof larg-est magnitude in an array of complex numbers.

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Solution:

Our function will make use of a complex arrayAdimension-ed size N. In addition, the complex absolute value library function is used. Otherwise no new concepts are involved. Let z = x +iywherex,yare real numbers and i2 = -1. The modulus or absolutevalue of z is given by \vertz\vert = \surd(x2 + y2) . Since x, y are real, there are an infinite number of complex numbersqfa givenmagnitude. COMPLEX FUNCTIONMAX(N,A) COMPLEXA(N) MAX =A(1) IF (N. EQ. l) RETURN DO 7 I = 1,N IF (ABS(A(I)).LE.ABS(MAX)) GO TO 7 MAX =A(I) 7CONTINUE RETURN END

Question:

Debug the following FORTRAN program for finding the roots of a quadratic equation: $ JOB CPROGRAM TO CALCULATE ROOTS OF QUADRATIC EQN. CASSUME POSITIVE DISCRIMINANT CREAD INPUT DATA 1)READ (5, 100), A, B, C 2)100FORMAT (3E12. 5) CCALCULATE ROOTS 3)DISC = (B\textasteriskcentered\textasteriskcentered2 - 4. \textasteriskcenteredA\textasteriskcenteredC)\textasteriskcentered\textasteriskcentered. 5 4)X1 = (- B + DISH) / 2. \textasteriskcenteredA 5)X2 = (- B - DISH) / 2. \textasteriskcenteredA CWRITE OUTPUT 6)WRITE (6, 300) A, B, C, X1, X2 7)200FORMAT (5 (5X, E12. 5)) 8)STOP 9)END

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Solution:

The two most common kinds of errors that a programmer can commit are 1) syntax errors. 2) logical errors. Syntax errors are detected by the compiler when translating from the source language to the machine language. Thus the data will not be processed and no results will be obtained. On the other hand, if a program is free of syntax errors but contains logical errors, the machine will process the program. The output obtained will be an incorrect solution to the problem the program was designed to solve. Going through the program, note the following syntax errors: Statement (1) contains an extra comma after (5,100). Statement (4) defines the discriminant as DISH instead of DISC (presumably what the programmer meant.) However, this is not recognized as an error by the compiler, which interprets DISH as a correct variable name. Statement (6) is the WRITE statement with the format label 300, while statement (7) is the corresponding format statement labelled 200. This error can be corrected by changing the value of either of those two labels to be identical to the other one. Statement (7) has the word FORMAT incorrectly spelled as FORNAT, which the compiler cannot decode. If any of these errors remain uncorrected (except for the one in statement (4)) the compiler cannot finish its task of translating the FORTRAN program into the machine language. Assuming there are no syntax errors, the program will be processed by the compiler and passed to memory from where the central proces-sing unit will manipulate the data according to the program instruc-tions. It will then transmit the results of these manipulations to the external output device. If logical errors are present, the output is suspect. The given program contains the following logical errors: (4)DISH has not been defined anywhere. Thus XI cannot be evaluated. (5)Note that according to the statement X2 = ((-B - DISC) / 2.)\textasteriskcenteredA This gives the wrong result since we want X2 = (-B - DISC) / (2.\textasteriskcenteredA). This error can be easily detected after a few sample runs of the program. However, if the calculated roots formed part of the input to a minor subroutine in a huge program, extensive logical debugging would be required to locate the error.

Question:

The drive shaft of an automobile rotates at 3600 rev/min and transmits 80 hp from the engine to the rear wheels. Compute the torque developed by the engine.

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Solution:

By definition, the infinitesimal amount of work done by the torque \Gamma (provided by the drive shaft) in turning the wheel through an infinitesimal angular dis-placement ∆\texttheta is ∆W = \Gamma ∆\texttheta Dividing both sides by ∆t ∆W/∆t = \Gamma (∆\texttheta/∆t) Taking the limit as ∆t \rightarrow 0 dW /dt= \Gamma \omega where \omega is the angular velocity of the wheel. But power P is defined as P =dW/dt= \Gamma \omega Hence\Gamma = P/\omega = (80 hp)/(3600 rev/min) But 1 hp = 550ft\bulletlb/s and 1 rev/min = (2 \pirad)/(60 s) and\Gamma = [(8) (550)ft\bulletlb/s] / [{(3600)(2\pi)rad} / {60 s}] = 117lb\bulletft \Gamma = P/\omega = [(44,000ft\bulletlb/sec) / (120\pirad/sec)] = 117 lb\bullet ft.

Question:

Discuss the fullword multiplication (M or MR) and the divi-sion (D or DR) instructions of IBM/360-370 computers. Can these two instructions be said to have anything in common?

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Solution:

The Multiplication Instruction (M or MR): Both types of this instruction have a register for the first operand. But the M-instruction is of register to in-dex (RX) format and has a fullword as its second operand. In contrast, the MR instruction is of register to register (RR) format and has a register as its second operand. Once more let us consider the first operand of the M and MR instructions. In writing the instruction a single register appears as the first operand, but in actuality we are employing two registers. In both cases the register designated as operand 1 must be an even (i.e. 2, 4, 6 etc.) register. Since two registers are employed, by IBM conven-tion, the even register designates the beginning of a "single" field. This 'single' field can be said to have resulted from a conceptual merger of the two 4-byte regis-ters to form a single 8-byte field. Since we perform mul-tiplication from right to left, the number to be multiplied is stored in the low order (right) bytes of the mergered field. Storing is done starting with the rightmost byte of the odd register since this register constitutes the 4 low order bytes of the 8-byte field. In doing multiplication the odd register is first loaded with the number to be mul-tiplied (the multiplicand). The resulting product replaces the multiplicand of the 8-byte field. In many cases the product is small enough to be contained in the 4 low order bytes of the 8-byte field. This is so since 4 bytes can contain 32 binary digits (bits) and each bit represents a power of two. As a result of this, a single register can contain a decimal number as large as four and a third bil-lion. These 4 bytes are of course the odd register. Thus, in storing the product at a convenient location in core, the contents of the odd register is stored. It must be noted that the procedure of storing the single register is done at the discretion of the programmer. Example 1 L7, MULCANSTORING NUMBER TO BE MULTIPLIED IN LOW 8 BYTES M6, MLTPLYRMULTIPLYING REGISTERS 6 AND 7 BY MULTPLYR (FULLWORD) STM6,7,DOUBLESTORING REGISTERS 6 AND 7 IN A DOUBLEWORD OR THE LAST STATEMENT CAN BE REPLACED BY: ST7, FULLSTORES ONLY THE SINGLE REGISTER. Interpretation of: M 6, MLTPLYR This instruction may be seen as: The Division Instruction (D or DR): The operands here are similar in form to those of the multiplication instruction. The D-instruction is of RX-format and takes a register for the first operand and a full-word for the second. The DR instruction is of RR format and has registers as both operands. The aforementioned con-ceptual merger also occurs. The number to be divided (the dividend) is also stored starting with the rightmost byte of the 8 byte field. After the calculation, the quotient is stored as a fullword in the odd register, and the remain-der, which takes the sign of the divider, is stored in the even register. If any of these results do not fit into the registers, a fixed-point exception occurs, and the pro-gram is usually cancelled. It is not difficult to see that the common feature of the IBM 360-370 multiply and divide instructions is that of the conceptual merger. It is this aspect of these instruc-tions that may provide problems for the programmer. If we see these instructions as operations on 8 byte fields de-signated by an even register we should find coding these instructions quite simple. The main idea behind the merger is to have a field large enough to contain the resulting data. It should be noted that the information in the regis-ters are stored in binary form although the examples indi-cate decimal digits.

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Question:

Assume that a gasoline company has a computer facility to take care of billing customers when they buy gasoline with credit cards. The computer is used to make out the check to the station, prepare a master list and create mailing lists and bills. Each customer's information is punched on the card, which contains the gas station number and name, the customer's number and name, and certain billing codes. Can you think of any things that could happen that would make the program useless?

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Solution:

Although this doesn't sound like a complicated computer problem, in reality programmers tend to forget about "trivial" details that often become troublesome. For example: a)What if the customer dies? The program must be able to accommodate uncollectable bills. b)What if a customer moves? The program should be able to alter addresses. c)What if a customer sends invalid checks? d)What if the customer loses his bill and wants a duplicate? e)What if a customer changes his or her name? Appropriate actions should be taken in these special cases. If the program doesn't take these possibilities into account, the output will be meaningless when these eventualities arise.

Question:

Write a BASIC program to obtain the relative frequencies of the three genotypes to be found in the next generation of a large Drosophila population. Assume the population is subject to the Hardy-Weinberg criteria. Use the general binomial equation (pA+qa)^2 to design a table for the number of individuals with AA,Aa, andaagenotypes.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G19-0474.htm

Solution:

The Hardy-Weinberg law states that a large population having completely randommatingsmaintains a constant genotypic frequency over every generation. This law is actually an idealized picture of reality, because it neglects the possibilities of natural selection, mutation, migration, and genetic drift. We want to design a program to simulate the genetic makeup of a suc-ceeding generation. First, recognize that (pA+qa)^2 = p^2AA + 2pqAa + q^2aa . Now, if we assume that the population contains 1000 individuals, choose the value for p-the frequency of a given allele A, to equal .7, and q-the frequency of the allele a, to equal .3, we can obtain three numbers which will describe the genotypes of the next generation. p2 will be the number of homozygous dominants, 2pq will be the number of hetero-zygous dominants, and q^2 will be the number of homozygous recessives. When each of these numbers are multiplied by the total population, the number of individuals of each genotype is obtained. AA is homozygous dominant, A is heterozygous dominant, and SMA is homozygous recessive. 1\OREM HARDY-WEINBERG EQUATION 2\OREAD P,Q 3\OPRINT ''P'', "Q", "HOMD.DOM.", "HETERO.DOM.", "HOUO.RBC." 4\OLET AA = (P\uparrow2) \textasteriskcentered 1000.0 5\OLET A = (2 \textasteriskcentered P \textasteriskcentered Q) \textasteriskcentered 1000.0 6\OLET SMA = (Q\uparrow2) \textasteriskcentered 1000.0 7\OPRINT P, Q, AA, A, SMA 8\ODATA \O.7\O, \O.3\O 9\OEND

Question:

Simplify the following expressions a) A = ST + VW + RST b) A = TUV + XY + Y c) A = F(E + F + G) d) A = (PQ + R + ST)TS g) A = (BE + C + F)C

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G03-0042.htm

Solution:

We need the following laws to simplify the expressions: Idempotent law :AA = A A + A = A Distributive law:A (B+G) = AB + AC A + BC = (A+B) (A+C) Law of Absorption :A(A+B) = A A + AB = A DeMorgan's law :AB=A+B A+B=AB a) A = ST + VW + RST is equal to ST + RST + VW by the associative law. A = ST(1 + R) + VW by distributive law but 1 + R = 1 by law of union. So, A = ST + VW b) A = TUV + XY + Y Use the law of absorption on XY and Y. A = TUV + Y c) A = F(E + F + G) Use the distributive law, A = FE + FF + FG Use the idempotent law on FF. A = FE + F + FG Use the law of absorption. A = F d) A = (PQ + R + ST)TS Use the distributive law A = TSPQ + TSR + STTS Use the idempotent law on STTS. A = TSPQ + TSR + TS Use the law of absorption. A = TS Use DeMorgan's law A = D + D +E Use the idempotent law on D + D. A = D +E A = Y(W + X + YZ)Z Use the distributive law A = YZW + YZX + YZYZ Use the idempotent law on YZYZ. A = YZW + YZX + YZ Use the law of absorption A = YZ g) A = (BE + C + F)C Use the distributive law A = CBE + CC + CF Use the idempotent law A = CBE + C + CF Use the law of absorption A = C Although the law of absorption enables us to solve problems somewhat faster, it generally takes the new reader some time to familiarize himself with it. In such cases, even though it may take an extra step or two, you will certainly find that it is less confusing to use the Law of distribution.

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Question:

A careful analysis shows that the energy of the photon in gamma radiation is slightly less than the difference between the energy of the two nuclear states. In the decay, the gamma-ray photon has momentum, which must be balanced by a recoiling nucleus. Therefore, the nucleus acquires a small portion of the available transition energy. Find what fraction of the transition energy is acquired by the ^43_19K nucleus in the emission of 0.22-MeV gamma photon. Assume that the nucleus was originally at rest.

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/Users/wenhuchen/Documents/Crawler/Physics/D33-1007.htm

Solution:

The momenta and the total energies of the particles before and after the decay are shown in the figure. The rest mass, M, of the potassium nucleus is 7.1 × 10^-26 kg. The energy of the photon, E_2 in MKS units is E_2 = (2.2 × 10^5eV) (1.6 × 19^-19 J/eV) = 3.5 × 10^-14 J The momentum of the photon is given by p^2 = (E_2/c) = [(3.5 × 10^-14 J) / (3 × 10^8 m/s)] The nucleus was initially at rest (see fig.) Therefore, the total momentum of the products after \Upsilon - emission must be zero, as a result of the conservation of total momentum. p_1^\ding{217} + p_2^\ding{217} = p^\ding{217}_initial = 0 orp_1 = p_2 The recoil of the nucleus can be treated non-relativistically since we don't expect it to acquire a large velocity because of its large mass. If the mass of the nucleus is M, and its recoil velocity is v, its momentum will be p_1 = Mv. Equating the momenta of the photon and the nucleus after the \Upsilon-emission, we get mv = p_2 orv = (p^2/m) = [(1.17 × 10^-22 kg m/s) / (7.1 × 10^-26 kg)] = 1.6 ×10^3 m/s The kinetic energy of the nucleus is K = (1/2) mv^2 = (1/2) (7.1 × 10^-26 kg)(1.6 × 10^3 m/s)^2 = 9.1 × 10^-20 J which corresponds to an energy of 0.6 eV. The fraction f of energy acquired by the nucleus is f = [(energy of the nucleus) / (total energy)] = [K / (K + E_2)] \approx (K/E_2) Since E2 > > K = [(6 × 10^-1 eV)/(2.2 × 10^5 eV)] = 2.7 × 10^-6

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Question:

A skier whose mass is 70 kg, takes off from a ski jump with a velocity of 50 m/sec at an unknown angle to the horizontal and lands at a point whose vertical distance below the point of take-off is 100 m. (as shown in the figure). What is his speed just before landing, if the friction of the air can be ignored?

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Solution:

In the absence of friction the law of conservation of energy demands that the sum of the kinetic energy and the potential energy remain constant. At take-off the kinetic energy of the skier is E_k =(1/2)m v^2 _0 where v_0 = 50 m/sec is the skier's speed as he leaves the ramp. During the skier's flight he rises to a maximum height of (h + y) meters above the ground and then falls to the ground. However, the skier's net loss of potential energy during his flight is mgh and not mg(h + y). From our study of projectile motion, all of the skier's original kinetic energy is converted to potential energy when he reaches his maximum height, but this added potential energy is converted back to kinetic energy when the skier falls back to the height at which he started (point A in the diagram). At point A, the skier has total energy equal to that at the beginning of his flight. Therefore, the skier's loss of potential energy between point A and the landing point, is equal to his net loss of potential energy over the entire flight. The net amount of potential energy he loses and gains as kinetic energy is given by: E_p = mgh Thus, the skier's final kinetic energy is equal to the sum of his ori-ginal kinetic energy and his net loss in potential energy: (1/2)mv2_f = (1/2)mv2_0 + mgh where v_f is the skier's speed as he hits the ground v_f = \surd(v2_0 + 2gh) = \surd{(50 m/sec)^2 + 2(9.8 m/sec^2)(100m)} = \surd4460(m/sec)^2 = 66.8 m/sec Just before landing the skier has a speed of 66.8 m/sec. The power of the law of conservation of energy lies in the fact that we can derive this result without knowing the angle of take-off, the horizontal distance traveled, or the exact nature of the curved tra-jectory. We also note that our result is independent of the mass of the skier.

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Question:

The following is a graph of the mechanical response, the twitch, of a muscle cell to a single action potential. Describe the main features of the isotonic twitch shown.

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/Users/wenhuchen/Documents/Crawler/Biology/F19-0470.htm

Solution:

When a muscle is given a single stimulus, such as an electric shock, it responds with a single, quick twitch. The single twitch shown above can be separated into three phases. The first phase (A) is known as the latent period, and represents the interval between the application of the stimulus and the beginning of the visible shortening of the muscle. The next phase (B) is the contraction period which represents the time during which the muscle shortens. The last of the three phases (C) is the relaxation period during which the muscle returns to its original length. Muscle fibers, like nerve fibers, have a refractory period, that is, a very short period of time immediately following one stimulus, during which they will not respond to a second stimulus. The refractory period in skeletal muscle is so short (about .002 second) that muscle can respond to a second stimulus while still contracting in response to the first. The result of this is the summation of contractions, which leads to a greater than normal shortening of the muscle fiber.

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Question:

Compare photosynthesis and cellular respiration in plants.

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/Users/wenhuchen/Documents/Crawler/Biology/F09-0220.htm

Solution:

Photosynthesis in plants is restricted to those cells containing chlorophyll, such as the cells of the leaf. Respiration occurs in all living cells, whether they be the non - chlorophyll - containing root cells or the chlorophyll - containing leaf cells. Photosynthesis requires water and carbon dioxide as raw materials, and sunlight as an energy source. It produces oxygen and simplehexosemolecules, such as glucose. Cellular respiration, on the contrary, consumes oxygen and organic molecules and releases water and carbon dioxide. Photosynthesis is an energy absorbing process, the energy ultimately coming from the sun; respiration releases energy: some is lost as heat and some is stored as chemical bond energy in ATP. Photosynthesis, as the name implies, occurs only when light is present, but respiration takes place continuously, during both night and day. Photosynthesis occurs in the chloroplasts, which contain the necessary enzymes and electron carriers; respiration occurs in the mitochondria, which contain the oxidative enzymes of thetricarboxylicacid cycle (also known as the Krebs cycle) and components of the electron transport system. Finally, photosynthesis results in an increase in the weight of the plant, as more organic substances are manufactured. Respiration results in a decrease in the weight of the plant, as organic materials are broken down for the synthesis and release of energy.

Question:

Suppose a particle of mass m, initially travelling with velocity v^\ding{217}, collides elastically with a particle of mass m initially at rest. Prove that the angle between the velocity vectors of the two particles after the collision is 90

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/Users/wenhuchen/Documents/Crawler/Physics/D06-0324.htm

Solution:

Whenever we are confronted with a collision problem, we may apply the law of conservation of momentum if no external forces act on the system. External forces are forces which act upon the system being considered (in our case, the system consists of the 2 masses) that are due to the environment external to the system, (i.e. friction). Because no external forces are acting in this problem, we may use this conservation law, which is written mathematically as p_10^\ding{217} + p_20^\ding{217} = p_(1)f^\ding{217} + p_(2)f^\ding{217}(1) In this equation, p_10^\ding{217} and p_20^\ding{217} are the initial moment of par-ticles 1 and 2,respectively,and p_(1)f^\ding{217} and p_(2)f^\ding{217} are the final moment of particles 1 and 2, respectively. Note two things: first, momentum is defined as the product of the mass of a particle and its velocity (i.e. p^\ding{217} = mv^\ding{217}) and, secondly, because v^\ding{217} is a vector (i.e. it has direction and magnitude, as in 50 mph EAST) p^\ding{217} is also a vector. Applying equation (1), and noting that v^\ding{217}_20 is 0 because particle 2 is initially at rest, we may write p_10^\ding{217} = p_(1)f^\ding{217} + p_(2)f^\ding{217}(2) Now, we have still not used one last bit of in-formation. The collision is elastic, and, whenever this is the case, kinetic energy is conserved. The law of kinetic energy conservation may be written mathematically as (1/2)m_1 v10^2 + (1/2)m_2 v20^2 = (1/2)m_1 v_(1)f ^2 + (1/2)m_2 v(2)f^2(3) where v_10, v_(1)f, v_20, v_(2)f have the same meaning as previ-ously. Using this equation, we obtain (1/2)m_1 v10^2 = (1/2)m_1 v(1)f^2 + (1/2)m2v(2)f^2(4) But m_1 = m_2 = m, hence (1/2)m v_10 ^2 =(1/2)mv(1)f^2 + (1/2)m v(2)f^2(5) v_10^2 = v(1)f^2 + v(2)f^2(6) Rewriting equation (2), we find mv_10^\ding{217} = mv_(1)f^\ding{217} + mv_(2)f^\ding{217}(7) Dividing both sides of (7) by m, v_10^\ding{217} = v_(1)f^\ding{217}+ v_(2)f^\ding{217}(8) We now want to express equation (8) in terms of magnitudes. To find the magnitude of a vector, we multiply it by itself, using the dot product, and take the square root. The dot product of 2 vectors is defined as A^\ding{217} \textbullet B^\ding{217} = AB cos \texttheta where A is the magnitude of A^\ding{217}, B is the magnitude of B^\ding{217}, and \texttheta is the angle between A^\ding{217} and B^\ding{217}. Note that A^\ding{217} \bullet A^\ding{217} = (A) (A) = A2 because, in this case, \texttheta = 0\textdegree and cos 0\textdegree = 1. Now, dotting each side of equation (8) into itself, we find v_10^\ding{217} \bullet v_10^\ding{217} = (v_(1)f^\ding{217}_ + v_(2)f^\ding{217}) \bullet (v_(1)f^\ding{217}_ + v_(2)f^\ding{217})(9) v_10^\ding{217} \bullet v_10^\ding{217} = v_(1)f^\ding{217} \bullet v^\ding{217}(1)f+ 2v_(1)f^\ding{217} \bullet v_(2)f^\ding{217} + v_(2)f^\ding{217} \bullet v_(2)f^\ding{217} orv10^2 = v(1)f^2 + v(2)f2+ 2v_(1)f^\ding{217} \bullet v_(2)f^\ding{217}(10) Subtracting equation (6) from (10) v10^2\ding{217} = v_(1)f ^2 + v(2)f^2 + 2v_(1)f^\ding{217} \bullet v_(2)f^\ding{217} -v10^2= v(1)f^2 + v_(2)f ^2 0= 2v_(1)f^\ding{217} \bullet v_(2)f^\ding{217} orv_(1)f^\ding{217} \bullet v_(2)f^\ding{217} = 0(11) v_(1)f v_(2)f cos \texttheta = 0(12) But because v_(1)f, v(2)f\not = 0,this means that cos \texttheta = 0 or \texttheta = 90\textdegree,as was to be shown.

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Question:

What do the terms critical temperature and pressure mean? If you want to liquefy a gas, what physical properties must you consider? If one gas has a higher critical tem-perature than another, what can be said about the relative forces of attraction between like molecules?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E02-0075.htm

Solution:

The critical temperature is the temperature above which a gas may not be liquefied regardless of the pressure. The critical pressure is the pressure necessary to liquefy a gas at its critical temperature. Both temperature and pressure are factors in the condensation of a gas. Molecules of a particular compound have a lower kinetic energy in their liquid state than in their gaseous state. kinetic energy = mass × (velocity)^2 Therefore, kinetic energy is proportional to the square of the velocities of the particles. In the gaseous state the velocities of the molecules are greater, when the temperature of a system is decreased the velocities of the molecules decrease and, hence, the kinetic energies decrease. If the pressure of the system is increased, the particles are forced closer together. At a certain point, when the kinetic energy will be low and/or the particles will be close enough together, the gas will liquefy. If a gas has a high critical temperature, less energy is necessary to liquefy it. In the liquid state the molecules are much closer together than in the gas. When less energy is needed to force the molecules together, the inter - molecular forces of attraction are greater.

Question:

A car covers a distance of 30 miles in 1/2 hour. What is its speed in miles per hour and in feet per second?

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Solution:

v_average= s/t = (30 mi)/(1/2)hr = 60 mi per hr = (60 mi)/hr × (5280 ft)/mi × (1 hr)/(3600 sec) = 88 ft per sec This useful relation, that 60 miles per hour equals 88 feet per second, is one you should commit to memory.

Question:

Design a program to move a knight around a chessboard so that in 63 moves, it will have touched each square once on a regular 8 × 8 chessboard. It is assumed that there are no other pieces on the board during the movement.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G23-0562.htm

Solution:

This is a classical problem first demonstrated by Leonard Euler in the 18th century. We define this sequence of moves as a "knight's tour." The length of the tour is the total number of squares visited, which, including the starting square, we want to be 64. We need an 8 × 8 matrix to represent the chessboard. The tour begins from position (1,1) , and only the character-istic L-shaped move of the knight is allowed. We will use what is known as an heuristic strategy. The major problem in the tour occurs when the knight must move into the corner positions. To avoid getting trapped, we define an array of weights to give higher priority to the squares farthest from the center of the board. Thus, in Figure 1, we have the array of weights. 7 6 5 4 4 5 6 7 Figure 1 6 5 4 3 3 4 5 6 5 4 3 2 2 3 4 5 4 3 2 1 1 2 3 4 4 3 2 1 1 2 3 4 5 4 3 2 2 3 4 5 6 5 4 3 3 4 5 6 7 6 5 4 4 5 6 7 Notice that when the knight comes closer to the center, it can move in eight possible directions, as shown in Figure 2: Each time the knight is to move, we should check each of the potential moves starting from direction A and moving clock-wise. After looking at the eight possibilities, we save the positions of the directions having the highest weights. If two possibilities have equal weight, we introduce a random generator to break the tie between them. Of course, the ran-dom process means that several tours will have to be attempt-ed before a complete tour is realized. A successful knight's tour produced by the heuristic method is given in Figure 3: 1 38 3 18 35 40 13 16 Figure 3 4 19 36 39 14 17 34 41 37 2 57 54 59 42 15 12 20 5 62 43 56 53 60 33 49 44 55 58 61 64 11 26 6 21 48 63 52 27 32 29 45 50 23 8 47 30 25 10 22 7 46 51 24 9 28 31 Note that it is not the only possible solution. Other suc-cessful tours do exist. The program requires a 12 × 12 array to create a buffer zone around the board. Then we fill the border rows 1, 2, 11, and 12 with nonzero values, and the board squares are filled with zeros. This is done to insure that no moves go off the board. A flowchart will help you to visualize the actions taken during execution of the program: INTEGER IM (8), JM(8), BOARD (12,12) INTEGER HIST (64), WGT (12,12) DATA(IM(K), K = 1,8)/1,2,2,1,- 1,- 2,- 2, - 1/ DATA(JM(K), K = 1,8)/2,1,- 1,- 2,- 2,-1,1,2/ DATA (HIST (I), I = 1,64) /64\textasteriskcentered0.0/ X = 0.5 CDO WHILE M < 101 5IF (M.GE. 101) GO TO 99 DO 10 I = 1,12 DO 15 J = 1,12 BOARD(I,J) = 0 IF(I.LT.3. OR. I.GT.10) BOARD (I,J) = 99 IF(J.LT.3. OR. J.GT.10) BOARD (I,J) = 99 WGT (I,J) = (IABS (2\textasteriskcentered1 - 13) + IABS (2\textasteriskcenteredJ - 13))/2 15CONTINUE 10CONTINUE CINITIALIZE TOUR ITUR = 3 JTUR = 3 N = 0 20CONTINUE N = N + 1 IF(N.GT.64) GO TO 30 I = ITUR J = JTUR BOARD(I,J) = N CNOW CHECK OTHER POSSIBLE MOVES BY CGENERATING A RANDOM INTEGER BETWEEN C1 AND 8 AND MOVE CLOCKWISE CALL RAND(X) KRND = INT (8.0\textasteriskcenteredX) + 1 MWGT = 0 IMWGT = 0 JMWGT = 0 DO 25 K = 1,8 K1 = MOD (KRND + K - 2,8) + 1 ITUR = I + IM(K1) JTUR = J + JM(K1) IF(BOARD(ITUR,JTUR) .NE.0) GO TO 25 IF (WGT(ITUR,JTUR).LT.MWGT) GO TO 2 5 MWGT = WGT (ITUR,JTUR) IMWGT = ITUR JMWGT = JTUR 25CONTINUE ITUR = IMWGT JTUR = JMWGT IF(MWGT.NE.0) GO TO 20 30CONTINUE HIST (N) = HIST (N) + 1 COUTPUT COMPLETED BOARD IF (N.LT.64) GO TO 99 WRITE(6,100) 100FORMAT(7X,'THIS IS A COMPLETED TOUR') WRITE(6,101) ((BOARD(I,J), J = 3,10), I = 3,10) 101FORMAT(1X,8I4) M = M + 1 GO TO 5 99CONTINUE WRITE (6,102)M 102FORMAT (1X,'TOTAL NUMBER OF TRIALS = ',I4) WRITE (6,101) (HIST (I), I = 1,64) WRITE (6,103) 103FORMAT (7X, 'THIS IS THE TABLE OF WEIGHTS') WRITE (6,101) ((W (I,J), J = 3,10), I = 3,10) STOP END

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Question:

A symmetrical top is described by the fact that, its moments of inertia about 2 of its principal axes are equal (i.e., I_1 = I_2 \not =I_3). Assuming that no external torques act, derive and solve the equations of motion of this body.

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Solution:

The general motion of a rigid body is very complex. For cases in which the object doesn't rotate about a fixed axis, we can-not relate the angular momentum (L) to the angular acceleration by L = I\alpha This relation only holds for rotations about a fixed axis. We must use N^\ding{217} = dL^\ding{217}/dt where N^\ding{217} is the net external torque on the body for these rotations. Another alternative is to use the Euler Equations I_1(d\omega_1/dt) + (I_3 - I_2)\omega_3\omega_2 = N_1 I_2(d\omega_2/dt) + (I_1 - I_3)\omega_1\omega_3 = N_2(1) I_3(d\omega_3/dt) + (I_2 - I_1)\omega_2\omega_1 = N_3 where the subscript 1 refers to the first principal axis of the rigid body (see fig. (A)),and similarly for the subscripts 2 and 3. Furthermore, \omega^\ding{217} is the angular velocity of rotation of the body with reference to an inertial frame. In this example, I_1 = I_2 \not = I_3 and N^\ding{217} = 0, whence I_1(d\omega_1/dt) + (I_3 - I_2)\omega_3\omega_2 = 0 I_2(d\omega_2/dt) + (I_1 - I_3)\omega_1\omega_3 = 0(2) I_3(d\omega_3/dt) = 0 Letting I_1 = I_2 = I, we obtain (d\omega_1/dt) + [(I_3 - I)/I]\omega_3\omega_2 = 0 (d\omega_2/dt) + [(I - I_3)/I]\omega_1\omega_3 = 0 (d\omega_3/dt) = 0 Defining \Omega \equiv [(I_3 - I)/I]\omega_3, we may write (d\omega_1/dt) + \Omega\omega_2 = 0 (d\omega_2/dt) + \Omega\omega_1 = 0(3) (d\omega_3/dt) = 0 A solution of (3) is given by \omega_1 = A cos \Omegat; \omega_2 = A sin \Omegat , where A is a constant. We see that the component of the angular velocity perpendicular to the figure axis (axis 3) (see figure (A)) of the top rotates with a constant angular velocity \Omega. The component \omega_3 of the angular velocity along the figure axis is constant. There-fore the vector \omega rotates uniformly with angular velocity \Omega about the figure axis of the top. In other words, a top which spins about its figure axis with angular velocity \omega_3 in force-free space will wobble with the frequency \Omega. For the earth I_3 is not exactly equal to I_1 because the earth is not exactly a sphere. The wobble is actually very well observed, giving rise to what is called the variation of latitude. The wobble is so interesting that the International Latitude Service maintains a number of observatories just for the purpose of measuring it.

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Question:

Find the decrease of air pressure with elevation under the assumption that the atmosphere is a homogenous ideal gas at a uniform temperature. (Neglect the variation of the gravitational acceleration g, with elevation).

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/Users/wenhuchen/Documents/Crawler/Physics/D14-0510.htm

Solution:

For this purpose, consider an air column or cross section A and height h, as shown in the figure. In the equilibrium state, the pressure at any elevation is produced solely by the weight of the gas above it. The differential volume element of height dh in the figure increases the force on the lower cross-section by dW = (Adh) \rhog where \rho is the density of the air, and (Adh)\rho is the mass of gas in the volume Adh. The resulting decrease in pressure is dP = - dW/A= -\rho g dh(1) (pressure decreases as height increases). Then the ideal gas law, written for one mole of the gas, is PV_M = R T. If M is the molar mass of the air, then \rho is \rho = M/V_M = MP/RT(2) Substituting (2) in (1) dP = -(MP/RT) g dh dP/P = -(Mg/RT) dh. Integration of this equation gives P(h) = P_0 e^-(Mg/RT)h where P_0 is the pressure at the earth's surface.

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Question:

Write a flowchart describing how one would design a computer program to print a multiplication table. You need only print 12 rows of products.

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Solution:

To accomplish this task, you will need a nested loop. The inner, or nested, loop outputs one number for each time the loop is executed. Because there are 12 × 12 = 144 numbers in the table, the nested loop is used 144 times. The outer loop, or nest, is executed 12 times: each execution of this loop causes one row to be printed. This flowchart is not well-refined. Although it does describe thoroughly the actions that must take place in order to obtain the table, it does not detail the calculations necessary to produce the products. The point is that if the calculations were exceedingly com-plex, you might waste a lot of time worrying about the calculation and forget about the printing of the table. This is a familiar case in programming, in which the student, to put it colloquially, cannot see the forest for the trees. If you can understand how to print a table, it is a relatively simple matter to adjust the calculations of the elements to be put into it. The flowchart is mostly self-explanatory. The dotted line indicates the boundaries of the nested loop.

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Question:

Describe briefly the Darwin-Wallace theory of natural selection. What is meant by "survival of the fittest"? Do you thinkthis applies to human populations today?

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Solution:

The Darwin-Wallace theory of natural selection states that a significantpart of evolution is dictated by natural forces, which select for survivalthose organisms that can respond best to them. Since more organismsare born than can be accommodated by the environment, there mustbe chosen among the large numbers born the limited number to live andreproduce. Variation is characteristic of all animals and plants, and it isthis variety which provides the means for this choice. Those individuals whoare chosen for survival will be the ones with the most and best adaptivetraits. These include the ability to compete successfully for food, water, shelter, and other essential elements, also the ability to reproduce andperpetuate the species, and the ability to resist adverse natural forces, which are the agents of selection. Essential to the theory of natural selection are the ideas of the "strugglefor existence" and "survival of the fittest". Because the resources ofthe environment are naturally limited, individuals must struggle among themselvesfor food and space. Those individuals with the traits that are bestsuited for the given environment will survive and multiply, while otherswill slowly decline in number or disappear completely. Since the naturalforces are constantly operating and changing, struggle for survival goeson forever and the fittest of the competitors survive. Survival of the fittest holds true for human popu-lations, but it is less obviousamong human beings than lower organisms. Whereas it is commonto see wild dogs fighting each other for a piece of food, it is less commonto see two human beings fighting for the same reason. Among humanbeings, the problems of food and shelter have been conquered successfully, such as by farming or by raising cattle. Actual physical strugglebetween human beings for food has become less important. The traitsof strong muscles, great agility and alertness, and other fighting abilitiesthat were once important to the cavemen have more or less lost theirsignificance today.

Question:

A ball is released from rest at a certain height. What is its velocity after falling 256 ft?

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Solution:

Since the initial velocity is zero, we use y = v_0t + (1/2)at^2 = (1/2)gt^2 taking \textquotedblleftdown\textquotedblright as the positive y-direction. Solving for the time to fall 256 ft, we have t= \surd(2y/g) = \surd[(2 × 256 ft)/(33 ft/sec^2)] = \surd(16 sec^2) = 4 sec The velocity after 4 sec fall is v= v_0 + at =gt= (32 ft/sec^2) × (4 sec) = 128 ft/sec.

Question:

A satellite of mass m moves around the Earth as shown (actual-ly, the path is an ellipse). Which instantaneous velocity is greater, v_p (at point P) or v_A ( at point A)?

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Solution:

Consider the Earth as a fixed object and neglect the in-fluence of the Sun and other planets. The angular momentum of the satellite around the earth L, is given by L^\ding{217} = r^\ding{217} × mv^\ding{217} where r^\ding{217} is the vector from the earth to the satellite, and v^\ding{217} is the velocity of the satellite. Since v^\ding{217} and r^\ding{217} are perpendicular L = mvr However,T = dL/dt(1) where the torque T is defined as T^\ding{217} = r^\ding{217} × F^\ding{217} F^\ding{217} is the gravitational force on the satellite keeping it in its orbit. (It is due to the mass of the Earth). Since the angle between F^\ding{217} and the radius vector r^\ding{217} is 0\textdegree we have T = Fr sin 0\textdegree = 0 Therefore, by equation (1), L of the satellite is constant in time. At time t_1 the particle is at A and at time t_2 it is at P. Hence, the angular momentum at the two points must be the same. Or L = mv_Ar_A = mv_Pr_P Since r_p < r_A. we must then have v_P > v_A . The velocity is greatest when the satellite is nearest the Earth; this point is called the perigee (labeled P in the diagram). The velocity is least at the farthest point from the Earth - the apogee (A) of the orbit.

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Question:

An object is 12 feet from a concave mirror whose focal length is 4 feet. Where will the image be found?

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Solution:

We first construct a ray diagram. The image is a real image (inverted) between the center of curvature and the focal point. It is smaller than the object. We now solve mathematically: D_0 is 12 feet and f is 4 feet. Substituting in (1/D_0) + (1/D_I) = (1/f) (1/12 ft) + (1/D_I) = (1/4 ft) whence D_I = 6 ft.

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Question:

Each of the following is an incorrect name for an alkane. Write a structural formula for each and provide the correct systematic name: (a) 2-ethylbutane; (b) 2-isopropylpentane; (c) 1, 1-dimethylpentane; (d) 2, 2-dimethyl-4-ethylpentane.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E20-0726.htm

Solution:

First, one writes the formula starting with the parent molecule. Next, one locates the number of the carbon on the parent molecule that is attached to a particular group by the number in front of that particular group in the compound's name. Thus, (a) 2-ethylbutane becomes (b) 2-isopropylpentane becomes (c) 1, 1-dimethylpentane becomes (d) 2, 2-dimethyl- 4 -ethylpentane Each name is incorrect in the same respect; the named parent molecule is not the longest chain of the compound. If a compound twists or bends in free space it must be taken into account when naming the compound. Thus, the correct names are (a) 3-methylpentane, since the longest chain has 5 carbons and a methyl group attached to carbon number 3. (b) 2, 3-dimethylhexane since the longest chain has 6 carbons and methyl groups are attached to carbon numbers 2 and 3. (c) 2-methylhexane, since the longest chain has 6 carbons and a methyl group is attached to carbon number 2. (d) 2, 2, 4-trimethylhexane, since the longest chain has 6 carbons and three methyl groups attached to carbons 2, 2 and 4; 1 methyl group attached to carbon 4 and 2 methyl groups attached to carbon 2.

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Question:

Explain briefly the background and applications of FORTRAN Computer Programming Language.

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Solution:

In 1957, a revolutionary new concept of scientific program-ming became available to users of the IBM-704 computer. It was called "FORTRAN" (FORmulaTRANslation), or, more formally, the FORTRAN Auto-matic Coding System for the IBM-704. Since its Introduction, features have been steadily added to the language. The latest version is FORTRAN V. FORTRAN is a digital computer programming language which resembles elementary algebra, augmented by certain English words such as DO, GO TO, READ, WRITE, IF, etc. Because of its similarity to ordinary algebra, the FORTRAN language is particularly well suited for solving problems in science, mathematics, and engineering. However, use of the language is by no means restricted to these areas. FORTRAN is also applied to a wide variety of problems in business, economics, psychology, medicine, library science - virtually any area which might require extensive manipulation of numerical data or character information. Such a huge number of areas of application, together with simplicity of learning and usage, make FORTRAN, probably, the most popular digital computer language. Although more powerful and flexible languages exist, the popularity of FORTRAN will probably continue for many years to come.

Question:

Calculate the solubility product constant of pure PbSO_4 in water. The solubility of PbSO_4 in water at 25\textdegreeC is 1.25 × 10-^4 moles/liter.

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Solution:

The solubility of a substance is the number of moles of that substance which will dissolve in one liter. Since PbSO_4 contains one Pb^2+ ion and one SO^2- ion per molecule in aqueous solution, the con-centration of each of these ions is equal to the solubility. Hence, [Pb^2+] = [SO_4 ^2-] = 1.25 × 10-^4 M . Thesolubilizationreaction for PbSO_4 is PbSO_4 (s) \rightleftarrows Pb^2+ + SO_4 ^2- . and the solubility product constant is K_sp= [Pb^2+] [SO_4 ^2-] . Substituting the values for [Pb^2+] and [SO_4 ^2-] gives K_sp= (1.25 × 10-^4 M ) × (1.25 × 10-^4 M ) = 1.6 × 10-^8moles^2 /liter^2 .

Question:

An observer on earth finds that the lifetime of a neutron is 6.42 × 10^2 s. What will the lifetime of a neutron be as measured by an observer in a spacecraft moving past the laboratory at a speed of 2 × 10^8 m/s?

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Solution:

If we measure the time required for the neutron to decay, first from the laboratory, and then relative to the neutron, we do not record the same time. The decay of the neutron takes longer for the observer in the lab. It is known that the faster the observer moves with respect to an event the slower the event seems to take place. This phenomenon is called time dilation and is a result of the laws of relativistic kinematics. The lifetime of an un-stable (i.e. decaying) particle will therefore be longer when viewed from the rocket. The formula relating thelifetime t_R observed from the rocket to one t_L, observed in the laboratory is t_R = [1 / {1 - (v2/ c^2)} ] t_L where v is the speed of the rocket with respect to the laboratory. Hence, for t_R we get t_R= 1 / \surd{1- [(2 × 10^8 m/s) / (3 × 108m/s)]^2} × 6.42 × 10^2 s = 1.34 × 6.42 × 102s = 8.62 × 10^2 s .

Question:

How does the compound eye of crustaceans and insects differ from man's eyes? What advantages do insects' eyes have that man's do not?

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Solution:

Compound eyes consist of a varying number of units or ommatidia (see Figure). Each ommatidium has its own lens, cornea, and group of neurons. Some arthropods, such as ants, have only a few ommatidia while others, such as dragonflies, have as many as 10,000. Each ommatidium, in gathering light from a narrow sector of the visual field, projects a mean intensity of the light from that sector onto the total retinal field. All the points of light from the various ommatidia form an image, called a mosaic pic-ture. The nature of a mosaic image can be understood by comparing it to the picture of a TV screen or a newspaper photo. The picture is a mosaic of many dots of different intensities. The clarity of the image depends upon the number of dots per unit area. The greater the number, the better the picture. The same is true of the compound eye. Each ommatidial receptor cell with its accompanying nerve-fiber projects a segment of the total picture. The compound eye may contain hundreds of thousands of neurons, clustered in groups of seven or eight per ommatidium. The human eye also forms a retinal picture composed of numerous points, each point corresponding to a rod or cone cell with its accompanying neuron. Considering the fact that the human eye contains approximately 125,000,000 rods and 6,500,000 cones, the concentration of corresponding points is much higher than in the compound eye, and the image formed is thus much finer. In addition, while the compound eye of the arthropod is composed of numerous lenses, the human eye has only one lens\textasteriskcentered for its entire corneal field. Hence, the insect has no structure strictly analogous to the human retina? their critical surface for vision is the outer surface of the compound eye itself, composed of the many closely packed individual lenses. Many arthropods have very wide visual fields? in some crustaceans, the corneal surface covers an arc of 180\textdegree. However, the anthropod cornea and lens are developed from the skeleton, and the eyes are fixed in place. The human eye, has a much smaller visual field, but compensates for this to some extent with movability. Compound eyes are particularly sensitive to motion. This is because the ommatidia recover very rapidly from a light impulse, making them receptive to a new impulse in a very short time. Compound eyes thus can detect flickers at extremely high frequency. Flies detect flickers of a frequency up to 265 per second, as compared to man's limit of about 53 per second. Because flickering light at higher frequency is seen as a continuous light by man, motion pictures are seen as smooth movement, and 60-cycle bulbs give off steady light. The ability of the compound eyes to detect high frequency light changes enables them to rapidly detect motion and facilitates the capture of prey and the avoidance of enemies. Compound eyes are sensitive to a broader range of light wavelengths than human eyes. Insects can see well into the ultraviolet range. Colors that appear identical to man may reflect ultraviolet light to different degrees, and appear strikingly different to insects. Compound eyes are also able to analyze the plane of polarization of light. A sky appearing evenly blue to man,reveals different patterns in different areas to insects, because the plane of polarization varies. Honeybees use this ability as an aid to navigation.

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Question:

An electron jumps from the orbit for which n = 3 to the orbit for which n = 2, thereby emitting some visible, red light. What is the magnitude of this quantum of energy?

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Solution:

In the hydrogen atom, the energy of the atom with its electron in quantum level n is E = [(2.18 × 10^-18 joule)/n^2] The energy of the hydrogen atom with its electron in the n = 2 energy level is E_2 = [(2.18 × 10^-18 joule)/2^2] The energy of the hydrogen atom with its electron in the n = 3 energy level is E_3 = [(2.18 × 10^-18 joule)/3^2] Then the magnitude of the quantum of energy is E_2 - E_3 = 2.18 × 10^-18 joule × [1/(2^2) - 1/(3^2)] = 3.03 × 10^-19 joule.

Question:

A flywheel, in the form of a uniform disk 4.0 ft in diameter, A flywheel, in the form of a uniform disk 4.0 ft in diameter, weighs 600 lb. What will be its angular acceleration if it is acted upon by a net torque of 225 lb-ft? (The rotational inertia of a wheel is I = (1/2)mR^2, where m is the wheel's mass and R is its radius.)

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Solution:

The flywheel is a massive wheel whose use is the "storing"of kinetic energy. The problem is one of applying the formula \cyrchar\cyrt^\ding{217} =I\alpha^\ding{217} (analogous to F^\ding{217} = ma^\ding{217}), in which \cyrchar\cyrt^\ding{217} is applied torque, and \alpha^\ding{217} is the resulting angular acceleration. We are given \cyrchar\cyrt^\ding{217}, we can determine I, and solve for \alpha^\ding{217}. Since the weight (W) of any object is W = mg where m is its mass and g is the acceleration due to gravity, we can find m m = W/g = (600 lb)/(32 ft/sec^2) = 18.8 slugs I= (1/2)mR^2 = (1/2)(18.8 slugs)(2.0 ft)^2 = 38 slug-ft^2 Therefore, substituting into \cyrchar\cyrt =l\alpha, 225 lb - ft =(38 slug-ft^2)\alpha \alpha = 5.9rad/sec^2 In radian measure the angle is a ratio of two lengths and hence is a pure number. The unit "radian" therefore does not always appear in the algebraic handling of units.

Question:

Locate the center of mass of the machine part in the figure consisting of a disk 2 in. in diameter and 1 in. long, and a rod 1 in. in diameter and 6 in. long, con-structed of a homogeneous material.

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Solution:

By symmetry, the center of mass lies on the axis and the center of mass of each part is midway be-tween its ends. The volume of the disk is: v_d = \pir^2h = \pi(1 in)^2 (1 in) = \pi in^3. The volume of the rod is: v_r = \pir^2h = \pi[(1/2) in]^2 (6 in) = (3/2) \pi in^3 Since the disk and the rod are both constructed of the same homogeneous material, the ratio of their masses will equal the ratio of their volumes: (mass of disk)/(mass of rod)= (m_d)/(m_r) = (\rhov_d)/(\rhov_r) = (\pi in^3)/[(3/2) \pi in^3] = (2/3) where \rho is their common density. The formula for the distance of the center of mass from a given origin 0 is: x = (m_d x_d + m_r x_r)/(m_d + m_r) where x_d and x_r are the distances of the centers of mass of m_d and m_r respectively from 0. Take the origin 0 at the left face of the disk, on the axis. Then x_d = 1/2 in., and x_r = 4 in. Since m_d = 2/3 m_r x= [(2/3)m_r (1/2 in) + m_r (4in)]/[(2/3) m_r + m_r] = 2.6 in. The center of gravity is on the axis, 2.6 in. to the right of 0.

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Question:

A saturated solution of the strong electrolyte Ca(OH)_2 is prepared by adding sufficient water to 5.0 × 10^-4 mole of Ca(OH)_2 to form 100 ml of solution. What is the pH of this solution?

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Solution:

We need to find [H_3O^+] in order to find thepH.[H_3O^+] is determined by substituting into the ex-pression for the water constant, K_W = [H_3O^+][OH^-] (K_W = 10^-14 mole^2/liter^2) , or [H_3O^+] = K_W/[OH^-] The hydroxyl concentration, [OH^-], is contributed by the dissociation of Ca(OH)_2, which is given by the equation Ca(OH)_2 \rightarrow Ca^2+ + 2OH^-. Ca(OH)_2 is a strong electrolyte, hence we assume that it dissociates completely. Since two hydroxyl ions are formed for every Ca(OH)_2 that dissociates, the con-centration of OH^- is equal to twice that of Ca(OH)_2, or [OH^-] = 2 × (0.0005 mole/100 ml) = 2 ×(0.0005 mole/0.10 liter) = 2 × 0.005 mole/liter = 0.010 mole/liter = 10^-2 mole/liter. The hydroniumion concentration is then [H_3O^+] =(K_W/OH^-) = [(10^-14 mole^2/liter^2)/(10^-2 mole/liter)] = 10^-12 mole/liter. The pH is defined as pH = - log [H_3O^+], or pH = - log [H_3O^+] = - log (10^-12) = - (- 12) = 12.

Question:

CF_4, is used as a low-temperature refrigerant under the name Freon-14. Its normal boiling point is - 127.8\textdegreeC and its heat of vaporization is 12.62 kJ/mole. At what pressure should you pump liquid Freon-14 to set a tem-perature of -135\textdegreeC?

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Solution:

Assuming that CF_4, vapor obeys the ideal gas law, and that the heat of vaporization ∆H_vap is in-dependent of temperature, then the following equation can be used to calculate the vapor pressures at two differ-ent temperatures. Log [P_2/P_1] = {[∆H_vap (T_2 - T_1)] / [2.303 R T_1T_2]}, where T_1 and T_2 are absolute temperatures, R = gas constant, and p_1 and p_2 are press-ures. Any units of pressure may be chosen as long as the same units are used for both pressures, and any units of heat may be used as long as ∆H_vap and RT have the same units of energy. Thus, substituting into the above equation the known data P_1 = 1 atm, T_1 = - 127.8\textdegree + 273\textdegree = 145.2\textdegreeK T2 = - 135\textdegree + 273\textdegree = 138\textdegreeK, ∆H_vap = 12620 J/mole R= 8.314 J/mole \textdegreeK Log [(P_2)/(1 atm)] = {[(12620 J/mole) (138\textdegreeK - 145.2\textdegreeK)] / [2.303 (8.314 J/mole \textdegreeK) (145.2\textdegreeK) (138\textdegreeK)]} = [(12620) (-7.2)] / [383662.77] = - [(90864) / (383662.77)] = -.2368 P_2 = (1 atm) 10^-.2368 = 0.58 atm.

Question:

Solder is a lead-tin alloy. 1.00 gram of the alloy treated with warm 30 % HNO_3 gave a lead nitrate solution and some insoluble white residue. After drying, the white residue weighed 0.600 g and was found to be SnO_2. What was the percentage tin in the sample? Sn + 4HNO_3 + SnO_2 + 4N_2 + 2H_2O

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Solution:

From thestoichiometryof this equation, 1 mole ofSnproduces 1 mole of SnO_2.Thus, if the number of moles of SnO_2 is known, the number of moles ofSnis also known (it is equivalent to the number of moles of SnO_2). Once the number of moles ofSnis known, the weight ofSncan be calculated and thus the percentage ofSnby weight in 1.00 gram of solder alloy can be determined. The number of moles of SnO_2 (MW = 151 g / mole) is [(0.600 g of SnO_2) / (151 g / mole)] = 0.00397 moles. The weight of 0.00397 moles ofSn(at. wt. 119 g / mole) is (0.00397moles) (119 g / mole) = 0.472 g. Thus, the percent-age ofSnin the alloy is [(0.472 gsn) / (1.00 g alloy)] × 100 = 47.2 %.

Question:

The hormone which brings about metamorphosis of the cecropia moth is secreted by a gland in the thorax. However, no metamorphosis takes place if the head is cut off, even though the larva continues to live. Explain.

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Solution:

In this organism, there is a two-lobed gland at the front of the head which plays a role in meta-morphosis. When this gland was transplanted into the body of a headless caterpillar, metamorphosis took place. A theory was proposed that the gland in the head of the moth produces a hormone which stimulates the gland in the thorax to secrete its hormone. When either the head or the thoracic gland alone was transplanted into the excised posterior half of a cecropia moth, there was no metamorphosis. If both glands were transplanted, the metamorphosis took place. This indicates that both the gland in the head and the gland in the thorax of the cecropia moth are necessary for the process of metamorphosis. An investigation into the nature of the action of these hormones showed that the thoracic hormone stimulates the development of the cytochromes in the body cells. The cytochromes play an important role in the energy release of the cell since they are acceptors of hydrogen in the electron transport system. An increase in the cytochromes results in an increase in the metabolic rate. It is this increase in metabolism which supplies the additional energy required for metamorphosis.

Question:

A converging lens of 5.0 cm focal length is used as a simple magnifier, producing a virtual image 25 cm from the eye. How far from the lens should the object be placed? What is the magnification?

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Solution:

The image produced is virtual, on the same side of the lens as the object. Therefore, the distance q of the image from the lens is negative. Using the general lens equation 1/p + 1/q = 1/f ; 1/p = 1/f - 1/q. Solve for p and substitute values. p = fq / (q - f) = {5.0cm (-25cm)} / (-25cm - 5.0cm) = 4.2cm. Using absolute values to find the magnification, M = │q/p│ = │-25 cm/4.2 cm│ = 5.9. Since the image is on the same side of the lens as the object, the light has been focused in a way that produces an erect image. In a real image, which is seen on the opposite side of the lens in relation to the object, the object is inverted.

Question:

Pick out the errors in the following FORTRAN statements and explain briefly why they are incorrect:

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Solution:

a) FORTRAN variable names cannot begin with numbers, so 2ERR is an invalid variable. b) In an IF statement, comparisons are made by using relational operators . An equality sign, which is used only in FORTRAN assignment statements , should be replaced by .EQ. TERM. c) CAPITAL is a seven-letter word. FORTRAN allows a maximum of six characters in each variable name. d) In FORTRAN, variables beginning with letters I through N are integer unless specified otherwise. Since WIDTH and AREA are real variable names , LENGTH cannot be used in order to avoid mixed mode multiplication .

Question:

An unknown compound consists of 82.98 % potassium and 17.02 % oxygen. What is the empirical formula of the compound?

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Solution:

The empirical formula of any compound is the ratio of the atoms that make up the compound by weight. It is the simplest formula of a material that can be derived solely from its components. Therefore, we must determine the ratio of gram-atoms of potassium (K) to the number of gram-atoms of oxygen (O). The number of gram-atoms of a substance equals the weight of the substance in grams divided by the weight per gram-atom of the substance. In other words, number of gram-atoms = (weight in grams) / (weight per gram-atoms) In this problem, we are given the percentages of the elements that make up the compound. These percentages are, in reality, the weight in grams, since in the de-finition of weight, we imply percentage. The weight per gram atom is the atomic weight of the element.which can be found in the periodic table of elements. Therefore, the number of gram-atoms for potassium is {82.98 g (wt of K)} / {39.10(atomic weight)} = 2.120 moles. For oxygen , the number of gram-atoms is {17.02 g(wt of O)} / {16.00(atomic wt)} = 1.062 moles. Recall, the empirical formula is the ratio of the elements by weight. Consequently, the ratio of potassium to oxygen is 2 : 1, since the gram - atom ratios are respectively 2.120 : 1.062. Therefore, the empirical formula is K_2O.

Question:

Analyze the reaction where two neutral pions are created from the annihilation of a neutron-antineutron pair. Assume that the neutron and Its twin were initially at rest (see figure).

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/Users/wenhuchen/Documents/Crawler/Physics/D36-1055.htm

Solution:

We wish to analyze the annihilation of a neutron-antineutron pair. We may relate the energies and momenta of the products of the annihilation to the energies and momenta of the neutron-antineutron pair by the laws of conservation of energy and momentun. Initially, the neutron and antineutron are at rest, and therefore, have only rest energy. Furthermore, the neutron and antineutron have equal masses, and 2m_n c^2 = E_\pi(1) + E_\pi(2)(1) where "n", "\pi_1", "\pi_2" refer to the neutron and 2 pions respectively. Applying conservation of momentum, 0 = p^\ding{217}_\pi(1) + p^\ding{217}_\pi(2) ,(2) since the neutron and antineutron are initially at rest. But, relativistically , P^\ding{217} = (m_0v^\ding{217}) / \surd{1 - (v^2 /c^2)} where v^\ding{217} is the velocity of the particle whose momentum we wish to measure, and m_0 is its rest mass. By (2) (m_0\pi(1)v_1^\ding{217}) / \surd{1 - (v^2_1 /c^2)} = (- m-_0\pi(2) v_2^\ding{217}_ ) / \surd{1 - (v^2_2 /c^2) } (3) Butm_0\pi(1) \approx m0\pi(2), and v_1^\ding{217} / \surd{1 - (v^2_1 /c^2) } = - v_2^\ding{217}/ {\surd1 - (v^2_2 /c^2) }(4) Taking the dot product of each side of (4) with itself, we have (since v^\ding{217} \textbullet v^\ding{217} = (v) (v) cos 0\textdegree= v^2) v^2_1 / { 1 - (v^2_1 /c^2) } = v^2_2 / {1 - (v^2_2 /c^2)} v^21= [v^2_2 {1 - (v^2_1 /c^2)}] / {1 - (v^2_2 /c^2)} v^2_1 {1 - (v^2_2 /c^2)} = v^2_2 {1 - (v^2_1 /c^2)} v^2_1 (c^2 - v^22) = v^22(c2- v^2_1) (v^2_1 c^2 ) - (v^21v^2_2)_ = (v^22c^2) - (v^21v^2_2) v^21c^2= v^22c2 v^21= v^2_2(5) Therefore, the magnitudes of v_1^\ding{217} and v_2^\ding{217} have the same value, say v. Using (2) and (3) 0 =[(m_0\pi(1) v^\ding{217}) / {\surd1 - (v^2 / c^2) }] + [ (m-_0\pi(2) v^\ding{217}) / {\surd1 - (v^2 /c^2)}](6) Hence the momenta of the 2 pions are equal and opposite. From (1), we may calculate the kinetic energies of \pi_1(T_\pi(1)) and \pi_2(T_\pi(2)). Noting that the energy of any particle equals the sum of its rest energy and kinetic energy, we may write E = m_0c^2 + T . With (1) 2m_nc^2 = m_\pi(1) c2+ T_\pi(1) + m_\pi(2) c2+ T_\pi(2) Butm\pi(1)= m\pi(2)and 2m_nc^2 \therefore 2m_\pic^2 = T_\pi(1) + T_\pi(2)(7) The mass-energy of the two neutrons is E_nc= 2m_nc^2 = (2) (1.67 × 10^-27kg) (3 × 10^8 m/s)^2 = 3 × 10^-10 J The rest mass-energy of the two neutral pions is E_\pi =2m_nc2= (2) (2.4 × 10^-28kg) (3 ×10^8m/s)^2 = 4.3 × 10^-11 J Therefore , T_\pi(1)^ + T_\pi(2) = 3 × 10^-10 J - 0.43 × 10^-10 J T_\pi(1)^ + T\pi(2)= 2.57 × 10^-10 J(8) To find how this energy is divided among the pions, note that E = [m_0c^2 / \surd{1 - (v^2 / c^2)}] where v is the velocity of the particle whose energy we wish to measure. Hence E _\pi(1) = [m0\pi(1)c^2]/ [\surd{1 - (v^2_1 / c^2)}] E _\pi(2) = [m0\pi(2)c^2] / [\surd{1 - (v^2_2 / c^2)}] Butv_1 = v_2 ,andm_\pi(1) = m\pi(2). whence E _\pi(1) = E _\pi(2) Orm_\pi(1)c^2+ T_\pi(1) = m_\pi(2)c^2+ T\pi(2) AndT_\pi(1) = T_\pi(2) . Using T_\pi(1) = T_\pi(2) = 1.29 × 10^-10 J.

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Question:

Describe the chemical reactions that occur in the human retina bringing about visual sensation.

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/Users/wenhuchen/Documents/Crawler/Biology/F20-0498.htm

Solution:

Rhodopsin is a visual pigment in the human eye that is responsible for the perception of an enormous range of light intensities. Located in the rod cell of the retina, rhodopsin has a chromophore (retinal1) bound to a protein (opsin). Retinal is the group) of vitamin A, also called retinol (see Figure 1) Retinal can be synthesized from vitamin A via an oxi-dation reaction, catalyzed by the enzyme alcohol de-hydrogenase. Retinal exists in the cis-configuration before light hits the rhodopsin molecule. When light strikes the rhodopsin molecule, cis-retinal absorbs the light energy and is converted to the trans form (Figure 2). Simultaneously, the sensation of vision, the stimulation of nerve impulses by the rod cells, occurs. The conversion causes a rearrangement of the molecular structure, resulting in an unstable compound called lumirhodopsin. Lumirhodopsin rapidly decomposes into free trans-retinal and opsin. This is due to the inability of trans-retinal to fit onto the opsin molecule as does the cis form. Rhodopsin has to be resynthesized from cis-retinal and opsin to provide for further excitation of the rods. Trans-retinal is first reduced to form trans-retinol, which is subsequently enzymatically converted into cis- retinol. The cis-retinol then undergoes oxidation to regenerate cis-retinal, which then combines with opsin to yield rhodopsin. The series of reactions from the light-induced breakdown of rhodopsin to its resynthesis, collectively termed the visual cycle, is outlined in figure 3. Because vitamin A is the precursor of retinal, synthesis of rhodopsin is dependent on its presence. In vitamin A deficiency, vision is most conspicuously affected. This deficiency is responsible for a condition known as xerophthalmia ("dry eyes") and may cause blindness. An early sign of vitamin A deficiency is night blindness, which is a maladaptation of the eyes to the dark.

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Question:

Calculate the mass of the light isotope of helium from the following reaction, which has a Q-value of 3.945 MeV: _3Li^6 + _1H^1 = _2He^4 + _2He^3. The masses in amu of _1H1, _2He^4, and _3Li^6 are 1.00813, 4.00386, and 6.01692, respectively.

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/Users/wenhuchen/Documents/Crawler/Physics/D36-1057.htm

Solution:

In the reaction 3.945 MeV of energy is released. By Einstein's mass energy relation E = mc^2 , this energy is equivalent to a mass whose value is: m = E / c^2 = (3.945 × 10^6 eV ) / ( 9 × 10^16 m^2/s^2) Since1 eV = 1.6 × 10^-19 Joules m = {(3.945 × 10^6 eV ) (1.6 × 10^-19 Joules / eV) } / ( 9 × 10^16 m^2/s^2) m = 7.013 × 10^-30 kg Converting this to atomic mass units by noting that 1 kg = 6.024 × 10^26 amu we obtain m = (7.013 × 10^-30 kg ) (6.024 × 10^26 amu /kg) m = 4.23 × 10^-3 amu m = .00423 amu This must be the difference in mass between reactants and products, or m = mreact. - m _prod.(1) The sum of the masses of the initial particles is m_react. = 1.00813 + 6.01692 = 7.02505 amu(2) But m_prod. = m\alpha+ m_He(3) Using (3) in (1) m = m_react. - m_\alpha- mHe Hence,m_He = m_react. - m\alpha- m Using (2) m_He = 7.02505 amu - 4.00386 amu - .00423 amu mHe= .01696 amu This must be the mass of an atom of the light isotope of helium.

Question:

A man with diabetes mellitus has a sudden fit of anger over a minor traffic accident, and lapses into a coma. Explain the physiological events leading to the coma and what should be done to bring him out of the coma.

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/Users/wenhuchen/Documents/Crawler/Biology/F21-0537.htm

Solution:

Diabetes mellitus is a disease which results from a hyposecretion of insulin. Insulin stimulates glu-cose uptake by all cells with the exception of brain cells. Hyposecretion of insulin reduces the cellular uptake of glucose, leading to an accumulation of glucose in the blood. The blood glucose may rise to such a level that a good deal of it cannot be reabsorbed by the kidney tubules. When this happens the level of glucose excreted into the urine increases tremendously. The osmotic force exerted by glucose in the urine retains water and prevents its reabsorption. Thus, the volume of water excreted is also abnormally high in people with diabetes. This loss of wa-ter, if sufficiently great, can ultimately reduce blood pres-sure to such an extent that there is a marked decrease in blood flow to the brain, resulting in coma and possibly death. In our specific example, coma is induced by the afore-mentioned means. Emotional stress stimulates the release of glucose from the liver through the action of epinephrine, thereby resulting in a rapid increase in blood glucose.In the case of a person with diabetes, this increased blood glucose results in a large amount of water loss, which in turn leads to a decreased blood pressure. Blood flow to the brain is reduced owing to the drop in blood pressure. Coma usually results as blood supply to the brain cells becomes insufficient. Insulin should be given as treatment for this comatose person. Insulin will stimulate glucose uptake, thereby reducing its level in the blood. Glucose concentration in the urine subsequently declines, and water loss is diminished. The pressure of the blood is thereby restored to normal. It is important to note that the dosage of insulin injected is critical. If an overdose of insulin is ad-ministered, the situation is equally fatal had no insulin been injected. This is because insulin cannot enter the brain, and hence the brain cells will not alter their rate of glucose uptake. Thus they are greatly dependent on the blood supply for glucose to maintain their energy metabolism. When too much insulin is given to the diabetic, his blood stream is essentially depleted of glucose units which have entered most of the body cells excepting the brain cells. Hence the brain will starve from an insuffi-cient supply of glucose, and its functions will break down. A coma is usually what follows.

Question:

Compute the wavelength of the electron matter waves in a Davisson-Germer apparatus if the electron velocity is 4.43 × 10^6 m/s.

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/Users/wenhuchen/Documents/Crawler/Physics/D33-1003.htm

Solution:

According to the de Broglie wave theory of matter, every particle has wave-like characteristics. To every particle there is associated a wave with wavelength given by: \lambda = (h/p) = (h/mV). For an electron (of mass 9.1 × 10^-31 kg.) and velocity 4.43 × 10^6 m/s we then have \lambda = [(6.6 × 10^-34 Js) / {(9.1 × 10^-31 kg)(4.43 × 10^6 m/s)}] = 1.64 × 10^-10 m The wavelength for a "matter wave" is now called the particles's de Broglie wavelength. The Davisson-Germer apparatus is used to ex-hibit the interference effects of the electron "matter wave."

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Question:

The mass of an electron is 9.109 × 10^-28 g. What is the atomic weight of electrons?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E19-0706.htm

Solution:

Atomic mass is defined as the weight of one mole of particles. There are 6.02 × 10^23,or Avogadro's number, of particles in 1 mole. Therefore, the atomic weight of electrons is equal to 6.02 × 10^23 times the mass of one electron. atomic weight of electron = 9.109 × 10^28 g × 6.02 × 10^23 / mole =5.48 × 10^-4 g/mole.

Question:

A single circular loop of wire is rotated in a uniform field of magnetic induction B^\ding{217}. We can suppose that the magnetic field is in the +z-direction, and the axis of rotation of the loop is in the x-direction along a diameter. If the speed of rotation \omega is constant, find the induced EMF in the coil.

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/Users/wenhuchen/Documents/Crawler/Physics/D22-0755.htm

Solution:

The flux through the circular loop is, since B^\ding{217} is constant, \varphi M = B \bullet A = BA cos \texttheta Where \texttheta is the angle between B and A (as shown in the Figure.) The angular velocity \omega, for constant speed of rotation, is \omega = (\texttheta/t)\texttheta = \omegat Therefore, \varphi_M = BA cos \omegat According to Faraday's law of induction, the EMF induced in a coil, due to a changing flux, is given by \epsilon = - [(d\varphi_M )/(dt)] where the negative sign indicates that the EMF is induced in such a direction as to oppose the change in flux which created it. Substituting for \varphi_M and differentiating, we then have \epsilon = \omegaBA sin \omegat = \omegaBA sin \texttheta In this case the induced EMF varies with time in a sinusoidal way.

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Question:

A baseball pitcher throws a ball weighing 1/3 pound with an acceleration of 480 feet per second^2. How much force does he apply to the ball?

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/Users/wenhuchen/Documents/Crawler/Physics/D04-0106.htm

Solution:

Newton's Second Law tells us that F = ma. However, we do not have the mass of the ball, but its weight which has the units of force. Since W = mg m = W/g where W is weight, m is mass, and g is the acceleration due to gravity. Therefore, m = [{(1/3) lb} / {32 ft/sec^2}] = (32/3) slugs Since the pitcher accelerates an object of mass (32/3) slugs with an acceleration of 480 ft/s^2, the force is F = (32/3) {(lb - s^2)/32 ft} × 480(ft/s^2) = 5 lbs. of force.

Question:

When did the first mammals appear? Compare the monotremesand marsupials.

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/Users/wenhuchen/Documents/Crawler/Biology/F28-0738.htm

Solution:

The first mammals arose in the earliest period (the Triassic) of the Mesozoic Era which dates to some 230 million years ago. They were warm-bloodedanimals which managed to adapt to the colder climate becauseof the ability to maintain a constant body temperature. Mammals thenproliferated and have become the dominant animal form of modern ages. Themonotremesare the earliest known mammals. Today, the survivorsof themonotremesare the duck-billed platypus and the spiny anteaterof Australia. Both have fur and suckle their young, but lay eggs liketurtles. The marsupials are another group of mammals, whose ancestorsevolved into existence during the Jurassic Period, some 181 millionyears ago. The marsupials are more advanced than the monotremesbecause they bring forth their young alive rather than laying eggs. They are, however, considered among the most primitive of modern mammalsbecause their young are underdeveloped and must remain for severalmonths in a pouch of the mother's abdomen, which contains the nipples.

Question:

In what way does Lamarck's theory of acquired characteristicsnot agree with present evidence?

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/Users/wenhuchen/Documents/Crawler/Biology/F27-0713.htm

Solution:

Lamarck, like most biologists of his time, believed that organisms areguided through their lives by innate and mysterious forces that enable themto adapt to and overcome adverse environmental forces. He believed that, once made, these adaptations are trans-mitted from generation to generation - that is, that acquired characteristics are inherited . In Lamarck's belief, then, new organs can arise in response to demands of the environment, their size being proportional to their use. Changes in the size of such organs are inherited by succeeding generations . An example of this, according to Lamarck was the giraffe's long neck. He claimed that it resulted when an ancestor of the present giraffe took to eating the leaves of trees, instead of grass. Since the giraffe had to continually reach up to eat, its neck became stretched. The longer neck that developed was then inherited by the giraffe's descendants. Lamarck's theory of the inheritance of acquired traits, though attractive, is unacceptable. Overwhelming genetic evidence indicates that acquiredcharacteristics cannot be inherited. Many experiments have been performedin attempts to demonstrate the inheritance of acquired traits, all endingin failure. Geneticists now firmly believe that acquired traits cannot beinherited because they are not translated into a genetic message and in-corporatedinto the genetic apparatus - the chromosomes - of cells. In particular, the genes of the sex cells are not changed. In other words, acquiredtraits involve changes only in the phenotype, not in the genotype; forthis reason they are not inherited, and thus, are of no evolutionary significance.

Question:

A water pump motor has a horsepower rating of 11.3 hp. If it is to pump water from a well at the rate of 1 ft^3/sec (water has a weight density of 62.4 lbs/ft^3), what is the maximum depth the well may have? Neglect friction in your calculations.

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/Users/wenhuchen/Documents/Crawler/Physics/D05-0285.htm

Solution:

The performance of a motor can be measured by the rate at which it does work. This rate is called the power P and is defined as P = ∆W/∆t where ∆W is the work done in the time interval ∆t. One unit of power is the horsepower, and is defined as 550 ft-lbs/sec. Therefore, the maximum power the motor can provide is P_max = (11.3 hp)[(55 ft-lbs/sec)/(hp)] = 6215 ft-lbs/sec The work the pump must do is just equal to the water's change in poten-tial energy, which is due only to gravity. It is equal to ∆W =mgh where h is the height through which the water is raised and is equal to the well's depth. In one second, the pump lifts one ft^3 of water or 62.4 lbs. Then ∆W/∆t = (62.4h lbs)/(1 sec) and the maximum depth is found from P_max= 6215 ft-lbs/sec = 62.4h_maxlbs/sec h_max= [(6215)/(62.4)] ft \approx 100 ft.

Question:

Find the current through the filament of a light bulb with a resistance of 240 ohms when a voltage of 120 volts is applied to the lamp.

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/Users/wenhuchen/Documents/Crawler/Physics/D20-0643.htm

Solution:

Since we wish to find the current, we use Ohm's Law in the form I = (E/R) . E = 120 volts,R = 240 ohms Therefore I = [(120 volts)/(240 ohms)] = 0.5 ampere .

Question:

The weight of the fruit in one variety of squash is determined by three pairs of genes. The homozygous dominant condition, AABBCC, results in 6-pound squashes, and the homozygous recessive condition, aabbcc, results in 3-pound squashes. Each dominant gene adds 1/2 pound to the minimum 3-pound weight. When a plant having 6-pound squashes is crossed with one having 3-pound squashes, all the offspring have 4 1/2-pound fruit. What would be the weights of the F_2 fruit, if two of these F_1 plants were crossed?

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/Users/wenhuchen/Documents/Crawler/Biology/F25-0663.htm

Solution:

This problem deals with polygenic inheritance; that is, the situation in which two or more independent pairs of genes have similar and additive effects on the same characteristic. Examples of such inheritance are height and skin color in man, and commercially important characteristics in animals and plants, such as the amount of eggs and milk produced, the size of fruit, and so on. In the cross between a 6-pound squash plant and a 3-pound squash plant : We are told that each dominant gene adds (1/2) pound to the weight. The presence of 3 dominant genes in F_1 (A, B, and C) would increase the weight by (1/2) + (1/2) + (1/2) or 1 (1/2) lbs. Therefore each squash from F_1 weighs 3 + 1 (1/2)or4 (1/2) pounds. In the cross between two F_1 plants, we must first determine the possible gametes from each plant. The number of possible gametes is obtained using the 2^n rule, where n is the number of heterozygous traits and 2n is the total number of different gametes formed. For each parent, n is equal to 3. Therefore the number of gametes is 2^3 or 8. The genotypes of these gametes can be obtained by dichotomous branching, which ensures that all possible combinations are considered. Each possible gene from each allelic pair is matched to every possible gene combination of the other two pairs as follows : Gametes ABC B C ABc A c AbC b C Abc c aBC B C aBc a c abC b C abc c Looking at the cross : Separating the trihybrid cross into three monohybrid crosses, and then using dichotomous branching to determine all possible combinations of the results of these crosses, one obtains: Summarizing:1/64weighs 6 pounds Summarizing:1/64weighs 6 pounds 6/64weighs 51/2 pounds 15/64weighs 5 pounds 20/64weighs 41/2 pounds 15/64weighs 4 pounds 6/64weighs 31/2 pounds 1/64weighs 3 pounds It is important to be able to associate quantitative characteristics with polygenic inheritance. The cross itself is not difficult but may at times be tedious (Note: this problem could also have been done using the Punnett square, but dichotomous branching is more frequently used in crosses involving three or more traits).

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Question:

At a certain temperature, the half-life for the decomposition of SO_2CI_2 is 4.00 hours, (a) In what period of time would the concentration of SO_2CI_2 be reduced to 10% of the origi-nal? (b) Starting with 100 mg of SO_2Cl_2 what weight would be left at the end of 6.50 hours? The reaction is first order, therefore the equation ln[A]/[A_0] = - kt can be used. [A] is the concentration of A at time t, [A_0 ] is the original concentration of A and k is rate constant.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E13-0464.htm

Solution:

(a) When the concentration of A is 10% of the original, the ratio of [A] / [A_0] equals .10. Therefore, one can use the equation given to solve for t after first determining k for the reaction, k is determined from the half-life in a first order reaction. The relation between k and t_1/2 (half-life) is t_1/2 = 0.693/k One is given the t_1/2. Solving for k: k = 0.693 / t1/2= 0.693 / 4.0 hrs = .173 hr^-1 One can now solve for t: {ln[A]} / [A_0] = -kt ln .10 = - .173 hr^-1 t t = ln.10 / -.173 hr^-1 = - 2.303/ -.173 hr^-1 = 13.3 hrs. (b)One can solve for the [A] after 6.5 hours by use of the equation: ln[A] / [A_0] = - kt One is given that there is originally 100 mg of A. 1 g = 1000 mg, thus 100 mg = 0.1 g, [A_0] must be in moles/liter. One finds the number of moles of SO_2CI_2 in 0.1 g by dividing 0.1 g by the molecular weight of SO_2CI_2(MW = 135). [A_0] = (no. of moles) / (liter) = (0.1g) / [(135 g/moles) / (liter)] = 7.41 × 10^-4 (moles / liter) Solving for [A] using the value of k found in part a: ln{[A] / (7.41 × 10^-4 moles / liter)}= - (.173 hr^-1) (6.5 hr) ln[A] -ln[7.41 × 10^-4 moles] = - (.173 hr^-1) (6.5 hr) ln [A] = - (1.12)+ln(7.41 × 10^-4 ) ln [A] = -1.12 - 7.21 ln [A] = - 8.33 [A] = 2.41 × 10^-4 moles One can find the number of grams by multiplying the number of moles by the molecular weight. weight = 2.41 × 10^-4 moles × 135 g/mole = .0326 g = 32.6 mg

Question:

A uniform circular disk of radius 25 cm is pivoted at a point 20 cm from its center and allowed to oscillate in a vertical plane under the action of gravity. What is the period of its small oscillations? Take g as \pi^2 m.s^-2.

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/Users/wenhuchen/Documents/Crawler/Physics/D09-0359.htm

Solution:

The moment of inertia of a uniform circular disk of radius R and mass M about an axis through its center perpendicular to its plane is 1/2MR^2. By the parallel-axes theorem, the moment of inertia about a parallel axis a distance h from the first is I = 1/2 MR^2 + Mh^2. The disk is acting as a physical pendulum, and hence its period for small oscillations is given by T = 2\pi \surd(1/Mgh) where I is the moment of inertia of the physical pendulum about its axis of suspension. Hence, T = 2\pi \surd[(1/2 MR^2 + Mh^2)/Mgh] = 2\pi \surd[(1/2 ×0.25^2 m^2 + 0.02^2 m^2)/(\pi^2 m\bullets^-2 × 0.2 m)] = 2 \surd(0.35625 s) = 1.193 s.

Question:

There are three processes which together enable the kidney to remove wastes while conserving the useful components of the blood. What are these processes and where do they occur?

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/Users/wenhuchen/Documents/Crawler/Biology/F18-0444.htm

Solution:

Blood flowing to the kidneys first undergoes glomerular filtration. This occurs at the junction of the glomerular capillaries and the wall of Bowman's capsule. The blood plasma is filtered as it passes through the capillaries, which are freely permeable to water and solutes of small molecular dimension yet relatively im-permeable to large molecules, especially the plasma pro-teins. Water, salts, glucose, urea and other small species pass from the blood into the cavity of Bowman's capsule to become the glomerular filtrate. It has been demonstrated that the filtrate in Bowman's capsule contains virtually no protein and that all low weight crystalloids (glucose, protons, chloride ions, etc.) are present in the same concentrations as in plasma. If it were not for the process of tubularreabsorption, the composition of the urine would be identical to that of the glomerular filtrate. This would be extremely wasteful, since a great deal of water, glucose, amino acids and other useful substances present in the filtrate would be lost. Tubular reabsorption is strictly defined as the transfer of material from the tubular lumen back to the blood through the walls of the capillary network in intimate contact with the tubule. The principal portion of the tubule involved in reabsorption is the proximal convoluted tubule. This tubule is lined with epithelial cells having many hair-like processes extending into its lumen. These processes sure the chief sites of tubular reabsorption. As the filtrate passes through the tubule, the epithelial cells reabsorb much of the water and virtually all the glucose, amino acids and other substances useful to the body. The cells then secrete these back into the blood-stream. The secretion of these substances into the blood is accomplished against a concentration gradient, and is thus an energy consuming process - one utilizing ATP. The rates at which substances are reabsorbed, and therefore the rates at which wastes are excreted (because what is not re-absorbed is eliminated), are constantly subjected to physiological control. The ability to vary the excretion of water, sodium, potassium, hydrogen, calcium, and phos-phate ions, and many other substances is the essence of the kidney's ability to regulate the internal environment. Reabsorption also occurs in the distal convoluted tubules, where sodium is actively reabsorbed under the influence of aldosterone, a hormone secreted by the adrenal cortex. When this occurs, chloride passively follows due to an electrical gradient; water is also reabsorbed because of an osmotic gradient established by the reabsorption of sodium and chloride. In addition, reabsorption of water takes place in the distal convoluted tubule and collecting duct, stimulated by the posterior pituitary hormone vaso-pressin, also known as antidiuretic hormone (ADH) . ADH in-creases the permeability of the distal convoluted tubule and collecting duct to water, allowing water to leave the lumen of the nephron and render the urine more concentrated. The kidney also removes wastes by means of tubular secretion. This process involves the movement of addi-tional waste materials directly from the bloodstream into the lumen of the tubules, without passing through Bowman's capsule. Tubular secretion may be either active or passive, that is, it may or may not require energy. Of the large num-ber of different substances transported into the tubules by tubular secretion, only a few are normally found in the body. The most important of these are potassium and hydrogen ions. Most other substances secreted are foreign substances, such as penicillin. In some animals, like the toadfish, whose kidneys lack glomeruli and Bowman's capsules, secre-tion by the tubules is the only method available for excre-tion.

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Question:

Write a program which uses a bubble sort procedure to put given strings in lexical order.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G15-0393.htm

Solution:

The function of BUBBLE sort is based on the function of a simple exchange sort, i.e., when an element, compared to the previous, or next one, is found to be out of order, it is moved to a preceding or following place respectively. The described procedure is continuously performed until that element is properly placed. The solution of this problem is as follows: \textasteriskcenteredBUBBLE SORT PROGRAM &TRIM = 1 DEFINE ('SORT(N)I') DEFINE('SWITCH(I)TEMP') DEFINE('BUBBLE(J)') \textasteriskcentered1:F(ERROR) N = INPUT A = ARRAY(N) \textasteriskcentered2 READI = I +1 A = INPUT:F(GO)S(READ) \textasteriskcentered3 GOSORT(N) \textasteriskcentered4 M = 1 PRINTOUTPUT = A:F(END) M = M + 1:(PRINT) \textasteriskcentered5 SORTI = LT(I,N-1) I +1:F(RETURN) LGT(A,A+ 1>):F(SORT) SWITCH(I) BUBBLE(I):(SORT) SWITCHTEMP =A A = A A =TEMP:(RETUN) BUBBLEJ = GT(J,1) J- 1:F(RETURN) LGT(A,A):F(RETURN) SWITCH(J):(BUBBLE) END Any statement in SNOBOL IV, starting with an asterisk in column 1, is a comment (or remark) statement. These statements are not executable and are usually used to explain the functions of different parts of the program. There are three functions used in this program, which are defined in the beginning of the program. Their procedures are given right after the comment numbered 5. The first card of the input data for this program must contain the number of strings to be sorted, otherwise the program stalls. The part of the program between the comments numbered 1 and 2 reads that number, assigns it to N, and sets up an array of N members. The next program segment (between comments 2 and 3) fills the array with the input items. Notice that counter I is not initialized to zero, because in SNOBOL IV the initial value of the variables is a null string. The statement between the comments 3 and 4 calls and executes the SORT function. Since the entry point label is omitted in the DEFINE statement, the computer looks for the statement labeled SORT, as the initial statement of the function procedure, and finds it right after the comment 5. The function LGT(X, Y), included in the second statement of the SORT function procedure, compares lexically strings X and Y, and succeeds if X follows Y alphabetically. Thus, LGT (SPACE, BLANK) will succeed, while LGT (CAT, DOG) will fail. The next two statements of the SORT function procedure call and execute functions SWITCH and BUBBLE. Function SWITCH makes the two strings switch their places, while the BUBBLE function compares the string with the preceding one and switches them around, if needed, continuously, until the first string is finally properly placed. Thus, the major function SORT works as follows: The first statement (I = LT(I,N-1) I + 1) checks if a currently considered string is numbered 1 through N-1 (there is no need to specially consider the last string N, because, when the string N-1 is considered, it is compared with the string numbered N). The next statement checks the alphabetical order of that string and the one following it in the array. If the order is correct, the next string is considered. If the order is wrong, the fol-lowing string moves up one place, is compared with the next string, moves up again, if needed, and so on, until it is properly placed. When all the input items are considered, the program execution returns to the main part, i.e., to the section between comments 4 and 5 and one by one prints all the input strings in a just arranged alphabetical order. For the input:3 COMPUTER ARRAY BRANCH THE OUTPUT IS:ARRAY BRANCH COMPUTER

Question:

The limits of the visible spectrum are approximately 400 nm to 700 nm. (a) Find the angular breadth of the first-order visible spec-trum produced by a plane grating having 15,000 lines per inch, when light is incident normally on the grating, (b) Show that the violet of the third-order spectrum overlaps the red of the second-order spectrum:

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Solution:

(a) The diffraction grating, shown in Fig. 1 has a grating spacing d given by d = [(2.54 cm)/(in.)] / [(15000 lines) (in.)] = 1.69 × 10^-4 cm. The nth order angular deviation for wavelength \lambda is sin \texttheta = n\lambda/d therefore the angular deviation of the violet is sin \texttheta_v= (\lambda_v)/(d)n = [(4 × 10^-5 cm)]/[(1.69 × 10^-4 cm)] = 0.237, \textthetav= 13\textdegree 40'. The angular deviation of the red is sin \textthetaR= (R)/(d)n = [(7× 10^-5 cm)] / [(1.69 × 10^-4 cm)] = 0.415, \texttheta_R= 24\textdegree30'. The first order deviations (n = 1) for the highest and lowest fre-quencies (violet and red respectively) of the visible spectrum will have an angular difference \texttheta_R - \texttheta_v. Hence, the first order visible spectrum includes an angle of \textthetaR- \textthetav= 24\textdegree30' - 13\textdegree40' = 10\textdegree50'. (b) For n = 3, violet light has the angular deviation sin \texttheta_v= (3\lambda_v)/(d) = [3 × (4 × 10^-5 cm) /[d] The second - order deviation for the red light is sin \texttheta_R = (2\lambda_R)/d = [2 × (7 × 10^-5 cm)] / [d] The first angle \texttheta_v is smaller than the second angle \texttheta_R, therefore whatever the grating spacing, the third order will always overlap the second, as can be seen in Fig. 2.

Question:

Write a program in PL / I to use the Gauss-Seidel method for solving linear equations. Specialize to 3 equations with 3 unknowns.

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Solution:

Consider the system a_11x_1 + a_12x_2 + a_13x_3 = b1 a_21x_1 + a_22x_2 + a_23x_3 = b_2 a_31x_1 + a_32x_2 + a_33x_3 = b_3 The solution algorithm is based on rewriting this system to a form a_11x_1 = b_1 - (a\textasteriskcentered_11x_1 + a_12x_2 + a_13x_3) a_22x-_2 = b_2 - (a_21x_1 + a\textasteriskcentered_22x_2 + a_23x_3) a_33x_3 = b_3 - (a_31x_1 + a_32x_2 + a\textasteriskcentered_33x_3) Let us introduce a P matrix associated with the A matrix, but modified so that the diagonal elements of the matrix A are replaced by 0. Using the new matrix P, one can rewrite the system as a_11x_1 = b_1 - (P_11x_1 + P_12x_2 + P_13x_3) a_22x-_2 = b_2 - (P_21x_1 + P_22x_2 + P_23x_3) a_33x_3 = b_3 - (P_31x_1 + P_32x_2 + P_33x_3) or in contact summation notation a_iix_i= bi- \sum_jP_ijx_jfori= 1, 2, 3. In PL/I notation this last equation becomes X (I) = [B (I) - SUM (P (I, \textasteriskcentered)\textasteriskcenteredX) / A (I, I). Note that we SUM over the subscript \textasteriskcentered corresponding to j. Also note that P_ij =a_ij,i\not = j;P_ii= 0. For this program we read in the dimension N of the square matrix A, and the value of K -the number of iterations. The program looks as follows: START: PROCEDURE; /\textasteriskcenteredGAUSS-SEIDEL\textasteriskcentered/ RPT:GET LIST (N, K) ; BEGIN; DECLARE A (N, N), B (N), X (N) ,P (N, N); GET LIST (A\textasteriskcenteredB); P = A; X = 0; DO I = 1 TO N; P (I, I) = 0; END; DO J = 1 TO K; DO I = 1 TO N; X (I) = [B (I) - SUM (P (I,\textasteriskcentered)\textasteriskcenteredX)) / A (I, I); END; PUT SKIP EDIT (X) (F (10,4)); END; END; PUT SKIP (2); GO TO RPT; END START;

Question:

Two masses of 200 gm and 300 gm are separated by a light rod 50 cm In length. The center of mass of the system serves as the origin of a Cartesian coordinate system. The rod lies in the xy plane and makes an angle of 20\textdegree with the y axis. Find the inertial coefficients I_xx and I_xy with respect to the center of mass.

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Solution:

Since we must calculate the moment of inertia of the rod about an axis through the center of mass, we must first locate the center of mass. Let us find the distance of the center of mass from the 200-gm mass. By definition of center of mass R_c.m. = (M_1x_1 + M_2x_2)/( M_1 + M_2) where x_1, x_2 are the distances of M_1 and M_2 from the origin (in our cases, the 200 gm. mass). Hence, using figure (A), R_c.m. = [(200 gm)(0) + (300 gm)(50cm)]/(500 gm) R_c.m. = 30 cm Looking at figure (B), the Cartesian coordinates of the 200 gm mass (denoted as M_1) are referred to the center of mass as origin, x_1 = (30 cm) sin 20\textdegree = (30 cm)(.342) \approx 10.3 cm y_1 = (30 cm) cos 20\textdegree = (30 cm)(.940) \approx 28.2 cm(1) z_1 = 0 The Cartesian coordinates of the 300 gm mass (denoted as M_2) are x_2 = (-20 cm) sin 20\textdegree \approx -6.8 cm y_1 = (-20 cm) cos 20\textdegree \approx -18.8 cm(2) z_2 = 0 Using these values of the coordinates, we proceed to evaluate the in-ertial coefficients defined by the equations I_xx = \sum_i m_i (y^2_i + z^2_i) I_xy = -\sum_i m_i x_i y_i For our problem, these reduce to I_xx = M_1 (y^2_1 + z^2_1) +M_2 (y^2_2 + z^2_2) I_xx = -M_1 x_1 y_1 - M_2 x_2 y_2 From (1) and (2) I_xx = (200 gm)[(28.2 cm)^2 + 0] + (300 gm) [(-18.8 cm)^2 + 0 ] I_xx = 265080 gm - cm^2 = 2.65 × 10^5gm \bullet cm^2 I_xy = -(200 gm)(10.3 cm)(28.2 cm) - (300 gm)(-6.8 cm)(-18.8 cm) I_xy = -96444 gm - cm^2 = -.96 × 10^5 gm - cm^2 Now suppose that the rod rotates about the x axis with angular velocity \omega. Find the components of J. In general, J_x I_xxI_xyI_xz \omega_x J_y I_yxI_yyI_yz \omega_y J_z I_zxI_zyI_zz \omega_z Where J_x, J_y, J_z, are the vector components of J^\ding{217}, and \omega_x, \omega_y, \omega_z are the componentsof \omega^\ding{217}. The quantities I_xx, I_xy, etc., are the products and moments of inertia of the system we are studying. In our problem, \omega_z = \omega_y = 0 and I_zx = 0. And J_x I_xxI_xyI_xz \omega_x J_y I_yxI_yyI_yz 0 J_z 0I_zyI_zz 0 Therefore, by the definition of matrix multiplication J_x I_xx \omega_x00 J_y I_yx \omega_x00 J_z 000 andJ_x= I_xx \omega_x J_y= I_yx \omega_x J_z= 0. Then(J_y)/(J_x) = (I_yx)/(I_xx) = (-.96 × 10^5)/(2.65 × 10^5) = -.363

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Question:

A thin rigid rod of weight W is supported horizontally by two props as shown in Figure A. Find the force F on the remaining support immediately after one of the supports is kicked out.

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Solution:

The moment the support is kicked out, the rod starts rotating about the other support as the free end of the support falls (Figure B). Let x be the displacement of the center of mass of the rod. Immediately after the kicking of the support, x is very small and is vertical. In this case, (d^2x)/(dt^2) becomes the downward acceleration of the center of mass: m[(d^2x)/(dt^2)] = W - F(1) where m is the mass. The torque on the rod about the remaining support is \cyrchar\cyrt = W(L/2) = I[(d^2\texttheta)/(dt^2)] where I is the moment of inertia with respect to the axis of rotation. The moment of inertia of a thin rod with respect to an end is known to be (1/3)mL^2 , hence (1/2)WL = (1/3)mL^2 [(d^2\texttheta)/(dt^2)] or (d^2\texttheta)/(dt^2) = (3/2)(W/mL) For small x, x \approx (L/2)\texttheta or(d^2x)/(dt^2)= (L/2)(d^2\texttheta)/(dt^2). Form (1) and (2), m(L/2)[(d^2\texttheta)/(dt^2)]= m(L/2)[(3W)/(2mL)] = W - F (3/4)W= W - F givingF= W - (3/4)W = (1/4)W.

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Question:

A capacitor of capacitance 0.1 \muF is charged until the difference in potential between its plates is 25 V. Then the charge is shared with a second capacitor which has air as the dielectric between its plates; the potential difference falls to 15 V. The experiment is repeated with a dielectric between the plates of the second capacitor, and the final potential difference is 8 V. What is the dielectric coefficient of the dielectric used?(See figures (a), (b) and (c)).

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Solution:

The situation is illustrated in figures (a) through (c). The charge possessed by the first capacitor initially, is, by the definition of capacitance (see figure (a)) Q = C_1 V_1 = 10^-7 F × 25 V = 2.5 \muC. This charge is shared with a second capacitor of capacitance C_2, no charge being lost. Hence Q = Q_1 + Q_2 where Q_1 is the charge on C_1 and Q_2 is the charge on C_2 (see figure (B)). Therefore Q = (C_1 + C_2 ) V_2or C_2 = (Q/V_2 ) - C_1 = [(2.5 x 10^-6 C)/(15 V)] - 10^-7 F = (2/3) × 10^-7 F. If a dielectric of coefficient \kappa is placed between the plates of capacitor C_2 and the experiment is re-peated, the capacitance of C_2 is increased to C_3 = \kappa C_2, where \kappa is the dielectric coefficient. In this case, Q is shared by C_3 and C_1. Hence Q = (C_1 + \kappa C_2) V_3 . \therefore\kappa = (1/C_2 ) [(Q/V_3 ) - C_1 ] = (3/2) × 10^7 F^-1 [{(2.5 × 10^-6 C)/(8 V)} - 10^-7 F] = 3/2 × 10^7 × 17/8 × 10^-7 = 3.2. In this analysis, we have used the fact that \kappa for air is equal to \kappa for vacuum, or \kappa_air = \kappa_vac = 1.

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Question:

(a) How much heat energy Is produced by a 5-kg rock that falls a vertical distance of 10 m before it strikes the surface of the earth? Assume that the rock was initially at rest. (b) How much would the temperature of 1 kg of water be raised by the rock striking the earth's surface? (4.19 × 10^3 J of heat energy is required to raise the temperature of 1 kg of water 1\textdegreeK.)

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/Users/wenhuchen/Documents/Crawler/Physics/D13-0467.htm

Solution:

(a) For this motion the kinetic energy is initially zero since the rock was released from rest. Just before colliding with the earth, the kinetic energy is a maximum. The conservation of energy requires that the increase in kinetic energy of the rock must equal the change in its potential energy, U - U_0 =mgh- mgh_0 = mg(h - h_0) Since h - h_0 = 0m - 10m, m = 5 kg, and g = 9.8 m/s^2 , the change in the potential energy is U - U_0 = (5 kg)(9.8m/s^2)(-10 m) = -4.9 × 10^2 J The minus sign means that the potential energy has decreased. The in-creased kinetic energy is then 4.9 × 10^2 J . This kinetic energy is all converted into heat upon collision with the earth. Consequently, the heat energy Q of the rock and earth is Q = 4.9 × 10^2J (b) Since 4.19 × 10^3 J of heat energy will raise the temperature of 1 kg of water 1\textdegreeK, 4.9 × 10^2 J of heat energy will cause a temperature rise of T = (4.9 ×10^2 J)/( 4.19 × 10^2 J/\textdegreeK) = 1.2 × 10^-1 \textdegreeK in 1 kg of water.

Question:

An object of mass 100 g is at rest. A net force of 2000 dynes is applied for 10 sec. What is the final velocity? How far will the object have moved in the 10-sec interval?

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/Users/wenhuchen/Documents/Crawler/Physics/D04-0102.htm

Solution:

Since the force is constant, the acceleration is also. To find a, apply Newton's second law of motion. a = F/m a = F/m = (2000dynes) / (100g) = (2000dynes) / (100g) = 20 cm/sec^2 = 20 cm/sec^2 . The kinematics equations for constant acceleration can be used to find the final velocity. The initial velocity is zero in this problem, since the object is initially at rest. v= v_0 + at v= at = (20 cm/sec^2) × (10 sec) = 200 cm/sec. To find the distance traveled in 10 seconds, use s = v_0t + (1/2)(at^2), and since v_0 = 0 s= 1/2(at^2) = 1/2(20 cm/sec^2)×(10 sec)^2 = 1000 cm = 10 m.

Question:

Let the magnitudes and directions of the emf's, and the Let the magnitudes and directions of the emf's, and the magnitudes of the resistances in the figure be given. Solve for the currents in each branch of the network.

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/Users/wenhuchen/Documents/Crawler/Physics/D20-0682.htm

Solution:

We assign a direction and a letter to each unknown current. The assumed directions are entirely arbitrary. Note that the currents in source 1 and re-sistor 1 are the same, and require only a single letter I_1. The same is true for source 2 and resistor 2; the current in both is represented by I_2. At any branch point we may apply Kirchoff's current law, which states that the current entering a branch point must equal the current leaving a branch point. There are only two branch points, a and b. At point b, \sumI = I_1 + I_2 + I_3 = 0(1) Since there are but two branch points, there is only one independent "point\textquotedblright equation. If the point rule is applied at the other branch point, point a, we get \suml = - I_1 - I_2 - I_3 = 0(2) which is the same equation with signs reversed. We now apply Kirchoff's voltage law to the two loops of the figure. This law states that the sum of the voltage drops around any closed loop of a circuit roust be zero. Hence, for loop defc, v_1 - I_3 r_3 + I_1 r_1 = 0(3) and for loop aefb v_2 - I_3 r_3 + I_2 r_2 = 0(4) We first solve for I_3. Solving (1) for I_1, I_1 = - I_2 - I_3(5) Substituting this in (3) v_1 - I_3 r 3 + (- I 2 - I_3 )r_1 = 0 orv_1 - I_3 r_3-I2r_1-I_3 r_1=0 - 2 r - = v_1 - I_3 (r_1 + r_3) - I_2 r_1 = 0(6) Solving this for I 2 I_2 r_1 = v_1 - I_3 (r_1 + r_3) I_2 = [{v_1 - I_3 (r_1 + r_3)} / (r_1 )](7) Substituting (7) in (4) v_2 - I_3 r_3 + [{v_1 - I_3 (r_1 + r_3)} / (r_1)] r_2 = 0 Solving for I_3 v_2 - I_3 r_3 + [(v_1 r_2)/(r_1)] - I_3 [(r_1 + r_3)/(r_1 )] r_2 = 0 orI_3 [r_3 + {(r_1 + r_3)/(r_1 )} r_2 ] = v_2 + [(v_1 r_2 )/(r_1 )] Hence I_3 = [(v_2 + {(v_1 r_2)/(r_1)}] / [r3 + {(r_1 + r_3)/(r_1)} r_2 ](8) Now, we solve for I_2 by substituting (8) in (4) v_2 - [([(v_2 + {(v_1 r_2)/(r_1)}] r_3 ) / (r_3 + [{(r_1 + r_3) r_2 } / (r_1 }])] + I_2 r_2 = 0 I_2 r_2 = [v_2 + {(v_1 r_2 )/(r_1 )} r_3] / (r_3 + [{(r_1 + r_3) r_2 } / (r_1 )]) - v_2 I_2 = [{[v_2 + {(v_1 r_2)/(r_1)}] (r_3 /r_2)} / {r_3 + [{(r_1 + r_3) r_2} / (r_1)]}] - (v_2/r_2 )(9) Substituting (9) and (8) in (5), we may solve for I_1 I_1 = (v_2 /r_2 ) - [{[v_2 + {(v_1 r_2)/(r_1)}] (r_3 /r_2)} / {r_3 + [{(r_1 + r_3) r_2} / (r_1)]}] - [[v_2 + {(v_1 r_2)/(r_1)}] / {r_3 + [{(r_1 + r_3) r_2} / (r_1)]}] Hence I_1 = (v_2 /r_2 ) -[[v_2 + {(v_1 r_2)/(r_1)}] / {r_3 + [{(r_1 + r_3) r_2}/(r_1)]}] [(r_3/r_2 ) + 1]

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Question:

A cubic foot of sea water at the surface weighs 64.0 lb. What volume of water weighs 100 lb at the sea bed, where the water pressure is 4000 lb\bulletft^-2? The compressi-bility of sea water is 36 × 10^-7 in^2\bulletlb^-1.

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Solution:

Using the definition of weight density D D = (Weight)/(Volume) thenV = W/D = (100 lb)/(64 lb/ft^3) = 1.5625 ft^3 where V is the volume 100 lb of water occupies at the surface. Where the water pressure P is 4000 lb \textbulletft^-2 = 4000 lb \textbulletft^-2 × 1 ft^2/144 in^2 = 4000/144 lb \textbulletft^-2, the volume will have decreased by ∆V due to this additional pressure. The compressibility k of sea water is defined by 1/K = (Stress)/(Strain) = P/(∆V/V) We are given that k = 36 × 10^-7 in^2\bulletlb^-1 \therefore ∆V/V = kp = (4000/144) lb \textbulletin^-2 × 36 × 10^-7 in^2\bullet lb^-1 = 10^-4. \therefore 1 - (∆V/V) = (V -∆V)/V = 1 - 10^-4or V - ∆V = (100/64) (1 - 10^-4) ft^3. = 1.5623 ft^3 Thus the volume occupied at the lower level is 1.5623 ft^3.

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Question:

When a flat plate of glass and a lens are placed in contact, a dis-tinctive interference pattern, known as Newton's Rings, is observed. (See figure A). Derive a formula giving the location of the fringes of the interference pattern relative to the center of the lens. (See figure B).

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Solution:

Destructive interference will result when the waves reflected from the apparatus shown in figure (B) are 180\textdegree out of phase. Let us trace the path of an incident ray. The ray will be partially reflected and partially transmitted at surface (1). When a ray of light is trams- mitted from a region of low refractive index to a medium of higher re-fractive index, it undergoes a phase change of 180\textdegree . Hence, the ray reflected at surface (1) is 180\textdegree out of phase with the incident ray. The transmitted ray next encounters surface (2). At this surface there is no phase change, since the light leaves an area of high refractive index and enters a region of low refractive index. In addition, part of this light is reflected at surface (2). The light transmitted at surface (2) next encounters surface (3) and is reflected with a 180\textdegree or \lambda/2 phase change. (\lambda is the wavelength the light). Hence, the ray reflected at (2), and the ray reflected at (3) are 180\textdegree out of phase. Now, the ray reflected at (3) travels a distance 2t greater than the ray reflected at (2). We will see destructive interference whenever 2t is an integral number of wavelengths, since the additional \lambda/2 re-quired for destruction is provided by the phase change due to reflection. Hence 2t = n\lambdan = 0, 1, 2,...destructive interference(1) 2t = [n + (1/2)]\lambdan = 0, 1, 2,..constructive Interference We must now find the location of the interference fringes in terms of the geometry of figure (B). from figure (B), t = R - \surd(R^2 - d^2)(2) t = R - R\surd(1 - d^2/R^2) t = R {1 - \surd(1 -d^2/R^2)}(3) But d << R and d/R << 1. (This means that the radius of curvature of the lens is large). We may therefore approximate the square root in (3) by the binomial theorem. Therefore, \surd(1 - d^2/R^2) \approx 1 - d^2/R^2(4) Substituting (4) in (3) t = R(1 - 1 + d^2/2R^2) t = d^2/2R d = \surd(2tR) Using (1)d = \surd(n\lambdaR)n = 0, 1, 2,...destructive interference d = \surd[{n+(1/2)}\lambdaRn = 0, 1, 2,...constructive interference The first equation locates the dark rings relative to the center of the lens, and the second equation locates the bright rings.

Question:

A shunt generator has a terminal potential difference of 120 volts when it delivers 1.80 kw to the line. Resistance of the field coils is 240 ohms, and that of the armature is 0.400 ohm. Find the EMF of the generator and the efficiency.

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Solution:

An electric circuit containing a shunt generator (that is, a generator whose field resistance R_F is in parallel with the generator) can be described schematically as shown above. The potential difference at the terminals where a load (or resistance) is connected is the terminal voltage, V. The load, when connected will draw a line current I_L, from the circuit. Therefore, the power deliver-ed to the load is P = I_L V. Solving for I_L = (1800 watts/120 volts) = 15.0 amp Using Ohm's Law,V = I_f R_f,(1) orI_F = (V/R_F) = (120 volts/240 ohms) = 0.500 amp By Kirchoff's current law, the sum of the currents entering a junction equals the sum of the current leaving the junction. Therefore, at point Q I_A = I_F + I_L = 0.50 amp + 15.0 amp = 15.5 amp. By Kirchoff's voltage law, the sum of the voltage drops about a loop is zero. Applying this law to loop (1), we obtain \epsilon = I_fR_f + I_AR_A Using (1) \epsilon = V + I_AR_A \epsilon = 120 volts + (15.5 amp)(.4 ohms) \epsilon = 126 volts The input power P_i = I_A \epsilon = 15.5 amp × 126 volts = 1950 watts and it is delivered by the generator. The output power is the power delivered to the line, and it is given as 1800 watts. Hence, Efficiency = (P_o/P_i) = (1800 watts/1950 watts) = 0.923 = 92.3%.

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Question:

For the following voltaic cell, write the half reactions, For the following voltaic cell, write the half reactions, designating, which is oxidation and which reduction. Write the cell reaction and calculate the voltage (E\textdegree) of the cell from the given electrodes. The cell is Cu; Cu^+2││ Ag^+2 ; Ag.

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Solution:

In a voltaic cell, the flow of electrons creates a current. Their flow is regulated by 2 types of reactions occur concurrently, oxidation and reduction. Oxidation is a process where electrons are lost and reduction where electrons are gained. The equation for these are the half-reactions. From the cell diagram, the direction of the reaction is always left to right. Cu \rightarrow Cu^+2 + 2e^-oxidation Ag^+2 + 2e^- \rightarrow Agreduction . Therefore, the combined cell reaction is Cu + Ag^+2 \rightarrow Cu^+2 + Ag. To calculate the total E\textdegree, look up the value for the E\textdegree of both half--reactions as reductions. To obtain E\textdegree for oxidation, reverse of the sign of the reduction E_O . Then, substitute into E\textdegree_cell = E\textdegree_red + E\textdegree_ox.If you do this, you find E\textdegree_cell= -(E\textdegree_redCu) + E\textdegree_red Ag^+2 = -.34 + .80 = .46 volt .

Question:

The atomic weight of iron is 55.847amu. If one has 6.02 g of iron, how many atoms are present?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E17-0608.htm

Solution:

A mole is defined as the weight in grams of a substance divided by its atomic weight: mole = amount in grams / atomic weight If one calculates the number of moles of iron, then the number of atoms present can be calculated. There are 6.02 × 10^23 atoms per mole. no. of moles of Fe = 6.02 / (55.847 g / mole) = 1.08 × 10-1moles no. of Fe atoms present= (1.08 × 10^-1 moles) × (6.02 × 10^23 atoms / mole) = 6.49 × 10^22 atoms.

Question:

A string of freight cars weighs 200 tons. The coefficient of rolling friction is 0.005. How fast can a freight engine of 400 hp pull the string of cars along?

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Solution:

Power is defined as Work divided by time. P = W/t and W = Fs (where F and s are in the same direction and are force and distance respectively). Then, P = Fs/tand since s/t = v P = Fv The force of friction can be calculated from F =\muN, where \mu is the coefficient of rolling friction and N is the normal force: F = 0.005 × (200 ton × 2000 lb/ton) = 2000 lb Using P = Fv from above and the conversion factor [(550 ft-lb/sec)/(1 hp)] to convert 400 hp to power in units ft-lb/sec: 400 hp (550 ft-lb/sec/hp) = 2000 lb × v v = 110 ft/sec = 75 mi/hr.

Question:

What is the magnetic field intensity at a point in air 30 cm away from and on the extension of the line of a bar magnet 10 cm long whose magnetic moments is 400 poles cm, if the point in question is nearer the north pole of the magnet?

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/Users/wenhuchen/Documents/Crawler/Physics/D21-0699.htm

Solution:

Draw a diagram. Recall that field Intensity H is a vector quantity, which with respect to a pole strength m is given by the expression H = (m/\mur^2 ) which follows directly from Coulomb's Law F = (mm')/(\mur^2) where, by definitionH = f/m' where m' is a test pole with a north polarity. Hence H = (m/\mur^2 ) . In this problem, the pole strength can be evaluated because the nagnetic moment is given by the defining equation M = mL, where M is 400 poles cm, and L is 10 cm. m = (M/L) = (400/10) = 40 poles . It follows that due to the north pole of the magnet, the field intensity H_1 at P is H_1 = (m/\mur^2) = [{40 poles}/{(1) (30cm)^2}] = (40/900) (poles/cm^2 ) = .044 oersteds to the right. Also, the field intensity H_2 due to the south pole is H_2 = [40 poles/{1 (40cm)^2}] = (1/40)(poles/cm^2 ) ThereforeH is the vector sum of H_1 and H_2 or H = .044^\ding{217} - .025^\ding{217} = .019 oersteds to the right. This problem is completely analogous to the problem of finding the electric field intensity at point p, due to two charges, one positive and one negative, at a distance 30 cm and 40 cm from the point P. Coulomb's law for the electrostatic force between charges would be the electro-static analogue for Coulomb's law for the magnetic force between poles.

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Question:

Render the following "nonsense"pseudocodeinto FORTRAN, including appropriate comments.Name the law ofpropositional logic you need to render the DO-WHILE and IF-THEN-ELSE constructs: inputL, J, KOUNT, N I \leftarrow 1 dowhile L < J\textasteriskcentered\textasteriskcentered3 and I < KOUNT KOUNT \leftarrow KOUNT + 2 ifJ < 0 or L < 0 thenJ\leftarrow J + 1 elseL \leftarrow L + 1 endif I \leftarrow I + 1 enddo-while outputKOUNT, J, L endprogram

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G08-0178.htm

Solution:

The name of the law requiredIsoften referred to asDeMorgan's Law, which is stated thus: not(P and Q) \equiv not P or not Q Actually,DeMorgan'slawIncludesthe Boolean equation P or Q \equiv not [not P and not Q] Here, the two constructs use the former equivalence. INTEGER L, J, KOUNT,N READ (5,100) L,J,KOUNT,N 100FORMAT (4 I 5) I =1 . CDO WHILE L LESS THAN J\textasteriskcentered\textasteriskcentered3 AND CI IS GREATER THAN KOUNT 10IF (L. GE.J\textasteriskcentered\textasteriskcentered3.OR.I. GE. KOUNT) GO TO 20 KOUNT = KOUNT + 2 CIF J LESS THAN 0 OR L GREATER CTHAN ZERO THEN DO IF (.NOT. (J.GE.0.AND.L.LE.0)) GO TO 30 CELSE DO L = L + 1 GO TO 40 30J = J + 1 40CONTINUE CENDIF I = I + 1 GO TO 10 CEND DO-WHILE 20CONTINUE STOP END

Question:

How does the Watson-Crick model of DNA account for its observed properties?

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/Users/wenhuchen/Documents/Crawler/Biology/F24-0612.htm

Solution:

By 1950, several properties of DNA were well established. Chargaff showed that the four nitrogenous bases of DNA did not occur in equal proportions, however, the total amount ofpurinesequaled the total amount ofpyrimi-dines(A+G = T + C). In addition, the amount of adenine equaled the amount of thymine (A = T), and likewise for guanine and cytosine (G = C). Pauling had suggested that, the structure of DNA might be some sort of an \alpha-helix held together by hydrogen bonds. The final observation was made by Franklin and Wilkins. They inferred from x-ray diffraction studies that the nucleotide bases (which are planar molecules) were stacked one on top of the other like a pile of saucers. On the basis of these observed properties of DNA, Watson and Crick in 1953 proposed a model of the DNA molecule. The Watson and Crick model consisted of a double helix of nucleotides in which the two nucleotide helices were wound around each other. Each full turn of the helix measured 34 \AA and contained 10nucteotidesequally spaced from each other. The radius of the helix was 10 \AA. These measurements were in agreement with those obtained from x-ray diffraction patterns of DNA. To account for Pauling's observation, Watson and Crick proposed that the sugar-phosphate chains of DNA should be on the outside, and the purinesandpyrimidineson the inside, held together by hydrogen bonds between bases on opposite chains. When they tried to put the two chains of the helix together, they found the chains fitted best when they ran in opposite directions.Moveover, because x-raydiffragtionstudies specified the diameter of the helix to be 20 \AA, the space could only accommodate onepurineand onepyrimidine. If twopurinebases paired, they would be too large to fit into the helix without destroying the regular shape of the double stranded structure. Similarly, twopyrimidineswould be too small. Moreover, for onepurineto be hydrogen-bonded with onepyrimidine properly, adenine must pair with thymine and guanine with cytosine. An A -T base pair is almost exactly the same in width as a C - G base pair , accounting for the regularity of the helix as seen in x-ray diffraction pictures. This concept of specific base pairing explained Chargaff's observation that the amounts of adenine and thymine in any DNA molecule are always equal, and the amounts of guanine and cytosine are always equal. Two hydrogen bonds can form between adenine and thymine, and three hydrogen bonds between guanine and cytosine (Fig. 2). The specificity of the kind of hydrogen bond that can be formed provides for correct base pairing during replication and transcription. The two chains are thus complementary to each other, that is, the sequence of nucleotides in one chain dictates the sequence of nucleotides in the other. The strands are alsoantiparallel, i.e., they extend in opposite directions and have their terminal phosphate groups at opposite ends of the double helix.

Question:

An air-core toroid of cross-sectional area A and of mean cir-cumferential length l is closely wound with N turns of wire. If N = 100, A = 10 cm^2 , l = 0.50 m, find its self inductance, L.

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/Users/wenhuchen/Documents/Crawler/Physics/D22-0744.htm

Solution:

The magnetic flux density B throughtout the coil will not deviate appreciably from its value at the mean radius r_0 of the torus, if its width is much smaller than l = 2\pir_0 . Applying Ampere's Law around the circle with radius r_0, we get \oint B.dl = \mu_0 NI orBl = \mu_0 NI B = [(\mu_0 NI)/l] and the flux is \varphi_B = BA = [(\mu_0 NAI)/l] . If a current I sets up a magnetic flux, \varphi_B a coil, then \varphi_B and I are related through the self inductance of the coil as follows, L = [(N\varphi_B )/I] where N is the number of turns. Hence L = [(\mu_0 N^2 A)/l] orL = [{4\pi × 10^-7 (w/a-m)} × (100)^2 × {1m^2 /(10^4 m^2 ) × 10 cm^2 }] \textbackslash(0.50 m) \approx 25 × 10^-6 henry .

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Question:

Two trains moving along parallel tracks in opposite directions converge on a stationary observer, each sounding its whistle of frequency 350 cycles. s^-1. One train is traveling at 50 mph. What must be the speed of the other if the observer hears 5 beats per second. The speed of sound in air is 750 mph.

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/Users/wenhuchen/Documents/Crawler/Physics/D26-0832.htm

Solution:

When a source is moving toward a stationary observer, the latter hears a frequency for the emitted note which is related to the frequency of the source by the expression f =uf_s/(u -v_s) where u is the speed of sound inair,v_sis the speed of the source andf_sis the frequency of the sound emitted by the source. If one of the trains is moving toward the observer with v_1- and the other with speed v_2, then since both have the same frequency whistle, f_1 = (uf_s)/( u - v_1)and f_2 = (uf_s)/( u - v_2) = (750 mph × 350 s^-1) / [(750 - 50) mph] = 375 cycles \bullet s^-1 But the observer hears 5 beats per second. This corresponds to a frequency difference f_1 = f_2 = \pm 5 cycles \textbullet sec^-1. Hence, f_1 = 370 cycles \textbullet s^-1 or 380 cycles \textbullet s^-1 \therefore f_1 = (uf_s)/( u - v_1) = 370 s^-1or380 s^-1 \therefore u - v_1 = (750 mph × 350 s^-1) / (370 s^-1)or (750 mph × 350 s^-1) / (380 s^-1) = 709.5 mphor690.8 mph \thereforev_1 = 40.5 mphor59.2 mph.

Question:

Complete the equations in Figure A. Name the reactants and the products.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E21-0777.htm

Solution:

Compound I is an acid chloride and compound II is an anhydride. Both of these substances are derivatives of carboxylic acids. Such species undergo nucleophilic substitution reactions. In these reactions, a stronger base displaces a weaker base. Thus, to complete these reactions, investigate whether H_2O is a stronger base than the Cl- ion or the carboxylate anion of the acid halide and the anhydride, respectively. A stronger base will replace a weaker base in a compound. Cl- ion and carboxylate anion are weak bases, so that the following occurs The stronger base OH- ion, replaced the weak CI- ion base. Reacton 2 is shown in Figure B. Again, OH- replaced the weaker carboxylate anion base.

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Question:

How much water must be added to 5 gallons of 90% ammonia solution to reduce the solution to a 60% solution?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E33-0944.htm

Solution:

The volume of the original solution is 5 gallons. If we let x represent the number of gallons of water added to the original solution, then the final solution will have a volume of (5 + x) gallons. No ammonia is added so the only ammonia in the final solution is the ammonia in the original solution. We can now record our given information in tabular form: Volume Percent (Gallons) Percent Concentration Gallons of 100%Ammonia Original solution 5 90 4.5 Final solution 5 + x 60 4.5 Note: The gallons of 100% ammonia for the original solution was obtained in the following way: multiply the total volume of the original solution by the percent concentration of the original solution. Therefore, 5 × 90% =5× (90/100_20) = (90/20) = (9/2) = 4.5 gallons of 100% ammonia. Also, the gallons of 100% ammonia for the final solution is the same as the gallons of 100% ammonia for the original solution; that is, 4.5 gallons, since no ammonia was added. The formula for the amount of substance in solution for the final solution is (total volume) × (percent concentration) = gallons of 100% ammonia. Therefore: (5 + x) × 60%= 4.5 or(5 + x) (60/100)= 4.5 or(5 + x).60= 4.5 or.60 (5 + x)= 4.5 3 + .60x= 4.5 .6x= 4.5 - 3 .6x= 1.5 x= (1.5) / .6 = 2.5 gallons. Therefore, 2.5 gallons of water must be added to reduce the concentration to 60%. Check: If 2.5 gallons of water are added to the original solution, we will have 7.5 gallons of solution, The amount of 100% ammonia remains the same 4.5 gallons. The concentration equals (4.5 / 7.5) × 100 = (450 / 7.5) = 60%.

Question:

To a good approximation, the force required to stretch a spring is proportional to the distance the spring is extended. That is, F = kx where k is the so-called spring constant or force constant and depends, of course, on the dimensions and material of the spring. Many elastic materials, if not stretched too far, obey this simple relationship - called Hooke's law after Robert Hooke (1635-1703), a contemporary of Newton. Suppose that it requires 100 dynes to extend a certain spring 5 cm. What force is required to stretch the spring from its natural length to a length 20 cm greater? How much work is done in stretching the spring to 20 cm?

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/Users/wenhuchen/Documents/Crawler/Physics/D08-0350.htm

Solution:

The force constant is k = F_1/x_1 = (100 dynes)/(5 cm) = 20 dynes/cm Therefore, the force required to extend the spring to 20 cm is F_2 = kx_2 = (20 dynes/cm) × (20 cm) = 400 dynes The average force required to stretch the spring to 20 cm is F= (F_2 + F_0)/2 Where: F_0 = initial force exerted on spring to keep its equilibrium length = kx_0 = k \bullet 0 = 0 F_2 = force required to stretch the spring to 20 cm = kx_2 ThenF= (F_2 + 0)/2 = kx_2/2 and the work expended is the average force multiplied by the distance: W =Fx_2 = (1/2) kx^2_2 = (1/2) × (20 dynes/cm) × (20 cm)^2 = 4000 dynes - cm = 4000 ergs since 1 erg = 1 dyne - cm.

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Question:

What are the fourextraembryonicmembranes foundin animals which lay their eggs on land, and what are their functions ?

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/Users/wenhuchen/Documents/Crawler/Biology/F23-0596.htm

Solution:

The development ofextraembryonicmembranes freed early reptiles from an obligatory embryonic de-velopment in water and set the scene for the flourishing of the land-dwelling vertebrates. The mammals have capitalized on their inheritance ofextraembryonicmembranes from the reptiles by utilizing them in intra-uterine development. Thechorion, orserosa, is an outer covering which, in reptiles, rests in contact with the inner surface of the shell, and which in mammals is established next to the cells of the uterine wall. This layer is the site where exchange of substances, largely gases, foods and wastes takes place between embryonic tissue and the maternal environment. To facilitate the exchange , various adaptations have been acquired. In the pig, the chorion elongates tremendously. Its great surface area affords abundant contact with the uterine wall. In most mammals, thechoriondevelops complex outgrowths known as chorionicvilli. These penetrate into the tissue of the uterine wall and greatly increase the area and intimacy of contact between fetal and maternal tissues. Usually there is a specialized region of thechorionwhere exchange between the mother and embryo is most efficient. Such an area is called the placenta. The humanchorion also secretes human chorionicgonadotropin(HCG), a hormone which functions to maintain the corpusluteum. The amnion is a thin fluid-filled, membranous sac which surrounds the embryo and provides it with protection from pressure, abrasion, irritation , and loss of water. The fluid within the amnion is the amniotic fluid . During labor of humans and other mammals, the pressure of the contracting uterus is transmitted via the amniotic fluid and helps to dilate the neck of the uterus. Later, shortly before the fetus is born, the amnion normally ruptures, re-leasing about a liter of amniotic fluid. This is referred to as the breaking of the water bag. Sometimes it fails to burst and the child is born with the amnion still enveloping its head. The amnion is essentially anoutfoldingof the body wall of the embryo consisting of ectodermal andmesod-ermaltissues. The inner part of this fold forms the amnion and the outer part forms thechorion. A thirdextraembryoniclayer found in reptiles, birds, and mammals is the yolk sac, which is an out-growth of the digestive tract of the embryo. This sac is richly endowed with blood vessels and functions to store and supply food. The human egg cell contains minimal yolk, but the sac has other functions. In addition, the yolk sac is the first source of blood cells and blood vessels of the embryo, and also performs biochemical functions which are later taken over by the liver. Theallantois, like the yolk sac, is an outgrowth of the digestive tract . It grows between the amnion andchorionand it fills almost all the space between the two. In reptiles and birds, theallantoisserves as a depot for nitrogenous wastes. The products of nitrogen metabolism are excreted as uric acid by the kidney of the developing embryo. The poorly soluble uric acid is deposited as crystals in the cavity of theallantoisand is discarded along with theallantoiswhen the young hatch out of the egg shells . In birds and reptiles, theallantoisfuses with thechorionto form a compound chorioallantoicmembrane, across which gas exchange takes place . Recall that in mammals both food and gases are exchanged across the chorion. In reptiles and birds, only gas is exchanged across this membrane which is here associated with theallantois. The egg shell is porous and oxygen coming in through it is picked up by blood vessels in the chorioallantois. In the human, wastes are transported through the placenta into the mother. Theallantoisis thus less important as an exchange site. But theallantoicstalk becomes the umbilical cord, linking the chorionto the embryo.

Question:

One electron is removed from O_2 and one from N_2. The bonding in O_2 is strengthened, while the bonding in N_2 is weakened. 1) Explain these findings, and 2) predict what happens if an electron is removed from NO.

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Solution:

There are two types of molecular orbitals: bonding and antibonding. A chemical bond is strengthened by electrons in bonding orbitals and weakened by electrons in antibonding orbitals. For every bonding orbital, there is a corresponding antibonding orbital. Each orbital can hold 2 electrons. Bond order measures bond strength by giving an indication of the number of electrons in bonding versus antibonding orbitals. Bond order is defined as one half the number of electrons in the bonding orbital less one half the number of electrons in the antibonding orbital. Thus, the higher the bond order, the stronger the bond. To find the original bond order, consider the valence electrons, the outermost electrons, since they are the only ones that participate in bonding. For N, Z = 7. Its electron configuration is 1s^2 2s^2 2p^3. The outer-most electrons are in 2p^3. This means that in N^2 there are a total of six valence electrons. There exist 3 bonding p orbitals. They can accommodate the six valence electrons. Since no electrons need be in antibonding orbitals, bond order = (1/2) (6) - (1/2) (0) = 3. When an electron is removed, the bonding orbitals have only 5 electrons. Thus, the bond order becomes (1/2) (5) - (1/2) (0) = 2.5. Since the bond order went from 3 to 2.5, the bond is weakened by removing the electron from N_2. For O, Z = 8. Since its electron configuration is 1s^2 2s^2 2p^4, each O atom has 4 valence electrons. O_2 has a total of 8 valence electrons. The three bonding p orbitals can hold only 6 .of these electrons. Thus, two electrons are in antibonding orbitals. The bond order is (1/2) (6) - (1/2) (2) = 2. When one removes an electron, it is removed from the antibonding orbitals, if electrons exist in such orbitals. Therefore, the O_2, after removal of the electron, has only 1 electron in an antibonding electron. This means the bond order becomes (1/2) (6) - (1/2) (1) = 2.5. The bond increased from 2 to 2.5, which means that bond strength increases when one removes an electron from O_2. To predict what happens to NO bond strength, consider the bond order before and after the electron removal. Recall, an O atom has 4 valence electrons and N has 3 valence electrons. The total in NO is seven. The 3 p bonding orbit-als can hold six of these electrons. This means 1 electron is in on antibonding orbital. Bond order = (1/2) (6) - (1/2) (1) = 2.5. If one removes one electron, the antibonding orbitals contain zero electrons. Thus, the bond order becomes (1/2) (6) - (1/2) (0) = 3. The bond increased from 2.5 to 3, which means the bond strength increases. Therefore, one can pre-dict the chemical bond strength of NO increases when an electron is removed.

Question:

You are handed a stack of IBM cards, each of which is supposed to contain a number between 0 and 50. Write a FORTRAN program to sum the results of the following operations on these numbers: a) Multiply numbers between 0 and 10 inclusive by 3 b) Divide numbers between 11 and 30 inclusive by 2 c) Subtract 5 from all numbers from 31 to 50 inclusive. Also have the program sum the number of errors in the input deck, i.e., how many cards contain numbers outside the specified range. The program should terminate when the value 99.9 is encountered in the deck.

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Solution:

A flowchart outlines the structure of the algorithm to be used: The idea is to check each number to find out the range in which it lies. The specified operation is performed upon every number; each answer (stored in the variable TEMP) is added to the final answer - the sum of all the adjusted numbers. Control passes to the output stage when the number 99.9 is reached. DATA ERRSUM, SUM, TEMP/0.0, 0.0, 0.0/ 10READ (2,100) X 100FORMAT (F3.1) IF (X.EQ. 99.9) GO TO 98 IF (X. LT.O. .OR.X.GT.50.) GO TO 20 20ERRSUM = ERRSUM + 1 GO TO 10 IF (X.GE.0. AND.X.LE.10.) GO TO 30 30TEMP = X{_\ast}3.0 SUM = SUM + TEMP GO TO 10 IF (X.GE.11.AND.X.LE.30.) GO TO 40 40TEMP = X/2.0 SUM = SUM + TEMP GO TO 10 TEMP = X - 5.0 SUM = SUM + TEMP GO TO 10 98WRITE (6,101) SUM, ERRSUM 101FORMAT (1X ,' SUM =', F8.4,' ERRORS IN DECK = ',F4.0) STOP END

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Question:

A lead ball, initially at rest, is dropped from a certain height. Assume it takes the ball 5 seconds to reach the ground. Develop a program segment to give position and velocity at one-second intervals.

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Solution:

The initial velocity v_0 zero. The formula for the change of position is y = (1/2)gt^2 , where g = -9.8 meters/sec^2. We include the negative sign since the ball travels in the negative y-direction. Now, with a constant acceleration, the instantaneous velocity is V_n =V_o + at , where n = 1,2,3,4,5. By iteration, we can simulate the ball 's^-motion. WRITE (5,100) 100FORMAT (1X,'POSITION', 5X, 'VELOCITY') DO 20 NT = 1,5 Y = 0.5 {_\ast} -9.8 {_\ast}(NT{_\ast}{_\ast}2) V = -9.8 {_\ast} (NT) WRITE (5,101) Y,V 101FORMAT (1X, F10.4, 5X, F10.4) 20CONTINUE

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Question:

A coil of resistance 10 ohm and inductance 0.1 H is in series with a capacitor and a 100\rule{1em}{1pt}V 60\rule{1em}{1pt}cycle \textbullet s^\rule{1em}{1pt}1 source. The capacitor is adjusted to give resonance in the circuit. Calculate the capacitance of the capaci-tor and the voltages across coil and capacitor.

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Solution:

The (complex) impedance Z (seen looking into the terminals a\rule{1em}{1pt}b) of the series R\rule{1em}{1pt}L\rule{1em}{1pt}C circuit (see figure) is Z^\ding{217} = (V^\ding{217}/I^\ding{217}) = R + j (X_L - X_C) where the reactances X_L = 2\pifL = \omegaL and X_C = 1/2\pifC = 1/\omegaC. f is the frequency of variation of voltage. The negative sign is introduced above because the inductive and capacitive reactances tend to cancel one another's effect. In the series combination, the voltages tend to cancel because they are in phase opposition. The (complex) voltage V^\ding{217} = \vertv^\ding{217}\vert e^j\texttheta1 where \vertv^\ding{217}\vert is the magnitude of the voltage and \texttheta_1 is the phase angle of the voltage relative to the current I^\ding{217} (= \vert I^\ding{217} \vert e^jo = \vert I^\ding{217} \vert). The magnitude and phase of the impedance are respectively \vert Z^\ding{217} \vert = \surd{R^2 + (X_L - X_C)^2}\texttheta = arctan {(X_L - X_C)/R} ThenZ^\ding{217} = \vert Z^\ding{217} \vert e^j\texttheta At resonance, X_L = X_C . (1/\omegaC) = \omegaL \therefore C = [1/(\omega^2 L)] = {1/(4\pi^2 × 60^2 s^\rule{1em}{1pt}2 × 0.1 H)} = 70.4 \muF. Further, the circuit becomes purely resistive for \vert Z^\ding{217} \vert = \surd(R^2 + O^2) or \vert Z^\ding{217} \vert = R, and \texttheta = arctan 0 = 0\textdegree. Thus \vert I^\ding{217} \vert = (\vert V^\ding{217} \vert/R) = (100 V/10\Omega) = 10 A. The voltage difference across the inductor is V^\ding{217}_L = I^\ding{217}Z^\ding{217}_L where Z_L is the impedance looking into the terminals a \rule{1em}{1pt} c (see the figure). However, the only impedance present is the inductive reactance X_L . Then \vert Z^\ding{217}_L \vert = X_L = 2\pifL \vert V^\ding{217}_L \vert = \vert I^\ding{217} \vert X_L = 10A × 2\pi × 60 s^\rule{1em}{1pt}1 × 0.1 Henry = 377 V. The voltage difference across the capacitor is, similarly, V^\ding{217}_C = I^\ding{217} Z^\ding{217}_C where Z_L is the impedance looking into the terminals c \rule{1em}{1pt} d (see the figure). But \vert Z^\ding{217}_C \vert = X_C = 1/(2\pifC), hence \vert V^\ding{217}_C \vert = \vert I^\ding{217} \vert X_C = 10 A × {1/(2\pi × 60s^\rule{1em}{1pt}1 × 70.4 × 10^\rule{1em}{1pt}6 F)} = 377 V. Thus, \vertV^\ding{217}_C \vert = \vert V^\ding{217}_L \vert. However V^\ding{217}_C and V^\ding{217}_L are in phase opposition or \vert \texttheta_C - \texttheta_L\vert = \pi, hence, their effects cancel. Therefore, at any time, V_ad = V_ac + V_cd = 0 and the voltage of the source appears across the resistor only. Notice that \vert V^\ding{217}_C \vert and \vert V^\ding{217}_L \vert , individually, may have values greater than source voltage.

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Question:

The equilibrium constant and standard Gibbs free energy change for ammonia synthesis at 400\textdegreeC or 673\textdegreeK are 1.64 × 10^-4 and 11, 657 cal/mole, respectively. The equation for this reaction is (1)N_2 (g) + 3H_2 (g) \rightarrow 2NH_3 (g) Calculate the equilibrium constants and standard free energy changes for (2)(1/2)N_2 (g) + 3/2 H_2 (g) = NH_3 (g)and (3)2NH_3 (g) \rightarrow N_2 (g) + 3H_2 (g).

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/Users/wenhuchen/Documents/Crawler/Chemistry/E15-0545.htm

Solution:

This problem illustrates the necessity of knowing the balanced chemical equation in order to determine the numerical value of the equilibrium constant. For ideal gases, the equilibrium constant,K_p, is the product of the partial pressure of the product gases, each raised to the power that corresponds with itsstoichiometriccoefficient in the balanced equation, divided by the product of the partial pressure of the reacting gases, each raised to the power that corresponds with itsstoichiometricco-efficient in the balanced equation. The quantity ∆G\textdegree is the standard change in Gibbs free energy. It is the difference between the Gibbs free energy of the products and the Gibbs free energy of the reactants. The Gibbs Free Energy and the equilibrium constant are related in the equation ∆G = - RT InK_p where R is the gas constant, T the absolute temperature andK_pthe equilibrium constant calculated using partial pressures. The equilibrium constant for reaction 1 is The equilibrium constant for reaction 1 is K_1 = [(P^2_NH3)/(P_N2 P^3_H2 )] = 1.64 × 10 K_1 = [(P^2_NH3)/(P_N2 P^3_H2 )] = 1.64 × 10 -4 Thus, for (2) (1/2)N_2 (g) + (3/2)H_2 (g) = NH_3 (g) Thus, for (2) (1/2)N_2 (g) + (3/2)H_2 (g) = NH_3 (g) K_2 = [(P_NH3)/(P^1/2_N2 P^3/2_H2 )] = K^1/2_1 = (1.64 × 10^-4 )^1/2 = 1.28 × 10^-2 By definition, at equilibrium, ∆G\textdegree = - RT In K_2 = - (1.987 cal\textdegreeK^-1 mole^-1) (673\textdegreeK) (2.303) × log(1.28 × 10^-2) = 5829 cal/mole. And for (3) 2NH_3 (g) = N_2 (g) + 3H_2 (g) K_3 = [(P_N2 P^3_H2 )/(P^2_NH3)] = (1/K_1) = [(1)/(1.64 × 10^-4 )] = 0.6098 × 10^4 ∆G\textdegree_3 = - RT In K_3 = - (1.987 cal\textdegreeK^-1 mole^-1) (673\textdegreeK) (2.303) × log (0.6098 × 10^4) = - 11, 657 cal/mole. If a reaction is reversed, the equilibrium constant becomes the reciprocal of that for the first reaction; the standard Gibbs free energy change has the same magnitude but the opposite sign.

Question:

Explain the features of the following program. EXAMPLE: PROCOPTIONS(MAIN); DCL BIT_STRINGBIT(4) VARYING; GETLIST(BIT_STRING); IF BIT_STRING \lnot=0 THEN PUT LIST (BIT_STRING, 'NOT ALL BITS ARE ZERO'); ELSE PUTLIST(BIT_STRING,'ALL BITS ARE ZERO'); END EXAMPLE,

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Solution:

The program illustrates the features of bit strings inPL/I. In the declare statement BIT_STRING is the name of a bit string variable. A bit string variable can take bit string data as values, just as arithmeticvariables take on numeri-cal values. A bit string constant consists of a sequence of one or more of the digits0 and 1. The constant is enclosed within quotes. The terminating quotessymbol is immediately followed by the letter B indicating a bit string. e.g.'111'B means that 111 is a bit string. (3)'101'B means that 101101101 is a bit string. ''Bmeans a null length bit string. Any bit string variable must always be explicitly de-clared as in the givenexample. The bit attribute consists of the keyword BIT followed by aninteger enclosed in paren-theses indicating the length of the bit string. Hence (4) in the DCL statement specifies the length of the variable named asBIT_STRING. The attribute VARYING specifies that the variable BIT_STRING is of varying length, and its maximum length is 4. The bit stringvariable can also be initialized using the attribute INITIAL. e.g. DECLARE INTEREST VARYING BIT(5) INITIAL(' 011 'B); initializesthe variable INTEREST to the bit string value 011. The second statement in the given program is a value ac-quisition statement. The rules regarding value acquisition are that all data on the datacard (input stream) be in the form of bit string constants without repetitionfactors. If the value in the input is longer in length than the declaredlength, truncation occurs and the rightmost excess digits are deleted. e.g. If on the data card we have '001100'B then, the value acquired will be BIT_STRING=0011. Thus, the last two bits are truncated. Similarly, if the value specified in the input is shorter in length than thedeclared length, padding, by means of zero bits occurs at the right. e.g. If on the data card the value is '01'B then, the value acquired by the BIT_STRING is equal to 0100. No padding occurs if the VARYING attribute is used. The third statement in the given program is of decision making, usingbbit strings. The variable which is named as BIT_STRING is tested, b inorder to arrive at a certain decision. For decision-making, the PL/I machine accepts a bit string of arbitrarylength. The True path is taken if the bit string contains at least onebit with the value 1. The False path is taken if none of the bits in the stringhas the value one. The null string ' 'B does not contain any bit; the False path is taken if a null string is present. In the same third statement the second part involves value disposition(PUT statement). It prints out the value of BIT_STRING and alsoprints out the remark depending on the decision which is made. Thus, bit-string processing is quite like character-string processing, andhas the facility of using the same built-in functions (SUBSTR, LENGTH, INDEX), concatenation and comparison operators as for character-strings.

Question:

Discuss the chemical composition of DNA.

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Solution:

Deoxyribonucleic acid or DNA is one type of nucleic acid present in the nucleus of all cells. Nucleic acids are rich in phosphorus, and contain carbon, oxygen, hydrogen, and nitrogen. DNA is composed of nitrogenous bases, a five-carbon sugar called deoxyribose, and phosphate groups. There are four kinds of nitrogenous bases: adenine, guanine, cytosine, and thymine. Adenine, and guanine belong to the class of organic compounds called purines, and cytosine and thymine belong to the pyrimidines. Each nitrogenous base is attached to deoxyribose molecule via a glycosidic linkage, and the sugar is attached to the phosphate by an ester bond. This combination of base-sugar-phosphate comprises the fundamental unit, termed a nucleotide, of nucleic acid. Four kinds of deoxyribonucleotides are found in DNA, each containing one of the four types of nitrogenous bases. The nucleotides are joined by phosphate ester bonds into a chain. Two complementary chains of nucleotides form a DNA molecule.

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Question:

A chemist has an E. coli bacterium cell of cylindrical shape. It is 2\mu (microns) long and 1\mu in diameter and weighs 2 × 10^-12 g. (a) How many lipid molecules are present, assuming their average molecular weight to be 700 and lipid content to be 2%? (b) If the cell contains 15,000 ribosomes, what percent of the volume do they occupy? You may assume them to be spherical with a diameter of 180 \AA; also 10^4 \AA = 1 micron.

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Solution:

(a) Because you are told the molecular weight of an average lipid molecule, you need only compute their total weight in the E. coli cell; and then find the number of moles present. A mole is defined as weight in (grams / mo-lecular weight) (MW) . From this, you use the fact that in one mole of any substance, (6.02 × 10^23) molecules (Avogadro's number) exist. The lipid content is given as 2%. If the total weight of the cell is (2 × 10^-12) g, the lipid molecules must have a total weight of (.02) (2 × 10^-12) or 4 × 10^-14 grams. The average molecular weight of a lipid molecule is given as 700 (g / mole). Thus, you have 4 × 10^-14 / 700 = 5.71 × 10^-17 moles of lipid molecules. If 6.02 × 10^23 mo-lecules are in a mole, then in 5.71 × 10^-17 moles, there are 6.02 × 10^23 5.71 × 10^-17 = 3.44 × 10^7 lipid molecules. (b) Calculate the volume of both the E. coli cell and the total volume occupied by the ribosomes. Volume of the E. coli cell: It is of cylindrical shape. The volume of a cylinder can be found from the product of its length and the area of its base. The area of the base, a circle. is \pir^2 , where r = radius. Because the diamater = 2r,the radius of the cylindrical E. coli cell is 1\mu divided by 2 or 0.5 \mu. This is converted to 5 × 10^3 \AA, by the conversion factor 10^4 \AA = 1 micron. Similarly, 2\mu = (2 × 10^4 \AA) = length of E. coli cell. Its total volume = (area of circle) (length) = [\pi(5 × 10^3)^2 (2 × 10^4] = 1.57 × 10^12. Volume of ribosomes: Each ribosome has a diameter of 180 \AA or a radius of 90 \AA. It is given that it is a sphere. The volume of a sphere is given by the formula (4 / 3 \pi^3), where r = radius. Substituting the volume of one ribosome = (4 / 3)\pi (90)^3 3.05 × 10^6. If there exists 15,000 ribosomes, their total volume must be (15000) (3.05 × 10^6) = 4.58 × 10^10. Therefore, the percentage volume it occupies = (total volume of ribosomes) / (volume of E. coli cell) × 100 = [(4.58 × 10^10) / (1.57 × 10^12)] × 100 = 2.9%.

Question:

Catalase, the enzyme that assists the breakdown of hydrogen peroxide into oxygen and water, has a turnover number of 2 × 10^8. In one hour, what weight of hydrogen peroxide could be decomposed by one molecule of catalase?

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Solution:

Enzymes may be defined as catalysts. A substrate is the compound that the enzyme acts upon to form the product. One aspect of enzymatic reactions is the rate at which the enzyme-substrate complex is formed and shifted to product. One can express this rate as the turnover number. It is defined as the number of molecules of substrate converted to product by one molecule of enzyme in one minute at optimum conditions. In this problem, you are given one molecule of enzyme, and excess substrate. The turnover rate is given as 2 × 10^8. You are asked for the weight of the substrate decomposed in one hour. To find the weight, you need the number of molecules decomposed. According to the definition of turnover number, 2 × 10^8 molecules, decompose in one minute. Thus, in 1 hour, 60 minutes × 2 × 10^8 = 1.2 × 10^10 molecules of hydrogen peroxide (H_2O_2) is decomposed. To calculate the total weight of these molecules, one uses Avogadro's number. It states that there are 6.02 × 10^23 molecules in one mole. Thus, 1.2 × 10^10 molecules represents 1.2 × 10^10 / 6.02 × 10^23 = 1.99 × 10^-14 moles of H_2O_2 decomposed. A mole is defined as weight in grams/mo-lecular weight. The molecular weight of H_2O_2 = 34. Thus, weight of H_2O_2 decomposed = (1.99 × 10^-14 moles) (34 g/mole) = 6.766 × 10^-13 grams.

Question:

Design a FORTRAN function to simulate trials according to the geometric distribution f(x) =pq^x, where p is the probability of success on any trial, and q = 1 - p is the probability of failure.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G19-0476.htm

Solution:

This perfectly binary probability structure is derived from the notion of Bernoulli trials, in which the only two possible outcomes are "success" and "failure". If a sequence of these independent trials is performed, the probability of obtaining X failures before reaching a success can be exhibited by the geometric distribution. There are two approaches to this problem, both depending on the value of p. If p is close to 0 or 1, we can generate a series of inde-pendent trials, and count the number of trials necessary before a suc-cess appears. As q = 1 - p, we can generate a series of random numbers between 0 and 1 and can call the series S. Let Si be thei-thterm of the series S. As we generate a new term Si, we compare its value with q. As soon as Si > q at any point during iteration, procedure stops, andi- 1 turns out to be the value of the geometric variable X. The alternate approach, when p is not close to 0 or 1, uses the formula [i= log r / log q], in whichiis the geometric variable, r is a random number such that 0 \leq r \leq 1, q = 1 - p, and the brackets indicate that only the integer part of the result is desired. This method is unreliable when p is close to 0 or 1 because the ALOG library function is computed with a power series that, on most machines, is imprecise when calculating values near the parameters of the chosen interval. The function procedure is described below. Comments highlight the two different methods of solution. INTEGER FUNCTION BERN(P) CRETURN TO THE MAIN PROGRAM IS CONTINGENT CUPON FINDING THE FIRST SUCCESS IF (P.LE. .\O5 .OR. P. GE. .95) GO TO 1\O CLOGARITHMIC METHOD X = RNDM (NUM) BERN = INT(1.\O + (ALOG 1\O (X) / ALOG 1\O(1.\O - P))) RETURN CTHIS IS THE COUNTING METHOD, WHICH CAVOIDS THE AFOREMENTIONED PRECISION DIFFICULTIES 1\ODO 2\O J = 1, 1\O\O\O X = RNDM (NUM) IF (X.LT.P) GO TO 3\O 2\OCONTINUE CIF NO SUCCESSES, BERN = 1\O\O\O\O BERN = 1\O\O\O\O RETURN CSUCCESS! 3\OBERN = J RETURN END Some FORTRAN compilers contain random number generation function as a library function. For those which do not, the function has to be defined and called. In this program, the random number generation function is called by means of X = RNDM (NUM) statement and has the following (generally used) procedure: FUNCTION RNDM (NUM) IF (NUM. GT. 0) INUM = NUM INU = 65539 \textasteriskcentered INUM IF (INU.LT. 0) INU = INU + 2147483647 + 1 RNDM = .4656613E - 9 \textasteriskcentered FLOAT (INU) INUM = INU RETURN END

Question:

Differentiate between fission, budding, and fragmentation as means of asexual reproduction.

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/Users/wenhuchen/Documents/Crawler/Biology/F22-0567.htm

Solution:

Most protozoans reproduce asexually by fission, which is the simplest form of asexual reproduction. Fission involves the splitting of the body of the parent into two approximately equal parts, each of which becomes an entire, new, independent organism. In this case, the cell division involved is mitotic. Hydras and yeasts reproduce by budding, a process in which a small part of the parent's body separates from the rest and develops into a new individual. It may split away from the parent and take up an independent existence or it may remain attached and maintain an independent yet colonial existence. Lizards, starfish, and crabs can grow a new tail, leg, or arm if the original one is lost. In some cases, this ability to regenerate a missing part occurs to such an extent that it becomes a method of reproduction. The body of the parent may break into several pieces, after which each piece regenerates its respective missing parts and develops into a whole animal. Such reproduction by fragmentation is common among the flatworms, such as the planaria.

Question:

Write a FORTRAN program that calculates the coefficients of a linear Diophantine equation of the form MX + NY = (M,N).

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Solution:

A Diophantine equation of this type is a polynomial with integer coefficients for which only integer solutions are permitted. The set (M,N) is called the solution of the equation. The simplest Diophantine equation is the one-variable, first-degree equation of the form MX = p where M, p are integers, and neither is equal to 0. By definition, this equation has an integer solution if and only if p/M is an integer. Two-variable linear Diophantine equations may be solved by applying Euclid's algorithm for finding the greatest common divisor (GCD). The program used incorporates Euclid's algorithm into the following main algorithm: 1) Input coefficients M and N. 2) Check integers; if M = 0, exit, else, continue. 3) Initialize: MO = N1 = 1, and NO = Ml = 0. 4) Divide M by N; store quotient in Q and remainder in R. 5) Calculate two new coefficients for M and N, which are, respectively M2 = MO - Q\textasteriskcenteredM1 N2 = NO - Q\textasteriskcenteredN1 6) Redefine the variables according to the following schedule: MO = M1 NO = N1 M1 = M2 N1 = N2 7) Redefine the dividend as the previous divisor and the divisor as the previous remainder (this is Euclid's strategy). 8) Loop back to step (4). 9) The final nonzero remainder is (M,N); if the first remainder is zero, (M,N) = │N│. Output M,N,M1,N1 and (M,N) in the form M \textbullet M1 + N \textbullet N1 = (M,N) One more operation must be performed. The absolute values of M and N must be stored in MABS and NABS, respectively, so that the signs of the solutions can be included in the output. Thus, the program is presented below. Comments are added to make it easier to follow the prose description of the algorithm. The program is terminated when 0 is entered as the value of M. Also note, that the program assumes M and N \leq 99999. INTEGER Q,R 5READ (5,100) M,N 100FORMAT (215) CDO WHILE M NOT EQUAL TO ZERO IF (M.EQ.O) GO TO 99 CINITIALIZE VARIABLES MO = 1 M1 = 0 NO = 1 N1 = 0 CFIND ABSOLUTE VALUES TO OBTAIN SIGNS OF M,N MABS = IABS(M) NABS = IABS(N) MSIGN = M/MABS NSIGN = N/NABS CCALCULATE QUOTIENT AND REMAINDER 20Q = MABS/NABS R = MABS - NABS \textasteriskcentered Q CIF R GREATER THAN ZERO, REAPPLY EUCLID'S CALGORITHM IF (R.LE.O) GO TO 30 M2 = MO - Q \textasteriskcentered M1 N2 = NO - Q \textasteriskcentered N1 CREDEFINE VARIABLES MO = M1 NO = N1 M1 = M2 N1 = N2 MABS = NABS NABS = R GO TO 20 CGIVE COEFFICIENTS OF M AND N WITH CTHEIR PROPER SIGNS 30Ml = MSIGN \textasteriskcentered M1 N1 = NSIGN \textasteriskcentered N1 COUTPUT RESULTS; NABS IS THE FINAL NON- CZERO REMAINDER WRITE (6,101) M,M1,N,N1,NABS 101FORMAT (1X,'(' ,I6, ') \textasteriskcentered (', I6,') + (', I6, ') \textasteriskcentered (', 1 I6, ') = ', I5) GO TO 5 99STOP END

Question:

Write a PL/I program to determine whether a given integer be-tween0 and 1000 (exclusive) is prime or composite.

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Solution:

As we know a prime is any positive integer p greater than 1 whose onlypositive divisors are 1 and p. Integers greater than 1 which are not primeare called com-posite. According to the fundamental theorem of arithmetic, every composite number n has at least two prime factors p and qsuchthat 11 has noprime factor less than or equal to \surdn, then n is prime." Thus we have analgorithm to determine whether a given number n is prime or composite: Divide n successively by the primes 2, 3, 5,7, ...p_n, wherep_ndenotes the largestprime that does not exceed \surdn, if none of these primes divides n, thenn is prime, other-wise n is composite. In our example to test integers less than 1000 we will need to divide by the primesless than \surd1000 = 31.6. These are 2, 3, 5, 7, 11, 13, 17, 19, 23,29 & 31.We can now de-tail the steps that our program is to perform. 1. Place the eleven primes into memory. 2. Read an integer n into memory. 3. (a) If 11000 THEN PUT LI ST ('OUT OF RANGE'); ELSE; /\textasteriskcentered CHECK THAT N IS WITHIN RANGE \textasteriskcentered/ RN = N; /\textasteriskcentered CONVERT N TO REAL NUMBER \textasteriskcentered/ NROOT=SQRT(RN); /\textasteriskcentered CALCULATE SQUARE ROOT \textasteriskcentered/ DOI = 1 TO 11; IF(NPR(I) -NROOT)>0 THEN PUT LIST(N,'THE NO IS PRIME'); This statement is followed by comment /\textasteriskcentered IS CURRENT PRIME GREATER THAN SQUARE ROOT \textasteriskcentered/ ELSE IF (N/NPR(I) \textasteriskcenteredNPR(I) -N)=0 THEN PUT LIST (N, 'THE NO IS COMPOSITE'); /\textasteriskcentered WE CHECK IF THE CURRENT PRIME DIVIDES N \textasteriskcentered/ ELSE; /\textasteriskcentered THIS IS A NULL ELSE AND SIGNIFIES THAT IF NEITHER OF THE ABOVE POSSIBILITIES HOLD, THEN THE LOOP IS TO BE REPEATED AND A NEW VALUE OF NPR.IS TO BE OBTAINED \textasteriskcentered/ GOTO LABEL; END PRIME;

Question:

The thymus gland is a two-lobed, glandular-appearing structure located in the upper region of the chest just behind the sternum. What are the two principal functions that have been attributed to this gland?

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Solution:

It is thought that one of the functions of the thymus is to provide the initial supply of lymphocytes for other lymphoid areas, such as the lymph nodes and spleen. These primary cells then give rise to descendent lines of lymphocytes, making further release from the thymus unnecessary. This first function of the thymus is non-endocrine in nature. The second function attributed to the thymus is the release of the hormone thymosin which stimulates the differ-entiation of incipient plasma cells in the lymphoid tissues. The cells then develop into functional plasma cells, capable of producing antibodies when stimulated by the appropriate antigens. To summarize, full development of plasma cells requires two types of inducible stimuli. They are: (1) stimulation by thymosin which initiates differentiation of all types of incipient plasma cells, and (2) stimulation by a specific antigen, which affects the functional maturation of only those cells with a potential for making antibodies against that particular antigen.

Question:

A biochemist performed several experiments on rats. In one, heused two groups of young rats. GroupArats were fed on adiet of purified casein (cheese protein), starch, glucose, lard, minerals, and water. Group B rats were fed the same dietwith the addition of 3 cm^3 of milk daily. It was observed thatGroup A rats stopped growing and lost weight, while Group B rats gained steadily in weight and size. After eighteendays the milk was given to Group A rats and removedfrom the diet of Group B. Group A rats now resumedgrowth and gained in weight, while Group B rats stoppedgrowing and lost weight. The investigator knew that 3 cm^3 of milk has an insigni-ficantfood value in terms of carbohydrates, fats, proteinsand minerals. Consider yourself in his position, confrontedwith only the information given so far. a) What conclusions could you reach about the presencein milk of substances other than these four foods, thepossible functions of the hypothetical substances, and thequantities needed by the rats? b) Why was it necessary to transfer the milkfrom Group B rats to Group A rats half-way through the experiment? c) The rats received only one type of protein (casein) . Why can protein starvationberuled out as a possible cause ofgrowth inhibition and weight loss?

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Solution:

One can conclude that some factors responsible for growth and weightincrease are present in milk. Such factors are called vitamins. Vitamins are complex chemicals which are required in minute amounts in thediet of all heterotrophic organisms. Autotrophic organisms, such as greenplants, also need to procure certain essential vitamins from the environmentfor their growth. Vitamins have no energy value. They play a vital role in many chemicalreactions in metabolism. Most vitamins function as coenzymes. Lack of any vitamin in the diet causes the reaction in which it takes part to slowdown, and since most metabolic reactions are part of a long sequenceof events, an alteration in the rate of any one of them can have widespreadeffects in the body. Since 3 cm^3 of milk is enough for normal growthin rats, it is shown that only small quantities of vitamins are needed. b) By interchanging the diet of Group A and Group B a control experimentis performed, demonstrating that the lack of growth is due to thediet alone and not to any constitutional defects in the rats.I.e, when normallygrowing rats in Group B are fed with the vitamin-deficient diet, theystop growing. c) Casein, a milk protein, is an example of an adequate protein. An adequateprotein is one which contains all the essential amino acids necessaryfor growth. Since the essential amino acids are present, we can ruleout protein starvation, a deficiency in essential amino acids, as a causeof the observed stunted growth.

Question:

Why is it that alveolar air differs in its composition from atmospheric air? Of what significance is this fact?

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Solution:

The respiratory tract is composed of conduct-ing airways and the alveoli. Gas exchange occurs only in the alveoli and not in the conducting airways. The maximum alveolar volume is about 500 ml. Let us consider then what takes place during expiration. Five hundred milliliters of air is forced out of the alveoli, and into the conducting airways of the respiratory tract. Of this air, 150 ml. remains in the respiratory airways following expiration, while 350 ml. of air is exhaled from the body. During the next inspiration 500 ml. of air are taken up by the alveoli, but 150 ml. of this air is not atmospheric, but rather is the air that remained in the tubes following the previous expiration. One can see, then, that only 350 ml. of fresh atmospheric air enters the alveoli during each inspiration. At the end of inspiration, 150 ml. of fresh air also fills the airways but does not reach the alveoli. Hence no gas exchange between this air and the blood can occur. This fresh air will be expelled from the body during the next expiration, and will be replaced again with 150 ml. of alveolar air, thus completing the cycle. From this cycle it can be seen that of the 500 ml. of air entering the body during inspiration, 150 ml. of it never reaches the alveoli, but instead remains in the con-ducting tubes of the respiratory system. The term "anatomical dead space" is given to the space within the conducting tubes because no gas exchange with the blood can take place there. The question arises then, as to the significance of this dead space. Because the tubes are not completely emptied and filled with each breath, "new" air can mix with "old" air. Consequently, alveolar air contains less oxygen and more carbon dioxide than atmospheric air. In addition, the air that remains in the alveoli following expiration (the residual volume) can, to a limited extent, mix with the incoming air and thereby alter its composition.

Question:

One of the first problems in linear programming was the so-called diet problem. The problem was to minimize the cost of eating three meals a day while meeting minimum daily requirements of certain essential nutrients. Suppose we consider only the contribution of vege-tables to the nutritional requirements. How many times should each vegetable be served during the next week in order to minimize cost while satisfying the nutritional and taste requirements? Use the following information: Units per serving Vitamin Vegetable Iron Phosphorus A B Niacin Cost/ serving Green beans Carrots Broccoli Cabbage Beets Potatoes 0.45 0.45 1.05 0.4 0.5 0.5 10 28 50 25 22 75 415 9065 2550 75 15 235 8 3 53 27 5 8 0.3 0.35 0.6 0.15 0.25 0.8 5\textcent 5\textcent 8\textcent 2\textcent 6\textcent 3\textcent Minimum daily requirements from vegetables 6.032517500(USP)245 mg5.0 mg In addition, cabbage cannot be served more than twice during the week, and the other vegetables cannot be served more than four times each during the week. A total of 14 servings of vegetables are required during the week.

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Solution:

A linear programming problem is said to be in canonical form when itis in the form: Min a_1 x1+ a_2 x_2 +.....+a_nx_n= C(1) subjectto b_11x_1, + b_12 x_2 +.........+ b_1n,x_n\geq C_1,(2) b_21x_1 + b_22 x_2 +.........+ b_2nx_n\geq C_2 ::::: : : : : ::: : : : ::::: : : : : ::: : : : b_m1x_1 + b_m2 x_2 +....... .+b_mnx_n\geq C_m andx_i \geq 0i= 1,2,....,n(3) To convert the given problem into canonical form, let x_1= number of times to serve green beans x_2= number of times to serve carrots x_3 = number of times to serve broccoli x_4 = number of times to serve cabbage x_5 = number of times to serve beets x_6 = number of times to serve potatoes The cost of one serving of green beans is $0.05. Thus x_1 servings will cost 5x_1 (in cents) Similarly, x_4 servings of cabbage will cost 2x_4. If we multiply the number of servings for a given vegetable by the cost per serving and sum over all vegetables, we will obtain the cost function (1). Thus, we have Min 5x_1 + 5x_2 + 8x_3 + 2x_4 + 6x_5 + 3x_6 Next, we hunt for the constraints. The minimum daily re-quirements from vegetables of iron, phosphorus, vitamin A, vitamin B and Niacin must be met or exceeded. Thus, for iron, .45x_1 + .45x_2 + 1.05x_3 + .4x_4 + .5x_5 + .5x_6 \geq 42.0 Here the minimum daily requirement of iron, 6.0 is multi-plied by 7 to give the weekly requirement. Proceeding with the other nutrients in a similar manner , we obtain the con-straint inequalities Min 5x_1 + 5x_2 + 8X_3 + 2x_4 + 6x_5 + 3x_6 subject to Iron .45x_1+ .45x_2 + 1.O5x_3 + .4x_4 + .5x_5 +5X_6 \geq 42.0 Phos. 10x_1 + 28x_2 + 50 x_3 + 25x_4 + 22x_5. + 75x_6 \geq 2275 Vitamin A415x_1 + 9065x_2 + 255OX_3 + 75x_4 + 25x_5 + 235x_6 \geq 122500 Vitamin C8x_1, + 3x_2 + 53x_3 + 27x_4 + 5x_5 + 8x_6 \geq 1715 Niacin. 3x1+.35x_2 + .6x_3 + .15x_4 + . 25x_5 + .8X_6 \geq 35 x_1\leq4 x_2\leq4 x_3\leq4 x_4\leq2 x_5\leq4 x_6\leq4 x_1 \geq 0x_2 \geq 0x_3 \geq 0x_4 \geq 0. The next step is to convert the program into a form suitable for computer manipulation . We add slack var-iables S_i to convert all inequalities into equalities . The simplex algorithm is then applied to the resulting system of equations to obtain an optimal solution. The simplex algorithm is available as a packaged pro-gram in most installations .

Question:

It has been found that the following sequence can be used to prepare sodium sulfate, Na_2 SO_4 : S (s) + O_2 (g)\rightarrow SO_2 (g) 2SO_2 (g) + O_2 (g)\rightarrow 2SO_3 (g) SO_3 (g) + H_2 O(l)\rightarrow H_2 SO_4 (l) 2NaOH + H_2 SO_4\rightarrow Na_2 SO_4 + 2H_2 O If you performed this sequence of reactions, how many moles of Na_2 SO_4 could possibly be produced if you start with 1 mole of sulfur? How many moles are possible with 4.5 g of water?

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Solution:

If you had the general equation, aA+bB\rightarrowcC+dD, a moles of A react with b moles of B to produce c moles of C and d moles of D. Thus, if you want to know how many grams of D will be produced, and you know how much A you have, calculate the number of moles of A present. From this, you can determine how many moles of D can be generated, since you know that a moles of A will produce d moles of D. With this in mind, you can proceed to answer these questions. From the first equation, you see that if you start with 1 mole of sulfur (S), 1 mole of SO_2 can be generated (since all the coefficients are one, although 1 is not written). Now that 1 mole of SO_2 is generated, you proceed to the second equation. It states that for every 2molesof SO_2, 2 moles of SO_3 are generated. Thus, in keeping with this relative ratio, 1 mole of SO_3 can be generated from one mole of SO_2. In the third reaction you again have 1 mole of SO_3 yielding 1 mole of H_2 SO_4. At this point, you have 1 mole of N_2 SO_4 (from 1 mole of starting sulfur). The last equation shows 1 mole of H_2S0_4 producing 1 mole of Na_2 SO_4. Therefore, if you where to start with 1 mole of S, you would obtain 1 mole of Na_2 SO_4. Now let us consider H_2 O. Water does not enter the sequence until the third equation. This means you start with SO_3 (g) + H_2 O (l) \rightarrow H_2 SO_4 (l). If you have 4.5 g of H_2 O (MW = 18 g/m), you possess 4.5 g/18 g/mole = .25 moles H_2 O. From the reaction, 1 mole of H_2 O yields 1 mole of H_2 SO_4. Thus, .25molesof H_2 SO_4 are generated from .25 moles of water. You have the same 1 : 1 ratio. The last equation is also a 1 : 1 ratio. Therefore, if you start with 4.5 grams of H_2 O, .25molesof Na_2 SO_4 will be produced.

Question:

Why is the flexor muscle of the crayfish well developed?

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Solution:

In order to answer this question, it is ne-cessary to understand the body plan of the crayfish. (Refer to figure in last solution.) Crustaceans have a head and trunk which is usually composed of a thorax and abdomen. The head and thoracic segments of the decapods are fused together dorsally by a carapace, which hangs over the sides of the animal and encloses the gills. Lobsters and crayfish have a large segmented abdomen, with appendages on each segment. The fourth thoracic appendages are chelipeds, walking legs that function as claws. The next four pairs of thoracic appendages are smaller, thinner walking legs. Lobsters, crayfish, and burrowing shrimps generally crawl with these walking legs. These animals are also capable of swimming backward rapidly. This is accomplished by flexing the abdomen ventrally. The flexor muscle of the crayfish is well-developed and powerful. It enables the animal to flex its abdomen rapidly in swimming. Many non-swimming decapods have a greatly reduced abdomen that is folded up under the thorax. This is often the case in crabs, as most crabs cannot swim. The blue crab, however, is a powerful and agile swimmer. Its last pair of legs are broad flat paddles that act as propellers. Shrimps are generally bottom-dwellers, i.e., they dwell at the bottom of a body of water. They swim intermittently. The abdominal appendages, pleopods, are the principal swimming organs. The pleopods are large and often fringed. Shrimps utilize ventral flexion of the abdomen for quick backward darts and for vertical steering.

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Question:

When ammonium nitrate (NH_4NO_3) is heated, laughing gas (N_2O) and water are produced. If the laughing gas is collected at STP (Standard Temperature and Pressure) and one starts with 10 grams of NH_4NO_3, how many liters of laughing gas could be collected?

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Solution:

When any gas is collected at STP (0\textdegreeC and 1atm), one mole of it will occupy 22.4 liters. The balanced equation for this chemical reaction is NH_4NO_3 \ding{217} N_2O + 2H_2O. One mole of NH_4NO_3 yields one mole of N_2O. If the number of moles of NH_4NO3that reacted is known, the number of moles of laughing gas produced is also known. The molecular weight of ammonium nitrate = 80 g / mole. 10 g of ammonium nitrate (10 g / 80 g / mole) = 0.125 moles of NH_4NO_3 produce 0.125 moles of laughing gas. Recall that 1 mole of any gas (at STP) occupies 22.4 liters. To find out what volume 0.125 moles occupy, set up the following proportion: (1 mole / 22.4 liters)= (0.125 / x), where x = the volume. Solving x= 2.8 liters = volume of laughing gas collected.

Question:

What is the equivalent capacitance between the points A and B in diagram (A)? The charge on the 6-\muF capacitor is 90 \muC. What potential difference exists between the points A and R?

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Solution:

In the arrangement shown in diagram (A) there are two sets of capacitors in parallel. Since the net capacitance of n capacitors in parallel is C_net = C_1, + C_2 + . . . + C_n the arrangement is equivalent to that shown in diagram (B). In each branch there are now two capacitors in series. If we have n capacitors in series, their net capacitance is (1/C_net) = (1/C_1) + (1/C_2) + . . . + (1/C_n) For each branch, then, (1/C_net) = (1/C_1) + (1/C_2). For the top branch (1/C_net ) = (1/12 \muF) + (1/4 \muF) = [(4 \muF + 12 \muF)/(48 \muF^2 )] C_net = 3 \muF For the lower branch (1/C_net ) = (1/4 \muF) + (1/2 \muF) = [(2 \muF + 4 \muF)/(8 \muF^2 )] C_net = (4/3) \muF Thus, the arrangement is equivalent to that shown in diagram (C). Here we have two capacitors in parallel. Their capacitances add. Hence, the equivalent capacitance between A and B is C = [3 + (4/3)] \muF = 4 (1/3) \muF. The charge on the 6-\muF capacitor is 90 \muC. Hence, the potential across it is, by definition of capacitance V_DS = [(Q_DS )/(C_DS )] = [(90 × 10^-6 C)/(6 × 10^-6 F)] = 15 V. This potential is common to all the capacitors in the top branch between A and S since they are all in parallel. Hence the total charge on the equivalent 12-\muF capacitor is Q_AS = C_AS V_AS = C_AS V_DS = 12 × 10^-6 F × 15 V = 180 \muC. The conductors connected to S are isolated and initially must have been uncharged. If a charge of - 180 \muC appears on the negative plate of the equivalent 12-\muF capacitor connected to S, a corresponding charge of + 180 \muC must be induced on the positive plate of the 4-\muF capacitor connected to S, leaving the net charge at S zero. A corresponding - 180-\muC charge is induced on the negative terminal of the 4- \muF capacitor, and the voltage between its plates is thus V_SB = (Q_SB /C_SB ) = [(180 × 10^-6 C)/(4 × 10^-6 F)] = 45 V. The total potential difference between A and B is thus V = V_AS + V_SB = (15 + 45) V = 60 V. This is the potential difference between A and B by either branch. Referring to diagram (C), the charge on the equivalent 4/3-\muF capacitor is thus Q_AE = C_AE V_AE = C_AE V = (4/3) × 10^-6 F × 60 V = 80 \muC. This is the charge on the plate attached to A in either the equivalent circuit or the original circuit, since the two produce identical effects. Hence the equivalent 4-\muF capacitor between A and R has charges of \pm 80 \muC on each of its plates. Hence the potential difference across it is V_AR = Q_AR /C_AR = [(80 × 10^-6 C)/(4 × 10^-6 F)] = 20 V.

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Question:

A man of mass 80 kg is standing on the rim of a stationary uniform circular platform, of mass 140 kg and diameter 8 m, which is free to rotate about its center. The man throws to a companion on the ground in a direction tangential to the rim a package of mass 1 kg at a speed of 20 m\bullets^-1 rela-tive to the ground. What angular velocity of the man and platform is produced in consequence? The man then walks so as to bring him to a position halfway between the rim and the center of the platform. What is the new angular velocity of the system?

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Solution:

The package is thrown tangentially at a speed of 20 m.s^-1 relative to the ground and consequently has an initial angular momentum about the center of the platform form given by \vertL\ding{217}\vert = \vertr\ding{217}× mv\ding{217}\vert = \vertmvrsin\texttheta\vert = mvr = 1 kg × 20 m\bullets^-1 × 4m \ding{217} \ding{217} \ding{217} = 80 kg\bulletm^2\bullets^-1. where \texttheta(=90\textdegree) is the angle between r\ding{217}and v\ding{217}. \ding{217} \ding{217} Since the platform is free to rotate about its center, there are no external (e.g. in the form of friction) forces acting on the man-package- platform system. There are no external torques acting on this system, but \Gamma\ding{217}= dL\ding{217}/dt \ding{217} \ding{217} where \Gamma\ding{217}is the sum of the external torques acting on the system. Thus \ding{217} 0 = dL\ding{217}/dt \ding{217} L\ding{217}= constant \ding{217} That is, the angular momentum of the system remains the same at all instances of time when no external force acts on the system. This is the principle of conservation of angular momentum. Since the man-package- platform system was initially at rest, L\ding{217}= 0. After the package is thrown, \ding{217} L\ding{217}must still be zero, thus the man-platform must acquire a momentum \ding{217} Equal and opposite to the momentum of the package, to satisfy the condition that the L\ding{217}= 0 for the man-platform-package system at all times. \ding{217} The plat- form is a distributed mass, as opposed to the localized masses of the man and package (which to a very good approximation may be treated as point masses). The fol-lowing expression involving the moment of inertia I of the platform must then be used. (The moment of inertia accounts for this mass distribution). I\cyrchar\cyromega = L(1) where I is the angular velocity of the platform due to the throwing of the package. (This is the rigid body analogue of linear momentum mv = p, where I corresponds to m, v corresponds to \cyrchar\cyromega and L corresponds to P.) I for the uni-form circular platform is I_1 = (1/2) m_1R2 where m_1 is the mass of the platform and R is its radius.For the man, the definition of moment of inertia for adiscrete particle gives I_2 = m_2R^2,where m_2 is the man'smass. Then, since the man also revolves about the center with angular velocity \cyrchar\cyromega, equation (1) becomes (I_1 + I_2)\cyrchar\cyromega = L_1 + L_2 = Lpackage [m_2R^2 + (1/2)m_1R^2)\cyrchar\cyromega = Lpackage (80 × 4^2 + 1/2 × 140 × 4^2) kg\bulletm^2 × \cyrchar\cyromega = 80 kg\bulletm^2\bullets^-1, \cyrchar\cyromega = (80 s^-1)/(80 × 16 + 70 × 16) = 1/30 rad\bullets^-1 . If the man walks toward the center, his moment of inertia about the center decreases. At the halfway posi-tion, his moment of inertia is m_2(R/2)^2 = (80 × 2^2) kg.m^2. But the angular momentum must stay the same, and so the angular velocity must increase to \cyrchar\cyromega', and (m_2R^2)\cyrchar\cyromega = L = [m_2(R/2)^2]\cyrchar\cyromega' (80 × 4^2 + 1/2 × 140 × 4^2)kg\bulletm^2 × \cyrchar\cyromega = (80 × 2^2 + 1/2 × 140 × 4^2) kg\bulletm^2 x \cyrchar\cyromega' . \therefore\cyrchar\cyromega' = (150/90)\cyrchar\cyromega = (1/18) rad\bullets^-1.

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Question:

A collimated beam of silver atoms emerges from a furnace at 1500\textdegreeK, passes through a circular hole, and falls on a screen 1m away. Assume that all the atoms travel with the same speed. What size hole is likely to give the smallest spot size on the screen? The mass of a silver atom is 1.8 × 10^-25 kg. (See figure).

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Solution:

If the hole has a radius a, the uncertainty in the z-co-ordinate of an atom passing through the hole is 2a. Thus \Deltaz = 2a . But if p_z is the momentum of the atom in the z-direction, then, at the hole, the Heisenberg Uncertainty Relationship yields \Deltap_z \Deltaz \approx h where \Deltap_z is the uncertainty in the z momentum. But \Deltap_z = m\Deltav_z where \Deltav_z is the uncertainty in the z velocity of the silver atoms. Hence m\Deltav_z \approx (h/2a) or \Deltav_z \approx (h/2am). Thus the atoms have an uncertainty in velocity in the z-direction of the order of h/2am. But the velocity in the z-direction is clas-sically zero. Hence, the atoms, by quantum-mechanical arguments, have velocities around the value h/2am in the z-direction. In the y-direction the atoms have a velocity which is assumed for simplicity to be the same for all atoms. The average kinetic energy possessed by an atom at temperature T is 3/2 kT, where k is Boltzmann's constant and T is given in degrees Kelvin. Thus the silver atoms of mass m have velocities v_y given by v_y = \surd3kT/m. They traverse the distance y in time t. Since no force acts on the silver atoms in the y direction y = v_y t or t = y/v_y In the same time the maximum distance achieved in the z-direction due to the z-velocity of the atoms is, similarly, a = \Deltav_z × t = (y\Deltav_z/v_y) = (hy/2am)\surd(m/3kT) = (hy/2a) \surd[1/(3mkT)] = [(6.6 × 10^-34 J \textbullet s × 1 m)/2a] × \surd[1 / (3 × 1.8 × 10^-25 kg × 1.38 × 10-23 J \textbullet k deg^-1 × 1500 k deg)] = (3.12/a) × 10^-12 m^2. Thus the largest possible distance from the axis at which a silver atom can strike the screen is found by observing the trajectory of the atom at the top of the hole. The furthest it will travel from the center of the hole is r = a + z = a+ (3.12/a) × 10^-12 m^2 . (dr/da) = 1 - (3.12/a^2) ×10^-12 m^2 , and for a minimum of r this quantity must be zero . It can be verified that this is a minimum by a second differentiation. Thus the radius of the hole which gives the smallest spot on the screen is that for which (3.12/a^2) × 10^-12 m^2 = 1 or a = 1.77 × 10^-6 m. In classical physics the spot size would be decreased indefinitely by reducing the size of the hole. Because of the wave nature of the electron, quantum mechanics does not supply the same answer. After a certain point, diminishing the size of the hole increases the size of the spot because of diffraction effects produced by the hole on the silver atoms. A minimum size of spot therefore results for a finite hole size.

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Question:

Diffusion is too slow a process to account for the observed rateof transport of materials in active cells. Explain why.

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Solution:

The cytoplasm of a living cell is a watery medium separated by the plasmamembrane from the external environment. The cytoplasm resemblesa loose gel which is somewhat denser than pure water. In order tomove from one part of a cell to another, materials have to move through thisrelatively dense gel. Thus diffusion of materials through a cell is necessarilya much slower process than diffusion of materials through water. Diffusion is a relatively slow method of transport of materials also becausethe molecules in motion are constant-ly colliding with other molecules. This causes them to bounce off in other directions and to take azig-zagpath instead of a straight one. Therefore molecules have to travela longer distance, which requires more time, in order to get to their destinations. Diffusion is only a passive process: it does not re-quire energy. Its ratedepends on the viscosity (thickness) of the medium, the difference in concentrationsof molecules in the two regions, the distance between the tworegions, the surface area of the two regions if they are in contact, and thetemperature. Often the transport of materials in living cells occurs at so rapida rate that it is not possible to be accounted for by simple (passive) diffusionalone. Instead, some active processes involving expen-diture of energyare believed to be responsible.

Question:

In the figure, a fine wire, having a positive charge per unit length \lambda, lies on the y-axis. Find the electric intensity set up by the wire at point P.

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Solution:

Let the wire be subdivided, in imagination, into short elements of length dy. The charge dq on an element is then \lambda dy. Let r represent the perpendicular distance from P to the wire and s^\ding{217} the vector from dq to P. If we view dq as a point charge, then it sets up a field dE^\ding{217} and P given by dE^\ding{217} = k [(\^{s} dq)/s^2] = k [(\^{s}\lambdadq)/s^2], dE^\ding{217} = k [(\^{s} dq)/s^2] = k [(\^{s}\lambdadq)/s^2], where \^{s} is a unit vector in the direction of s^\ding{217}, and k = 9 × 10^9 N \textbullet m^2/c^2, and the resultant intensity E^\ding{217} is the sum of all these infinitesimal fields dE^\ding{217} along the y-axis, or E^\ding{217} = k \int[(\^{s}\lambdadq)/s^2]. The unit vector \^{s} lies in the yz-plane, so its x- component is zero. The magnitude of its y-component is sin \texttheta and that of its z-component is cos \texttheta. The vector equation above is then equivalent to the three scalar equations E_x = 0 E_y = k\lambda ^+\infty\int_-\infty [(sin \texttheta dy)/s^2 ] E_z = k^+\infty\int_-\infty [(cos \texttheta dy)/s^2 ]. The wire is considered to be sufficiently long so that the limits of integration are from - \infty to + \infty. Let us change the integration variable from y to \texttheta, so y = r tan \texttheta, s = r sec \texttheta and \texttheta varies from - (\pi/2) to \pi/2 as y varies between -\inftyand + \infty. Since y = r tan \texttheta, dy = r sec^2 \texttheta d\texttheta. Hence, E_y = k\lambda\pi/2\int_-\pi/2 [{(sin \texttheta)(r sec^2 \texttheta) d\texttheta}/s^2 ] Since sec \texttheta = s/r, E_y = k\lambda\pi/2\int_-\pi/2 [{(sin \texttheta)(s^2/r)d\texttheta}/s^2 ] = k\lambda/r\pi/2\int_-\pi/2 sin \texttheta d\texttheta = 0. This result is not unexpected since in the figure we see that for each dq at y = Y, there is another dq at y = - Y. The y-components of the electric fields of these two charges cancel, whereas the z-components add up. E_z = k\lambda\pi/2\int_-\pi/2 [(dy cos \texttheta)/s^2] = k\lambdar^\pi/2\int_-\pi/2 [{(cos \texttheta)(sec^2 \texttheta) d\texttheta}/s^2 ] = k\lambdar^\pi/2\int_-\pi/2 [{(cos \texttheta)(s^2/r^2)}/s^2 ] = k\lambda/r^\pi/2\int_-\pi/2 cos \texttheta d\texttheta = 2k\lambda/r

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Question:

A person seated on a swivel chair is rotated rapidly. The rotation is suddenly stopped and the person is told to stand up. He complains that he feels dizzy. How can you account for his dizziness?

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Solution:

The labyrinth of the inner ear has three semicircular canals, each consisting of a semicircular tube connected at both ends to the utricle. Each canal lies in a plane perpendicular to the other two. At the base of each canal, where it leads into the utriculus, is a bulb-like enlargement (the ampulla) containing tufts of hair cells similar to those in the utriculus and sacculus, but lacking otoliths. These cells are stimulated by movements of the fluid (endolymph) in the canals. When a person's head is rotated, there is a lag in the movement of the endolymph in the canals. Thus, the hair cells in the ampulla attached to the head rotate, in effect, in relation to the fluid. This movement of the hair cells with respect to the endolymph stimulates the former to send impulses to the cerebellum of the brain. There, these impulses are interpreted and a sensation of dizziness is felt.

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Question:

The same system which enables starfish to rocky substrates is useless on sandy beaches. Explain.

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Solution:

Starfish (also called sea star) locomotion occurs by means of tube feet. The tube feet are part of a hydraulic system found only in echinoderms called the water vascular system. The under surface of each arm of a sea star has hundreds of pairs of tube feet, which are hollow, thin-walled, sucker-tipped, muscular cylinders. Internally, the tube foot is connected to a muscular sac, an ampulla, at its base. The tube feet are connected via short lateral canals to a radial canal which extends the length of the arm. The radial canals are joined at the central disc of the sea star by a ring canal. The ring canal leads via a short tube called the stone canal to a sieve-like plate, called a madreporite, on the surface of the animal. The entire system is filled with watery fluid. In order to extend the tube foot, the ampulla contracts, forcing fluid into the tube foot. A valve prevents the fluid from flowing into the radial canals. The tube foot attaches to the sea floor (or substratum) by a sucker at its tip, which acts as a vacuum. Longitudinal muscles cause the tube foot to shorten in length; this pulls the sea star forward, and water is forced back in to the ampulla. The tube foot is released from the substratum, and then again. The rapid repetition of this cycle of events enables sea stars to move slowly. One can see that the use of suction in the tube feet would be useless to the sea star in moving on sandy beaches. On a soft surface, the tube feet are employed as legs. Locomotion becomes a stepping process, involving a backward swinging of the middle portion of the podia followed by a contraction, shoving the animal forward. Certain sea stars have tube feet without suckers at their tips? instead the tips are pointed. These sea stars live on soft bottoms, and can feed on buried animals. Starfish are slow-moving carnivorous animals. Their prey are generally sedentary or slow moving animals, such as clams and oysters.

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Question:

If the last operation performed on a computer with an 8 bit word was an addition in which the operands were 2 and 3, what would be the values of the following flags? (a) Carry (b)Zero (c)Overflow (d)Sign (e)Even Parity (f)Half-carry What if the operands were -1 (two's complement) and +1?

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Solution:

The addition operation of 2 and 3 is: +2:0000 \vert 0010 +3:0000 \vert 0011 +5:0000 \vert 0101 bit #7654 \vert 3210 (a)The Carry flag is 1 if there is a carry from the most significant bit position (bit 7). Since there is no carry from bit 7 the contents of the Carry flag is 0. (b)The Zero flag is 1 if the result is zero. Since the result is not zero the contents of the Zero flag is 0. (c)The Overflow flag is 1 if the two operands in an addition have the same sign but the result has the oppo-site sign. As an example: +64:01000000 +64:01000000 10000000 This answer is incorrect if the numbers are signed because the answer is negative. Thus, this result is too large for 8-bit two's complement notation and the Overflow flag would be set to 1. Another example: -96:10100000 -64:11000000 01100000 This answer is also incorrect because it is positive and thus the Overflow flag would be set to 1. In the addi-tion of 2 and 3 the sign bit is correct so the contents of the Overflow flag is 0. (d)The Sign flag is 1 if the most significant bit (bit 7) is 1 (corresponding to a negative number) . Thus the contents of the Sign flag is 0 since bit 7 is 0. (e)The Even Parity flag is set to 1 if the result of the last operation contained an even number of 1's. The addition of 2 and 3 produced two l's and hence the con-tents of the Even Parity flag is 1. (f)The Half-carry flag is set to 1 if the last opera-tion produced a carry from the lower half word (bits 0-3). The addition of 2 and 3 produced no carry from bit 3 so the contents of the Half-carry flag is 0. The addition of -1 and +1 is: -1:1111 \vert 1111 +1:0000 \vert 0001 0:10000 \vert 0000 bit #7654 \vert 3210 (a)A carry was generated from bit 7 so the content of the Carry flag is 1. When binary numbers of different signs are being added, the computer will indicate a carry and not an overflow, if the result extends outside of its storage field, as is the case in this example. (b)The addition of -1 and +1 produced zero so the con-tents of the Zero flag is 1. (c)There is no overflow so the contents of the Over-flow flag is 0. (d)The sign bit (bit 7) is 0 so the contents of the Sign flag is 0. (e)There are no 1's in the result so the contents of the Even Parity flag is 1 (zero is an even number). (f)A carry was generated from bit 3 so the contents of the Half-carry flag is 1.

Question:

If we project a body upward with speed 1000 cm/s, how high will it rise?

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/Users/wenhuchen/Documents/Crawler/Physics/D05-0228.htm

Solution:

We use the principle of energy conservation to find the height h. Assume that the level of pro-jection is the position of zero potential energy. Then at the point of projection the total energy is purely kinetic E = 0 + (1/2) Mv^2 = (1/2)M ×(10^6 cm^2/s^2) At maximum height v = 0, and the total energy is now purely potential, hence E =Mgh. By equating the two expressions for E, we have Mgh= (1/2)M × (10^6 cm^2/s^2) h = [(1/2) × 10^6 cm^2/s^2] / g = [10^6 cm^2/s^2] / [2(980 cm/s^2] h = .51 × 10^3 cm h = 510 cm.

Question:

What weight of ice could be melted at 0\textdegreeC by the heat liberated by condensing 100 g of steam at 100\textdegreeC to liquid. Heat of vaporization = 540 cal/g, heat of fusion = 80 cal/g.

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Solution:

The quantity of heat necessary to convert 1 g of a liquid into a vapor is termed the heat of vaporization. For water, 540 calories are necessary to change liquid water at 100\textdegreeC into vapor at 100\textdegreeC. In this problem vapor is condensed, thus 540 cal of heat are evolved for each gram of liquid condensed. no. of cal evolved in condensation = 540 cal/g × weight of vapor Here 100 g of vapor is condensed. Thus, no. of cal evolved = 540 cal/g × 100 g = 54000 cal When ice melts, heat is absorbed. About 80 calories of heat are required to melt 1 g of ice, the heat of fusion. Here, 54000 cal are evolved in the condensation. Therefore, this is the amount of heat available to melt the ice. Because 80 cal are needed to melt 1 g of ice, one can find the number of grams of ice that can be mfelted by 54000 cal, by dividing 54000 cal by 80 cal/g. no. of grams of ice melted = (54000 cal) / (80 cal/g) = 675 g.

Question:

When is a condition expression used in COBOL? Give the rules for comparison of numeric or nonnumeric items.

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Solution:

Conditional Expressions: Conditional expres-sions are expressions used in situations where need of test-ing any condition arises. The outcome of a test determines the next logical step to be performed. The conditional expression reduces to a single value, "true" or "false." The truth or falsity is determined by a relational test. This test is based on a comparison of characters. Rules of Comparison: Characters are compared and evaluated on the basis of a computer collating sequence in which the characters have a specified order of magnitude. This collating sequence varies from machine to machine. As an example, the ordering scheme of fig. 1 will be used in the following discussions. Items declared numeric in the DATA DESCRIPTION section cannot be compared with items declared to be nonnumeric. Comparison of Numeric Items: The comparison of numeric items is based on respective values of the item considered purely as algebraic values. The item length, in terms of the number of digits is not itself significant. e.g., comparison of +00003 with +03 results in an equal condition, and +01 is greater than -155. Comparison of Nonnumeric Items: For two nonnumeric items, possibly containing numeric characters, a comparison results with respect to the ordered character set. There are two cases: a) Equal Length Items: If items are of equal length, comparison proceeds by comparing characters in corresponding character positions. Starting from the high-order end, com-parison continues until either a pair of unequal characters is encountered or the lower order end of the item is reached. As an example, consider : 1ABCD and ABBD results in inequality ABBD > ABCD . 2SAM and SAM results in equality. 3NAME 1 and NAME 2 results in inequality, NAME 1 < NAME 2. b) Unequal Length Items: If the items are of unequal length, comparison proceeds as above. If the process exhausts the characters of the shorter item without detection of differ-ence, the shorter item is Less Than the longer item. e.g.,comparingABC & ABCD, every character A, B and C of the first item equals every character of the later item but due to the shorter length of ABC the result is ABC < ABC.

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Question:

Describe how J-K, T and D type flip-flops can be derived starting from the S-C type flip-flop, and obtain excita-tion equations for them from the excitation equation of the S-C type flip-flop.

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Solution:

a) J-K type flip-flop can be obtained from an S-C type flip-flop by adding two AND-gates as shown in fig. 1. The input constraint S\bulletC = 0 of the S-C flip-flop is automatically achieved in the J-K flip-flop, as shown in figure 1, because of the AND-gates. And thus, the J and K inputs need not worry about any input constraints. Both J and K can be = 1. The excitation equation of the J-K flip-flop is obtained by substitution in the excitation equation for the S-C flip-flop as follows: y\rightarrow= SY' + c'Y \rightarrow y\rightarrow= (JY')Y' + (KY)'Y (substitution) \rightarrow y\rightarrow= JY'Y' + (K' +Y')Y (by DeMorgans law) \rightarrow y\rightarrow= JY' + K'Y + Y'Y (distributive law) \rightarrow y\rightarrow= JY' + K'Y \rightarrow b) A T type flip-flop is obtained by shorting together the J and K inputs of the J-K flip-flop obtained in (a) as shown, in fig. 2. The excitation equation of the T flip-flop is obtained from the excitation equation of the J-K flip-flop by putting J = K = T \therefore Y\rightarrow= JY' + K'Y(Now put J = K = T) \rightarrow \therefore Y\rightarrow= TY' + T'Y \rightarrow = T + Y c) A D type flip-flop can be obtained from an S-C type flip-flop by giving mutually inverted inputs to the S and C inputs, as shown in Figure3: The inverter automatically ensures that both S and C will never be equal to 1 at the same time. The excitation equation is obtained from the excitation equation of the S-C flip-flop as follows: y\rightarrow= Sy' + C'y = Dy' + (D')'y \rightarrow \therefore y\rightarrow= Dy' + Dy = D (y' + y) = D \textbullet 1 = D. \rightarrow

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Question:

Indicate how you would prepare (1) acetone from acetic acid and (2) acetic acid from ethyl alcohol.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E21-0779.htm

Solution:

This problem is concerned with synthesis of organic compounds. In such problems you want to write out the structures of the initial and final products involved, so that you can identify functional groups. From this, you employ reactions that change the functional groups until the desired product is obtained. If you had calcium acetate, Ca, you could obtain acetone by heating it, since calcium acetate forms calcium carbonate and acetone upon the addition of heat. To obtain calcium acetate, add Ca(OH)_2, which is a base. Because it is a base, the acid reacts with it to produce the calcium acetate, which is the salt, and water. Thus, the sequence of reactions becomes Ethyl alcohol : CH_3CH_2OH. To prepare acetic acid from ethyl alcohol, you must know one important fact; Ethyl alcohol is a primary alcohol, i.e., one where the OH group is located on a carbon atom bonded to only one other carbon atom. Primary alcohols can be directly oxidized to the corresponding carboxylic acid by the addition of KMnO_4, potassium permanganate. Thus, you perform the following reaction:

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Question:

Represent the exclusive-OR function, using a) NAND gates only b) NOR gates only Sketch the circuit realizations in each case.

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Solution:

a) Exclusive-OR realization using NAND gates only: The exclusive-OR function can be written as A × B = A \bullet B' + A' \bullet B [Using the Sum of Products form] = [(A\bulletB'+A' \bulletB)' ]' [Complementing Twice] = [(A\bulletB') ' \textbullet (A'\bulletB) ' ]' [Applying deMorgan's Rules on the inner primed bracket] = [(A\uparrowB')\uparrow(A'\uparrowB)] [Using the Shaeffer stroke notation]. We assume that both unprimed variables (A,B) and primed variables (A', B') are available as inputs. Note that the outer bracket represents a NAND function of the two inner brackets. Each inner bracket is a NAND function of two variables each. The circuit realization is shown in figure 1. b) Exclusive-OR realization using NOR gates only. We can write: A × B = (A+B) \textbullet (A'+B')[Using a Product of Sums form] = [ ((A+B) \textbullet (A '+B'))' ]'[Complementing twice ] = [ (A+B)' + (A'+B')']'[Applying DeMorgan's Rules on the inner primed bracket] = [(A\downarrowB')\downarrow(A'\downarrowB')][Using the Pierce Arrow notation]. Observe that in the last two steps the outer square bracket represents a NOR function of the two inner brackets, each of which is a NOR function of two variables. The circuit realization using NOR gates only is shown in figure 2.

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Question:

Design a buffer system that will function in the low, near neutral, and high pH levels, using combinations of the three sodium phosphates (Na_3PO_4, Na_2HPO_4, NaH_2PO_4) and phosphoric acid (H_3PO_4). The equilibrium constants for this polyproticacid are k_1 = 7.5 × 10^-3, k_2 = 6.2 × 10^-8, and k_3 = 4.8 × 10^-13. With this information, calculate the pH at both extremes of the buffer, assume 10 : 1 and 1 : 10 ratios, and at the mid-range of the buffer, assume a 1 : 1 ratio. Assume, also, that the acid-salt ratio, in making up the buffer is equal to HA/A^- in solution.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E10-0375.htm

Solution:

Solutions that contain appreciable amounts of a weak acid, such as phosphoric acid, and its salt or salts, such as sodium phosphates, are called buffers. Their utility is in maintaining a relatively constantpH.This problem requires one to find various pH values for this buffer system. To do this, note that the pH of a buffer is given by pH =pk_a+ log ([A^-] / [HA]) =pk_a+ log ([salt] / [acid]). One knows that to find the pH, one must determine [H_3O^+], since pH = - log [H_3O^+]. To do this consider the ionization of thepolyproticacid, phosphoric acid, which can dissociate to produce 3 protons, k_1, the first ionization constant, corresponds to H_3PO_4 + H_2O \rightleftarrows H_3O^+ + H_2PO_4^- so that k_1 ={[H_3O^+][H_2PO_4^-]} / [H_3PO_4]. k_2, the second ionization constant, corresponds to H_2PO_4^- + H_2O \rightleftarrows H_3O^+ + HPO_4^2- so that k_2 = {[H_3O^+][ HPO_4^2-]} / [H_2PO_4^-]. k_3, the third ionization constant, corresponds to HPO_4^2- + H_2O \rightleftarrows H_3O^+ + HPO_4^3- and k_3 = {[H_3O^+][HPO_4^3-]} / [HPO_4^2-]. In general, the k_a of any acid is given by the expression k_a = {[H_3O^+][A^-]} / [HA]or(H_3O^+) = k_a ([HA] / [A^-]) . It is from this equation that the [H_3O^+] must be calculated, and thus yield the pH value wanted. The three systems: Buffer system with k_1 Consider the ratios of 10 : 1, 1 : 1, and 1 : 10 for the acid : salt ratios, given that the ratio equals [HA]/[A^-]for the low, neutral and high pH values, re-spectively. This means that [H_3O^+] = 10 k_a , k_a and 0.1 k_a for low, neutral, and high, respectively. Here, k_a = k_1 which is = 7.5 × 10^-3.pk_1 = 2.12. Since [H_3O^+] = 10 k_a , k_a , and 0.1 k_a ; pH = - log [H_3O^+] = (pk_a- 1),pk_aand (pk_a+ 1) for low, middle, and high pH values, re-spectively. Thus, with pk_1 = 2.12, pH = 1.12 (low), 2.12 (middle) and 3.12 (high). Buffer system with k_2 One still has pH = (pk_a- 1),pk_a, andpk_a+ 1. But now k_a = k_2 = 6.2 × 10^-8 pk_2 = 7.21, Thus, pH = 6.21 (low), 7.21 (middle), and 8,21 (high). Buffer system with k_3 One still has pH = (pk_a- 1),pk_a, andpk_a+ 1. Now, k_a = k_3 = 4.8 × 10^-13 pk_3 = 12.32. Thus, pH = 11.32 (low), 12.32 (middle), and 13.32 (high).

Question:

From collision theory, it is found that when the rate- determining step involves collision of two A molecules, the rate will be proportional to the square of the A concentra-tion. Explain why.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E13-0453.htm

Solution:

Collision theory makes the assumption that for a chemical reaction to take place, particles must collide. The rate will depend upon the number of collisions per unit time between the particles involved and the fraction of these collisions that are effective. If you increase [A], there are more A molecules which can be hit. Also, the average A molecule gets hit more often. As such, the number of A with A collisions varies as the square of concentra-tion due to this double effect.

Question:

How many amperes will a 720-watt electric iron take from a 120-volt line?

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/Users/wenhuchen/Documents/Crawler/Physics/D24-0783.htm

Solution:

P = 720 watts,E = 120 volts P = E × I Then I = (P/E) = (720 watts/120 volts) = 6 amp

Question:

An object is 24 inches from a convex lens whose focal length is 8 inches. Where will the image be?

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Solution:

The use of a ray diagram is helpful here. We can see that the image is real (on the far side of the lens and inverted). We then solve mathematically: D_0 is 24 inches and f is 8 inches. Substituting in 1/D_0 + 1/D_I = 1/f 1/24 in. + 1/DI= 1/8 in. Solving D_I = 12 in. Hence D_I = 12 in. which means that the image is 12 inches from the lens on the side away from the object.

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Question:

Why is the mitochondria referred to as the "powerhouse of the cell " ?

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/Users/wenhuchen/Documents/Crawler/Biology/F02-0036.htm

Solution:

Mitochondria are membrane-bounded organelles concerned principally with the generation of energy to support the various forms of chemical and mechanical work carried out by the cell. Mitochondria are distributed throughout the cell, because all parts of it require energy. Mitochondria tend to be most numerous in regions of the cell that consume large amounts of energy and more abundant in cells that require a great deal of energy (for example, muscle and sperm cells). Mitochondria are enclosed by two membranes. The outer one is a continuous delimiting membrane. The inner membrane is thrown into many folds that extend into the interior of the mitochondrion. These folds are called cristae. Enclosed by the inner membrane is the ground substance termed the matrix (see accompanying diagram). Many enzymes involved in the Krebs cycle are found in the matrix. Enzymes involved in the generation of ATP by the oxidation of NADH_2 , or, the electron transport reactions, are tightly bound to the inner mitochondrial membrane. The enzymes for the specific pathways are arranged in sequential orders so that the products of one reaction do not have to travel far before they are likely to encounter the enzymes catalyzing the next reaction. This promotes a highly efficient energy production. The reactions that occur in the mitochondria are all related in that they result in the production of ATP (adenosine triphosphate), which is the common currency of energy conversion in the cell. Some ATP is produced by reactions that occur in the cytoplasm, but about 95 percent of all ATP produced in the cell is in the mitochondria. For this reason the mitochondria are commonly referred to as the powerhouse of the cell.

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Question:

It is estimated that 3 × 10^5 tons of sulfur dioxide, SO_2, enters the atmosphere daily owing to the burning of coal and petroleum products. Assuming an even distribution of the sulfur dioxide throughout the earth's atmosphere (which is not the case) , calculate in parts per million by weight the concentration of SO_2 added daily to the atmosphere. The weight of the atmosphere is 4.5 × 10^15 tons. (On the average, about 40 days are required for the removal of the SO_2 by rain).

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/Users/wenhuchen/Documents/Crawler/Chemistry/E01-0013.htm

Solution:

Here, one is asked to find the number of tons of SO^2 per 10^6 tons, i.e. per million, of atmosphere. This is done by using the following ratio: Let x = no. of tons of SO_2 per 10^6 tons of atmosphere. (3.0 × 10^5 tons SO_2 ) / (4.5 × 10^15 tonsatm) =( x / 10^6 tonsatm) x = (3.0 × 10^5 tons SO_2 ×10^6 tonsatm) / (4.5 × 10^15 tonsatm) x = 6.67 × 10\Elzbar5tons SO_2 or 6.67 × 10\Elzbar5ppmSO_2 .

Question:

What is the significance of hydrophobic and hydrophilic moieties in biological systems?

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/Users/wenhuchen/Documents/Crawler/Biology/F01-0026.htm

Solution:

All cells possess a cell membrane, which serves to regulate the materials entering and leaving the cell, and to give the cell shape and protection. The regulation of the movement of materials across the cell membrane (which materials and how much of each) is referred to as selective permeability. Since the cell's internal and external environment are mainly aqueous, the cell membrane must have a unique structure to be able to allow for the passage of water-soluble as well as water- insoluble materials. Water-soluble substances are called hydrophilic ("love of water") and water-insoluble substances are called hydrophobic ("fear of water"). The cell membrane is composed of a double layer of phospholipid with protein molecules dispersed throughout the membrane. Phospholipids and proteins are molecules that have both a hydrophilic and a hydrophobic portion. It is precisely these qualities which bestow the selective permeability property to the cell membrane. The hydrophilic portions (the polar or charged portions) are arranged so that they face the inside or outside of the cell. The hydrophobic portions are arranged so that they lie within the cell membrane. The polar phosphate moiety of the phos-pholipid is hydrophilic and thus faces the inner or outer surface. The fatty acid tails, the hydrophobic regions of the phospholipid, are oriented towards the middle of the bilayer. Membrane proteins are classified as peripheral (extrinsic) or integral (intrinsic) Peripheral proteins are on the inner or outer surface, exposed to the aqueous solution. Integral proteins may span the entire membrane or just the bilayer. Only the polar sections face the surfaces. Note: The above model is not a monomeric unit of the cell membrane. Cell membranes have diversity both externally and internally.

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Question:

Suppose a body of mass 0.5 kg slides down a track of radius R^\ding{217} = 1 m, like that in the Fig., but its speed at the bottom is only 3 m/sec. What was the work of the frictional force acting on the body?

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Solution:

Note that we cannot calculate the work done by friction from the definition of work because W_friction = \intF^\ding{217}_friction \bullet ds^\ding{217} and we do not know the functional form of F^\ding{217}_friction. Since work is a form of energy, our next thought may be to try to calculate the work done by the frictional force by using the principle of conservation of energy. At position 1, the mass is at rest, and its energy is equal to its potential energy E = mgR(1) In sliding down the track, some of this energy will be dissipated by friction and be transferred into heat energy Q and some will be transformed into the kinetic energy of the mass. Hence, at position 2, E = Q + (1/2)mv^2(2) Note that the potential energy of the mass at position 2 is zero, since this is our reference position for potential energy. Combining equations (1) and (2) mgR = Q + (1/2)mv^2 andmgR - (1/2)mv^2 = Q or(.5 kg) (9.8 m/s^2) (1 m) - (1/2)(.5 kg) (9 m^2/s^2) = Q 4.9 kg \textbullet (m^2/s^2) - 2.25 kg \textbullet (m^2/s^2) = Q Q = 2.65 Joules This is the heat energy produced by friction, and it is positive because heat energy has been gained by the system. Therefore, the work done by friction must be negative because this energy was lost to heat. Hence W_f = - Q = - 2.65 Joules

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Question:

Compute the r.m.s. speed of the molecules of oxygen at 76.0 cm. Hg. pressure and 0\textdegree C , at which temperature and pressure the density of oxygen is 0.00143 gm/cm^3.

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Solution:

The pressure exerted on a wall by an ideal gas can be calculated as follows. In Fig. (a), the elastic collision of a molecule with the wall is shown. If the force exerted on the mole-cule during the collision is perpendicular to the wall, then the x and y components of the molecule's velocity are not affected. The collision is assumed to be elastic, therefore the z component of the velocity is reversed in direction but remains unchanged in magnitude v'_x= vx v'y= v_y v'_z = \rule{1em}{1pt}vz The z-component of momentum of the molecule changes by an amount ∆p_z = p'_z - p_z = m(v'_z - v_z) = - 2mv_z. Now, we need to know the number of molecules that strike the wall with Velocities v^\ding{217}, per unit time per unit area. Let us consider an in-finitesimal area dA on the wall and the molecules incident on it with velocities parallel to v^\ding{217}. These molecules will be included in the tube shown in Fig. (b). Molecules in the tube that strike the wall with a velocityv^\ding{217} in time dt must be at most a distance l = vdt from the wall. Therefore, if we restrict the length of the tube to vdt, all the molecules inside it with velocity v^\ding{217} will strike the wall within the time dt. The molecules lying outside this tube with the same velocity v^\ding{217} will not strike dA during the same period. Since the molecules move in random directions, we expect that the number of molecules per unit volume, moving with velocity v^\ding{217}, does not depend on the direction of v^\ding{217} but rather on its magnitude v. If the density of the molecules with speed v is n(v), the total number of the molecules in the tube with speed v is = n(v) × volume = n(v) × vdt dA cos \texttheta = n(v) v_z dt dA . Because of the randomness of the motion, the average number of mole-cules moving toward the wall with velocity v^\ding{217} is equal to the average number moving away from the wall with velocity -v^\ding{217}. Therefore, the number of the molecules in the tube striking the wall is half the number we found for those with speed v; dN(v^\ding{217}) = (1/2)n(v) v_z dt dA. Therefore, the total number of particles incident on an area dA per unit time is [dN(v^\ding{217})]/[dt] = (1/2)n(v) v_z dA. The total rate of change of momentum is the number of collisions per unit time multiplied by the momentum change in each collision. Using Newton's Second Law. dF_z = [(dp_z)/(dt)] = (1/2) n(v) v_z (\rule{1em}{1pt}2mv) dA = - mv_z^2 n(v) dA where dF_z is the element of force which the area dA exerts on the molecules which strike it. The reaction force exerted on the wall by the molecules, using Newton's third law, is dF = -dFz thus the pressure on the wall becomes p = -(dF_z/dA) = mv_z^2 n(v). Now we have to find the total average pressure, since not all mole-cules have the same v^\ding{217}. For this, we re-express the average of the quantity v^2_z n(v) as < v^2_z n(v) >_av =v^2_zn wherenis the average number of particles per unit volume; n = (total number)/(total volume) = N/V. The average of v^2, for a gas, is the sum of the averages of its com-ponents v^2 =v^2_x+v^2_y+v^2_z. The motion is completely random, therefore all the directions of motion are equally probable. Therefore we may take v^2_x=v^2_y=v^2_z= (1/3)v_2 which gives p = (1/3)mv^2(N/V). In this problem, the pressure of oxygen is the weight of a 0.76 m. mercury column with unit area; p = volume × density × g = h\rho_Hg^g = 0.76 m × 13.6 × 10^3 kg/m^3 × 9.8 m/sec^2 = 1.01 × 10^5 nt/m^2. Pressure can be expressed as p = (1/3)v^2(mN/v) = (1/3)v^2[(M_total)/(v)] = (1/3)v^2\rho_ox givingv^2= 3p/\rho_ox Hence, the r.m.s speed is v^2= (3 × 1.01 × 10^5 nt/m^2)/(1.43 kg/m^3) = 2.12 × 10^5 m^2/sec^2 v_rms = \surd(v^2) = 461 m/sec.

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Question:

A current-carrying wire in the form of a semicircle lies in a plane at right angles to the direction of a uniform magnetic induction. Show that the force on the wire is the same as that experienced by a straight wire lying along the diameter between the ends of the semicircle.

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Solution:

Let the semicircle carrying a current i have radius r, and let the magnetic induction have magnitude B. Consider Fig. A, in which a vector element d1^\ding{217} of the current-carrying wire is shown in a direction from the center 0 of the semicircle making an angle \texttheta with the radius of symmetry. The force on that element due to the magnetic field is dF^\ding{217} = i(d1^\ding{217} × B^\ding{217}). Here d1^\ding{217} and B^\ding{217} are at right angles, and thus dF^\ding{217} lies perpendicular to both d1^\ding{217} and B^\ding{217}, or along the radius from d1^\ding{217} to 0. Also dF^\ding{217} has components, in the x- and y- directions shown, of - i d1B sin \texttheta and - i d1B cos \texttheta, respectively(see figure B). But there is a corresponding element d1^\ding{217} in a direction from 0 making an angle - \texttheta with the radius of symmetry. This element is subjected to a force which has x- and y-components of + i d1B sin \texttheta and - i d1B cos \texttheta, re-spectively. The x-components of the forces of these two elements thus cancel out, and this is true of all pairs of elements chosen at all possible angles \texttheta on the semicircle. It follows that the total force F^\ding{217} on the semicircle has a y-component only. Now dl/r = d1/r = d\texttheta radians or d1 = rd\texttheta. Also, when 1= 0, \texttheta = \pi/2 andwhen 1 = unity, 0 = + \pi/2. Therefore, F = ^1\int_0dF_y= F = ^1\int_0 - i d1B cos \texttheta = ^(\pi/2)\int_(-\pi/2)- ird\textthetaBcos \texttheta = - iBr[ sin \texttheta ^(\pi/2)]_(-\pi/2) = - 2i Br . Consider the straight wire; the force F^\ding{217}' on it is F^\ding{217}' = i (2r^\ding{217} × B^\ding{217}). Thus F^\ding{217}' has magnitude 2iBr and points in the negative y-direction. This is the same force as that which acts on the semicircle. One may arrive at this result and a more general result by noting that a closed current-carrying loop lying in a plane at right angles to the magnetic field experiences no net force. If the semicircular loop is closed by allowing the current to return along a wire occupying the vacant diameter (or by any loop whatsoever lying in the plane and joining the two ends of the semicircle), the complete circuit so formed experiences no net force. It follows that the forces on the semicircle and on the return wire are equal and opposite. Since reversing the direction of the current reverses the magnetic force on it, it imme-diately follows that the force on the semicircle is equal to the force on any current-carrying conductor lying in the plane and having the same endpoints.

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Question:

Outline briefly the events occurring in each stage of mitosis. Illustrate your discussion with diagrams if necessary.

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/Users/wenhuchen/Documents/Crawler/Biology/F25-0642.htm

Solution:

Mitosis refers to the process by which a cell divides to form two daughter cells, each with exactly the same number and kind of chromosomes as the parent cell. In a strict sence, mitosis refers to the division of nuclear material (karyokinesis). Cytokinesis is the term used to refer to the division of the cytoplasm. Although each cell division is a continuous process, in order for it to be studied, can be artificially divided up into a number of stages. We will describe each stage separately, beginning with interphase. 1) Interphase:This phase is called the resting stage. However, the cell is "resting" only with respect to the visible events of division in later phases. During this phase, the nucleus is metabolically very active and chromosomal duplication is occurring. During interphase, the chromosomes appear as vague, dispersed thread-like structures, and are referred to as chromatin material. 2) Prophase:Prophase begins when the chromatin threads begin to condense and appear as a tangled mass of threads within the nucleus. Each prophase chromosome is composed of two .identical members resulting from duplication in interphase. Each member of the pair is called a chromatid. The two chromatids are held together at a dark, constricted area called the centromere. At this point the centromere is a single structure. The above events occur in the nucleus of the cell. In the cytoplasm, the centriole (a cytoplasmic structure involved in division) divides and the two daughter centrioles migrate to opposite sides of the cell. From each centriole there extends a cluster of raylike fila-ments called an aster. Between the separating centrioles, a mitotic spindle forms, composed of protein fibrils with contractile properties. In late prophase the chromosomes are fully contracted and appear as short, rod-like bodies. At this point individual chromosomes can be distingushed by their characteristic shapes and sizes. They then begin to migrate and line up along the equatorial plane of the spindle. Each doubled chro-mosome appears to be attached to the spindle at its centromere. The nucleolus (spherical body within the nucleus while RNA synthesis is believed to occur) has been undergoing dissolution during prophase. In addition, the nuclear envelope breaks down, and its disintegration marks the end of prophase. 3) Metaphase:When the chromosomes have all lined up along the equatorial plane, the dividing cell is in metaphase. At this time, the centromere divides and the chromatids become completelty separate daughter chromo-somes. The division of the centromeres occurs simulta- neously in all the chromosomes. 4) Anaphase:The beginning of anaphase is marked by the movement of the separated chromatids (or daughter chromosomes) to opposite poles of the cell. It is thought that the chromosomes are pulled as a result of contraction of the spindle fibers in the presence of ATP. The chromosomes moving toward the poles usually assume a V shape, with the centromere at the apex pointing toward the pole. 5) Telophase:When the chromosomes reach the poles, telophase begins. The chromosomes relax, elongate, and return to the resting condition in which only chromatin threads are visible. A nuclear membrane forms around each new daughter nucleus. This completes karyokinesis, and cytokinesis follows. The cytoplasmic division of animal cells is accom-plished by the formation of furrow in the equatorial plane. The furrow gradually deepens and separates the cytoplasm into daughter cells, each with a nucleus. In plants, this division occurs by the formation of a cell plate, a partition which forms in the center of the spindle and grows laterally outwards to the cell wall. After the cell plate is completed, a cellulose cell wall is laid down on either side of the plate, and two complete plant cells form.

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Question:

A rectangular tank 6.0 by 8.0 ft is filled with gasoline to a depth of 8.0 ft. The pressure at the surface of the gasoline is 14.7 lb/in^2. (The density of gasoline is 1.325 sl/f^3). Find the pressure at the bottom of the tank and the force exerted on the bottom.

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/Users/wenhuchen/Documents/Crawler/Physics/D10-0410.htm

Solution:

The total pressure at the tank's bottom is the sum of the air pressure at the surface of the fluid and the pressure due to the gasoline above the tank bottom: P_air = 14.7 lb/in^2 Since 1 lb/in^2 = 144 lb/f^2 P_air = 14.7 lb/in^2 = (14.7 ) (144 lb/f^2) = 2120 lb/f^2(1) To find the pressure on the bottom of the tank due to the gasoline, we note that the pressure is equal to the force on the bottom of the tank divided by the area of the bottom P_gas = F/A P_gas = F/A But F_gas = \rhogV where \rho is the density of gasoline, g is 9.8 m/s^2, and V is the volume of the gasoline in the tank. Hence P_gas = \rhogV/A But V = hA, where h is the height of thegasoline in the tank. Therefore P_gas = \rhogh = (1.313 sl/f^3)(32 f/s^2)(8 f) P_gas = 336 lb/f^2(2) Hence, using (1) and (2) P_total = (336 + 2120)lb/f^2 P_total/Noting that = 2456 lb/f^2 P_total = F_total/A andF_total = P_total A = (2456 lb/f^2)(48 f^2) = 117888 lb

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Question:

When we hold our head upright, there is very little problem keeping our balance. Yet when our head is turned upside down , as in a head stand, we still manage to maintain our equilibrium . How is balance achieved in the latter case? Discuss your answer in relation to the ear.

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/Users/wenhuchen/Documents/Crawler/Biology/F20-0502.htm

Solution:

In the inner ear, there are two small, hollow sacs known as the utriculus andsacculus. Each sac is lined with sensitive hair cells, upon which rest small crystals of calcium carbonate (otoliths). Normally when the head is upright, the pull of gravity causes theotolithsto press against particular hair cells, stimulating them to initiate impulses to the brain via sensory nerve fibers at their bases. Changes in the position of the head cause these crystals to fall on and stimulate some other hair cells, thereby giving the brain a different set of signals. The relative strength of the signals to the brain indicates the position of the head at any given moment . When a head stand is being done, theotolithsin theutriculusand sacculus are reoriented by gravity and come to rest on other hair cells. These hair cells are stim-ulated and send impulses to our brain, informing us that now our head is in a new position. The brain will then transmit the appropriate response to the effectors so that we can maintain our balance even when our head is upside down. It is important to realize that equilibrium in humans depends upon the sense of vision, stimuli from theproprio-ceptors, and stimuli from cells sensitive to pressure in the soles of the feet in addition to the stimuli from the organs in the inner ear. This is why in certain types of deafness, in which the cochlea as well as the equilibrium organs of the inner ear are impaired , the sense of equi-librium is still present.

Question:

Ethyl chloride can be chlorinated to give two isomers, A and B. A can be chlorinated to give two isomers of C_2H_3Cl_3, while B gives only one isomer of C_2H_3Cl_3 upon further chlorination. Indicate the equations for the chlorination of ethyl chloride and identify A and B structurally.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E21-0754.htm

Solution:

You are told about isomerism in the compounds A and B, and their chlorinated products. When two compounds are isomers they have the same molecular formula, but differ in the sequence of linkages of atoms. These are structural isomers. Ethyl chloride has the structural formula The equation for chlorination is its reaction with Cl_2 in the presence of light. Note that there exist two possible carbons on which the chlorines can bond. These are the two isomers A and B. Thus you can write The question is which one is A, as the other must be B. You are told that A gives 2 isomers, while B yields 1, if they are chlorinated again. Let's start with If you chlorinate it, you would obtain These are two different compounds in the arrangement of atoms. They are structural isomers. Thus, isomer A must be If this is correct, then the chlorination of B should yield only one product. Let's see if this is true. and is the same as which means B yields only 1 isomer.

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Question:

Coal drops at the rate of 25 slugs per second from a hopper onto a horizontal moving belt which transports it to the screening and washing plant. If the belt travels at the rate of 10 ft per second, what is the horsepower of the motor driving the belt? Assume that 5% of the energy available is used in overcoming friction in the pulleys. (See figure.)

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/Users/wenhuchen/Documents/Crawler/Physics/D06-0312.htm

Solution:

We take the belt and hopper as our system. By using the principle of conservation of momentum, we can relate the net external force on the system (provided by the motor) to the time rate of change of the system momentum P^\ding{217},or (dP^\ding{217})/(dt) = F^\ding{217}_ext(1) We examine a small element of mass, \Deltam, leaving the hopper and falling on the belt. (See figure.) At t = 0, the momentum of \Deltam in the x direction is P_0^\ding{217} = (\Deltam) (0) and, at t = \Deltat, it is P_f^\ding{217} = \Deltam u^\ding{217} Hence\DeltaP^\ding{217} = P_f^\ding{217} - P_0^\ding{217} = \Deltam u^\ding{217} or\DeltaP^\ding{217}/ \Deltat = u^\ding{217}[(\Deltam/\Deltat)] Taking the limit as \Deltat \rightarrow 0, (dP^\ding{217}/dt) = u^\ding{217} [(dm/dt)](2) Comparing (2) with (1) F_ext^\ding{217} = u^\ding{217} [(dm)/(dt)] The power provided by the motor must be the time rate of change of the work, W, done on the belt by the motor. Hence, P = dW/dt = d/dt [\int F_ext^\ding{217} \bulletds^\ding{217}] where ds^\ding{217} is an element of path traversed by the belt. Assuming F_ext^\ding{217} = constant, and noting that ds^\ding{217} /dt = u^\ding{217}, we obtain P = (d/dt) (F_ext^\ding{217} \bullet \int ds^\ding{217}) = (d/dt) (F_ext^\ding{217} \bullet s^\ding{217}) P = F_ext^\ding{217} \bullet (ds^\ding{217}/dt) = F_ext^\ding{217} \bullet u^\ding{217} Since F_ext^\ding{217} and u^\ding{217} are parallel, we find, using(2) P = (F_ext)(u) = u^2 (dm/dt) P = (10 ft/s)2(25 s^1/s) = 2500 1b \bulletft/s(3) This power must be supplied by the motor to keep the belt moving at a uniform rate, assuming that none of it is dissipated in friction. Now, if 5% of the power supplied by the motor is used in overcoming friction, only 95% remains to power the belt. Since the power needed to move the belt is given by (3), we obtain, P = 95% P' orP' = (100/95) P = (100/95) (2500 lb \bullet ft/s) P' = 2631.6 ft \textbullet lb/s Since1 hp = 550 ft \textbullet lb/s P' = 4.8 hp Taking friction into account, the rate of working of the motor is 4 Taking friction into account, the rate of working of the motor is 4 .8 hp.

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Question:

Consider the collision of 2 particles of mass M_1 and M_2 , that stick together after colliding. Let M_2be at rest initially, and let v^\ding{217}_1 be the velocity of M_1 before the collision. (1) Describe the motion of M = M_1 + M2after the collision. (2) What is the ratio of the final kineticenergy to the initial kinetic energy? (3) What is the motion of the center of mass of this system before and after collision? (4) Describe the motion before and after the collision in the reference frame in which the center of mass is at rest.

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/Users/wenhuchen/Documents/Crawler/Physics/D06-0332.htm

Solution:

The basic principle used in solving a collision problem is the law of conservation of total momentum. This principle may be applied to any collision, so long as there are no external forces (forces due to the outside environment) acting on the system. From figures (1) and (2), we see that the initial momentum is M_1v^\ding{217}_1, and the final momentum is (M_1 + M_2)v^\ding{217}, and we obtain M_1v_1^\ding{217} = (M1+ M_2) v^\ding{217} v^\ding{217} = [M1/ (M_1 + M_2)] v_1^\ding{217} Hence, (M_1 + M_2) moves with velocity v^\ding{217}, parallel to v_1^\ding{217}. (2) The kinetic energy k_f after the collision is K_f = (1/2) (M1+ M_2)v^2 = (1/2) (M_1 + M_2) [(M1^2v1^2) / (M_1 + M_2)^2] Kf= [(M_1^2 + v_1^2)] / [2(M_1 + M_2)] The initial kinetic energy, k_i is (1/2) M_1v1^2, hence kf/k_i = (M_1^2v_1^2)/[2(M_1 + M_2)] \bullet [1/(1/2 M_1V_1^2)] = [M_1/(M_1 + M_2)] k_f/k_i = [1 - {M_2/(M_1 + M_2)}] The difference, k_i - k_f, is lost to increased internal motion in the (M_1 + M_2) system (i.e. internal excitations and heat). When a meteorite (M_1) strikes and sticks to the earth (M_2),essentially all the kinetic energy of the meteorite will be lost to heat in the earth. This follows from the fact that if M_2 >> M_1, k_f/k_i = [M_1/(M_1 + M_2)] = [1/{1 + (M_2/M_1)}] \approx0 Hence, k_f \approx0 and all the initial kinetic energy is transformed into heat. (3)The center of mass of a system of particles is a fictitious point whose motion is supposed to describe the trajectory of an imaginary bag which contains all the particles in that system. As the particles move around, the shape and the volume of this bag changes but not its momentum. The interactions of particles among themselves cannot result in a net-resultant force or change in momentum of the bag as a result of the action- reaction principle (for each force exerted on one particle by the others, there is an equal and opposite force exerted by this particle on the others). Furthermore, the collisions of particles with each other conserve the momentum of the colliding particles in each collision and cannot change the total momentum of the bag. In this way, we can view the motion of the center of mass as represent-ing the net effect of only the external forces on the system. If there are no external forces, then the center of mass will not change its velocity, irrespective of the final velocities of the particles. The position of the center of mass is given by R_cm^\ding{217} = (Sum of all m_iv_i^\ding{217})/(M_t) where m_i and r_i^\ding{217} are the masses and the positions of individual particles, M_t is the total mass. In our problem, let the collision take place at the origin of our coordi-nate system and at time t = 0. After the collision, the center of mass will just coincide with the mass (M_1 + M_2) and we have R_cm^\ding{217} = vt^\ding{217} = [M_1/(M_1 + M_2)] v_1^\ding{217}_ t The center of the mass velocity is v_cm^\ding{217} = (sum of all p_i^\ding{217})/(M_t) where p_i^\ding{217} are the individual moment. For our problem v_cm^\ding{217} = v^\ding{217}=[M_1/(M_1 + M_2)] v_1^\ding{217} This expression for the velocity of center of mass will be true for all times, i.e. also before the collision (for which t < 0). (4) In this reference frame v_cm^\ding{217} = 0. (This reference frame is attached to the center of mass of the system.) The new velocity of M_1 with respect to this observerwill be the velocity of M_1 in the old frame minus thevelocity of the c.m. frame with respect to the old frame or u_1^\ding{217} = v_1^\ding{217} - v_cm^\ding{217} = [M_2/(M_1+ M_2)] v_1^\ding{217} and, for M2 u_2^\ding{217} = v_2^\ding{217} -v_cm^\ding{217}= - v_cm^\ding{217} = - [M_1/(M_1 + M_2)] v_2^\ding{217} This result for u^\ding{217}_2 could be guessed right away.When the observer was in the old frame,M_2 was station-ary. As the observer moves with v^\ding{217}_cm With respect to this old frame, then with respect to this observer M_2 will appear to move in the opposite direction with equal speed: u_2^\ding{217} = -v_cm^\ding{217} . The total momentum p^\ding{217}_total in this system should add up to zero since v_cm^\ding{217} is zero; indeed we have p\ding{217}_total = M_1u_1^\ding{217} + M_2u_2^\ding{217} = [(M_1M_2)/(M_1 + M_2)] v^\ding{217}_1 - [(M_1M_2)/(M_1 + M_2-)] v_1^\ding{217} = 0 The advantage of the center of mass frame is that the total momentum in it is zero.

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Question:

A uniform, spherical bowling ball is projected without initial rotation along a horizontal bowling alley. How far will the ball skid along the alley before it begins to roll without slipping? Assume that the ball does not bounce. (See figure).

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/Users/wenhuchen/Documents/Crawler/Physics/D04-0209.htm

Solution:

The ball is projected along the alley at point A. When it hits the alley at point B, the ball will roll and slip. At point C, the ball has begun to roll without slipping. In order to find the distance at which skidding (or slipping) stops, we must find the acceleration of the ball, and then solve for its position. Applying Newton's Second Law to the horizontal direction of motion of the ball, we obtain ma_x = - f(1) where a_x is the horizontal acceleration of the ball of mass m. Note that we take a_x to be positive in the positive x-direction. Applying the Second Law to the vertical direction ma_y = N - mg But a_y , the y acceleration of the ball, is zero since the ball doesn't bounce. Therefore N = mg(2) In order to calculate torques,\cyrchar\cyrt^\ding{217}, about 0, we use \cyrchar\cyrt^\ding{217} \cyrchar\cyrt^\ding{217}=I\alpha^\ding{217}(3) \cyrchar\cyrt^\ding{217} \alpha^\ding{217} where\alpha^\ding{217}is the angular acceleration of the ball. But, the only torque acting \alpha^\ding{217} on the ball is due to f. Hence \cyrchar\cyrt^\ding{217}= r\ding{217}×f\ding{217} \cyrchar\cyrt^\ding{217} \ding{217} × \ding{217} where r\ding{217}is the location of the point of application of f\ding{217}. Since r\ding{217}and f\ding{217} \ding{217} \ding{217} \ding{217} \ding{217} are perpendicular (see figure), \cyrchar\cyrt = Rf From (3) Rf = I\alpha(4) \alpha where\alphais positive in a direction pointing into the plane of the figure.Now, \alpha Now, if f is constant, (1) tells us that a_x is constant, since a_x = (-f)/m - Hence, we may use the kinematics equations for constant acceleration to find v_x = ^vx_0 [(-f)/m] t(5) (-f)/m] t(5) - x = x_0 + ^vx_0 t [(-f)/(2m)]t^2(6) - where x_0 andvx_0are the initial position and velocity of the ball. Similarly, v x_0 we use (4) to solve for the angular velocity of the ball. \alpha= Rf/I \alpha But\alpha= d\omega/dt where\omegais the angular velocity of the ball. Therefore \alpha \omega \omega d\omega/dt = Rf/I \omega \omega\int_\omega_0 d\omega= Rf/I ^t\int_0 dt \omega \int_\omega \omega \int where\omega_0 is the angular velocity at t = 0. \omega \omega=\omega_0 + [Rf/I]t(7) \omega \omega While the ball is skidding, (7) and (5) are completely independent relations. However, when the ball starts rolling without slipping, they are related by v_x =\omegaR(8) \omega Substituting (7) and (5) in (8) vx_0-(f/m) t' =\omega_0R + [(R^2f)/I] t' v x_0-(f/m) t' =\omega_0R + [(R^2f)/I] t' - \omega But\omega_0 = 0. Solving for t' \omega vx_0 = t'[{(R^2f)/(I)} + (f/m)] v x_0 = t'[{(R^2f)/(I)} + (f/m)] t' = (^vx_0)/[ {(R^2f)/(I)} + (f/m) ](9) At t = t', slipping stops. To find the position of the ball when slipping stops, we substitute (9) in (6) x = x_0 -vx_0 [(^vx_0)/{(R^2f / I) + (f/m)}] - ( f/2m) [(^vx_0^2)/{(R^2f / I) + (f/m)}] v x_0 [(^vx_0)/{(R^2f / I) + (f/m)}] - ( f/2m) [(^vx_0^2)/{(R^2f / I) + (f/m)}] Furthermore, x_0 = 0 (see figure), and x =(^vx_0^2) / [f{(R^2/I) + (I/m)}]-[-f(^vx_0^2)]/[2mr^2{(R^2/I) + (I/m)}^2](10) (^vx_0^2) / [f{(R^2/I) + (I/m)}]-[-f(^vx_0^2)]/[2mr^2{(R^2/I) + (I/m)}^2](10) - - The frictional force law is f = u_kN(11) where N is the normal force of the alley on the ball, and u_k. is the coefficient of kinetic friction. (We use the coefficient of kinetic friction because the ball skids from t = 0 to t = t'.) Substituting (2) in (11) f = u_kmg Substituting this in (10), x = (^vx_0^2) / [u_kmg{(R^2/I) + (1/m)}]-(^vx_0^2) / [2u_km^2g{(R^2/I) + (1/m)}^2] - x = (^vx_0^2) / [u_kmg{(R^2/I) + (1/m)}]-[1 - (1 / [2m{(R^2/I) + (1/m)}])] - For a sphere I = 2/5 mR^2 and R^2/I = R2/ (2/5 mR^2) = 5/2m then x = (^vx_0^2) / [u_kmg{(5/2m) + (1/m)}]-[1-(1 / [2m{(5/2m) + (1/m)}])] - - x = [(^2vx_0^2) / (7u_kg)] [1-(1/7)] =(^12vx_0^2)/(49u_kg) - This is the position at which slipping stops.

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Question:

Of the terrestrial arthropods, the insects are among the best adapted for the prevention of water loss. Explain.

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/Users/wenhuchen/Documents/Crawler/Biology/F12-0306.htm

Solution:

Insects are a large group of mostly terrestrial arthropods. Their success of survival .on land is partly due to the adaptation of water conserving mechanisms. Their epicuticle, the outermost exoskeletal layer, is impregnated with waxy compounds. This serves to prevent loss of water by evaporation. Their wastes are excreted as uric acid, a crystalline substance, thus minimizing water loss due to protein metabolism. Uric acid is ex-creted via Malpighian tubules. Malpighian tubules lie more or less free within the hemocoel (blood sinus) , with the proximal end attached to the digestive tract at the junction of the midgut and hindgut (rectum). There may be 2 to 250 tubules. If there are only a few, the tubules are usually long, slender, and convoluted. If there are many, the tubules may be short and grouped in bunches. The cells lining the tubule lumen are cuboidal epithelial cells. The outer layer of the tubule wall is composed of elastic connective tissue and muscle fibers. The tubules are capable of peristalsis. Uric acid, formed by the tissues and released into the blood, is selectively absorbed by the tubule cells from the blood within the hemocoel. Salt, water, and amino acids are also absorbed. Together, the substances are discharged from the Malpighian tubules into the hindgut. Nutritive substances and some of the salts are reabsorbed by the rectal epithelium and are returned to the blood. Much water is also reabsorbed to conserve water inside the body. The uric acid is excreted with the waste products of digestion. There are several other methods of excretion in insects. Some excess salts and other substances are deposited in the cuticle and shed when the insect molts. Some particulate wastes are picked up and degraded intra-cellularly by nephrocytes, cells located on or near the heart.

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Question:

What is wrong with each of the following gate representa-tions ?

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G04-0068.htm

Solution:

a) This is an example of the incorrect assumption that a single variable may be controlled simultaneously from two or more elements whose outputs are inconsistent. Even though the definition of a gate structure is fulfilled \rule{1em}{1pt}\rule{1em}{1pt} a set of elements with given behaviors and a set of couplings between elements \rule{1em}{1pt} \rule{1em}{1pt} the contradictions inherent in the structure make its analysis meaningless. For example, if X_1 = X_2 = 0 and X_3 = X_4 = 1, then Y could be either 1or 0. b) Although a circuit may contain dependency loops, this circuit contains a closed loop of inconsistent dependency in which a variable Y depends upon another variable X in such a way that two different values of Y are required si-multaneously. This example fails for Y = 0 and X = 1 be-cause Y must also equal 1 now. c) Let us represent the output in terms of a pair of simul-taneous Boolean equations: Y_1 = X_1 \textbullet Y_2 Y_2 = X_2 + Y_1 When solving these equations for Y_1 and Y_2 by substitution, we obtain either of two sets of equations: Y_1 = Y_2 \textbullet X_1 Y_2 = X_2(a) or Y_1 = X_1 Y_2 = X_2 \textbullet X_1 If we begin with X_2 = 0 and X_1 = 1, set (a) gives Y_1 = Y_2 = 0, but set (b) gives Y_1 = Y_2 = 1. With this state, the response of the circuit is ambiguous.

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Question:

An experiment to measure Avogadro's number involves the collection of a beam of alpha particles to which electrons are added to give neutral helium. Suppose that for a particular run it takes a number of electrons equivalent to 6.40 × 10^-5 coulomb/sec to neutralize the beam, and that after 24 hours of collection 1.14 × 10^-4 g of neutral helium are collected. From this data, determine Avogadro's number. The charge on one electron is 1.602 × 10^-19 coulombs.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E04-0114.htm

Solution:

Remembering that it takes two electrons to neutralize one alpha particle, i.e., one positively charged He atom, we will calculate the total charge required during the 24 hour run and, from this, the number of electrons. The number of alpha particles neutralized, or the number of helium atoms produced, is then equal to one-half the number of electrons. Using the mass of helium collected and the atomic mass of helium (4.0026 g/mole), we will calculate the corresponding number of moles. Dividing the number of atoms by the number of moles will give the number of atoms per mole, or Avogadro's number. 24 hours is equivalent to 24 hours × 60 min/hr × 60 sec/min = 86,400 sec. Thus, in 24 hours it took 6.40 × 10^-5 coulomb/sec × 86,400 sec = 5.530 coulomb to neutralize the beam. Since the charge on one electron is 1.602 × 10^-19 coulomb, dividing this number into the total number of coulombs gives the equivalent number of electrons, or number of electrons = coulombs/ coulombs of electron = 5.530 coulombs / (1.602 × 10^-19 coulomb / electron) = 3.452 × 10^19 electrons . It takes two electrons to neutralize one alpha particle, hence the number of alpha particles neutralized by this number of electrons is number of alpha particles = number of electrons / 2 = (3.452 × 10^19) / 2 = 1.726 × 10^19 alpha particles. But one alpha particle produces one helium atom, so that the beam produced 1.726 × 10^19 helium atoms. The mass of helium corresponding to this number of helium atoms produced by the beam is given as 1.14 × 10^-4 g. Dividing by the atomic mass of helium, we convert this mass to moles: mass helium = mass helium / atomic mass helium = (1.14 ×10^-4 g) / ( 4.0026 g/mole) = 2.84 × 10^-5 mole. Hence, 1.726 × 10^19 helium atoms correspond to 2.84 × 10^-5 mole of helium. Dividing these two numbers gives Avogadro's number: Avogadro's number = number of atoms / moles = (1.726 × 10^19 ) / (2.84 × 10^-5 mole) = 6.08 × 10^23 atoms/mole. Comparing this with the currently accepted value of 6.02 × 10^23 atoms/ mole, we see that the estimate calculated is in error by [{(6.08 × 10^23) - (6.02 × 10^23)} / (6.02 × 10^23)] × 100% = [{( 6.08- 6.02 ) / (6.02 )} × 100% ] = (0.06 / 6.02) × 100% = 1% .

Question:

Write a program that finds, for integers m and n, the quotient and remainder when m is divided by n.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G22-0546.htm

Solution:

The Fundamental Theorem of Arithmetic states that if m and n are any integers with n \not = 0, then there exist unique integers q and r, 0 \leq r < │n│ such that m =qn+ r. As examples: 1) m = 43 n = 5; using the division algorithm q = 8 r = 3 and m =nq+ r, or 43 = 8(5) + 3. 2) m = -43 n = -5 q=9 r = 2 m =nq+ r, or -43 = 9(-5) + 2 Here r = 2 < │n│ = │-5│ = 5 . 3) m = 8 n = 32; 8 = 0(32) + 8 . The following computer program accepts any integers m, n (n \not = 0) and computes the quotient and remainder of m/n. CINPUT INTEGERS M AND N. IF N IS CZERO, READ ANOTHER PAIR. 5READ (5,10) M,N IF (N.EQ.0) GO TO 5 10FORMAT (2I6) CFIND THE QUOTIENT IQ = INT (M/N) CFIND THE REMAINDER IR = M - IQ\textasteriskcenteredN COUTPUT RESULTS WRITE (6,60) M,N 60FORMAT (1HO, 10HDIVIDEND =, I6, 10H DIVISOR =, I8) 70WRITE (6,80) IQ, IR 80FORMAT (1H, 10HQU0TIENT =, I6, 12H REMAINDER =, I6) 90STOP END

Question:

What is the magnetic induction B at the center of a Circular cable consisting of 100 turns of wire having a common radius of 20 cm carrying 15 amperes?

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/Users/wenhuchen/Documents/Crawler/Physics/D22-0740.htm

Solution:

Consider the circular contour about the circular loop shown in The figure. By the Biot-Savart law the contribution to the magnetic induction at the center of the circular loop, due to an infinitesimal element, dl^\ding{217}, of the loop, is given by dB = (\mu_0 /4\pi) [(NI dl sin \texttheta)/a^2] where \mu_0 = 4\pi × 10^-7 Weber/A \bullet m is the constant of permeability, \texttheta is the angle between dl^\ding{217} and the radius vector a^\ding{217} (\texttheta = 90\textdegree), and NI is the total current through the loop. At the center of the loop, B is directed per-pendicular to the plane of the loop. B = \int dB = (\mu_0 /4\pi) (NI/a^2 ) \int dlfor sin 90\textdegree = 1 B = (\mu_0 /4\pi) (NI/a^2 ) 2\pia = (\mu_0 /2) (NI/a) We are given that N =100 turns, I = 15 amp and a = 20 cm × 1 m/100 cm then B = {2\pi × 10^-7 [W/(A-m)] × 100 × 15A}/(0.2 m) = 1.5 × 10^-4 \pi weber

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Question:

Can you distinguish between two gene loci located on the same chromosome that have 50 percent crossing over and two gene loci each located on different chromosomes?

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Solution:

When two linked genes have 50 percent crossing over between them, it means that 50 percent of the progeny will be recombinants. For example, a heterozygous parent genotypically will form four will form four types of gametes if a cross-occurs between A and B: Now if the recombinant types occurred 50% of the time, this means that they are produced in numbers equal to the parental types. This could happen only if crossover occurred between the genes during every meiosis in every individual. In other words, if the genes are so far apart that it is certain that crossover will occur between them, then in any meiotic event, recombinant types will always be formed, and in equal numbers with parental types. Therefore, the four types of gametes occur in a 1:1:1:1 ratio. When testcrossed, the progeny will also occur in a 1:1:1:1 phenotypic ratio. This means that the ratio of the progeny from parents having linked genes showing 50% crossing over is indistinguishable from that of parents with genes that are not linked. Thus when genes are separated by 50 map units, it is impossible to differentiate between linkage and non-linkage. Note that no more than 50% recombination is ever expected between any two loci because only two of the four chromatids are involved in crossing over. In fact, the ratio is usually less than 50% because crossover may not always occur and because double or multiple cross-- overs can reduce the apparent number of crossover events, as shown below.

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Question:

Explain briefly the background and applications of BASIC computer language.

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Solution:

BASIC (Beginner's All-purpose Symbolic Instruction Code). In the early 1960's, a project was undertaken at Dartmouth College to make the computer more accessible and easy to use by both students and faculty. One of the objectives was to develop a language that could be quickly learned but powerful enough to be used in solving most small to medium scale problems in any discipline. The result of this effort was BASIC, and it more than met its objective. Since BASIC provides the opportunity for someone with little or no previous computer experience to quickly begin writing programs and using the computer, it is commonly available on time sharing systems. BASIC's statements and operands are, probably, the closest to actual English language and algebra respectively and the easiest to comprehend, which ensures its huge popularity in almost every field of science, business, and engineering.

Question:

Calculate the number of grams of NH_4Cl that must be added to 2 l of 0.1 M NH_3 to prepare a solution of pH = 11.3 at 25\textdegreeC. The K_b for NH_3 = 1.8 × 10^-5. The equilibrium equation is: NH_3 + H_2O \rightleftarrows NH_4^+ + OH^-.

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Solution:

To determine how many grams of NH_4Cl must be added to this solution to maintain a pH of 11.3, one must first calculate the [NH_4^+] needed, because [NH_4^+] = [NH_4Cl] in the solution. This is done by using the ex-pression that describes the dissociation, NH_3 + H_2O \rightleftarrows NH_4^+ + OH^-. The dissociation constant (K_b) for this reaction is 1.8 × 10^-5. K_b is equal to the product of the concentrations of the substances formed divided by the concentration of the reactants. These concentrations are raised to the power of their coefficients in the equation for the reaction. [H_2O] is not included because it is assumed to remain constant. K_b = {[NH_4^+] [OH^-]} / [NH_3] = 1.8 × 10-^5 . To solve for [NH_4^+], one must know [OH^-] and [NH_3] first. One is given that NH_3 solution is .1 M, thus [NH_3] = 1. One can find [OH^-] by using the definition ofpOH.pOH= 14.0 - pH = - log [OH^-]. The pH of this solution is 11.3, thereforepOH= 14.0 - 113. = 2.7. One solves for OH^- by using the relationpOH= - log [OH^-]. Solving for [OH^-] 2.7 = - log [OH^-] 10^-2.7 = [OH^-] 2.0 × 10^-3 = [OH^-] One can now use the expression for K_b to solve for [NH_4^+] 1.8 × 10^-5 = {[NH_4^+] [OH^-]} / [NH_3] = {[NH_4^+] [2.0 × 10^-3]} / [.1] [NH_4^+] = {(.1) (1.8 × 10^-5)} / (2.0 × 10^-3) = 9.0 × 10^-4 M One is given that 2 l of this solution has been made up. Because, Molarity= number of moles / volume in liters; number of moles =molarity× volume = [9.0 × 10^-4 (mole/liter)] (2 liters) = 18 × 10^-4 moles. 18 × 10^-4 moles of NH_4Cl must be added to the solution. Because moles = weight in grams/molecular weight weight in grams = moles × M.W. (MW of NH_4Cl = 53.5) = 18 × 10^-4 moles (53.5 g/mole) = 0.096 g of NH_4Cl to be added.

Question:

Determine the radii of the first and second quantum orbits in the hydrogen atom.

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Solution:

We will derive the Bohr theoretical expression for the radius of the quantum orbit of an electron. Consider an electron of mass m and charge e moving at a constant distance r from the nucleus with a tangential velocity v, as shown in the accompanying figure. In order for this orbit to be stable, the Coulombic force attracting the electron to the nucleus (e^2/r^2) must be balanced by the centrifugal force on the electron (mv^2/r), which, if not compensated for, would compel the electron to leave its orbit. Hence, the magnitude of these two forces must be equal. (e^2 /r^2) = (mv^2/r) Solving for r, we obtain r = (e^2/mv^2) Bohr introduced quantization of the electron`s orbit by assuming that the electron's angular momentum (mvr) is given by: mvr = n (h/2\pi), where h is Planck's constant and n is an index, called the principal quantum number, which takes on the values n = 1 for the first orbit, n = 2 for the second orbit, and so on. Solving for v, v = [(nh)/(2\pimr)] We can now obtain an expression for r which includes quantization and is in terms of known quantities. Using the above expressions for v and r, r = (e^2/mv^2) = (e^2/m) [(2\pimr)/(nh)] = [(4\pi^2 e^2 mr^2)/(n^2 h^2)], r = [(n^2 h^2)/(4\pi^2 e^2 m)] The radius of the first orbit (n = 1), is called the Bohr radius. Solving for r : e = 4.80 × 10^-10 esu and m = 9.11 × 10^-28 g. r_1 = [(1^2 × h)/(4\pi^2 e^2 m)] = [{1 × (6.626 × 10^-27 erg-sec)^2} / {4\pi^2 × (4.80 × 10^-10 esu)^2 × 9.11 × 10^-28 g)}] = 5.29 × 10^-9 cm = 0.529 \AA. Let the radius of the nth orbit be denoted r_n. Then r_n = [(n^2 h^2)/(4\pi^2 e^2 m)] = n^2 [(h^2)/(4\pi^2 e^2 m)] = n^2 r_1 = n^2 × 0.529 \AA. Hence, the radius of the second quantum orbit is r_2 = n^2 × 0.529 \AA = (2)^2 × 0.529 \AA = 2.12 \AA.

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Question:

Describe between an exoskeleton and an endoskeleton.

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Solution:

In most of the largemulticellularanimals, some kind of skeleton is found that serves for protection, support and locomotion. The skeleton of an animal may be located outside or inside the body. If the former is the case , the skeleton is termed an exoskeleton, and is usually composed of a semiflexible substance known as chitin. If the latter is the case, the skeleton is termed an endoskeleton, and is composed of either carti-lage bone , or both. The hard shells of lobsters, crabs, oysters,clanis, and snails are examples of exoskeletons. In fact, themolluscand arthropod phylums are charac-terized by the presence of an exoskeleton. The advantage of an exoskeleton as a protective device is obvious, but a serious disadvantage is the attendant difficulty of growth. Snails and clams solve this difficulty by secret-ing additions to their shells as they grow. Lobsters and crabs have evolved a complicated process, called molting, whereby the outer shell is first softened by the removal of some of its salts . A new similarly soft cuticle is secreted beneath the old shell. The old shell is then split down the back. Following this split, the animal crawls out of the old shell, grows rapidly for a short time, and thenredepositsthe removed mineral salts in the new shell, hardening it. During molting, the animal usually remains well-hidden, because this is a time when it is defenseless and easily killed by its enemies. Human beings and all other chordates have an endo-skeleton. The skeleton of sharks and rays is made of cartilage. The human skeleton consists of approximately 200 bones; the exact number varies at different periods of life as some bones, at first distinct, gradually become fused. The skeleton in man acts as a supporting framework for the other organs of the body. Since the skeleton is internal and grows with the organism, no problem such as that encountered in the growth of lobsters and crabs occurs.

Question:

Trace the movement of the integers 4, 3, 5, 1, 2 as they are sorted in ascending order with the bubble sort. Can you give a general formula for the time needed for sorting N numbers in the best case? In the worst case?

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Solution:

The bubble sort compares each pair of adjacent elements in a list. If in the proper order, the pair is left alone, and the next pair is examined. If out of order, the items are switched, and the next pair is examined. In this fashion, the larger elements are "bubbled\textquotedblright down the list, while the smaller elements move to the head of the list. Of course, if the list is to be sorted in descending order, the opposite process will occur. Let us illustrate the steps necessary to arrange the integers in ascending order: 1st comparison3, 4, 5, 1, 2Compare 3 and 4,no switch 2nd comparison3, 4, 5, 1, 2Compare 4 and 5,no switch 3rd comparison3, 4, 1, 5, 2Compare 5 and 1,switch them 4th comparison3, 4, 1, 2, 5Compare 5 and 2,switch - end of 1st pass 5th comparison3, 1, 4, 2, 5Compare 4 and 1,switch 6th comparison3, 1, 2, 4, 5Compare 4 and 2,switch - end of 2nd pass 7th comparison1, 3, 2, 4, 5Compare 3 and 1,switch 8th comparison1, 2, 3, 4, 5Compare 3 and 2,switch - end of 3rd pass In the best case, i.e., when the data are already order, the bubble sort makes only one pass through the list. No sorting method can do any better than this. However, in the worst case, the bubble sort makes a maximum of N (N - 1) / 2 comparisons, just like other sequential sorting methods. The worst case is realized when the data are in the opposite order from the order you wish to have. For example, if we wish to sort the integers 5, 4, 3, 2, 1 into ascending order, the bubble sort will make 10 switches. 4, 3, 2, 1, 54 switches 3, 2, 1, 4, 53 switches 2, 1, 3, 4, 52 switches 1, 2, 3, 4, 51switches 10 switches total

Question:

Explain the process of gaseous exchange in the leaf.

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/Users/wenhuchen/Documents/Crawler/Biology/F10-0255.htm

Solution:

A leaf that is actively carrying on photosynthesis must have a high rate of gas exchange with the environment to remove the excess oxygen and to acquire an adequate supply of carbon dioxide. As carbon dioxide is utilized in photosynthesis, its con-centration within the cell decreases and carbon dioxide diffuses into the cell from the film of water surrounding the cell. Other carbon dioxide molecules pass from the air spaces of the spongy layer of the leaf and dissolve in the water film. They will then diffuse into the cell as the cell rapidly uses up carbon dioxide. Still other carbon dioxide molecules diffuse from the air outside the leaf through the stomata and into the air spaces within the leaf. In short, carbon dioxide moves into the cell along a concentration gradient. Carbon dioxide molecules diffuse from a region of higher concentration such as the atmosphere, to a region of lower concentration, in this case, the cells of the leaf . In a similar manner but opposite direction, oxygen produced within the cells as a result of photo-synthesis passes from cell to water film to air spaces and through the stomata to the exterior by diffusion along a concentration gradient.

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Question:

Calculate the molecular weight of a pure isoelectric protein if a 1% solution gives an osmotic pressure of 46 mm of H_2O at 0\textdegreeC. Assume that it yields an ideal solution.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E22-0812.htm

Solution:

The relationship between molecular weight and osmotic pressure is M = (C/ \pi)RT, where M is the molecular weight, c the concentration in grams per liter, R the gas constant (0.082 liter atm deg-1 mole^-2) , T the absolute temperature, and \pi the osmotic pressure in atmospheres. One is given these quantities in the problem, but they are expressed in different units than those needed to solve for molecular weight. Converting these quantities: Concentration: The solution is 1% by weight protein. One liter of water weighs 1000 g; 1% of 1000 is 10 so that c = 10. Osmotic pressure: The osmotic pressure is 46 mm H_2O. To convert to atmospheres, 46 mm H_2O must be converted to mm Hg. mmHg= 46mm × [(.9970 density of H_2O) / (13.534 density of Hg)] =3.39. There are 760 mm Hg in 1 atm. Therefore, the number of atmospheres is equal to (3.39 mm Hg) / (760 mm Hg/atm) = 4.45 × 10^-3 atm Temperature: T = 0\textdegreeC + 273 = 273\textdegreeK. Solving for M: M = [(10 g / litre)] / [4.45 × 10^-3 atm] ×.082 (liter-atm/\textdegreek-mole) × 273\textdegreek = 50,300 g/mole.

Question:

Explain the action of the following procedure which in-cludes asubroutine procedure within itself. VERIFY_TEST_VALUES: PROCEDURE; DCL LENGTH FIXEDDEC(3); CALL GET_AND_TEST_INPUT; \textbullet \textbullet \textbullet \textbullet \textbullet CALL GET_AND_TEST_INPUT' \textbullet \textbullet \textbullet \textbullet \textbullet GET_AND_TEST_INPUT:PROCEDURE; AGAIN:GETLIST(LENGTH); IF LENGTH = 0 THEN GOTO L; IF LENGTH<0 \vert LENGTH>90 THEN DO; PUTLIST('ERROR', LENGTH);GOTOAGAIN; END; /\textasteriskcentered END OF DO GROUP \textasteriskcentered/ END GET_AND_TEST_INPUT; \textbullet \textbullet \textbullet \textbullet \textbullet CALL GET_AND_TEST_INPUT; L:ENDVERIFY_TEST_VALUES;

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Solution:

From three different points in the program, the subroutine procedureis called. This is called a subroutine procedure because no re-sult value is produced. This is what differentiates a sub-routine procedure from a functionprocedure, which initiates an evaluation of a result to be later usedby the main pro-gram. Each time the subroutine is invoked in the above ex-ample, it acquiresa new value length from the data card. Then, the subroutine checksif each length value is between 0 and 90. If the value is <0 or >90 then the PUT LIST statement is executed, andthe control goes to the statementlabelled'AGAIN'. Thus, a new value forlength is obtained. If a value of zero is obtained for length, it is an indicationthat data is over and the program returns to the statement labelledFINISH in the calling program. But, if the value of length lies between0 and 90, then the program does not go to the statementlabelled FINISH, nor does it enter the DO group. Instead, the program control goes tothe next statement after the DO group, which happens to be the last statementof the subroutine procedure. From here, control is transferred backto the calling procedure, which con-tinues with the other steps of the program.

Question:

What is the equilibrium concentration of oxygen gas in fresh water at 25\textdegree C when exposed to air at a pressure of 760 torr? The Henry's law constant for oxygen in water at 25\textdegreeC is 3.30 × 10^7torrs. Express your answer in milligrams of oxygen per liter.

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Solution:

Solving this problem requires the use of Henry's law. It is stated P_2 = X_2 , K_2 , where P_2 is the partial pressure,X_2 is the mole fraction, and K_2 is the Henry's law constant. The subscript, 2, indicates the solute (or component at lower concentration). Since P_2 = 760torrand K_2 = 3.30 × 10^7torrsare known, we can solve for X_2 or the mole fraction of O_2 in water. Thus, X_2 = (P_2 ) / (K_2 ) = (760torr) / (3.3 × 10^7torrs) = (moles of O_2 ) / [(moles of O_2 ) + (moles of H_2O)] The volume is assume to be 1 liter or 1000 grams of H_2O . The number or moles of H_2O is [(1000g) / (18.02g/mole)] 55.49 moles H_2O . In addition, the number of moles of O_2 is negligible in comparison with the number of moles of H_2O, thus the number of moles of H_2O is [(760torr) (55.49 moles H_2O)] / (3.3 × 10^7torrs) , = 1.28 × 10-^3 moles O_2 . The weight in milligrams of O_2 is (1.28 × 10-^3 moles ) (32g/mole) (10^3 mg/g) = 41 mg. Since there is 1 liter of H_2O , then the concentration of O_2 is 41mg/l .

Question:

Baking powder consists of a mixture of cream of tartar (potassium hydrogen tartrate, KHC_4 H_4 O_6, molecular weight = 188 g/mole) and baking soda (sodium bicarbonate, NaHC0_3, molecular weight = 84 g/mole). These two components react according to the equation KHC_4 H_4 O_6 + NaHCO_3 \rightarrow KNaC_4 H_4 O_6 + H_2 O + CO_2. How much baking soda must be added to 8.0 g of cream of tartar for both materials to react completely?

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Solution:

From the equation, we know that one mole of NaHCO_3 reacts with one mole of KHC_4 H_4 O_6. Hence, if we con-vert 8.0 g of KHC_4 H_4 O_6 to moles, we know how manymoles of NaHCO_3 must be added. Finally, all we need to do is to convert moles of NaHCO_3 to grams of NaHCO_3. In order to convert from grams to moles, we use the relationship moles = mass/molecular weight. The number of moles of KHC_4 H_4 O_6 in 8.0 g is moles= [(mass) / (molecular weight)] = [(8.0 g) / (188 g/mole)] = 4.3 × 10^-2 mole. Hence, we must add 4.3 × 10^-2 mole of NaHCO_3. Using the relationship mass = moles × molecular weight, we find that 4.3 × 10^-2 mole of NaHCO_3 corresponds to 4.3 × 10^-2 moles × 84 g/mole = 3.6 g. Hence, 3.6 g of baking soda must be added.

Question:

lf a muscle had to rely on its stored supply of ATP for contraction, it would be completely fatigued within a few twitches. Therefore, if a muscle is to maintainitscontractile activity, molecules of ATP must be made avail-able as rapidly asthey are broken down. What are the sources of ATP in a skeletalmuscle? What is oxygen debt?

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Solution:

There are three sources of ATP in muscle cells: 1)creatine phosphate, 2) substrate levelphos-phorylationduringglycolysisand the TCA cycle, and 3) oxidativephosphorylationin the mitochondria. In an exercised muscle, the increased levels of ADP and inorganic phosphateresulting from the breakdown of ATP ultimately act as positive modulatorsfor oxidativephosphorylation. However, a short time elapses beforethesemultienzymaticpathways begin to deliver newly formed ATP ata high rate. It is the role ofcreatinephosphate to provide the energy for ATP formation during this inter-val.Creatinephosphate can transfer its phosphategroup to a molecule of ADP, converting it to ATP and leaving a creatinemolecule behind. A single enzyme catalyzes this reversible reaction. During moderate exercise, the muscle cell is able to derive ATP fromsources other thancreatinephosphate. The source of ATP during moderateexercise comes from the complete oxidation of carbohydrate (e. g. glycogen) to carbon dioxide and water viaglycolysis, the Krebs (TCA)cycle,and the electron transport chain. During the unavoidable delaybefore adjustments of the respiratory and circulatory systems increasethe oxygen supply to the active muscles, some of the oxygen for aerobicrespi-ration may come fromoxymyoglobin.Myoglobinis a compoundin muscle which is chemically similar to hemoglobin. It forms a loosecombination with oxygen while the oxygen supply is plentiful and storesthe oxygen until the demand for it increases. During heavy muscularactivity, a number of factors begin to limit the cell's ability to replaceATP by oxidativephosphorylation: the delivery of oxygen to the muscle, the availability of substrates such as glucose, and the rates at whichthe enzymes in the metabolic pathways can process the substrates. Any of these may become rate-limiting under various conditions. Since oxidativephosphorylationdepends upon oxygen as the final electron acceptor, the continued formation of ATP depends upon an adequate deliveryof oxygen to the muscle. Should the oxygen deliverybe insufficient, lactic acid (homolactic) fermentation occurs. The muscles obtainthe extra energy they need from this anaerobic process and lactic acid, the end product, accumulates. The muscle thus incurs what physiologistscall oxygen debt. When the violent activity is over, the musclecells consume large quantities of oxygen as they convert the lactic acidintopyruvicacid.Pyruvicacid is oxidized via the TCA and the electrontransport processes. The cells utilize the energy thus obtained to resynthesizeglycogen from the lactic acid that remains. In this manner, theoxygen debt is paid off, and the lactic acid is removed. This is why one continuesto breathe hard for some time after one has stopped high levels ofactivity.

Question:

Determine the free energy change, ∆G, for transforming liquid water at 100\textdegreeC and 1 atm. to vapor at the same conditions. ∆H = 9720 cal (molar enthalpy of vaporization) .

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Solution:

This problem deals with the energies of reactions: enthalpy, entropy and free energy. The entropy of a system is a measure of the state of randomness. As more thermal energy is added to a system, the more random it becomes. Entropy can be measured quantitatively. Entropy or ∆S =q_rev/T, whereq_revis the amount of heat added to the system that can be recovered if the reaction were reversed. For example, the energy needed to melt ice can be recovered by freezing the liquid back to ice. Enthalpy may be defined as the heat content of a system. The concept of energy and entropy may be combined to define the driving force of a reaction. This driving force is referred to as Gibbs free energy (G) where ∆G = ∆H - T∆S, where ∆H is the enthalpy. This energy must be equal to the heat content of the system (∆H) minus the energy necessary to increase the randomness, or entropy, of the same system. Therefore, to solve this problem, you need to sub-stitute into this equation the following values: T = 100\textdegreeC = 373\textdegreeK q = ∆H = 9720 cal Therefore, ∆S = (q/T) = [(9720 cal)/(373\textdegreeK )] Now, ∆G = 9720 cal - 373 \textdegreeK [(9720 cal)/(373\textdegreeK)] = 0. The answer is zero. The system can derive no usable energy to perform work from this change. For any system when ∆G = 0, the system is at equilibrium.

Question:

Briefly discuss some abnormalities of the respiratory tract.

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Solution:

The respiratory tract is subject to various infections and damages which may interfere with normal functioning. Tuberculosis is caused by the bacteriaMycobacterium tuberculosisand results in the destruction of lung tissue. It was once a major cause of death. In-fectious bacteria or viruses can bring about infections in the nasal cavities, the sinuses, the throat, the larynx (laryngitis), the bronchial tubes (bronchitis), and the lungs (pneumonia). Pneumonia, which may result from infection by bacteria or viruses, results in the filling of the alveoli with fluid. As the infection spreads through-out the lungs, the surface area for gas exchange is reduced. Allergic reactions may also occur in the respiratory tract. They occur when a person becomes sensitized to certain substances in the air, such as pollution, smoke or pollen. A person who is allergic to these substances has a condition known as hay fever. In such people, allergic reactions are initiated in the nasal cavities. Should these substances pass into the bronchioles and not be filtered in the nasal cavities, allergic asthma may result. The bronchiole tubes swell and become clogged with mucus, and the person has respiratory difficulty. It has long been known that cigarette smoking can cause cancer in the lungs. It is not as widely known, however, that it can also result in emphysema. Emphysema results from a thickening of the lung tissues, which causes a loss of elasticity and a loss in the ability to absorb oxygen. In extreme cases the vital capacity is so reduced that breathing becomes a major effort. Since exercise requires increased oxygen consumption, physical activity is greatly reduced in people with emphysema. Alveolar cells secrete a detergent-like chemical called surfactant, which intersperses with the water molecules in the alveoli. The air within each alveolus is separated from the alveolar membrane by an extremely thin layer of fluid. Surfactant markedly reduces the water molecules cohesive force, thereby decreasing the surface tension. If the surfactant were absent, this tension would be strong enough to cause the alveoli to collapse. In some newborn infants, predominantly those born prematurely, the lungs do not secrete an adequate amount of surfactant for their inspiratory forces to over-come the surface tension. Hence their alveoli are unable to expand when they inhale. Without treatment, these infants die of suffocation. The disease associated with insufficient surfactant production is referred to as respiratory distress syndrome. The recent discovery that the administration of hormones from the adrenal cortex enhances maturation of the surfactant-synthesizing cells may provide an important means of combating this disease.

Question:

Suppose we have two charges, q_1 = + 4stateCand q_2 = \rule{1em}{1pt}6 stateC, with an initial separation of r_1 = 3 cm. What is the change in potential energy if we increase the separation to 8 cm?

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Solution:

The potential energy of two point charges, q_1 and q_2 , separated by a distance r_1, is given V_12 = q_1q_2 / r_1 . Therefore as the separation is increased from r_1 to r_2 the change in potential energy is ∆V_12 = q_1q_2[(1 / r_2) \rule{1em}{1pt} (1 / r_1)] = (4statC)(\rule{1em}{1pt}6statC) × [(1 / 8cm) \rule{1em}{1pt} (1 / 3cm)] = (\rule{1em}{1pt}24 statC^2)(\rule{1em}{1pt}5cm / 24cm^2) = +5[(statC)^2 / cm] = +5 ergs In this case, there is a net increase in the electrostatic potential energy (that is, ∆V_12 > 0) because work was done by an outside agent against the attractive electrostatic force.

Question:

Write a FORTRAN program segment to find a root of the continuous func-tion f(x) via the bisection method.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G21-0514.htm

Solution:

The bisection method starts with two values of X, called HI and LO, for which the values of f(x) have opposite signs. The object is to find X when f(x) = 0. We do this by halving the in-terval between HI and LO, alternately from the negative side (LO) and from the positive side (HI). As we continue the process, we can approach X until we reach an approximation with any desired degree of accuracy. [Note that rounding errors can alter the accuracy of the approximation.] The program logic is given below, using the function X3 + 3X - 5 (Figure) as a sample f(x). We have chosen HI = 2.0, so f(HI) = (2)3 + 3(2) - 5 = 9 , and LO = 0.0, so f(LO) = (0)3 + 3(0) - 5 = -5. REAL F, X, HI, LO, ERROR CCHOOSE SOME INITIAL HI AND LO VALUES, CSUCH AS HI = 2.0 AND LO = 0.0. CALSO CHOOSE ERROR TO BE, SAY, = .002. DO 20 J = I, 1000 IF ((HI - LO).LE. ERROR) GO TO 30 X = (LO + HI)/2.0 F = (X \textasteriskcentered\textasteriskcentered 3) + (3.0 \textasteriskcentered X) - 5.0 IF (F.GT.O.O) GO TO 15 LO = X GO TO 20 15HI = X 20CONTINUE 30WRITE (5,\textasteriskcentered) X [Note: This WRITE statement includes an asterisk, a feature provided on some machines. It simply indicates that only variables are to be outputted, without any literals. Its purpose is to save time by eliminating a FORMAT statement.]

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Question:

A satellite of mass M_s is placed in a stable circular orbit of A satellite of mass M_s is placed in a stable circular orbit of radius R around the earth. What is its angular momentum about an axis through the earth perpendicular to the plane of its orbit? Assume that R > > radius of satellite.

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Solution:

In the figure, the earth's gravitational pull on the satellite accounts for the satellite's centri-petal acceleration: (M_sv^2)/R = (GM_sM_e)/(R^2) Where M_e is the mass of the earth. Thus:v = \surd[(GM_e)/(R)] Its angular velocity is therefore: \omega = v/R = \surd[(GM_e)/(R^3)] Since R > > radius of satellite, when calculating the satellite's moment of inertia, with respect to the axis A-A', we can take all of its mass to be at a distance R from the axis. This reduces to the case of finding the moment of inertia of a point mass with respect to an axis. This is: I = M_sR^2 The angular momentum L of the satellite is therefore: L = I\omega = M_sR^2\surd[(GM_e)/(R^3)] = M_s \surd(GM_eR)

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Question:

A chemist has tert-butane. If he adds Br_2 and light, what are the monobromo substitution products expected? Which one will predominate and why? How would you prepare the predominating compound by an addition reaction?

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Solution:

The first question concerns substitution in alkanes. You are performing a halogenation substitution reaction and are asked to predict the monohalogenation products expected for tert-butane. This necessitates the writing of its structure and analyzing its hydrogens, since they will be replaced by the bromine atoms. Tert butane can be written All the hydrogens on the methyl groups are equivalent. This means that if you substitute a bromine atom for one hydrogen atom from a methyl group, it is the same as sub-stituting for any other hydrogen on one of the other methyl groups. The other hydrogen (the one attached to the 3\textdegree carbon) is not equivalent to the hydrogens bound to the 1\textdegree carbons. Thus you expect two monobromo products. They are You could have placed the Br on the other two methyls and the same molecule would be obtained, for Example, When the Br is attached to the 3\textdegree carbon tert-butyl bromide is obtained. This product predominates greatly. To see why, consider the mechanism of this reaction. The mechanism is as follows (let x = a halogen). x_22x \textbullet(radicals denoted by dots) x \textbullet + alkaneHx + alkane \textbullet alkane \textbullet + x_2alkyl halogen + x \textbullet You see that the reacton proceeds by radical formation. To form isobutyl bromide, a 1\textdegree radical. must be formed first. T-butyl bromide goes through a 3\textdegree radical, i.e., A 3\textdegree radical is more stable and, thus, has a longer lifetime than the 1\textdegree radical. Since the life time of the 3\textdegree radicals is longer, there will be more 3\textdegree radicals to react with the bromine radicals, and the amount of t-butyl bromide formed exceeds the amount of iso-butyl bromide formed. You can prepare t-butyl bromide, by an addition re-action. HBr can be added across the double bond of iso-butylene as shown in the following reaction. It has been shown that when an inorganic acid, such as HBr, adds across a double bond in an olefin the hydrogen ion will add to the carbon with the most hydrogen atoms already bound to it. This is called Markovnikov's rule.

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Question:

A spaceship from earth enters a circular orbit 22,000 km above the surface of Mars at the equator, which allows it to rotate at the same speed as Mars and thus to stay always above the same point on the planet's surface. The Martians set up a guided missile directly beneath it and fire it vertically upward in an attempt to destroy the intruder. With what minimum velocity must the missile leave the surface in order to suc-ceed in its mission? Mars is a sphere of radius 3400 km and of mean density 4120 kg\bulletm^-3.

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Solution:

The missile will just reach the spaceship if its velocity at the latter's position is zero. We must therefore calculate the missile's velocity as a function of position. Using Newton's Second Law, we ob-tain F = ma(1) where m is the missile's mass,ais its acceleration and F is the net force acting on it. Note that F = -GmM_m/r^2(2) where M_m is the mass of Mars and r is the radial distance of the missile from the center of Mars. Assuming that the missile travels in a radial direction, we may write a =dv/dt= (dv/dr)(dr/dt) = v(dv/dr) Substituting (2) and (3) in (1) -GmM_m/r^2 =mv(dv/dr) orv(dv/dr) = -GmM_m/r^2 vdv= -GM_m(dr/r^2) ^v\int_v0 (vdv) = -GM_m^R\int_R0 (dr/r^2) where v = v at r = R, and v = v_0 at r = R_0. (R_0 is the position of the surface of Mars relative to its center). (v^2 - v^2 _0)/2 =GM_m(1/R - 1/R_0) v2_0 = v^2 + 2GM_m(1/R_0 - 1/R) The minimum value of v_0 is found by requiring that themissle'svelocity at the spaceship's position be zero (v = 0). Hence v2_0 _min = 2GM_m(1/R_0 - 1/R)(3) The mass of Mars is its volume times its density (\rho_m). M_m = (4/3)\piR3_m×\rho_m= (4/3)\pi× (3.4 × 10^6m)^3 × 4120\bulletm^-3 = 0.678 × 10^24kg. orv^2 = 2 × 6.67 × 10^-11N\bulletm^2 \bulletkg^-2 × 0.678 × 10^24kg [{1/(3.4 ×10^6m)} - {1/(25.4 × 10^6m)}] = 2.30 × 10^7 m^2\bullets^-2. ThereforeV = 4.8 km\bullets^-1. Note that R in (3) is the position of the spaceship from the center of Mars, orR = R_0 + 2.2 × 10^6m R = 25.4 × 10^6m

Question:

Enzyme behavior may be described by the lock-and-key theory. Explain and describe this theory.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E23-0829.htm

Solution:

Millions of chemical reactions occurring in the body require catalysts to make them go to completion. Enzymes serve as these catalysts. Their behavior and/or function is described by the lock-and-key theory. According to this theory, enzymes have definite three-dimensional structures. They are arranged so that the substrate (the substance upon which the enzyme acts) fits into the enzyme's structure. Only a specific kind of substrate can fit into a given enzyme. When the substrate and enzyme unite, i.e., form an aggregate, the substrate is exposed for a reaction to occur. In Fig. A, the three- dimensional shape of the enzyme is arranged to accommodate the substrate. The active site is that portion of the enzyme that catalyzes the reaction. In Fig. B, the sub-strate is held in position by the enzyme while the re-action occurs. In Fig. C, the product(s) of the reaction leaves the enzyme, thus, freeing the enzyme to catalyze another reaction.

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Question:

How can we deduce information about prehistoric man from the fossil teeth, jaws, and craniums found?

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Solution:

Of all the fossils discovered relevant to the evolution of man, the craniums, jaws, and teeth have been the most numerous and useful. Teeth have been found in large numbers at archaeological sites because their enamel coating gives them great durability. Teeth are complex structures with many distinctive features. The size and shape of the cusps, and the root size of the fossil teeth can be compared with modern man's dental anatomy and to that of modern apes. In this manner, a fossil man can be placed in a chronological and evolutionary relationship to otherhominiod-like forms, whether they be extinct or extant. Much can also be said about our ancestor's diet by studying the structure of a fossil's teeth. For example, there is an evolutionary trend towards reduction in size of the canine teeth in man's evolution (see Figure). Man and apes are charac-terized by larger canines. Since the canines are used in tearing raw meat, the reduction in their size would seem to parallel the evolution of man's mode of feeding and his diet. A major problem encountered by anthropologists is the absence of teeth from many of the jaws that are found. It becomes difficult to determine the exact number and size of the teeth when the jaw is not complete. As a general rule, the size of the jaw is directly related to the size of the teeth. For instance, man has charac-teristically both smaller teeth and smaller jaws than his predecessors. The shape of the jaw alone is also an important evolutionary indicator. The dental arch, or portion of the jaw where the teeth are attached, is rectangular in the apes and parabolic in man. This indi-cates that in the apes, the back teeth on either side of the jaw are parallel to each other, with the incisors in a straight row. In man and some of his fossil ancestors, parabolic dental arches are present wherein the distance between the canines on either side of the jaw is smaller than that between the molars. The structure and size of the teeth, the ridges on the jaw, and the dimensions and shape of the dental arch indicate how the face may have looked, and the nature of the diet. If the fossil man ate hard, rough objects that required teeth with effective chewing and grinding surfaces, we would most likely observe huge molars that have undergone much wear, and jaws with large ridges for attachment of strong chewing muscles. Enlarged chewing muscles are also accompanied by large ridges on the sides and top of the skull, since thetemporalismuscles used for chewing originate in the skull region in hominoids, although not always in the same place. Huge neck muscles would also be accompanied by large ridges on the skull. This would indicatequadrupedalism. If the animal was quadruped, huge neck muscles would be needed to support the neck, and obvious muscle ridges would be present in the skull. If the animal was bipedal, the neck muscles and ridges would be reduced since the head would now be balanced on the spinal column. The centered position of the head indicatingbipedalismcan be further determined by the position of the foramen magnum.Bidepalorganisms like man have the foramen magnum located underneath the center of the brain, while in apes, the foramen magnum is located toward the rear of the skull. The overall dimensions of the braincase are im-portant in determining an animal's cranial capacity. If the skulls of the various fossil men are put into a chronological sequence, a trend toward increasing brain size is noticed as well as a general thinning of the skull bones. A great deal of the information indicated above is still to a large degree speculative, since skulls, jaws, or teeth are rarely found in their original form without great amounts of weathering and fragmentation.

Question:

A 3000-lb automobile at rest at the top of an incline 30 ft high and 300 ft long is released and rolls down the hill. What is its speed at the bottom of the incline if the average retarding force is 200 lb?

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Solution:

The potential energy at the top of the hill is available to do work against the retarding force F^\ding{217} and to supply kinetic energy. The work done by the retarding force F^\ding{217} is W = \int F^\ding{217} \bullet ds^\ding{217} where the integral is evaluated over the path of motion of the auto. If F^\ding{217} is constant, W = F^\ding{217} \bullet \int ds^\ding{217} = F^\ding{217} \bullet s^\ding{217} s^\ding{217} being the vector displacement of the auto. Hence, by the principle of energy conservation, mgh = F^\ding{217} \textbullet s^\ding{217} + (1/2) mv^2 where v is the velocity at the bottom of the incline. Since F^\ding{217} and s^\ding{217} are parallel and in the same direction, we may write Wh = Fs + (1/2) mv^2 (1) Since the height of the incline is much less than the length of the base, we may use 300 feet as an approxima-tion to the length of the hypotenuse (see figure). Then, substituting into (1) 3000 lb × 30 ft = 200 lb × 300 ft + (1/2) × 94 slugs × v^2 9.0 × 10^4 ft-lb - 6.0 × 10^4 ft-lb = (1/2) × 94 slugs × v^2 v^2 = (3.0 × 10^4 ft-lb)/(47 slugs) = 640 ft^2/sec^2 v = 25 ft/sec.

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Question:

Two long, straight wires, each carrying a current of 9 A in the same direction are placed parallel to each other. Find the force that each wire exerts on the other when the separation distance is 1 × 10^-1 m.

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Solution:

The currents carried bv the wires i_a = 9 A, i_b = 9 A, and the distance between them, d = 1 × 10^-1 m, are given. The first current carryingwire (a) produces a field of induction B at all nearby points around it. The magnitude of B is B = (\mu_0 /2\pi) (i_a /d)(1) where i_a is the current through wire a, d is the distance a separating the two wires and the permeability constant \mu_0 = 4\pi × 10^-7 W/(A \bullet m). The other wire will find itself immersed in the field due to the first wire. The magnetic field it creates, that is its own self field, has no influence on its behavior. The magnetic force on this second current carrying wire (of length l) is F^\ding{217} = i_b l^\ding{217} × B^\ding{217} where i_b is the current through the second wire. Since the wire length is perpendicular to the magnetic field vector B, we then have F = i_b lB(2) F is directed inward toward the first wire. It is perpendicular to both B and to the length vector. There-fore, combining equations (1) and (2) we have F = (\mu_0 /2\pi) [(i_a i_b)/d] l The force per unit length on a wire is then (F/l) = (\mu_0 /2\pi) [(i_a i_b )/d] = 2 × 10^-7 (N/A^2) [(i_a i_b)/d] = 2 × 10^-7 (N/A^2) [{(9A) (9 A)} /(1 × 10^-1 m)] = 1.62 × 10^-4(N/m)

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Question:

What basic nutritional requirements do all livingorganisms have in common? Comparephototrophsandchemotrophs, autotrophs andheterotrophs.

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Solution:

Organisms, ranging from bacteria to man, share a set of nutritional requirements necessary for normal growth. These requirements must be known in order to cultivate microorganisms in the laboratory, and purecul- tures may be obtained through the preparation of appropriate selective growth media. All organisms require a source of energy. Green plants and some bacteria can utilize radiant energy (from the sun) and are thus called phototrophs . Animals and non-photosyn-thetic bacteria must rely on the oxidation of chemical compounds for energy and are thus called chemotrophs . All organisms require a carbon source. Plants and many bacteria require only carbon dioxide as their carbon source. They are termed autotrophs . Animals and other bacteria require a more reduced form of carbon , such as an organic carbon compound. Sugars and other carbohydrates are examples of organic carbon compounds. Organisms which have this requirement are termed theterotrophs. They depend upon autotrophs for their organic form of carbon, which they use as both a carbon source and an energy source. All organisms require a nitrogen source. Plants utilize nitrogen in the form of inorganic salts such as potassium nitrate (KNO_3), while animals must rely on organic nitrogen- containing compounds such as amino acids. Most bacteria utilize nitrogen in either of the above forms, although some bacteria can use atmospheric nitrogen. All organisms require sulfur and phosphorous. While phosphorous is usually supplied by phosphates, sulfur may be supplied by organic compounds , by inorganic compounds, or by elementary sulfur. All organisms need certain metallic elements, and many require vitamins . The metallic elements include sodium, potassium, calcium, magnesium , manganese, iron, zinc, and copper. While vitamins must be furnished to animals and to some bacteria, there are certain bacteria capable of synthesizing the vitamins from other nutrient compounds. Finally, all organisms require water for growth. For bacteria and plants , all the above nutrients must be in solution in order to enter the organism . Bacteria show considerable variation in the specific nutrients required for growth. For example, all heterotrophic bacteria require an organic form of carbon, but they differ in the kinds of organic compounds they can utilize. Different bacteria utilize nitrogen in its various forms. Some require several kinds of amino acids and vitamins, while others require only inorganic elements.

Question:

A heavy particle of mass M collides elastically with a light particle of mass m (see the figure below) . The light particle is initially at rest. The initial velocity of the heavy particle is v_h^\ding{217} = u_h\^{\i}, the final velocity is w_h^\ding{217}. If the particular collision is such that the light particle goes off in the forward (+\^{\i}) direction, what is its velocity w_1^\ding{217}? What fraction of the energy of the heavy particle is lost in this collision?

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Solution:

This problem can be solved using the principle of conservation of linear momentum. The linear momentum before collision must equal the linear momentum after collision. The initial momentum p_i of the system is: p_i = Mv_h + mv_1 = Mv_h since the smaller mass m is initially at rest. The final momentum p_f of the system is: p_f = Mw_h + mw_1 By the conversation of linear momentum, p_i = p_f : Mv_h = Mw_h + mw1 Thus:mw_1 = Mw_h - Mwh w_1 = (M/m) (v_h - w_h) The energy of the heavy particle before the collision is(1/2) Mvh^2. After the collision the kinetic energy is (1/2) Mwh^2. The fraction of its original kinetic energy that the heavy mass M retains after the collision is: (E_f)/(E_i) = [{(1/2) Mv_h^2} / {(1/2) Mw_h^2}] = (v_h/w_h)^2 where E_i and E_f are the initial and final kinetic energies respectively. The fraction of the energy of the heavy particle that is lost in the collision is Fractional energy loss = (E_i - E_f) / (E_i) = 1 - (Ef/ E_i) = 1 - (vh/ w_h)^2 .

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Question:

Estimate the minimum voltage needed to excite K radiation from an x-ray tube with a tungsten target, and calculate the wavelength of the K_\alpha line.

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Solution:

Moseley found experimentally that the x-ray line spectrum varies in a regular way from element to element, unlike the irregular variations of optical spectra. This regular variation occurs because characteristic x-rays are due to transitions involving the innermost electrons of the atom. Because of the shielding of the other elec-trons the inner electron energies do not depend on the com-plex interactions of the outer electrons, which are respon-sible for the complicated optical spectra. Also, the inner electrons are shielded from the interatomic forces respon-sible for the binding of the atoms in solids, According to the Bohr model of the atom, the energy of an electron in the first orbit is proportional to the square of the nuc-lear charge. Mosely therefore, reasoned that the energy, and thus the frequency, of a characteristic x-ray photon should vary as the square of the atomic number of the element. He plotted the square root of the frequency of particular x-ray lines versus the atomic number Z of theelement. He found that his curves could be described by the general equation, equation, f = A^2 (Z - b)^2 f = A^2 (Z - b)^2 Where f is the frequency of the x-ray and A and b are constants for each line. One family of lines, called the K-series, has b = 1. If the bombarding electron in an x-ray tube knocks an electron from the n = 1 inner orbit of a target atom out of the atom, photons are emitted corres-ponding to transitions of other electrons to the vacancy in the n = 1 orbit. The lowest frequency line corresponds to the lowest energy transition, from n = 2 to n = 1. This line is called the K_\alpha line. Bohr's relation for the frequency of a photon emitted from a one electron atom is, f = Z^2 [(2\pi^2m e^4) / {(4\piE_0)^2 h^3}] [{1/(n_2^2)} - {1/(n_1^2)}](2) The energy E of a photon is given by hf. Using Z - 1 in place of Z in eq (2) so as to fit the form of Moseley's equation (1) for the K-series, we find, E = hf = [(2\pi^2m e^4) / {(4\piE_0)^2 h^2}] [{1/(n_2^2)} - {1/(n_1^2)}](z - 1)^2(3) For K radiation to occur, an electron must be knocked from the n_1 = 1 orbit to infinity (n_2 = \infty). This is the minimum energy that must be transmitted to the atom for K radiation and is, E_ionization = [(2\pi^2me^4) / {(4\piE_0)^2 h^2}] (Z - 1)^2 [(1/\infty) - 1] = 2.18 × 10^-18 (Z - 1)^2joules = 13.6 ev (Z - 1)^2 For tungsten, Z = 74, and the energy input required is (13.6 ev) (73)^2 = 72,800 ev. So the anode must be at least 72,800 volts above the cathode potential. The first line in the K series of tungsten has a wavelength given by E = hf. But f\lambda_K\alpha = c and E = hf = hc/(\lambda_K\alpha) = 13.6 ev (Z - 1)^2 [(1/1^2) - (1/2^2)] \lambda_K\alpha = [{6.625 × 10^-34 joule-sec (3.0 × 10^8 m/sec)} / {(13.6 × 1.60 × 10^-19 joule) (73)^2 (1 - 1/4)}] = 0.21 × 10^-10m = 0.21 A

Question:

An iron steam pipe is 200 ft long at 0\textdegreeC. What will its increase in length when heated to 100\textdegreeC? (\alpha = 10 x 10^\rule{1em}{1pt}6 percelsiusdegree).

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Solution:

The change in length, \DeltaL, of a substance due to a temperature change is proportional to the change, \DeltaT, and to the original length,L_0,of the object: \DeltaL = \alpha L_0 \DeltaT where \alpha is the proportionality constant and is called the coefficient of linearexpension. L_0 = 200 ft,\alpha = 10 × 10^\rule{1em}{1pt}6 per C\textdegree, T = 100\textdegreeC,T_0 = 0\textdegreeC. Increase in length= \DeltaL = \alpha L_0\DeltaT = (10 × 10^\rule{1em}{1pt}6) (200) (100) = 0.20 ft.

Question:

A particle of mass m is released from rest at point a in Fig. 1 falling parallel to the (vertical) y-axis. (a) Find the torque acting on m at any time t, with respect to origin O. (b) Find the angular momentum of m at any time t, with respect to this same origin, (c) Show that the relation \tau = d1/dt yields a correctresult when applied to this problem.

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Solution:

(a) The torque is given by \tau = r × F, its magnitude being given by \tau = rF sin\texttheta. r = sin\texttheta = b and F = mg so that \tau = mgb = a constant Note that the torque is simply the product of the force (mg) times the moment arm (b). The right-hand rule shows that \tau is directed perpendicularly into the figure. (b) The angular momentum is given by 1 = r × p, its magnitude being given by 1 =rp sin\texttheta. r sin\texttheta = b and p = mv = m(gt) so that 1 = mgbt. The right- hand rule shows that 1 is directed perpendicularly into the figure, which means that 1 and \tau are parallel vectors. The vector 1 changes with time in magnitude only, its direction always remaining the same in this case. (c) Since d1, the change in 1, and \tau are parallel, we can replace the vector relation \tau = d1 / dt by the scalar relation \tau = d1 / dt . Using the expression for \tau and 1 from (a) and (b) above mgb = (d/dt) (mgbt) = mgb, Which is an identity. Thus the relation \tau = d1 / dt yields correct results in this case. Canceling the constant b out of the first two terms above substituting for gt the equiva-lent quantity v, mg = (d/dt) (mv). Since mg = F and mv = p, this is the familiar result F = dp / dt. Thus, relations such as \tau = d1/ dt, though often vastly useful, are not new basic postulates of classical mechanics but are rather the reformulation of the Newtonian laws for rotational motion. Note that the values of \tau and 1 depend on choice of origin, that is, on b. In particular, if b = 0, then \tau = 0 and 1 = 0.

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Question:

Summarize the roles of the various enzymes involved in digestion.

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Solution:

Enzymes in Digestion Enzyme Source Optimum pH Product(s) Salivary amylase Salivary glands Neutral Maltose Pepsin Stomach Acid Peptides Rennin Stomach Acid Coagulated casein (A milk protein) Trypsin Pancreas Alkaline Peptides Chymotrypsin Pancreas Alkaline Peptides Lipase Pancreas Alkaline Glycerol, Fatty acids, Mono- andDiglycerides Amylase Pancreas Alkaline Maltose Ribonuclease Pancreas Alkaline pentoses , nucleotides and nitrogen bases Deoxyribonuclease Pancreas Alkaline pentoses , nucleotides and nitrogen bases Carboxypeptidase pancreas Alkaline Free amino acids Aminopeptidase Intestinal Glands Alkaline Free amino acids Maltase Intestinal Glands Alkaline Glucose Sucrase Intestinal Glands Alkaline Glucose and fructose Lactase Intestinal Glands Alkaline Glucose andgalactose

Question:

Explain briefly the background and applications of Pascal.

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Solution:

Pascal is a modern high-level programming langu-age developed in the early 1970's in Switzerland and named in honor of 17th century French mathematician and inventorBlaisePascal. Originally intended as a language for teach-ing structured programming, it is now used for a variety of scientific and commercial applications. An important boost for Pascal has come from its widespread implementation on microcomputers. Being one of the newest general purpose programming languages, Pascal features modern control structures for se-quence, selection, and iteration , which makes it particularly, well-suited for structured programming , makes programs read-able, and allows for implementation of various data-structures.

Question:

[(A + B) + (A + C) + (A + D)] \textbullet [A \textbullet \textasciitilde B](1) The + operator denotes the inclusive OR function and the \bullet operator denotes the AND function. NOT is represented by \textasciitilde \bullet (b) Change statement (1) into its equivalent switching circuit. (c) Show the gate representation of statement (1).

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G03-0050.htm

Solution:

(a) Set up the truth table by specifying all possible combinations of A,B,C, and D and evaluating the terms of the state-ment for each combination. Since there are 4 variables, the truth table has 2^4 = 16 rows of entries as shown in figure 1. First, evaluate the innermost terms: A B \backsimB C D A + B A\bullet\backsimB A + C A + D 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 0 0 0 0 1 1 1 1 0 0 0 0 1 1 1 1 1 1 1 1 0 0 0 0 1 1 1 1 0 0 0 0 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 1 1 1 1 0 0 0 0 0 0 1 1 0 0 1 1 1 1 1 1 1 1 1 1 0 1 0 1 0 1 0 1 1 1 1 1 1 1 1 1 Fig. 1 Now, transfer the results in columns 6, 8, and 9 to a second truth table, shown in figure 2 which evaluates the first bracketed term. A + B A + C A + D (A + B) + (A + C) + (A +D) 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 0 0 1 1 0 0 1 1 1 1 1 1 1 1 1 1 0 1 0 1 0 1 0 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 Fig. 2 Finally, transfer the result in the last column of this truth table and the result in column 7 of the first truth table to a third truth table, shown in figure 3, which evaluates the entire statement (1). A \bullet \backsim B (A + B) + (A + C) + (A +D) Statement (1) 0 0 0 0 0 0 0 0 1 1 1 1 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 1 1 1 1 0 0 0 0 Fig. 3 The entry in each row of column 3 is the result of performing the operation on the entries in columns 1 and 2 of that row. (b) The switching circuit equivalent to statement (1) is obtained by connecting, in parallel, variables which are operands of the + operation and connecting, in series, variables which are operands of the \bullet operation. First, find equivalent switching circuits for the innermost terms. These are shown in figure 4. The switching circuit equivalent to the first bracketed term is a parallel connection of the three parallel circuits A + B, A + C, A + D, as shown in figure 5. The switching circuit equivalent to statement (1) is a series con-nection of the parallel circuit [(A + B) + (A + C) + (A + D)] and the series circuit [A \textbullet \backsim B]. The complete switching circuit is shown in figure 6. [(A + B) + (A + C) + (A + D)] \textbullet [A \textbullet \textasciitilde B] Note that the circuit conducts if and only if A is closed and B is open, regardless of the settings of C and D. This agrees with the truth table, since statement (1) has value if and only if A = 1 and B = 0. (c) Variables which are operands of the + operation are inputs to an OR gate. Variables which are operands of the \bullet operation are inputs to an AND gate. A variable that is the operand of the \textasciitilde operation is the input to a NOT gate. The gate representation of the terms in parentheses are shown in figure 7. The composition of 2 operations is done by plugging the output of 1 gate into the input of another, so that the gate representation of [A \textbullet \backsim B] is shown in figure 8. Finally, by the same reasoning, the complete gate representation of [(A + B) + (A + C) + (A + D)] \textbullet [A \textbullet \textasciitilde B] is shown in figure 9.

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Question:

The barometric pressure on the lunar surface is about 10^-10 torr. At a temperature of 100\textdegreeK, what volume of lunar atmosphere contains (a) 10^6 molecules of gas, and (b) 1 millimole of gas?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E02-0052.htm

Solution:

This problem is an application of the ideal gas equation, PV = nRT, where P = pressure, V = volume, n = number of moles, R = gas constant, and T = absolute temperature. Solving for V, V = nRT/P, in the first part of this problem, we use the definition n = [(number of molecules)/(Avogadro's number)] = (N/A). We must then substitute into the formula V = nRT/P = (N/A) (RT/P) in order to obtain the volume corresponding to N molecules. In the second part of this problem we can use V = nRT/P directly, remembering that 1 millimole = 10^-3 mole. Hence, for the first part of the problem, V = (N/A) (RT/P) = [(10^6 molecules)/(6.02× 10^23 molecules/mole)] × [(0.0821 l -atm/\textdegreeK - mole × 100\textdegreek)/(1.316 × 10^-13 atm)] = 1.04 × 10^-4 l, where we have used P = 10^-10 torr = 10^-10 torr × [(1 atm)/(760 torr)] = 1.316 × 10^-13 atm. For the second part of the problem, V = (nRT/P) = [(10^-3 mole × .0821 l -atm/\textdegreeK - mole × 100\textdegreeK) / (1.316 × 10^-13 atm)] = 6.24 × 10^10 l.

Question:

All the silicon and most of the oxygen in the earth's crust exists in the form of silicate minerals, such as granite. If the basic chemical unit of all silicates is the Sio_4^4- ion, which consists of an Si^4+ ion and four O^2- ions, write the electronic configurations of the Si^4+ and O^2- ions in silicate materials.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E27-0907.htm

Solution:

An electronic configuration of an element (or ion) gives a picture of the arrangement of electrons that surround a nucleus. The atomic number gives the total number of electrons of the atom (notthe ion). With this in mind, write the electron configuration of the oxygen and silicon atoms. The electronic configuration of oxygen with atomic number 8 is 1s^22s^22p^4 and for silicon with atomic number 14 it is 1s^22s^22p^63s^23p^2. To find the electronic configurations of the ions Si^4+ and O^2- note each ion's charge. The charge of 4+ on Si indicates that it lost 4 electrons (electronsare negatively charged). Thus, if the silicon atom had a configuration of1s^22s^22p^63s^23p^2, and it loses 4 electrons, the configuration becomes 1s^22s^22p^6 (ions will form by losing only high energy electrons). The oxygen atom configuration was 1s^22s^22p^4. The fact that the charge was 2- indicates that the atom gained two electrons. Consequently, the configuration becomes 1s^22s^22p^6.

Question:

The electron of a hydrogen atom, revolving in the first Bohr orbit can be thought of as being equivalent to a current loop, as shown in the figure. Find the magnetic moment \mu_B of this loop (which is called one Bohr magneton).

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/Users/wenhuchen/Documents/Crawler/Physics/D21-0710.htm

Solution:

The average charge, per unit time, passing a point of the orbit, or the average current I is e/t, where T is the period of rotation. The angular mo-mentum of the first Bohr radius is given by L = mvr = (h/2\pi) , orv = [h/(2\pimr)] . Then, the period and average current are T = (2\pir)/v = (4\pi^2 r^2 m)/h . I = (eh)/(4\pi^2 r^2 m) . The magnetic moment of the loop is \mu_B = IA, where A is its area, \pir^2 . Therefore, \mu_B = (h/4\pi) \bullet (e/m) = [(6.62 × 10^-34joule - sec × 1.76 × 10^11 coul/kq) / (12.57)] = 9.27 × 10^-24 amp \textbullet m^2 .

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Question:

A rocket blasts off from Earth at 12 noon. The original mass of the rocket is M_0 , and the exit velocity of the fuel jet isV_e_. Assume that fuel is burned at a constant rate, dm/dt. How high will the rocket be at 12:15 PM? (assuming a straight-line trajectory). Write a FORTRAN program to find the height at intervals of 0.5 minutes.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G12-0302.htm

Solution:

Force may be expressed as the change of momentum, as given by Newton's second law: F =dp/dt. Now, we establish that positive y is the vertical direction upwards. If M_0 is the total mass of rocket and fuel before the launch, and m is the mass of the exiting fuel jet, then the rate of change in momentum is given by: F = (d/dt)(MV) + (V -V_e) (dm/dt) We must also take into account the force of gravity, which is acting against the motion of the rocket. The formula will not be derived step-by-step here; a text on differential equations will have the thorough explanation. However, all variables will be explained, after presenting the general equation. The initial condition v = 0 at t = 0 yields the equation: V = -gt-V_eln[(m_0 -ut) / u] then, by integrating, the equation of motion is obtained, which expresses the height y as a function of time t: y = -(1/2)gt2+V_e[t + {( m_0 -ut) / u} 1n {(m_0 -ut) / m_0}] The variables are as follows: y = vertical height; -(1/2)gt^2 = the effect of gravity pulling the rocket downward; hence the use of the negative sign. V_e= the exit velocity of the fuel jet M_0 = original mass of the rocket plus theunburaedfuel u = dm/dt= the rate of change of the mass of the unburned fuel Note that as the rocket rises, the mass of the fuel, and ultimately, the total mass of the rocket is decreasing. Also notice that in the program, a decision should be made upon appropriate values for M_0, \mu ,V_e. It is assumed that these have already been read in. Also, since g is defined in meters per sec^2 , it may be best to use the MKS system of units. CSIMULATE IHE VERTICAL ASCENT OF AN CUNCONTROLLED ROCKET FOR 15 MINUTES CAT 30-SECOND INTERVALS Y = 0.0 G = 9.80616 WRITE (5,100) 100FORMAT (1X,'TIME', 5X, 'HEIGHT') DO 20 I = 1,30 T = 0.5 {_\ast} FLOAT(I) Y = -0. 5{_\ast}G{_\ast} (T{_\ast}{_\ast}2) + (VE{_\ast} (T + (RMD - (RMU{_\ast} 1T))RMU){_\ast}L0G((RMO - RMU){_\ast}T/RMO)) WRITE (5,101) T,Y 101FORMAT (2X, F5.2, 4X, F12.4) 20CONTINUE STOP END SAMPLE OUTPUT: TIMEHEIGHT 0.50 1.00 1.50 2.00

Question:

Calculate ∆m, the difference in mass between the final and initial nuclei in grams per mole for the emission of a \gamma ray, ^19_8O\textasteriskcentered (excited state)\rightarrow^19_8O (ground state) + \gamma (∆E = 1.06 × 10^8 Kcal/mole).

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/Users/wenhuchen/Documents/Crawler/Chemistry/E19-0713.htm

Solution:

One solves for ∆m by using Einstein's Law, which relates ∆m and ∆E. ∆E = ∆mc^2, where ∆E is the energy, ∆m is the change in mass, and c is the speed of light, 3 × 10^10 cm/sec. When using Einstein's Law, the ∆E, in Kcal/mole, must be first conver-ted to ergs/mole. There are 4.18 × 10^10ergsin one Kcal, therefore one converts Kcal to ergs by multiplying the number of Kcal by 4.18 × 10^10 ergs/Kcal. no. of ergs= 4.18 × 10^10 ergs/kcal × 1.06 × 10^8 kcal/mole = 4.43 × 10^18 ergs/mole. One can now solve for ∆m. ∆E = ∆mc^2∆E= 4.43 × 10^18 ergs/mole c= 3 × 10^10 cm/sec ergs= (g cm^2) / (sec^2) 4.43 × 10^18 {(g cm^2) / (sec^2 mole)} = ∆m[3 × 10^10 (cm/sec)]^2 [(4.43 × 10^18) / (3 × 10^10)^2] [g/mole] = ∆m 4.92 × 10^-3 g/mole = ∆m When a \gamma particle is emitted, one knows that the final state weighs less than the initial state, therefore, ∆m = - 4.92 × 10^-3 g/mole.

Question:

If we have a list of tabular values represented by a one-dimensionalar-ray, then the first differences of the list areformed by subtracting each element except the last [sincenothing follows the last element] from the element immediatelyfollowing it. Suppose we have a one--dimensionalarray named X that contains 50 elements. Compute the 49 elements of another array named DX from DX(I) = X(I + 1) - X(I) I = 1,2,...,49. Write a program segment to perform this calculation.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G21-0511.htm

Solution:

Dimension both arrays according to the given dimensions and then writea simple DO loop to define the difference array DX. Thus we get: DIMENSIONX(50), DX(49) DO 50 I = 1, 49 50DX(I) = X(I + 1) - X(I) STOP END

Question:

A laboratory analyst puts Fehling solution in a test tube and then adds the urine of a diabetic which contains some sugar. On heating, what red copper compound precipitates?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E32-0935.htm

Solution:

Copper forms two series of compounds: Copper (I) (cuprous), and copper II (cupric) compounds. Copper (I) oxide (Cu_2O) occurs naturally as the mineralcuprite. This red oxide may also be obtained by careful oxidation of copper in air or by precipitation of Cu+ ions from solutions with sodium or potassium hydroxide. Copper (I) hydroxide is not known. Copper (I) oxide is also obtained as a re-duction product of Fehling's solution, a reagent used in the diagnosis of diabetes to test for the presence of reducing sugars in the urine. Fehling's solution consists of an aqueous solution of copper (II) sulfate, sodium hydroxide, and sodium potassiumtartrate. Thetartrate forms a complex with Cu^2+. As a result, so few free Cu^2+ are left in solution that no precipitate of copper (I) hydroxide is obtained when the sodium hydroxide is added. Reducing sugars reduce Cu^2+ to Cu+, and Cu_2O separates as a reddish precipitate.

Question:

How has the role of internal body secretions in behavior been demonstrated in silk worms?

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/Users/wenhuchen/Documents/Crawler/Biology/F31-0804.htm

Solution:

Although much of the stimuli which trigger specific behavioral responses have an external origin, many responses occur as a result of internal stimulation. Much of this internal stimulation is initiated by hormones. The silkworm larva must spin its cocoon at a certain time in life, just following its larval stage. The stimulus for the spinning behavior comes from the disappearance of a specific substance, juvenile hormone. This is secreted by the corporaallata, a pair of glands located behind the brain of the silkworm. The secretion of juvenile hormone prevents the spinning of the cocoon. As the larva grows, the level of juvenile hormone drops, allowing for the spinning reaction to occur. No external sign stimuli is necessary, only internal hormonal levels. The cocoon-spinning behavior of the larvae of silk worms is an example of a behavior under the control of hormones.

Question:

100 ml of 0.1 M NaOH is mixed with 100 ml of 0.15 M HOAC. K_a = 1.75 × 10^-5 for acetic acid, what is the pH of this mixture?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E12-0421.htm

Solution:

To find the pH of the mixture, first determine the H_3O^+ concentration, since pH = - log [H_3O^+]. To find H_3O^+, consider what happens when acetic acid is added to NaOH, a base. When an acid is added to a base, a neuralization reaction occurs such that water and a salt are produced. Thus, This reaction does not show the presence of any H_3O^+ ions, which may lead one to incorrectly conclude that its con-centration was zero. There are 100 ml of 0.15 M HOAC. M = Molarity = moles/liter. Thus, 0.15 = moles/0.10 liters (1 liter = 1000 ml). Therefore,(0.15) (0.10) = 0.015 moles of HOAC. For NaOH, M = 0.10 = moles/0.10 liters. Thus, the number of moles of NaOH = (0.10) (0.10) = 0.010. From the previously written neutralization equation, one mole of acid reacts for every mole of base. Only 0.01 moles of HOAC could react with NaOH. Thus, 0.015 - 0.010 = 0.005 moles of HOAC was NOT neutralized by the base, NaOH. This excess 0.005 moles of HOAC is, in the presence of its salt NaOAC. Thus, it is a buffer solution. Solutions that con-tain a weak acid, such as HOAC, and its salt are called buffers. Therefore, consider the dissociation of acetic acid; it is in excess and is the source of the hydronium ion concentration. HOAC + HOH \rightleftarrows H_3O^+ + OAC^-. The equilibrium constant for this reaction measures the ratio of the concentration of the products to reactants, each raised to a power based on its coefficient in the chemical equation. Thus, K_a , the equilibrium constant, = {[H_3O^+] [OAC^-]} / [HOAC]. Water is not included since it is a constant. One is solving for [H_3O^+] to find pH, let x = [H_3O^+]. From the dissociation equation x = [OAC^-] also, since they are formed in equimolar amounts. Recall, however, that the neutralization reaction produced OAC^- also. There were 0.010 moles each of NaOH and HOAC which reacted. From the equation, 0.010 moles of OAC^- should have been produced. The volume was 200 ml. Thus, M = concentration = (0.01 moles) / (200 ml / 1000 ml/l) =0.05 M. The total concentration of OAC^-, from the acid dissociation and neutralization, is x + .05. The acetic acid concentration is the initial amount minus the amount that dissociated. Recall, there were 0.005 moles of acetic acid in 200 ml. to give (0.005 moles)/ (200 ml / 1000 ml/l) = 0.025M. x moles/liter of it dissociated. Thus, the final concentration is (0.025 - x). K_a = {[H_3O^+] [OAC^-]} / [HOAC]. 1.75 × 10^-5 = {[H_3O^+] [OAC^-]} / [HOAC]. Substituting final concentrations, 1.75 × 10^-5 = {(x) (x + 0.5)} / (0.025 - x). The excess OAC^-, from the dissociation of HOAC, x, is small, Thus, {(x) (x + 0.05)} / (0.025 - x) \approx (0.050 x) / (0.025) = 1.75 × 10^-5. Solving for x, one finds [H_3O^+] = x = 8.8 × 10^-6 M. Thus, pH = - log [H_3O^+] = - log [8.8 × 10^-6] = 5.06.

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Question:

Write a program in PL/I to calculate a table of returns on an investment of original amount A which is compounded yearly at P percent return, for Y years.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G18-0445.htm

Solution:

After Y years the original amount A will grow to T =A(1 + P/100)^Y The program below reads in values of A,P, and Y. Then it calculates the corresponding value of T and prints out A, P, Y, and T. /\textasteriskcenteredCOMPOUND INTEREST\textasteriskcentered/ ON ENDFILE (SYSIN) GO TO L60 L10:GET LIST (A, P, Y); L20:T = A\textasteriskcentered(1 + P/100) \textasteriskcentered\textasteriskcenteredY; L30:PUT EDIT (A, P, Y, T) (F (16,4)) ; L40:PUT SKIP; L50:GO TO L10; L60:END

Question:

If 1.588 g of nitrogen tetroxide gives a total pressure of 1 atm when partially dissociated in a 500-cm^3 glass vessel at 25\textdegree, what is the degree of dissociation, \alpha? What is the value of K_p ? The equation for this reaction is N_2O_4\rightarrow 2NO_2 .

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/Users/wenhuchen/Documents/Crawler/Chemistry/E09-0324.htm

Solution:

The determination of the density of a partially dissociated gas provides one method for measuring the extent to which the gas dissociates. When a gas dissociates, more molecules are produced, and at constant temperature and pressure the volume increases. The density at constant pressure then decreases, and the difference between the density of the undissociated gas and that of the partially dissociated gas is directly related to the degree of dis-sociation. If \alpha represents the fraction of gas dissociated, then 1 - \alpha represents the fraction of gas undissociated.Assuming one starts with 1 mole of gas the number of moles of gaseous products in the balanced chemical equation is 1 +\sumv_i . (where \sumv_i is the summation of all the coefficients; positive coefficients for the products and negative co-efficients for the reactants) . Therefore, the number of moles of gas present at equilibrium is (1 - \alpha) + (1 + \sumv_i )\alpha = 1 + \alpha\sumv_i The density of an ideal gas at constant pressure and temperature is inversely proportional to the number of moles for a given weight and the ratio of the density \rho_1 of the undissociated gas to the density \rho_2 of the partially dissocciated gas is given by the expression \rho_1 / \rho_2 = 1 + \alpha \sumv_i \alpha = (\rho_1 - \rho_2 ) / (\rho2\sumv_i) If there is no dissociation, then \alpha = 0 and \rho_1 = \rho_2 , if dissociation is complete, then \alpha = 1, \rho2\sumv_i = \rho_1 - \rho_2 and \rho_1 = (1 + \sumv_i ) \rho_2 . Molecular weights are proportional at constant tem-perature and pressure to the gas densities. Therefore, molecular weights can be substituted for the densities: \alpha = ( M_1 - M_2 ) / (M_2 \sumv_i) where M_1 is the molecular weight of the undissociated gas, and M_2 is the average molecular weight of the gases when the gas is partially dissociated. The dissociation of nitrogen tetroxide proceeds by the following equation: N_2O_4= 2NO_2andK_p = (P^2 _N(O)2) / (P_(N)2(O)4) The degree of dissociation is \alpha; (1 - \alpha) is pro-portional to the number of moles of undissociated N_2O_4; 2\alpha is proportional to the number of moles of NO_2 ; and (1 - \alpha) + 2\alpha or 1 + \alpha is proportional to the total number of moles. If the total pressure of N_2O_4plus NO_2 is P, the partial pressures are: P_N2O4 = { (1 - \alpha) / (1 + \alpha) } P and P_(NO)2 = { 2\alpha / (1 + \alpha) } P Then,K_p = [{2\alpha / (1 + \alpha)} P]^2 / [{ (1 - \alpha) / (1 + \alpha) } P] = (4 \alpha^2 P) / (1 - \alpha)^2 M_1 = 92.02 g/mole. Using the Combined gas Law, PV = n RT and setting n, the number of moles of substance, equal to the mass of substance, m, divided by its molecular weight, M_2 , one obtains PV = (m / M_2 )RT Rearranging and solving for M_2 , M_2 = (RT/P) × (m/V) M_2 = (RT/P)(m/V) = {(0.082 liter atm \textdegreeK-^1 mole-^1 ) (298\textdegreeK) (1.588 g)} / {(1 atm ) (0.500 liter)} = 77.68 g/mole \alpha = ( M_1 - M_2 ) / (M_2 \sumv_i) where \sumv_i for the reaction equals [(coefficient of NO_2 ) - (coefficient of N_2O_4) = (2 - 1) = 1] and M_1 = 92.02 and M_2 = 77.68. \alpha= (92.02 - 77.68) / 77.68 = 0.1846 K_p = (4 \alpha^2 P) / (1 - \alpha^2) = { 4 (0.1846)^2 (1 atm) } / {1 - (0.1846)^2 } = 0.141.

Question:

As part of a certain program, there is a declare statement as follows: DCL (A,C) FIXED(5,2), B CHAR(5) VAR, D FIXED(2,1), E CHAR(4), F BIT(7) VAR, G FIXED (3,2); Also, as part of the same program, there is an EDIT-directed output statement as follows: PUT EDIT(A,B,C,D,E,F,G) (LINE(40), COL(15), F(7,2), PAGE, A(3), F (6,2), LINE(10), F(7,3), X (3), A(5), COL(30), B(6), SKIP(7), X(5) , F(3,1) ) ; Show what will be printed out, if the contents of the memory storage are as follows: Name of Memory location Contents A - 123\textbullet45 B B2GLb b C 123\textbullet45 D - 1\textbullet8 E CARS F 1011bbb G - 7\textbullet94 Note:bis a symbol used to denote a blank space.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G13-0325.htm

Solution:

The PUT EDIT statement states that the values for the variables A, B, C, D, E, F and G are to be printed out. The format in which these are to be printed out is given in the second set of parentheses ( ). The first format instruc-tion is LINE(40). Hence, the printer skips to line (40) on the current page. This assumes that the printer did not pass line 40 already. If the printer was indeed past line 40, e.g., on line 45, then, the printer skips the rest of the page and goes to line 40 on the next page. Now, the next format in-struction says COL(15), which is a short form for COLUMN(15). This is an instruction to the printer to go to column 15 on the current line, which is line 40. However, if in case column 15 has already been passed as a result of carrying out some previous format instruction, the printer then skips to col. 15 on the next line. That is, it would skip to column 15 on line 41 in this case if the present position of the printer was, for example, column 59 on line 40. Thus, after having moved to the correct page, line and column, the computer reads the next format instruction which specifies F(7,2). This specification of F(7,2) corresponds to the first element A in the data list of the PUT EDIT statement. Therefore, the computer retrieves the value for A from its memory. This value is - 123\textbullet45, and it agrees with the specification of the F(7,2) format, which requires a number of the type XXXX\textbulletXX. Hence, starting from column 15, of the current line (say, line 40), the printer prints out - 123\textbullet45. After this printing, the printer is on column 22 (15 + 7 = 22). The minus sign and the decimal point also count as one column position in - 123\textbullet45. Now the computer wants to print out the next data item, which is B. Therefore, it checks the format list and sees that the next format instruction says PAGE. This causes the printer to leave blank the rest of the current page and go to the first line, first column of a new page. Next comes the instruction A(3), which means, that the data item B has a character field of 3 characters to be printed. Thus, the value of B, which is, B2GLbis retrieved from memory and as only 3 characters are to be printed, the printer prints B2G. The L is cut off, and the printer now has come to column 4 of the first line of the new page. The next format instruction F(6,2) is for the third item C from the data list of the PUT EDIT statement. There-fore the value of C is fetched from memory as 123\textbullet45. The format specification requires a form XXX\textbulletXX. Hence, 123\textbullet45 is printed out from column 4 onwards, and the printer comes to column 10. Next comes the turn for item D. For this, the first format instruction specifies LINE(10). Therefore the printer jumps from the current LINE(1) to LINE (10). The next format instruction is F(7,3). The value for D from memory is - 1\textbullet8 now, F(7,3) requires a form XXX\textbulletXXX. Therefore - 1\textbullet8 is con-verted to 0 - 1\textbullet800. Now, the print out is: 0 - 1\textbullet800, starting at column 1. Note that the leading zero is suppressed. This brings the printer to column 8. Now, the next item on the data list, which is, E is to be printed out. The first format instruction for E says X(3). This means that 3 columns are to be skipped. The printer now skips from the present column 8 to column 11. Now, a character field A(5) is specified. The value fetched for E from memory is CARS, which has only 4 characters, therefore a blank is added at the end of CARS to make up the 5 characters specified by the format. The printer now prints out CARSb(note that the 5th character is a blank). b The printer thus comes as a result to column 16 of line 10. Next, the item F is to be printed. For this, the format first specifies COL(30); as a result the printer jumps from column 16 to column 30 of the same line 10. Then, the computer reads the specification B(6), which means that a Binary bit string is to be printed out from memory. The value for F is obtained as 1011bbb. This string has 4 characters, and 3 blanks, but the format specification, B(6), requires 6 bits to be printed out. Hence, zeroes are filled in to make up the required string length, and the printer prints out 101100. Now, the last item on the data list, which is G is to be printed out. The format specification for G specifies first a SKIP(7). This causes the printer to jump 7 lines from the present line 10 to line 17. The next format specification says X(5). Hence, the printer jumps 5 columns from column 1 of line 17 to column 6 on the same line 17. The next format instruction specifies F(3,1) for G. The value for G as obtained from memory is -7.94. But the format specification requires a form X\textbulletX. Hence, the value of G is truncated at both ends, and the printer prints out 7\textbullet9. Note that the minus sign and the digit 4 have been truncated. The truncation of the decimal place (digit 4) is not very serious. At most is affects the accuracy of the number. However, the truncation of the minus sign from the left is very serious. It changes a negative number into a positive number. Trunca-tion from the left can also cut off the higher order digits. Hence, the computer is programmed to regard truncation of a number from the left as an error condition. This error is called a SIZE condition. Also, the printer prints out a warning that a SIZE condition has occurred. To summarize, the print-out for this problem would be as shown in fig. 1.

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Question:

A string under a tension of 256newtonshas a frequency of 100 vibrations per second. What is its frequency when the tension is changed to 144newtons?

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/Users/wenhuchen/Documents/Crawler/Physics/D25-0816.htm

Solution:

The tension is only (144/256) as much as it was at first. The square root of (144/256) is (12/16), or (3/4) . Since the frequency is directly proportional to the square root of the tension, the new frequency is (3/4) × 100, or 75vps.

Question:

Using FORTRAN rules what would be the value of sum in each case.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G12-0275.htm

Solution:

The rules adopted by the creators of FORTRAN state that evaluation of arithmetic operations is conducted from left to right, except when the succeeding operation has a higher "binding power" than the one currently being considered. The binding power of "+" and "-" is the lowest in strength; operators "{_\ast}" and "/" are of middle strength, and raising to a power ({_\ast}{_\ast}) has the highest priority. The concept of parenthesizingsubexpressionswas added to the rules to achieve a sequence in operations different from the rules of binding power of operators. Therefore, using this knowledge we find the solutions to be : to 3. The result is 7.

Question:

The square ABCD has a side of 10 cm length. Equal positive charges of +50 statcoulombs are placed at A and B and equal negative charges of\rule{1em}{1pt}100 statcoulombs are placed at C and D. Calculate the electric field at P, the center of the square.

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/Users/wenhuchen/Documents/Crawler/Physics/D18-0591.htm

Solution:

P is the point where the two diagonals AC and BD cross. Let AP = BP = CP = DP = R. Then, applying Pythagoras' theorem to the right angle triangle APB, AP^2 + BP^2 = AB^2 2R^2 = 10^2 cm^2 R^2 = 50 cm^2 R = 7.07 cm The charge q_A at A produces a contribution E_A to the field at P given by definition as EA= q_A / R^2 = 50 statcoul /50 cm^2 Since 1 (statcoul / cm^2 ) = 1 (dyne / statcoul) E_A = 1 dyne per statcoulomb Since the charge at A is positive, E_A points directly away from A, (see figure) . Similarly, the charge at B produces a contribution E_B = 1 dyne per statcoulomb which points directly away from B. The negative charge q_C at C produces a contribution E_C = q_C /R^2 = (100 statcoul) /(50 cm)^2 = 2 dyne per statcoulomb Since q_C is negative this contribution points directly toward C. Similarly, the charge at D produces a contribu-tion E_D = 2 dyne per statcoulomb which points directly toward D. Adding together E_A and E_C, which point in the same direction, we get a vector of magnitude 3 dyne per stat-coulomb pointing toward C. Adding E_B to E_D we get a vector of magnitude 3 dyne per statcoulomb pointing toward D. This is shown in the figure. This figure also gives the triangle of vectors PQR used to add the two vectors shown. Here we are using the parallogram law of vector addition. The resultant field E =PQ^\ding{217} is obviously vertically downward and its magnitude is obtained from Pythagoras' theorem PQ^2 = PR^2 + RQ^2 E^2 = (3^2 + 3^2)dyne^2 / statcoul^2 = 18 dyne^2/statcoul^2 E = 4.24 dyne per statcoulomb The resultant field at P is therefore 4.24 dyne per statcoulomb pointing in the directionAD^\ding{217}.

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Question:

A 0.96-lb ball A and a 1.28-lb ball B are connected by a stretched spring of negligible mass as shown in the diagram. When the two balls are released simultaneously, the initial acceleration of B is 5.0 ft/sec^2 westward. What is the initial acceleration of A?

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/Users/wenhuchen/Documents/Crawler/Physics/D04-0124.htm

Solution:

If we take the balls and spring as our system, as shown in the figure, no external forces act. The only forces acting are then internal, and consist of the force of m_A on m_B, and the force of m_B on m_A,(f\ding{217}_ABand f\ding{217}_BA, respectively). But by Newton's Third Law, f\ding{217}_AB = -f\ding{217}_BA(1) at each instant of time. By Newton's Second Law, f\ding{217}_AB = m_B a\ding{217}_B f\ding{217}_BA = m_A a\ding{217}_A(2) since f\ding{217}_AB is the only force on m_B, and similarly for f\ding{217}_BA and m_A. Using (2) in (1) m_B a\ding{217}_B = -m_A a\ding{217}_A orm_B g a\ding{217}_B = -m_A g a\ding{217}_A(3) Butm_B g = W_B m_A g = W_A where W_A and W_B are the weights of A and B. Hence, (3) yields W_B a\ding{217}_B = -W_A a\ding{217}_A Solving for a\ding{217}_A, a\ding{217}_A = -(W_B/W_A) a\ding{217}_B(4) To find a\ding{217}_A at t = 0, we use the value of a\ding{217}_B at t = 0 in (4) giving a\ding{217}_A (t = 0) = -{1.28 lb/.96 lb}(5 ft/s^2 westward) a\ding{217}_A (t = 0) = -6.7 ft/s^2 westward a\ding{217}_A (t = 0) = 6.7 ft/s^2 eastward since eastward is the opposite of westward.

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Question:

The process of ovulation continues until a period of the female's life called the menopause. What are the changes that occur during this period?

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/Users/wenhuchen/Documents/Crawler/Biology/F21-0564.htm

Solution:

The menopause usually comes between the ages of 45 and 50. It is apparently attributable to two reasons: (1) the cessation of secretion of LH by the pituitary as a result of declining activity of the LH- stimulating center in the hypothalamus, and (2) the declining sensitivity of the ovaries to the stimulatory activity of gonadotropins. In the absence of LH, secretory activity in the ovaries falls off. Neither estrogen nor progesterone can be produced in significant quantities. Ovulation and menstruation become irregular and ultimately cease completely. Some secretion of estrogen generally continues beyond these events but gradually diminishes until it is inadequate to maintain the estrogen-dependent tissues, most notably the breasts and genital organs, which become atrophied. Sexual drive, however, is frequently not diminished and may even be heightened. Severe emotional disturbances are not uncommon during menopause and are generally ascribed not to the direct effect of estrogen deficiency, but to the disturbing nature of the wholeperiod-the awareness that reproductive potential has ended. Woman during menopause often experience hot flashes. These result from dilation of the skin arterioles, causing a feeling of warmth and marked sweating. The reason for this is unknown. Many of the symptoms of the menopause can be reduced by the administration of small amounts of estrogen. Just recently, another aspect of menopause was discovered, that being the relationship between plasma estrogen and plasma cholesterol. Estrogen significantly lowers the level of plasma cholesterol. This explains why women have a significantly lower incidence of arte-riosclerosis (a disease positively correlated to the level of cholesterol in the plasma) than men until the menopause, when the incidence becomes similar in both sexes. Furthermore, decreased levels of estrogens are implicated in osteoporosis since estrogen normally stimulates osteoblasts to form new bone.

Question:

The unaided eye can perceive objects which have a diameterof 0.1 mm. What is the diameter in inches?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E01-0002.htm

Solution:

From a standard table of conversion factors, one can find that 1 inch= 2.54 cm. Thus, cm can be con-verted to inches by multiplying by 1 inch/2.54cm. Here, one is given the diameter in mm, which is .1 cm.Milli-- metersare converted to cm by multiplying the number of mm by .1 cm/1 mm. Solving for cm, you obtain: 0.1 mm × .1 cm/1 mm = .01 cm. Solving for inches: .01 cm × [(1 inch) / (2.54 cm)] = 3.94 ×10\Elzbar^3 inches.

Question:

One group of compounds that may become quite important in birth control is the prostaglandins. What are the advantages of prostaglandins in birth control?

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/Users/wenhuchen/Documents/Crawler/Biology/F21-0561.htm

Solution:

The compounds known as prostaglandins have been identified as cyclic, oxygenated, 20-carbon fatty acids. There are now 16 known natural prostaglandins. The prostaglandins, which are secreted by the seminal vesicles and other tissues, appear to mediate hormonal action by influencing the formation of cyclic AMP. The end result of their actions is varied. Though they are expected to be valuable in treating many ailments, from nasal congestion to hypertension, their most significant use may be in birth control. Prostaglandins induce strong contractions in the uterus, and hence they may be powerful agents for abortion. As abortive agents, the prostaglandins efficient in almost all of the cases where they are used. When more is known about the side effects of pros-taglandins, it is believed that their application in abortion would be safer than mechanical means. The pros-taglandins are also being considered as a once-a-month replacement for birth control pills. The powerful uterine contractions induced by prostaglandins prevent the fertilized ovum from being implanted. Since they can be introduced directly into the uterus through the vagina, the prostaglandins will probably be safer contraceptives than the oral chemical agents which must travel through the blood and may affect other parts of the body.

Question:

\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2Three forces acting at a point are F_1\ding{217} = 2i˄ - j˄ + 3k˄, F_2\ding{217} \"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2 \"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2 \"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2= - i˄ + 3j˄ + 2k˄, and F_3\ding{217} = - i˄ + 2j˄ - k˄. Find the direction \"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2 \"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2and magnitudes of F_1\ding{217} + F_2\ding{217} + F_3\ding{217}, F_1\ding{217} - F_2\ding{217} + F_3\ding{217}, and \"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2 \"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2F_1\ding{217} + F_2\ding{217} - F_3\ding{217}. \"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2 \"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2 \"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2 \"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2

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/Users/wenhuchen/Documents/Crawler/Physics/D01-0002.htm

Solution:

\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2When vectors are added (or subtracted) , their components in the \"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2 \"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2directions of the unit vectors add (or subtract) algebraically. Thus since \"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2 \"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2F_1\ding{217} = 2i˄ - j˄ + 3k˄, F_2\ding{217} = - i˄ + 3j˄ + 2k˄, F_3\ding{217} = - i˄ + 2j˄ - k˄, \"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2 \"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2then it follows that 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2i˄ + 0j˄ + 6k˄ 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\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2 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\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2 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It therefore has a magnitude of \surd(4^2 + 4^2) units = 4 \surd2 units \"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2 \"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2= 5.66 units, and lies in the y- z plane, making an angle \texttheta with the \"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2 \"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2y - axis, as shown in figure (a), where tan \texttheta = 4/4 = 1. \"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2 \"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2Thus \texttheta = 45\"{\i}\textquestiondown1⁄2. \"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2 \"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2Similarly, F_1\ding{217} - F_2\ding{217} + F_3\ding{217} has a magnitude of 2 \surd2 units = 2.82 \"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2 \"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2units, and lies in the x - y plane, making an angle \textphi with the x-axis, as \"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2 \"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2shown in figure (b), where tan \textphi = + 2/-2 = - 1. Thus \textphi = 315\"{\i}\textquestiondown1⁄2. \"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2 \"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2Also, F_1\ding{217} + F_2\ding{217} - F_3\ding{217} has a magnitude of \surd(2^2 + 6^2) units = 2 \surd10 \"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2 \"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2 \"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2units = 6.32 units, and lies in the x - z plane at an anglexto the x-axis, as \"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2 x \"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2shown in figure (c), where tanx= 6/2 = 3. Thusx= 71\"{\i}\textquestiondown1⁄234\"{\i}\textquestiondown1⁄2. \"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2 x x

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Question:

The transformation of a spermatid into a functioning sperm is a process of differentiation. It varies in detail in different species, but the basic features are the same. Outline this process of differentiation in humans.

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/Users/wenhuchen/Documents/Crawler/Biology/F22-0574.htm

Solution:

Within the cytoplasm of the human spermatid, one can find proacrosomic granules, centrioles, and mitochondria. During differentiation of the spermatid, a process called spermiogenesis, the proacrosomic granules, which were formed from the Golgi complex, gather toward one pole of the cell and condense into the form of a bead. This acrosomal bead contains appreciable amounts of carbohydrates and several enzymes of lysosomal nature. This region of the differentiating spermatid eventually becomes the acrosome, and is chiefly concerned with enzymatic penetration of the egg. The two centrioles of the spermatid move toward the opposite side of the nucleus. Here the centriole nearest the nucleus spins a filament that grows outward and becomes the axis of the sperm's tail or flagellum. This flagellum shows the usual 9 + 2 tubule pattern, and is often accompanied by additional fibrous structures that run alongside the tubules. (It is interesting to note that whereas basal bodies usually form flagella, here the centriole forms the flagellum.) The centrioles then move apart. The proximal one remains close to the posterior surface of the nucleus. The distal one forms a ring around the axial filament and recedes a short distance. Some of the mitochondria now gather around the axial filament in the space between the two centrioles and form a spiral in an area called the middle piece. Most of the cytoplasm pinches away and is discarded, but a little remains as a delicate sheath over the head area. Thus the resultant sperm consists of 3 pieces. The head con-tains an anterior acrosome, a posterior nucleus, and little cytoplasmic material, and by virtue of its lytic enzymes, functions in penetration of the egg. The midpiece contains the mitochondria necessary for energy-requiring motility. Finally, the tail with a 9 + 2 microtubule arrangement, contains the contractile machinery for whiplike movements. It was shown that before fertilization, sperm in the vicinity of an ovum in the oviduct undergo a sequence of structural changes called the acrosome reaction. The outer membrane of the acrosome fuses at multiple points with the overlying plasma membrane of the sperm head to create numerous openings, through which the enzyme-rich contents of the acrosome are liberated in a process not unlike the release of secretory products from a glandular cell. The released enzymes help the sperm to digest its way through the protective coating of the egg. The result is the entry of the sperm into the egg cytoplasm.

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Question:

An electron initially at rest is accelerated though1cm by an electric field of 3×10^4 V/m. What is the terminal speed?

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/Users/wenhuchen/Documents/Crawler/Physics/D19-0621.htm

Solution:

In order to solve for kinematical variables such as position, velocity, and acceleration, we must use Newton's Second Law to relate the net force on our system to its acceleration. We then integrate this equation to obtain v(t) and x(t). In this problem, the only force acting on the electron is that due to the electric field E^\ding{217}. By definition E^\ding{217} = F^\ding{217}/q where F^\ding{217} is the force acting on q. Hence, F^\ding{217} = qE^\ding{217} Since the electron charge is - e, where e = 1.6 x 10^\rule{1em}{1pt}19 coul. we may write F^\ding{217} = -eE^\ding{217} But, by the second law, F^\ding{217} = ma^\ding{217} where m and a^\ding{217} are the electron mass and acceleration, respectively. Therefore, ma^\ding{217} = \rule{1em}{1pt} eE^\ding{217} ora^\ding{217} = (\rule{1em}{1pt} e / m) E^\ding{217} For our problem, E^\ding{217} is constant. Since e and m are also constants, the same must be true of a^\ding{217}. Looking at the figure, we see that E^\ding{217} is in the negative x direction. If \^{\i} is a unit vector in the positive x direction E^\ding{217} = E\^{\i} anda^\ding{217}= (\rule{1em}{1pt} e/m) (\rule{1em}{1pt} E\^{\i}) = (eE / m)\^{\i}(1) Noting that a^\ding{217} is only in the x direction we may drop the vector notation in (1) and resort to scalar notation. a = eE/m(2) where a is positive when in the positive x direction. But we want to know v in terms of x. Using the fact that a= dv/dt = [dv/dx] [dx/dt] =[dv/dx] (v) we obtaina = v(dv / dx) oradx = vdv Using (2), (eE/ m) dx = vdv (eE/ m) ^(x)f\int_(x)0 dx =^(v)f\int(v)0vdv(3) where the subscripts "f\textquotedblright and "\textdegree" indicate final and initial values, respectively. Integrating(3) (eE/m) (xf\rule{1em}{1pt} x_0) = 1/2 v2(v)f│_(v)0 = 1/2 (v2f\rule{1em}{1pt} v2_0) Solving for vf (eE/m) (x_f \rule{1em}{1pt}x_0) = 1/2 (v2f\rule{1em}{1pt} v2_0) (2eE/ m) (x_f \rule{1em}{1pt} x_0) + v2_0 = v2f vf= \surd[(2eE/m) (x_f \rule{1em}{1pt} x_0) + v^2 _0 ] But v_0 = 0 since the electron is initially at rest, and x_f \rule{1em}{1pt} x_0 = 1 cm. Therefore, v_f = \surd[{2(1.6 × 10^\rule{1em}{1pt}19 coul) (3× 102v/cm) (1cm)} / (9.11 × 10^\rule{1em}{1pt}31 kg)] v_f = \surd(1.054 × 10^14 m/s^2 ) v_f = 1.03 × 10^7 m/s

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Question:

Find the angular width of the central maximum of a Fraunhofer dif-fraction pattern if the light used has wavelength \lambda = 6000\textdegreeA and the slit width is .1 × 10^-3 m.

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/Users/wenhuchen/Documents/Crawler/Physics/D30-0904.htm

Solution:

The angular location of the minima of a Fraunhofer diffrac-tion pattern is given by sin \psi = m\lambda/a(m = 1,2,...) where a is the slit width. The central maximum is flanked on either side by the first order minima (m = 1). The angular position of this minimum is sin \psi = [{(1) (6000\textdegreeA)} / {(.1×10^-3 m)}] Since1\textdegreeA = 10^-10 m sin \psi = (6 × 10^3 × 10^-10 m) / (1 × 10^-4) m sin \psi = 6 × 10^-3 = .006 \psi = 1/3\textdegree Hence, the angular breadth of the central maximum is (see figure) 2\psi = 2/3\textdegree.

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Question:

An elevator is accelerated upward at 2 ft/sec^2. If the elevator weighs 500 lb, what is the tension in the supporting cable?

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/Users/wenhuchen/Documents/Crawler/Physics/D04-0110.htm

Solution:

The net force acting on the elevator is \sumF = T - mg where T is the cable tension, and mg is the elevator's weight. (Note that the positive direction is taken as upward). By Newton's Second Law, this must equal the product of the elevator's mass and acceleration, whence T - mg = ma Solving for T(1) T = m(g + a) We don't know the mass, m, of the elevator, but we do know its weight W. Since W = mg (1) becomes T = (W/g) (g + a) Using the data provided T = [(500 lb) / (32 ft/s^2)](32 ft/s^2 + 2 ft/s^2) T = (34/32) 500 lb T = 531.2 lb.

Question:

atom, isotope, ion. Could a single particle of matter be all three si-multaneously?

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/Users/wenhuchen/Documents/Crawler/Biology/F01-0005.htm

Solution:

An atom is the smallest particle of an element that can retain the chemical properties of that element. It is composed of a nucleus, which contains positively charged protons and neutral neutrons, around which negatively charged electrons revolve in orbits. For example, a helium atom contains 2 protons, 2 neutrons, and 2 electrons. An ion is a positively or negatively charged atom or group of atoms. An ion which is negatively charged is called an anion, and a positively charged ion is called a cation. Isotopes are alternate forms of the same chemical element. A chemical element is defined in terms of its atomic number, which is the number of protons in its nucleus. Isotopes of an element have the same number of protons as that element, but a different number of neutrons. Since atomic mass is determined by the number of protons plus neutrons, isotopes of the same element have varying atomic masses. For example, deute-rium (^2H) is an isotope of hydrogen, and has one neutron and one proton in its nucleus. Hydrogen has only a proton and no neutrons in its nucleus. A single particle can be an atom, an ion, and an isotope simultaneously. The simplest example is the hydrogen ion H+ . It is an atom which has lost one electron and thus developed a positive charge. Since it is charged, it is therefore an ion. A cation is a positively charged ion (i.e. H+) and an anion is a negatively charged ion (i.e. Cl-). If one compares its atomic number (1) with that of deuterium (1) , it is seen that although they have different atomic masses, since their atomic numbers are the same, they must be isotopes of one another.

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Question:

Explain how the metabolism of a fermenting yeast cell differs from that of a fatigued muscle cell.

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/Users/wenhuchen/Documents/Crawler/Biology/F03-0077.htm

Solution:

We shall first examine the anaerobic metabolism of yeast. When yeast is placed in an environ-ment devoid of oxygen, their cells are able to extract energy from glucose in the absence of molecular oxygen via glycolysis, which occurs in both aerobic and anaerobic organisms. There are no hydrogen acceptors due to the absence of molecular oxygen, the final acceptor. When O_2 is not available, all intermediates of the respiratory chain are reduced, and the TCA cycle and electron transport system are turned off. Pyruvate, the degradation product of glucose, cannot therefore be oxidized to acetyl CoA and enter the TCA cycle for further energy extraction. Instead, pyruvate itself accepts the electrons (via acetaldehyde) and is reduced to ethyl alcohol (ethanol). This process, known as alcoholic fermentation, converts each glucose molecule to two ethanol molecules and two CO_2. Alcoholic fermentation also serves to regenerate oxidized NAD+, so that glycolysis can continue. Two net ATP's are formed per glucose molecule in alcoholic fer-mentation. In man and other organisms, anaerobic respiration can supplement aerobic respiration for short periods of time. In a fatigued muscle cell, pyruvic acid, produced via glycolysis, again serves as an electron acceptor since oxygen is unavailable. However, unlike yeast, human cells produce lactic acid instead of ethanol and carbon dioxide: This process, sometimes called homolactic fermentation, occurs during strenuous exercise when muscle cells cannot obtain enough oxygen to support aerobic respiration. Instead, the energy expended comes from glycolysis (2 ATP per glucose molecule) with the build-up of lactic acid, the cause of muscle fatigue. Although both processes, alcoholic and homolactic fermentation, enable energy production, they are not as efficient as aerobic respiration. A yeast cell or skeletal muscle cell can theoretically obtain 18 times as much ener-gy by oxidizing a glucose molecule completely to carbon di- oxide and water, instead of stopping at ethanol, if it em-ploys the aerobic type of respiration.

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Question:

What is the volume in liters of a rectangular tank which measures2.0 m by 50 cm by 200 mm?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E01-0009.htm

Solution:

One liter is equal to 1000cc; therefore, one should find the volume ofthe tank in cubic centimeters first and then convert to liters. This method isbest for this problem because the sides of the tank are given in units whichcan quickly be converted to centimeters. There are 100 cm in 1 m.Thus ,2 m is equal to 2.0 m × 100 cm/1 mor 200 cm. There are 10 mm in 1cm,therefore 200 mm is equal to 200 mm×1 cm/10 mm or 20 cm. Solving for the volume of the tank in cubic centimeters: volume= 200 cm × 50 cm × 20 cm = 2.0 × 10^5 cc. To convert from cubic centimeters to liters multiply by 1 liters/1000cc. volumein liters = 2.0 × 10^5 cc × 1 liters/1000cc = 200 liters.

Question:

Define the operating system, and describe the resource man-agement functions it carries out.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G01-0019.htm

Solution:

An operating system is a set of programs that: a) Provides an interface between the users and the hardware. This generally means creating a more hospitable environment for the users. Examples are interactive use of the computer, convenient execution of the programs, and receiving meaningful error messages when programs fail to run. b) Manages the sharing of computer resources. This means the management of hardware, software, and information reliably and efficiently, as discussed next. OS Resource Management Functions: The operating system (also called the executive, monitor, master, supervisor, etc.) manages the following resources: 1. CPU or Multiple Processing Elements (PEs) : This is done by multiprogramming and multitasking when there are several PEs. Multitasking refers to loading (assigning tasks to) of process-ing elements in a balanced manner. The CPU scheduling is used to assign and reclaim CPU to processes. 2. Primary Memory and Cache Memory : This is done through vari-ous allocation anddeallocationof memory to user programs. One popular method is demand paging, where fixed-size program sec-tions (pages) are allocated to the equal size physical memory regions (frames) on demand. Cache memories (faster than regular primary memory) are used to keep track of allocation of pages to frames. 3. Secondary Storage and Files: The disk space is managed through contiguous, linked, or indexed allocation methods. Some-times, the amount of free disk space is kept track of using bitmaps (sequence of bits indicating whether a disk block is free or used) . The disk space management is part of the file management functions. File management allows creation, move, copy, and deletion of files. In addition, it supports file op-erations such as open, read, write, position, append, close. 4. I/O and Communication Devices : The Operating System (OS) keeps track of the I/O devices such as tapes, terminals, and printers and communication devices such as channels, multiplex-ers, and modems. The OS assigns these devices to user processes in a fair, reliable, and efficient manner; and reclaims them when they are no longer needed. 5. Data and Information : The operating system also keeps track of and allows efficient sharing of data such as system buffers and stacks contents, the messages stored in mailboxes (shared communication buffers), and the like. The information management functions include the accounting for CPU usage, and system status information such as the time, date, directory contents, available disk space, free memory space, etc.

Question:

Discuss the separate half-reactions and net cell reaction that occur in a flashlight battery. If you have a current of .01 amp, how long can the current flow if the battery contains 25.0g of a paste of MnO_2 , NH_4Cl, and ZnCl_2 of which 20% is MnO_2 ?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E16-0580.htm

Solution:

A standard flashlight dry cell, or Leclanche cell, is composed of a graphite rod in a moist MnO_2 - ZnCl_2- NH_4Cl paste, all in a zinc wrapper. The zinc wrapper serves as the electrode for the oxidation half-- cell (the anode), while the graphite rod is the electrode for the reduc-tion half-cell (the cathode). Thus, the probable half-reactions are: at the anode: Zn(s) \rightarrow Zn2++ 2e^- and at the cathode: 2MnO_2(s) + 2H_3O^+ + 2e^- \rightarrow 2MnO(OH) + 2H_2O . The net reaction for the overall process is Zn(s) + 2MnO_2 \rightarrow ZnO \bullet Mn_2O_3 . To determine how long the current can flow, calculate the number of moles of MnO_2 present and the number of Faraday's required to reduce it per mole. You are told that 20% of 25 grams of paste is MnO_2.Thus, you have (.20)(25g) = 5.0g of MnO_2 .The molecular weight of MnO_2 = 86.94g . Thus, you have (5g) / (86.94g /mole) = .0575 moles of MnO_2 . From the net equation, you see that 2 electrons are transferred. A Faraday = 96,500 coul and is capable of reducing one equivalent posi-tive charge, which means Avogardro's number of individual unit charges (one mole of electrons). Since 2 electrons were transferred per 2 moles, it takes 2 Faraday's to reduce 2 moles of MnO_2 . According to the net equation 2 moles of MnO_2 are reduced for every mole of ZnO \bullet Mn_2O_3 obtained. Thus, you can write that the current will last for (.0575 moles MnO_2) (2F / 2moles MnO_2) (96,500 coul / F) (amp /coul/sec) (1 / .01 amp) = 154 hours.

Question:

A dc series motor operates at 120 volts and has a resistance of 0.300 ohm. When the motor is running at rated speed, the armature current is 12.0 amp. What is the counter EMF in the armature?

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/Users/wenhuchen/Documents/Crawler/Physics/D24-0787.htm

Solution:

An electric circuit containing a series motor (that is a motor whose coil resistance is in series with the motor) is drawn schematically above. The motor draws current, and uses it to produce the mechanical motion of a shaft. By Kirchoff's voltage law, the sum of the voltage drops around the circuit must be zero. Using this fact along with Ohm's Law (V = IR) we obtain V = \epsilon + IR or\epsilon = V \rule{1em}{1pt} IR = 120 volts \rule{1em}{1pt} 12.0 amp × 0.300 ohm = 116 volts.

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Question:

Distinguish between a hormone and a vitamin, and a hormone and an enzyme.

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/Users/wenhuchen/Documents/Crawler/Biology/F21-0531.htm

Solution:

A hormone is an organic substance synthesized by a specific organ or tissue and secreted into the blood, which carries it to other sites in the body where the hormonal actions are exerted. Vitamins are chemical compounds, a small quantity of which is essential for the survival of certain organisms. Unlike hormones, most vitamins cannot be synthesized by the body but must be obtained from external sources (i.e. diet). Like hormones, vitamins are necessary in only small quantities to ensure proper body functioning. Vitamins ordinarily function as coenzymes, which are non-protein portions of enzymes that help in binding the enzyme to the substrate. Enzymes are organic compounds that function as bio-chemical catalysts produced by living organisms, and are primarily constituted of protein. Unlike hormones which may travel considerable distances to target sites, enzymes generally are used at, or very near to, their site of production. For example, pancreatic enzymes exert their effect in the duodenum, which is nearby the pancreas. Protein hormones that act through the adenylcyclasesystem generally elicit their responses by activating certain enzymes in their target cells. Whereas enzymes are always made up of protein, hormones may consist of amino acids or steroids. Also, enzymes are not released into the blood as are hormones.

Question:

A wire 0.5 mm in diameter is stretched along the axis of a cylinder 5 cm in diameter and 25 cm in length. The wire is maintained at a temperature of 750\textdegreeK by passing a current through it, the cylinder is kept at 250\textdegreeK, and the gas in it has a thermal conductivity of 6 × 10^-5 cal\bulletcm^-1\bulletC deg^-1\bulletsec^-1. Find the rates at which the heat is dissipated both by conduction through the gas and by radiation, if the wire is perfectly black.

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/Users/wenhuchen/Documents/Crawler/Physics/D16-0549.htm

Solution:

The heat flow due to conductivity through a hollow cylinder is given by H = [2\piKL(t_2 - t_1)] / [1n (R_2/R_1)] where R_1 and R_2 are the inner and outer radii of the cylinder, L is its length, and K is the coefficient of thermal conductivity of the material of which the cylinder is composed. Also, t_1 is the cylinder temper-ature at its inner radius, and similarly for t_2. In this case, therefore H = [2\pi × 6 × 10^-5cal \bullet cm^-1 \bullet C deg^-1 \bullet s × 25 cm × (750-250)C deg] / [1n (2.5 cm/0.025 cm)] = 1.02 cal \bullet s^-1. The rate of emission of energy per unit area by radiation is the net outflow according to Stefan's Law. Thus, in the usual notation, R = e \sigma (T^4 - T^4_0) where T is the absolute temperature of the outer cylinder surface, and similarly for T_0, e is the emissivity of the cylinder's surface, and \sigma is a constant. However, H' = RA = e \sigma A(T^4 - T^4_0) where A is the area of the surface which emits the radiation. Since the wire is emitting the energy, and it is to be considered a black body, e = 1 andA = 2\pi R_1 L H' = [5.67 × 10^-12W \bullet cm^-2 \bullet Kdeg^-4 × 2\pi × 0.25 cm × 25 cm(750^4 - 250^4)(K deg)^4] / [4.186 J\bullet cal^-1] = 1.67 cal \bullet s^-1.

Question:

How many ounces of silver alloy which is 28% silver must be mixed with 24 ounces of silver which is 8% silver to produce a new alloy which is 20% silver?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E33-0945.htm

Solution:

Let x = number of ounces of 28% silver to be used. The relationship used to set up the equation is Volume of 28% silver + Volume of 8% silver = Volume of silver in mixture. .28x + .08(24) = .20(x + 24) 28x + 8(24)= 20 (x + 24) 8x= 288 x= 36 ounces of silver Check: Volume of 28% silver= (.28) (36) = 10.08 Volume of 8% silver= (.08) (24) = 1.92 Total amount of silver = 12 ounces The total mixture contains 24 + 36 = 60 ounces, and 12 ounces is 20% of 60 ounces.

Question:

The analysis of a sample of soil weighing 0.4210 g yields a mixture ofKClandNaClweighing 0.0699 g. From theKCl, 0.060 gAgClis precipitated out. Calculate the percent of KCl in the soil.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E27-0892.htm

Solution:

To solve this problem (1)setup the equation for the reaction ofKCltoAgCl (2)determinethe number of moles ofAgClprecipitated (3)fromthestoichiometryof equation determine how many moles of KClwere consumed (4)calculatethe weight ofKClconsumed (5)calculatethe percentage ofKClin the soil by dividing the weight ofKClby the total weight of soil and multiplying by 100. The equation for the reaction is KCl + Ag^+ \ding{217}AgCl+ K+ The number of moles ofAgClis (weightofAgCl) / (MW ofAgCl) = [(0.60 g) / (143.5 g / mole)] = 4,18× 10^-4 moles From thestoichiometryof the equation, 4.18 × 10^-4 moles ofKClis consumed . The weight ofKClconsumed is the number of moles ofKCltimes its molecular weight. Thus, weightofKCl= (4.18 × 10^-4 moles) (74.6 g / mole) = 0.0312 g The percentage ofKClin the soil is [(weight ofKCl) / (weight of soil)] × 100 = (0.0312 gKCl) / (0.4210 g Soil) = 7.41 %.

Question:

Two men, each of whom weighs 150 lb, stand opposite each other on the rim of a small uniform circular platform which weighs 900 lb. Each man simultaneously walks clockwise and at a fixed speed once around the rim. The platform is free to rotate about a vertical axis through its center. Find the angle in space through which each man has turned. (The platform's moment of inertia is I = (1/2)MR^2 , where M is its mass).

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Solution:

The figure (parts (a) through (c)) illustrates various posi-tions of the 2 men, and the platform, relative to an observer on the ground. Our ultimate goal is to find an equation, for each man, de-scribing his angular position, \texttheta, as a function of time, relative to his initial position figure (a)). It is equally acceptable to find an equation for the angular velocity \omega^\ding{217} of each man, since this may be integrated to find \texttheta. Rather than jumping right into a dynamic analysis of this problem, it might be worth our while to see if we can use any conservation relations in solving this problem. Note that the angular momentum of the system consisting of the 2 men and platform is constant in time, since no external torques (i.e., friction) act on the system. Furthermore, using angular momentum as our conserved quantity will give us a relation between the angular kinematical variables (\alpha,\omega,\texttheta) at 2 times during the motion of the system. We take these 2 times to be as illustrated in figures (a) and (b). The initial angular momentum of the system, L_0, is zero, since figure (a) shows the system at rest. The final angular momentum, L_F^\ding{217}, is due to the angular momentum of each man(L_1^\ding{217}.L_2^\ding{217}) plus the platform's angular moment-um, L_p^\ding{217}. Hence L_f^\ding{217} = L_1^\ding{217} + L_2^\ding{217} + L_p^\ding{217}(1) But, by the definition of the angular momentum of a particle about point C, L_1^\ding{217} = r_1^\ding{217} × p_1^\ding{217} L_2^\ding{217} = r_2^\ding{217} × p_2^\ding{217} where p_1^\ding{217} and p_2^\ding{217} are the linear momentum of each men, and r_1^\ding{217} and r_2^\ding{217} are as illustrated in figure (a). Also, form figure (b) r_1^\ding{217} = -r_2^\ding{217} p_1^\ding{217} = -p_2^\ding{217} since each man walks with the same velocity. Then L_1^\ding{217} = r_1^\ding{217} × p_1^\ding{217} andL_2^\ding{217} = r_1^\ding{217} × p_1^\ding{217}(2) Furthermore, the angular momentum of the platform is L_p^\ding{217} = I \omega^\ding{217}(3) where \omega^\ding{217} and I are its angular velocity and moment of inertia relative to C. (See figure (a)). Using (2) and (3) in (1) Lf^\ding{217} = 2r_1^\ding{217} × p_1^\ding{217} + I \omega^\ding{217}(4) But, the momentum of particle l(p_1^\ding{217}) is related to its angular velocity (\omega_1^\ding{217}) relative to an external observer by p_1^\ding{217} = m_1\omega_1^\ding{217} × r_1^\ding{217} Therefore, r_1^\ding{217} × p_1^\ding{217} = m_1r_1^\ding{217} ×(\omega_1^\ding{217} × r_1^\ding{217})(5) Since, for any vectors A^\ding{217}, B^\ding{217} and C^\ding{217} A^\ding{217} × (B^\ding{217} × C^\ding{217}) = (A^\ding{217} \bullet C^\ding{217})B^\ding{217} - (A^\ding{217} \bullet B^\ding{217})C^\ding{217} we obtain, using (5) r_1^\ding{217} × p_1^\ding{217} = m_1r_1^\ding{217} ×(\omega_1^\ding{217} × r_1^\ding{217}) = m_1r^2_1 (\omega_1\ding{217}) - m_1(r_1\ding{217}\bullet \omega_1\ding{217})r_1^\ding{217}(6) \ding{217} \ding{217} \ding{217} ^\ding{217} ^\ding{217} Now (\omega_1^\ding{217}). is perpendicular to the plane of the platform, which contains r_1^\ding{217}. Then \omega_1^\ding{217} \bullet r_1^\ding{217} = 0 and (6) becomes r_1^\ding{217} × p_1^\ding{217} = m_1r_1^\ding{217} × (\omega_1^\ding{217} × r_1^\ding{217}) = m_1r^2_1 \omega_1^\ding{217}(7) Using (7) in (4) L_f^\ding{217} = 2m_1r^2_1 \omega_1^\ding{217} + I\omega^\ding{217}(8) By the principle of conservation of angular momentum L_f^\ding{217} = L_0^\ding{217} = 0 and (8) yields 2m_1r^2_1 \omega_1^\ding{217} + I\omega^\ding{217} = 0 or \omega_1^\ding{217} = [(-I)/(2m_1r^2_1)] \omega^\ding{217}(9) Since each man travels with the same speed, \omega_1^\ding{217} = \omega_2^\ding{217} = \omega'^\ding{217}(10) \omega_1^\ding{217} = \omega_2^\ding{217} = \omega'^\ding{217}(10) Also, the masses of the 2 men are equal, and m_1 = m_2 = m.(11) Noting that r_1 equals r_2, and that each equals the platform's radius, R, we obtain r_1 = r_2 = R.(12) Using (10) through (12) in (9) \omega'^\ding{217} = [(-I)/(2mR^2)] \omega^\ding{217}(11) Now \omega = (d\texttheta)/(dt) and \omega' = (d\texttheta')/(dt) by definition, where \texttheta and \texttheta' are the angular positions of the plat-form and either man relative to an outside observer. Then (d\texttheta')/(dt) = [(-I)/(2mR^2)][(d\texttheta)/(dt) or \int^\texttheta'_\texttheta'0 d\texttheta'= [(-I)/(2mR^2)] \int^\texttheta_\texttheta0 d\texttheta where\texttheta' = \texttheta'_0 and \texttheta = \texttheta_0 at t = 0. (figure (a)). Finally \texttheta' - \texttheta'_0 = [(-I)/(2mR^2)](\texttheta - \texttheta_0) This relates the net angle traversed by each man (relative to an observer on the ground) to the net angle which the platform rotates through. Using the given data \texttheta' - \texttheta'_0 = -[{(1/2)(MR^2)}/(2mR^2)](\texttheta - \texttheta_0) = (-M)/(4m)(\texttheta - \texttheta_0) \texttheta' - \texttheta'_0 = [(-Mg)/(4mg)](\texttheta - \texttheta_0) = [(-900 lb)/(600 lb)](\texttheta - \texttheta_0) \texttheta' - \texttheta'_0 = -(3/2)(\texttheta - \texttheta_0)(12) Each man makes 1 revolution (or traverses 2\pi radians) relative to the disc. (See figure (c) ). Then, relative to an outside observer, the man traverses an angle (\texttheta' - \texttheta'_0) equal to 2\pi; plus the angle the disc turns through relative to him (\texttheta - \texttheta_0). Hence, (\texttheta' - \texttheta'_0) = (\texttheta - \texttheta_0) + 2\pi(13) Using (12) in (13) (\texttheta' - \texttheta'_0) = -(2/3)(\texttheta' - \texttheta'_0) + 2\pi (\texttheta' - \texttheta'_0) = -(2/3)(\texttheta' - \texttheta'_0) + 2\pi (5/3)(\texttheta' - \texttheta'_0) = 2\pi \texttheta' - \texttheta'_0 = (6\pi)/(5) = 216\textdegree Thus \texttheta - \texttheta_0 = -(2/3)(\texttheta' - \texttheta'_0) = -144\textdegree The negative sign indicates that the men move in a direction opposite to the direction of motion of the disc.

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Question:

An incident wavefront of light makes an angle of 60\textdegree with the surface of a pool of water. The speed of light in water is 2.3 × 10^8 m/s. What angle does the refracted wavefront make with the surface of water?

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Solution:

The angle between \texttheta_i the incident ray and the normal to the surface (as shown in the figure),equals the angle between the incident wavefront and the water surface, and \texttheta_i = 60\textdegree. Snell's Law, relating the angle of incidence, \texttheta_i , to the angle of refraction, \texttheta_r of the light, is n_1 sin \texttheta_i = n_2 sin \texttheta_r where n_1 and n_2 are the refractive indices of air and water, respectively. Hence sin \texttheta_r = (n_1/n_2) sin \texttheta_i But n_1 = [{speed of light (vacuum)}/{ speed of light (air)}]n_2 = [{speed of light (vacuum)}/{ speed of light (water)}] Hence sin \texttheta_r = [(2.3 × 10^8 m/s)/(3 × 10^8 m/s)] sin 60\textdegree = .664 or\texttheta_r = 42\textdegree. \texttheta_r also equals the angle the refracted wavefront makes with the water surface.

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Question:

Anucleusof mass M emits a \Upsilon ray of energy E_\Upsilon^\ding{217} . The nucleuswas initially at rest . What is its recoil energy after the emission?

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/Users/wenhuchen/Documents/Crawler/Physics/D36-1044.htm

Solution:

From momentum conservation we have 0 =p_\Upsilon+p_nuc In addition we know that E=h\Elzpscrv=hc/\lambda and by thedeBroglierelation p = h/\lambda \textbullet Therefore we have E = pc and p_\Elzpbgam = E_\Elzpbgam/c. Hence, ▏p_nuc ▏= E_\Upsilon / c . The recoil energy is given by K_nuc = 1/2 Mv^2 = (M^2v^2)/ (2M) = (Mv)2/ 2M= (p^2_nuc) / 2M \textbackslash = E_\Elzpbgam^2/ 2Mc^2 The last equality involves substituting E_\Elzpbgam/c for P_nuc . We have assumed that the recoil velocity may be treated as nonrelativistic.

Question:

Calculate the percent composition by weight of ether (C_2H_5)_2O.

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Solution:

To understand this problem, one must first realize that the subscript 2 outside the brackets means that all elements inside the brackets are multiplied by 2. Carrying this procedure out, one obtains the formula Cu_4H_10O for ether. Percent composition is found by dividing the individual weights of the components by the molecular weight. Therefore, first find molecular weight by using the table below. Elements Number of Atoms Atomic Weight Total AtomicWeight C 4 12 48 H 10 1 10 O 1 16 16 M.W. = 48 + 10 + 16 = 74 g/mole. Percent composition is then calculated from the formula % composition = [(total atomic weight) / (M.W.)] × 100 % C = (48/74) × 100 = 64.86 % % H = (10/74) × 100 = 13.51 % % O = (16/74) × 100 = 21.63 %. To double check, make sure that the total percent equals 100.

Question:

esterification, saponification, elastomer and nitration.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E21-0772.htm

Solution:

Esterification is the process that occurs during the reaction of an alcohol (R'-OH) with a carboxylic acid (R-COOH) to produce an ester . It can be illustrated as follows in the formation of the ester ethyl acetate Saponification is the term used to describe those reactions of an esterwith a base to produce soft and hard soaps. The process can be illustrated by the formation of sodium stearate (C_17H_35COO)_3C_3H_5 + 3NaOH \rightarrow 3C_17H_35COONa + C_3H_5(OH)_3 Ester (tristearin)BaseSoap (sodiumGlycerol stearate) Elastomer is the general name of all rubber sub-stitutes. They tend to be better than natural rubber. For example, most elastomers are more oil resistant and some are less permeable than rubber. Nitration:This is the name of the reaction in which nitric acid (HNO_3) is mixed with sulfuric acid (H_2SO_4) and added to a benzene ring, such that an NO_2 is substituted for a hydrogen atom attached to the ring to form nitro-benzene (fig. A).

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Question:

What is the de Broglie wavelength of an electron with a kineticenergy of 1eV?

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Solution:

When calculating the de Broglie wavelength, it is important to know whetheror not to use relativistic formulas in order to calculate the quantitiesappearing in the de Broglie formula. One way of deciding this is torealize that, when the velocity v is very small compared with c, (and, hence, the problem is non-relativistic) the kinetic energy is small when comparedto the rest mass energy, m_0c^2 . The rest mass energy of an electronis 0.512 × 10^6eVwhich is large compared with 1eV. We need notuse relativistic formulas in the present example. In the formula for the de Broglie wavelength, \lambda = (h/mv)(1) itis permissible to insert the rest mass, m_0, for m, m\approx m_0 = 9.11 × ^10-28 gram. The kinetic energy is (1/2)m_0v^2= 1eV. Because 1eV= 1.6 × 10^-12 erg (1/2)m_0v^2 = 1.6 × 10^-12 erg. Therefore v =\surd[(2 × 1.6 × 10^-12 erg)/m_0] =\surd[(2 × 1.6 × 10^-12 erg)/(9.11 × 10^-28 g)] = 5.92 × 10^7 cm/sec. Notice that this is, in fact, very much less than the velo-city of light (3 × 10^10 cm/sec). Inserting these values of m and v into equation (1), \lambda = [(6.625 × 10^-27 erg sec)/{(9.11 × 10^-28 × 5.92 × 10^7)g \textbullet cm/sec}] = 1.23 × 10^-7 cm. This is about ten times larger than the diameter of an atom.

Question:

Give examples of valid variable names in SNOBOL IV. List the basic statements and briefly explain their operations.

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Solution:

The variable names in SNOBOL IV must begin with a letter which may be followed by any number of letters, digits, periods and underscores. Some examples of valid variable names are listed below: A, B1, VARIABLE, D987654321, Y0U.OR.ME, K6.J2 The basic statements are as follows: variable: variable = INPUT - Assigns the first read in string to a variable name. OUTPUT = string or variable - Prints out a given string, PUNCH = string - Punches the given string, variable = string - Assigns the given string to a variable.

Question:

TheRiverdellEnglewoods, a little League team of the future, is offering a prize to anyone who can write a FORTRAN program to keep track of the box scores of their games. An underfinanced team, theEnglewoodshave an ancient card-punching machine in the dugout. Each time an Englewood comes to bat, the scorekeeper prepares a card with the following items: 1)the player's number (2 digits) 2)the result of his at-bat (0 for out, 1 for a single, etc., up to 4 for a home run) 3)whether the player scored a run (either 0 or 1) 4)the number of runs batted in The manager wants the program to output the player's number, the number of at-bats, hits, runs scored, and runs batted in for each game. After the last player has been listed, compute and output the team totals.

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Solution:

We assume that there is a total of 20 players on the team numbered 1 through 20. Thus, we need a loop to compute the required statistics for each player. After each computation, we store the elements in arrays so that team totals may be computed at the end. Variable names are chosen to match closely the baseball terminology. The program stores each player's totals until a card containing all zeros is encountered. This signifies the end of the deck, whereupon the output procedure begins, completed with appropriate headings. CBOX SCORES FOR ENGLEWOODS INTEGER PLAYNO, EFORT, RUNSCO, RBI INTEGER ATBAT, HITNO, RUNNO, RBINO INTEGER TOTBAT, TOTHIT, TOTRUN, TOTRBI DIMENSION ATBAT(20), HITNO(20), RUNNO(20), RBINO(20) CINITIALIZE ALL VARIABLES TO ZERO DATA PLAYNO, EFORT, RUNSCO, RBI/0,0,0,0/ DATA ATBAT, HITNO, RUNNO, RBINO/20{_\ast}0 , 20{_\ast}0, 20{_\ast}0 , 20{_\ast}0/ DATA TOTBAT, TOTHIT, TOTRUN, TOTRBI/0,0,0,0/ CREAD CARDS 10READ (2,100) PLAYNO, EFORT, RUNSCO, RBI 100FORMAT (12,3I1) CDO WHILE PLAYNO IS NOT EQUAL TO ZERO IF (PLAYNO. EQ. 0) GO TO 20 ATBAT (PLAYNO) = ATBAT (PLAYNO) + 1 IF (EFORT. NE. 0) HITNO (PLAYNO) = HITNO (PLAYNO) + 1 RUNNO (PLAYNO) = RUNNO (PLAYNO) + RUNSCO RBINO (PLAYNO) = RBINO (PLAYNO) + RBI GO TO 10 CPRINT HEADINGS WRITE (6,101) 101FORMAT(1X, 'PLAYER NUMBER',5X, 'AT BATS' ,5X, 1'HITS',5X,'RUNS',5X,'RBIS') 20DO 25 J = 1,20 IF (ATBAT (J) .EQ. 0) GO TO 25 WRITE (6,102) J, ATBAT (J), EFORT, RUNSCO, RBI 102FORMAT (7X,I2,13X,I1,11X,I1,2(8X,I1)/) TOTBAT = TOTBAT + ATBAT (J) TOTHIT = TOTHIT + HITNO (J) TOTRUN = TOTRUN + RUNNO (J) TOTRBI = TOTRBI + RBINO (J) 25CONTINUE WRITE (6,103) TOTBAT, TOTHIT, TOTRUN, TOTRBI 103FORMAT (1X, 'TEAM TOTALS',9X, I2,11X, I2,2 (8X, 12) ) STOP END

Question:

Certain substances are maintained in the blood at a high threshold while others are held at a very low threshold. What factor determines what the kidney threshold for a substance will be? Which substances characterize each level?

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Solution:

The amounts of the individual substances dissolved in the blood exhibit wide variation. Some substances are high threshold substances, which means that they are maintained in the blood at a high concentration relative to their concentration in the urine. Substances maintained at such levels include glucose, amino acids, fatty acids, vitamins, and hormones. These are filtered out of the blood at theglomerulusand almost fully reabsorbed by the tubules. Recall that it is through active transport by the cells of the proximal convoluted tubule thatreabsorptiontakes place. The threshold level ofreabsorptionis thus a function of the rate of active transport across the tubule surface. Only if the concentration of a high threshold substance in the blood is very great will the amount filtered surpass the amount reabsorbed and cause the substance to spill into the urine. Substances with high threshold values are essential nutrients or other compounds needed in body metabolism. Low threshold substances are those having a high concentration in the urine relative to their concentration in the blood. Only a small fraction of these substances is reabsorbed in the tubules. Low threshold substances are mostly waste products, such as urea; the concentration of urea in the urine may be up to 70 times that in the blood. Various drugs, such as aspirin and antibiotics, are also included in this group. This is not always advanta-geous, for the kidneys may remove a drug so fast that continued application is needed for maximal effectiveness. This problem is solved in some cases by mixing the medication with a binder so that it is released slowly over an extended period of time. Substances having medium thresholds include inorganic mineral ions and certain organic substances. Such substances are roughly equally reabsorbed and excreted. Important ions in this group are the sodium and chloride ions. These ions are involved in the maintenance of the pH of the blood (7.3) by their association with hydrogen and hydroxyl ions. The cellular metabolism of food particles most often results in an excess of hydrogen ions; these are then excreted along with chloride ions, resulting in a slightly acidic urine.

Question:

Describe the evolution of the reptilian excretory system to account for the transition from an aquatic to a terrestrial habitat.

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Solution:

The reptilian excretory system has evolved in a way that enables the animal to conserve most of its water. The conservation of body fluid is a necessary characteristic of a terrestrial animal. Without this feature, the body tissues would become dessicated in the dry environment. Reptiles conserve water by having a coarse, dry, horny skin. In addition, the glomeruli of the reptilian kidney have diminished in size so that less water is filtered from the blood. Another modification is a greater degree of reabsorption of water from the glomerular filtrate by the kidney tubules. This occurs as the rep-tiles have evolved kidney tubules with two highly coiled regions and a long loop of Henle extending deep into the medulla of the kidney. These long portions of the tubule function in the reabsorption of water. Their ability to produce a concentrated hypertonic excretory product is an important adaption to land life. The conservation of water in the reptile has affected the nature of the excretory product. Nitrogenous wastes are excreted as uric acid in reptiles, as opposed to urea in amphibians (whose ancestors gave rise to primitive reptiles). Uric acid is excreted as a watery paste in reptiles. It is less toxic and less soluble in water than urea. Therefore, it is not necessary for reptiles to use water to dissolve their nitrogenous wastes. In fact, only a comparatively very small amount is required to simply flush the uric acid out of the excretory system. In this way, the reptile conserves most of its body water and prevents dehydration.

Question:

Evaluate 16^\rule{1em}{1pt}3/4.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E33-0953.htm

Solution:

16^\rule{1em}{1pt}3/4= 1/(16^3/4) = [1/(^4\surd16)^3 Note that2^4= 2\bullet2\bullet2\bullet2 = 16, hence ^4\surd16 = 2. Thus, 16^\rule{1em}{1pt}3/4. = 1/2^3 = 1/(2\bullet2\bullet2) = 1/8.

Question:

An E. coli cell is 1.0\mu in diameter and 2.0\mu long and may be assumed to be cylindrical. It contains 80% water. If the intracellular pH is 6.4, calculate the number of H^+ ions in a single cell.

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Solution:

One can find the number of H^+ ions in one E. coli cell from the dimensions of the cell and the pH, because pH = -log [H^+].[H^+] is found in moles per liter and the volume of the cell can be calculated in liters. Solving for [H^+]: 6.4 = - log [H^+] [H^+] = 10^-6.4 = 3.98 × 10^-7 moles / liter To determine the number of ions in the cell, the number of moles present must be determined first. This can be found from the volume. The volume of a cylinder is equal to l\pir^2, where l is the length and r is radius of the cross-section, 1\mu = 10^-4 cm. Volume of the cell = (2 × 10^-4 cm) (\pi) (0.5 × 10^-4 cm)^2 = 1.57 × 10^-12 cm^3 = 1.57 × 10^-15 l If 80% of the cell is H_2O, 80% of the volume equals the volume of H_2O present. The H^+ ions can only exist in the aqueous phase of the cell. Volume of H_2O = .80 × 1.57 × 10^-15 l = 1.26 × 10^-15 l To solve for the number of moles of H^+, multiply [H^+] by the volume of H_2O present. No. of moles of H^+ = 1.26 × 10^-15 l × 3.98 × 10^-7 moles / l = 5.01 × 10^-22 moles. There are 6.02 × 10^23 ions per mole. No. of H^+ ions = 6.02 × 10^23 (ions / mole) × 5.01 × 10^-22 mole = 3.02 × 10^2 ions = 302 H^+ ions.

Question:

Compare the processes of digestion in the hydra, flatworm and earthworm.

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/Users/wenhuchen/Documents/Crawler/Biology/F17-0419.htm

Solution:

Hydra is a member of the coelenterate phylum. Hydra is a multicellular, sac-like organism whose body wall is two cells thick. Many of the cells are semi-specialized, differing in function from their immediate neighbors. The hydra's digestive system, the gastrovascular cavity, has only one opening-the mouth (refer to Figure 1) . The tenta-cles of the hydra have special cells, called nematocysts, that spear tiny organisms swimming near the hydra. The ten-tacles then draw the prey into the mouth. The cells lining the gastrovascular cavity begin digestion by secreting en-zymes. This mode of digestion, occuring outside cells, is extracellular digestion. Extracellular digestion is followed by intracellular digestion in the hydra. When small food fragments come into contact with the cells lining the cavities, these fragments are phagocytized and digestion is completed inside the cell. The indigestible material is expelled through the mouth. Thus, the hydra utilizes extracellular digestion to break down relatively large organisms into small pieces which are then digested intracellularly by individual cells. Flatworms are small animals that are often parasitic. Parasitic flatworms have lost most of their ability to procure food and are provided with nutrients by their host. However, some flatworms such as planaria are non-parasitic and are free-living organisms. The mouth of the planaria is on the ventral side, in the middle of the body (refer to Figure 2). It opens into a tubular muscular pharynx that can protrude through the mouth directly onto the prey. The pharynx opens into a gastrovascular cavity similar to those other coelenterates but much more branched. The extensive branching brings nutrients to all parts of the body and helps increase the absorptive surface area of the cavity. Some extracellular digestion occurs in the cavity but most food is phagocytized and digested intracellularly in the cells lining the cavity. An earthworm has a complete digestive tract. It has two openings, the mouth and anus, (refer to Figure 3), with wastes expelled through the anus and food ingested through the mouth. Various regions of the tract are specialized: the pharynx to suck in food, the esophagus to transport food to the crop, the crop to store the food, the gizzard to break the food up, and the intestine to digest and absorb the food. Most digestion is extracellular, utilizing the enzymes secreted by cells lining the tract. The digestive tract is not widely branched, but is rather like a simple gastrovascular cavity. However, large numbers of folds are found in the tract lining, which greatly increase the absorptive surface area. Since the earthworm is a land animal, it must also conserve its body fluids by reabsorbing water from the indigestible wastes. This occurs in the posterior region of the intestine, analogous to the human large intestine.

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Question:

In an electrolytic cell, a liter of a 1M aqueous solution of MnO_4^- is reduced at the cathode. Determine the number of faradays required for each of the following to be made a) a solution that is .01 M MnO_4^2- ; b) 1 gram of MnO_2 ; c) 1 gram-equivalent of Mn metal.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E16-0565.htm

Solution:

For each part, you need to write the reaction that occurs. The masses of substances produced during the process are proportional to their equivalent weights. The unit electrical equivalent = 1 Faraday or 1F. One Faraday = 96,500 coul and is capable of reducing one equivalent positive charge, an equivalent positive charge contains Avogadro's number of individual unit charges. You proceed as follows: reaction is MnO_4^- + e^- \rightarrow Mn_4^2-.One electron is being transferred. You want [MnO_4^2-] = .01 M . Since the solution has a volume of 1 liter, .01 M = .01 moles, since M = molarity = moles / liter. Due to the fact that only 1 electron is transferred, only one Faraday is needed to reduce a mole of MnO_4^-, since it would require only a mole of electrons. You need only, however, .01 moles of MhO_4^2- . Since a mole of MnO_4^2- is formed from every mole of MnO_4^- oxidized you need to reduce .01 moles of MnO_4^2- . To do this, requires, therefore, .01 F . Mn0_4^- + 4H^+ + 3e^- \rightarrow MnO_2 + 2H-_2O . Here, 3 electrons are required to reduce MnO_4^- to MnO_2 . Thus, one Faraday can only reduce (1 / 3) of a mole of MnO_4^-,. Recall, one Faraday is Avogadro's number of electrons and this reaction requires three moles, or 3F per mole of MnO_4^-. You want lg of MnO_2 (atomic wt. = 86.94 g / mole), which is (1g) / (86.94g / mole) or (.0115) moles of MnO_2 . Since one mole of MnO_2 will be produced for every mole of MnO_4^- , you must reduce .0115 moles of MnO_4^- . One faraday will reduce (1 / 3) of a mole of MnO_4^- . To reduce all of it, you want 3 Faraday's / mole. Thus, number of Faraday's required = (3F / mole)(.0115 mole) = .0345F . MnO_4^- + 8H^++ 7e^- \rightarrow Mn + 4H_2O to occur. 7 electrons are to be transferred. Thus, 1 Faraday can only reduce (1 / 7) mole of material. However, you want 1 gram-equivalent of material, which is the amount deposited by 1F. Hence, 7 equiv. of Mn = 1 mole Mn. Thus, to deposit this amount, you use (1.00 equiv. Mn)(F / equiv. Mn) = 1.00F.

Question:

A coil of 600 turns is threaded by a flux of 8.0 × 10^-5 weber. If the flux is reduced to 3.0 × 10^-5 weber in 0.015 sec, what is the average induced EMF?

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/Users/wenhuchen/Documents/Crawler/Physics/D22-0741.htm

Solution:

We note that there is a change in magnetic flux. This immediately implies the use of Faraday's Law, which relates a change in magnetic flux to an induced E.M.F. (electromotive force). The flux linking the coil will induce an E.M.F. in the coil. Faraday's Law states: E.M.F. = [(-Nd\varphiB)/dt](1) where N is the number of turns in the coil, and \varphi_B is the magnetic flux linking the coil. We may write this using average values as E.M.F. = - N [(∆\varphiB)/∆t](2) where ∆\varphi_B is the change in magnetic flux over the interval ∆t, andE.M.F. is the average value of the resulting E.M.F. Substituting the values given in the statement of the problem into equation (2) , we obtain E.M.F. = - [(600) (8 × 10^-5 weber - 3 × 10^-5 weber)]/(.015 s) = - 2 [(weber)/(sec)] But1 weber = 1 [(Joule - sec)/(coul)] Therefore, E.M.F . = - 2 [(Joule - sec)/(coul - sec)] = - 2 Volts

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Question:

Four moles of PCl_5 are placed in a 2-liter flask. When the following equilibrium is established PCI_5\rightleftarrows PCI_3 + CI_2 , the flask is found to contain 0.8 mole of Cl_2 . What is the equili-brium constant?

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Solution:

The equilibrium constant (Keq) for this reaction is: Keq= {[PCI_3 ] [CI_2]} / [PCI_5] where [ ] indicate concentrations. One is given that, at equilibrium, there are 0.8 mole of Cl_2 present. From the equation for the re-action, it follows that there is 0.8 mole of PCl_3 present also. This means that 0.8 moles of PCl_5 has reacted. Originally 4 moles of PCI_5 were present, thus, at equilibrium, 4.0\Elzbar 0.8 moles or 3.2 moles are present. These components are present in a two-liter flask. Therefore, the concentrations will be expressed in moles/2 liters. One can now solve for the equilibrium constant. Keq= {[PCI_3] [CI_2] } / [PCI_5][PCI_3] = 0.8 moles/2 liters [CI_2]= 0.8 moles/2 liters [PCI_5] = 3.2 moles/2 liters Keq={(0.8/2) (0.8/2)} / {(3.2/2)} = {(0.4M) (0.4M) = {(0.4M) (0.4M) } / {(1.6M)} = (0.16M^2) / (1.6M) = 0.1M.

Question:

A spacecraft, with a mass m = 2000kg, rotates with an angular velocity of \cyrchar\cyromega = (0.05 rad / s)\^{\i} + (0.15 rad / s)\^{\j}. Two small jets, located at points A and B, are turned on in a direction paral-lel to the z-axis. Each jet has a thrust of 25N. The radii of gyration of the spacecraft are k_x = k_z = 1.5m and k_y = 1.75m. Determine the required time of operation for each jet, so that the angular velocity of the spacecraft reduces to zero.

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Solution:

Choose a coordinate system fixed in the body such that the rockets at A and B lie in the x, y plane and the origin is at the center of mass. The positions of A and B are r^\ding{217}_1 and r^\ding{217}_2 respectively. Qualitatively, what happens is both the rockets oppose the original motion about the y axis but B will increase the original motion about the x- axis while A will oppose it. Thus, A will have to be fired longer in order to compensate for the increase about the x- axis due to B. The torque, \tau^\ding{217}, produced by A and B is \tau^\ding{217}_B = r^\ding{217}_1 × F^\ding{217}_1 = (y_1F_1i^\ding{217}) - (x_1F_1y^\ding{217})(1) \tau^\ding{217}_A = r^\ding{217}_2 × F^\ding{217}_2 = -(y_2F_2i^\ding{217}) - (x_2F_2j^\ding{217}). From Newton 's Second Law, \tau^\ding{217} = l\alpha^\ding{217} where I is the moment of inertia and \alpha^\ding{217} is the angular acceleration. In components this becomes \tau^\ding{217} = I_x\alpha_xi^\ding{217} + I_y\alpha_yj^\ding{217} + I_z\alpha_zk^\ding{217}(2) where I_x , I_y and I_z are the moments of inertia about the x, y, z axes. Comparing equations (2) and (3) gives \alphaA_x = [(d\cyrchar\cyromegaA x )/(dt)] = -[(y_2F_2) / I_x] = -c_1; [(d\cyrchar\cyromegaA_y)/(dt)] = -[(x_2F_2 ) / I_y] = -c2;; [(d\cyrchar\cyromegaA_z) / (dt)] = 0(3) [(d\cyrchar\cyromegaB_x)/(dt)] = -[(y_1F_1 ) / I_x] = c_3; [(d\cyrchar\cyromegaB_y)/(dt) = -[(x_1F_1 ) / I_y] = -c_4 ; [(d\cyrchar\cyromegaB_z) / (dt)] = 0 These equations can be integrated directly to give ∆\cyrchar\cyromegaA_x = -c_1t_A ; ∆\cyrchar\cyromegaA_y = -c_2t_A ; \cyrchar\cyromegaA_z = constant(4) ∆\cyrchar\cyromegaB_x = c_3t_B ; ∆\cyrchar\cyromegaB_y = -c_4t_B ; \cyrchar\cyromegaB_z = constant where ∆\cyrchar\cyromega = \cyrchar\cyromega(t) - \cyrchar\cyromega(0) is the change in angular velocity about each axis due to firing the rockets. Since initially \cyrchar\cyromega^\ding{217} = 0.2 (rad / sec)i^\ding{217} + 0.1 (rad /sec)j^\ding{217}; w_z(0) = 0, the constants in equation (4) must also be zero. Since we want the final angular velocity to be zero, the sum of the initial angular momentum plus the additional in-crements about each axis must be zero. So, \cyrchar\cyromega_(x)0 + ∆\cyrchar\cyromegaA_x + ∆\cyrchar\cyromegaB_x = 0 \cyrchar\cyromega_(y)0 + ∆\cyrchar\cyromegaA_y + ∆\cyrchar\cyromegaB_y = 0. Substitution from equation (3) gives (C_1t_A) - (C_3t_B) = \cyrchar\cyromega(x)0 (C_2t_A) + (C_4t_B) = \cyrchar\cyromega_(y)0 These are two simultaneous equations in the unknowns t_A and t_B. From the theory of equations, the solutions are t_A = (1 / ∆)\mid\cyrchar\cyromega_(x)0-c_3\mid= (1 / ∆)[(c_4\cyrchar\cyromega_(x)0) + (c_3\cyrchar\cyromega_(y)0)](5) \mid\cyrchar\cyromega_(y)0c_4\mid t_B = (1 / ∆)\midc_1\cyrchar\cyromega_(x)0\mid= (1 / ∆)[(c_1\cyrchar\cyromega_(y)0) - (c_2\cyrchar\cyromega_(x)0)] \midc_2\cyrchar\cyromega_(y)0\mid where ∆ =\midc_1-c_3\mid= (c_1c_4) + (c_2c_3). \midc_2c_4\mid For this problem x_1 = 2.5 m, y_1 = 4 m, x_2 = 4 m, y_2 = 2.5 m m = 2 × 10^3kg, F_1 = F_2 = 25 n, k_x = k_z = 1.5 m and k_y = 1.75 m. From equation (3) the C values are c1= (y_2F_2) / (I_x) = [(2.5 m 25 n) / {2 × 10^3 kg(1.5 m)^2}] = 0.0139(rad /sec^2) c2= (x_2F_2) / (I_y) = [(4 m 25 n) / {2 × 10^3 kg(1.75 m)^2}] = 0.0163(rad /sec^2) c3= (y_1F_1) / (I_x) = [(4 m 25 n) / {2 × 10^3 kg(1.5 m)^2}] = 0.0222(rad /sec^2) c4= (x_1F_1) / (I_y) = [(2.5 m 25 n) / {(2 × 10^3 kg(1.75 m)^2}] = 0.0102(rad /sec^2) ∆ = (c_1c_4) + (c_2c_3) = 0.000503 (rad / s^4). Substitution in eq. (5) gives the times t_A = [(c_4\cyrchar\cyromega_(x)0) + (c_3\cyrchar\cyromega_(y)0)] / (∆) = [0.0102 × 0.05 + 0.0222 × 0.15] / (0.000503) = 7.634 sec. t_B = [(c_1\cyrchar\cyromega_(y)0) - (c_2\cyrchar\cyromega_(x)0)] / (∆) = [0.0139 × 0.15 - 0.0163 × 0.05] / (0.000503) = 2.525 sec. Thus, it is necessary to fire A approximately three times as long as B in order to stop the capsule's rotation. This agrees with our initial qualitative remarks regarding the motion.

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Question:

Which of the following would maintain a constant body temperature in spite of changes in the environmental temperature: frog, robin, fish, dog, or lizard?

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Solution:

Animals that can regulate their internal body temperature so that it remains constant in varyingenvirinmentaltemperatures are known as homeotherms. Of the phylumChordata, only mammals and birds can thermregulate- i.e., they possess temperature-regulating sys-tems. They are considered to be warm-blooded animals. From the list of animals given, two of them arehomeotherms. The robins, since it is a bird, belongs to the class Aves, and the dog is a member of the class Mammalia. Hence these two animals arehomeothermicand will maintain a constant body temperature. The remaining animals arepoikilotherms. They possess no temperature-regulating system and as a result, their internal body temperature varies directly with the environmental temparature. Thispoikilothermiccharacteristic places strict limitations on the environment exploited by the animal, which in turn affects its behavioral patterns. The frog is an amphibian whereas the lizard is a reptile. A fish belongs to thesuperclasspisces. These three animals are poikilothermsand hence do not maintain a constant body temperature independent of changes in environmental temperature.

Question:

Explain how a change of one base on the DNA can result in sickle-cellhemoglobin rather than normal hemoglobin.

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Solution:

If one base within a portion of DNA comprising a gene is altered, andthat gene is transcribed, the codon for one amino acid on the mRNA willbe dif-ferent. The polypeptide chain which is translated from that mRNA maythen have a different amino acid at this position. Although there are hundredsof amino acids in a protein, a change of only one can have far-reaching effects. The human hemoglobin molecule is composed of two halves of protein, each half formed by two kinds of polypeptide chains, namely an alphachain and a beta chain. In persons suffering from sickle-cell anemia,.amutation has occurred in the gene which forms the beta chain. As a result, a single amino acid substitution occurs, whereglutamicacid is replacedbyvaline. Onecodonforglutamicacid is AUG, while one for valineis UUG. (A = adenine, U = uracil, G = guanine). If U is substituted forA in thecodonforglutamicacid, thecodonwill specifyvaline. Thus, it iseasy to see how a change in a single base has resulted in this amino acidsubstitu-tion. This change is so important, however, that the entire hemoglobinmolecule behaves differently. When the oxygen level in the blooddrops, the altered molecules tend to form end to end associations andthe entire red blood cell is forced out of shape, forming a sickle- shaped body. These may aggregate to block the capillaries. More importantly, thesesickledcells cannot carry oxygen properly and a person havingthem suffers from severe hemolytic anemia, which usually leads to deathearly in life. The condition is known as sickle-cell anemia. In a normal hemoglobin molecule, theglutamicacid residue in question , being charged, is located on the outside of the hemoglobin moleculewhere it is in contact with the aqueous medium. When this residueis replaced byvaline, a non-polar residue, the solubility of the hemoglobinmolecule decreases. In fact, the hydrophobic side chain of valinetends to form weak vanderWaals bonds with other hydrophobic sidechains leading to the aggregation of hemoglobin molecules.

Question:

Delivery trucks which operate by making use of the energy stored in a rotating flywheel have been in use for some time in Germany. The trucks are "charged up" before leaving by using an electric motor to get the flywheel up to its top speed of 6000 rev\bulletmin^-1. If one such flywheel is a solid homogeneous cylinder of weight 1120 lb and diameter 6 ft, how long can the truck operate before returning to its base for "recharging\textquotedblright, if its average power requirement is 10 hp?

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Solution:

The angular speed of the flywheel is \omega = 6000 rev\bulletmin^-1 × 2\pi rev\bulletmin^-1 × 1/60 min\bullets^-1 = 200\pi rev\bullets^-1 The kinetic energy stored in the flywheel is given by E_k = (1/2) I\omega^2 where I is the moment of inertia. For a disc of radius r and mass M, I = (1/2) Mr^2. Therefore, E_k = (1/2) x (1/2) Mr^2 x \omega^2 = (1/4) Mr^2\omega^2 = (1/4) × (1120/32) slugs x 9 ft^2 x 4\pi^2 x 10^4 s^-2 = (63\pi^2)/2 x 10^5ft\bulletlb. The truck consumes this energy at a rate of P = 10 hp = 10 x 550 ft\bulletlb-s^-1, Thus, assuming that the frictional losses are negligible, the truck can work for t = (E_k)/p = (63\pi^2 x 10^5ft\bulletlb)/(2 x 5500 ft\bulletlb\bullets^-1) = [(63\pi2x 10^5)/(2 x 5500 x 60)] min = 94.2 min. The flywheel must therefore be "recharged" before this time has elapsed.

Question:

Describe the light reactions of photosynthesis. What are the products of these reactions?

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Solution:

The essential requirements for the energy - requiring dark reactions of photosynthesis are ATP and NADPH _2. These molecules are produced via the light reac-tions of photosynthesis, which occur entirely within the chloroplasts. This was proven by Daniel Arnon in 1954 who showed that intact chloroplasts carefully isolated from spinach leaves were able to carry out the complete photo-synthetic reaction. Scientists now agree that there are two reaction pathways occurring in the chloroplasts which generate ATP molecules from light energy, one of the pathways being designated cyclic photophosphorylation, the other noncyclic photophosphorylation. Both reactions require the presence of light. Each pathway is a photsystem. A photosystem is an energy-trapping center which receives packets of en-ergy from excited chlorophyll molecules and converts them into a stream of high - energy electrons which can then travel down the electron transport chain. Photosystem I is called P700 due to the wavelength at which chlorophyll a absorbs light (700 nm). Photosystem II is similarly designated P680 due to the response by chlorophyll b. In cyclic photophosphorylation, light striking a chlorophyll a molecule excites one of the electrons to an energy level high enough to allow it to leave the molecule. The chlorophyll a+ molecule, having lost an electron, is now ready to serve as an electron acceptor because of its net positive charge. However, the ejected electron does not return ti its ground state and the chlorophyll a^+ molecule directly; instead it is taken up by ferredoxin and passed along an electron transport chain. As the electron passes from ferredoxin to the cytochromes, to plastocyanin and finally back to chlorophyll a+ , two ATP molecules are produced. In this way, light energy is con-verted into chemical bond energy, and not lost as heat as it would have been had the excited electron returned di-rectly to its ground state. In noncyclic photophosphorylation, oxygen is produced by the photolysis of water. NADPH and ATP molecules are formed. The excitation of chlorophyll a by light at Photosystem I ejects high energy electrons. These electrons pass to ferredoxin and then to NADP^+, reducing it to NADPH. To restore chlorophyll a+ to its original state, another set of chlorophyll molecules known as chlorophyll b come into play at Photosystem II. When light energy is absorbed by chlorophyll b, it also ejects high energy electrons. These pass to an electron ac-ceptor called Q and then via plastoquinone, cytochromes and plastocyanin to chlorophyll a+. During these steps, ADP is phosphorylated to ATP. The electrons necessary to restore chlorophyll b+ to its ground level come from the splitting of water due to the high affinity that chlorophyll b+ has for electrons. The water molecule is thus split into component protons, electrons, and oxygen. The oxygen, after combining with another oxygen atom, is released in gaseous form as O_2. The two elec-trons are picked up by cytochrome b and the hydrogen ions are used in the conversion of NADP^+ to NADPH.

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Question:

What was the Appalachian Revolution? When did itoccur andwhat were its effects?

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Solution:

The Appalachian Revolution refers to the general folding of the earth'scrust which raised the great Appalachian mountain chain running fromNova Scotia to Alabama. It occurred at the end of the Permian Period, the final period of Paleozoic Era, some 280 million years ago.The Appalachian Revolution also brought into existence some of the mountain rangesof Europe. Because of a greatglaciationwhich affected most of thelands existing at that time, the climate became colder and drier. Many ofthe Paleozoic forms of life were unable to adjust to the climatic changes andbecame extinct. Many marine forms were lost during the Appalachian Re-volution, due to the cooling of the water and the decrease in the amountof available space caused by the drying up of the shallow seas. Other new forms of organisms arose, and the Paleozoic Era gave way to the Mesozoic Era.

Question:

One end of a fingernail file is clamped in a vise and the other end is given a to-and-fro vibration. The motion of the free end is approximately S.H.M. If the frequency is 10 vibrations per second and the amplitude is 4 millimeters, what is the velocity when the dis-placement of the free end is 2 millimeters ?

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Solution:

The problem states that the motion is S.H.M. Therefore, we know that the displacement of the file is x = A sin (\omegat+ \alpha)(1) where A is the amplitude and \alpha is a constant.\omegais the angular frequency of the vibration. If f is the frequency of the motion \omega = 2\pif. The velocity of the end of the file is, differentiating (1), v = A \omegacos(\omegat+ \alpha)(2) We need the velocity when x = 2 mm. At this position, using (1) 2 mm= 4 mm sin (\omegat+ \alpha) sin(\omegat+ \alpha)= 1/2 whence (\omegat+ \alpha)= 30\textdegree Hence, using (2), v= A \omegacos(30\textdegree) orv= A (2\pif)cos(30\textdegree) v= (4 mm) (6.28) (10 per sec) {\surd(3/2)} v= [(40 mm)/(sec)] (6.28) (.866) = 218 mm/sec.

Question:

What is the gravitational force on a person whose mass is 70 kg when he is sitting in an automobile that accelerates at 4 m/s^2?

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Solution:

The mass, m = 70 kg, and the acceleration, g = 9.8 m/s^2, are the known observables. The gravitational force on the person is given by Newton's Second Law, F = mg where g is the acceleration due to gravity. F = mg = (70 kg)(9.8 m/s^2) = 6.86 × 10^2N. Although the person will experience the force causing the acceleration, 4 m/s^2, his weight is unaffected by the car's motion. This occurs since the acceleration of the auto-mobile is perpendicular to that caused by gravity and has no effect upon the person in the downward direction.

Question:

Given log_102 = 0.3010, find log_1032.

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Solution:

Note that, 32 = 2 \bullet 2 \bullet 2 \bullet 2 \bullet 2 = 2^5. Thus, log_1032 = log_102^5. Recall the logarithmic property, log_bx^y = y log_bx. Hence,log_1032 = log_102^5= 5 log_102 = 5 (0.3010) = 1.5050

Question:

In microscopy, small spherical bodies are often seen attached to the network of endoplasmic reticulum. What are these bodies? What function do they serve in the cell? The role of the ribodome in protein synthesis.

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/Users/wenhuchen/Documents/Crawler/Biology/F02-0038.htm

Solution:

The small spherical bodies that we see studding the endoplasmic reticulum - more accurately, the rough endoplasmic reticulum - are the ribosomes. The rough endoplasmic reticulum owes its rough appearance to the presence of ribosomes. The smooth endoplasmic reticulum appears smooth because it lacks ribosomes. Ribosomes consist of two parts, a large subunit and a small subunit. Both the large and small subunits are made of proteins and ribonucleic acid (RNA). However, the two subunits differ both in the number and in the type of proteins and RNA they contain. The large subunit contains large and more varied RNA molecules than the small subunit. It also has more protein molecules than the smaller one. An interesting point to note is that when we put together all the chemical components of a ribosome, under favorable conditions, these parts will rearrange themselves and come together, without direction from pre- existing ribosomes, to form a functional assembly. This ability to self- assemble may provide us with a clue to the origin of living things. Ribosomes are the sites of protein synthesis in the cell. Messenger RNA (mRNA), which carries genetic in-formation from the nucleus, associates with the small ribosomal subunit first and then binds to the large sub-unit as a prelude to protein synthesis. This association of mRNA to ribosomes holds the components of the complex system of protein synthesis together in a specific manner for greater efficiency than if they were dispersed freely in the cytoplasm. The mRNA then pairs with complementary molecules of transfer RNA (tRNA), each carrying a specific amino acid. The linking up of tRNA molecules into a chain complementary to mRNA brings together amino acids which bind with each other to form a highly specific protein molecule. Thus ribosomes are the sites where proteins are synthesized under genetic control.

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Question:

How and where is seminal fluid produced in man?

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Solution:

Semen containing seminal fluid and sperm is transferred to the vagina during sexual intercourse. There are three pairs of glands that together contribute to pro-duce the two to five ml of seminal fluid per ejaculation. These are the paired seminal vesicles, the prostate gland, and Cowper's glands. While the contents of semen have been determined, the contributions from each gland are only suggested below. The seminal vesicles secrete mucus and nutrients (such as fructose) into the ejaculatory duct. The single prostate gland secrets a milky, alkaline fluid which increase the pH of the semen. At the base of the erectile tissue of the penis lies a third pair of glands, called Cowper's glands, which contribute more mucus to the seminal fluid. The seminal fluid that results contains glucose and fructose which are metabolized by the sperm for energy, acid-base buffers to protect the sperm from the acidic secretions of the vagina, and mucous materials that lubricate the passages through which the sperm travel.

Question:

Liquid helium boils at 4\textdegreeK. What is the boilingtemperature onthe Fahrenheit scale?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E01-0020.htm

Solution:

The temperature in Kelvin is the temperature in degrees Centigrade addedto 273. In this problem, the boiling point is given in \textdegreeK. Hence, the temperatureshould be converted to \textdegreeC and, then, to Fahrenheit using the relation \textdegreeF = 9/5 \textdegreeC + 32 The boiling point of helium can be converted to \textdegreeC by subtracting 273 from theboiling point in \textdegreeK. \textdegreeC = \textdegreeK\Elzbar 273 \textdegreeC = 4\textdegreeK\Elzbar 273 =\Elzbar 269\textdegreeC. After the temperature is converted to the Centigrade scale, the temperatureon the Fahrenheit scale can be determined. \textdegreeF = 9/5 \textdegreeC + 32 \textdegreeF = 9/5 (\Elzbar 269\textdegreeC) + 32 =\Elzbar 452\textdegreeF The boiling point of helium on the Fahrenheit scale is\Elzbar 452\textdegreeF.

Question:

Assume that 90% of the electron density is representative of the volume of the atom. The following atomic radii have been obtained for 7 elements across the second period of the periodic table and for six elements down through the first family: H 0.37 LiBeBCNOF 1.23.89.80.77.74.74.72 Na 1.57 K 2.03 Rb 2.16 Cs 2.35 Explain the observed trends of these atomic radii.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E18-0655.htm

Solution:

Atomic radii increase going down a family, but decrease going across a period. To explain this, one must consider the change in the elements in going across a period or down a family and see how it could affect atomic radii. In moving across a period, the atomic number increases, which means, the nuclear positive charge increases. But the principal quantum number of the outside electrons remains constant. This means that the outer electrons, assuming all other factors, are constant, will remain at the same distance from the nucleus. Not all factors are constant, though. With increasing atomic number, the nuclear positive charge increases, and, thus, the force pulling the electrons towards the nucleus increases. Therefore, the atomic radii decreases as one moves across a period. Moving down a family presents a similar case of increasing atomic number. But, here, the principal quantum number is also increasing, which means the outermost electrons are farther and farther away from the nucleus. The principal quantum number has a stronger effect than atomic number in determining atomic radii. Thus, the radii increase going down a family.

Question:

Compare the processes by which glucose and fatty acids are activated.

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/Users/wenhuchen/Documents/Crawler/Biology/F03-0092.htm

Solution:

Before glucose and fatty acids can be oxidized, they must first be activated in the first steps of their respective pathways. A fatty acid is activated by reaction with ATP and coenzyme A to yield a fatty acyl Co A: R-COOH + ATP + CoA - SH \textemdash\textemdash> R - CO \sim S - CoA + AMP +PPi fatty acylfatty acyl CoA A series of enzymes, called acyl - CoA- synthetases are involved in the catalysis of this reaction. The bond formed between the fatty acid and coenzyme A is a high energy thioester bond. The energy for its formation comes from the hydrolysis of ATP to AMP and pyrophosphate. While a fatty acyl - AMP intermediate is formed, the final activated compound is a thioester. The activated form of glucose, on the other hand, is phosphorylated. Only ATP is involved in the activation, which yields glucose - 6 - phosphate and ADP Unlike fatty acid activation, ATP is hydrolyzed to ADP, not AMP, and only after glucose is converted to pyruvate does coenzyme A serve as an activator, enabling pyruvate to enter the TCA cycle.

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Question:

A ring of charge with radius 0.5 m has a 0.02 m gap (see figure (a)). Compute the field at the center if the ring carries a charge of +1 coulomb.

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/Users/wenhuchen/Documents/Crawler/Physics/D17-0573.htm

Solution:

The field can be found by superposition of the fields of an imaginary ring of charge, and a negative charge segment located where the gap would be ( see figure (b) ). We must first calculate the linear charge density of the incomplete ring. \sigma = q/(length of ring) = q/(2\piR-0.02) = [1 coulomb] / [{2\pi(0.5) - 0.02}m] = 0.321 coulombs/m where \sigma is the linear charge density and R = 0.5 is the radius. We can approximate the arc length of the gap with the linear distance between the ends since this chordal length is small compared with the radius. The field at the center due to the complete ring is zero. Since all charge elements are diametrically opposite to the center, all field elements must cancel. Thus the field is entirely due to the negative charge segment. Both the ring and the segment must have charge den-sities whose absolute values (since the segment is negative) equal the charge density of the incomplete ring. Thus, the charge on the segment is: q' = (0.02m)(\rule{1em}{1pt}0.321 coulombs/m) = 6.42 × 10^-3 coulombs The field at the center is: E = k(q'/r^2) = 9 × 10^9 (N - m^2/C^2) [(6.42 ×10^-3 C)/(0.5m)^2 ] = 2.31 × 10^8 N/C where we have treated the segment as if it were a point source.

Question:

IfDNAaseis added to a bacterial cell, the DNA is hydrolyzed , the cell cannot make any more proteins and eventually dies. IfDNAaseis added to RNA viruses, they continue to produce new proteins. Explain.

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/Users/wenhuchen/Documents/Crawler/Biology/F05-0155.htm

Solution:

By means of electron microscopy and x-rays diffraction studies, much has been discovered about the structure and composition of viruses. The infective ability of viruses is due to their nucleic acid composition. An individual virus contains either DNA or RNA but not both as is true for cells . Therefore since the virus proposed in the question contains RNA, it is not affected byDNAase. The RNA replicates, forming a complementary RNA strand which acts as messenger RNA in order to code for the synthesis of new viral proteins. To produce new viral RNA, the viral RNA first synthesizes a complementary strand and thus becomes double- stranded . The double-stranded RNA serves as a template for synthesis of new viral RNA. The virus could have been tobacco mosaic virus, influenza virus or poliomyelitis virus. These are all viruses which contain single- stranded RNA as their genetic material. There are at least two groups of RNA viruses in which the RNA is normally double-stranded, and assumes a double-helical form. DNA viruses, such as the smallpox virus, SV 40 (a tumor- inducing virus), and certain bacterial viruses, such as bacteriophages T_2, T_4, and T_6, contain double-stranded DNA. Yet there are somebacteriophageswhich have asingle-strandedDNA molecule. It does not matter whether the genetic information is contained in DNA or RNA, or if it exists as a single strand or as a double helix. For viruses, the important point is that the genetic message is present as a sequence of nucleotide bases.

Question:

Suppose that all of the electrons in a gram of copper could be moved to a position 30 cm away front the copper nuclei. What would be the force of attraction between these two groups of particles?

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Solution:

The atomic mass of copper is 63.5. Therefore, 1 g of copper contains a number of atoms given by Avogadro's number divided by the mass of 1 mole (that is, 63.5 g): No. atoms = [(6.02 × 10^23 atoms / mole) / 63.5 g / mole ] = 0.92 × 10^22 atoms / g The atomic number of copper is 29; in other words, each neutral cop-per atom contains 29 electrons. Therefore, the number of electrons in 1 g of copper is No. electrons in 1 g = 29 × 0.92 × 10^22 = 2.7 × 10^23 electrons Thus, the total change on the group of electrons is q_e = 2.7 × 10^23 ×(\rule{1em}{1pt}e) = 2.7 × 10^23 × (\rule{1em}{1pt}4.8 × 10^\rule{1em}{1pt}10 stateC) = \rule{1em}{1pt}1.3 × 10^14 stateC A similar positive charge resides on the group of copper nuclei. Hence, the attractive electrostatic force (in the OGS system) F_E = (q_1q_2) / (r^2) where r is the distance between charges q_1 and q_2 . Therefore, since the 2 groups of charges with which we are concerned both have magnitude q_e. F_E = (q2_e) / (r^2) = (\rule{1em}{1pt}1.3 × 10^14 stateC)^2 / (30cm)^2 = 1.9 × 10^25 dyne Because the nuclei and electrons have opposite charges, this force is attractive. It is as great as the gravitational force between the Earth and the moon.

Question:

Two species form a predator-prey pair. Their relation is expressed by the rule: The rate of predator population Increase is proportional to an excess in prey population and the rate of decrease of the prey pop-ulation is proportional to the increase of its enemy's population. Let N_1(t), N_2(t) be the prey and predator populations, respectively and A_1, A_2, are positive constants for natural (i.e., in absence of other species) rates of birth (per unit of population per unit of time). Write a FORTRAN program which uses the modified Euler method to simulate this system from time t = 0 to t =t_fif N_1(0) = N_10 and N_2(0) = N_20.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G24-0587.htm

Solution:

Proceeding directly to the mathematical model, one can express the predator-prey relation as: Ṅ̇_1 = -A_2N_2(1) Ṅ̇_2 = A_1N_1(2) The program applies the modified Euler method first to equation (1) and then to equation (2). N1 and N2 are kept in a COMMON block so that the new computed value of N1 could be used in determining N2. A logical flag, TWO, is used to determine which equation is being integrated: REALT/0.0/,TFIN,DT,REALN,ACCUR,N1,N2,N10,N20,A1,A2 INTEGER I,N LOGICAL TWO/.FALSE./ COMMON TW0,N1,N2,A1,A2 READ,N,TFIN,ACCUR,NIO,N20,A1,A2 PRINT,T,NIO,N20 REALN = N DT = TFIN/REALN N1 = N10 N2 = N20 DO 10 I = 1,N T = T + DT CALL MEULER(T,N1, ACC UR, DT) TWO = .TRUE. CALL MEULER (T,N2,ACCUR,DT) TWO = .FALSE. PRINT, T,N1,N2 10CONTINUE STOP END FUNCTION G(W) REAL G,W,N1,N2,A1,A2 LOGICAL TWO COMMON TWO,N1 ,N2 ,A1 ,A2 G = (-A2)\textasteriskcenteredN2 IF (TWO) G = A1\textasteriskcenteredN1 RETURN END Note, that if equation (1) is differentiated, the result is: Ṅ̇= -A_2Ṅ_2(3) Substituting equation (2) into equation (3) one gets: Ṅ̇_1 = -A_2A_1N_1(4) Differentiating equation (2) and substituting equation (1) into the result yields: Ṅ̇_2 = -A_2A_1N_2(5) Equations (4) and (5) are of the form ӱ = -Ay where A = -A_2A_1 is constant. The conclusion can be made that this is a free harmonic oscillating system.

Question:

Explain the properties of NAND gates, and use NAND gates to generate AND, OR, and NOT operations. Draw the symbol diagrams, and give truth tables for each operation.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G03-0051.htm

Solution:

The NAND gate is a combination of an AND gate followed by a NOT gate. In fig. 1 the structure of a NAND gate is shown. The key to understanding a NAND gate is to notice that the output is 0 if and only if its inputs are simultaneously 1. The Truth table of the NAND gate is given in fig. 2. ab ab f(a,b) 00 01 10 11 0 0 0 1 1 1 1 0 Fig. 2 F(a,b) =ab NAND gates are usually represented as in fig. 3 when used in circuit diagrams; Several other interesting properties of NAND gates are; f(a , b) =ab=a+b(DeMorgan's Law) f(a , a) =a+a=a A NAND gate with both of its inputs the same acts like a NOT gate. This idea is illustrated in fig. 4. Truth table for the gate illustrated in fig. 4 is shown in fig. 5. a f(a) 0 1 1 0 Fig. 5 Use of NAND gates to generate AND operations; Two NAND gates connected as shown in fig. 6 generate an AND operation at the output. Since AND gates have the truth table shown in fig.7. a b f(a, b) 0 0 1 1 0 1 0 1 0 0 0 1 Fig. 7 It can easily be verified that the output in fig. 4 has the sane operation. When inputs a and b are both 0, output of the first NAND gate is 1. This output is the input of the second HAND, gate, therefore both inputs to the second NAND gate are 1. The output of this HAND gate then is found to be 0, from the definition of NAND gates. The OR gate can be obtained from three NAND gates, connected as shown in fig. 8. When inputs a and b are both 0, the input to the third NAND gate is 11 and its output is 0. In all the other input combinations the output will be 1. The truth table of the OR gate is given in fig. 9. a b f(a, b) 0 0 1 1 0 1 0 1 0 1 1 1 Fig, 9

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Question:

In writing a basic assembler language program, our first and foremost concern is that of housekeeping. What is housekeep-ing and why is it necessary?

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G09-0188.htm

Solution:

The first six or so instructions of the IBM/360- 370 assembly language are the housekeeping instructions. In almost all programs, there is some housekeeping to be done. These instructions provide standard linkage between the program and the operating system. In general, these instructions are: LABELSTART\varphi \varphi \varphi SAVE(14, 12) BALR3, 0 USING\textasteriskcentered, 3 ST13, SAVE+4 LA13, SAVE These housekeeping instructions have three basic func-tions: 1. They provide standard operating system (OS) program linkage. This is done by means of the SAVE, ST and LA in-structions. 2. They load and identify the base register for the program. The BALR and USING statements perform this task. The third function is not of fundamental concern to the BAL programmer. This step involves preparation of the units and materials needed by the main processing part of the program. In most instances, the task is to open input and output files, a task which the system itself usually fails to perform. Let us now consider the basic functions of each instruc-tion: LABEL START\varphi: The label is usually the name of the program. \varphi START\varphigives the system a location in memory where the program \varphi should be stored. The address coded is almost never obtained so a zero is substituted, later to be re-placed by the assembler. SAVE (14, 12): Here, the contents of 15 (Register 13 excluded) of the system's 16 general purpose registers are saved. It is important to save the contents of these re-gisters since they may hold information pertinent to the system's operation. The save area generally has a length of 18fullwords, and saving of the registers starts in the fourth word. Let us now discuss the 'ST 13, SAVE+4' and 'LA 13, SAVE' instructions. Because of some system operations which the programmer doesn't need to concern himself with present-ly, it was found to be convenient to store the contents of Register 13 in the secondfullword, (SAVE + 4), of the SAVE area. The 'LA 13, SAVE' simply places the address of the SAVE area in Register 13. This is done so that the area can be referred to, and the contents of the register re-stored at the end of the program. Now, going back, we can look at the 'BALR 3,0' and the 'USING \textasteriskcentered,3' instructions. The BALR sets up the base regis-ter for the program in Register 3. It then stores the base address (or just the address) of the program. The 'USING \textasteriskcentered, 3', in effect confirms that Register 3 is the base regis-ter of the program. It tells the computer that starting at that point (indicated by \textasteriskcentered-asterick) the base register of this program is Register 3. The way these two instructions are coded indicates the natural order of these two instruc-tions in most assembly language housekeeping procedures. The choice of the base register, of course, is left up to the programmer. From the above discussion we can clearly see why house-keeping is of foremost concern to the programmer. It allows the computer to preserve, for later use, the present con-tents of its registers, and it enables the assembler to properly place the users program in core.

Question:

What is the binding energy of the lithium nucleus?

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Solution:

The binding energy is the energy released when a nucleus is formed; it is what holds the nucleus together. We would have to add this energy again in order to break up the nucleus. The source of this energy, released when the nucleus was destroyed, is in the form of a mass loss in the nucleons.Therefore, by comparing the "standard" mass of all the nucleus, (protons and neutrons) and the actual mass of the nucleus, we can calculate the mass defect - the amount of mass converted into energy in the formation of the atom. Einstein's theory tells us that mass and energy are equivalent. Since there are 931 Mev (energy) per 1 amu (mass) we can claulate the energy released in the destruction of the nucleus. In the isotope L_1^7_13 there are 3 protons and 4 neutrons (7 - 3). So z = 3,A = 7, M = 7.01822 where z is the atomic number, A the mass number, and M the atomic mass of Lithium. Therefore, mass defect = [3(1.00814) + 4(1.00898) - 7.01822] = [(3.02442 + 4.03592) - 7.01822] amu = [7.06034 - 7.01822] = 0.04212 amu Converting mass to energy : binding energy in Mev = 931 × mass defect = 931 × 0.04212 in amu = 39.2Mev

Question:

Determine the entropy change that takes place when 1 mole of ammonia (a) Passes from the liquid state to the gaseous state at its boiling point, - 33\textdegreeC; ∆H_VAP = 5570 cal/mole (b) as a gas at - 33\textdegreeC comes to room temperature, 25\textdegreeC. Assume heat capacity is constant at 8.9 cal/deg - mole for this range.

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Solution:

Entropy, S, may be defined as an indication of the randomness of a system. The change in entropy for any change in the state of a system may be written as ∆S =S_final-S_initial_. .For a reversible process, ∆S = [(q_rev)/(T)], whereq_rev= amount of heat involved in a reversible process and T = the temperature in \textdegreeK at which the amount of heat is reversibly absorbed or evolved. ∆S can also be defined in terms of heat capacity (C_P) and temperature ranges, say T_1 \rightarrow T_2, by the following equation: ∆S = 2.303 C_P log T_2/T_1. With this in mind, proceed as follows: (a)1 mole of NH_3 (ammonia) goes from liquid to gas at - 33\textdegreeC. The ∆H_VAP is the heat necessary to vaporize 1 mole of liquid material at its boiling point at constant temperature. A gas can also be liquefied, thus, this change of state is reversible. Recalling that ∆S =q_rev/T, You can say, therefore, that ∆S = ∆H_VAP /T. Since ∆H_VAP is the only heat involved,q_rev= ∆H_VAP. By substitution into ∆S = ∆H_VAP /T, you obtain ∆S = [(5570 cal/mole)/(240\textdegreeK)]or∆S = 23.2 cal/deg - mole. (b)This question uses heat capacity, which demands use of the expression ∆S = 2.303 C_P log T_2/T_1. Since all the values are known, you can substitute them into the above equation: ∆S = 2.303 C_P log (T_2 /T_1 ) = (2.303) (8.9) log [(298\textdegreeK)/(240\textdegreeK)] = 20.5 log 1.24 = (20.5) (.0934) = 1.91 cal/deg - mole.

Question:

In the figure below, C_1 = 6 \muf, C_2 = 3 \muf and V_ab = 18 v. In the figure below, C_1 = 6 \muf, C_2 = 3 \muf and V_ab = 18 v. Find the charge on each capacitor. What is the value of the Find the charge on each capacitor. What is the value of the equivalent capacitance (see figure)? equivalent capacitance (see figure)?

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Solution:

The charge Q on a capacitor having capacitance C is Q = VC, where V is the potential difference across the capacitor. Then, Q_1 = V_ab C_1 = (18v) × (6 × 10^-6 f) = 108 × 10^-6 C Q_2 = V_ab C_2 = (18v) × (3 × 10^-6 f) = 54 × 10^-6 C The equivalent capacitance must carry the same total charge as the original system, since charge is conserved (none leaks out of the system). Hence Q_net = Q_1 + Q_2 = 162 × 10^-6 C Then, the equivalent capacitance is C_e = (Q_net /V_ab) = [(162 × 10^-6 C)/(18 V)] = 9 × 10^-6 f

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Question:

What is differential reproduction? What factors bring about differential reproduction?

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/Users/wenhuchen/Documents/Crawler/Biology/F27-0715.htm

Solution:

When the percentage of viable offspring carrying a certain genotype cannot be accounted for by purely random mating, differential reproduction is said to have occurred. Differential reproduction can result from nonrandom mating, differential fecundity, or dif-ferences in either zygote viability or offspring fertility. Nonrandom mating may occur due to genotypic variaton, genetic mutations, or both. There are well established behavior patterns of courtship and mating in many species, which are primarily genetically controlled. For example, in many species of birds and fish, a brightly colored spot on the male serves to stimulate the female for copu-lation. Males having this characteristic will succeed in finding a mate easily and more frequently than males with-out this trait, hence producing in their lifetime more offspring carrying their genes. Males lacking this trait will find it difficult to pass their genes on to succeeding generations. Mutations that lead to, formation of bigger or brighter spots will benefit the males with the mutated gene by rendering them more attractive to females. Conversely, mutations that lead to disappearance of these spots or formation of smaller, duller ones, will have a negative selective value. These mutations are inherited by the progeny, perpetuating differential reproduction through nonrandom mating. Success in mating, however, can neither guarantee successful fertilization nor assure large numbers of off-spring. Such factors can also be important in differen-tial reproduction. Differences in the number of gametes produced by different individuals and the proportion of their gametes that will successfully unite with others to form zygotes can be collectively termed differential fecundity. Differential fecundity may be determined ge-netically. Within a species, the individuals producing large numbers of gametes or having a large percentage of successful matings (resulting in fertilization) will nec-essarily contribute the greatest percentage of genes to the next generation. This depends, of course, on the via-bility of the zygotes formed. In environments which result in low zygote viability, individuals having high fecundity will tend to reproduce in larger numbers. In fish, for example, zygotes must be fertilized and also develop in an external environment; those species producing large numbers of both gametes and zygotes will tend to be preserved. Mammals, on the other hand, whose zygotes develop internally, cannot and need not produce such large numbers of gametes or zygotes. In fact, large numbers of zygotes may actually be a disadvantage, since this limits the amount of care and feeding available to each offspring. Another factor affecting the reproducing ability of a given individual or individuals is the fertility of the offspring. The offspring produced as a result of a given mating may be viable, but unless they are themselves fertile, the result is the same as if they were not viable. Thus, it is the interaction of a variety of factors which is responsible for the reproductive capacity of a given individual or species. Any one of these factors could be responsible for differential reproduction.

Question:

Two equal conducting spheres of negligible size are charged with 16.0 × 10^\rule{1em}{1pt}14C and \rule{1em}{1pt}6.4 × 10^\rule{1em}{1pt}14 C, re-spectively, and are placed 20 cm apart. They are then moved to a distance of 50 cm apart. Compare the forces between them in the two positions. The spheres are connected by a thin wire. What force does each now exert on the other?

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Solution:

The equation giving the force between the spheres, which may be considered as point charges, is by Coulomb's law, F = (1 / 4\pi\epsilon_0)(q_1q_2) / r^2 where q_1, q_2 are the charges on the spheres, and r is their separation. Thus F_1 = (1 / 4\pi\epsilon_0)[(q_1q_2) / (0.2)^2m^2 ] and F_2 = (1 / 4\pi\epsilon_0)[(q_1q_2) / (0.5)^2m^2 ] \therefore F_1 / F_2 = (0.5)^2 / (0.2)^2 = 6(1/4) If the spheres are joined by a wire, the charges, which are attracted to one another, can flow in the wire under the influence of the forces acting on them. The charges will neutralize as far as possible and [16.0 × 10^\rule{1em}{1pt}14 + (\rule{1em}{1pt} 6.4 × 10^\rule{1em}{1pt}14)] = 9.6 × 10^\rule{1em}{1pt}14 C will be left distributed over the system. Neglecting the effect of the wire, by symmetry 4.8 × 10^\rule{1em}{1pt}14 C will reside on each sphere. The force between the two spheres is now F = (1 / 4\pi\epsilon_0)(q^2 / r^2) = 9 × 10^9 N.m^2.C^\rule{1em}{1pt}2 × [(4.8 × 10^\rule{1em}{1pt}14)^2C^2 ] / (0.5)^2m^2 = 8.29 × 10^\rule{1em}{1pt}17N.

Question:

How much work is required to raise a 100-g block to a height of 200 cm and simultaneously give it a velocity of 300 cm/sec?

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Solution:

The work done is the sum of the potential energy, PE = mgh, and the kinetic energy, KE = (1/2) mv^2: PE= mgh = (100 g) × (980 cm/sec^2) × (200 cm) = 1.96 × 10^7 g-cm^2/sec^2 = 1.96 × 10^7 ergs KE= (1/2) mv^2 = (1/2) × (100 g) × (300 cm/sec)^2 = 4.5 × 10^6 g-cm^2/sec^2 W= PE + KE = 1.96 × 10^7 ergs + 0.45 × 10^7 ergs = 2.41 × 10^7 ergs = 2.41 J

Question:

What activation energy should a reaction have so that raising the temperature by 10\textdegreeC at 0\textdegreeC would triple the reaction rate?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E13-0468.htm

Solution:

The activation energy is related to the tem-perature by the Arrhenius equation which is stated k = Ae ^-E/RT where A is a constant characteristic of the reaction; e is the base of natural loragithms, E is the activation energy, R is the gas constant (8.314 J mol^-1 deg^-1) and T is the absolute temperature. Taking the natural log of each side: ln k = ln A - E/RT For a reaction that is 3 times as fast, the Arrhenius equation becomes 3 k = Ae-E/R(T + 10\textdegree) Taking the natural log: ln 3 + ln k = ln A - E/R(T + 10\textdegree) Subtracting the equation for the final state from the equation for the initial state: lnk =lnA - E/RT - (ln3 + ln k = ln A - E/R(T + 10)) -ln3 = - E/RT + E/R(T + 10) Solving for E: -ln3 = - E/RT + E/R(T + 10)R = 8.314 J/mole \textdegreeK T = 0 + 273 = 273 -ln3 = - E/(8.314 J/mole-K) (273K) + E/(8.314 J/mole-K)(283K) -1.10 = - E/2269.72 J/mole + E/2352.86 J/mole (2269.72 J/mole)(2352.86 J/mole) × - 1.10 = (- E/2269.72 J/mole + E/2352.86 J/mole)(2269.72 J/mole) (2352.86 J/mole) - 5.874 × 10^6 J2/ mole^2 = (- E) (2352.86 J/mole) + E (2269.72 J/mole) - 5.874 × 10^6 J2/ mole^2 = - 8.314 × 10^1 J/mole × E 7.06 × 10^4 J/mole = E

Question:

If a wire carrying 2 amperes lies perpendicularly across a uni-form magnetic field of flux density 5 × 10^-2 weber/meter^2 in such a manner that 15 cm of the wire are subjected to the field (i.e., if the pole pieces of the magnet are 15 cm across), how much side thrust is experienced by the wire, and which way does it act?

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/Users/wenhuchen/Documents/Crawler/Physics/D22-0738.htm

Solution:

By the right-hand rule, it is seen that the thrust is down-ward. The magnitude of the thrust is given by F^\ding{217} = IL^\ding{217} × B^\ding{217} where B^\ding{217} is the magnetic induction, and I is the current in a wire of length L. Since the angle between L^\ding{217} and B^\ding{217} is 90 o , then F = BLI thereforeF = 5 × 10^-2 w/m^2 × 15cm × (1m/100cm) × 2 amp = 15 × 10^-3 n = .015 nt where we have used the fact that 1 cm = (1/100) n .

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Question:

What is the critical angle between carbon disulfide and air? (The index of refraction for Carbon Disulfide is 1.643.) Critical Angle

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/Users/wenhuchen/Documents/Crawler/Physics/D27-0860.htm

Solution:

Carbon disulfide is a more optically dense mater-ial than air. Therefore, as a beam of light passes from carbon disulfide to air, the angle of refraction is larger than the angle of incidence. There is an angle of incidence smaller than 90\textdegree for which the angle of refraction is equal to 90\textdegree, meaning that the beam of light emerges parallel to the boundary between the two mediums. This angle of incidence is called the critical angle. If the angle of incidence is greater than this value, the light will not escape from the carbon disulfide. It will be reflected back into the carbon disulfide following the regular law of reflection. Solving for the critical angle \texttheta_1, let \texttheta_2 be 90\textdegree. The index of re-fraction for carbon disulfide is 1.643 and for air it is 1.00 Using Snell's law n_1 sin \texttheta_1 = n_2 sin \texttheta_2 1.643 sin \texttheta_1 = 1.00 sin 90\textdegree sin \texttheta_1 = (1.00/1.643) = 0.608 \texttheta_1 = 37.4\textdegree.

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Question:

Explain why penicillin is effective only against actively growing bacteria. Describe the mode of action of some other antimicrobial agents.

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/Users/wenhuchen/Documents/Crawler/Biology/F05-0123.htm

Solution:

An antimicrobial agent is one that interferes with the growth and activity of microorganisms. Knowledge of the mode of action of a particular agent makes it possible to determine the conditions under which it will act most effectively. Among the known sites of action of antimicrobial agents are the cell wall, cell membrane, protein structure and synthesis and enzyme activity. The cell walls of some gram-positive bacteria are attacked by the enzyme lysozyme, which is normally found in tears, mucous secretions, and leucocytes (white blood cells). Lysozyme breaks down the peptide linkages in the cell wall complex. Some bacteria secrete enzymes which, degrade cell walls of other bacteria or prevent cell wall formation. Without a cell wall to provide support for the bacterium, it will soon lyse and die. The antimicrobial effect of penicillin is attributed to its inhibition of cell wall synthesis. Penicillin prevents the incorporation of the amino sugar, N-acetylmuramic acid into the mucopeptide structure that comprises the cell wall. This is why penicillin only works on actively growing bacteria. If cell wall formation is complete, penicillin has no effect. The cell membrane helps contain the cellular con-stituents and provides for selective transport of nutrients into the cell. Damage to this membrane thus inhibits growth or causes death. The bactericidal (kills bacteria) action of phenolic compounds, such as hexachlorophene, is attributed to their effect on cell permeability. This results in leakage of cellular constituents and eventual death. Proteins are essential to the cell for both structure and enzymatic activity. Protein denaturation (alteration of their natural configuration) causes irreparable damage to the cell. High temperatures, acidity, and alcohol denature proteins. Streptomycin combines with the ribosomes of sensitive bacteria and disturbs protein synthesis. Many agents inhibit enzymes involved in the energy- supplying reactions of the cell. For example, cyanide inhibits cytochrome oxidase in the electron transport chain, fluoride inhibits glycolysis, and dinitrophenol uncouples oxidative phosphorylation. All of these inhibit ATP synthesis. A familiar antibiotic, sulfanilamide, works by blocking the synthesis of folic acid, a necessary substrate for certain reactions in the cell. A precursor of folic acid is para-aminobenzoic acid (PABA) whose structure is very similar to that of sulfanilamide: Sulfanilamide, by effectively competing with this precursor for the binding site on an enzyme involved in the pathway, inhibits folic acid synthesis. Any compound, such as sulfanilamide, that interrupts synthetic processes by substituting itself for a natural metabolite is called an antimetabolite or metabolic analogue.

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Question:

A force of 2000 dynes produced an acceleration of 500 cen-timeters per second^2. What was the mass of the object accelerated?

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/Users/wenhuchen/Documents/Crawler/Physics/D04-0097.htm

Solution:

Here, we can apply Newton's Second Law, F = ma. In this case F = 2000 dynes and a = 500 cm/sec^2. Then 2000 dynes= M × 500 cm/sec2 whence M= 4 gm

Question:

Starch (amylose), (C_6H_10O_5)x, is a polymer of the sugar glucose, C_6H_12O_6, linked by - O - bonds; each starch mo-lecule may contain 200 to 1000 glucose units. A mixture of amino acids added to a solution of amylose produces no detectable reaction. One molecule of the enzyme amylase, containing the same amino acids, however breaks (hydrolyzes) 4000 -O- bonds per second, (a) Estimate the mass in picograms of glucose produced by an amylase molecule in 1 day. (b) The heat in calories liberated to a cell by the oxidation of the glucose: C_6H_12O_6 (aq) + 6O_2 (g) \rightarrow 6O_2 (g) + 6H_2O (l). ∆H = - 686 Kcal. How many calories does the cell obtain from the mass of glucose calculated in part (a)?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E23-0831.htm

Solution:

(a) Five steps can be used to solve for the number of picograms of glucose liberated from the amylose by 1 amylase in a day. (1) find the number of seconds in one day (2) calculate the number of glucose molecules formed per day (3) determine the weight of one molecule of glucose (4) solve for the number of grams of glucose formed per day (5) convert from grams to picograms. Solving: (1) There are 60 seconds in a minute,60 minutes in an hour and 24 hours in a day. One can solve for the number of seconds in a day by multiplying these factors together The units will cancel out. 60 (sec/min) × 60 (min/hour) × 24 (hours/day) = (8.64 × 10^4) (sec/day) (2) The problem states that one amylase molecule liberates 4000 molecules per second. Therefore the number of molecules liberated per day can be found by multiplying the number of seconds in a day by the number of molecules formed per second. no. of glucose molecules formed per day = = (8.64 × 10^4)(sec/day) × 4000 (molecules/sec) = (3.456 × 10^8) (molecules/day) (3) The weight of one molecule of glucose is found by dividing the molecular weight of glucose by the number of molecules per mole. (MW of C_6H_12O_6 = 180 g/mole, Avogadro's Number = 6.02 × 10^23 molecules/mole) weight of one molecule = (180 g/mole) / (6.02 × 10^23 molecule/mole) =2.99 × 10^-22 g/molecule (4) After one knows the number of molecules formed per day and the weight of one molecule, one can find the weight in grams of the glucose formed in one day by one amylase. no. of grams of glucose formed per day = = 2.99 × 10^-22 g/molecule × 3.456 × 10^8 molecules/day = 1.03 × 10^-13 g/day. (5) There are 10^12 picograms in one gram. Grams can be converted to picograms by multiplying the number of grams by the conversion factor 10^12 pg/g no. of pg produced = 1.03 × 10^-13 g/day × 10^12 pg/g = 1.03 × 10^-1 pg/day. (6) From the equation given one knows that 686 Kcal are liberated when 1 mole of glucose is oxidized. One can find the number of calories liberated by 1) determining the number of moles of glucose present 2) solving for the number of Kcal produced 3) converting Kcal to calories. Solving: 1) One has solved for the number of molecules produced per day previously (part (a)(2)). Therefore one can find the number of moles produced by dividing the number of mo-lecules produced by the number of molecules in one mole. (3.456 × 10^8 molecules/day) / (6.022 × 10^23 molecules/mole) = 5.74 × 10^-16 moles/day. 2) There are 686 Kcal liberated per mole of glucose oxidized, therefore the number of Kcal formed when 5.74 × 10^-16 moles are oxidized is equal to 5.74 × 10^-16 moles/day × 686 Kcal/mole = 3.94 × 10^-13 Kcal/day. 3) There are 1000 cal in 1 Kcal, thus Kcal can be converted to cal by multiplying by the conversion factor 1000 cal/Kcal no. of cal produced = 3.94 × 10^-13 Kcal/day × 1000 cal/Kcal = 3.94 × 10^-10 cal/day.

Question:

A 500-kg ore bucket is raised and lowered in a vertical mine shaft using a cable. Determine the upward force exerted by the cable when (a) the upward acceleration is 4 m/s^2 and (b) the bucket is moving upward with a constant velocity.

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/Users/wenhuchen/Documents/Crawler/Physics/D04-0111.htm

Solution:

(a) The known quantities are the mass of the ore bucket, m = 5 × 10^2 kg, and the acceleration, a = 4 m/s^2. The two forces acting on the bucket are the weight and the tension of the cable, as illustrated in the figure. To accelerate the bucket upward, the tension, f, must be larger than the weight of the bucket. This is a consequence of Newton's first law. Newton's first law states that an ob-ject will maintain a constant velocity (or remain at rest) when no net force acts on the object. If the object accelerates, a net force must be acting on it. By vector addi-tion, the resultant upward force, F, is equal to f + w, where w = mg. Since f and w are oppositely directed, the magnitude of their vector sum is equal to the difference of their magnitudes. F = f - w. Using Newton's second law, f - w= ma(1) f= ma + w The weight of the bucket is w= mg andf= ma + mg. Substituting the known observables, f= (5 × 10^2 kg)(4 m/s^2) + (5 × 10^2 kg)(9.8 m/s^2) = 6.9 × 10^3 N. (b) If the bucket moves upward with constant velocity (uniform motion), the acceleration is zero. (This follows from Newton's first law). Equation (1) becomes f = w. The upward force (tension) exerted by the cable equals the weight of the bucket, 4.9 × 10^3 N.

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Question:

It is known that both metallic and ionic crystals are good conductors of electricity, yet many of their other proper-ties differ significantly. Explain why.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E06-0216.htm

Solution:

The similarities and differences of the two will be seen after an investigation of their space lattices, i.e., those patterns of points describing the atomic or molecular arrangements of points in a unit of a crystal. For something to conduct electricity, it must have charges. In an ionic crystal, the lattice points are occupied by positive or negatively charged ions, each of which occupies a position exactly equivalent to every other species of like charge. The electric conductivity of ionic crystals is low but increases with Temp, because more electrons are excited into the conduction band of the crystal. Ionic crystals are held together by electro-static forces between the ions. In a metallic crystal, you have discrete atoms, not ions, at the lattice points. It seems that the species should not conduct electricity. It is important to realize, however, the valence or outer electrons of the metallic crystal are distributed over the crystal as a whole rather than being localized on each atom. The mobile electrons, therefore, which are able to move in an applied electric field, account for the electrical conductivity. This mobility of electrons also explains why a metallic crystal is generally strong, lustrous and malleable. In the ionic crystal, you have strong electro-static forces between the ions, which accounts for the high melting points. However, these crystals also tend to be hard and brittle because the crystal consists of parallel sheets of positively and negatively charged ions, lateral displacement may bring ions of like charge into the vicinity of each other, resulting in electro-static repulsion of like charges. This, then, helps to explain why they possess a facile crystal fracture.

Question:

Explain the Header Card method of stopping a computer job, i.e., a program.

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Solution:

In this, the first executable statement in the program is a statement whichreads in the value of some var-iable, say N, the value of which denotesthe number of data items which will follow. If the program is punchedout on cards, the value of the variable N is punched on the first cardof the data section of the deck of cards. This method is useful when duringdifferent re-runs of the same program, the set of data that is to be usedin the program changes. We have no advance knowledge, while writing the program, how muchdata will be in each set of data, and what will be their range of values. Hence, while writing the program, we just use the variable N. Later, just before running the pro-gram, we punch out a card with the value ofN for the parti-cular set of data we have. We call this card the 'Header Card', and place it as the first card of the data cards. Below is a program to illustrate the use of the Header Card Method. SUM: PROCOPTIONS(MAIN); DCL (N,X,Y)FIXED(3)INIT(0),SUM FIXED(4)INIT (0); GETLIST(N); /\textasteriskcentered WE READ THE VALUE OF N FROM THE HEADER CARD \textasteriskcentered/ LOOP: GET LIST(X) /\textasteriskcenteredWEREADTHE VALUE OF X FROM A DATA CARD \textasteriskcentered/ SUM = SUM+X; Y = Y + 1; /\textasteriskcenteredY IS INCREMENTED BY 1 \textasteriskcentered/ IF Y = N THEN GOTO FINISH; GOTO LOOP; /\textasteriskcentered IF Y IS NOT EQUAL TO N THEN WE GO BACK TO LOOP AND GET ONE MORE VALUE OF X \textasteriskcentered/ FINISH: PUTLIST(SUM); END SUM;

Question:

Army engineers have thrown a pontoon bridge 10 ft in width over a river 50 yd wide. When twelve identical trucks cross the river simultaneously the bridge sinks 1 ft. What is the weight of one truck? The density of water is 1.94 slugs\bulletft^-3.

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/Users/wenhuchen/Documents/Crawler/Physics/D10-0400.htm

Solution:

When the trucks are on the bridge, the extra volume of the bridge immersed is 1 ft deep, 10 ft wide, and 50 yd = 150 ft long, i.e., 1500 ft^3. The up thrust on this extra volume immersed in water is the weight of an equal volume of water according to Archimedes' principle. Thus U =v\rho_wg = 1500 ft^3 × 1.94 slugs\bulletft^-3 × 32 ft\bullets^-2 = 9.312 x 10^4 lb. where \rho_w is the density of water, and g is the gravitation-al acceleration. But this up thrust just balances the weight of the twelve trucks. Hence one truck has a weight W = U/12 = (9.312 × 10^4 lb)/12 = 7760 lb.

Question:

Find the weight density of water at a pressure of 4000 lb/in.^2, taking the weight density at normal atmospheric pressure as 62.4 lb/ft^3. (The bulk modulus of water is 2.97 × 10^5 lb/ft^2).

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Solution:

The change ∆V in the volume of water as a result of a change ∆P in the pressure acting on it is given by -∆V/V = ∆P/B where B is the bulk modulus of water. It is defined as stress/strain = ∆P/(∆V/V) . Initially, the pressure is that of air and it is increased to 4000 lb/in.^2 Therefore, ∆P = 4000 lb/in2- P_air- = (4000 - 15) lb/in.^2 and-(∆V/V) = [(4000 - 15) lb/in^2]/[(144 in^2/ft^2) × (2.97 × 10^5 lb/ft^2)] = 9.35 × 10^-5 where we have converted the bulk modulus to lb/in^2. The weight density of water is mg/V, hence the fractional change in the density of water is ∆D = [(Mg)/(V + ∆V)] - [(Mg)/(V)] = -[(Mg)/(V)] [1 - {1/(1 + ∆V/V)}] where the weight Mg is unchanged. We find that ∆D = - D [(∆V/V)/(1 + ∆V/V)] and, since ∆V/V is much less than 1, we get ∆D/D \approx - (∆V/V) = 9.35 × 10^-5 The final density of water is D_f = D + ∆D = [62.4 + 9.35 × 10^-5 × 62.4] lb/ft^3 = 62.406 lb/ft^3. Since the compressibility of water is very small, there is not an appreciable increase in density.

Question:

When a 2\rule{1em}{1pt}V cell is connected in series with two electrical elements, the current in the circuit is 200 mA. If a 50 cycle. s^\rule{1em}{1pt}1 , 2-V ac source replaces the call, the current becomes 100 mA. What are the values of the circuit ele-ments? Suppose that the frequency is increased to 1000 cycles \textbullet s^\rule{1em}{1pt}1. What is the new value of the current?

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Solution:

Since a dc current can flow through the elements, neither of them can be a capacitor, which offers an infinite resistance to direct current. The resistance of the elements is R = (V/I) = {(2 V)/(200 × 10^\rule{1em}{1pt}3 A)} = 10 \Omega Since the current changes when alternating current is supplied, the circuit must also contain an inductor of inductance L. The reactance (the ac analogue of resist-ance) of an inductor is X = \omegaL where \omega = 2\pin. n is the frequency of variation of voltage. The impedance Z (i.e, the ratio of V to I at a set of terminals of a circuit as shown in Figure A) of the series combination of the resistor and inductor (Fig. B) is Z^2 = R^2 + X^2 Z = \surd(R^2 + \omega^2L^2) = (V'/I') = {2V/(100 × 10^\rule{1em}{1pt}3 A)} = 20 \Omega. \therefore100 \Omega^2 + \omega^2L^2 = 400 \Omega^2orL^2 = (300 \Omega^2/\omega^2). \thereforeL = {(17.32 \Omega)/(2\pi × 50 s^\rule{1em}{1pt}1)} = 0.055 H. When the frequency is increased to 10^3 cycles. s^\rule{1em}{1pt}1 , I' = (V'/Z') = (2V)/{\surd[100 \Omega^2 + (2\pi × 10^3 × 0.055)^2 \Omega^2]} = (2/346) A = 5.77 mA. We see that when the frequency n = 0, I = 200 mA. When n = 50 sec^\rule{1em}{1pt}1, I = 100mA, and when n = 1000 sec^\rule{1em}{1pt}1 , I = 5.77 mA. The inductor does then act as a resistor. Its opposition to current (the reactance) is directly proportional to frequency as is seen by the relation X = (L)\omega = (2\piL)n = Kn where K is a constant.

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Question:

A 100-keV X ray is Compton scattered through an angle of 900 . What is the energy of the X ray after scattering?

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Solution:

When X-rays pass through a crystal, some of the X-rays knock electrons out of the crystal. The scattering of electromagnetic waves (light, X-rays, etc.) by the electrons in solids can be considered to be a collision between a photon of energy hѵ and an electron (Fig. 1). The electron absorbs some of the energy of the incident photon and the scattered photon therefore has less energy and thus a lower frequency ѵ'. The collision is elastic, hence total energy as well as total momentum is conserved during the collision (Fig. 2). Hence, E^ph _i + Ee1_i = Eph_f + Ee1_f P^\ding{217}ph _i = P\ding{217}ph_f + P\ding{217}e1_f . The expressions for energies and momenta are P_i = hѵ/c, P_f = hѵ'/c ; P_e = m_ev / [\surd(1 - \beta^2)] E^Ph _i = P_ic, Eph_f = p_fc ; Ee1_i = m_ec^2 ,Ee1_f = m_ec^2 / [\surd(1 - \beta^2)] where m_e is the rest mass of the electron, \beta = v/c . Conservation of momentum is shown schematically in Fig. 2. Using the law of cosines for the triangle in the figure, p2_e = p2_i + p2_f - 2p_ip_f cos \texttheta ; adding and subtracting the quantity 2p_ip_f on the right hand side, we have P2_e= (p_i - pf)^2 + 2p_ip_f(1 - cos \texttheta). The expression for the electron energy is modified to eliminate \beta from the expression for convenience; (E^e1_f)^2 =m^2_ec^4 / (1 - \beta^2) or(E^e1_f)^2 - m^2_ec^4 = [m^2_ec^4 \beta^2]/ (1 - \beta^2)] = [m_ev / \surd(1 - \beta^2)]^2 c^2 therefore(E^e1_f)^2 - m^2_ec4+ p^2_ec^2. From conservation of energy p_ic + m_ec^2 = p_fc + \surd(m^2_ec^4 + p^2_ec^2) or\surd(m^2_ec4+ p^2_ec^2) = (P_i - P_f) c + m_ec^2. Squaring both sides, we obtain another expression for p^2_e ; P^2_e = (P_i - P_f)^2 + 2m_ec(P_i - P_f). Equating the two expressions we obtained for P^2_e ; (P_i - P_f)^2 + 2P_iP_f(1-cos \texttheta) = (P_i - P_f)^2 + 2m_ec(P_i - P_f) orP_iP_f(1 - cos \texttheta) = m_ec(P_i - P_f). Now, we substitute hѵ/c and hѵ'/c for p_i and p_f respectively; ѵѵ'(1 - cos \texttheta) = [m_ec^2 / h] (ѵ - ѵ') , and divide both sides by ѵѵ' to obtain Compton's equation 1/hѵ' = [1/hѵ] + [1/ m_ec^2] (1 - cos \texttheta) Using the values m_ec^2 = 511 keV and cos 900 = 0, we find 1/hѵ' = [1/(100 keV)] + [1/ 511 keV] or hѵ' = [{(100 keV) × (511 keV)} / {(100 keV)+(511 keV)}] = [51100/611] keV = 84 keV The electron, of course, carries off the remainder of the incident energy: KE = 100 keV - 84 keV = 16 keV

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Question:

In a certain species of flies, the male presents the female witha silken balloon before mating. Explain the evolutionary derivationof this behavior.

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Solution:

Animals with beneficial adaptive behavior patterns usually survive longerand successfully raise more offspring than those without. These behaviorpatterns will often be continued through successive generations. As they are passed on, they evolve. Thus, new behavior patterns can be seento arise from ancestral behavior patterns. The presentation of a silkenballoon to the female fly (familyEmpididae) before mating is the resultof the evolution of a behavior pattern. Different species of the fly family,Empididae, ex-hibit slightly differentcourtship behaviors. These dif-ferent behaviors may represent the differentstages of the same behavior as it evolved from the ancestral behaviorpattern. The male of the most primitiveEmpididaespecies mates withthe female without giving her anything. This often results in the femalecapturing and eating the male. In more advanced species, the male will first capture an insect and presentit to the female. While the female is occupied with eating the insect, the male mates with her. In even further advanced species, the malecaptures an insect, wraps it in silk and presents this to the female. The female attempts to open the silken wrapping while the male mates withher. In the most advanced species ofEmpididaethe male presents anempty balloon of silk to the female. The female pro-ceeds to unravel the silkenballoon while the male mates with her. At this stage in the evolution ofthis behavior, the silken balloon (not the insect) is the important ele-ment. By studying different species, a clue to the deriva-tion of behavior patternscan be obtained. It is impor-tant to realize that many seemingly oddbehaviors per-formed by animals have an evolutionary origin and have helpedthat animal adapt to its environment.

Question:

Suppose that a satellite is placed in a circular orbit 100 miles above the earth's surface. Determine the orbital speed v and the time t required for one complete revolution of the satellite.

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Solution:

The radius R of the circular path is determined as follows (see the figure), R= R_e + 100 miles where R_e , the earth's radius, is 6.378 × 10^6m; a distance of 1 mile is equal to 1,609m; therefore, 100 miles = 1.61 × 10^5m R = 6.378 × 10^6m + 1.61 × 10^5m = 6.539 × 10^6m. For our purposes, it is sufficient to retain only the first two digits: R = 6.5 × 10^6m The gravitational pull on the satellite is F_g = mg. This pull provides the centripetal force for the circular motion, therefore F_center = mv^2/R = mg orv^2 = gR = (9.8m/s^2)(6.5 × 10^6m) = 63.7 × 10^6m^2/s^2 \approx 64 × 10^6m^2/s^2 v = 8 × 10^3m/s This orbital speed is only approximately correct because it has been assumed that the effect of gravity 100 miles above the earth is the same as at the earth's surface. The gravitational "pull" weakens as one recedes from the earth's surface, but at 100 miles above the earth it is only slightly different from the value g, so the calculation above Is reason-ably accurate. To determine the time interval required for one revolu-tion of the satellite, the distance the satellite travels in one revolu-tion must be calculated. This is just the circumference C of a circle of radius R: C= 2\piR = (2)(3.14)(6.5 × 10^6m) = 4.1 × 10^7m The period of the motion is t= C/v = (4.1 × 10^7m)/(8 × 10^3m/s) = 5.1 × 10^3s t= (5.1 × 10^3s)/(60s/min) = 85 min The time required for one complete revolution of the satellite is about 85 min.

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Question:

A G.T.O. has a 22 gal. cooling system. Suppose you fill it with a 50-50 solution by volume of (CH_2OH)_2, ethylene glycol, and water. At what temperature would freezing become a problem? Assume the specific gravity of ethylene glycol is 1.115 and the freezing point depression constant of water is (1.86 C\textdegree / mole).You might have placed in methanol (CH_3OH) instead of (CH_2OH)_2. If the current cost of ethylene glycol is 12 cents / lb and the cost of methanol is 8 cents / lb, how much money would you save by using CH_3OH? And yet, ethylene glycol is the more desirable antifreeze. Why? density = (.79 g / ml) for CH_3OH and 3.785 liter = 1 gallon.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E07-0251.htm

Solution:

To determine the freezing point of the water- ethylene glycol solution, you employ the following equation ∆T_f= [(1000gK_f(N_2)] / [M_1(N_1 + N_2)] where ∆T_fis the change in the freezing point of the solvent, water,K_fthe freezing point depression constant of the solvent, M_1 is the molecular weight of the solvent, N_1 the number of moles of solvent and N_2 the number of moles of solute. You are givenK_ffor water and the molecular weight is 18; hence, you can solve for ∆T_fafter determining N_1 and N_2. You can calculate the number of moles of water and of ethylene glycol present by using their densities and volumes. Use the following steps: 1) Convert 22 gal (the volume of the tank) to liters, volume = (22 gal × 3.785 l / gal) = 83.3 l 2) You are told that the mixture is 50-50 by volume or one half water and one half ethylene glycol. Hence there are (83.3 / 2)l or 41.68l of each. 3) Use the density of each compound to find the number of grams of each present. (density of H_2O = (1.0 and 1000 cm^3 = 1l) mass of H_2O = (1.0 g / cm^3) × (1000 cm^3 / l) × (41.68 l) = 41.68 × 10^3 g H_20 mass of (CH_2OH)_2 = (1.115 g / cm^3) × (1000 cm^3 / l) × (41.68l) = 46.47 × 10^3 g of (CH_2OH)_2 4) The number of moles of each is found by dividing its mass by its molecular weight. (MW of H_2O = 18, MW of (CH_2OH)_2 = 62). moles of H_2O = (41.68 × 10^3 g) / ( moles 18 g / mole) = 2.32 × 10^3 moles of (CH_2OH)_2 = (46.47 × 10^3g) / (62g / mole) = 7.50 × 10^2 moles You can now solve for ∆T_fby substituting into the above equation ∆T_f= [1000 g (1.86 \textdegreeC / mole) (7.50 × 10^2mole)] / [(18 g / mole) (2.32 × 10^3 + 7.50 × 10^2 moles)] The normal freezing point of water is 0\textdegreeC, after adding the ethylene glycol, the freezing point becomes (0\textdegree-25.2\textdegreeC) or (-25.2\textdegreeC). Determine the amount of money it is possible to save by using a methanol-water mixture by calculating how much 41.68l (half of the volume of the tank) of methanol costs and subtracting this value from the cost of 41.68l of ethylene glycol. To calculate the cost of each 1) 1lb = 454 g. Since the costs are given in (cents / 1b) weight of determine the weight in pounds of 41.68 l of each. Earlier the weight of 41. 68 l of ethylene glycol was found to be (46.47 × 10^3g). Converting this to pounds, weight of (CH_2OH)_2 = 46.47 × 103g × (1 lb / 454 g) = 102.41b weight of (CH_3OH)= density × volume = (.79 g / ml) × (10^3 ml / l ) × 41.68l = 33.07 × 10^3 g Converting to pounds weight of CH_3OH = 33.07 × 10^3 g ×(1lb / 454 g) = 72.8 lb. The cost of each is found by multiplying the price per pound by the weight of each. cost of (CH_2OH)_2 = 102.4 lb × $.12 / lb = $12.29 cost of (CH_3OH) = 72.8 lb × $.08 / lb = $5.82 The difference in cost is $12.29 - 5.82 = $6.47. Hence, you save $6.47 by using methanol. You have found that a 50-50 mixture of water and ethylene glycol protects the car to a temperature of -25.2\textdegreeC. The most desirable antifreeze will protect the car at the lowest temperature, hence, to determine whetherehtyleneglycol is more desirable than methanol, calculate the lowest temperature at which the car will be protected by methanol. This is accomplished by using the above equation for freezing point depression. ∆T_f= [(1000 g (K_f) (N_2)] / [(M_1)(N2+ N_1)] Previously, one has determinedK_f, N_1and M_1 to be (1.86\textdegreeC / mole), (2.32 × 10^3) moles and (18 g / mole), respectively. Thus, to solve for ∆T_f determine N_2 first. As shown, the weight of the methanol is 33.07 × 10^3 g. Find N_2 by dividing this weight by the molecular weight of methanol, 32. No. of moles of methanol (N_2) = (33.07 × 10^3 g) / (32 g / mole) = 1.03 × 10^3 moles. Solving for ∆T_f: ∆Tf= [(1000 g)(1.86 \textdegreeC / mole)(1.03 × 10^3 moles)] / [(18 g / mole)(1.03 × 10^3 + 2.32 × 10^3 moles)] = 31.8\textdegreeC. The methanol will protect the car at temperatures as low as (0\textdegreeC-31.8\textdegreeC) or (-31.8\textdegreeC). This is fine while the car is not in motion and is cold, but when the car is driven the engine becomes very warm. Methanol has a boiling point of 60\textdegreeC and will evaporate while the engine is operating, whereas the ethylene glycol has a much higher boiling point and will not evaporate. Thus, if methanol is used as an anti-freeze it will have to be replaced after each use of the car while the ethylene glycol will last all winter. In the end it is far more expensive and troublesome to use methanol.

Question:

Calculate the magnetic induction B along the axis of a very long 2.5 × 10^-1 meters solenoid of 1000 turns of wire if the radius of the coil is 2 × 10^-2meters andthe current in the wire is 15 amperes.

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Solution:

Consider the rectangular contour (C) about the solenoid as shown in the figure. It encloses a total current NI where I is the current through the wire and N is the number of turns. By Ampere's Law \oint_c B^\ding{217} \textbullet dl^\ding{217} = \mu_0 (NI) where the constant of permeability \mu_0 = 4\pi × 10^-7 Weber/amp-m and dl^\ding{217} is the vector infinitesimal ele-ment of contour length. Assume there is no magnetic field outside the solenoid, and neglect fringing effects. Then, \oint B^\ding{217} \textbullet dl^\ding{217} = ^b\int_a B^\ding{217} \bullet dl^\ding{217} + ^c\int_b B^\ding{217} \bullet dl^\ding{217} + ^d\int_c B^\ding{217} \bullet dl^\ding{217} + ^a\int_d B^\ding{217} \bullet dl^\ding{217} = BL cos 0 o + Brcos 90 o + (0)L + Br cos 90 o = BL Hence, Ampere's law for the solenoid becomes BL = \mu_0 NI B = [(\mu_0 NI)/(L)] If N = 1000 turns, I = 15 amp, and L = 2.5 × 10^-1 m then B = {4\pi × 10^-7 [W/(a-m)] × 1000 × 15a}/(2.5 × 10^-1 m) = 7.54 × 10^-2 (weber/m^2 ) = .0754 (W/m^2 ) The radius of the solenoid does not appear in any of the calculations and is extraneous information.

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Question:

How would you define ecology? Differentiate between autecologyandsynecology.

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Solution:

Ecology can be defined as the study of the interactions between groupsof organisms and their environment. The termautecologyrefers to studiesof individual organisms or populations, of single species and their interactionswith the environ-ment.Synecologyrefers to studies of various groupsof organisms which are associated to form a functional unit of the environment. Groups of organisms are characterized by three levels of organization- populations,communititesandecocystems. A population is agroup of organisms belonging to the same species which occupy a given area. A community is a unit composed of a group of populations living in a givenarea. The community and the physical environment considered togetheris an ecosystem. Each of these designations may be applied to a smalllocal entity or to a large widespread one. Thus a small group of sycamoretrees in a park may be regarded as a population, as could be thesycamore trees in the eastern United States. Similarly, a small pond andits inhabitants or the forest in which the pond is located may be treatedas an ecosystem. From these examples, we see that the limit of an ecosystemdepends on how we define our ecosystem. However each ecosystemmust consist of at least some living organisms inhabiting a physicalenvironment. Various ecosystems are linked to one another by biological, chemicaland physical processes. The entire earth is itself a true ecosystembecause no part is fully isolated from the rest. This global ecosystemis usually referred to as the biosphere.

Question:

Use the principle of conservation of mechanical energy to find the velocity with which a body must be projected vertically upward, in the absence of air resistance, to rise to a height above the earth's surface equal to the earth's radius, R.

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Solution:

Let the center of the earth be the origin. Then the initial distance of the body is r_1 = R and its final position is r_2 = 2R. Let v_1 be the initial velocity. v_2, the final velocity, of the body of mass m is zero since 2R is the maximum height the body rises. Using conservation of energy, we have KE_1 + PE_1= KE_2 + PE_2 (1/2)mv^2_1 - G(mm_E/r_1) = (1/2)mv^2_2 - G(mm_E/r_2) wherem_E, is the earth's mass. Note that potential energy is negative. Substitution yields (1/2)mv^2_1 - G(mm_E/R) = 0 - G (mm_E/2R), v^2_1 =Gm_E/R.

Question:

What is meant by regeneration? Explain regeneration by usingplanariaas an example.

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Solution:

Regeneration is defined as the repair of lost or injured parts. Planaria , like hydra, has great powers of regeneration. Because of its larger size, theplanariahas been studied the most. If a planarian is cut in two pieces, each part will regenerate into a whole worm; that is, a new head will form on the tail-piece and a new tail will form on the headpiece. If a slice is taken out from the animal immediately posterior to the head, this slice will be seen to regenerate two heads. A larger section cut out from the abdomen will regenerate a head from one end and a tail from the other. It is also possible to cut the head region down the middle, keeping the cut surfaces apart, and have two heads regenerated. The capacity for this regeneration is related to cells calledneoblasts, which are part of the mesoderm, that retain the capability to proliferate and differentiate into the different tissues found inplanaria. The striking degree of regeneration, such as seen in hydra and planaria , becomes restricted in more complex animals due to greater cellular differentiation. A star-fish may regenerate its arms and an earthworm can re-generate a new head, but in man and other warm blooded vertebrates, the adult's capacity for regeneration is restricted to the healing of wounds.

Question:

For the reaction CO_2 (g) + H_2 (g) \rightleftarrows CO(g) + H_2O (g), the value of the equilibrium constant at 825\textdegreeK is 0.137. If 5.0 moles of CO_2 , 5.0molesof H_2 , 1.0 mole of CO , and 1.0 mole of H_2O are initially present, what is the composition of the equilibrium mixture ?

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Solution:

This problem is an application of the expression for the equilibrium constant . From thestoichiometryof the reaction, one mole of CO and one mole of H_2 O are produced for one mole of CO_2 and one mole of H_2 O that are reacted. Hence, if x moles of CO are produced at equili-brium, then x moles of H_2 O are produced, x moles of CO_2 are consumed, and x moles of H_2 are consumed. Therefore, at equili-brium, there are 1+x moles of CO, 1 + x moles of H_2 O, 5x moles of CO, and 5 \rule{1em}{1pt} x moles of H_2 . If we let v denote the volume of the reaction vessel, the equilibrium concentrations are [CO] = [H_2 O] = (1 + x)/v, and [CO_2] = [H_2] = (5 \rule{1em}{1pt} x)/v . Substituting these values into the expression for the equilibrium constant gives K= 0 137 = { [CO] [H_2 O] } / { [CO_2 ] [H_2 ] } = {[(1 + x)/v] [(1 + x)/v] } / {[(5 \rule{1em}{1pt} x)/v] [(5 \rule{1em}{1pt} x)/v] } = { (1+x)^2 v^2} / { (5\Elzbarx)^2 / v^2 } = { 1+2x+x^2 } / { 25\Elzbar10x + x^2 } . or 0.137(25) \rule{1em}{1pt} 0.137(10)x + 0.137x^2 = 1 + 2x + x^2 0.863x^2 + 3.370x \rule{1em}{1pt} 2.425 = 0 . Using the quadratic equation, x =[\Elzbar3.370 \pm \surd{(3.370)^2\Elzbar 4(0.863) (-2.425)}] / 2(0.863) or x = 0.62 , x =\Elzbar 4.52 . The second of these is nonphysical and is therefore discarded. The equilibrium concentrations are then [CO] = 1 + x = 1 + 0.62 = 1.62 moles [H_2 0] = 1 + x = 1 + 0.62 =1.62 moles [C0_2] = 5 \rule{1em}{1pt} x = 5 \rule{1em}{1pt} 0.62 = 4.38 moles [H_2] = 5 \rule{1em}{1pt} x = 5 \rule{1em}{1pt} 0.62 = 4.38 moles .

Question:

Which of the following organic compounds would you predict to be associated liquids: (a) CH_3OH; (b) CH_3OCH_3; (c) CH_3CI; (d) CH_3NH_2.

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Solution:

To answer this question, one should first understand the term, associated liquids. Liquids whose molecules are held together by hydrogen bonding are associated liquids. Hydrogen bonding is an especially strong kind of attraction between a hydrogen atom (that is bound to a highly electronegative atom) and another highly electronegative atom. This type of bond is formed by purely electrostatic forces between the positive end of one polar molecule and the negative end of another polar molecule. For hydrogen bonding to be important, the electronegative atoms must be F, O, or N. Thus, once the structures of the species are known, one can tell if they are associated liquids. (a) CH_3OH, (methanol); It is an associated liquid because oxygen is bonded to the hydrogen atom. And the hydrogen atom thus forms a bond with another oxygen molecule of methane. Liquid (b), CH_3OCH_3 (dimethyl ether); is not an associated liquid; the oxygen is not bonded directly to any hydrogen atom and therefore cannot form any hydrogen bonds. Liquid (c), CH_3CI, (methyl chloride); is not an associated liquid because Cl is not electro-negative enough to form a stable hydrogen bond. Liquid (d), CH_3NH_2, (methyl amine); is an associated liquid, N is highly electronegative and the hydrogen can form hydrogen bonds with nitrogen atoms of other CH_3NH_2 molecules.

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Question:

What are the functions of roots? What are the two typesof rootsystems?

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/Users/wenhuchen/Documents/Crawler/Biology/F10-0241.htm

Solution:

Roots serve two important functions: one is to anchor the plant in thesoil and hold it in an upright position; the second and biologically more im-portantfunction is to absorb water and minerals from the soil and conductthem to the stem. To perform these two functions, roots branch andrebranchextensively through the soil resulting in an enormous total surfacearea which usually exceeds that of the stem's. Roots can be classifiedas a taproot system (i.e. carrots, beets) in which the primary (first) root increases in diameter and length and functions as a storage placefor large quantities of food. A fibrous root system is com-posed of manythin main roots of equal size with smaller branches. Additional roots that grow from the stem or leaf, or any structure otherthan the primary root or one of its branches are termed adventitious roots. Adventitious roots of climbing plants such as the ivy and other vines attachthe plant body to a wall or a tree. Adventitious roots will arise from thestems of many plants when the main root system is removed. This accountsfor the ease of vegetative propagation of plants that are able to produceadventitious roots.

Question:

What factors affect the rate of transpiration?

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/Users/wenhuchen/Documents/Crawler/Biology/F10-0257.htm

Solution:

A number of different factors affect the rate of transpiration. The most important of these is temperature. Because temperature is a measurement of the average molecular kinetic energy, a high temperature heating up the plant body will provide the water molecules within the leaves with increased internal energy, and many of them may soon gainadequate kinetic energy to enter the gaseous phase and escape into the atmosphere . The higher the temperature rises, the greater will be the rate of transpiration, all other factors remaining constant. Plants of the desert, thus , must adapt some mechanisms to slow down water loss due to rapid evapo-ration by the hot desert air. Humidity is a second important factor. Water is lost much more slowly into air already laden with water vapor. This is why plants growing in shady forests, where the humidity is generally high, typically spread large luxuriant leaf surfaces since their chief problem is getting enough sunlight , not losing water. In contrast, plants of grasslands or other exposed areas often have narrow leaves and relatively little total leaf surface , since they are able to get sufficient light but are constantly in danger of excessive water loss. Air currents also affect the rate of transpiration. Wind blows away the water vapor from leaf surfaces, allowing more vapor molecules to escape from the leaf surface. If the air is humid, wind may actually decrease the transpiration rate by cooling the leaf, but if the breeze is dry, evaporation is increased and the trans-piration rate is greatly enhanced. Leaves of plants growing in windy, exposed areas are often hairy. These hairs are believed to protect the leaf surface from wind action by trapping a layer of air saturated with water vapor over the surface of the leaf. As before , humid air reduces the rate of evaporation, thus preventing excessive loss of water.

Question:

Write a FORTRAN function subprogram that determines whether a particular item is contained in an ordered binary tree. All items in the tree are assumed to be integers.

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Solution:

A binary tree is said to be ordered if the follow-ing properties hold: For each node in the tree, all numbers contained in the left subtree are less than the number at the node being considered. All numbers contained in the right subtree of a node are greater than the number at that node. When a search is begun, we start at the root, turning left if the number being searched is smaller, or turning right if the number is greater. Consider the tree below: We must set up a linked list, complete with two pointers at each node, one for each branch. We also require a pointer for the root node. To implement this in FORTRAN, we use a two-dimensional array called TREE (100,3). We are searching the tree for an integer between 0 and 100, hence the size of the first parameter. The second parameter refers to the data for each node and the two pointers which check, respectively, the left and right subtrees. The following table shows the initial status of the array: I TREE(I, 1) TREE(I, 2) TREE(I, 3) 1 3 0 0 2 15 1 4 3 19 0 0 4 20 3 5 5 22 0 0 6 25 2 8 7 29 0 0 8 39 7 9 9 40 0 10 10 73 0 0 Array Component Node Data Left Subtree Right Subtree The values of the pointers represent row numbers in the array. We take the root of the tree to be 6. The strategy used involves logical variables. We de-clare the function to return a value of .TRUE, if the element is in the tree, and to return a value of .FALSE, if the ele-ment is not found. Nodes are checked in succession. If the value is not found at a particular node, the left branch is investigated first. Then the right branch is checked. If the node is terminal (i.e., having no branches), we exit from the program. The value of K is used to indicate the node under current consideration. The steps are given as follows. Comments will illumi-nate the modular structure of the program. LOGICAL FUNCTION SEARCH (TREE, ROOT, ITEM) INTEGER TREE (100,3), ROOT, ITEM, K K = ROOT CLOOP BEGINS HERE 10CONTINUE CIF NODE IS FOUND, SET SEARCH = .TRUE. AND EXIT IF (TREE (K,1).NE. ITEM) GO TO 20 SEARCH = .TRUE. GO TO 99 20CONTINUE CTRY SEARCHING THE BRANCHES, STARTING WITH CTHE LEFT IF (TREE(K,1).LT. ITEM) GO TO 30 CIF THIS NODE IS TERMINAL,SET SEARCH = .FALSE. AND EXIT IF (TREE (K,2) . NE. 0) GO TO 40) SEARCH = .FALSE. GO TO 99 40CONTINUE K = TREE (K,2) GO TO 98 CNOW TRY THE RIGHT BRANCH; IF NODE IS CTERMINAL, SET SEARCH\textasteriskcentered.FALSE. AND EXIT IF (TREE(K,3).NE.0) GO TO 50 SEARCH = .FALSE. GO TO 99 50CONTINUE K = TREE (K,3) 98CONTINUE GO TO 10 CEND LOOP 99CONTINUE RETURN END

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Question:

Suppose the change HC_2 O_4 ̄ + Cl_2 CO_3 ^2 ̄ +Cl ̄ is to be carried out in basic solution. Starting with 0.10 mole of OH ̄, 0.10 mole of HC_2 O_4 ̄, and 0.05 mole of Cl_2, how many moles ofCl ̄ would be expected to be in the final solution?

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Solution:

The equation for this reaction is HC_2 O_4 ̄ + Cl_2 + 5OH ̄ \rightarrow 2Cl ̄ + 2CO_3 ^2 ̄ + 3H_2 O From this equation one knows that 1 mole of HC2O_4 ̄, 1 mole of Cl_2 / and 5 moles of OH ̄ are needed to form 2 moles ofCl ̄. One needs 5 times as much OH ̄ as H_C O ̄ or Cl_2. This means that if one wishes to react 0.10 moles of HC_2 O_4, or Cl_2 there must be 0.50 moles of OH ̄ present. There are 0.10 moles of OH ̄ present, thus one needs only .02molesof each HC_2 O_4 ̄ and Cl_2. Some of these reactants will re-main unchanged. From the equation, one notes that there are 2 moles ofCl ̄ formed for each Cl_2 reacted. If .O_2 moles of Cl_2 are used then .04 moles ofCl ̄ will be formed. OH ̄ is called the limiting reagent.

Question:

Define the terms \textquotedblleftrange\textquotedblright and \textquotedblleftcenter of origin.\textquotedblright

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/Users/wenhuchen/Documents/Crawler/Biology/F28-0728.htm

Solution:

The range of a given species is defined as that portion of earth in whichit is found. The range may be small or large. Thekangeroos, for example, are found only in Australia, while man is found almost all over theworld. The range of a species does not neces-sarily include its center oforigin. The center of origin of a species is that particular place where it originated, since each species of plant or animal originated only once. It is nowknown that a new species arises in a particular geographic area. If thespecies is favored for survival, it spreads out until it is stopped by a barrierof some kind - physical, such as an ocean; environmental, such as anadverse climate; or biological, such as the absence of food, danger frompredators, the failure to compete with native members, or the inability to produce fertile young.

Question:

What are the chief components and structures of a eucaryotic cell?

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Solution:

Membranes are a crucial component of both animal and plant cells. The plasma membrane surrounds the cell and serves to separate the internal living matter from the external environment. Plant cells have in addition a cell wall external to the plasma membrane. It is composed mainly of cellulose and is fairly rigid but permeable. Selectivity of materials entering the plant cells is not a function of the cell wall but of the plasma membrane. Membranes are also present inside the cell, dividing the cell into compart-ments distinctive in both form and function. All membranes are composed basically of lipids and proteins. For this reason the plasma membrane and the membranes inside the cell are termed unit membranes. Subcellular structures in the cytoplasm are known as organelles. When these are surrounded by membranes, they are termed membrane-bounded organelles. Each organelle performs some specific functions. The nucleus is the controlling center of the cell. The nucleus is surrounded by two layers of unit membrane which form the nuclear envelope. Within the nucleus are found the chromosomes, the substances of inheritance. Chromosomes are composed chiefly of deoxyribonucleic acid (DNA) and protein. Genes located within the chromosomes, direct cellular function and are capable of being replicated in nuclear division. The nucleolus is a specialized region in the nucleus involved in the synthesis of ribosomal RNA , the material making up the ribosomes. Mitochondria and chloroplasts are membrane-bounded organelles involved in energy production in the cell. Since energy can neither be created nor destroyed, these organelles actually convert one form of energy to an-other. Chloroplasts found in plant cells, convert solar energy to chemical energy contained in organic substances; the latter are oxidized in the mitochondria to yield energy useful to the cell in the form of ATP. Cells which utilize food for energy must take food into the cell and degrade it. Lysosomes are membrane- bound organelles whose digestive enzymes break down organic substances to simpler forms, which can be used by the cell to yield energy for its life-sustaining process. Most cells contain membrane-delimited bodies called vacuoles. Small vacuoles may be termed vesicles. These structures may contain the ingested materials taken from the cell exterior or materials to be released by the cell. Mature plant cells usually contain a single large fluid- filled vacuole. This vacuole aids the cell in maintaining an internal pressure and thus rigidity. Organelles involved in the synthesis and transport of cellular components are the ribosomes, endoplasmic reticulum, and Golgi apparatus. Ribosomes are involved in protein synthesis. The endoplasmic reticulum is a system of membranes providing transport channels within the cell. It transports the synthesized protein to other parts of the cell. The Golgi apparatus is involved in the packaging of cellular products before they can be released by the cell to the outside. Organelles involved in the maintenance of cellular shape and in movement are the microfilaments and micro-tubules, also known as the skeleton of the living cell. Microfilaments are involved in the connection of adjacent cells for intercellular communication. They also function in the transport of products within the cell. The micro-tubules are the basic substance in the cilia and flagella of motile cells. By the contracting action of micro-tubules, flagella and cilia beat and propel the cell. The microtubules have also been identified as the fundamental substance in the spindle apparatus during cell division. Both microfilaments and microtubules are protein structures.

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Question:

(a) What is the magnitude of the electric field at a distance of 1A\textdegree (= 10^-8 cm) from a proton? (b) What is the electrostatic potential at this distance? (c) What is the potential difference in volts between positions 1A\textdegree and 0.2A\textdegree from a proton? (d) A proton is released from rest at a distance of 1A\textdegree from another proton. What is the kinetic energy when the protons have moved infinitely apart? If one proton is kept at rest while the other moves, what is the terminal velocity of the moving proton? (e) A proton is accelerated from rest by a uniform electric field. The proton moves through a potential drop of 100 volts. What is its final kinetic energy? (Note that 100 volts \cong 0.33statvolts.)

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Solution:

(a) From Coulomb's law E = e/r^2 \cong (5 × 10^-10esu)/(1 × 10^-8cm)^2 = 5 × 10^6statvolts/cm StatVolt/cm is the dimension of electric field. The field is directedradially outward from the proton. (b) Electrostatic potential is given by ^\infty\int_r E^\ding{217} \bulletds^\ding{217}. By convention, we shall assume that the potential is 0 at infinity, and we shall set our limits of integration accordingly. \varphi(r) = ^\infty\int_r(e/r^2)dr= \rule{1em}{1pt}(e/r) ^\infty\vert_r = e/r \cong (5 × 10^-10esu)/(1 × 10^-8 cm) \cong 5 × 10^-2statvolts. (c) The potential difference between two points is ^p2\int_p1 E^\ding{217} \bulletds^\ding{217} where p_1 and p_2 are the two points. Therefore, ^p2\int_p1 E^\ding{217} \bulletds^\ding{217} = ^p2\int_p1 (e/r^2)dr= ^(1×10)-8cm\int_(0.2×10)-8cm (e/r^2)dr = \rule{1em}{1pt}(e/r) ^(1×10)-8cm\vert_(0.2×10)-8 cm \cong \rule{1em}{1pt} [(5 × 10^-10esu) / (1 × 10^-8 cm)] + (5 × 10^-8esu) / (0.2 × 10^-8 cm) \cong 0.2statvolts (d) By conservation of energy we know that the kinetic energy must equal the original potential energy. Potential energy is given by \intF^\ding{217} \textbulletds^\ding{217}, where potential energy is zero at infinity. The force is given byqE. Therefore, ^\infty\int_(1×10)-8 cmqE\bulletds= ^\infty\int_(1×10)-8 cm (e^2/r^2) \bulletdr = \rule{1em}{1pt} (e^2/r)^\infty\vert(1x10)-8 cm = (4.8 × 10^-10esu)^2 / (1 × 10^-8 cm) \cong 23 × 10^-12 erg If one proton is kept at rest while the other moves, the terminal velocity of the moving proton is given by (using conservation of energy) (1/2) Mv^2 \cong 23 × 10^-12 erg v^2 \cong (2 × 23 × 10^-12 erg) / (1.67 × 10^-24 gm) \cong [2 × 23 × 10^-12 (gm - cm^2/sec^2)] / [1.67 × 10^-24 gm] \cong 27 × 10^12(cm^2/sec^2) v\cong 5 × 10^6 cm/sec (e) The kinetic energy will be equal to the change in potential energy. This is equal to ^p2\int_p1 F \textbulletds. Here F =qEand energy is^p2\int_p1qE\bulletds = q^p2\int_p1 E \bulletds= q\varphi_12, or (4.8 × 10^-10statcoul) (0.33statvolt) \cong 1.6 × 10^-10 erg.

Question:

Why would it be disadvantageous for a host-specificparasite tokill its host?

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/Users/wenhuchen/Documents/Crawler/Biology/F30-0767.htm

Solution:

In the course of their evolution, parasites usually develop special featuresof behavior and physiology that make them better adjusted to the particularcharac-teristics of their host. This means that they often tend to becomemore and more specific for their host. Where an ancestral organismmight have parasitized all species in a particular family, its variousdescendants may parasitize only one species of host at each stagein its development. This means that many parasites are capable of livingonly in one specific host species and that if they should cause the extinctionof their host, then they themselves would also become extinct. A dynamicbalance exists where the host usually survives without being seriouslydamaged and at the same time allowing the parasites to moderatelyprosper. Probably most long-established host-parasite relationshipsare balanced ones. The ideally adapted parasite is one that canflourish without reducing its host's ability to grow and reproduce. For everyone of the parasitic species that cause serious disease in man and otherorganisms, there are many others that give their hosts little or no trouble. It is generally true that the deadliest of the parasites are the ones thatare the most poorly adapted to the species affected. Relationships thatresult in serious disease in the host are usually relatively new ones, or onesin which a new and more virulent form of the parasite has recently arisen, or where the host showing the serious disease symptoms is not theprimary host of the parasite. Many examples are known where man is onlyan occasional host for a particular parasite and suffers severe diseasesymptoms, although the wild animal that is the primary host showsfew ill effects from its relationship with the same parasite.

Question:

Compute the equivalent resistance of the network in Figure 1, and find the current in each resistor.

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/Users/wenhuchen/Documents/Crawler/Physics/D20-0684.htm

Solution:

Successive stages in the reduction to a single equivalent resistance are shown in Figs. 1 and 2. The 6\Omega and 3\Omega resistors in Fig. 1 are in parallel, hence they are equivalent to the [(6\Omega × 3\Omega)/(6\Omega + 3\Omega)] = 2\Omega resistor in Fig. 2. The series combination of this 2\Omega resistor with the 4\Omega resistor gives the 6\Omega resistor in Fig. 3. The total current I_1, in the circuit is I_1 = [(18v)/(6\Omega)] = 3 amp. The current I_2 through the 60 resistor in Fig. 1 is I_2 = [(V_cb )/(6\Omega)] where V_cb is the potential across the 6\Omega and 3\Omega resistors. Similarly, the current I_3 through the 3\Omega resistor is I_3 = [(V_cb )/(3\Omega)] V_cb is obtained by considering the loop equation for Fig. 2; V_cb = I_1 × 2\Omega = 3 amp × 2\Omega = 6v. Therefore I_2 = [(6V)/(6\Omega)] = 1 amp, I_3 = [(6V)/(3\Omega)] = 2 amp.

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Question:

A building society lends money to house purchasers subject to a monthly payment of one per cent of the amount borrowed . This payment covers both capital repayment and interest due, interest being charged at a rate of eight percent per annum, calculated monthly. Write a program which reads in the amount of a loan, tabulates the series of payments required, as a four column table showing the payment number, the interest.duethat month, the capital repayment that month, and the outstanding capital balance. The table should also show the final non-standard payment required to complete the payment.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G16-0417.htm

Solution:

PROGRAM LOANPAYMENT (INPUT, OUTPUT); (\textasteriskcentered This program reads in the amount of a loan \textasteriskcentered) (\textasteriskcentered and prints the repayment table showing \textasteriskcentered) (\textasteriskcentered the payment number, the monthly interest, \textasteriskcentered) (\textasteriskcentered the interest repayment, and the residual \textasteriskcentered) (\textasteriskcentered balance. \textasteriskcentered) CONST ANNUALRATE =8;(\textasteriskcentered per cent \textasteriskcentered) VARNUMBER :1 . . MAXINT; INTEREST, REPAYMENT, LOAN, PAYMENT, RESIDUE, MONTHRATE:REAL; BEGIN READ (LOAN); WRITE ('AMOUNT BORROWED =' ,LOAN: 12:2); PAYMENT: = LOAN/100; WRITELN (' ':10, 'MONTHLY REPAYMENT =', PAYMENT:10:2); WRITELN; WRITELN; WRITELN ('NUMBER INTEREST REPAYMENT RESIDUE'); WRITELN; MONTHRATE: =ANNUALRATE/100/12; NUMBER: = 1; RESIDUE: = LOAN; REPEAT INTEREST: = MONTHRATE\textasteriskcenteredRESI DUE; REPAYMENT: = PAYMENT - INTEREST; RESIDUE: = RESIDUE - REPAYMENT; WRITELN (NUMBER: 7, INTEREST: 10:2, REPAYMENT; 10:2, RESIDUE:10:2 ); NUMBER: = NUMBER +1 UNTIL RESIDUE+RESI DUE \textasteriskcenteredMONTHR ATE <= PAYMENT; WRITELN; WRITELN; WRITELN ('LAST PAYMENT=', RESIDUE+RESIDUE \textasteriskcenteredMONTHRATE: 10:2) END.

Question:

Why does a physician advise reduced salt intake in cases of hypertension?

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/Users/wenhuchen/Documents/Crawler/Biology/F15-0389.htm

Solution:

Hypertension or high blood pressure, is caused by abnormalities in arterial pressure regulation. The mean arterial pressure (M.A.P.) is the average pressure throughout the pressure pulse cycle. It is not equal to the average of the systolic and diastolic pressure, since the heart remains in diastole twice as long as in systole. M.A.P. = [(SBP + 2DBP) / 3] = [{120 + 2(80)} / 3] = 93.3 Thus mean arterial pressure is closer to the diastolic than to the systolic blood pressure. Arterial pressure is determined by two factors: cardiac output and total peripheral resistance. The relationship between them is: Pressure = Cardiac Output × Total Peripheral Resistance Cardiac output is the amount of blood pumped by each ventricle per unit time, and is determined by both the heart rate and the stroke volume. Total peripheral resistance is the resistance of the entire systemic circulation. Any factor that causes an increase in cardiac output or total peripheral resistance will cause an increase in mean arterial pressure. One such factor is the secretion of the hormonealdosteroneby the adrenal cortex.Aldosteroneacts on the tubules of the kidney to increase thereabsorptionof sodium. This increased retention of sodium causes more water to be retained, thereby increasing the volume of the blood. This increased blood volume increases the stroke volume and hence the cardiac output. Arterial blood pressure increases. When excessive aldosteroneis secreted (such as by a tumor in the adrenal cortex), hypertension results. Salt, or sodium chloride, is a source of sodium. A decreased salt intake causes less water to be retained, decreasing blood volume, and in turn lowering blood pressure. Blood pressure can also be increased by vasoconstriction (constriction of the blood vessels). Constriction causes an increased resistance to flow (since a decrease in vessel radius increases vascular resistance). A protein calledangiotensinII causes both vasoconstriction and stimulation ofaldosteroneproduction and secretion.AngiotensinII is producedenzymaticallyfromangiotensinI, which is converted from angiotensinogen, a glycoprotein made in the liver. The enzyme rennin, formed in the kidney, catalyzes this conversion. Kidney damage can cause anoversecretionof rennin, which causes an increase of angiotensinII, and thus hypertension. Slight hypertension usually accompanies arteriosclerosis, the deposition of fatty material (such as cholesterol) in the arterial walls. The arteries become fibrous and calcified. This buildup of deposits inhibits the flow of blood, increasing resistance to flow and thus increasing the blood pressure needed to overcome the resistance. Oversecretionof rennin andaldosteroneand arteriosclerosis can thus cause hypertension, but in most cases (90%), the cause is unknown. These cases, known as \textquotedblleftessential\textquotedblright hypertension (because of unknown etiology) appear to be hereditary. Hypertension is harmful because it increases the work load of the heart and damages the arteries by subjecting them to excessive pressure. Limiting the amount of salt in foods may help to alleviate hypertension.

Question:

For underground electrical transmission cables using insu-lating materials, the power loss in the insulator is propor-tional to V^2, where V is the potential applied to the cable. Compare the power loss per meter for two standard transmission voltages, 69,000 V and 345,000 V.

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/Users/wenhuchen/Documents/Crawler/Physics/D24-0793.htm

Solution:

The available power is given by the relation-ship, P = VI(1) where I is the current through the cable. I can be re-placed by I = (V/R), by ohm's Law, where R is the cable re-sistance. Then, from (1), P = (V^2/R) If R is constant, P \propto V^2(2) For the transmission cable, power loss is proportional to the available power and, by (2), is therefore also proportional to V^2 . If the same cable is used for both poten-tials, the following ratio provides a comparison: (power loss of 345,000-V line/ power loss of 69,000-V line) = {(3.45 × 10^5V)^2/(6.9 × 10^4 V)^2} = 25

Question:

Assume X and Y are two arrays, each already sorted in ascending order, with M and N elements respectively. Merge X and Y into a single sorted array Z.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G07-0145.htm

Solution:

One method is to compare the first elements of each array. The smaller of the two will become the first member of the new array, while the greater of the two will come next. This picture will illustrate the idea: The flow chart is as follows: INTERIOR L,M,N,K,NX,NY,X(M) ,Y(N) L = M + N K = 1 NX = 1 NY = 1 20IF (X(NX).LT.Y(NY)) GO TO 50 CTHIS SECTI0N EXECUTES ALL STATEMENTS C\ON THE RIGHT SIDE 0F THE FL0W CHART. Z (K) = Y (NY) K = K + 1 NY = NY + 1 IF (NY.LE.N) GO TO 20 D\O 30 IY = K, L Z (I) = X (NX + IY - K) 30C\ONTINUE G\O T\O 9\O CTHIS SECTION EXECUTES ALL STATEMENTS ON THE CLEFT SIDE OF THE FLOWCHART. 50Z (K) = X (NX) K = K + 1 NX = NX + 1 IF (NX.LE.M) G\O T\O 2\O D\O 40 IX = K, L Z (I) = Y (NY + IX - K) 40C\ONTINUE 90 ST\OP END [Note: Merge sorts of this type are used frequently in business applications, in which large files must be updated with new material.]

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Question:

Let us assume that the conduction electrons in a metal are completely free except for a retarding force which is pro-- portional to the speed, a) Find the variation of the electron velocity with time when a constant electric field is set up in a metal, b) How long does it take for the electrons to reach maximum velocity?

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/Users/wenhuchen/Documents/Crawler/Physics/D19-0632.htm

Solution:

a) The acceleration of an electron due to an electric field is given by m_e (dv/ d_t)1=Ee where v is the velocity of the electron, and e is the electronic charge. The retarding force gives rise to a deceleration given by m_e (d_v / d_t)_2 = \rule{1em}{1pt}bv where b is some constant. The equation of motion is then m_e (d_v / d_t) = Ee \rule{1em}{1pt}bv The solution of this equation is (see the figure) v(t) = eE / b(1\rule{1em}{1pt} e\rule{1em}{1pt}t / T) where T = m_e / bis called the relaxation time. We see that as t becomes large, a terminal velocity v_t = eE / b is established b)If we write v_t as v_t = (eE / m_e ) (me/ b) = (eE / m_e )T it turns out to be just that velocity which an electron would attain in a time T with acceleration eE / me, v_t = aT = (eE / M_e )T.

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Question:

Define stack, queue, anddeque. What characteristics of these three list structures make each of them useful for different applications?

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G07-0137.htm

Solution:

A queue is a type of list structure with the lim-itation that items may be added or deleted at either end, but not in the middle. The term queue is also used in con-nection with computer memory to mean a group of items chained together sequentially by addresses. But here, we will dis-cuss the queue as a way of entering data at the macro level. Queue is a general term which describes several types of lists. One type of queue is known as a FIFO queue, which stands for "first-in-first out." Items that are added to a FIFO queue at one end are removed at the other. The other type of queue is a LIFO queue, which stands for "last-in- first-out." This type of queue is known as a stack. Stacks have the property that the items are removed in the reverse of the order in which they are entered. In addition, a stack allows insertions and deletions to take place only at one end. The termdequeis an acronym for "double-ended queue", in which items may be added to, or deleted from, either end. This type of list may be constructed with a double set of pointers, one to keep track of items moving up the list, and one to keep track of items moving down the list. One sample application of queues is in the simulation of customer arrival at a ticket window. New customers coming up to the window can be placed at one end of a queue which represents the line of prospective ticket buyers. When the tickets are bought, the customer is removed from the list at the opposite end. Using this concept, the owner of the ticket booth can simulate the flow of customers and plan his business activities accordingly. Stacks are especially useful in processing data struc-tures such as trees. Stacks can also be used when converting expressions in algebraic notation to Polish expressions. Recursive procedures, sometimes found in PL/1 programs, use the notion of stacks to allocate additional memory space for themselves. Deques, since they can be constructed with two sets of pointers, are ideal for two-way, or undirected, lists. This allows the user to search a list back and forth from either end.

Question:

In man, the kidney performs the bulk of the excretion of wastes from the body. Outline the structure of the human kidney and urinary system.

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/Users/wenhuchen/Documents/Crawler/Biology/F18-0441.htm

Solution:

Located on each posterior side of the human body just below the level of the stomach are the bean-shaped kidneys. Each kidney is about 10 cm long, and consists of three parts: an outer layer called the cortex, an inner layer called the medulla, and a sac-like chamber called the pelvis (see figure 1) . The functional unit of a kidney is the nephron; there are about a million nephrons per kidney. A nephron consists of two components: a tubule for conduct-ing cell-free fluid and a capillary network for carrying blood cells and plasma. The mechanisms by which the kidneys perform their functions depend on both the physical and physiological relationships between these two components of the nephron. Throughout its course, the kidney tubule is composed of a single layer of epithelial cells which differs in structure and function from one portion of the tubule to another. The blind end of the tubule is Bowman's capsule, a sac embedded in the cortex and lined with thin epithelial cells. The curved side of Bowman's capsule is in intimate contact with the glomerulus, a compacttuft of branching blood capillaries, while the other opens into the first portion of the tubular system called the proximal con-voluted tubule. The proximal convoluted tubule leads to a portion of the tubule known as the loop of Henle. This hairpin loop consists of a descending and an ascending limb, both of which extend into the medulla. Following the loop, the tubule once more becomes coiled as the distal convoluted tubule. Finally, the tubule runs a straight course as the collecting duct. From the glomerulus to the beginning of the collecting duct, each of the million or so nephrons is completely separate from its neighbors. However, the collecting ducts from separate nephrons join to form common ducts, which in turn join to form even longer ducts, which finally empty into a large central cavity, the renal pelvis, at the base of each kidney. The renal pelvis is con-tinuous with the ureter, which empties into the urinary bladder where urine is temporarily stored. The urine remains unchanged in the bladder, and when eventually excreted, has the same composition as when it left the collecting ducts. Blood enters the kidney through the renal artery, which upon reaching the kidney divides into smaller and smaller branches. Each small artery gives off a series of arterioles, each of which leads to a glomerulus. The arterioles leading to the glomerulus are called afferent arterioles. The glomerulus protrudes into the cup of Bowman's capsule and is completely surrounded by the epithelial lining of the capsule. The functional signi-ficance of this anatomical arrangement is that blood in the capillaries of the glomerulus is separated from the space within Bowman's capsule only by two extremely thin layers: (1) the single-celled capillary wall, and (2) the one- celled lining of Bowman's capsule. This thin barrier permits the filtration of plasma (the non-cellular blood fraction) from the capillaries into Bowman's capsule. Ordinarily, capillaries recombine to form the begin-nings of the venous system. However, glomerular capil-laries instead recombine to form another set of arterioles, called the efferent arterioles. Soon after leaving the region of the capsule, these arterioles branch again forming a capillary network surrounding the tubule. Each excretory tubule is thus well supplied with circulatory vessels. The capillaries eventually rejoin to form venous channels, through which the blood ultimately leaves the kidney.

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Question:

What is meant by the term muscle fatigue?

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/Users/wenhuchen/Documents/Crawler/Biology/F19-0471.htm

Solution:

A muscle that has contracted several times, exhausting its stored supplyof organic phosphates and glycogen, will accumulate lactic acid. (Review chapter 3, cellular metabolism.) It is unable to contract any longer andis said to be "fatigued". Fatigue is primarily induced by this accumulationof lactic acid, which correlates closely with the depletion of themuscle stores of glycogen. Fatigue, however, may actually be felt by theindividual before the muscle reaches the exhausted condition. The spot most susceptible to fatigue can be demonst-rated experimentally. A muscle and its attached nerve can be dissected out and thenerve stimulated repeatedly by electric shock until the muscle no longercontracts. If the muscle is then stimulated directly by placing electro-deson the muscle tissue, it will contract. With the proper device for detectingthe passage of nerve impulses, it can be shown that upon fatigue, the nerve leading to the muscle is not fatigued, but remains capableof con-duction. Thus, since the nerve is still conducting impulses andthe muscle is still capable of contracting, the point of fatigue must be atthe junction between the nerve and the muscle, where nerve impulses initiatemuscle contrac-tion. Fatigue is then due in part to an accumulation oflactic acid, in part to depletion of stored energy reser-ves, and in part to breakdownin neuromuscular junction transmission. In contrast to true muscle fatigue, psychological fatigue may cause anindividual to stop exercising even though his muscles are not depleted ofATP and are still able to contract. An athlete's performance depends not onlyupon the physical state of his muscles but also upon his will to perform.

Question:

Determine themolarityof an H_2SO_4, solution, 15.0 ml of which is completely neutralized by 26.5 ml of 0.100 M NaOH.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E10-0366.htm

Solution:

Acids and bases react with each other to produce salts and water. Such reactions are called neutralizations. To find themolarityof the neutralized acid solution, first write out the balanced reaction. H_2SO_4 + 2NaOH \rightarrow Na_2SO_4 + 2HOH. When neutralization is complete, there is no longer any acid or base left. It is given that 26.5 ml of 0.100 MNaOHis used. From this, one wants to compute how many moles ofNaOHwere present. It is known from the neutralization reaction that the number of moles of acid must be 1/2 the number of moles of base for a complete neutralization. From the number of moles of base present, the number of moles of acid needed are determined. Sincemolarity= moles/liter and one is given the volume, themolaritycan be calculated. Proceed as follows: There exists 26.5 ml of 0.100 MNaOH. There are 1,000 ml in one liter. This means that 26.5 ml is equal to 0.0265 liters. Recalling the definition ofmolarity. 0.1 = [(no. of moles ofNaOH)/(0.0265 l)] . Solving moles = 0.00265 ofNaOH. This, then, must be twice the number of moles of acid. For the acid, therefore,molarity= [(.00265)(1/2)]/liters. One is told that the volume of the H_2SO_4 solution is 15 ml, or 0.015 liters (1000 ml = 1 liter).Molarityof the acid becomes (0.001325/0.015) = 0.088 M.

Question:

What kind of marine life is found in the different regions of the ocean and what are these organisms' adaptations to their environment?

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/Users/wenhuchen/Documents/Crawler/Biology/F30-0788.htm

Solution:

The neritic, or coastal, waters support a greater population of marine life forms than do those of the open ocean. This abundance results from the rich supply of nitrates, phosphates, and other nutrients dumped into coastal waters by rivers and streams or brought to the surface by upwellings and turbulence. These substances are required by the producers who are the photosynthetic organisms that form the base of the food chain for animal life. Also, the ocean floor receives much more energy from sunlight here and as a result there is a great luxury of both plants and animals in the neritic waters. More is known about the communities of the intertidal and subtidal regions than about those of the lower neritic region. On a typical sandy shore (Fig. 1) the dominant animals in the intertidal region might include ghost shrimps, burrowing isopods, burrowing amphipods, mole crabs, polychaete worms, and beach clams, and the dominant forms in the subtidal region might include hermit crabs, sand dollars, borrowing shrimp, ascidians, and copepods. In contrast, on a rocky shore (Fig. 2) the dominant animal forms in the intertidal region might include barnacles, oysters, and mussels, with such forms as sea anemones, sea urchins, and corals in the subtidal region. On mud bottoms, various kinds of burrowing clams and snails are found. Many of the bottom dwellers are adapted to feed on plankton and dead organism. The organisms of the intertidal region are subjected to alternating periods of exposure to the air and to the strong action of waves. They are adapted to these en-vironmental factors in a number of ways. For example, the seaweeds have tough outer coverings and pliable bodies. Such animals as crabs, barnacles, mollusks and starfish have hard calcerous shells, whereas others like the sea anemones have a leathery outer covering. Also, many animals are able to burrow in the sand to escape exposure to air. Neritic pelagic communities include a great variety of fish species, larger crustaceans, turtles, seals, whales, and others. Most of the fish are plankton feeders - the adult fish feed mostly on zooplankton and are thus secondary consumers. Some fish, such as sharks, are dangerous predators. The euphotic zone of the oceanic region is inhabited by planktonic organisms and a variety of nektonic forms, including many kinds of fish. Mackeral and herring are among the commercial fish found in this oceanic region. Whales may also be present, as well as many kinds of birds. The bathyal and abyssal zones are incomplete ecosystems since there can be no producers where light cannot reach. The food of the inhabitants consists partly of the dead bodies of plants and animals that rain down from above, and partly of each other. A variety of species of fish, crustaceans, echinoderms, worms and mollusks are found in these zones and on the ocean floor. Many of the inhabitants posses light-producing structures which are significant adaptation in this darkened habitat. The deep sea forms are adapted to withstand great pressure by keeping the pressure uniform inside their bodies and on the outside. In the hadal zone at the great depths of the Pacific, organisms are subjected to pressures greater than 1000 atmospheres. At a depth of 10,500 meters, sea cucumbers, sea anemones, bivalve mollusks and crustaceans have been found. This is close to the deepest part of the ocean yet discovered, which is 10,860 meters.

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Question:

In humans, the liver is one of the largest and most active organsin the body. Describe its functions.

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Solution:

One of the most important functions of the liver is to regulate the levelof glucose in the blood stream. The maintenance of a constant level ofblood glucose is essential because specialized cells, such as the brain cells, are easily damaged by slight fluctuations in glucose levels. Excess glucoseis stored in the liver as glycogen while declining blood glucose canbe restored by conversion of liver glycogen into glucose. In addition to theformation of glycogen, the liver can convert glucose into fat, which is theneither stored in the liver itself or in special fat-deposit tissues known asadipose tissues. The production of bile occurs in the liver. Bile is of major importancein aiding the digestion of fats. It is secreted by the liver cells intoa number of small ducts which drain into a common bile duct. Bile is storedin the gallbladder, which releases its contents upon stimulation by thepresence of fats in the duodenum. When proteins are metabolized, ammonia, a toxic sub-stance, is formed. The cells of the liver are able to detoxify ammonia rapidly by transformingit into the relatively inert substance urea. Urea can then be excretedfrom the body in the urine. The liver also produces a lipid which has been frequently associatedwith heart diseases - cholesterol. In fact, a diet with reduced cholesterolcontent may not be effective in lowering cholesterol level in the bodybecause the liver responds by producing more. The liver plays some important roles in blood clotting. It is the site ofproduction ofprothrombinand fibrinogen, substances essential to the formationof a blood clot. In addition, bile secreted by the liver aids in the absorptionof vitamin K by the intestine. Vitamin K is an essential cofactor inthe synthesis ofprothrombinand the plasma clotting cofactors. Vitamin D, produced by the skin, is relatively inactive and requires biochemicaltransformation, first by the liver and then by the kidneys, beforeit is fully able to stimulate absorption of calcium by the gut. The mechanismof activating vitamin D is not yet fully understood. However, thepresence of liver and kidney enzymes is essential to the process.

Question:

Simulate a game of draw poker between a player and the computer. Use the Basic language.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G23-0554.htm

Solution:

In the game, both players (i.e. the computervsthe operator) start with a fixed stake (say $200 each). The com-puter accepts IOUs for certain personal items. 10PRINT \textquotedblleftPOKER 20DIM A (50) , B (5) 30DEF FNA (X) = INT (10\textasteriskcenteredRND (1)) 40DEF FNB (X) = X - 100 \textasteriskcentered INT (X/100) 50PRINT \textquotedblleftWELCOME TO THE TABLE. I WELL OPEN THE\textquotedblright ; 60PRINT \textquotedblleftBETTING BEFORE THE DRAW; YOU OPEN AFTER. WHEN\textquotedblright ; 70PRINT \textquotedblleftYOU FOLD, BET 0; TO CHECK, BET .5\textquotedblright 80PRINT : LET O = 1 90LET C = 200: LET S = 200 95LET P = 0 100PRINT: IF C < = 5 THEN 3340 110PRINT \textquotedblleftTHE ANTE IS $5. I WILL DEAL\textquotedblright 120PRINT: IF S > 5 THEN 140 130GOSUB 3490 140LET P = P + 10: LET S = S - 5: LET C = C - 5 150FOR Z = 1 TO 10: GOSUB 1460: NEXT Z 160PRINT "YOUR HAND :" 170N = 1 180GOSUB 1580 190N = 6: I = 2 200GOSUB 1830 210PRINT: IF I <> 6 THEN 330 220IF FNA (0) < = 7 THEN 250 230LET X = 11100 240GOTO 300 250IF FNA (0) < = 7 THEN 280 260LET X = 11110 270GOTO 300 280IF FNA (0) > = 1 THEN 320 290X = 11111 300I = 7: Z = 23 310GOTO 430 320Z = 1: GOTO 365 330IF U > = 13 THEN 390 340IF FNA (0) > = 2 THEN 360 350GOTO 300 360Z = 0 365K = 0 370PRINT \textquotedblleftI CHECK\textquotedblright 380GOTO 465 390IF U < = 16 THEN 420 400Z = 2 410IF FNA (0) > = I THEN 430 420Z = 35 430V = Z + FNA (0) 440GOSUB 3130 450PRINT \textquotedblleftI OPEN WITH\textquotedblright V 460K = V 465GOSUB 2700 470GOSUB 490 480GOTO 630 490IF I <> 3 THEN 580 500PRINT: PRINT \textquotedblleftI WIN\textquotedblright 510C = C + P 520PRINT "NOW I HAVE $ "C" AND YOU HAVE $" S 530PRINT "DO YOU WISH TO CONTINUE"; 540INPUT H$ 550IF H$ = "YES" THEN 95 560IF H$ = "NO" THEN 3750 570GOTO 530 580IF I <> 4 THEN 620 590PRINT: PRINT "YOU WIN" 600S = S + P 610GOTO 520 620RETURN 630PRINT: PRINT "NOW WE DRAW. HOW MANY CARDS DO YOU WANT ?\textquotedblright; 640INPUT T 650IF T = 0 THEN 780 660Z = 10 670IF T < 4 THEN 700 680PRINT "YOU CAN'T DRAW MORE THAN THREE CARDS" 690GOTO 640 700PRINT "WHAT ARE THEIR NUMBERS?" 710FOR Q = 1 TO T 720INPUT U 730GOSUB 1450 740NEXT Q 750PRINT "YOUR NEW HAND:" 760N = 1 770GOSUB 1580 780Z = 10 + T 790FOR U = 6 TO 10 800IF INT (X/10 \uparrow (U - 6)) <> 10\textasteriskcenteredINT(X/10\uparrow (U - 5)) THEN 820 810GOSUB 1450 820NEXT U 830PRINT: PRINT "I AM TAKING" Z - 10 - T "CARD"; 840IF Z = 11 + T THEN 860 850PRINT "S": PRINT: GOTO 860 860N = 6: V = I: I = 1 870GOSUB 1830 880B = U: M = D 890IF V <> 7 THEN 920 900Z = 28 910GOTO 1060 920IF I <> 6 THEN 950 930Z = 1 940GOTO 1060 950IF U > = 13 THEN 1000 960Z = 2 970IF FNA (0) <> 6 THEN 990 980Z = 19 990GOTO 1060 1000IF U > = 16 THEN 1050 1010Z = 19 1020IF FNA (0) <> 8 THEN 1040 1030Z = 11 1040GOTO 1060 1050Z = 2 1060K = 0 1070GOSUB 2700 1080IF T <> .5 THEN 1170 1090IF V = 7 THEN 1130 1100IF I <> 6 THEN 1130 1110PRINT "I'LL CHECK" 1120GOTO 1180 1130V = 2 + FNA (0) 1140GOSUB 3130 1150PRINT "I BET" V 1160K = V: GOSUB 2710 1170GOSUB 490 1180PRINT: PRINT "NOW WE COMPARE HANDS" 1190J$ = H$ 1200K$ = 1$ 1210PRINT "MY HAND:" 1220N = 6 1230GOSUB 1580 1240N = 1 1250GOSUB 1830 1260PRINT: PRINT "YOU HAVE"; 1270K = D 1280GOSUB 3360 1290H$ = J$ 1300I$ = K$ 1310K = M 1320PRINT "AND I HAVE"; 1330GOSUB 3360 1340IF B > U THEN 500 1350IF U > B THEN 590 1360IF H$ = "A FLUSH" THEN 1420 1370IF FNB (M) < FNB(D) THEN 590 1380IF FNB(M) > FNB(D) THEN 500 1390PRINT "THE HAND IS DRAWN" 1400PRINT "ALL $" P "REMAINS IN THE POT" 1410GOTO 100 1420IF FNB (M) > FNB (D) THEN 500 1430IF FNB (D) > FNB (M) THEN 590 1440GOTO 1390 1450Z = Z + 1 1460A (Z) = 100 \textasteriskcentered INT (4\textasteriskcenteredRND (1)) + INT (100\textasteriskcenteredRND (1)) 1470IF INT (A(Z)/100) > 3 THEN 1460 1480IF A(Z) - 100 \textasteriskcentered INT(A(Z)/100) > 12 THEN 1460 1490IF Z = 1 THEN 1570 1500FOR K = 1 TO Z - 1 1510IF A(Z) = A(K) THEN 1460 1520NEXT K 1530IF Z < = 10 THEN 1570 1540N = A (U) 1550A (U) = A (Z) 1560A(Z) = N 1570RETURN 1580FOR Z = N TO N + 4 1590PRINT Z "- - \textquotedblright; 1600GOSUB 1660 1610PRINT "OF"; 1620GOSUB 1750 1630IF Z/2 <> INT (Z/2) THEN 1640 1640PRINT: NEXT Z: PRINT 1650RETURN 1660K = FNB(A(Z)) 1670IF K <> 9 THEN 1690 1680PRINT "JACK"; 1690IF K <> 10 THEN 1700: PRINT "QUEEN"; 1700IF K <> 11 THEN 1710: PRINT "KING"; 1710IF K <> 12 THEN 1720: PRINT "ACE"; 1720IF K > = 9 THEN 1740 1730PRINT K + 2; 1740RETURN 1750K = INT (A(Z)/100) 1760IF K <> 0 THEN 1780 1770PRINT "CLUBS", 1780IF K <> 1 THEN 1800 1790PRINT "DIAMONDS", 1800IF K <> 2 THEN 1810: PRINT "HEARTS", 1810IF K <> 3 THEN 1820: PRINT "SPADES", 1820RETURN 1830U = 0 1840FOR Z = TO N + 4 1850B(Z) = FNB (A(Z)) 1860IF Z = N + 4 THEN 1890 1870IF INT (A (Z)/100) <> INT (A(Z + 1)/100) THEN 1890 1880U = U + 1 1890NEXT Z 1900IF U <> 4 THEN 1970 1910X = 11111 1920D = A (N) 1930H$ = "A FLUS" 19401$ = "H IN" 1950U = 15 1960RETURN 1970FOR Z =N TO N + 3 1980FOR K = Z + 1 TO N + 4 1990IF B(Z) < = B(K) THEN 2050 2000X = A (Z) 2010A (Z) = A (K) 2020B (Z) = B(K) 2030A (K) = X 2040B(K) = A(K) - 100 \textasteriskcentered INT(A(K)/100) 2050NEXT K 2060NEXT Z 2070X = 0 2080FOR Z = N TO N + 3 2090IF B (Z) <> B (Z + 1) THEN 2130 2100X = X + 11 \textasteriskcentered 10 \uparrow (Z - N) 2110D = A (Z) 2120GOSUB 2420 2130NEXT Z 2140IF X <> 0 THEN 2280 2150IF B (N) + 3 <> B (N + 3) THEN 2180 2160X = 1111 2170U = 10 2180IF B (N + 1) + 3 <> B (N + 4) THEN 2280 2190IF U <> 10 THEN 2260 2200U = 14 2210H$ = "STRAIG" 2220I$ = "HT" 2230X = 11111 2240D = A (N + 4) 2250RETURN 2260U = 10 2270X = 11110 2280IF U > = 10 THEN 2350 2290D = A (N + 4) 2300H$ = "SCHMAL" 2310I$ = "TZ," 2320U = 9 2330X = 11000 2340GOTO 2400 2350IF U <> 10 THEN 2380 2360IF I = 1 THEN 2400 2370GOTO 2410 2380IF U > 12 THEN 2410 2390IF FNB (D) > 6 THEN 2410 2400I = 6 2410RETURN 2420IF U > = 11 THEN 2470 2430U = 11 2440H$ = \textquotedblleftA PAIR\textquotedblright 2450I$ \textquotedblleftOF\textquotedblright 2460RETURN 2470IF U <> 11 THEN 2570 2480IF B (Z) <> B (Z - 1) THEN 2530 2490H$ = \textquotedblleftTHREE\textquotedblright 2500I$ = \textquotedblleft\textquotedblright 2510U = 13 2520RETURN 2530H$ = \textquotedblleftTWO P\textquotedblright 2540I$ = \textquotedblleftAIR,\textquotedblright 2550V = 12 2560RETURN 2570IF U > 12 THEN 2620 2580U = 16 2590H$ = \textquotedblleftFULL H\textquotedblright 2600I$ = \textquotedblleftOUSE,\textquotedblright 2610RETURN 2620IF B (Z) <> B (Z - 1) THEN 2670 2630U = 17 2640H$ = "POUR" 2650I$ = "" 2660RETURN 2670U = 16 2680H$ = "FULL H": I$ = "OUSE," 2690RETURN 2700G = 0 2710PRINT "WHAT IS YOUR BET?"; 2720INPUT T 2730IF T - INT(T) = 0 THEN 2790 2740IF K <> 0 THEN 2770 2750IF G <> 0 THEN 2770 2760IF T = .5 THEN 3060 2770PRINT "NO SMALL CHANGE, PLEASE" 2780GOTO 2710 2790IF S - G - T > = 0 THEN 2820 2800GOSUB 3490 2810GOTO 2710 2820IF T <> 0 THEN 2850 2830I = 3 2840GOTO 3030 2850IF G + T > = K THEN 2880 2860PRINT "IF YOU CAN'T SEE MY BET, THEN FOLD\textquotedblright 2870GOTO 2710 2880G = G + T 2890IF G = K THEN 3030 2900IF Z <> 1 THEN 3070 2910IF G > 5 THEN 2950 2920IF Z > = 2 THEN 3000 2930V = 5 2940GOTO 3070 2950IF Z = 1 THEN 2970 2960IF T < = 25 THEN 3000 2970I = 4 2980PRINT "I FOLD" 2990RETURN 3000IF Z = 2 THEN 3080 3010PRINT "I'LL SEE YOU" 3020K = G 3030S = S - G 3040C = C - K 3050P = P + G + K 3060RETURN 3070IF G > 3\textasteriskcenteredZ THEN 3000 3080V = G - K + FNA (0) 3090GOSUB 3130 3100PRINT "I'LL SEE YOU, AND RAISE YOU" V 3110K = G + V 3120GOTO 2710 3130IF C - G - V > = 0 THEN 3330 3140IF G <> 0 THEN 3170 3150V = C 3160RETURN 3170IF C - G > = 0 THEN 3010 3180IF (O/2) <> INT(O/2) THEN 3260 3190PRINT "WOULD YOU LIKE TO BUY BACK YOUR SHIRT?"; 3200PRINT "AND COAT FOR $50?"; 3210INPUT J$ 3220IF J$ = "NO" THEN 3260 3230C = C + 50 3240O = O/2 3250RETURN 3260IF O/3 <> INT (0/3) THEN 3340 3270PRINT "WOULD YOU LIKE TO BUY BACK YOUR GOLD"; 3280PRINT "CHAIN FOR $50?"; 3290INPUT J$ 3300IF J$ = "NO" THEN 3340 3310C = C + 50 3320O = O/3 3330RETURN 3340PRINT "I'M BUSTED. CONGRATULATIONS!" 3350STOP 3360PRINT H$; I$; 3370IF H$ <> "A FLUS\textquotedblright THEN 3410 3380K = INT (K/100) 3390GOSUB 1760 3400PRINT: RETURN 3410K = FNB (K) 3420GOSUB 1670 3430IF H$ = "SCHMAL" THEN 3450 3440IF H$ <> "STRAIG" THEN 3470 3450PRINT "HIGH" 3460RETURN 3470PRINT "S" 3480RETURN 3490PRINT: PRINT "YOU CAN'T BET WHAT YOU DON'T HAVE" 3500IF O/2 = IN (O/2) THEN 3620 3510PRINT "WOULD YOU LIKE TO SELL YOUR SHIRT AND COAT?"; 3520INPUT J$ 3530IF J$ = "NO" THEN 3620 3540IF FNA(0) > = 7 THEN 3580 3550PRINT "I WILL GIVE YOU $75 FOR IT" 3560S = S + 75 3570GOTO 3600 3580PRINT "THIS COAT LOOKS MOTHEATEN - I'LL GIVE YOU $25" 3590S = S + 25 3600O = O\textasteriskcentered2 3610RETURN 3620IF O/3 <> INT (O/3) THEN 3740 3630PRINT "WILL YOU PART WITH YOUR GOLD CHAIN?"; 3640INPUT J$ 3650IF J$ = "NO" THEN 3730 3660IF FNA (0) > = 6 THEN 3700 3670PRINT "YOU ARE NOW $100 RICHER" 3680S = S + 100 3690GOTO 3720 3700PRINT "THIS IS ROLLED GOLD. $25" 3710S = S + 25 3720O = O\textasteriskcentered3 3730RETURN 3740PRINT: PRINT "YOU' RE BUSTED. HAVE A REVOLVER" 3750END

Question:

A small object lies 4 in. to the left of the vertex of a concave mirror of radius of curvature 12 in. Find the position and magnifica-tion of the image. Concave Mirror

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/Users/wenhuchen/Documents/Crawler/Physics/D27-0852.htm

Solution:

The mirror equation for a concave mirror is (1/p) + (1/q) = (2/R) with a positive R. The object distance is p = +4 in. and R is +12 in. Then, (1/4 in.) + (1/q) = (2/12 in.) q = - 12 in., Magnification m is given by m = - (q/p) = - (-12 in./4 in.) = 3 The image is therefore 12 in. to the right of the vertex (q is negative), is virtual (q is negative), erect (m is positive), and 3 times the height of the object. See the figure.

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Question:

Write a program, incorporating the function TAN, to compute tangents of angles 0,10,20,....,360 degrees.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G16-0414.htm

Solution:

PROGRAM TABLEOFTANS (OUTPUT); CONST PI = 3.1415926536; VAR DEGREES: INTEGER; FUNCTION TAN (x:REAL):REAL; BEGIN TAN:=sin(x) /cos( x) END; (\textasteriskcenteredTAN\textasteriskcentered) BEGIN( \textasteriskcenteredTABLEOFTANS\textasteriskcentered) DEGREES: = o; WHILE DEGREES < = 360 DO BEGIN WRITE (DEGREES); IF DEGREES MOD 180 = 90 THEN WRITELN ('INFINITY') ELSE WRITELN (TAN (DEGREES\textasteriskcenteredPI/180)); DEGREES: DEGREES + 10 END END.(\textasteriskcenteredTABLEOFTANS\textasteriskcentered).

Question:

A person whose diet is deficient in B vitamins is likely to be listlessand lacking in energy for normal activities Explain.

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/Users/wenhuchen/Documents/Crawler/Biology/F03-0098.htm

Solution:

The vitamin B complex includes (among others) thiamine (B1), riboflavin(B2) pyridoxine (B6), biotin, niacin or nicotinic acid,cobalamin (B12) andpantothenicacid.All these vitamins function as parts of coenzymesused in the oxidation of glucose to form ATP. Thiamine is an essential constituent of thiamine pyrophosphate, the coenzymeused in the oxidativedecarboxylationofpyruvicand \alpha- ketoglutaricacids in the TCA cycle. Characteristics of mild deficiencies are: fatigue, loss of appetite and weakness. Severe deficien-cies enhance thesesymptoms and cause degeneration of nerves and muscle atrophy, a conditionknown as beriberi. Niacin is a component of two coenzymes, NAD+ and NADP+, which serveas hydrogen acceptors in reactions ofglycolysisand the TCA cycle. Deficiency of niacin causes stunted growth and the disease pellagra. Riboflavin forms part offlavinadeninedinucleotide(FAD), a coenzymeofsuccinicdehydrogenasein the TCA cycle. Deficiency results instunted growth, cracks in the corner of the mouth, and cornea damage. Pantothenicacid is especially important since it forms part of coenzymeA, and thus used in a number of reac-tions in the metabolism of carbohydrates, fats and proteins. An expected symptom ofpantothenic aciddeficiency is growth failure. A person with any of these vitamin deficiencies would show signs of slowATP production since these vitamins are necessary providers of coenzymesinvolved in metabolic pathways leading to ATP formation. Since ATP is needed for energy-requiring reactions, a lack of ATP would causethe vitamin-deficient person to be sluggish and to exhibit slower growth.

Question:

Fe(s) + Co^2+(.5M) \rightarrow Fe^2+(1.0M) + Co(s) . Fe^2+ + 2e^- \leftrightarrows Fe(s) with E\textdegree = -.44e and Co^2+ + 2e^- \leftrightarrows Co(s) with E\textdegree = -.28, find the standard cell potential ∆E , the cell potential ∆E and the concentration ratio at which the potential generated by the cell is exactly zero.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E16-0594.htm

Solution:

Assume that the reaction proceeds spontaneously. This means, therefore, that the reaction must have a positive value for ∆E\textdegree. With this in mind, you proceed as follows: You are given two half-reactions: Fe^2+ + 2e^- \leftrightarrows Fe(s)E\textdegree = -.44 Co^2++ 2e^- \leftrightarrows Co(s)E\textdegree = -.28ev Both reactions represent reduction (gain of electrons). But, the over-all reaction in a cell is a combination of both a reduction and an oxidation reaction. Thus, you must reverse one, keeping in mind that the ∆E\textdegree must be a positive value. Recall, also, that ∆E\textdegree = E\textdegree_oxid + E\textdegree_red . You can write Fe^2+ + 2e^- \leftrightarrows Fe^2+;E\textdegree = (-.44) = -44ev Co^2+ + 2e^- \leftrightarrows Co(s) : E\textdegree = -.28ev Fe(s) + Co^2+ \leftrightarrows Fe^2+ + Co(s) (overall reaction) with ∆E\textdegree = .44 - .28 = .16 ev. Notice: By reversing the Fe reaction and combining it with the other, you obtained the overall reaction with a ∆E\textdegree = .16 ev, a positive value, which indicates that the reaction proceeds spontaneously. To find ∆E, use the Nernst equa-tion, which states ∆E = ∆E\textdegree -(.059 / n) log K, where n = number of electrons transferred and K = equilibrium constant of reaction. In this problem, the number of electrons trans-ferred is 2, so that n = 2. K = [Fe^2+] / [Co^2+], i.e., the ratio of the concentrations of products to reactants, each raised to the power of its respective coefficient in the chemical equation. Note, Co(s) and Fe(s) are omitted, because they are solids and, thus, considered constants them-selves. You are given [Fe^2+] and [Co^2+] and you have calculated ∆E\textdegree . Therefore, ∆E = ∆E\textdegree -(.059 / n)log [(Fe^2+) / (Co^2+) ] = .16 -(.059 / n)log[(1.0) / (.5)] =.16 - .0295 log 2 = .16 - .01 = .15 ev. To find the concentration ratio, K, when ∆E = 0, use the Nernst equation; ∆E = ∆E\textdegree -(.059 / n) log K. ∆E = 0, ∆E\textdegree = .16, n = 2 and K = [Fe^2+] / [Co^2+] , so that 0 = .16 - (.059 / n) log [Fe^2+] / [Co^2+] or 0 = .16 - .0295 log [Fe^2+] / [Co^2+]. [Fe^2+] / [Co^2+] = antilog(.16 / .0295) = antilog 5.4 = 2.5 × 10^5 thus, [Fe^2+] / [Co^2+] = 2.5 × 10^5, when ∆E = 0.

Question:

Determine the current in each of the resistors in figure A.

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/Users/wenhuchen/Documents/Crawler/Physics/D20-0678.htm

Solution:

We find the resistance between points B and C (R_BC) by using the relation for resistors R_1, R_2 , . . . in parallel (see figure (A)). (1/R_Total ) = (1/R_1 ) + (1/R_2 ) + . . . (1/R_BC ) = (1/6.0 ohms) + (1/9.0 ohms) + (1/18.0 ohms) = [(6.0 )/(18.0 ohms)] R_BC = [(18.0)/(6.0)] ohms = 3.0ohms Using the formula for resistors in series, R_Total = R_1 + R_2 + . . . , we find the resistance between points A and C in the circuit, (see figure (B)) R_AC = R_AB + R_BC = 4.0 ohms + 3.0 ohms = 7.0 ohms The current I_t in the circuit is obtained from Ohm's Law I_t = [(V_AC )/(R_AC )] = [(35 volts)/(7.0 ohms)] = 5.0 amp . The current through each individual resistor is, by Ohm's Law, equal to the voltage across the resistor divided by its resistance. Hence I_FG = [(V_FG )/(R_FG )] = [(15 volts)/(6.0 ohms)] = 2.5 amp I_HK = [(V_HK )/(R_HK )] = [(15 volts)/(9.0 ohms)] = 1.7 amp I_LM = [(V_LM )/(R_LM )] = [(15 volts)/(18.0 ohms)] = 0.83 amp Note that I_t = I_FG + I_HK + I_LM by Kirchoff's Current Law. I_t = (2.5 + 1.7 + 0.8) amp = 5.0 amp.

Image

Question:

Write two FORTRAN programs which will compute n!intwo differentways: (a) First according to the standard formula for a positive integern \geq 1, n! =n(n-1) \textbullet\textbullet\textbullet (3)(2)(1) (b)thenuseStirling'sapproximation formula n! \cong (2\pi)1^/2n^n^+(1/2) e^-n

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G21-0520.htm

Solution:

We will formulate the answers as two function routines which will be usedin later work. Denote these two functions byFACT(N) and FACTST(N). Then by applying the above formulas directly we get the programsbelow: INTEGER FUNCTIONFACT(N) IPROD = 1 IF (N.LE.1) GO TO 200 DO 100 I = 2,N IPROD = IPR0D\textasteriskcenteredI 100CONTINUE FACT = IPROD GO TO 300 200FACT = 1 300RETURN END FUNCTIONFACTST(N) Z1 =EXP(-FLOAT(N)) Z2 = (FLCAT(N))\textasteriskcentered\textasteriskcentered(FLOAT(N) + 0.5) CFLOAT CONVERTS INTEGERS TO REALS Z3 = SQRT (2\textasteriskcentered3.14159) FACTST = Z1\textasteriskcenteredZ2\textasteriskcenteredZ3 RETURN END

Question:

Illustrate using flowcharts the need for debugging when dealing with improper loop sequences.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G08-0186.htm

Solution:

Most beginners have some difficulty understanding and applying the rules dealing with loops. Consequently, many of their debugging concentrates on the proper construction of a loop. Consider the flowchart of figure 1 that develops an algorithm for finding the factors of a number. Now, suppose the instruction NUMBER \leftarrow K is omitted from the inner loop, leaving the (partial) flowchart of figure 2. Suppose NUMBER is 12. The loop begins with J = 2. Since 2 divides 12, K will be an Integer. The computer prints 2 IS A. FACTOR and returns to the start of the loop. Since the value of NUMBER has not been changed, the computer repeats exactly the same computations on the second pass around the loop. 2 IS A FACTOR is printed and the computations repeated. The computer is in a loop for which the normal exit condition will never be satisfied. The program must be aborted. The error here can be easily recognized since 2 or 3 pages of '2 IS A FACTOR' indicate that the value of NUMBER is not changing. Now, suppose that the print statement has also been omitted. In this case there is no output to tell the user that he has constructed an infinite loop. He will not receive any output and so may assume that the system is 'down'. In fact, the computer is slaving away trying to exit out of an infinite loop. In this case, the system will also abort the program before it is complete. The lack of output tells the programmer that he should include some PRINT instructions in his loops to find what caused the program to abort.

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Question:

For HF,K_diss= 6.7 × 10^-4 what is the H_3O^+ and OH^- concentrations of a 0.10M HF solution? Assume K_W = 1 × 10^-14 .

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/Users/wenhuchen/Documents/Crawler/Chemistry/E12-0419.htm

Solution:

To answer this question, you must set up the equations for the equilibriainvolved. HF is an acid, which means it donates pro-tons (H+). The only base present is H_2O. They react as shown in the following equation: HF + H_2O \rightleftarrows H_3O^+ + F^- . This is not the only source of H_3O^+ . Recall that water also can dissociate into ions. H_2O + H_2O \rightleftarrowsH_3O^+ + OH^- . This means that in the consideration of the H_3O^+ concentration you must look at both sources. Because you are asked the concentrations, you must employ equilibrium constants,K_dissandK_w. Let x = the number of moles/liter of H_3O^+ from the reaction of HF with H_2O, then x is also equal to the number of moles/liter of F^- produced. They have the same mole ratio, as indicated by theequilibria. Since HF is the stronger acid, some of its protons ionize and then join with H_2O to form H_3O^+ . Thus, the concentration of HF is (0.1 - x). If you let y = moles/liter of OH^-, then y = moles/liter of H_3O^+ formed by the dissociation of water. These variables that represent the concentrations at equilibrium must satisfy the two equilibrium conditions: K_diss= {[H_3O^+] [F^-]} / [HF] = 6.7 × 10^-4 = {(x+y)(x)} / (0.1 - x) K_w= [H_3O^+] [OH^-] = 1 × 10^-14 = (x+y)(x) . For both conditions, the H_3O^+ concentration is represented by (x+y) since, at equilibrium, the concentration must be the sum from both sources of H_3O^+ production. Solve these equations simultaneously for x and y, you can avoid cumbersome calculations by noting which contribution is the dominant one. In this way, the other can be neglected to give a fairly good approximation. A general rule for determining which is dominant is to compare the dissociation constants. The larger the dissociation constant, the more it dom-inates the final equilibrium state. From inspection of these con-stants, you find that the contribution of H_3O^+ from HF dominates over that from water. You need consider only HF + H_2O \rightleftarrows H_3O^+ + F^-, therefore, K_diss= {[H_3O^+] [F^-]} / [HF] = 6.7 × 10^-4 = {(x+y)(x)} / (0.1 - x) . Solving, you obtain x = 7.7 × 10^-3M = [H_3O^+]. To find y, go back to K_W = [H_3O^+] [0H^-] = 1 × 10^-14 = (x+y) (y) and substitute in your value for x. Assuming that y is negligible in comparison to x, the answer becomes (1.0 × 10^-14) / (7.7 × 10^-3) = 1.3 × 10^-12M = [OH^-].

Question:

A two-person barbershop has five chairs to accommodate waiting customers. Potential customers arrive at the average rate of 3.7634/hr. and spend an average of 15 min. in the barber's chair. Assume M = 7 (total number of persons in the system including the two barbers.) \lambda = 3.7634/hr. \mu = 4/hr. S = 2 servers. a) What is the probability of a customer getting directly into the barber's chair upon arrival? b) What is the expected number of customers waiting for a haircut? c) What is the effective arrival rate assuming Pn = 1? d) How much time can a customer expect to spend in the barbershop? e) What fraction of potential customers are turned away?

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G18-0457.htm

Solution:

a) The probability of a customer getting directly into the barber's chair upon arrival is the same as in the case of no customers in the system. This is computed as follows: b) The expected number of customers waiting for a haircut: c) The effective arrival rate S = number of servers M = maximum number of persons in the system Therefore, since n \leq S, Pn = (1/ n!) (\lambda/\mu)^n Po = (1/1) [(3.7643)/4]^1 Po = (1/1) [(3.7643)/4] (.36133) = .2457 \lambdaeff= \mu [S - ^S-1\sumn = 0(S-n) Pn] = 4[ 2- 2 (0.36133) - 0.339957] = 3.7495 d) The time a customer is expected to spend in the barbershop, W = (L/ \lambdaeff) = {L_q + (2 - 2po - p^1)} / \lambdaeff = (1.183083) / (3.7495) = 0.03155 hr \cong 19 min e) Fraction of potential customers that will be turned away P (7 in system) = P_7 = [1/ (S!S^n-S)] (\lambda/\mu)nPo Since S < n \leq M = 2 < 7 = 7 = [1 / (2!2^7-2)] (3.7634/4)7(0.36133) = 0.00368 = 0.3 percent Thus, only 0.3 percent of the potential customers will be turned away. The Fortran program for the problem follows: \textasteriskcentered\textasteriskcentered INFINITE SOURCE, FINITE QUEUE, MULTIPLE SERVERS \textasteriskcentered\textasteriskcentered \textasteriskcentered\textasteriskcentered QUEUING PROGRAM \textasteriskcentered\textasteriskcentered WE ASSUME ARRIVALS TO SYSTEM ARE COMPLETELY AT RANDOM - POISSON INPUT ARRIVALS FORM A SINGLE QUEUE INFINITE SOURCE AND FINITE QUEUE FIFO QUEUE DISCIPLINE THERE ARE MULTIPLE SERVERS WITH EXPONENTIAL SERVICE TIME DEPARTURES FROM SYSTEM OCCUR COMPLETELY AT RANDOM TO CALCULATE AND PRINT LAVERAGE # OF CUSTOMERS IN THE SYSTEM LQAVERAGE # OF CUSTOMERS IN QUEUE PNPROBABILITY OF N CUSTOMERS IN SYSTEM AT ANY POINT IN TIME PZEROPROBABILITY OF NO CUSTOMERS IN THE SYSTEM WAVERAGE TIME CUSTOMER SPENDS IN SYSTEM WQAVERAGE TIME CUSTOMER SPENDS IN QUEUE LAMEFFOVERALL EFFECTIVE ARRIVAL RATE FOR FINITE QUEUE WITH INFINITE SOURCE REAL\textasteriskcentered4 TITLE (20) , LAMBDA, MU, N, M, L, LQ, LAMEFF 5READ (5, 10, END = 2000) TITLE 10FORMAT (20A4) WRITE (6, 11) TITLE 11FORMAT ('1', 20A4, //) READ (5, 20) LAMBDA, MU 20FORMAT (3F10.0) READ (5, 20) N, M, S C\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered C\textasteriskcenteredCALCULATE PZERO\textasteriskcentered C\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered IS = S+1 SUM = 0.0 DO 405 NK = 1, IS SK = NK-1 CALL FACT (SK, TFACT) SUM = SUM+ (1/TFACT) \textasteriskcentered (LAMBDA/MU) \textasteriskcentered\textasteriskcenteredSK 405CONTINUE SUM2 = 0.0 NM = M DO 410 NK = IS, NM SUM2 = SUM2+ (LAMBDA/ (MU\textasteriskcenteredS)) \textasteriskcentered\textasteriskcentered (NK-S) 410CONTINUE CALL FACT(S, SFACT) PZERO = 1/ (SUM+ (1/SFACT) \textasteriskcentered ( (LAMBDA/MU) \textasteriskcentered\textasteriskcenteredS) \textasteriskcenteredSUM2) C\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered C\textasteriskcenteredCALCULATE PN AND LQ\textasteriskcentered C\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered IF(N.LE.S) GO TO 415 IF (N.GT.M) GO TO 420 PN= (1/ (SFACT\textasteriskcenteredS\textasteriskcentered\textasteriskcentered (N-S))) \textasteriskcentered ((LAMBDA/MU) \textasteriskcentered\textasteriskcenteredN) \textasteriskcenteredPZERO GO TO 425 415CALL FACT(N, TFACT) PN= (1/TFACT) \textasteriskcentered ( (LAMBDA/MU) \textasteriskcentered\textasteriskcenteredN) \textasteriskcenteredPZERO GO TO 425 420PN=0.0 425LQ= (((LAMBDA/MU) \textasteriskcentered\textasteriskcentered (S+1) \textasteriskcenteredPZERO) \textasteriskcentered (1- (LAMBDA/(MU\textasteriskcenteredS)) \textasteriskcentered\textasteriskcentered (M-S) - (M-S) \textasteriskcentered ( (LAMBDA/(MU\textasteriskcenteredS)) \textasteriskcentered\textasteriskcentered (M-S) ) \textasteriskcentered (1-LAMBDA/ (MU\textasteriskcenteredS ) ) ) )/(S\textasteriskcenteredSFACT\textasteriskcentered(1-LAMBDA/(MU\textasteriskcenteredS))\textasteriskcentered\textasteriskcentered2) C\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered C\textasteriskcenteredCALCULATELAMEFF, L, W, WQ\textasteriskcentered C\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered NS=S SUM=0 DO 560 NK=1,NS KK=NK-1 SKK=KK CALL FACT(SKK,AFACT) SUM=SUM+ (S-KK) \textasteriskcenteredPZERO\textasteriskcentered (1/AFACT) \textasteriskcentered (LAMBDA/MU) \textasteriskcentered\textasteriskcenteredKK 560CONTINUE LAMEFF=MU\textasteriskcentered(S-SUM) L=LQ+LAMEFF/MU W=L/LAMEFF WQ=LQ/LAMEFF C\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered C\textasteriskcenteredPRINT RESULTS\textasteriskcentered C\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered WRITE (6,810) LAMBDA, MU, L, LQ, N, PN, PZERO, W, WQ, LAMEFF 810FORMAT(' LAMBDA =',F10.5/' MU = F10.S/' L =',F10.5/\textasteriskcentered LQ =',F10.5/' N =',F10.5/' PN =',F10.5/\textasteriskcentered PZERO =',F10.5/\textasteriskcentered WQ =',F10.5/' LAMEFF =',F10.5) GO TO 5 2000STOP END SUBROUTINE FACT(P, PROD) NUM=P PROD=1. IF (NUM.EQ.0) GO TO 20 DO 10 K=1, NUM 10PROD=PROD\textasteriskcenteredK 20RETURN END /DATA WITH N=1 3.76344. 1 .7.2. 1 .7.2. WITH N=1 LAMBDA= 3.76340 MU= 4.00000 L= 1.18309 LQ= 0.24571 N= 1.00000 PN= 0.33996 PZERO= 0.36133 W= 0.31553 WQ= 0.06553 LAMEFF= 3.74953

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Question:

The dorsal root of a nerve leading to the leg of a rat is severed. How will this affect the rat? How would the effect vary if the ventral root were severed instead?

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/Users/wenhuchen/Documents/Crawler/Biology/F20-0508.htm

Solution:

A cross-section of the spinal cord (see problem 26-16) reveals that the paired dorsal roots are the junctions where sensory fibers from the peripheral receptor areas enter, while the paired ventral roots are where axons of the motor neurons leave the spinal cord to effectors, such as muscles. When the dorsal root of a spinal nerve is cut, the sensory fibers are severed and afferent impulses (that is impulses to the nervous system) can no longer reach the spinal cord and thence the brain. If a hot iron is brought to touch the leg of a rat, in which the dorsal root of the nerve leading to the leg is severed, the rat will not feel any burn sensation, as evidenced by its lack of attempt to avoid the stimulus. Since the receptor cells on the skin of the leg are not impaired, they do respond to the stimulus by firing impulses along the sensory fibers to the spinal cord. However, the impulses cannot reach the central nervous system and there-fore cannot be translated and interpreted in the brain. Also, simple reflex action such as the rapid withdrawal of the leg cannot occur in the rat, as impulses from the sensory neurons cannot be transmitted to the motor neurons in the reflex arc. As a consequence, the rat will not feel the stimulation of the hot iron, nor would it perform reflexive withdrawal of the leg from the iron. In higher animals such as man in which memory and experience play an important role in behavior, severing the dorsal root may cause loss of sensation and loss of reflex action, but the somatic response to the same stimulus may be present. For instance, by past experience one has learned that the hot iron causes an unpleasant sensation. Although by lesion of the dorsal root he does not sense the burning sensation, he may remove his hand or leg from the hot iron based on this information stored in the memory center of his brain. If instead of the dorsal root, the ventral root were severed, the injury would be inflicted on the motor neurons. In this instance, the rat would sense the stimu-lation of the hot Iron, but would find itself unable to withdraw the leg since the impulses generated by the motor neurons cannot reach theeffectormuscles of the leg. However, the rat may demonstrate other responses, for example, it may squeak in pain.

Question:

How much power is required to service a city of 100,000 homes, each of which draws an average current of 20 A at a potential of 120 V? If the transmission lines that carry power were capable of carrying a current of 1,000 A, what potential would be required to transmit the power from the source?

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/Users/wenhuchen/Documents/Crawler/Physics/D24-0791.htm

Solution:

The total power which must be supplied is equal to the total power dissipated in all the homes. Since power equals the product of the current used (or supplied) by a device and the voltage across the device, we calculate, P = IV for each home. Here I is current and V is voltage. Each home uses power, P = (20A) (120V) \rule{1em}{1pt} 2400 watts But there are 10^5 homes and P_net = (2.4 × 10^3 watts) (10^5) P_net = 2.4 x 10^8 watts If the current is 1,000 A, the required voltage is V = (Pnet/I)=[(2.4 × 10^8Watts)/(1.3 × 10^3 A)] = 2.4 × 10^5V

Question:

An aluminum calorimeter of mass 50 g contains 95 g of a mixture of water and ice at 0\textdegreeC. When 100 g of aluminum which has been heated in a steam jacket is dropped into the mixture, the temperature rises to 5\textdegreeC. Find the mass of ice originally present if the specific heat capacity of aluminum is 0.22 cal/g\bulletCdeg.

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/Users/wenhuchen/Documents/Crawler/Physics/D13-0482.htm

Solution:

The heat lost by the cooling aluminum must equal the heat gained by the calorimeter and contents. If a mass y of ice were originally present, the total heat gained would have to include the heat acquired by the ice in melting, the heat gained by the 95 g of water in rising in temperature, and the heat gained by the calorimeter in doing likewise. The heat Q absorbed by a mass m of specific heat c as its temper-ature rises an amount ∆t is: Q = m c ∆T Also one gram of ice absorbs 80 calories of heat in changing into water. Thus, heat gained by the aluminum calorimeter = (50 g)(0.22 cal/g\bulletCdeg) (5 - 0)\textdegreeC heat gained by ice as it melts = (y) (80 cal/g) heat gained by water= (95 g) (1 cal/g\bulletCdeg) (5 - 0)\textdegreeC heat lost by chunk of aluminum = (100 g) (0.22 cal/g\bulletCdeg) (100 - 5)\textdegreeC since steam has a temperature of 100\textdegreeC. Thus 100 g × 0.22 cal/g\bulletCdeg × (100 - 5)\textdegreeC = y × 80 cal/g + 95 g × 1 cal/g\bulletCdeg× (5 - 0)\textdegreeC + 50 g × 0.22 cal/g\bulletCdeg × (5 - 0)\textdegreeC \therefore80 y cal/g = [0.22(9500 - 250) - 95 × 5] cal. \thereforey = 1560/80 = 19.50 g .

Question:

What are the basic structural and functional differences between white blood cells and red blood cells? Describe two different ways that white blood cells act to protect the body from foreign agents such as microorganisms.

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/Users/wenhuchen/Documents/Crawler/Biology/F14-0358.htm

Solution:

In addition to red blood cells, human blood contains five types of white blood cells, or leukocytes. Unlike the red blood cells, all types of leukocytes contain nuclei and do not contain hemoglobin. Leukocytes are generally larger than erythrocytes and are far less numerous. There are approximately 5,400,000 red cells per cubic millimeter of blood in an adult male (4.6 million in females), while only about 7,000 leukocytes per cubic millimeter. There are five different types of leukocytes (see Figure). Leukocytes are classified as granular (eosino-phils, neutrophils and basophils) or angular (monocytes and lymphocytes). The granulocytes have granules in their cytoplasm, have lobed nuclei, and are produced in red bone marrow. The agranulocytes are produced in organs such as the lymph nodes, spleen and thymus. Neutrophils make up over 60% of the leukocytes present in the body. Like other leukocytes, they can squeeze through the pores of blood vessels and enter the tissue spaces. They then move by amoeboid movement to sites of infected tissue. For example, when bacteria enter a certain tissue of the body, they can either attack cells or produce damaging toxins. The blood vessels in the in-fected region dilate and allow more blood flow to the site, causing the heat and redness characteristic of inflammation. Blood vessel permeability also increases, causing fluid to enter the tissue and swelling to result. Neutrophils and monocytes pass through the blood vessels and engulf, ingest and destroy the bacteria. Foreign particles and dead tissue can also be engulfed by this process of phagocytosis. These leukocytes are chemically attracted to the inflamed site by products released from the damaged tissues. The monocytes (5.3%) usually appear after the neutro-phils and are more important in fighting chronic infections. Monocytes can phagocytize bacteria but more often, they en-large and become wandering macrophages, which can move more quickly and engulf more bacteria. Eosinophils (2.3%) are weakly phagocytic, and are involved in aller-gic reactions (by releasing antihistamines) and in fight-ing trichinosis (an infection caused by the parasitic worm, trichinella). Basophils ( .5%) liberate the anti-coagulant heparin which combats the coagulative pro cesses that sometimes occurs in prolonged inflammation. The second means by which leukocytes protect the body from invasion is carried out by the lymphocytes. Lymphocytes (30%) function in the process of immunity. Lymphocytes are involved in the production of anti-bodies, proteins made by the body in response to speci-fic foreign substances. These foreign substances are subsequently attacked and destroyed by the anitbodies. Immunity and its agents will be discussed in further questions. The main functions of the leukocytes are thus protective: phagocytosis and immunity. The major function of the erythrocytes is to carry hemoglobin for gaseous transport and exchange, which is essential to the continuation of cellular metabolic processes.

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Question:

Distinguish between messenger RNA (mRNA), ribosomal RNA (rRNA), and transfer RNA (tRNA). What is the role of each in protein synthesis?

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/Users/wenhuchen/Documents/Crawler/Biology/F24-0624.htm

Solution:

These three types of RNA are all single- stranded and are transcribed from a DNA template by RNA polymerase in the nucleus. However, the function of each type after they leave the nucleus is what determines their differences. Messenger RNA carries the genetic information coded for in the DNA and is responsible for the translation of that information into a polypeptide chain. Each set of 3 bases of the mRNA comprises a codon which directs the incorporation of a specific amino acid into the polypeptide chain. Messenger RNA binds reversibly to the smaller subunit of the ribosome, where protein synthesis is initiated. It can dissociate from the ribosome without jeopardizing the integrity of the ribosome. Degradation of mRNA by ribo-nucleases allows the cell to control its metabolism. Messenger RNA in E. coli has a lifetime of about 2 minutes. Degradation of mRNA is in the 5' \textemdash> 3' direction and may proceed even as ribosome's are translating the mRNA. Ribosomes initiate protein synthesis by binding to the 5' end of the mRNA and reading the codons in the 5' \textemdash>3' direction. If degradation were 3' \textemdash>5', there would be synthesis of incomplete protein chains. Messenger RNA is heterogeneous in size due to the different lengths of polypeptide chains which a cell needs to synthesize, there-fore, it is the most variable type of RNA. While mRNA can dissociate from the ribosomes, rRNA is an integral part of the ribosome and its removal results in the destruction of the ribosome. The 30 S subunit of a bacterial ribosome contains 21 proteins and a 16 S fragment of rRNA. The 50 S subunit is composed of 35 proteins and 2 fragments of rRNA with Svedberg constants of 23 S and 5 S, when placed together, the ribosomal proteins and rRNA undergo spontaneous self-assembly into a fully functional ribosome. This suggests that rRNA interacts with the ribo-somal proteins and helps maintain the characteristic 3-D shape of the ribosome. While the function of rRNA is not completely understood, regions of unpaired bases may be involved in the binding of tRNA and mRNA to the ribosome. Ribosomal RNA has an average molecular weight of 5 × 10^5 to 5 × 10^6 daltons and is the least variable type of RNA, having only the three types mentioned above (5S, 16S, 23S). The single strand of both mRNA and rRNA is able to fold back on itself with subsequent formation of hydrogen bonds between complementary regions, giving rise to apparent "double -stranded" helical regions. (see figure 1) Transfer RNA is the smallest type of RNA. With approximately 80 nucleotides, its molecular weight is 2.5 × 10^4 daltons. While there are at least 20 different types of tRNA's (one specific for each amino acid), the sequence of bases in the different tRNA molecules is highly conserved. This leads to the sharing of the same basic shape among the different types (see figure 2 a). The 3' end always ends in the sequence CCA-3' and the 5' end is a guanine. The amino acid residue is bound to the 3' adenine. Like mRNA and rRNA, tRNA folds back upon itself. Unlike the other two RNA's, tRNA, forms loops and double -stranded sections (see figure 2b). In one of these loops is located the anticodon, which is what distinguishes the different types of tRNA. The function of tRNA is to insert the amino acid specified by the codon on mRNA into the polypeptide chain, and it is through the complementation of anticodon and codon that the appropriate amino acid is incorporated. Hydrogen bonds form between the complementary codons and anticodons, thereby orienting the amino acid for peptide bond formation. A distinguishing feature of tRNA is the presence of unusual bases such as methyl inosine, dimethylguanosine, and ribothymidine. It is the presence of these alkylated bases which protects tRNA from digestion by ribonucleases.

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Question:

Using the Periodic Table of Elements, find the following for sodiumdihydrogenphosphate, NaH_2PO4: (a) formula weight, (b) percent composition of oxygen, (c) weight in grams of 2.7 moles, and (d) percentage composition of oxygen in 2.7 moles.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E04-0149.htm

Solution:

This problem encompasses work in chemical stoichiometry. With this in mind, you proceed as follows: (a)The formula weight = molecular weight, which is the sum of atomic masses of all the atoms in the substance. Na = 22.98, H = 1.008, O =15.9999, and P = 30.97. Thus, formula weight = Na + 2H + P + 4O = 22.98+ 2(1.0080) + 30.97 + 4(15.9999) = 119.9.(b) Percentage composition of oxygen is (total weight of oxygen in compound) / (total weight of compound) = [(4 oxygen atoms) × (15.9994 mass/oxygen atom)] /(119.9) = .5334 g of oxygen or 53.34 % by weight in NaH_2PO4. (C) Mole = (mass of substance) / (molecular weight) . You are given that there are 2.7 moles and you calculated the molecular weight. Thus, the weight in grams of 2.7 moles = (119.9 g/mole) (2.7moles) = 324 g of NaH_2PO_4 . (d) Following a procedure similar to the one in part (b), Percent composition = [{(no. of moles of O) × (MW of O)} / ( weight of comp/no. of moles)] × 100 = {(4 moles × 15.9 g/mole ) / (324 g / 2.7moles)} × 100 = 53.34% The percent composition of any element in any compound does not change when the amount of the compound present is changed.

Question:

For which of the following two reactions should ∆S be more positive? Why?

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Solution:

If the heat gained by a system is equal to that lost by the surroundings, then the entropy change for the surroundings is the negative of the entropy change for the system; for both the system and surroundings taken together, ∆S is zero if the transfer of heat is carried out reversibly. The entropy of a system increases with the increasing randomness of the molecules. Thus, for reaction (a), ∆S is positive owing to the loss of rigidity or increasing randomness of reactants. For reaction (b), ∆S is negative owing to the loss of independent translational motion of the reactants. In other words, the product is more stable or more ordered than the reactants and, thus, the entropy decreases for the reaction.

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Question:

Find log_10 100.

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Solution:

The following solution presents 2 methods for solving the given problem. Method I. The statement log_10 x = y is equivalent to 10^y = x, hence log_10 100 = x is equivalent to 10^x = 100. Since 10^2 = 100, log_10 100 = 2. Method II. Note that 100 = 10 × 10; thus log_10 100 = log_10 (10 × 10). Recall: log_x (a × b) = log_x a + log_x b, therefore log_10 (10 × 10)= log_10 10 + log_10 10 = 1 + 1 = 2.

Question:

An elevator and its load weigh a total of 1600 lb. Find the tension T in the supporting cable when the elevator, origin-ally moving downward at 20 ft/sec, is brought to rest with constant acceleration in a distance of 50 ft. (See fig.)

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/Users/wenhuchen/Documents/Crawler/Physics/D04-0125.htm

Solution:

The mass of the elevator is m = w/g = (1600 lb) / (32 ft/sec^2) = 50 slugs where w is the weight of the elevator and its load. From the equations of motion with constant acceleration, v^2 = v2_0 + 2ay,a = (v_2 - v^2 _0)/2y. Let the upward direction be positive and the origin (y = 0) be at the point where the deceleration begins. Then the initial velocity v_0 is -20 ft/sec, the final velocity v is zero, and its displacement during this Interval is y = -50 ft. Therefore a = {0 - (- 20 ft/sec)}^2 / (-2 × 50 ft) = 4 (ft/sec^2). The acceleration is therefore positive (upward). From the free-body diagram (Fig. ) the resultant force is \sumF = T - w = T - 1600 lb. Hence, from Newton's second law, \sumF = ma, T - 1600 lb = 50 slugs × 4 (ft/sec^2) = 200 lb, T = 1800 lb.

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Question:

Find the minimum energy of a simple harmonic oscilla-tor using the uncertainty principle\Deltax\Deltap\geqh.

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Solution:

Consider a particle of mass m on a spring of force-constant k. The kinetic and potential energies, respectively, are, K = (1/2) mv^2 = (p^2/2m) U = (1/2) kx^2 = (1/2) m\omega^2 x^2 where p =mvand \omega^2 = (k/m). The total energy is constant (conservation of energy) and is therefore equal to its average value. E = = (< p^2 >/2m) + (1/2) m\omega^2 (1) where < > indicates the time average. As a result of symmetry with respect to the equilibrium point, < p > and are zero. The uncertainty in x and p are defined as, (\Deltax)^2 = < (x - )^2 > = - 2< x> + ^2 = - ^2 = - 0 = and (\Deltap)^2 = < (p -

^2 > = - 2

+

^2 = -

^2 = - 0 =

^2 Therefore (1) can be written as, E = (1/2m) (\Deltap)^2 + (1/2) m\omega^2 (\Deltax)^2(2) Using the uncertainty relation \Deltap\Deltax\geqh or (\Deltap)^2 \geq [h^2/(\Deltax)^2] the total energy (2), becomes, E - (1/2) m\omega^2 (\DeltaX)^2 = (1/2m) (\Deltap)^2 \geq [h^2/{2m(\DeltaX)^2}] or E \geq (h^2/2m) [1/(\DeltaX)^2] + (1/2) m\omega^2 (\Deltax)^2 The value of\Deltaxfor which \DeltaE is a minimum is given by the condition, [dE/{d(\Deltax)}] = 0 Hence, - 2(h^2/2m) (\Deltax)^-3 + 2 (1/2) m\omega^2\Deltax= 0 (\Deltax)^4 = (h^2/m^2\omega^2) (\Deltax)^2 = (h^2/m\omega) E_min= (h^2/2m) (m\omega/h) + (m\omega^2/2) (h/m\omega) =h\omega

Question:

A chemist dissolves 10 g of an unknown protein in a liter of water at 25\textdegreeC. The osmotic pressure is found to be 9.25 mmHg. What is the protein's molecular weight. Based upon the number of moles in 10 g of protein, what would the freezing point depression and boiling point elevation be? Assume R = Universal Gas Constant = .0821 (liter- atm / mole\textdegreeK), k_f = (1.86\textdegreeC / m), and k_b = (.52\textdegreeC / m).

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Solution:

The osmotic pressure (\pi) can be related to the molar concentration (C), Universal Gas Constant (R) and temperature (T) in degrees Kelvin of a solution via the formula \pi = CRT. After solving for C one can determine the molecular weight of the protein .Recall, C = (moles / liter) = (N / V), where N = moles and V = volume in liters. Hence, \pi = (N / V) RT. But N = moles = No. of (grams / molecular) wt = (g / M.W.) Rewriting, \pi = [(q / M.W.) /(V)]RTorM.W. = (gRT / \pi V). Substituting for the known values M.W = [(10g)(.821)(298\textdegreeK)] / (9.25mmHG) × [(760mm) / (atm)] = 20,100 (g / mole). Note, you assumed a volume of 1 liter and multiplied by the conversion factor (760 mm / atm), since R is in atm. and \pi in mmHg. Now that you know that the molecular weight is 20,100, the number of moles in the 10 g of protein is (10 / 20),100 = 4.98 × 10^-4 moles. To find the freezing point depression, ∆T_f, and boiling point elevation, ∆T_b, you must find the molality, since ∆T_b = k_bm and ∆T_f = k_fm, where k_f = molal freezing point depression constant, k_b = molal boiling point elevation constant and m = molality. Molality = moles of solute per 1 kg solvent. Thus, molality = .000498 (mole / 1 kg), since in 1 liter of water you have 1000 ml, which weighs 1kg. Therefore, ∆T_f = (1.86\textdegreeC) (.000498) = 9.3 × 10^-4 \textdegreeCand ∆T_b = (.52\textdegreeC) (.000498) = 2.59 × 10^-4\textdegreeC.

Question:

Given that the average speed of an O_2 molecule is 1,700 km/h at 0\textdegreeC, what would you expect the average speed of a CO_2 molecule to be at the same temperature?

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Solution:

The speed of a molecule is related to the mass and temperature by the equation (1/2) mv^2 = 3/2 kT where m is the mass, v is the speed, k is Boltzmann's constant, and T is the absolute temperature. Since k and T are constant 3/2 kT is constant for this system. Thus 1/2 m_(O)2v^2_(O)2 = 1/2 m_C(O)2v^2_C (O)2 where m_(O)2 is the mass of O_2, v_(O)2 is the speed of O_2, m_C(O)2 is the mass of CO_2, and v_C(O)2 is the speed of CO_2. The masses are equal to the molecular weights here. Solving for v_C(O)2 : v_C(O)2 = \surd[{(1/2) m_(O)2v^2_(O)2} / {(1/2) m_C(O)2}]m_(O)2 = 32 m_C(O)2----_--- = 44 v_(O)2 = 1700 km/h. v_C(O)2= \surd[{(1/2) (32) (1700)^2} / {(1/2) (44)}] = 1700 \surd[32/44] = 1700 × (.85) = 1450 km/h.

Question:

What is the molecular weight of a substance, each molecule of which contains 9 carbon atoms and 13 hydrogen atoms and 2.33 × 10^-23 g of other components?

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Solution:

The molecular weight of a compound is the sum of the weights of the components of the compound. It is the weight of one mole of the substance, thus this compound weighs the sum of the weight of 9 moles of C, 13 moles H and (6.02 × 10^23) × (2.33 × 10^-23 g) . Because each molecule of the third substance weighs 2.33 × 10^-23 g, one mole of it weighs (6.02× 10^23) × (2.33 × 10^-23 g) . There are 6.02 × 10^23 molecules of this other substance in one mole of the compound (Avogadro's number). molecular weight=(9 × MW of C) + (13 × MW of H) + (2.33 × 10^-23g) (6.02 × 10^23 /mole) = (9 × 12.01 g/mole) + (13 × 1.00 g/mole) + (14.03g/mole) = 108.09 g/mole \div 13.0 g/mole + 14.03 g/mole = 135.12 g/mole.

Question:

How many calories of heat are required to raise 1,000 grams of water from 10\textdegreeC to 100\textdegreeC?

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Solution:

The temperature rise is 90 degrees centigrade. The number of calories needed is 1000 × 900 = 90,000 calories, since one calorie is required to raise the temperature of one gram of water one degree centigrade.

Question:

The density of KF is 2.48 g/cm^3. The solid is made up of a cubic array of alternate K^+ and F^- ions at a spacing of 2.665 × 10^-8 cm. between centers. From these data, calculate the apparent value of the Avogadro number.

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Solution:

Avogadro's number is the number of particles in one mole of a substance. Here, one is given the dimensions of a cubic lattice made of (1/2)K^+ ions and (1/2)F^- ions. Due to the fact that KF crystallizes likeNaCl, it can be seen that there are eight ions or 4 formula units of KF in the cube. To calculate Avogadro's number from the data, one should find: (1) the volume of one cube 2) the weight of one cube using the density 3) the number of cubes in one mole of KF 4) Avogadro's Number by multiplying the number of cubes by 4, the number of formula units (KF) per cube. Solving for Avogadro's Number: 1) The volume of the cube is found by cubing the length of the edge. From the crystallization of KF, it is seen that the length of the edge of cube is twice the spacing between ion centers. This is due to the fact that three ions make up an edge, and, as such, 2spacingsbe-tween the ions exist. Therefore, length of edge = 2.665 × 10^-8 cm × 2 = 5.33 × 10^-8 cm volume of cube = (5.33 × 10^-8)^3 = 1.51 × 10^-22cm^3. 2) Using the density, one knows that 1 cm^3 weighs 2.48 g. Therefore, the weight of one cube is equal to the density times the volume of a cube. weight of one cube = 2.48 g/cm^3 × 1.51 × 10^-22 cm^3 = 3.76 × 10^-22 g/cube 3) One mole of KF weight the sum of the weights of one mole of K^+ and one mole of F^-. (MW of K = 39.10, MW of F = 19.00.) MW of KF = 39.10 + 19.00 = 58.10 g/mole The number of cubes in one mole is then found by dividing 58.10 g/mole by the weight of one cube. no. of cubes in one mole = (58.10 g/mole) / (3.76 × 10^-22 g/cube) = 1.55 × 10^23 cubes/mole. 4) One finds the number of KF particles in one mole (Avogadro's Number) by multiplying 1.55 × 10^23 cubes/mole by 4 formula units/cube, because there are 4 KF formula units in each cube. Avogadro's Number = 4 [formula units/cube] × 1.55 × 10^23 [cubes/mole] = 6.2 × 10^23 formula units/cube.

Question:

A body is whirled in a vertical circle by means of a cord attached at the center of the circle. At what minimum speed must it travel at the top of the path in order to keep the cord barely taut, i.e., just on the verge of collapse? Assume radius of circle to be 3 ft.

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Solution:

At the top of the circle, the net force on the body of mass m is \sumF = mg + T wherethe positive direction is taken downward, and T is the cord tension. Since the motion is circular, the net force is centripetal and \sumF = (mv^2)/R where v is the body's velocity and R is the circle's radius. Hence mg +T = (mv^2)/R andT = (mv^2)/R - mg If the cord is just on the verge of collapse, T = 0 and (mv^2)/R - mg = 0 Whencev = \surdgR Using the given datav = \surd[(32 ft/s^2)(3 ft)] = \surd[96 ft^2/s^2] v = 9.8 ft/s

Question:

What weight of sulfur must combine with aluminum to form 600 lbs of aluminum sulfide?

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Solution:

In this problem one wants to find out how much reactant (sulfur) was needed to produce a given amount of product (aluminum sulfide). The balanced equation for this reaction is: 2Al + 3S \rightarrow AL_2 S_3 The first method for solving this problem is the pro-portion method. This procedure involves the ratios of weights and molecular weights. Set up the balanced equation showing the weights and molecular weights. Xlbs600 lbs 2 Al + 3S\rightarrowAL_2 S_3 3(32) lb150 lb M.W. of S is 32. M.W. of Al_2 S_3 is calculated below. Atom Number of Atoms Atomic Weight Total Atomic Weight Al 2 27 54 S 3 32 96 M.W. of Al_2 S_3 = 150. Use the proportion equation, [(weight _reactant ) / (moles × M.W. _reactant ) = [(weight _Product ) / (moles × M.W. _product )] Solve for weight_reaotant. weight _reactant = [(3) (32) (600)] / [150] = 384 lbs. sulfur. The second method for solving the problem is by the mole method. We see that 1 mole of Al_2 S_3 requires moles of S. Therefore, if one can calculate the actual number of moles present of 1 substance one can obtain the number of moles of the other substance by setting up a ratio. Then, knowing the number of moles, one can calculate the weight by using the equation moles = weight/M.W. One is given that 600 lbs. of Al_2 S_3 are present. Therefore, the number of moles is 600/150 = 4 pound moles of Al_2 S_3. Setting up the ratio, [(S) / (Al_2 S_3 )] = (3/1) = (x/4) it is seen that x = 12 lb. moles of S. Next, solve the mole equation for weight of S weight = moles × M.W. = 12 × 32 = 384 lbs.

Question:

A circus clown whose mass is 1 × 10^2 kg steps onto the outer edge of a large disk with a radius of 2.0 × 10^1 m and a mass of 2 × 10^3kg. As-sume that the disk is mounted on a frictionless bearing with a vertical axis of rotation and is initially at rest. If the clown now runs clock-wise around the edge of the disk at a speed of 2m/s, how fast does the disk turn and what is the angular momentum?

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Solution:

Since no external torques are acting on the clown-disk system, we may apply the principle of conservation of angular momentum to this system. The angular momentum of the system before the clown stepped on the disk was zero. Therefore, after the clown steps on the disk, the angular momentum of the disk-down system must remain zero, or J\ding{217}_system = 0 \ding{217} But J\ding{217}_system = J\ding{217}_disk + J\ding{217}_clown= 0 \ding{217} \ding{217} \ding{217} and J\ding{217}_disk =-J\ding{217}_clown \ding{217} - \ding{217} This last equation states that the magnitude of J\ding{217}_disk and J\ding{217}_clown are the \ding{217} \ding{217} same, but that J\ding{217}_disk and J\ding{217}_clown are opposite in direction. Hence, because \ding{217} \ding{217} the clown is running clockwise, the disk travels counter-clock- wise. If it is assumed that the clown can be represented as a particle, the orbital angular momentum of the clown, is J\ding{217}= r\ding{217}×p\ding{217} \ding{217} \ding{217} × \ding{217} where p\ding{217}is the linear momentum of the clown, and r\ding{217}is a vector from the \ding{217} \ding{217} axis of rotation to the clown. Because the clown is running in a circle r\ding{217}is \ding{217} perpendicular to p\ding{217}and \ding{217} \vertJ\ding{217}_clown\vert = rp_clown = mrv_clown \ding{217} = (1×10^2kg)(2 m/s) (2×10^1m) × × = 4×10^3J\bullets × \bullet which is also the magnitude of the disk's angular momentum. Now, \vertJ\ding{217}_dlsk\vert = I_disk\omega_disk \ding{217} \omega where I_disk is the moment of inertia of the disk about its axis of rotation. Therefore, since \vertJ\ding{217}_dlsk\vert = \vertJ\ding{217}_clown\vert, \ding{217} \ding{217} \omega_disk = (\vertJ\ding{217}_clown\vert)/(I_disk)(1) \omega \ding{217} The only variable we don't know in (1) is I_disk. This we now calculate. By definition, I_disk =\intr^2 dm \int where r is the distance of a mass element dm from the axis about which we calculate I_disk (see figure), and the integral is taken over the mass of the disk. Now \rho= dm/dv \rho where\rhois the density of the disk and dv is a volume element. \rho Whence I_disk =\intr^2\rhodm I_disk =\intr^2\rhodm \int \rho If we consider the disk to be very thin, dv = ds, an element of area of the disk, and ds = r dr d\texttheta \texttheta I_disk =2\pi\int_0^R\int_0\rhor^3 dr d\texttheta I_disk =2\pi\int_0^R\int_0\rhor^3 dr d\texttheta \pi \int \int \rho \texttheta I_disk =2\pi\int_0\rho(R^4/4) d\texttheta I_disk =2\pi\int_0\rho(R^4/4) d\texttheta \pi \int \rho \texttheta I_disk = (\pi\rhoR^4)/(2) I_disk = (\pi\rhoR^4)/(2) \pi\rho where R is the disk radius. But\rho= M/\piR^2, M being the disk's mass. \rho \pi I_disk = (\piR^4)/(2) (M/\piR^2) = MR^2/2 \pi \pi Using (1), \omega_disk= (\vertJ\ding{217}_clown\vert)/(MR^2/2) \omega \ding{217} = [4×10^3J\bullets]/[(2×10^3kg/2)(4×10^2m^2)] × \bullet × × \omega_disk= [4×10^3J\bullets]/[4×10^5m^2\bulletkg] \omega × \bullet × \bullet = 10^-2 N\bulletm\bullets/kg\bulletm^2 \bullet \bullet \bullet \omega_disk= 10^-2 [(kg\bulletm^2/s^2\bullets)/(kg\bulletm^2)] \omega \bullet \bullet \bullet = 10^-2 s^-1 Furthermore, v_disk =\omega_disk R is the velocity of a point on the rim of the disk. \omega v_disk = (10^-2 s^-1)(2×10m) = 0.2m/s × The disk rotates much slower than the clown runs around the edge.

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Question:

What will be the final concentration of a solution of 50 ml of .5M H_2SO_4 added to 75 ml of .25M H_2SO_4 (the final volume is 125 ml)?

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Solution:

Two facts must be known to answer this question: (1) M =Molarity = the number of moles of solute per liter of solution and (2) that the number of moles of H_2SO_4 in the final solution is the sum of the number of moles of H_2SO_4 in each solution. molarity= [(no. of moles)/(liters of solution)] One can rearrange this equation to determine the number of moles present in each solution. no. of moles =molarity× liters of solution The total number of moles in the final solution is equal to the sum of the number of moles of each of the two initial solutions. total number of moles = (0.05l)(.5 moles/l) + (.075l) (.25 moles/l) = 0.025 moles + .019 moles = .044 moles. To obtain the final concentration, you return to the definition of molarity, supplying the known values. Thus,molarity(final concentration)= [(0.044moles) / (.125 liter)] = 35 M H_2SO_4 (It was given that the volume of the final solution equaled 125 ml = .125 liters.)

Question:

How many full strokes of a bicycle pump (chamber 4.0 cm diameter and 40.0 cm long) would you need to make in order to pump up an automobile tire from a gauge pressure of zero to 24 pounds per square inch (psi)? Assume tem-perature stays constant at 25\textdegreeC and atmospheric pressure is one atmosphere. Note, that gauge pressure measures only the excess over atmospheric pressure. A typical tire volume is about 25 liters.

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Solution:

One atmosphere equals 14.7 psi, therefore, the amount of pressure needed to fill the tire is 24 + 14.7 psi or 38.7 psi. Converting back toatm: atm. contained in inflated tire = 38.7 psi × 1atm/14.7 psi = 2.63 atm. When the tire is deflated the pressure is 1atmand the volume is 25 l, Eachatmof pressure occupies 25 l. Therefore, the volume of the tire when inflated is 25l × 2.63 atm. Volume of inflated tire = 25l, × 2.63 = 65.75l Since there is 25l present in the tire before in-flation, the volume that the pump must contribute is 65.75 - 25l or 40.75l. The volume of air forced into the tire at each stroke of the pump is equal to the volume of the pump. Volume of pump= \pir^2 h = \pi × (2 cm)^2 × 40 cm = 503 cm^3 /stroke There are 1000 cm^3 in 1 liter, therefore 40750 cm^3 of air must be pumped into the tire. If 503 cm^3 is pumped in per stroke the number of strokes necessary to fill the [(40750 cm^3 )/(503 cm^3 /stroke)] No. of strokes = [(40750 cm^3 )/(503 cm^3 /stroke)] = 81 strokes.

Question:

TheRydberg- Ritz equation governing the spectral lines of hydrogen is (1/\lambda) = R [(1/n_1 ^2) - (1/n_2 ^2)], where R is the Rydberg constant, n_1 indexes the series under consideration (n_1 = 1 for the Lyman series, n_1 = 2 for theBalmerseries, n_1 = 3 for thePaschenseries), n_2 = n_1 + 1, n_1 + 2, n_1 + 3, . . . indexes the successive lines in a series, and \lambda is the wave- length of the line corresponding to index n_2. Thus, for the Lyman series, n_1 = 1 and the first two lines are 1215.56 \AA (n_2 = n_1 + 1 = 2) and 1025.83 \AA (n_2 = n_1 + 2 = 3). Using these two lines, calculate two separate values of the Rydberg constant. The actual value of this constant is R = 109678 cm^-1.

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Solution:

The first thing to do is to convert the wave - lengths from \AA to more manageable units, i.e. centimeters.Using the relationship 1 \AA = 10^-8 cm, the first two Lyman lines are 1215.56 \AA = 1215.56 × 10^-8 cm for n_2 = 2, and 1025.83 \AA = 1025.83 × 10^-8 cm for n_2 = 3. Solving theRydberg- Ritz equation for R, one obtains : R = [\lambda {(1/n_1 ^2) - (1/n_2 ^2)}]^-1, For the first line, R = [\lambda {(1/n_1 ^2) - (1/n_2 ^2)}]^-1 = [1215.56 × 10^-8 cm {(1/1^2) - (1/2^2)}]^-1 = 109689 cm^-1 and for the second line, R = [\lambda {(1/n_1 ^2) - (1/n_2 ^2)}]^-1 = [1025.83 × 10^-8 cm {(1/1^2) - (1/3^2)}]^-1 = 109667 cm^-1 The first of these is 0.0100% greater than the true value, and the second is 0.0100% less than the true value.

Question:

Find Antilog_10 1.4850.

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Solution:

By definition, Antilog_10 a = N is equivalent to log_10N = a. Let Antilog_101.4850 = N. Hence, Antilog_10 1.4850 = N is equivalent to log_10N = 1.4850. The characteristic is 1. The mantissa is 0.4850. Therefore, the number that corresponds to this mantissa will be multi-plied by 10^1 or 10. The mantissas which appear in a table of common logarithms and are closest to the mantissa 0.4850 are 0.4843 and 0.4857. The number that corresponds to the mantissa 0.4850 will be found by interpolation. Set up the following proportion. (d/.01) = (.0007/.0014) cross-multiplying, .0014d= (.01) (.0007), or d =.01 (.0007/.0014) = (1 ×10^\rule{1em}{1pt}2) [(7 × 10^\rule{1em}{1pt}4) / (1.4 × 10^\rule{1em}{1pt}3)] = [(7 × 10^\rule{1em}{1pt}6) / (1.4 × 10^\rule{1em}{1pt}3)] = (7/1.4) × (10^\rule{1em}{1pt}6/10^\rule{1em}{1pt}3) = 5 × 10^\rule{1em}{1pt}6\rule{1em}{1pt}(\rule{1em}{1pt}3) = 5 × 10^\rule{1em}{1pt}3 = 5 ×.001 = .005 Hence,d= 0.005 x= d + 3.05 = 0.005 + 3.050 = 3.055 Hence, N = Antilog_10 1.4850 = 3.056 × 10 = 30.550 = 30.55 Therefore Antilog_10 1.4850 = 30.55.

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Question:

A Rowland ring, made of iron, of mean circumferential length 30 cm and cross section 1 cm^2 , is wound uniformly with 300 turns of wire. Ballistic galvanometer measurements made with a search coil around the ring show that when the current in the windings is 0.032 amp, the flux in the ring is 2 × 10^-6 weber(see figure). Compute (a) the flux density in the ring, (b) the magnetic intensity, (c) the perme-ability,(d) the relative permeability.

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/Users/wenhuchen/Documents/Crawler/Physics/D21-0733.htm

Solution:

a)The flux, \varphi, through the ring is defined as \varphi = \int B^\ding{217} \textbullet dA^\ding{217} where B is the flux density and dA^\ding{217} is an infinitesimal cross sectional element through which the flux passes. B is constant in the iron, and the angle between B^\ding{217} and the area vector is zero. Therefore \varphi = B \int dA = BA where A is the cross section of the iron. Hence B = (\varphi/A) = [(2 × 10^-6 w)/(1 cm^2 × 1 m^2 /10^4 cm^2 )] = 2 × 10^-2 w/m^2 . b)Magnetic intensityH = Ni/l = [(300 × .032)/(.30)] = 32 amp-turns/m. c)Permeability \mu = (B/H)(by definition) = [(2 × 10^-2 w/m^2 )/(32 amp/m)] = [6250 × 10^-7 w / amp-m.] d)Relative permeability K_m = (\mu/\mu_0 ) = [(6250 × 10^-7 )/(12.57 × 10^-7 )] = 498(no units).

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Question:

Write condensed structural formulas for all the alkynes, i.e. unsaturated compounds with triple bonds, with a molecular formula of C_5H_8.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E20-0734.htm

Solution:

A condensed structural formula provides all the information represented by other structural formulas (i.e. Lewis diagrams, bond diagrams), but it is not as cumbersome. To solve this problem, first write all the structural isomers with five carbons in a continuous chain: Second, write all the structural isomers with a four carbon parent molecule and one carbon in a branch: No other structural formulas are possible. Thus, there are only three structural formulas poss-ible for C_5H_8 . Next, one writes these formulas in the condensed form. Namely,

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Question:

Explain the types and effects of competition.

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Solution:

Competition is the active demand of two or more organisms for a commonvital resource. There are many ways for organisms to compete butbasically competition can be categorized in two ways. Contest competitionis the active physical confrontation between two organisms whichallows one to win the resource. Extreme contest competition involvesdirect aggression. Scramble competition is the exploitation of a commonvital resource by both species. In scramble competition, one organismis able to find and utilize vital resources more efficiently than another. The aggressive behavior of ant colonies is an example of contest competition. Actual colony warfare can establish territorial limits within and betweenspecies. Often, contest competition takes the form of threatening gestureswhich involves safer means to achieve the same goal. Scramble competition can best be demonstrated with observations ofthe fruit fly Drosophila. Combat and aggressivebehavior havenothing todo with the type of competition these flies demonstrate. It has been observedthat when the food supply was limited, there were four conditionsfor survival for the fly larvae. Those that survived were the quickestfeeders, best adapted for the particular medium, heaviest at the beginningof competition, and most resistant to changes due to population increase. Every species occupies a particular niche that to a large extent helpsavoid competition. But niches do overlap in many areas and competition can never be clearly eliminated.

Question:

An E. coli culture will grow to a limiting concentration of about 10^9 cell per cm^3. At this concentration what percentage of the total volume of the culture medium is occupied by the cells? Assume that an E. coli cell is cylindrical and that it is 2.0\mu long and has a 1.0\mu diameter.

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Solution:

This percentage can be found by first determining the volume that one E. coli cell occupies and then the volume that 10^9 cells occupy. This volume divided by 1 cm^3 multiplied by 100 equals the percent of the total volume that the cells occupy. The volume of a cylinder is equal to its length multiplied by the area of its cross-section. Area of a circle = \pir^2, where r = radius of the circle. 1\mu = 10^-4 cm Volume of 1 E. coli cell= (2.0 x 10^-4 cm) (\pi) × (0.5 × 10^-4cm)2 = 1.57 × 10^-12 cm^3 There are 10^9 cells per cm^3 of solution, therefore the volume that the cells occupy in 1 cm^3 of solution is 10^9 × 1.57 × 10^-12 cm^3 or 1.57 × 10^-3 cm^3. Percent of total volume that the E. coli cells occupy = (1.57 × 10^-3 cm^3)/ (1.0cm^3 × 100) = .157%.

Question:

Compute ther.m.s. speed of 0_2 at room temperature.

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/Users/wenhuchen/Documents/Crawler/Physics/D14-0491.htm

Solution:

Ther.m.s. speed of a gas molecule is v_r.m.s . = \surd[(3RT)/(M)] where M is the molar mass of the gas, T is its temper-ature in degrees Kelvin, and R is the gas constant. Hence, v_r.m.s = \surd{[(3)(8.31 joule/mole\textdegreeK)(273\textdegreeK)]/(32 × 10^-3 kg/mole)} Here, we have used the fact that 0\textdegreeC = (0 + 273)\textdegreeK. v_r.m.s = \surd(2.13 × 105m^2/g^2) v_r.m.s = 4.61 × 10^2 m/s.

Question:

(a) Explain what is meant by error detection through redundancy. For each of the following words, determine if it is a "legal" code word for the (b) binary-coded decimal (BCD) system. (c) excess-3(X-3) code system. (i)0 1 0 1 (ii) 0 1 0 0 (iii) 1 1 0 0

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G06-0126.htm

Solution:

(a) All information within a computer is represented as bits of 1's and 0's. The flow of this information within, to and from the computer occurs along some form of transmission line. A basic physical characteristic of any transmission line is the presence of disturbances which introduce the possibility of errors occurring in the information transmitted on this line. Let us represent a trans-mission system by letting the transmitted information be the binary n-tuple (x1 ,x2 ,...,x_n ) which moves along n parallel data lines, forming n independent networks. Thus, one faulty line can influence only a single bit, and therefore, at worst, introduce the error of complementing one of the n-tuple components in the output: This independent error representation of transmission lines character-izes data buses (bidirectional communication lines between the CPU, memory, and I/O devices, which carry instructions and transfer data). A simplifying assumption that will be useful is that, in a data word consisting of n bits, only one bit will be incorrect. Probability theory can be used to show that for an error rate of one in 1000 trans-mitted bits, the odds of having two erroneous bits in a 5-bit word (n = 5) are less than one in 100,000. So the single-error approxima-tion is a reasonable, though simplistic, assumption. Computers are supplied with automatic error detectors. An error can be detected by redundancy, which is the inclusion of extra information with each transmission. This extra information helps the receiver to decide if the received message has been altered in transmission. The simplest error detecting system, conceptually, would be duplicate transmission of each data unit. Thus, if the first transmission doesn't match the second transmission, the receiver can signal that an error has occurred. If we are willing to surrender a degree of certainty in exchange for efficiency, better methods are available. One such method is the binary- coded decimal (BCD) code system. In the BCD system, four bits are used to code the decimal digits 0-9 in their binary representation. Decimal BCD 0 0000 1 0001 2 0010 3 0011 4 0100 5 0101 6 0110 7 0111 8 1000 9 1001 From these 10 digits any positive decimal Integer can be constructed. In BCD the binary representations for the numbers 10-15 are not used and, if they show up at a receiver, they can be detected as "illegal" by checking the received datum with a "legal word" list: (b) Since the legal BCD word list consists of the binary representa-tions of the decimal digits 0-9, any received 4 bit word whose decimal equivalent is greater than 9 is illegal. (i) Since binary 0101 = decimal 5, 0101 is a legal BCD code word, (ii) Since binary 0100 = decimal 4, 0100 is a legal BCD code word, (iii) Since binary 1100 = decimal 12 is greater than 9, 1100 is not a legal BCD code word. (c) The excess-3 (X-3) code is obtained by adding binary 0011 to the BCD codes: Decimal BCD Excess-3 0 0000 0011 1 0001 0100 2 0010 0101 3 0011 0110 4 0100 0111 5 0101 1000 6 0110 1001 7 0111 1010 8 1000 1011 9 1001 1100 Note that X-3 does not allow 0000 or 1111. Thus, every legal X-3 word contains at least one 0 and at least one 1, which provides the receiver with the information that the data channel is active and transmitting. Again, the error-detection algorithm involves table look-up and comparison. Since 0101, 0100, and 1100 are all on the legal word list, they are all legal X-3 code words.

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Question:

With the aid of a diagram, describe the structures found in the human ear.

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Solution:

Three parts of the human ear can be distinguished: the outer ear, the middle ear, and the inner ear. The outer ear consists of the skin- covered cartilaginous flap, or pinna, and a channel known as the external auditory meatus. At the inner end of this canal is a membrane called the tympanic membrane (eardrum). On the other side of the tympanic membrane is a small chamber, the middle ear, which contains three tiny bones. These three bones are the malleus (hammer), incus (anvil) and stapes (stirrup) and are arranged in sequence across the middle ear from the tympanic membrane to another membrane, the oval window, which separates the middle ear from the inner ear. The middle ear is connected to the pharynx via the Eustachian tube, which serves to equalize the air pressure between the outer and middle ear. The inner ear consists of a complicated labyrinth of interconnected fluid-filled chambers and canals. One group of chambers and canals is involved with the sense of equilibrium. There are two small chambers - the sacculus and utriculus - and three semicircular canals arranged perpendicularly to one another. The utricle and saccule are small, hollow sacs lined with sensitive hair cells and containing small ear stones, or otoliths, made of calcium carbonate. The semicircular canals contain hair cells and are filled with a fluid known as the endolymph. Movement of the endolymph over the hair cells stimulates the latter to send impulses to the brain. Besides the structures involved in maintaining equilibrium, the inner ear also houses the organ for hearing, the cochlea. The cochlea is a coiled tube made up of three canals - the vestibular canal, the tympanic canal and the cochlear canal, each separated from each other by thin membranes. The oval window is linked to the vestibular canal while the tympanic canal is sealed by a membrane, the round window, which leads to the middle ear. These two canals are connected with each other at the apex of the cochlea and are filled with a fluid known as the perilymph. The cochlear canal, filled with endolymph, contains the actual organ of hearing, the organ of Corti. The organ of Corti consists of a layer of epithelium, the basilar membrane, on which lie five rows of specialized receptor cells extending the entire length of the coiled cochlea. Each receptor cell is equipped with hairlike projections extending into the cochlear canal. Overhanging the hair cells is a gelatinous structure, the tectorial membrane, into which the hairs project. The hair cells of the organ of Corti initiate impulses via the fibers of the auditory nerve to the brain.

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Question:

What will be the relative concentrations of Zn^+2 and Cu^+2 when the cell Zn; Zn+2II Cu^+2; Cu has completely run down? (E\textdegree = 1.10 volts).

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/Users/wenhuchen/Documents/Crawler/Chemistry/E16-0601.htm

Solution:

The key phrase to understand before solving this problem is "has completely run down". That means there is no apparent current, leading us to assume a value of zero for the electrode potential, E. Next, one should realize that whenever concentrations are involved one should apply the Nernst Equation, which states E = E\textdegree - (RT / nF)In Q , where Q is the equilibrium constant expression, R is the gas constant (8.314 joules per degree), F is 96,500 coulombs, n is the number of moles of electrons transferred and T is 25\textdegreec (or 298\textdegree K). Since the overall cell reaction is Zn + Cu^+2 - Zn^+2 + Cu , 2 moles of electrons are transferred and log Q equals log {[Zn^+2] / [Cu ]}. Remember that solids are omitted from all concentration equations. This yields (via substitution): E = E\textdegree - (RT / nF) In Q and 0 = 1.1 volts - (.059 / 2) log {[Zn^+2] / [Cu^+2]} , which means log {[Zn^+2] / [Cu^+2]} = 37.3 . Solving, [Zn^+2] / [Cu^+2] = 2 × 10^37 . This means that the concentration of Zn^+2 is far greater than Cu^+2 .

Question:

How is an ARRAY declared inPL/I? Explain using examples.

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Solution:

Every identifier to be used as an array name requires explicit declaration. The declaration must also de-fine the subscripting to be used whenthe array elements are referenced. For declaration of an array variable, the declar-ation has the following format: DCL ARRAY NAME (DIMENSION ATTRIBUTES) DATA ATTRIBUTES; Array names can be any variable names and every element within thearray is referenced by the array name declared. The use of the dimension attribute in an array declara-tion is governedby the following rules. 1. The dimension attribute has the form of one or more boundary specificationsenclosed within a pair of parentheses. If two or more boundaryspecifications are used in the list, they are separated by commas. The number of boundary speci-fications in the dimension attributeestablishes dimensional-ity of the array variable. 2. A boundary specification is composed of two, possib-ly signed, decimalinteger constants, separated by a colon. The first integer value specifiesthe lower bound and the second the upper bound of that particulardimension. For example DECLAREA(-1:2, 8:10); In the above, A is the name of the array. It could have been any other nametoo, e.g., NANCY, PETE, POOR, etc.. But the same name cannot beused to refer to more than one data struc-ture or data item. Also note thatin the above, A is a two dimensional array as there are two boundary, specifications. The first subscript may assume any one of the integer values-1, 0, 1, or 2. The second subscript may assume any one of the integervalues in the range of 8 through 10. Thus, in the first subscript thereare 4 elements and the second sub-script has 3 elements so the arrayA contains 4 × 3=12 elements. 3. In a boundary specification, the lower bound must have its value lessthan or equal to the value of the upper bound. The dimension attributes, e.g.,(3:5), (-3:0, 0:3), (9:9, 1:1), arewell formed. But, (5:3) is not well formed because the lower bound is greater thanthe higher bound. If the lower bound is not specified it is assumed to havea numeric value of one. For example, the dimension attributes (17,26,5) and (1:17, 1:26, 1:5) areequivalent. 4. The dimension attribute must always appear as the first attribute followingthe variable name declared. Using the above 4 rules a well-formed declare statement can be written. E.g., consider the following declare state-ment: DECLAREB(5,12) FIXED DEC(8,3); The above statement is a well-formeddeclarestatement. It establishes a twodimensional array B, with upper bounds 5 and 12. The elements are fixeddecimal numbers with preci-sion of the type XXXXX.XXX. However, thefollowing declare statement is wrong. DECLARE BFIXED(5,12) DEC(8,3); The machine would be unable to distinguish between dimension and precisionattributes when written as above. The declara-tions for more than onearray can sometimes be condensed. For example, consider the following: DECLARE (A, B, MATRIX) (10, 1:9, -2:5) FLOAT; DECLARE (X FIXED, Y FLOAT) (17, 20); This is how the array structure is declared. In the above, A, B and MATRIX are3-dimensional arrays with 10×9×8=720 elements in each. All theelements have a Float-ing Point number representation. X and Y are both 2-dimensional arrays, each of 17 × 20 = 340 elements. But, the elements of X have a Fixed Point repre-sentation, and the elements of Y have a Floating Point re-presentation.

Question:

What is the acceleration of a block on a frictionless plane What is the acceleration of a block on a frictionless plane in-clined at an angle \texttheta with the horizontal?

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/Users/wenhuchen/Documents/Crawler/Physics/D04-0122.htm

Solution:

In order to find the acceleration, a, of the block, we must calculate the net force, F, on the block, and relate this to its acceler-ation via Newton's Second Law, F = ma. (Here m is the mass of the block). The only forces acting on the block are its weight mg and the normal force N exerted by the plane (see figure). Take axes parallel and perpendicular to the surface of the plane and resolve the weight into x- and y-components. Then \sumF_y = N - mg cos \texttheta \sumF_x = mg sin \texttheta. But we know that the acceleration in the y direction, a_y = 0, since the block doesn't accelerate off the surface of the inclined plane. From the equation \sumF_y = ma_y we find that N = mg cos \texttheta. From the equation \sumF_x = ma_x, where a_x is the acceleration of the block in the x direction, we have mg sin \texttheta = ma_x, a_x = g sin \texttheta. The mass does not appear in the final result, which means that any block, regardless of its mass, will slide on a frictionless inclined plane with an acceleration down the plane of g sin \texttheta. (Note that the velocity is not necessarily down the plane).

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Question:

Using positive logic, determine the logic function represented by the diode circuit of figure 1.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G04-0063.htm

Solution:

The ideal voltage-current characteristic of a diode is shown in fig. 2. The diode conducts when V_D is positive and acts as an insulator when V_D is negative. Hence, the diode is similar to a valve which lets current flow in only one direction. As can be seen in fig. 1, when all inputs are high (greater than \rule{1em}{1pt}5V) the diodes conduct, there is a large positive current in R, and the output voltage is greater than \rule{1em}{1pt}5V. When only one or two inputs are high; those diodes corresponding to the high inputs conduct and those diodes corresponding to the low inputs (less than or equal to \rule{1em}{1pt}5V) act as insulators. The conducting diode(s) pass current to the resistor R, forming a voltage drop. Hence, the output voltage is greater than \rule{1em}{1pt}5V. When all inputs are low (less than or equal to \rule{1em}{1pt}5V) all the diodes act as insulators and there is no current in resistor R, hence, the output voltage is \rule{1em}{1pt}5V. From this information it is seen that the output is low only when all of the inputs are low. Tabulating this result, we get: V IN 1 V IN 2 V IN 3 V IN OUT L L L L L L H H L H L H L H H H H L L H H L H H H H L H H H H H It is evident that this table represents an OR gate.

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Question:

Calculate the pressure required to compress 2 liters of a gas at 700 mm pressure and 20\textdegreeC into a container of 0.1 liter capacity at a temperature of - 150\textdegreeC.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E02-0039.htm

Solution:

One is dealing with changing volumes, press-ures and temperatures of a gas. Therefore, this problem can be solved using the combined gas law. It states that as the pressure increases, the volume decreases and that as the temperature increases, the volume in-creases. These factors are related by the equation (P_1 V_1 )/T_1= (P_2 V_2 )/T_2 where P_1, V_1 and T_1 are the initial pressure, volume and temperature and P_2, V_2, and T_2 are the final values. For any problem dealing with gases, the first step always involves converting all of the temperatures to the degree Kelvin scale by the equation \textdegreeK = \textdegreeC + 273 For this question T_1 = 20\textdegreeC = 20 + 273 = 293\textdegreeK T_2 = - 150\textdegreeC = - 150 + 273 = 123\textdegreeK. This seems to indicate that the pressure would decrease. But one is also told that the volume de-creases, which would have the effect of increasing the pressure. Therefore, one cannot predict the final change in volume. For the sake of clarity, set up a table as given below. P_1 = 700 mmP_2 = ? V_1 = 2 litersV_2 = 0.1 liter T_1 = 293\textdegreeKT_2 = 123\textdegreeK Since one is given 5 of the 6 values, it is poss-ible to use the combined gas law equation to determine P_2 (P_1 V_1 )/T_1= (P_2 V_2 )/T_2 P_2 = [(T_2 V_1 P_1 )/(T_1 V_2 ) = [{123\textdegreeK (2 liters) (700 mm)}/{293\textdegreeK (0.1 liter)}] = 5877 mm.

Question:

The skin is much more than merely an outer wrapping for an animal; it is an important organ system and performs many diverse functions. What are some of the primary functions of the skin in man?

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Solution:

Perhaps the most vital function of the skin is to protect the body against a variety of external agents and to maintain a constant internal environment. The layers of the skin form a protective shield against blows, friction, and many injurious chemicals. These layers are essentially germproof, and as long as they are not broken, keep bacteria and other microorganisms from entering the body. The skin is water-repellent and therefore protects the body from excessive loss of moisture. In addition, the pigment in the outer layers protects the underlying layers from the ultraviolet rays of the sun. In addition to its role in protection, the skin is involved in thermoregulation. Heat is constantly being produced by the metabolic processes of the body cells and distributed by the bloodstream. Heat may be lost from the body in expired breath, feces, and urine, but approximately 90 per cent of the total heat loss occurs through the skin. This is accomplished by changes in the blood supply to the capillaries in the skin. When the air 'temperature is high, the skin capillaries dilate, and the increased flow of blood results in increased heat loss. Due to the increased blood supply, the skin appears flushed. When the temperature is low, the arterioles of the skin are constricted, thereby decreasing the flow of blood through the skin and decreasing the rate of heat loss. Temperature-sensitive nerve endings in the skin reflexively control the arteriole diameters. At high temperatures, the sweat glands are stimulated to secrete sweat. The evaporation of sweat from the surface of the skin lowers the body temperature by removing from the body the heat necessary to convert the liquid sweat into water vapor. In addition to their function in heat loss, the sweat glands also serve an excretory function. Five to ten per cent of all metabolic wastes are excreted by the sweat glands. Sweat contains similar substances as urine but is much more dilute.

Question:

The solar constant, or the quantity of radiation received by the earth from the sun is 0.14 W\bulletcm^-2. Assuming that the sun may be regarded as an ideal radiator, calculate the surface temperature of the sun. The ratio of the radius of the earth's orbit to the radius of the sun is 216.

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Solution:

To calculate the temperature of the sun, T, we use Stefan's Law R = e \sigma T4 Here, e is the emissivity of the radiator, \sigma is a constant, T is the temperature of the radiator in Kelvin degrees, and R is the rate of emission of radiant energy per unit area of the radiator. Hence, T^4 = R/e\sigma(1) Regarding the sun as an ideal radiator, e = 1. Furthermore, R = P/A where A is the surface area of the sun, and P is the power provided by the sun as a result of radiation. Using these facts in (1) T^4 = P/\sigmaA(2) Now, the power per unit area intercepted by the earth is P/A' = .14 W/cm^2 where A' is the surface area of a sphere having a radius equal to that of the earth's orbit. Hence, P = (.14 W/cm^2) A'(3) Using (3) in (2) T^4 = {(.14 W/cm^2)A'}/\sigmaA Now A'/A = (4\pir^2)/(4\piR^2), r where r and R are the radius of the earth's orbit and the radius of the sun, respect-ively. Then, T^4 = [(.14 W/cm^2)/\sigma] (r/R)^2or T^4 = [(0.14 W\bulletcm^-2)/(5.6 × 10^-12 W\bulletcm^-2\bulletK deg^-4)] × (216)^2 \therefore T = 5.84 × 10^3 \textdegreeK.

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Question:

Using the pedigree below, determine the method of inheritance of the trait, and as far as possible, fill in the genotypes of each individual.

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Solution:

One should immediately note that the trait is only exhibited by male members of the family. Although the number of individuals is small enough that this could have been a a chance occurrence, it strongly suggests that the trait is sex-linked. Proceeding on this assumption, we can deduce the possible genotype of each member. Letting r represent the recessive, sex-linked allele, we can assign the genotype X^rY to P_1, F_1,7, F_2,1 ,F_2,3 ,F_2,5 ,and F_3,1, since all are males expressing the trait. Since F_2,7 ,a male, does not express the trait, and there-fore does not carry the r allele, his genotype is X^RY, where R represents the normal allele. We know that F_3,1 must have received the r allele from his mother, F_2,6 since to a male offspring, the mother contributes the X while the father contributes the Y chromosome. The genotype of F_2,6 is then X^RX^r, since she does not express the trait. F_1,5 must also be a carrier (X^RX^r) since F_2,6 is a carrier, F _2,5 expresses the trait, and F1,4,who must be X^RY, could not have donated the recessive allele to his offspring. Since we have no information concerning the offspring of F_2,4 ,her genotype could be either X^RX^R or X^Rx^r, depending on whether she received a dominant or recessive allele from F_1,5 . Since F_1,5 is a carrier, and her father P_3 ,who does not express the trait, is X^RY, we can assign P_4, the geno-type X^RX^r , which shows that she is also a carrier. F_1,6 can be either X^RX^R or X^Rx^r, one dominant allele in either case being necessarily donated by P_3. F_1,3 a male not expressing the trait, is X^RY. Since no male offspring of P_1 and P_2 express the trait, it is likely though not necessary, that P_2 .is not a carrier. Her possible genotypes are X^RX^R and X^Rx^r. F_1,2 must have received the recessive allele from P_1 and be a carrier (X^Rx^r). F_1,1 must be X^RY since he does not express the trait. F_2,2 could be either X^RX^R or X^RX^r. Summarizing, we can see that our assumption of sex linkage is consistent with the inheritance pattern of the trait :

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Question:

Our eyes are the principal organs involved in seeing. Yet we are able to perform manual acts such as dressing or tying knots when our eyes are closed. Why?

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Solution:

There are many sense organs (receptors) in the body. Our pair of eyes is just one example. Although our eyes are extremely important to our perception of this world, we will not be totally and helplessly lost when we can no longer use our eyes. For instance, we are still aware of the relation of our body to the environment even if our eyes are closed. We know whether we are standing or sitting, we know where our limbs are, and we know where one part of our body is in relationship to another. Such perception without the use of our eyes in achieved with a different set of sense organs known asproprioceptors. Proprioceptorsare receptors found in muscles, tendons and joints, and are sensitive to muscle tension and stretch. They pick up impulses from the movements and positions of muscles and limbs relative to each other and relay them to the cerebellum for coordination. Impulses from the proprioceptors are extremely important in ensuring the coordinated and harmonious contraction of different muscles involved in a single movement . Without them, complicated skillful acts would not be possible. Proprioceptorshave other important functions. They help maintain the sense of balance and give the body general awareness of its environment .

Question:

Calculate the mass of the electron by combining the results of Millikan's determination of the electron charge q and J.J. Thomson's measurement of q/m.

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/Users/wenhuchen/Documents/Crawler/Physics/D31-0920.htm

Solution:

The charge of the electron, q = 1.6 × 10^-19 C, and the ratio of charge to mass, q/m = 1.76 ×10^11 C/kg, are known. q/m = 1.76 × 10^11 (C/kg)but q = 1.6 × 10^-19 C Therefore, m = [{1.6 × 10^-19C} / {1. 76 × 10^11 (C/kg)}] = 9.1 × 10^-31 kg The mass of the electron is about 9.1 × 10^-31 kg.

Question:

Bone always develops by replacement of a preexisting connective tissue. When bone formation takes place in preexisting cartilage it is called endochondral ossification. Describe this method of bone formation.

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/Users/wenhuchen/Documents/Crawler/Biology/F19-0478.htm

Solution:

Bones at the base of the skull, in the vertebral column, the pelvis and the limbs are called cartilage bones because they originate from cartilage. This cartilage, present in the infant is replaced with bone in later years by means of a process called endo-chondral ossification. This can best be studied in one of the long bones of an extremity. We first start with a cartilaginous shaft. This shaft begins to ossify, or harden into bone, around its midportion, due to the deposition of calcium by the cartilage cells (chondrocytes) . At the same time, blood vessels from the surrounding layer of connective tissue grow into the diaphysis. The calcified cartilage cells then die and are replaced by cells called osteoblasts, which form the bone matrix. The thinwalled blood vessels branch and grow toward either end of the cartilage model, forming capillary loops that extend into the blind ends of the cavities in the calcified cartilage. Cells are brought into the interior of the cartilage by these vessels. These cells later form bone marrow or bone matrix. In the continuing growth in length, the cartilage cells in the epiphyses become arranged in longitudinal columns. The epiphyses are, like the diaphyses before them, invaded by blood vessels and begin undergoing ossification. The expansion of these centers of ossifi-cation gradually replaces all of the epiphyseal cartilage except that which persists as the articular cartilage, and a transverse disk of longitudinal columns of cartilage between the original area of ossification and the epiphy-seal area of ossification, called the epiphyseal plate. The epiphyseal plate contains the cartilage columns whose zone of proliferation is responsible for all subsequent growth in length in long bones. Under normal conditions, the rate of multiplication of cartilage cells in this zone is in balance with their rate of replacement by bone. The epiphyseal plate, therefore, retains approximately the same thickness. Growth in length is the result of the cartilage cells continually growing away from the shaft and being replaced by bone as they recede. The net result is an increase in the length of the shaft. At the end of the growing period, proliferation of cartilage cells slows and finally ceases. The remaining cartilage becomes converted to bone and it is at this point that no further growth in length can occur. The growth in diameter of bone does not depend upon the calcification of cartilage but rather is the result of deposition of new bone by the periosteum.

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Question:

Of the following pairs, which member should exhibit the largest dipole moment. Use the data from the accompanying table, (a) H-O and H-N; (b) H-F and H-Br; (c) C-O and C-S.

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Solution:

A dipole consists of a positive and a negative charge separated by some distance. Quantitatively, a dipole is described by giving its dipole moment, which is equal to the charge times the distance between the positive and negative centers. The polarity of a bond is measured as the magnitude of the moment of the dipole. Thus, to find which member has the higher dipole moment, determine which bond has the greatest polarity. The polarity of the bond is indicated by the difference in electronegativities of the atoms (i.e., the difference in their tendency to attract shared electrons in a chemical covalent bond). The greater the difference in electronegativities, the greater the bond polarity, giving it the greater dipole moment. From the table, note that in part (a) the difference in electronegativities of H-O is 3.5 - 2.1 or 1.4. For H-N, the difference is 3.0 - 2.1 or 0.9. Therefore, H-O has the largest dipole moment. Using similar calculations, one can determine the bonds with the larger dipole moments in parts b and c. (c) H-F and H-Br For H-F: 4.0 - 2.1 = 1.9 For H-Br: 2.8 - 2.1 = 0.7. Thus, H-F has the larger dipole moment. (c) C-O and C-S For C-O: 3.5 - 2.5 = 1.0 For C-S; 2.5 - 2.5 = 0. Thus, C-O has the larger dipole moment here.

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Question:

Occasionally a boy is born withundescendedtestes (cryptorchidism), and he will be sterile unless they are surgically caused to descend into the scrotum. Explain the reason for this.

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Solution:

Human sperm cannot develop at the high temperature found within the body cavity. They can only develop in the testes which are at a slightly lower temperature since they are suspended from the body in the scrotum. When the weather is warm, and the body is radiating considerable heat, the skin of the scrotum loosens so that the testes hang away from the body. When the weather is cold, the skin of the testes contracts and draws the testes closer to the body, thus preventing excess loss of heat. During its de-velopment in the mother, the testes remain inside the body cavity of the male fetus. At birth, the testes descend into the scrotum. If there is a deficiency of male hormone or if the canal is too narrow, the testes will not descend. The result of this would be sterility, the sperm being unable to develop at the higher tempe-rature in the body cavity.

Question:

The volume of oil contained in a certain hydraulic press is 5 ft^3. Find the decrease in volume of the oil when sub-jected to a pressure of 2000 lb/in^2. The compressibility of the oil is 20 × 10^-6 per atm.

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Solution:

The volume decreases by 20 parts per million for a pressure increase of one atm. Since 2000 lb/in^2 = 136 atm., the volume decrease is 136 × 20 = 2720 parts per million. Since the original volume is 5 ft^3, the actual decrease is [(2720)/(1,000,000)] × 5 ft^3 = 0.0136 ft^3 = 23.5 in^3. or, the change in volume of the oil is proportional to the original volume and the pressure exerted on the oil. ∆V = - k V_0p The constant of proportionality, k, equals the compressibility of the oil. Therefore, the change in volume can be found using the above equation. ∆V = - kV_0p = - 20 × 10^-6 atm^-1 × 5 ft^3 × 136atm = - 0.0136 ft^3.

Question:

What advantages domulticellularorganisms haveover single-celledorganisms?

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Solution:

Single-celled organisms represent one of the great success stories ofevolution. They probably comprise more than half the total mass of livingorganisms, and have successfully colonized even the harshest environ-ments. Biochemically, many unicellular organisms, such as bacteria, are far more versatile than man being able to synthesize virtually everythingthey need from a few simple nutrients. With the development of eukaryotic cells, certain evolutionary breakthroughsoccurred. Eukaryotic cells are not only capable of being muchlarger than prokaryotes, they have the ability to aggregate into multicellularfunctional units. The cells ofmulticellularorganisms are specializedfor a variety of functions, and interact in a way that make the organismmore than the sum of its parts. Multicellularorganisms employ a high order of complexity. The constituentcells are specialized for the division of labor. The cells that makeup the body of a man are not all alike; each is specialized to carry outcertain functions. For example, red blood cells carry oxygen from the lungsto the various parts of the body while nerve cells are involved in the transmissionof impulses. This specialization allows each cell to function moreefficiently at its own task, and also allows the organism as a whole to functionmore efficiently. Also, injury or death to a portion of the organism doesnot nec-essarily inhibit the functioning and survival of the individual as awhole. Multicellularorganisms are better adapted to sur-vive in environmentsthat are totally inaccessible to unicellular forms. This is most strikingin the adapta-tion ofmulticellularorganisms to land. Whereasuni- cellularanimals and plants, such as theprotozoansand the one-celled algae, survive primarily in a watery en-vironment, the higher,multicellular organsisms, such as the mammals and angiosperms, are predominantly land-dwellers.Mutlicellularityalso carries the potential for diversity. Millions of different shapes, specialized organ systems, and patterns of behaviorare found inmulti-cellularorgansisms. This diversity greatly increasesthe kinds of environments that organisms are able to exploit.

Question:

How many kilograms of O_2 are contained in a tank whose volume is 2 ft^3 when the gauge pressure is 2000 lb/in^2 and the temperature is 27\textdegreeC?

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Solution:

Assume the ideal gas laws to hold. The molecular weight of O_2 is 32 gm/mole. The volume V is V = 2 ft^3 = 2 ft^3 × 28.3litres/ft^3 = 56.6 liters. Gauge pressure is the pressure above air pressure. Air pressure is 14.7 lb/in^2. Therefore P_abs = 2000 lb/in^2 + 14.7 lb/in^2 = 2015 lb/in^2 = [2015 lb/in^2]/[14.7atm/(lb/in^2)] = 137atm The temperature must be expressed in Kelvin degrees in order for us to be able to use the ideal gas laws. T = t + 273 = 300\textdegreeK. Hence the number of moles of gas in the tank is then n =P_absV/RT = (137atm× 56.6 liters)/(0.082liter\bulletatm/mole\bullet\textdegreeK × 300\textdegreeK) = 315 moles. The mass of this amount of gas is m = (315 moles)(32 gm/mole) = 10,100gm = 10.1kgm Note that we had to convert P and V to the appropriate units in order for them to be consistent with the units of the ideal gas law.

Question:

What is a sigma bond? What is a pi bond? what are their basic differences?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E18-0657.htm

Solution:

A molecular orbital that is symmetrical around the line passing through two nuclei is called a sigma (\sigma) orbital. When the electron density \sigma in this orbital is con-centrated in the bonding region between the two nuclei, the bond is called a sigma bond. The covalent bonds in H_2 and HF are sigma bonds. In the formation of the bonding orbital between two fluorine atoms, the 2p orbitals overlap in a head-to-head fashion to form a sigma bond. However, there is a second way in which half-filled p orbitals of two different atoms may overlap to form a bonding orbital. If the two p orbitals are situated perpendicular to the line passing through the two nuclei, then the lobes of p orbitals will overlap intensively sideways to form an electron cloud that lies above and below the two nuclei. The bond resulting from this sideways or lateral overlap is called a pi (\pi) bond; the bonding orbital is called a pi orbital. It differs from a sigma orbital in that it is not symmetrical about a line joining the two nuclei. Pi bonds are present in molecules having two atoms connected by a double or triple bond. The sigma bond has greater orbital overlap and is usually the stronger bond? a pi bond, with less overlap, is generally weaker.

Question:

Explain what you understand by the DECLARE statement andlist how it is used for Numeric and Non numeric Variables.

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Solution:

The DECLARE statement, abbreviated as DCL, in PL/I allows for properstorage allocation of programmers' variable. It lets the compiler knowin what form the variable will be stored by declaring its name followedby a list of its attri-butes . Numerical variables: PL/I has a wide variety of DCL state-ment forms for declaringnumeric variables. a) Fixed Decimal Point variables: -Thegeneral form is as follows: DECLARE name scale attribute base attribute (precision attribute); For example, DECLARE Y FIXED DECIMAL (6, 3); The above could be followed by a statement as follows: Y = 12 \textbullet 4; Thus, the stored result will be equivalent to 012\textbullet400. Zeros are added on both sides of 12\bullet4 because Y was de-clared to have6 digits in all, but with 3 digits after the decimal point. Similarly, if the variablehad a lesser to-tal number of digits declared, compared to the precisionand the value of the variable, a truncation of the number takes place. This truncation must be regarded as an error. As an example DECLARE A FIXED DECIMAL (5, 2); GET DATA (A); PUT DATA (A). If A = 5218\textbullet79, in view of the precision of (5, 2) declared above, A in storage = 218\textbullet79. Hence, the PUT DATA statement will lead to a print out of A = 218\textbullet79. A truncation of the left-most bit exceeding the precision attribute, has taken place. Similarly if A = 6179, then,Ain storage will be 179\textbullet00. In this case, thePUT DATA (A); statement will lead to a print-out of A=179\textbullet00; (Totally thereshould be five digits, with two digits after the decimal point.). b) Float decimal point variables: -Thebasic declara-tion reads as follows: DECLARE name FLOAT DECIMAL (W) ; The precision is specified by W, i.e., the total number of digits to be accomodatedin the floating point number. For example, DECLARE YFLOATDECIMAL (8); The above could be followed up with a statement as follows: Y = 112\textbullet4357; Hence the stored result will be as follows: 0\textbullet11243570E + 3. If the value of the variable is specified using more digits than the declarationprovides PL/I will truncate, just as for fixed point variables, as illustratedbelow: For example DCL A FLOAT DECIMAL (5); If A = 317\textbullet04 at input, then A = 3\textbullet1704E + 02 at output. If A = 0\textbullet0278 at input, then A = 2\textbullet7800E - 02 at output. If A = 8241\textbullet356 at input, then A = 8\textbullet2413E + 03 at output. c) Similar to declarations for fixed and floating point decimal variables, there are declarations for Binary varia-bles. For example, DECLARE name FIXED BINARY (w, d); e.g., DCL C BINARY FIXED (8, 3); PL/I requires a 'B' immediately following a binary con-stant, to distinguishit from decimal constants, because otherwise, 101\textbullet10 can be a decimalconstant as well as a bi-nary constant. Now consider the following: DCL C BINARY FIXED (8, 3); The above willaccomodatea value of C as follows, for ex-ample, C = 10011\textbullet000B;( =19 decimal). But, if B is not present, the PL/I compiler takes it to be 10011\textbullet000 decimal andwill then convert it to binary accord-ing to normal practice, producing a value10011100011011. (As can be verified by continuously dividing 10011 by 2, thus converting 10011 to a binary form).For the floating point representationof BINARY numbers we can have a declara-tion of the type illustratedbelow: DECLARE name FLOAT BINARY (W); d) Non-Numeric Variables: -Theallocation of stor-age for non- numericvariables must always be explicit. i) CHARACTER STRINGS: -Forcharacter strings, the declaration isas follows: DECLARE name CHARACTER (length); 'length' is a decimal integer specifying the number of char-acters to be accomodated. Character strings have to be shown between single quotationmarks for instance DECLARE X CHARACTER (5) X = 'PARIS' As a result of the above declaration, the compiler saves5 storage locationsfor X, & the five character 'PARIS' will be placed in X. When a stringlength differs from the stor-age area made available by the declaration, PL/I either pads or truncates the string, as necessary. For example, consid-er the following Declare statement: DECLARE B CHARACTER (7); Now, if B='VANbGOGH', then, the stored value for B =VANbGOGIf B = 'KUGEL', then, the stored value for B =KUGELbb. ii) BIT STRINGS: -Whena variable is declared with a statement as follows, viz., DECLARE name BIT (length); then, that variable is treated as a bit string, consisting of only ones and zeros. Bit strings are handled in the same way as character strings. Multiple Declarations: -Itis not necessary to allo-cate storage for- eachvariable in separate Declare statements. For example, consider the followingfour separate declare statements: DCL A FLOAT (6); DCL B FIXED BINARY (18); DCL C CHARACTER (12); DCL X FIXED (7,2); The above can be combined together as illustrated below: DCL AFLOAT(6), B FIXED(18), C CHARACTER(12), XFIXED(7,2); Factored Declarations: -Whenstorage is to be allocated for severalvariables that have a common set of attributes, it is possible to condensethem, as shown below: DCL AFIXED(5, 2), B CHAR(5), C FIXED(5, 2), D FLOAT(7); can be writtenas follows: DCL (A, C)FIXED(5, 2), B CHAR(5), D FLOAT(7); In the above, theFIXED(5, 2) attribute of both A and C has been factored out, leading to a condensed form of the de-clare statement. Note that no more factoring is possible now in the above statement.

Question:

How are the legs of a horse adapted for running? Compare the method of walking of a bear, a cat and a deer.

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/Users/wenhuchen/Documents/Crawler/Biology/F19-0488.htm

Solution:

Animals differ with respect to which part of the foot they put on the ground when walking and running. Men and bears walk on the entire sole of the foot. This method of locomotion is known asplatingrade. Animals such as dogs and cats, to increase their effective limb length and thus their running speed, have become adapted to running on their digits, or fingers. This type of locomotion is known asdigitigrade. Speed is increased still further in the horses, deer, and cattle, which walk on their hoofs, or nails. In these animals the lower limb bones are lengthened, raising the wrist and ankle farther off the ground. This type of locomotion is known asunguligrade. In the case of the horse, the leg is supported by one digit, terminating in a hoof, whereas in cattle, two digits are used for support, terminating in two hoofs. Those ungulates, such as the horse, which walk on one digit are known as thePerisso-dactyla. Those walking on two digits are known as theArtiodactyla.

Question:

Write a CSMP111 program for the following inventory control system: ẏ = u - V ẋ = V - S V = (Y/T_1) u = S + K(I - X) where X - Current inventory level Y - Outstanding level of orders placed with the suppliers u - Rate of ordering from suppliers V - Delivery rate from suppliers S - Sales rate I - Planned inventory level T_1- Average delivery time (= 3 days) Let the initial conditions be I= X = 0 Y = 12 S = 4 items/day.

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Solution:

The control of inventory requires a feedback system. A retailer tries to keep inventory at a level where he maintains what he views as a reasonable balance between the cost of holding goods in inventory and the penalty of losing sales if the inventory should become empty. A simple model of the inventory control system would need two levels and three rates defined as follows: X - Current inventory level Y - Outstanding level of orders placed with the supplier u - Rate of ordering from supplier V - Rate of delivery by supplier S - Rate of sales Also, assume that planned inventory level and average lead timeI andT_1respectively, are constant. A systems diagram for the above levels diagram for the above levels is: The rate of change of the level of orders is Ẏ = u - V while the time derivative of X is Ẋ - V - S. The rate of delivery is inversely proportional to the outstanding level of orders placed with the suppliers. The order rate is u = S + K(I - X) where K is the ordering constant. The systems diagram acts like a flowchart for the CSMPIII program. Assume that time is measured in days and let the initial conditions be, at t = 0: I= X = 20 Y= 12 S = 4 items/day The program looks as follows: TITLEINVENTORY CONTROL SYSTEM \textasteriskcentered PARAMK=(1.0, 0.5, 0.25, 0.125, 0.0625) \textasteriskcentered S=4.0+2.0\textasteriskcenteredSTEP(4.0) V=Y/T_1 u=S+K\textasteriskcentered(I-X) YDOT=U-V Y=INTGRL(YO,YDOT) XDOT=V-S X=INTGRL(XO,XDOT) \textasteriskcentered CONST I=20.0, YO=12.0, XO=20.0, T1=3. 0 \textasteriskcentered TIMER DELT=0.1 FINTIM=50.0 PROEL=1.0 PRINT u, v, X, Y END STOP

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Question:

Write the micro-operations for the execute cycles of the Memory reference instructions of Figure 2 using these registers and control signals. M - memory PC - program counter MAR - memory address register MBR - memory buffer register OPR - op-code register F, R - cycle identification registers E - carry bit register The cycle code is shown in Figure 1. Cycle ID Registers FR Cyclecode Cycledescription 00 C_0 Fetch cycle 01 C_1 Indirect cycle 10 C_2 Execute cycle 11 C_3 Interrupt cycle Fig. 1 Each cycle consists of four steps t_0, t_1, t_2, t_3. Instruction Symbol OP-Code Control Signal Effective Address Memory Word Symbolic Designation AND 000 q_0 M M AC\leftarrow[AC]\wedgeM \wedge ADD 001 q_1 m M EAC\leftarrow[AC]+M LDA 010 q_2 m M AC\leftarrow[M] STA 011 q_3 m M M\leftarrow[AC] BUN 100 q_4 m - PC\leftarrowm BSA 101 q_5 m M M\leftarrow[PC] PC+m+1 ISZ 110 q_6 m M M\leftarrowM+1, if M+1=0, then PC\leftarrow[PC]+1 Fig.2

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Solution:

At the beginning of the execute cycle the registers contain the following information: [PC] = The address of the instruction about to be exe-cuted; incremented by one. [MBR]_4-15 = The address of the operand (effective address). [OPR] = The op-code part of the instruction format (this was decoded into one of eight control signals q_0 - q_7) [F]= 1 }decoded to excite c_2 [R] = 0 } A step is specified if the control signal expression to the left of the colon is true (logic 1). AND:This instruction logically "ands" the contents of the accumulator and the operand. The result is stored back in the accumulator. Note:[] means "contents of"; < > means "addressed by." q_0c_2t_0 :MAR \leftarrow [MBR]_4-15\midFetch operand q_0c_2t_1 :MBR \leftarrow [M]\mid q_0c_2t_2 :AC \leftarrow [AC] \wedge [MBR]AND with AC, store result in AC q_0c_2t_3 :F\leftarrow0Go to Fetch cycle (c_0 is excited) ADD:This instruction adds the contents of the AC and the operand. The result is stored in the AC. If there is a carry, it is stored in E. q_1c_2t_0 :MAR \leftarrow [MBR]_4-15\midFetch operand q_1c_2t_1 :MBR \leftarrow [M]\mid q_1c_2t_2 :EAC \leftarrow [AC] + [MBR]Add to AC, store result in AC and E q_1c_2t_3 :F\leftarrow0Go to Fetch cycle LDA:This instruction loads the operand into the AC. q_2c_2t_0 :MAR \leftarrow [MBR]_4-15\midFetch operand, clear AC q_2c_2t_1 :MBR \leftarrow [M]; AC\leftarrow0\mid q_2c_2t_2 :AC \leftarrow [AC] + [MBR]Add to AC q_2c_2t_3 :F\leftarrow0Go to Fetch cycle STA:This instruction stores the contents of the AC into the memory location specified by [MBR]_4-15. q_3c_2t_0 :MAR \leftarrow [MBR]_4-15Store address in MAR q_3c_2t_1 :MBR \leftarrow [AC]Store [AC] in MBR q_3c_2t_2 :M\leftarrow[MBR]Load into memory q_3c_2t_3 :F\leftarrow0Go to Fetch cycle BUN(branch unconditionally):This instruction is similar to the JUMP or GO TO statement in FORTRAN. The address of the next instruction is specified by the address portion of the MBR. The execute cycle is: q_4c_2t_0 :PC \leftarrow [MBR]_4-15Transfer next address to PC q_4c_2t_1 :Do nothing q_4c_2t_2 :Do nothing q_4c_2t_3 :F\leftarrow0Go to Fetch cycle BSA(branch and save return address):This instruction stores the contents of the PC in the memory location specified by the effective address and then control branches to the next higher location of the effective address. q_5c_2t_0 :MAR \leftarrow [MBR]_4-15\mid MBR_4-15 \leftarrow [PC]\midLoad effective address into PC PC\leftarrow[MBR]4-15\mid q_5c_2t_1 :[M]\leftarrowMBRLoad old PC into effective address q_5c_2t_2 :PC\leftarrow[PC]+1Increment PC q_5c_2t_3 :F\leftarrow0Go to Fetch cycle The BUN and BSA instructions are very useful when branch-ing to and from subroutines. ISZ(increment and skip-if-zero):This instruction incre-ments the operand and then tests to see if the operand is zero. If the operand is not zero then the program flows normally. If the operand is zero then the next' instruction is skipped and control resumes at the instruction after the skipped instruction. q_6c_2t_0 :MAR \leftarrow [MBR]_4-15\midFetch operand q_6c_2t_1 :MBR \leftarrow [M]\mid q_6c_2t_2 :MAR\leftarrow[MBR]+1Increment operand q_6c_2t_3 :M\leftarrow[MAR]\midStore new operand; if [MBR] = 0 then PC=PC+1\midIncrement PC if \midoperand is zero; F\leftarrow0\midGo to Fetch cycle.

Question:

A typical bacterial cell has a cell wall and plasma membrane. If the cell wall is removed by treatment withlysozyme, an enzyme that selectively dissolves the cell wall material, will the bacteria live? Explain. What is the biochemicaldifference between the cell walls of bacteria, and plants?

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Solution:

The main function of the bacterial cell wall is to provide a rigid framework or casing which supports and protects the bacterial cell from osmotic disruption. Most bacteria live in a medium which is hypotonic relative to the bacterial protoplasm: i.e., the bacterial protoplasm is more concentrated than the medium, and water tends to enter the cell. If intact, the cell wall provides a rigid casing, which prevents the cell from swelling and bursting. If the cell wall were destroyed, water would enter the cell and cause osmoticlysis(bursting). If the bacterium is placed in a medium which has the same osmotic pressure as the bacterial cell contents, osmotic disruption would not occur when the cell wall was dissolved. A bacterium devoid of its cell wall is called a protoplast. Pro-toplasts can live in an isotonic medium, (medium having equal osmotic pressure as the bacterial protoplast).If the bacterium was placed in a hypertonic medium, waterwould leave the cell andplasmolysis(shrinkage) would ensue. The cell wall varies in thickness from 100 to 250 \AA (an angstrom, \AA, is 10^\rule{1em}{1pt}10 meter) and may account for as much as 40% of the dry weight ofthe cell. While the cell wall ineucaryotesis composed of cellulose, in bacteria , the cell wall is composed ofinsoluablepep-tidoglycan. Peptidoglycan consists of sugars (N-aceylglucosamineand N- acetylmuramic acid) and amino acids, includingdiaminopinelicacid, an amino acid unique to bacteria. In gram-negative bacteria, the peptidoglycan constitutes a much smaller fraction of the wall component than it does in gram-positive bacteria. The higher lipid content in the cell walls of gram-negative bacteria accounts for the differences in gram- staining . The plasma membrane, a thin covering immediately beneath the cell wall, is too fragile to provide the support needed by the cell. Instead, this semipermeablemembrane controls the passage of nutrients and waste products into and out of the cell.

Question:

A ship is being towed by a tug by means of a steel wire. If the drag on the ship is equivalent to 2 × 10^6 lb, and if the breaking strain of the wire is 0.025, what is the smallest permissible diameter of wire that may be used?

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Solution:

If the ship is being towed, it eventually settles down at a steady speed so that the force being exerted by the tow wire is equal and opposite to the drag on the ship. Now, the formula for Young's modulus is Y = (stress)/(strain) = (F_n/A)/(∆1/1_0) where the symbols have their usual significance. Thus A = F_n1_0/Y∆1, where F_n the normal force on the wire, is equal to the force exerted by the wire on the ship, which balances the drag force. (See the fig.) Further, the strain ∆1/1_0 must be less than 0.025, the strain at which the wire breaks. A table of Young's modulus for different materials shows that for steel, Y = 29 × 10^6 lb\bulletin^-2 Hence A \geq [(2 × 10^6 lb)/(29 × 10^6lb\bulletin^-2 × 0.025)] The cross sectional area of the wire (A) is \piR^2, where R is the radius of the wire. Since R = D/2 where D is the diameter of the wire, then A = \pi[(D^2)/4] and [(\piD^2)/4] \geq 80/29 in^2. \therefore D \geq \surd(320/29\pi) in. = 1.87 in., Hence, 1.87 in. is the smallest possible diameter for the tow wire.

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Question:

Write a PL/I procedure that reads an integer N > 1 and calculatesthe sequence Y =[ ^N\sum_i= 1{i(N -i)^2} ] / N The program should output N and the corresponding Y on eachline.

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Solution:

We will set up a DO loop to calculate the terms as in-dicated by the summationsymbol, (\sum). Program control passes to the end of the program whenthe value of N is less than or equal to zero. In this example, we will input N = 2, N = 10, and N = 25. SEQ: PROC OPTIONS (MAIN); DCL Y FIXED (10,3); LOOP: GET DATA (N); / \textasteriskcentered TEST FOR END OF DATA \textasteriskcentered / IF N < = 0 THEN GO TO FINISH; / \textasteriskcentered INITIALIZE Y AND COMPUTE THE SUM \textasteriskcentered / Y = 0; SUMLOOP: DO J = 1 TO N; Y = Y + J \textasteriskcentered (N - J) \textasteriskcentered (N - J); END SUMLOOP; Y = Y/N; / \textasteriskcentered PRINT ANSWERS \textasteriskcentered / PUT SKIP DATA (N,Y) ; GO TO LOOP; FINISH: END SEQ; Sample input: N = 2; N = 10; N = 25; N = 0; Sample output: N =2Y =0.500; N =10Y =82.500; N =25Y =1300.000;

Question:

A 1,000-horsepower steam engine produces about 750,000 W of mechanical power. If this engine operates a generator with an efficiency of 95 per cent, what is the electrical power produced ?

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/Users/wenhuchen/Documents/Crawler/Physics/D24-0790.htm

Solution:

Efficiency \epsilon is defined by \epsilon = (P_output/P_input)×100% The steam engine delivers 750,000 W to the generator. The generator operates at an efficiency of 95% whereP_outputis its Power output. For the generator , {(P_output)/(7.5 × 10^5 W)} = 95% = 9.5 × 10^\rule{1em}{1pt}1 The electric power delivered in this energy-transformation process is, P = (7.5 × 10^5 W) (9.5 × 10^\rule{1em}{1pt}1) = 7.125 × 10^5 W The 5 per cent difference (37,500 W) appears as heat that must be dissipated by either passing air through cooling fins or circulating water.

Question:

A Celsius thermometer indicates a temperature of 36.6\textdegreeC. What would a Fahrenheit thermometer read atthat temperature ?

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/Users/wenhuchen/Documents/Crawler/Physics/D12-0439.htm

Solution:

The relationship between the Celsius and Fahrenheit scales can be derived from a knowledge of their cor-responding values at the freezing and boiling points of water. These are 0\textdegreeC and 32\textdegreeF for freezing and 100\textdegreeC and 212\textdegreeF for boiling. The temperature change between the two points is equivalent for the two scales and a temperature difference of 100 Celsius degrees equals 180 Fahrenheit degrees. Therefore one Celsius degree is (9/5) as large as one Fahrenheit degree. We can then say \textdegreeF = (9/5)\textdegreeC + B where B is a constant. To find it, substitute the values known for the freezing point of water: 32\textdegree = (9/5) × 0\textdegree + B = B. We therefore have \textdegreeF = (9/5)\textdegreeC + 32\textdegree. For a Celsius temperature of 36.6\textdegree, the Fahrenheit temperature is F = (9/5) × 36.6\textdegree + 32\textdegree = 65.9\textdegree + 32\textdegree = 97.9\textdegree.

Question:

Write a PL/I program to integrate a general function D(X) overA \leq X \leq B. Illustrate for D(X) = (1 - X)-2 .

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G21-0529.htm

Solution:

We slice the interval [a, b] into n equal parts. The size of the slices ish = (b - a)/n . The area under the curve is given ac-cording to Simpson's ruleby S = (h/3) [f (a) + 4f (a+h) +2f(a+2h) + 4f(a+3h) + 2f(a+4h)+...+ f(b)] /\textasteriskcentered SIMPSON RULE INTEGRATION \textasteriskcentered/ RPT: GET LIST (A,B,N); H = (B - A)/N; X = A; S = 0; DO I = 1 TO N/2; S = S + H/3\textasteriskcentered (D(X) + 4\textasteriskcenteredD(X+H) + D(X+2\textasteriskcenteredH)); X = X + 2\textasteriskcenteredH; END; PUT SKIP EDIT (A,B,N,S) (2F(10,4)), F(10), F(10,4)); GO TO RPT; D:PROCEDURE (X); RETURN (1/1 - X)\textasteriskcentered\textasteriskcentered2); END D;

Question:

Why did Mendel succeed in discovering the principlesof transmissions of hereditary traits, whereas others who had conducted such investigations had failed?

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/Users/wenhuchen/Documents/Crawler/Biology/F25-0640.htm

Solution:

Mendelssuccess was a combination of good experimental technique and luck. He chose the garden pea for his studies because it existed in a number of clearly defined varieties; the different phenotypes were thus evident and distinguishable. In addition, he only studied the transmission of one trait at a time, and crossed plants differing with respect to just that one trait. Previous investigators had studied entire organisms at a time, and had crossed individuals differing in many traits. Their results were a conglomeration of mixed traits and interacting factors, making it impossible to figure out what was happening. Mendel was also careful to begin with pure line plants developed through many generations of natural self - fertilization. However, luck was involved in that the traits which Mendel randomly chose were coded for by genes that were located on different chromosomes, and thus assorted independently . If he had chosen traits whose genes were linked, his ratios would not have been understandable in terms of what he knew then.

Question:

With what speed is an electron moving if its kinetic energy is 1% larger than the value that would be calculated if relativistic effects were not present? What is the value of the kinetic energy of the electron?

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/Users/wenhuchen/Documents/Crawler/Physics/D32-0947.htm

Solution:

The relativistic expression for the kinetic energy of a particle of rest mass m_0 and mass m traveling with speed v is T = E - m_0c^2= {m_0c^2 / (\surd1 - v^2c^2)}- m_0c^2 m_0c^2[{1 / \surd(1 - (v2/ c^2)} - 1 ] = m_0c^2 [ {1 - (v2/ c^2)}^-1/2 - 1] If the error involved in using the classical formula for T is only 1%, the particle is traveling at a speed such that v^2/ c2<< 1 Hence, using a Taylor expansion for {1 - (v2/ c^2)}^-1/2 = (1 + v2/2c2+ 3v^4/ 8c^4 + .....) T = m_0c^2 {(1/2) ( v2/ c^2) (3/8) ( v4/ c^4) + .......} = [(1/2)m_0v2{1 + (3/4) ( v2/ c^2)}] , when we ignore terms of higher order than those in v^2 / c^2 . But (1/2)m^0v^2is the nonrelativistic expression for the kinetic energy of the particle, and thus, for the electron in the problem, (1/2)m_0v^2 {1 + (3/4) ( v2/ c^2)}] = (1/2)m_0v2+ (1%) [(1/2)m_0v^2] = 1.01[(1/2)m_0v^2] or 1 + ( 3v2/ 4c^2) = 1.01 ( 3v2/ 4c2) = .01 v^2 = .04c^2 / 3 = 4c^2 / 300 v = 0.1155c. T = (101/100) × (1/2)m_0v2 =(101/100) × (1/2) × 9.108 × 10^-31 kg × (4 / 300) × 2.998^2 × 10^16 m^2 \bullet s^2 = 5.512 × 10^-16 J Since 1 eV = 1.602 × 10^-19 J T = (5.512 × 10-16J) / (1.602 × 10^-19 J / eV) = 3442 eV .

Question:

Write a BASIC program to calculate the hypotenuse of a right triangle. Your program should read in values of the two sides A and B, and calculate the hypotenuse C according to the Pythagorean theorem. Illustrate the results when A = 3, B = 4; A = 6, B = 8; A = B = 1.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G10-0228.htm

Solution:

This program illustrates the use of Library Functions in BASIC. It executes function SQR (Square Root) for determining the hypotenuse. The hypotenuse C is calculated from the formula C = \surd(A^2 + B^2). The program follows: 1\O REM CALCULATES THE HYPOTENUSE 2\O REM OF A RIGHT TRIANGLE 3\O PRINT "SIDE A", "SIDE B", "HYPOTENUSE C" 4\O READ A, B 5\O LET C = SQR (A _\uparrow 2 + B _\uparrow 2) 6\O PRINT A,B,C 7\O GO TO 4\O 8\O DATA 3, 4, 6, 8, 1, 1 9\O END OUTPUT: SIDE A SIDE B HYPOTENUSE C 3 4 5 6 8 1\O 1 1 1.41...

Question:

Name the major organs of the human digestive tract and explain their functions.

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/Users/wenhuchen/Documents/Crawler/Biology/F17-0422.htm

Solution:

The human digestive system begins at the oral cavity. The teeth break up food by mechanical means, in-creasing the substrates' surface area available to the action of digestive enzymes. There are four types of teeth. The chisel-shaped incisors are used for biting, while the pointed canines function in tearing, and the flattened, ridged premolars and molars are used for grinding and crushing food. In addition to tasting, the tongue mani-pulates food and forms it into a semi-spherical ball (bolus) with the aid of saliva. The salivary glands consist of three pairs of glands. The parotid glands, located in the cheek in front of the ear, produce only watery saliva which dissolves dry foods. The submaxillary and sublingual glands, located at the base of the jaw and under the tongue, produce watery and mucous saliva which coagulates food particles and also lubricates the throat for the passage of the bolus. Saliva also contains amylases which break down starches. The tongue pushes the bolus into the pharynx which is the cavity where the esophagus and trachea (windpipe) meet (see Fig. 2). The larynx is raised against the epiglottis and the glottis is closed, preventing food from passing into the trachea. The act of swallowing initiates the movement of food down through a tube con-necting the mouth to the stomach called the esophagus. Once inside the esophagus, the food is moved by involuntary peristaltic waves towards the stomach. At the junc-tion of the stomach and the esophagus, is a special ring of muscles called the lower esophageal sphincter. These muscles are normally contracted, but when food from the esophagus reaches the sphincter, it opens reflexively and allows food to enter the stomach. The stomach is a thick muscular sac positioned on the left side of the body just beneath the ribs. The upper region of the stomach, closest to the heart, is called the cardiac region. Below that is the crescent part of the sac called the fundus. The pyloric region is tubular and connects the stomach to the small intestines. The wall of the stomach is made up of three thick layers of muscle. One layer is composed of longitudinal, one of circular, and the other of oblique (diagonal) fibers. The powerful contractions of these muscles break up the food, mix it with gastric juice and move it down the tract. Gastric juice is a mixture of hydrochloric acid and enzymes that further digest the food. Gastric juice and mucus are se-creted by the small gastric glands in the lining of the stomach. The mucus helps protect the stomach from its own digestive enzymes and acid. The partially digested food, called chyme, is pushed through the pyloric sphincter in-to the small intestine. The pyloric sphincter is similar in structure and function to the lower esophageal sphincter. The first part of the small intestine, called the duodenum, is held in a fixed position. The rest of the intestine is held loosely in place by a thin membrane called mesentery which is attached to the back of the body wall. In the duodenum, bile from the liver that has been stored in the gallbladder is mixed with pancreatic juice from the pancreas. The secretions of the pancreas and glandular cells of the intestinal tract contain enzymes that finish digesting the food. As digestion continues in the lower small intestine, muscular contractions mix the food and move it along. Small finger-like protrusions, called villi, line the small intestine facing the lumen. They greatly increase the intestinal surface area and it is through the villi that most of the nutrient absorption takes place. The small intestine joins the large intestine (colon) at the cecum. The cecum is a blind sac that has the appendix protruding from one side. Neither the appendix nor the cecum are functional. At the junction of the cecum and the lower small intestine (ileum.) is a sphincter called the ileocecal valve. Undigested food passes through this valve into the large intestine. The large intestine has the function of removing water from the unabsorbed material. At times there is an excretion of certain calcium and iron salts when their concentrations in the blood are too high. Large numbers of bacteria exist in the colon. Their function in man is not fully understood, but some can synthesize vitamin K which is of great importance to man's blood clotting mechanism. The last section of the colon stores feces until it is excreted through the anal sphincter.

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Question:

Discuss the adaptations for land life evident in the bryophytes.

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/Users/wenhuchen/Documents/Crawler/Biology/F07-0176.htm

Solution:

The bryophytes are considered one of the early invaders of the land. In order to survive on land, a plant must have certain structures which will enable it to exploit a terrestrial environment for food, water, gases and to afford it protection against environ-mental hazards. Although the bryophytes are far from being truly terrestrial and retain a strong dependency on a moist surrounding, they have accumulated important adaptations enabling them to live successfully on land. Because land plants are removed from a water environment, they face the potential danger of dehydration due to evaporation. To reduce water loss, most bryophytes have an epidermis. In the mosses, it is thickened and waxy, forming a cuticle. Unlike aquatic plants, which obtain and excrete gases dissolved in water, land plants require an efficient means to exchange gases with the atmosphere. The epidermis of the bryophytes is provided with numerous pores for the diffusion of carbon dioxide and oxygen. Diffusion through pores is a much faster and more efficient process for gaseous exchange than simple diffusion through membranes. Besides gases, land plants need to obtain water, which may be a limiting factor in a terrestrial environ-ment. Most bryophytes absorb water and minerals directly and rapidly through their leaves and plant axis, which may have a central strand of thin-walled conducting cells. Generally, the plant is attached to the substrate by means of elongated single cells or filaments of cells called rhizoids. In those mosses having a cuticle on their leaves, the rhizoids may function to some extent in water absorption. In most bryophytes, however, the rhizoids are not true roots and serve only in anchoring the plant. In addition to structures, the bryophytes have evolved a life cycle in which the developing zygote is protected within the female gametophyte. Here the zygote obtains food and water from the surrounding gametophytic tissue and is protected from drought and other physical hazards present in a terrestrial environment. The bryophyte sperm cell is flagellated and requires a moist medium for its transport. Fertilization, however, can occur after a rain or in heavy dew. The sperm cell swims, in a film of moisture, to the female gametophyte (archegonia) wheregametic union occurs. Because the bryophytes have acquired rhizoids, cutinized epidermis, a porous surface, and a reproductive style in which the embryo is protected within the female gametophyte, they are able to succeed in a terrestrial environment.

Question:

Define the following symbols as they are used in flowcharting:

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G08-0166.htm

Solution:

These symbols have very specific meanings for flowcharting procedures. The following explanations come from the American National Standards Institute: Input or output using a line printer or a terminal Two or more actions may follow from this point, based upon the response to the question raised Input or output using the standard 80- column card Commentary on a particular section of the algorithm, used to explain ambiguous actions Usually with a number inside the circle, it connects flowlines when going to or coming from another spot on the same page or another page

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Question:

In peas, yellow color is dominant to green. What will be the colors of the offspring of the following crosses? a) homozygous yellow × green b) heterozygous yellow × green c) heterozygous yellow × homozygous yellow d) heteroygous yellow × heterozygous yellow

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/Users/wenhuchen/Documents/Crawler/Biology/F25-0649.htm

Solution:

This problem offers practice in genetic crosses. First we must determine the alphabetic repre-sentation of the genes involved. Let Y be the gene for yellow color in peas and y be the gene for green color in peas. For each cross, we write down the genotypes of the parents, the possible types of gametes, and de-termine the F_1 offspring. If it helps we may use the Punnett Square to obtain the first generation. In the crosses then: All the progeny of F_1 are yellow because Y is dominant. Half the progeny is yellow (Yy) and half is green (yy). All the progeny will be phenotypically alike (yellow). Genotypically, half will be homozygous and half will be heterozygous. A ratio is obtained of 3 yellow progeny (1 YY + 2 Yy) to 1 green (yy).

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Question:

If the gold foil in a Rutherford type experiment is 1/10,000 in. thick, what is a probable minimum number of gold atoms an alpha particle passed through before hitting the fluorescent screen? 1 in = 2.54 cm, 1 cm = 10^8 \AA. The radius of a gold atom is 1.5 \AA.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E18-0665.htm

Solution:

In Rutherford's experiment, a beam of alpha particles is aimed at a gold foil of given thickness. Most of the alpha particles pass through undeflected. The deflections that occur suggest that the alpha particles collided with a solid body. This body is the gold nucleus. This experiment elucidated some important properties of the nucleus. One of these properties is the positive charge of the nucleus. If one knows the thickness of the foil and the atomic radius of the particles that compose the foil, one can calculate the number of particles which make up the width of the foil. If one assumes cubic packing in the foil, the sum of the lengths of the diameters of the gold atoms will be equal to the thickness of the foil. Proceed as follows: The radius, and therefore the diameter, of the gold atom is given in terms of angstroms. The thickness of the foil must be converted to the same units. The thickness = (10^-4 in) (2.54 cm/in) (1 × 10^8 \AA/cm) = 2.54 × 10^4 \AA. Diameter of gold nucleus = 2 × Radius = 2 × 1.5 \AA= 3 \AA. The minimum number of gold atoms passed by alpha particles is equal to [(2.54 × 10^4 \AA)/(3 \AA)] = 8.466 × 10^3

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Question:

It is desired to produce penetrating x-rays of wavelengths about 0.20 A. What is the minimum voltage at which the x- ray tube can be operated?

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/Users/wenhuchen/Documents/Crawler/Physics/D35-1034.htm

Solution:

An x-ray tube accelerates electrons through a potential difference. In order to see how this occurs, let us refer to the figure. The net force on an electron emitted from plate A is, by definition of electric field F^\ding{217} = eE^\ding{217} where e is the signed charge of an electron, and E^\ding{217} is the electric field intensity. Hence, the work done by this force (since it is the resultant force on the electron) is equal to the change in kinetic energy of the electron (1/2)m_e v2_b -(1/2)m_e v2_a = ^b\int_a F^\ding{217} \bullet dr^\ding{217}. Assuming that the initial velocity, v_a, of the electron is zero, (1/2)m_e v2_b = ^b\int_a eE^\ding{217} \bullet dr^\ding{217} (1/2)m_e v2_b = e ^b\int_a E^\ding{217} \bullet dr^\ding{217}. Replacing the signed number e by - \verte\vert (since e < 0), we may write (1/2)m_e v2_b = - \verte\vert ^b\int_a E^\ding{217} \bullet dr^\ding{217} = \verte\vert(V_b - V_a) Hence (1/2)m_e v2_b = \verte\vert (V_b - V_a). Assuming that V_a = 0 volts (1/2)m_e v2_b = \verte\vert V_b . If all the energy of the electron when it reaches point b is changed into the energy of an x-ray photon, we obtain (1/2)m_e v2_b = \verte\vert V_b = hѵ HenceV_bmin = hѵ/\verte\vert . V_b is the minimum voltage needed to produce x-rays of fre-quency ѵ. (Only those electrons which give up all their energy to the x-ray photons will produce x-rays). V_bmin = hѵ/\verte\vert Butѵ = c/\lambda where c = speed of light and \lambda is the x-ray wavelength. V_bmin = (hc)/(\verte\vert\lambda) V_bmin = (6.63 × 10^-34 joule-sec3.0 × 10^8 m/sec)/(1.60 × 10^-19 coul0.20 × 10^-10m) = 62,000 volts

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Question:

What is the possible function of nonsense codons in the genes?

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/Users/wenhuchen/Documents/Crawler/Biology/F24-0619.htm

Solution:

Nonsense codons are those which do not code. There are three such nonsense triplets for an amino acid : UAA, UAG, and UGA. They are found at the end of nucleotide sequences coding for given polypeptides, and code specifically for the termination of the chain. They signify that peptide elongation at the ribosome should stop. These codons are read by specific proteins known as release factors. These release factors are necessary to release the chain from the ribosome. Two release factors have so far been identified, each of which recognizes two codons. One is specific for UAA and UAG and the other for UAA and UGA. Nonsense codons may be employed in the following manner: the nucleotide sequence etc.- CUU AGG UAU AUA UAG GCC ACG - etc. on the mRNA might represent two genes, with the UAG triplet acting as a terminating codon between them :

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Question:

Explain the TRAILER CARD method of stopping a computer job, i.e., a program.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G13-0333.htm

Solution:

In this method, we use a certain unique number on the last card in thedata section of a program card deck. This last card is called the Trailer Card. The unique number selected is such that it is different in value, or numberof digits, etc., from all the other authen-tic data on the other precedingdata cards of the program deck. For example, if we know that allour data consists of positive numbers, we could use a negative number onthe Trail-er Card. Or, if we know that all the data is non-zero, we could usea value of zero on the Trailer Card. Then, in the body of the program weinsert a statement which tells the computer to end the job when the valuewhich we have put on the Trailer Card is encountered. The Trailer Card method is most useful in cases where different re- runsof a given program will have different amounts of data, which cannot beknown in advance. Hence, instead of modifying the program each time thedata changes, we use the same program, but take advantage of the Trailer Card to tell the computer when to end the program. We give below a program to illustrate the use of the Trailer Card method. SUM: PROCOPTIONS(MAIN); DCL(X,SUM)FIXED(4)INIT(0); LOOP:GET LIST (X) ;/\textasteriskcentered WE READ ONE DATA VALUE \textasteriskcentered/ SUM = SUM + X; IF X = 0000 THEN GO TO FINISH; /\textasteriskcentered WE READ A VALUE OF 0000 FOR X, WHICH IS ON THE TRAILER CARD, AND, AS THE TRAILER CARD IS THE LAST DATA CARD, WE GO OUT OF THE LOOP TO LABEL FINISH. HERE WE ASSUME NONE OF THE DATA NORMALLY HAS A VALUE '0000' \textasteriskcentered/ GOTO LOOP; /\textasteriskcentered IF THE TRAILER CARD IS NOT REACHED, WE REPEAT THE LOOP \textasteriskcentered/ FINISH: PUTLIST(SUM); END SUM;

Question:

A + B \rightarrow products. Consider the following reaction rate data and then determine the rate law for each case: (a) When [A] is doubled, the initial rates doubles; when [B] is doubled, the initial rate doubles, (b) When [A] doubles, initial rate doubles; when [B] doubles, the initial rate is halved, (c) Doubling [A], doubles initial rate; but doubling [B] leaves the rate unchanged.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E13-0452.htm

Solution:

The given reaction is one with two different molecules being converted to products. The rate of any such reaction is: r = rate = k [A]^x [B]^y,where k = a rate constant and [A] = concentration of A and [B] = concentration of B. x and y are the number of each A and B molecules, re-spectively, that are present in the activated complex or transition state. The determination of these exponents is the key to writing the rate law for each case. You proceed as follows: (a) You are told that when A or B is doubled, the rate is doubled. Thus, when A \rightarrow 2a,r \rightarrow 2r, which means 2r = k [2A]^x[B]^y. Now, if x = 1, then [2A]^x = [2A] and dividing by 2 you come back to r \rightarrow k[A][B]^y. In other words, r \rightarrow 2r if A \rightarrow 2A, implies rvaries directly as A; that is, x = 1. The same type of situation exists with B, so that y = 1. Thus, the rate law becomes r = k[A][B]. (b)Here, when A \rightarrow 2A, r \rightarrow2a, which means, again, A varies directly as r, and so x = 1. When B \rightarrow 2B, r \rightarrow (1/2)r , however, in this case, (1/2)r = [A][2B]^y. This can only come about if y = - 1. If y = - 1, then [2B]^-1 = 1/[2B], so that (1/2)r = [A] 1/[2B]. The halves can be cancelled to return a directionproportion. Thus , the rate law becomes r = k[A][B]. (c)Here, when [A] \rightarrow [2A], r \rightarrow 2r but when [B] \rightarrow[2B], r \rightarrow r. For A: you have, again, a direction proportion between A and r, which requires x = 1. For B: when B's concentration is doubled, nothing happens to r; it remains constant. This means r does not depend upon the [B]. For r not to depend upon [B]^y, y must equal zero, thus, r = k [A] [B]\textdegree = k [A] (1). As such, the rate law becomes r = k [A].

Question:

Figure 1 shows a flat strip of copper of width a and negligi-ble thickness carrying a current i. Find the magnetic field B at point P, at a distance R from the center of the strip, at right angles to the strip. .

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/Users/wenhuchen/Documents/Crawler/Physics/D37-1079.htm

Solution:

Subdivide the strip into long infinitesimal fila-ments of width dx, each of which may be treated as a wire carrying a current di given by i(dx/a). The field contribu-tion dB at point P in Fig. 1 is given, for the element shown, by the differential formula for a long cylindrical wire, which is dB = (\mu_0 / 2\pi)(di / r) = (\mu_0 / 2\pi) [{i(dx/a)} / (R sec \texttheta)]. Note that the vector dB is at right angles to the line marked r, Only the horizontal component of dB, dB cos \texttheta, is effec-tive, the vertical component being canceled by the contribution associated with a symmetrically located filament on the other side of the orgin. Thus B at point P is given by the (scalar) integral B = \intdB cos \texttheta = \int [{\mu_0i (dx / a)} / (2\piR sec \texttheta)] cos \texttheta = [\mu_0i / (2\piaR)] \int[(dx) / (sec^2\texttheta)] . The variables x and \texttheta are not independent, being related by x = R tan \texttheta or dx = R sec^2\texttheta d\texttheta. Bearing in mind that the limits on \texttheta are \pmtan^-1 (a/2R) and eliminating dx from this expression for B, B = [\mu_0i / (2\piaR)] \int[(R sec^2\texttheta d\texttheta) / (sec^2\texttheta)] = [\mu_0i / (2\pia)]^+tan-1(a/2R)\int^ _-tan-1(a/2R) d\texttheta = [\mu_0i / (\pia)] tan^-1(a/2R). At points far from the strip, (a/2R) is a small angle, for which tan^-1\alpha \cong \alpha. Thus, as an approximate result, B \cong [\mu_0i / (\pia)] (a/2R) = [\mu_0 / (2\pi)] (i / R). This result is expected because at distant points the strip cannot be distinguished from a cylindrical wire.

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Question:

A source emits sound waves at 1000 cycles/second (frequency f_0) and the speed of the wave with respect to the source is 1000 ft/second(v_w) rest with respect to the source (\lambda_0)? towards the listener, what frequency (f_sl) and wavelength (\lambda_sI) does he observe? listener, what frequency (f_s2) and wavelength (\lambda_s2) does he observe?

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/Users/wenhuchen/Documents/Crawler/Physics/D26-0840.htm

Solution:

Everyone has heard the drop in pitch of a passing train's whistle. The Doppler effect is the apparent change of wave frequency observed by a listener when he and the wave source are in relative motion. In this problem, we deal with sound waves and derive the change in observed frequency and wavelength for two kinds of relative motion: motion of the observer and motion of the source. For any observer, v = \lambdaf , which means that the velocity at which he observes the wave to move is equal to the product of its observed frequency and wavelength. This is no more than an application of the conventional velocity de-finition v = ∆s/ ∆t if we remember f = 1/T where T is the wave's period. Then v = \lambdaf = \lambda/T where \lambda is the observed displacement in time T. (a) Using v = \lambdaf, v_w = \lambda_0f_0, \lambda_0 = v_w/f_0 = (1000 ft/second)/(1000/second) = 1 ft (b) It is known that the velocity of waves in a medium depend only on the mechanical properties of that medium. When the source moves towards the listener (see L_1 in figure) a "bunching up" of waves is noted by the listener, as shown in the figure. We seek a mathematical relation to determine by what amount their wavelength is effectively reduced, and how the frequency is affected. In a time, t, the source emits t/T = f_0t waves. Along the line between listener and source these waves are spread over a distance v_wt - v_st. Thus, if the wavelength is smaller by a constant amount, it must now be \lambda_s1 = [(v_wt - v_st)/(f_0t)] = [(v_w - v_s)/(f_0)] v_wt - v_st f_0t v_w - v_s)/(f_0)] = [(1000 ft/sec - 100 ft/sec) / (1000/sec)] = [(900 ft/sec) / (1000/sec)] = 0.9 ft. As the speed of the wave is unchanged, the listener will observe an in-crease in the wave's frequency, according to v_w = \lambda_s1f_s1; f_s1 = v_w/\lambda_s1 = (1000 ft/sec)/(.9 ft) = 1111 sec^-1 (c) As the source recedes a listener observes the waves to occupy more space than those of a stationary source, as can be seen by looking at L_2 in the figure. In a time, t, f_0t waves are emitted, spread over a distance v_wt + v_st. As before, we find the wavelength \lambda_s2 = [(v_wt + v_st)/( f_0t)] = [(v_w + v_s)/(f_0)] = [(1000 ft/sec + 100 ft/sec) / (1000/sec)] = 1.1 ft And, as in (b) we have for the observed frequency, f_s2 = v_w/\lambda_s2 = (1000 ft/sec)/(1.1 ft) = 909/sec.

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Question:

2.000picogram(pg) of ^33P decays by ^0_-1\beta emission to 0.250 pg in 75.9 days. Find the half-life of ^33P.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E19-0719.htm

Solution:

The half-life is defined as the time it takes for 1/2 of the amount of a certain compound present to de-compose. For example, if a substance has half-life of day, after one day there will only be 1/2 of the original amount left. When given the original and final amount of a substance, after a given time has elapsed, the number of half-lives that have passed can be found. his is done by dividing the original amount by the final amount and determ-ining how many factors of 2 are present in the quotient. The half-life is then found by dividing the time elapsed by the number of half-lives. Solving for the half-life of ^33P: (original amount) / (final amount) = (2.000 pg) / (0.250) = 8 8 = 2 × 2 × 2, therefore 3 half-lives have passed. One is given that these half-lives elapse in 75.9 days. half-life = (75.9 days) / (3) = 25.3 days.

Question:

Find all three-digit numbers with non-zero first digit that are equal to the sum of the cubes of their digits.

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Solution:

The solution is based upon having the computer check each three-digit number HTU = 100\textasteriskcenteredH + 10\textasteriskcenteredT + U to see if the condition HTU = H3 + T3 + U3 is satisfied. [Note that H stands for hundreds, T stands for tens, and U-for units]. Since we want all possible combinations of H,T, and U where U = 0,1,2,...,9; T = 0,1,2,...,9; and H = 1,2,3,...,9 - there are 10 × 10 × 9 = 900 numbers to check. We do this checking with three nested DO loops. The DO LOOP parameters take on values 1,2,...,10; 1,2,...,10; and 1,2,...,9, progressing from the inner to the outer loops. Since 1 is the lowest possible Integer to be assigned to any DO-LOOP index, the program subtracts 1 from indexes TP1 and UP1 to initiate them at zero. Meanwhile, the final values of TP1 and UP1 are set to 10, which become 9's after the same subtraction. The program looks as follows: INTEGER H,T,U,TP1, UP1 DO 10 H = 1,9 DO 10 TP1 = 1,10 T = TP1 - 1 DO 10 UP1 = 1,10 U = UP1 - 1 IF(100\textasteriskcenteredH + 10\textasteriskcenteredT + U.EQ. H\textasteriskcentered\textasteriskcentered3 + T\textasteriskcentered\textasteriskcentered3 + U\textasteriskcentered\textasteriskcentered3) GO TO 1 GO TO 10 1WRITE (6,100) H,T,U 10CONTINUE 100FORMAT (3I1) STOP END

Question:

Explain how character strings are declared and manipulated in Pascal.Give examples.

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Solution:

Unlike some other programming languages, Pascal has no facilities for explicit declaration of character strings. Instead, Packed arrays of characters are used to define character strings. Since the length of an array must be known at compilation time, only strings of fixed length are allowed. Thus, if we define TYPE STRING = PACKED ARRAY[1..10] of char; VAR WORD: string; then the variable WORD can be assigned any value consisting of exactly ten characters. The following assignments are valid: WORD: =' HELLObbbbb'; WORD: =' GGODbBYEbb' ; WORD: = 'CHARACTERS'; WORD: = '#123456789' and the following are not: WORD: = 'HELLO' WORD: = 'WORD' WORD: = 'AVERYLONGWORD' When a variable declared as packed array of characters ap-pears in a WRITE or WRITELN statement, the entire array is written out. For example, the effect of these two state-ments WORD: =' HELLObbbbb'; write1n(word); will be HELLObbbbb However, to read in a character string, the only possibility is character by character, and then only if the length of the string is exactly equal to the length of the array hold-ing it. Thus, to read in a ten-character word, we would write FOR I:= 1 to 10 DO READ(word[I]); Character strings can also be compared. For example, the Boolean variable P will have value 'true' after the execu-tion of any of these statements. P: = 'ALEX' < 'JILL' P: =' JOHNbbb' < ' JOHNSON ' P: ='ZITO' > 'MARY' P: = 'JACK' ='JACK'

Question:

A sample of blood has freezing point of - 0.560\textdegreeC. What is the effective molal concentration of the solute in the serum? According to this concentration, what is the osmotic pressure in the blood serum at body temperature (37\textdegreeC)?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E26-0880.htm

Solution:

The molal freezing point depression and boiling point elevation for the ideal 1 molal solution are also called the cryoscopic and ebullioscopic constants. The mathematical expressions for the freezing point depression is ∆T_f = m K_f where K_f is the freezing point constant for water, 1.86\textdegreeC, m is the effective molality of the solute, and ∆T_f is the change in the freezing point. For this particular problem ∆T_f = 0.560\textdegreeC and K_f = 1.86\textdegreeC, so that m = ∆T_f / K_f = 0.560\textdegreeC / 1.86\textdegreeC = 0.30 m. Hence, 0.30 m is the effective molal concentration of solute in serum. Osmosis is the diffusion of a solvent through a semi- permeable membrane into a more concentrated solution. The osmotic pressure of a solution is the minimum pressure that must be applied to the solution, in excess of the pressure on the solvent, to prevent the flow of solvent from pure solvent into the solution. An empirical osmotic pressure equation for solutions is \pi = CRT, where \pi is the osmotic pressure, C is the concentration in molality or molarity, R is the gas constant (0.082 atm- liter/\textdegreeK-mole), and T is the temperature (in \textdegreeK). Thus, at 37\textdegreeC = 310\textdegreeK and at 0.30 m, blood serum has an osmotic pressure of \pi = CRT = (0. 30 m) (0.082 atm-liter/\textdegreeK mole) (310\textdegreeK) = 7.6 atm.

Question:

Balance the equation for the following reaction taking place in aqueous basic solution: MnO^-_4 + H_2O_2 \rightarrow MnO_2 + O_2

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Solution:

The equation in this problem involves both oxidation and reduction. When balancing it, you can use the following rules. Separate the net reaction into its two major components, the oxidation reaction (the loss of electrons) and the reduction reaction (the gain of electrons). For each half- reaction, balance the charges with H^+, if the medium is acidic, or OH^-, if the medium is basic. Next, balance the oxygens by the addition of H_2O. Balance total atoms by the addition of H atoms. Finally, combine the two half reactions, so that all charges from electron transfer cancel out. You employ these rules as follows: The net reaction is MnO^-_4 + H_2O_2 \rightarrow MnO_2 + O_2 The oxidation process is H_2O_2 \rightarrow O_2 + 2e^- The oxygen atoms in H_2O_2 go from -1 to zero in O_2. Thus, you have a loss of two electrons. To balance charges, add 2 OH^- to the left side, since there exist 2 negative charges on the right side. You obtain H_2O_2 + 2OH^- \rightarrow O_2 + 2e^- You now have 4 oxygens on the left, but only 2 on the right. Thus, add 2 water molecular ro the right, obtaining H_2O_2 + 2OH^- \rightarrow O_2 + 2e-+ 2H_2O There are the same number of H's on each side. Proceed, now, to reduction half-reaction. Here, MnO^-_4 + 3e^- \rightarrow MnO_2 Mn begins with a + 7 oxidation number and ends up with + 4 in MnO_2. Therefore, 3 electrons must be added to the left side of the equation. To balance the charges, add 4 OH^- ions to right, since you have a total of 4 negative charges on left. Rewriting the equation MnO^-_4 + 3e^- \rightarrow MnO_2 + 4OH^-. Add 2 water molecules to the left, so that oxygen atoms can be balanced, obtaining 2H_2O + MnO^-_4 + 3e^- \rightarrow MnO_2 + 4OH^-. The hydrogens are balanced. Thus oxidation:H_2O_2 + 2OH^- \rightarrow O_2 + 2e-+ 2H_2O reduction:2H_2O + MnO^-_4 + 3e^- \rightarrow MnO_2 + 4OH^-. To combine these two so that electrons cancel out. Select a multiple of 3 and 2, since these are the number of electrons involved in the half- reactions. This multiple is six. Multiply the oxidation by 3 and the reduction by 2, obtaining oxidation:3H_2O_2 + 6OH^- \rightarrow 3O_2 + 6e-+ 6H_2O reduction:4H_2O + 2MnO^-_4 + 6e^- \rightarrow 2MnO_2 + 8OH^- The net reaction is the total. Thus, adding you obtain: 3H_2O_2 + 4H_2O + 6OH^- + 2MnO^-_4 + 6e^- \rightarrow 3O_2 + 6e^- + 6H_2O + 8OH^- + 2MnO_2 Cancel the electrons, subtract OH^- ions and H_2O's to obtain: 2MnO^-_4 + 3H_2O_2 \rightarrow 2MnO_2 + 3O_2 + 2H_2O + 2OH^- which is the balanced equation.

Question:

A car starts from rest and reaches a speed of 30 miles per hour in 8 seconds. What is its acceleration?

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/Users/wenhuchen/Documents/Crawler/Physics/D03-0058.htm

Solution:

v = at for constant acceleration. We shall convert the velocity in miles per hour into feet per second. A useful conversion factor to remember is that 60 mph is about 88 ft. per second. Therefore, 30 mph is about 44 ft. per second. Substituting we have: a =v_final/t= (44 ft)/(sec × 8 sec) = 5.5 ft per sec per sec.

Question:

The difference of potential V_1 between the terminals of a resistor is 120 volts when there is a current I_1 of 8.00 amp in the resistor. What current will be maintained in the re-sistor if the difference of potential is increased to 180 volts?

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/Users/wenhuchen/Documents/Crawler/Physics/D20-0646.htm

Solution:

Ohm's law indicates that the resistance R will remain the same when the potential difference is increased; hence we can write R = [(V_1 )/(I_1 )] = [(120 volts)/(8.00 amp)] = 15.0 ohms I_2 = [(V_2 )/(R)] = [(180 volts)/(15.0 ohms)] = 12.0 amp .

Question:

Write an assembler language program in BAL to add 20 numbers from main storage.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G09-0200.htm

Solution:

Let AA denote the starting location of the num-bers A_i to be added together. We assume these numbers oc-cupy consecutive full words in main storage (i.e. 4 bytes per word). If the address of the A_i just added to the sum is kept in a register, the number 4 can be added to the con-tents of that register; the result is the address of the next number to be added. Suppose register 7 is initialized to AA, the address of A_0. Each time the adjustment step is executed, the quantity in register 7 is increased by 4. Hence the second time the instruction A 8,\varphi(7) is executed, the address \varphi in register 7 is 4 greater than the address of A_0; thus it is the address of A_1. The third time the body of the loop is executed, the address of register 7 has been increased again, so A_2 is growing to sum. The process of adding on A_i and adjusting the address continues until exe-cution of the loop is terminated. We exit from the loop when the contents of register 9 is 20. \textasteriskcenteredA SEGMENT TO ADD 20 NUMBERS FROM \textasteriskcenteredCONSECUTIVE FULL WORDS IN MAIN STORAGE \textasteriskcenteredINITIALIZATION LA8,\varphiSUM =\varphi \varphi \varphi LA9,\varphiI =\varphi \varphi \varphi LA7, AA ADDRESS OF FIRST A \textasteriskcenteredBODY: ADD A (SUBI) TO SUM LOOPA8,\varphi(7) \varphi LA9, 1 (9) INCREASE I LA7, 4 (7) GET ADDRESS OF NEXT A C9, = F '2\varphi' \varphi BNE LOOP BRANCH BACK IF I NOT YET 20, LV LOOP IF I = 2\varphi \varphi

Question:

Explain how a conditional branch instruction is specified in APL programs.

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Solution:

A conditional branch instruction is specified in APL by means of a right-pointing arrow followed on its right by an expression. For example: i) [4] \rightarrow (I \leq N) × 2 In the above, a test for I is performed. Is I \leq N? If I is indeed less than or equal to N, then the value of the test is true, i.e., 1. But if I is not less than or equal to N then the value of the test is false, i.e., 0. Now, if the result of the test is 1, the value of the expression (I \leq N) × 2 reduces to 1 × 2, equal to 2. This means that in the given example the statement [4] of the program re-duces to the following: [4] \rightarrow (I \leq N) × 2 i.e.,[4] \rightarrow 1 × 2 i.e.,[4] \rightarrow 2. Thus, a branching to statement 2 is performed. However, if the result of the test is 0, then the statement [4] of the example reduces to the following: [4] \rightarrow (I \leq N) × 2 i.e.,[4] \rightarrow 0 × 2 i.e.,[4] \rightarrow 0. This means that a branching to statement 0 is to be per-formed. However, there is no statement 0 in the program. Hence, the program execution halts. ii) [4] \rightarrow (I \leq N) × BOB. The above is similar to parti) explained before, except that instead of the statement number of the statement to which a branch off is required, a label is specified. La-bels have an advantage over numbers. If statements are added or removed from the program, the statement numbers change. This necessitates corrections to be made within the statements too, in order to branch to the proper state-ment according to the new numbering. But if labels are used, corrections of the statements are not necessary. A label takes on a value equal to the statement number of the statement which has the given label. In the example given above, the label is BOB. This means that if the test is true, the program will jump to a statementlabelledBOB. The program could be something like the following: [2] BOB: I \leftarrow I + 1 [4] \rightarrow (I \leq N) × BOBor,[4] \rightarrow (I \leq N) × BOB [7] BOB: I \leftarrow I + 1 In the above, if the test is false, statement [4] becomes: [4] \rightarrow 0 × BOB, i.e.,[4] \rightarrow 0. and the program halts. iii) In the above two cases, the program went to a halt if the test was false. However, at times it is not proper for the program to halt when the test is false. It is just required that the program goes to the next sequen-tial instruction if the test is false, and branch to the statement indicated if the test is true. Consider the following statement, for example: [4] \rightarrow 2 × ʅ(I \leq N) In the above, if the test of I \leq N is true, the value of (I \leq N) is equal to 1. Therefore the statement [4] reduces to 2 × ʅ(1) = 2 × 1=2. Thus, the program branches to statement 2. But, if the test is false, the statement [4] reduces to 2 × ʅ(0) Now, iota zero, i.e., ʅ 0, generates an empty vector, and therefore the statement [4] becomes meaningless. Hence, the program goes to the next sequential statement. iv) A label can be used, instead of a statement num-ber, with the iota function. For example, [4] \rightarrow BOB × ʅ(I \leq N). v) Another way to specify a branch instruction is by using the compression function, which has the symbol /. It is not necessary to know the meaning of the compression function in order to understand the branching. Consider the following illustration: [4] \rightarrow (I \leq N) / 2. In the above, if the test, that is, the quantity to the left of the compression function symbol is true, a transfer is made to the statement whose number is specified on the right of the compression function symbol. Thus, if I \leq N is true, the program jumps to statement 2. But, if the test is false, the program goes to the next sequential statement, i.e., to statement 5. vi) A label can also be used on the right side of the compression function symbol instead of a number. For ex-ample, [4] \rightarrow (I \leq N) / BOB.

Question:

Consider an elementary inventory control system where there is no delay between the ordering of goods and their receipt into inventory. Suppose the order rate is direct-ly proportional to the difference between desired inventory and actual inventory, with proportionality constant 1/A, where A is the time that would be required to correct the inventory if the order rate were constant. Write a FORTRAN program to simulate this system, using modified Euler's method (the method is fully explained, and the subroutine for it is given in the SIMULATION chapter) from time t=0 to t=t_f if I(t) is the inventory level at time t and I(0)=I_0.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G18-0464.htm

Solution:

Clearly, the goal of this system is to maintain the desired inventory, D. If \.{I} is the order rate (and hence, the rate of change in inventory, since there is no delay), we can write: \.{I} = 1/A(D - I)(1) with block diagram as follows: Note that this model "fills to capacity" using negative feedback control. That is, knowledge of the present system state is used to determine the future system state. If change in inventory is done continuously, the exact solution to equation (1) would be: I(t) = D - (D - I_\O)^e-t/A Note, that as t \rightarrow \infty, l(t) approaches D, which is the steady state solution, the program looks as follows: DATA T/_\O , \O/ COMMON D, A INPUT N, TFIN, ACCUR, D, A, I\O OUTPUT T, I SPACE = D - I\O REALN = N DT = TFIN/REALN I = I\O DO 1\O J = 1, N T = T + DT CALL MEULER (T, I, ACCUR, DT) EXACT = D - SPACE \textasteriskcentered EXP(-T/A) ERROR = ABS( (EXACT - I) /EXACT) \textasteriskcentered1\O\O. OUTPUT T, I, EXACT, ERROR 1\OCONTINUE STOP END FUNCTION G (W) COMMON, D, A G = (D - W)/A RETURN END

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Question:

CO_2 + H_2O +hv(light energy) (CH_2O) + O_2. It was generally assumed in the past that the carbohydrate came from a combination of the carbon atoms and water molecules and that the oxygen was released from the carbon dioxide molecule. Do you agree with this assumption? Explain your answer.

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Solution:

Superficially, the assumption appears sound, and was believed for years to be true. It is however, quite wrong, and was refuted in the 1930's by C.B. vanNielof Stanford University. VanNielwas studying photosynthesis in different types of photosynthetic bacteria. He observed that in their photosynthetic reactions, bacteria reduce carbon dioxide to carbohydrate, but do not release oxygen. The purple sulfur bacteria, which require hydrogen sulfide for photosynthesis, were found to accumulate globules of sulfur within the cells. Based on these observations VanNiel then wrote the following equation for photosynthesis in purple sulfur bacteria: CO_2 + 2H_2S \rightarrow CH_2O + H_2O + 2S This equation shows that it is unlikely that carbon dioxide is split during photosynthesis; if it were, oxygen would be formed. Moreover, carbohydrate could not have come from the combination of carbon atoms and hydrogen sulfide molecules. It would be reasonable to propose that hydrogen sulfide was the molecule to be cleaved rather than carbon dioxide. This led VanNielto propose the following general equation for photosynthesis, CO_2 + 2H_2A \rightarrow (CH_2O) + H_2O + 2A Here, H_2A represents anoxidizablespecies capable of donating its electrons upon being split. According to this hypothesis, in the photosynthesis of green plants, it is the water molecule, not carbon dioxide, that is split to release O_2. This brilliant speculation was not proven until many years later when advancement of radio - isotopic techniques made it possible to reveal the mechanism of the actual reaction. The use of isotopic oxygen, ^18O, allowed the successful tracing of the path of oxygen from water to free molecular oxygen: CO_2 + 2H_2^18O - CH_2O + H_2O + ^18O_2 Thus, it can readily be seen that molecular oxygen comes actually from water and not from carbon dioxide. It is now known that the splitting of water occurs during the light reactions of photosynthesis, releasing O_2 gas, electrons, which are energized by light to provide for ATP formation, and protons, which are used to form NADPH, the reducing power necessary for carbohydrate formation. Carbon dioxide is then reduced in the dark reactions of photosynthesis using both ATP and NADPH to form carbohy-drates.

Question:

What are the implications of the Hardy-Weinberg Law?

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Solution:

The Hardy-Weinberg Law states that in a population at equilibrium both gene and genotype frequen-cies remain constant from generation to generation. An equilibrium population refers to a large interbreeding population in which mating is random and no selection or other factor which tends to change gene frequencies occurs. The Hardy-Weinberg Law is a mathematical formulation which resolves the puzzle of why recessive genes do not disappear in a population over time. To illustrate the principle, let us look at the distribution in a population of a single gene pair, A and a. Any member of the popu-lation will have the genotype AA,Aa, oraa. If these genotypes are present in the population in the ratio of 1/4 AA : 1/2Aa: 1/4aa, we can show that, given random mating and comparable viability of progeny in each cross, the genotypes and gene frequencies should remain the same in the next generation. Table 1 below shows how the genotypic frequencies of AA,Aa, andaacompare in the population and among the offspring. Table 1 The Offspring of the Random Mating of a Population Composed of 1/4 AA, 1/2Aaand 1/4aaIndividuals Mating Frequency Offspring MaleFemale AA×AA 1/4 × 1/4 1/16 AA AA×Aa 1/4 × 1/2 1/16 AA + 1/16Aa AA×aa 1/4 × 1/4 1/16Aa Aa ×AA 1/2 × 1/4 1/16 AA + 1/16Aa Aa ×Aa 1/2 × 1/2 1/16 AA + 1/8Aa+ 1/16aa Aa ×aa 1/2 × 1/4 1/16Aa+ 1/16aa aa ×AA 1/4 × 1/4 1/16Aa aa ×Aa 1/4 × 1/2 1/16Aa+ 1/16aa aa ×aa 1/4 × 1/4 1/16aa Sum : 4/16 AA + 8/16Aa+ 4/16aa Since the genotype frequencies are identical, it follows that the gene frequencies are also the same. It is very important to realize that the Hardy-Weinberg law is theoretical in nature and holds true only when factors which tend to change gene frequencies are absent. Examples of such factors are natural selection, mutation, migration, and genetic drift.

Question:

What is the mole fraction of H_2SO_4 in a 7.0 molar solution of H_2SO_4 which has a density of 1.39 g/ml?

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Solution:

The mole fraction of H_2SO_4, is equal to the number of moles of H_2SO_4 divided by the sum of the number of moles of. H_2SO_4 and of H_2O. mole fraction of H_2SO_4 = [( moles of H_2SO_4 )/(moles of H_2SO_4 + moles of H_2O)] Since the solution is 7.0 molar in H_2SO_4, you have 7 moles of H_2SO_4 per liter of solution. If one knows how much 7 moles of H_2SO_4 weighs, and how much one liter of the solution weighs, the weight of water can be determined by taking the difference between the quantities. The weight of one liter of the solution can be calculated by multiply-ing the density by the conversion factor 1000 ml/1 liter. weight of 1 liter of the solution = 1.39 g/ml × 1000 ml/liter = 1390 g/l. weight of the water in a one liter solution = weight of the total solution - weight of 7 moles of H_2SO_4 7 moles of H_2SO_4 weigh 7 times the molecular weight of H_2SO_4 . weight of 7 moles of H_2SO_4 = 7 × 98 g = 686 g. Hence, weight of the water = 1390 g - 686 g = 704 g. The number of moles of H_2O is found by dividing its weight by the molecular weight of H_2O. no. of moles of H_2O = [(704 g)/(18 g/mol] = 39 moles The number of moles of both components of the system is now known, therefore, the mole fraction of H_2SO_4 can be obtained. mole fraction H_2SO_4 = [(moles of H_2SO_4)/(moles of H_2SO_4 + moles H_2O)] mole fraction of H_2SO_4 = [7 / (7 + 39)] = 0.15.

Question:

A proton, starting from rest, falls through a potential differenceof 10^6 V. What is its final kinetic energy andfinal velocity?

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Solution:

At its rest position, the proton has only potential energy and no kineticenergy. As it falls, it loses potential energy but gains kinetic energy. In its final position, the proton has only kinetic energy and no potential energy. However, the law of conser-vation of energy demands that at all times, the sum of the kinetic and potential energies remain constant. Consequently, the sum of the two energies at the rest position must equal thesum of the two energies at the final position. Or PE + 0 = 0 + KE By definition of potential V, V = Energy/Charge=E/q Then PE =eV wheree is the charge of the proton and V is the potential dif-ference throughwhich it falls. Therefore 10^6 elec. volts = 1/2m_pv^2 , wherean electron volt is the energy required to move a charge e througha potential of 1 volt. The final kinetic energy is then 10^6eV= 10^6eV×(1.602 X 10\rule{1em}{1pt}12erg/1eV) =1.602 × 10^\rule{1em}{1pt}6 erg = 1/2 m_pV2 To compute the final velocity, we use this relation, then v=\surd[{2×(1.60×10^\rule{1em}{1pt}6 erg)} / (1.67×10^\rule{1em}{1pt}24g)]. wherem_p = 1.67 ×10^\rule{1em}{1pt}24 g. Consequently, v =1.38 × 10^9 cm/sec

Question:

What is the elemental percent composition (by weight) of a mixture that contains 20.0 g of KAl(SO_4)2and 60.0 of K_2 SO_4 ?

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Solution:

The elemental percent composition by weight is equal to the weight of the element present divided by the total weight of the mixture multiplied by 100. percent composition = {(wt of elements) / (total wt) }× 100 The weight of the total mixture is equal to the sum of the weights of the KAl(SO_4)_2 and the K_2 SO_4 . total weight = 20.0 g + 60.0 g = 80.0 g The weights of the various elements present is found by dividing the molecular weight of the element multiplied by the number of atoms of that particular atom present in the molecule by the molecular weight of the compound, and then multiplying the quotient by the number of grams of the compound present. Solving for the weights of the elements in KAl(SO_4)2: MW = 258.12. For K = MW = 39.1 weight of K = {(30.1 g/mole × 1)/(258.12 g/mole)} × 20.0g = 3.03g For Al: MW = 26.98 weight of Al = {(26.98 g/mole × 1)/(258.12 g/mole)} × 20.0g = 2.09g. For S: MW = 32.06 weight of S = {(32.06 g/mole × 2 moles)/(258.12 g/mole)} × 20.0g = 4.97g. For O: MW = 16 weight of O = {(16 g/mole × 8moles)/(258.12 g/mole)} × 20.0g = 9.92g. Solving for the weight of the elements in KAl(SO_4)_2 : MW = 174.26 For K: MW = 39.1 weight of K = {(39.1g/mole × 2 moles)/(174.26 g/mole)} × 60.0g = 26.93 g For S; MW = 32.06 weight of S = {(32.06 g/mole × 1moles)/(174.26 g/mole)} × 60g = 11.04 g For O: MW = 16 weight of O = {(16 g/mole × 4 moles) / (174.26 g/mole)}× 60g = 22.04 g. One can find the total weights of the various elements by taking the sum of their weights from the two compounds. Total weight of K = 3.03 g + 26.93 g = 29.96 g Total weight of Al = 2.09 g Total weight of S = 4.97 g + 11.04 g = 16.01 g Total weight of O = 9.92 g + 22.04 g = 31.06 g. One can now determine the elemental percent composition of the mixture. % = {(weight of elements) / (total weight of mixture)} × 100 % of K = {(29.96g) / (80.0g)} × 100 = 37.45% % of Al = {(2.09g) / (80.0g)} × 100 = 2.61% % of S = {(16.01g) / (80.0g)} × 100 = 20.01% % of O = {(31.96g) / (80.0g)} × 100 = 39.95% .

Question:

You have 1 liter of an ideal gas at 0\textdegreeC and 10 atm press-ure. You allow the gas to expand against a constant ex-ternal pressure of 1 atm, while the temperature remains constant. Assuming, 24.217 cal/liter-atm, find q, w, ∆E and ∆H in calories, (a) in these values, if the expansion took place in a vacuum and (b) if the gas were expanded to 1 atm pressure.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E14-0506.htm

Solution:

The solutions to the parts of this problem require a combination of thermodynamics and ideal gas law theory. The gas expands because its pressure is greater than the external pressure. When the pressure of the gas falls to 1 atm, which is the pressure being applied ex-ternally, gas expansion will terminate. This allows for determination of the volume change using Boyle's Law. You want ∆V (volume change) , since w = work = P∆V. Boyle's law states that PV = constant for ideal gas. Thus, P_1V_1 = P_2 V_2. Originally, P_1 = 10 atm with V_1 = 1 liter. You end up with P_2 = 1 atm. Thus, V_2 = (P_1 V_1) / P_2 = [{(10 atm) (1 liter)} / {1 atm }] = 10 llters. ∆V = 10 liters (final) - 1 liter (original) = 9 liters. Therefore, w = P∆V = (1 atm)(9 liters) = 9 liters-atm. But, there are 24.217 cal per liter-atm, so that, in calories, w = 9 liter-atm × 24.217 cal/liter-atm = 218 cal. q = the heat absorbed by the system. To find its value, you employ the first law of thermodynamics, which says ∆E = q - w, where ∆E = change in energy of system and w = work performed. Since the gas is ideal, E = energy is only a function of temperature. As such, ∆E = 0, since the temperature is constant. You have, therefore 0 = q - w or q = w. But, you just found w, which means q = 218 cal. To find ∆H, the enthalpy change, remember that ∆H = ∆E + ∆(PV). You know that ∆E = 0. To find ∆(PV) , re-call that ∆(PV) = P_2 V_2 - P_1 V_1, as derived from Boyle's law. However, P_1 V_1 = P_2 V_2 thus; P_2 V_2 - P_1 V1= 0. This means ∆H = 0. In a vacuum, there is no external pressure. This means that no work can be done, so that w = 0. This is derived from the fact that work is defined as performing an action against a surrounding environment. If there is no environment, there cannot be any work. Since ∆E remains 0 and q = ∆E + w, then q = 0. Since there is zero pressure, PV = 0. Since ∆ (PV) = 0, ∆E = 0 and ∆H = ∆E + ∆(PV), then ∆H = 0. You now find the work (w) , required in going from zero pressure to 1 atm pressure; use the equation for the reversible expansion of a gas in terms of work. w = 2.303 n RT log V_2/V_1, where V_1 = initial volume, V_2 = final volume, R = universal gas constant, T = temperature in Kelvin (Celsius plus 273\textdegree) and n - number of moles. V_2 was calculated from Boyle's law. The rest is given. Thus, you substitute to obtain w = 2.303 (1 mole) (1.987) (273)log (10/1) = 1249 cal. ∆E\textdegree and ∆H\textdegree are still zero, but since q = ∆E + w, q = 0 + 1249 = 1249 cal.

Question:

One gram of water (1 cm^3) becomes 1671 cm^3 of steam when boiled at a pressure of 1 atm. The heat of vaporization at this pressure is 539 cal/gm. Compute the external work and the increase in internal energy.

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Solution:

The work done by the water in expanding from volume V_1 to volume V_2 is W = ^V2\int_V1 pdV where p is the pressure exerted by the water on its container. In our problem, p is constant, whence W = p(V_2 - V_1) = p(V_V - V_L) where V_V is the volume occupied by the steam, and V_L is the volume occupied by the water. Hence, W = p(V_V - V_L) = 1.013 × 10^6 dynes/cm^2 (1671 - 1)cm^3 Here we used the fact that 1atm= 1.013 × 10^6 dynes/cm^2 = 1.695 × 10^9 ergs Since 1 erg = 2.389 × 10^-8 cal W= (1.695 × 10^9 ergs)(2.389 × 10^-8 cal/erg) W= 41 cal The Law of Thermodynamics is ∆U = ∆Q - ∆W where ∆U is the change in internal energy of the water- steam system during the stated process. Also, ∆Q and ∆W are the heat added to and the work done by the system during the process, respectively. Denoting the gaseous state by the subscript V, and the liquid state by the subscript L, we obtain U_V - U_L = ∆Q - ∆W The heat, ∆Q, added to the system is the amount needed to vaporize the water, or ∆Q =mL= (1 g)(539 cal/g) = 539 cal ThenU_V - U_L= 539 cal - 41 cal = 498 cal. Hence the external work, or the external part of the heat of vaporization, equals 41 cal, and the in-crease in internal energy, or the internal part of the heat of vaporization, is 498 cal.

Question:

For the following PL/I program, CONDITION' arises. b) Show what will be printed out if the program is run. c) Write a corrected program to avoid errors due to this condition. d) Show what the print out will be from the corrected program . EXAMPLE:PROCOPTIONS(MAIN); DCL (B(6),C) CHAR(4), Z FIXED (2); GET LIST (Z,C); DO I=1 TO Z; B(I)=SUBSTR(C,1,I); PUTLIST(I,B(I)); END; END EXAMPLE; The DATA CARD holds data as follows:6,ROVE.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G13-0343.htm

Solution:

a) The purpose of the above program is to illus-trate a type of conditionknown as 'STRINGRANGE CONDITION'. In the program, the array B is declared as having 6ele-ments . The GET LIST statement gets the value of Z from the DATA CARD asZ = 6. Also, the value of C is obtained from the DATA CARD as C = 'ROVE'. In the next statement, viz., DO I = 1 TO Z, the value of Z = 6 gets substituted. Hence, the DO loop statement becomes DO I = 1 TO 6;. This means that the DO loop will be performed 6 times. Each time., thevalue of an array element B(I) is calculated. This value ofB(I) = SUBSTR(C,1,I) means that from C, i.e., from 'ROVE', starting from the first character, select I characters. These selectedI characters are to be assigned to B(I). Thus, for example, if 1 = 2, then SUBSTR(C,1,2) gives B (2) = 'RO'. In the given program, as Z = 6, the loop will be perform-ed 6 times. But, during the 5th loop, on substituting values ofIand C in the equation: B(I) = SUBSTR(C,1,I) , gives: B(5) = SUBSTR ('ROVE',1,5). The above statement tries to find a substring of 'ROVE', starting fromthe first character up to (and including) the fifth character. However, 'ROVE' has only 4 characters. Hence, the computer will just print the available characters. The programmermust be informed. Hence, the computer is pro-grammed to printout a message stating that the STRINGRANGE CONDITION had occurredin a certain program statement of the program. b) The printout of the program will be as follows: 1'R ' 2'RO ' 3'ROV ' 4'ROVE ' 5'ROVE ' 6'ROVE ' c) The program can be corrected as follows: The computer must be told thatit must look out for the STRINGRANGE CON-DITION. This is done by usinga word within parentheses, viz., (STRINGRANGE):, as a prefix to thefirst program state-ment, as shown below. Also, in the body of the program, a statement such as:. On STRINGRANGE GOTO .....,must be in-cluded. A good place for inserting this statement is just after the DCL statementso that the computer reads it be-fore it enters the loop. The corrected program can now be written as shown below: (STRINGRANGE) :EXAMPLE:PROC OPTIONS(MAIN); DCL(B(6),C) CHAR(4), Z FIXED(2) ; ON STRINGRANGEB(I) = '0000' ; GETLIST(Z ,C) ; DO I = 1 TO Z; B (I) = SUBSTR(C, 1,I) ; PUTLIST(I,B(I)); END; FINISH:END EXAMPLE; In the above program the program is made to go to the FINISH label, whichis the end of the program when the STRINGRANGE CONDITION arises. But as a matter of fact any valid instruc-tion could be given to the computerinstead of the 'GOTO FINISH' instruction. d) The print-out of the corrected program is as follows: 1'R ' 2'RO ' 3'ROV ' 4'ROVE ' 5'0000 ' 6'0000 ' The computer will also print out a STRINGRANGE CONDITION mes-sage. It is assumed that the user of the program knows that whenever '0000' is printedout it means that a STRINGRANGE CONDITION has occurred at that stage.

Question:

Write a BASIC program which calls for the input of an arbitrary number, and then prints out whether the number read in is positive, negative, or zero.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G10-0225.htm

Solution:

This problem represents an aspect of elementary decision making. The implementation of these decisions is best handled by use of several IF ... THEN clauses. 1\O\O REM DETERMINE IF X IS POSITIVE, NEGATIVE 11\O REM OR ZERO 12\O PRINT "WHEN I ASK, YOU ENTER A NUMBER AND"; 13\O PRINT "I WILL TELL YOU WHETHER YOUR NUMBER"; 14\O PRINT "IS POSITIVE, NEGATIVE, OR ZERO." 15\O PRINT 16\O PRINT "WHAT IS YOUR NUMBER?"; 17\O INPUT X 18\O IF X > \O THEN 19\O 181 IF X < \O THEN 21\O 182 IF X = \O THEN 23\O 19\O PRINT X; "IS POSITIVE" 2\O\O GO TO 15\O 21\O PRINT X; "IS NEGATIVE" 22\O GO TO 15\O 23\O PRINT X; "IS ZERO" 24\O GO TO 15\O 999 END

Question:

A biology student places one bacterium in a suitable medium and incubates it under appropriate conditions. After 3 hours and 18 minutes, he determines that the number of bacteria now present is 1000. What is the generation time of this bacteria?

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Solution:

Bacterial growth usually occurs by means of binary fission with one cell dividing into two and these two cells dividing into four, etc. The population at any time can be represented by a geometric progression (1, 2, 4, 8...). The time interval required for each division (or for the entire population to double) is called the generation time. Different bacteria have different generation times, ranging from 20 minutes forEscherichiacoli to 33 hours forTreponemapallidum(the bacterium responsible for syphilis). The generation time also differs for the same bacterial species under different environmental conditions. To determine the generation time (G) of a bacterial population, we must know the number of bacteria present initially (N\textdegree_0), the number of bacteria present at the end of a given time period (N), and the time period (t). We can determine the generation time by using some simple mathemati-cal expressions. If we start with a single bacterium , the total population (N) after the nth generation is 2^n: N = 1 × 2^n For example, at the end of three generations, we would have eight bacteria, (2^3 = 8) . However, since we usually start with many bacteria, we must modify the formula to account for more than one parental bacterium: N = N_0 × 2^n. Solving for n, the number of generations, we get log N = log N_0 + n log 2. By substituting .301 for the log 2 and rearranging, n = (log N - log N_0) / (.301) = 3.3 (log N - log N_0). Since the difference of the logarithms of two numbers is the logarithm of the quotient of the two numbers, n = 3.3 log (N / N_0) The generation time G is simply equal to the time elapsed between N_0 and N divided by the number of generations: G = (t / n) = [t] / [3.3 log (N / N_0)] In our example: N = 1000, N_0 = 1 and t = 198 minutes. Substituting into the general formula, G = (198) / (3.3 × log 1000) . G = (198) / (3.3 × 3) = 20 minutes The time interval required for this particular bacterial species to divide is twenty minutes. If we plot the number of bacterial cells against time, we get a bacterial growth curve. In our case, with a generation time of 20 minutes, we would obtain this curve: This curve does not represent the complete pattern of growth, but rather one selected portion of the normal growth curve. This portion is called the exponential or logarithmic phase of growth. Only at this phase does the population double at regular time intervals. Eventually the reproduction rate is checked because of the lack of nutrients or the accumulation of waste products. In our example, we assumed that the bacteria were growing at a logarithmic rate where both N_0 and N should be measured during the exponential growth phase.

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Question:

Describe the curve one would obtain by plotting pressure versus volume for an ideal gas in which the temperature and number of moles of gas are held constant.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E02-0053.htm

Solution:

This problem requires a plot of the ideal gas equation. This equation reads PV = nRT, where P = pressure of gas, V = volume of gas, n = number of moles of gas, R = gas constant, and T = absolute tem-perature of gas. Since R is a constant and n and T are held constant, the product nRT may be combined into a single constant, call it k. Then, PV = nRT = k, or PV = k which is the equation of a hyperbola. Plotting P on the abscissa axis and V on the ordinate axis, we obtain the following curve: Note: Even if we did not know PV = k is the graph of a hyperbola, we could still obtain this graph. For ex-ample, if k is a constant, P and V must vary inversely with each other. In other words, if P is large, v must be small for k to remain constant. Likewise, if V is large, P must be small.

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Question:

A racing car traveling in a circular path maintains a constant speed of 120 miles per hour. If the radius of the circular track is 1000 ft, what if any acceleration does the center of gravity of the car exhibit?

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Solution:

Since the car is traveling a circular path at constant speed, v, its acceleration is radial, and given by a = v^2/R Here, R is the radius of the circular path. Using the given data a = [(120 miles/hr)^2]/(1000 ft) = (14400 miles^2)/(1000 ft\bullethr^2) a = 14.4 miles^2/ft\bullethr^2 In order to keep units consistent, note that 1 hr = 3600 s and 1 mile = 5280 ft Then a = [14.4(5280 ft)^2]/[(1 ft)(3600 s)^2] a = 30.98 ft/s^2

Question:

Write a program in BASIC to find and print the sum of two complex numbers, inputted from the terminal. You may treat a single complex number as an ordered pair of real numbers.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G10-0223.htm

Solution:

If Z_1 and Z_2 are two complex numbers given by a + bi and c +di respectively wherei= \surd-1 , we define Z_1 + Z_2 = B (a + c) + (b + d)i. In terms of ordered pairs of real numbers, Z_1 = (a, b), Z_2 = (c, d), and Z_1 + Z_2 = (a+c,b+d), where a, b, c and d are the coordinates of the two points representing given numbers: a and c are on the real axis, b and d are on the imaginary axis The program below used the input of a particular number 999 to ter-minate the program. This is done by means of an IF .., THEN statement, which has the form IF variable = expression THEN label AND works as fol-lows: If the value of given variable equals the given expression, then the next statement to be executed is the one with the label following the word THEN. Otherwise, the computer goes to the next statement. 1\O PRINT "THIS PROGRAM ADDS TWO COMPLEX 15 PRINT "NUMBERS IN A, B FORM" 2\O PRINT 3\O PRINT "TYPE IN THE FIRST NUMBER" 4\O INPUT A, B 5\O IF A = 999 THEN 999 6\O PRINT "TYPE IN THE SECOND NUMBER" 7\O INPUT C,D 8\O PRINT "THE SUM Z1 + Z2 = (";A + C; " , " ;B + D; ")" 9\O GO TO 20 999 END

Question:

A capacitor is found to offer 25 ohms of capacitive reactance when connected in a 400-cps circuit. What is its capacitance?

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/Users/wenhuchen/Documents/Crawler/Physics/D23-0763.htm

Solution:

Capacitive resistance, X_C, is given by X_C = [1/(2\pifC)] Here X_C = 25 ohms and f = 400 cps. Then X_C = 25 ohms = {1/(2\pi × 400 cps × C)} orC = [1/(2\pi × 400 cps × 25 ohms)] Since 1 farad = 1 (coul/volt) = {1coul/(amp \textbullet ohm)} = 1 (sec/ohm) C = [1/(2\pi × 400 × 25) farad = 1.6 × 10 ^-5 farad = 16\muf

Question:

Lemon juice is very acidic, having a pH of 2.1. If we assume that the only acid in lemon juice is citric acid (represented by HCit), thatHCitismonoproticand that no citrate salts are present, what is the concentration of citric acid in lemon juice? The dissociation constant for citric acid HCit+ H_2O \rightleftarrows H_3O^+ + Cit^- is K_a = 8.4 × 10^-4 mole/liter.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E12-0417.htm

Solution:

The solution to this problem involves deter-mination of [HCit] from the expression for K_a of the reaction HCit+ H_2O \rightleftarrows H_3O^+ + Cit^-. By definition, K_a = {[H_3O^+] [Cit^-]} / [HCit], or,[Hcit] = {[H_3O^+] [Cit^-]} / K_a Since one mole of Cit^- is produced per mole of H_3O^+ produced, [Cit^-] = [H_3O^+]. [H_3O^+] can be determined from thepH.By definition, pH = - log [H_3O^+], hence [H_3O^+] = 10^-pH = 10^-2.1 mole/liter. Therefore, [H_3O^+] = [Cit^-] = 10^-2.1 mole/literand we have [HCit]= {[H_3O^+] [Cit^-]} / K_a = (10^-2.1 mole/liter×10^-2.1 mole/liter) / (8.4 × 10^-4 mole/liter) = [(10^-2.1)^2 / (8.4 × 10^-4)] mole/liter = [(10^-4.2) / (8.4 × 10^-4)] mole/liter \allequal [(6.3 × 10^-5) / (8.4 × 10^-4)] mole/liter = 7.5 × 10^-2 mole/liter = 7.5 × 10^-2 M.

Question:

A billiard ball is struck by a cue as in figure (a). The line of action of the applied impulse is horizont-al and passes through the center of the ball. The initial velocity v_0^\ding{217} of the ball after impact, its radius R, its mass M, and the coefficient of friction \mu between the ball and the table are all known. (a) How far will the ball move before it ceases to slip on the table and starts to roll? (b) What will its angular velocity be at this point?

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/Users/wenhuchen/Documents/Crawler/Physics/D04-0203.htm

Solution:

(a) Between the time the ball is struck by the cue, and the time it begins pure rolling, friction with the table decelerates it linearly, but simultaneously exerts a torque upon it about its center of mass. This causes the ball to undergo an angular acceleration. The ball begins pure rolling when its linear velocity and its angular velocity have been decreased and increased respectively to the point at which the relation v = R\omega holds. We recognize this as the definition of linear velocity with respect to angular velocity for pure rolling. The force of friction on the ball is by definition: F_f = \muN = \mumg where N = mg is the normal force between the ball and the table. The negative sign indicates that F_f is directed opposite to V_0. The ball's linear acceleration is: a = Ff/ M =- \mug Thus, its linear velocity at time t is given by: v(t) = v_0 + at = v_0 - \mugt The torque on the ball is (see figure (b)) : \cyrchar\cyrt = F_fR = \muMgR Since we know that the moment of inertia of a solid sphere about an axis passing through the center is I as 2/5 MR^2, we can calculate its angular accelera-tion: \cyrchar\cyrt = I\alpha,\alpha = \cyrchar\cyrt/I = (\muMgR)/[(2/5)MR^2] = (5/2)(\mug/R) The ball's angular velocity at time t is given by: \omega(t) = \omega_0 + \alphat,\omega_0 = 0 \omega(t) = \alphat = (5/2)(\mug/R) t To calculate the distance the ball will move before it begins pure rolling, we must first calculate how long it is after the ball has been struck that this occurs. Rolling begins when v(t) = R\omega(t) v_0 - \mugt = R[(5/2)(\mug/R)] t = (5/2) \mugt (7/2) \mugt= v_0 , t = (2/7)(v_0/\mug) The distance the ball travels is therefore, since the acceleration is constant, s = v_0t + (1/2)at^2 = v_0t - (1/2)\mugt^2 s = v_0[(2/7)(v_0/\mug)] - (1/2)\mug[(2/7)(v_0/\mug)]^2 = (2/7)[(v_0^2)/(\mug)] - (2/49) [(v^2_0)/(\mug)] = (12/49) [(v^2_0)/(\mug)] Here v_0 is the ball's initial velocity. (b) Its angular velocity at this point is: \omega(t) = \omega_0 + \alphat \omega(t) = \alphat = (5/2)(\mug/R) [(2/7) (v_0/\mug)] = (5/7)(v_0/R) Since the ball's initial angular velocity \omega_0- = 0.

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Question:

Given that 1 gram of sugar, starch, or other dry carbo-hydrate requires one liter of oxygen for complete oxi-dation and one gram of vegetable oil or other dry fat requires about 2 liters, answer the following: (a) Why is there a difference in the volume of oxygen required, (b) If mineral oil, paraffin wax, and hydrocarbons are foods would they have fewer calories per unit weight than vegetable oils? (c) Can hydrocarbons be a food source for man?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E23-0838.htm

Solution:

Consider what is meant by oxidation and the general chemical composition of the substances mentioned. Complete oxidation reactions may be defined as the addition of oxygen to an organic compound to produce water (H_2O) and carbon dioxide (CO_2) plus heat. The general formula of a fat may be written (RCO_2)_3C_3H_5 , where the R's can be the same or different molecules. In other words, a fat may be defined as an ester of glycerol and three long chain carboxylic acids. Carbo-hydrates are simple sugars or the substances that hydrolyze to yield simple sugars of general formula C(H_2O) x, where x may range from 3 to many thousands. Thus, you can proceed as follows: (a) In fats, from the general formula, you see the percentage of C and H in fats is greater than that in carbohydrates. The percentage of O is less in fats than in carbohydrates. The amount of oxygen required in combustion is a direct function of the C, H, and O percentages in the organic compound. The greater the C, H percentage, the more oxygen is required. Thus, fats should need much more oxygen for complete combustion than carbohydrates. (b) Mineral oil, paraffin wax, or hydrocarbons should yield more Calories per unit weight than vegetable oil. From (a) , you know that the amount of oxygen re-quired is proportional to the amount of heat produced upon oxidation; this is a direct function of C, H, and O percentage in the organic compound. Mineral oil, paraffin wax, and hydrocarbons have a higher C and H content than fats. (Hydrocarbons are 100% H and C whereas fats are part oxygen.) Therefore, mineral oil, paraffin and hydrocarbons yield more calories (or heat). (c) Hydrocarbonscan notbe used as a food source. Food is not useful to the body until the food particles have been degraded and oxidized by enzymes to yield heat. The human body possesses no enzyme systems that can perform the degradation of hydrocarbons.

Question:

Write a FORTRAN program to implement Gauss's elimination method for solving the systems of equations.

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Solution:

Consider the system of n linear equations with n unknowns: a_11x_1 + a_12x_2 + ... + a_1nx_n = b_1 a_21x_1 + a_22x_2 + ... + a_2nx_n = b_2(1) .. .. .. a_n1x_1 + a_n2x_2 + ... + a_2nx_n =b_n Gauss's elimination method is used to find a solution of (1) i.e., a set [x_1,...,x_n} such that when it is substituted into (1), all the equations are satisfied. The method is as follows: 1) Divide the first equation by a_11 to obtain x_1 + (a12/ a_11) x_2 + ... + (a1n/ a_11)x_n= b_1 / a_11.(2) 2) Now subtract a_21 times first equation from row 2, a_31 times the first equation from row 3,... , a_n1 times the first equation from row n to obtain x_1 + (a_12 / a_11) x_2 + ... + (a_1n / a_11)x_n= b_1 / a_11(3) w_22x_2 + ... + w_2nx_n = c_2 . . . w_n2x_2 + ... +w_nnx_n=c_n. where the result of adding - (a_j1 (a1j/ a_11)x_j) froma_ijx_ihas been written as w_ijx_j and the result of adding - (a_i1b_1 /a_11) to b_1 has been written asc_i (i= 2,...,n). The method is applied again to eliminate x_2 from the third, fourth,...,n-thequations. Repeated application yields thetridiagonal system: x_1 + v_12x_2 + ... + v_1nx_n = d_1 x_2 + v_23x_3 + ... + v_2nx_n = d_2(4) ...... x_n-1 + vn-1,nxn= d_n-1 x x_n=d_n Now the system can be solved by back-substituting. The X, Y and Z's need not be presented to solve such systems on a computer. The coefficients can be loaded into an array A(N,N+1) and solved in a similar manner. DIMENSION A (N, N+1) CAN EXTRA COLUMN IS NEEDED FIRST TO STORE. CTHE CONTINUOUSLY CHANGING VALUES (b_1....b_n) OF THE CEQUATIONS, AND THEN THE ROOTS OF THE SYSTEN READ (5,10)N DO 40 I = 1, N DO 40 J = 1, N+1 READ (5, 20) A (I, J) 40CONTINUE M = N - 1 L = N + 1 DO 60 K = 1, M K1 = K + 1 DO 60 J = K1, L DO 60 I = K1, N A (I, J) = A (I, J) - A(I, K) / A (K, K)\textasteriskcenteredA (K, J) 60CONTINUE I = N J = I + 1 DO 70 I1=1, N DO 70 J1 = 1, J A (I1, N + 1) = A (I1, N + 1) / A (I1, I1) 70CONTINUE 100I = I - 1 110J = J - 1 B (I, N + 1) = B (I, N + 1) + A (I, J)\textasteriskcenteredA (J, N + 1) IF (J.EQ.0) GO TO 120 GO TO 100 120I = N I = I - 1 J = J - 1 A (I, N + 1) = I / A (I, I)\textasteriskcentered(A (I1, N + 1) - B (I, N + 1)) 130IF (I.EQ.0) GO TO 140 GO TO 120 140WRITE (3, 30) (A (I, N +1), I = 1, N) STOP END Note: The formats 10, 20, 30 are not included because they depend on actual data.

Question:

What are some practical applications of plant hormones?

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Solution:

Auxins have a variety of practical uses and are of tremendous economic importance. Because auxins stimulate root growth, it is possible to propagate a plant vegetatively by obtaining a cut stem containing an apical meristem which can produce auxin. Root formation will occur on the cut stem because of the presence of auxin. A new plant will grow from the cut stem. By treating the female flower part with auxin, it is possible to produce parthenocarpic fruits (fruits formed without pollination and hence without seed, such as seedless oranges and grapes). Auxin-treated fruits grow larger, ripen faster, and remain on the plant until they can be harvested. Plants sprayed with synthetic auxins retain their leaves, branches, and flowers, as well as their fruits, for longer periods of time. The synthetic auxin 2, 4-D (2, 4-dichlorophenoxy acetic acid) is widely used as a weed killer. Most of the common weeds are dicotyledonous plants which are more susceptible to the auxin than are the monocots. The dicots take up the 2, 4-D and are stimulated to metabolize at such a high rate that they consume their own cellular constitu-ents and eventually die. Many of the monocots are important food sources, such as corn, oats, rye, barley and wheat. Auxins are especially useful since they destroy the dicotyledonous weeds, but leave the monocotyledonous crop untouched. Like auxins, gibberellins can cause the development of parthenocarpic fruits, including apples, currants, cucumbers, and eggplants. When treated with both auxin and gibberellin, these plants produce fruits more than twice as large as those obtainable by the application of either hormone alone. Application of gibberellin to some species of long-day (or short-night) plants induces flowering without appropriate long-day exposure. Gibberellins will also substitute for the dormancy-breaking requirement and induce germination by promoting the growth of the embryo and the emergence of the seedling. Treatment of a plant with cytokinin causes the loss of apical dominance and the plant grows bushier. It also prevents leaves from yellowing, and thus has particular commercial value in indoor, decorative plants.

Question:

Given the equilibrium H_2 S + H_2 O \rightleftarrows H_3 0^+ + HS\Elzbarand HS\Elzbar+ H_2 O \rightleftarrows H_3 O^+ + S\Elzbar2, find the final concentration of S\Elzbar2, if the final concentrations of H_3 O^+ and H_2 S are 0.3 M and 0.1 M, respectively. k_1 = 6.3 × 10\Elzbar8for H_2 S + H_2 O \rightleftarrows H_3 0^+ + HS\Elzbarand k_2 = 1 × 10\Elzbar14for HS\Elzbar+ H_2 O \rightleftarrows H_3 0^+ + S\Elzbar2.

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Solution:

This problem can be solved by writing the equilibrium constant expressions for the equilibrium. These expressions give the ratio of the concentrations of products to reactants, each raised to the power of its coefficient in the equilibrium equation. Therefore, for H_2 S + H_2 O \rightleftarrows H_3 0^+ + HS\Elzbar, we can write k_1 = 6.3 × 10\Elzbar8= {[H_3 0^+] [HS\Elzbar] } / [H_2 S] For HS\Elzbar+ H_2 O \rightleftarrows H_3 0^+ + S\Elzbar2, k_2 = 1 × 10\Elzbar14= {[H_3 0^+] [S\Elzbar2]} / [HS^\rule{1em}{1pt}] Note: water concentration is not included in the equi-librium expression since its concentration is assumed to be constant. One is not given any information concerning [HS\Elzbar]. [HS\Elzbar] is common to both k_1 and k_2 , so that if one solves one equation for [HS\Elzbar] and substitutes it into the other equation, [HS\Elzbar] is eliminated. Proceed as follows: If one solves for [HS\Elzbar] in k_1 one obtains [HS\Elzbar] = {k_1 [H_2 S] } / [H_3 0^+ ] = { 6.3 × 10\Elzbar8[H_2 S] } / [H_3 0^+ ] Substituting this into k_2 , one obtains k_2 = 1 × 10\Elzbar14= {[H_3 0^+] [S\Elzbar2]} / [HS^\rule{1em}{1pt}] = {[H_3 O^+] [S\Elzbar2]} / ({6.3 × 10\Elzbar8[H_2 S]} / [H_3 0^+ ]) Rewriting in terms of [S\Elzbar2] , [S\Elzbar2] = { (1 × 10\Elzbar14) (6.3 × 10\Elzbar8) [H_2 S] } / [H_3 0^+ ]^2 One is given that [H_2 S] and [H_3 O^+] equal 0.1 M and 0.3 M, respectively. Therefore, one can substitute these values into this equation to solve for [S\Elzbar2] . [S\Elzbar2] = {(1 × 10\Elzbar14) (6.3 × 10\Elzbar8) (0.1)} / [0.3]^2 = 7 × 10\Elzbar22M.

Question:

Differentiate clearly between "breathing" and "respiration".

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Solution:

Respiration has two distinct meanings. It refers to the oxidative degradation of nutrients such as glucose through metabolic reactions within the cell, resulting in the production of carbon dioxide, water and energy . Respiration also refers to the exchange of gases between the cells of an organism and the external environment. Many different methods for exchange are utilized by different organisms. In man, respiration can be categorized by three phases: ventilation (breathing), external respiration, and internal respiration. Breathing may be defined as the mechanical process of taking air into the lungs (inspiration) and expelling it (expiration). It does not include the exchange of gases between the bloodstream and the alveoli. Breathing must occur in order for respiration to occur; that is, air must be brought to the alveolar cells before exchange can be effective. One distinction that can be made between respiration and breathing is that the former ultimately results in energy production in the cells. Breathing, on the other hand, is solely an energy consum-ing process because of the muscular activity required to move the diaphragm.

Question:

Using conditioning, how can one determine the sensitivity of a dog's hearing to sounds of different frequencies?

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Solution:

Dogs can usually discriminate between sounds of different frequencies or pitches (measured in cycles per second, cps) . To determine how well they discriminate, one can use operant conditioning. By rewarding or punish-ing a particular act when it occurs, the probability that it will occur again is either increased or decreased. For example, one can expose the dog to a sound frequency of 1000 cps. The dog is then rewarded, whenever it responds to a sound of that frequency by barking. Soon the dog barks whenever it hears a sound of 1000 cps. In the same manner, the dog can be conditioned to raise its paw when-ever it hears sounds of another frequency. By making this frequency closer and closer to 1000 cps, and observing the dog's response, either a bark or a paw lift, one can determine the smallest differences in frequency between which the dog can discriminate. A point is eventually reached beyond, which the dog is unable to discriminate between a "bark\textquotedblright pitch and a "lift-paw" pitch. Dogs have been found to be capable of discriminating between two sounds that differ by only two cycles per second. Generalization is the response that is contradictory to discrimination. This occurs when an animal conditioned to one stimulus responds in the same manner to a similar, though different stimulus. For example, a bird conditioned to peck at a dark blue spot may peck at a violet spot when the blue one is unavailable.

Question:

Perform the following conversions (synthesis) using any inorganic reagents you require:

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Solution:

In any synthesis problem you can use a step by step approach. Try to find reactions that lead to inter-mediates from which the product can be obtained. In (A) you want to shift the hydroxyl group from the 1\textdegree carbon to the 2\textdegree carbon. You can do this through the correspond-ing alkene. Once the alkene is obtained, the addition of acid and water will generate the desired product. To obtain the intermediate, add acid to the starting material. Because it is an alcohol, it will dehydrate to the alkene. You have, then, In (B) you want to switch the CI from the 2\textdegree carbon to the 1\textdegree carbon. Convert the starting material to an alkene and from this the product can be obtained. You have, therefore, the following process: Potassium hydroxide has the ability to dehalogenate an alkyl halide to an alkene. With peroxides, the hydrogen from the HCI adds to the carbon with the least number of carbons attached. You are performing an addition reaction across a double bond. In (C), you are converting the OH group to a NH_2 group. You can convert the OH to CI, first, since upon addition of NH_3, the NH_2 group is obtained. To obtain the chlorine, dehydrate the alcohol with acid to form the alkene, then add HCI. You have the following sequence:

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Question:

In 1932, many cities showed higher sunlight levels than in the previous year. Explain this observation.

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Solution:

To answer this problem, consider the effect of atmospheric particles on the amount of sunlight received and what happened in 1932. By 1932 three years of slackened industrial activity- had passed. Then as now, industries released massive a- mounts of particulate matter, i.e. solid and liquid aerosols, into the atmosphere. These particles have the ability to scatter and absorb light. Because industrial activity was drastically reduced, fewer particles were released that could scatter or absorb the sunlight. The cities, therefore, received more sunlight in 1932 than in years when industry was thriving. It is interesting to note that because of particulate matter, cities receive about 15 to 20 % less solar radiation than do rural areas.

Question:

Gas expanding in a gas engine moves a piston in a cylinder and does 2000 ft-lb of work at a constant pressure of 3000 lb/sq ft. How much does the volume change?

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Solution:

The work done by a gas in expanding a piston a distancedsis dW= F^\ding{217} \textbulletds^\ding{217} where F^\ding{217} is the force exerted by the gas on the piston of area A. Since F^\ding{217} andds^\ding{217} are in the same direction, dW= Fds By definition of the gas pressure, p, we have p = F/A orF =pA Using this in the expression fordW, dW = p Ads= pdV wheredVis the change in volume of the gas during the expansion. The net work done by the gas, in ex-panding at constant pressure from volume V_1 to volume V_2, is W = ^V2\int_V1 p dV = p ^V2\int_V1 dV = p(V_2 - V_1) Solving for V_2 - V_1, we obtain V_2 \rule{1em}{1pt} V_1 = W/P = (2000 ft lb)/(3000 lb/ft^2) = (2/3) ft3 for the volume change of the gas.

Question:

Given the balanced equation 4NH_3 (g) + 5O_2 (g) \rightarrow 4NO (g) + 6H_2 0 (g), how many grams of NH_3 will be required to react with 80 g of O_2 ?

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Solution:

In this problem, one is asked to find the number of grams of NH_3 required to react with 80 g of O_2. From the equation, one can see that 4 moles of NH_3 react with 5 moles of O_2. This means that one must first determine the number of moles of O_2 in 80 g. One can then use the following ratio to determine the number of moles of NH_3 that will react with this number. This ratio holds because the ratio of NH_3 to O_2 is 4 : 5. (4/5) = [(number of moles of NH_3 ) / (number of moles of O_2 )] After one knows the number of moles of NH_3 required, one can determine the weight of the NH_3 by multiplying the number of moles by the molecular weight. To solve this problem: (1)calculate the number of O_2 in 80 g (2)calculate the number of moles of NH_3 that react with 80 g O_2 (3)find the number of grams of NH_3. Solving: (1)The number of moles of O_2 in 80 g is found by dividing 80 g by the molecular weight of O_2. The molecular weight of O is 16, which means the molecular weight of O_2 is twice that amount, or 32. number of moles = [(number of grams present) / (molecular weight)] number of moles = [(80 g) / (32 g/mole)] = 2.5 moles (2)The number of moles of NH_3 reacting with a certain number of moles of O_2 is in the ratio 4 : 5, as shown previously. (4/5) = [(number of moles of NH_3) / (number of moles of O_2)] (4/5) = [(number of moles of NH_3) / (2.5moles)] number of moles of NH_3 = [(4 × 5 moles) / (5)] = 2 moles 2 moles of NH_3 are necessary to react with 2.5 moles or 80 g of O_2. (3)The number of grams of NH_3 in 2.0 moles is found by multiplying 2.0 moles by the molecular weight of NH_3 MW of NH_3 = 17. number of grams = number of moles × molecular weight number of grams = 2.0 moles × 17 g/mole = 34 g.

Question:

what is the energy content of 1 gm of water?

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Solution:

If the mass of the gram of water was completely converted to energy, the amount of energy released would be E = mc^2 = 1 × 10^-3 × (3 × 10^8)^2 = 9 × 10^13 joules

Question:

Calculate the interatomic nonbonded distance between 2 bromine nuclei in a molecule of carbontetrabromide (CBr_4) . The C-Br distance is 1.94 \AA.

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Solution:

From diagram A, one can see that CBr_4 is in a tetrahedral conformation. Thus, one knows that the Br-C-Br angle is equal to 109.5\textdegree. As such, diagram B is then true. One is solving for the length ofBr_1Br_2. Angle Br_1-C-A is 109.5\textdegree / 2 or 54.75\textdegree, since one can let lineCAbe an angle bisector. To findBr_1\AA, note that sin 54.75 = B_1A / 1.94 A^o. Therefore, Br_1A= 1.94 \AA × sin 54.75 Br_1A= 1.94 \AA × .8166 = 1.58 \AA B_1B_2= 2 ×Br_1A= 2 × 1.58 \AA = 3.17 \AA, sinceBr_1A=Br_2AforCAis also a median.

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Question:

500 liters of a gas at 27\textdegreeC and 700torrwould occupy what volume at STP?

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Solution:

STP (Standard Temperature and Pressure) is defined as being 0\textdegreeC and 760torr, thus, in this problem, one is asked to find the new volume of a gas when the temperature and pressure are changed. One refers to the combined gas law in such a case. This law can be stated: For a given mass of gas, the volume is in-versely proportional to the pressure and directly pro-portional to the absolute temperature. Stated algebrai-cally (PV/T) = K where P is the pressure, V is the volume, T is the absolute temperature, and K is a constant. This means that if two of the variables are changed, the third changes so that the relation PV/T = K remains true. This means that one can now state that (P_1 V_1 )/T_1= (P_2 V_2 )/T_2 where P_1 is the original pressure, V_1 is the original volume, T_1 is the original temperature, P_2 is the new pressure, V_2 is the new volume, and T_2 is the new absolute temperature. In this problem, one is given the temperature on the Celsius scale, it must be converted to the absolute scale before using the combined gas law. This can be done by adding 273 to the temperature in \textdegreeC. T_1 = 27 + 273 = 300\textdegreeK T_2 =0 + 273 = 273\textdegreeK One knowsP_1, V_1, T_1, P_2, and T_2. One is asked to find V_2. Using the Combined Gas Laws: (P_1 V_1 )/T_1= (P_2 V_2 )/T_2P_1 = 700torr V_1 = 500 liters T_1 = 300\textdegreeK P_2 = 760torr V_2 = ? T_2 = 273\textdegreeK [(700torr× 500 liters)/(300\textdegreeK)] = [(760torr× V_2)/(273\textdegreeK)] V_2 = [(700torr× 500 liters × 273\textdegreeK)/(760torr× 300\textdegreeK)] V_2 = 419 liters.

Question:

Write Declare statements for the following variables or con-stants, making use of the rules of PL/I regarding attributes. Also explain the reasons why you selected the particularat-tributes. The variables are given as follows: a) A variable named PERCENT which takes values from 0\textbullet00 to 100\textbullet00.PERCENT is an array of 8 elements. b) A variable named AMOUNT which represents dollars and cents, from 0\textbullet00 to 1 million dollars. c) A variable named NUMBER which represents Avogadro's number= 6\textbullet022×10^23 d) A variable named TOTAL which represents integer numbersfrom 1 to 4000, to be stored in a binary form.

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Solution:

a) We can Declare PERCENT as follows: DCLPERCENT(1:8) FIXED (5,2) REAL DEC; Here, the parentheses (1:8) shows that PERCENT is an array. FIXED specifies the Scale. (5, 2) specifies the precision, which is of the form XXX\textbulletXX. REAL specifies the Mode DEC specifies the Base. b) We can Declare AMOUNT as follows: DCL AMOUNTFIXED(9, 2) DEC REAL; In the above, the words FIXED, DEC and REAL specify the scale, the Base and the Mode respectively.Also, (9, 2) can represent num-bers from -9,999,999\textbullet99 to +9,999,999\textbullet99.Thus, our range of 0\textbullet00 to 1,000,000\textbullet00 canbe easilyaccomodatedin this range. c) We can Declare NUMBER as follows: DCL NUMBERFLOAT(4) REAL DECIMAL; In the above, the value of the variable NUMBER is equal to 6\textbullet022 × 10^23. Hence, if we tried to use the FIXED Scale, we would have to specify its scaleand precision as follows: FIXED (26, 3). This huge number of almost all digits of zero, is very cum-bersome to use, store, etc. Hence, we use instead the Floating point scale as above. The specificationFloat(4) means that the number is of the form X\textbulletXXX, i.e., 4 digits, with the remaining multiplier of powers of ten implicitly provided. A printoutof the NUMBER would be: 6\textbullet022E + 23. d) We declare TOTAL as follows: DCL TOTALFIXED(12) BINARY REAL; In the above, we have specified the Scale as FIXED, the Mode as REAL, and, the Base as BINARY. The reason for specifying the precision as (12) isas follows: We are given that the variable TOTAL takes values from 0 to 4000. And we observe that a number 2^11 = 2048,2^12 = 4096, 2^13 = 8192. Hence, a precision of 2^12 is just right to represent a range from 0 to 4000.

Question:

An astronaut is to be put into a circular orbit a distance of 1.6 × 10^5 m (about 100 miles) above the surface of the earth. The earth has a radius of 6.37 × 106m. and mass of 5.98 × 10^24 kg. What is the orbital speed?

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Solution:

The force between the astronaut and the earth is: F = G (mM/R^2) = ma where G is the gravitational constant, m the mass of the astronaut and his ship, M the mass of the earth, and the letterathe centripetal acceleration of the ship. The term on the far right of the equation is just a state-ment of Newton's second law. We know that centripetal acceleration a, is equal to V^2/R where V is the instantaneous linear velocity of the orbiting body at any time, thus: G(mM/R^2) = m(V^2/R) V = [G (M/R)]^1/2 = [(6.67 × 10^-11 N - m^2/kg^2){(5.98 × 10^24 kg)/(6.37 × 10^6m)}]^1/2 = 7.91 × 10^3 m/sec [Note 1N = 1 {(kg - m)/(sec^2)}].

Question:

Suppose that an electron bounces back and forth along the x axis in a box of length 10^-10 m. (a) What is its lowest energy level? (b) What wavelength of radiation would be emitted if the electron changed from the n = 2 to the n =1 state?

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Solution:

(a) No external forces act on the electron except when it hits the walls and is reflected elastically. There-fore, it experiences no loss of energy and its momentum while it travels between the walls is constant in magnitude. Wilson and Sommerfeld stated a general rule prescribing the quantization of a system which undergoes periodic motion. This rule is \ointp_qdq = nh(1) where p_q is the momentum associated with the periodic coordinate q. (In our case, the x position of the particle is varying periodically. Hence, q = x.). Also, n is an integer, h is Planck's constant, and the integral in (1) is evaluated over 1 period of the variable q. (For this example, we evaluate (1) over 1 period of x). If we consider the electron to be at the left wall at the beginning of its cycle, then \ointp_xdx = ^L\int_0 mvdx + ^0\int_L (-mv) dx = 2mvL.(2) Note that the integral is split into 2 parts because the momentum changes direction in 1 cycle of motion. Using (2) in (1) 2mvL = nh orp = mv = (nh/2L).(3) Assuming the electron to be nonrelativistic, its energy is E = (1/2)mv^2 = (p^2/2m) = (n^2h^2) / (8mL^2). The lowest energy level occurs in the n = 1 state. E_1 = [h^2/(8mL^2)] = [{(6.625 × 10^-34)^2} / {8(9.11 × 10^-31) × 10^-20}]joule. = 6.02 × 10^-18 joule. Because 1 joule = 6.24 × 10^18 eV E_1 = (6.02 × 10^-18 joule) [6.24 × 10^18 (eV/joule)] E_1 = 37.5 eV where v is the photon frequency. (b) The energy of the photon is hv = E_2 - E_1 = [h^2/(8mL^2)] (2^2 - 1^2) hv = [{(6.625 × 10^-34 J \textbullet s)^2}/{(8) (9.11 × 10^-31 kg) (10^-20 m^2)}](4 - 1) = 18.1 × 10^-18 joule. Since 1 joule = 6.24 × 10^18 eV hv = (18.1 × 10^-18 joule) [6.24 × 10^18 (eV/joule)] hv = 113 eV = 18.1 × 10^-18 joule. From c = v\lambda, the wavelength \lambda is \lambda = (hc/E) = [{6.625 × 10^-34(3.0 × 10^8)} / (18.1 × 10^-18)]m = 1.10 × 10^-8 m = 110 A.

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Question:

GivenK_ifor acetic acid is 1.8 × 10^-5, calculate the percentage of ionization of 0.5 M acetic acid. The dis-sociation reaction is HC_2H_3O_2 \rightleftarrows H^+ + C_2H_3O_2^-.

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Solution:

K_iis the ionization constant and indicates to what degree acids and bases will dissociate in solution. Acetic acid is a weak acid becauseK_iis so small. The larger the value ofK_i, the greater the % dissociation and the stronger the acid. A useful range of values forK_iis given below. Strength Range Very strong greater than 1 × 10 ^3 strong 1 × 10 ^3 to 1 × 10^-2 weak 1 × 10^-2 to 1 × 10^-7 Very weak less than 1 × 10^-7 K_iis calculated from the ratio of products to re-actants, each raised to the power of its coefficient in the reaction equation; K_i= ({[products]^reaction ^moles ^(or ^coefficient)}/{[reactants]^ reaction ^moles ^(or ^coefficient)}) From the chemical reaction, K_i= {[H^+][C_2H_3O_2^-]}/[HC_2H_3O_2] = 1.8 × 10^-5 To solve this problem, one uses the following method. Before the reaction begins, no dissociation occurs, so that only acetic acid is present, and in its full concentration of 0.5 M. After dissociation occurs, b number of moles/liter has dissociated to the products in equal amounts. Or b moles/liter of each H^+ and C_2H_3O_2^- are produced since the coefficients of the reaction indicate that they are formed inequimolaramounts. If b moles/liter dissociate, then there are 0.5 - b moles/liter of acetic acid left. This can be summarized as follows: Before: 0.5 moles O O After: 0.5 - b b b This gives aK_iof, K_i= {[b][b]} / [0.5 - b] = 1.8 × 10^-5 . SinceK_iis so small, very few moles of acetic acid dissociate, which causes b to be insignificantly small. Therefore, 0.5 - b is approximately equal to 0.5, giving K_i= {[b][b]} / [0.5] = 1.8 × 10^-5 . Solving for b, one obtains: b^2 = 9.0 × 10^-6 b = 3.0 × 10^-3 mole of acetic acid ionized. Using the following equation to find % ionization: % ionization = [(number of moles ionized)/(original number of moles of acetic acid)] × 100 = [(3.0 × 10^-3) / (0.5)] × 100 = 0.60 %.

Question:

Define a fossil and distinguish between the two major kinds of fossils. Discuss their preservation, dating, and usefulness in tracing human evolution.

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Solution:

Fossils are remains or traces of a plant or animal found in the earth. There are basically two types of fossils: physical (anatomical) fossils and cultural fossils (archaeological artifacts). Physical fossils are those which reveal the form and structure of the animal or plant. Examples of physical fossils are bones, teeth and footprints. Cultural fossils differ in that they reveal information concerning the activities of an organism, or a group of organisms. Cultural fossils include tools and ceremonial artifacts. Successful preservation of any type of fossil depends on the material of which it is composed and the environment in which it was formed. Whereas soft tissues of the body deteriorate rapidly in the air and leave little trace, hard bones may remain preserved for millions of years. The most frequently preserved anatomical fossils are teeth, which are coated with dentin . This is one of the hardest substances found in the body. Also commonly found are the larger and thicker bones of the cranium, femur, jaw, and pelvis. Archaeological artifacts found are made of materials that are difficult to decompose. This explains why the obsidian tools (obsidian is a dark volcanic rock resembling glass) made by early man are found in large quantity, whereas no wooden tools have been uncovered from the same period. Preservation of the body of an organism is most effective when death occurs rapidly and an air supply is immediately eliminated. This may occur when an organism falls into a quicksand pit, or is trapped in an avalanche or in the lava flow of a recently erupted volcano. Once the animal has died and the soft tissues begin to decay, minerals from soil or rocks may start to infiltrate these tissues resulting in a well preserved, petrified specimen. However, this preservation of an entire organism trapped by some natural phenomenon is relatively rare. More often only fragments of fossilized teeth and bones are discovered, from which the nature of the unpreserved body parts must be deduced. Digging for fossils is usually a costly enterprise requiring a great deal of time and precision. It is therefore important that a logical site be chosen before the project has begun. Historic water sites, caves and natural shelters are suspicious sites for a digging. When fossils are found, precise maps are drawn relating their positions relative to one another and to the surroundings. Digging is done in levels since the lower a fossil is found the older it usually is. This may not be true in areas of extreme erosion or regions where mountain formation or other natural phenomena such as earthquakes have upset the ordered layering of sediment. The dating of fossils is essential in fitting together the pieces of the puzzle of man's history. The scheme of the evolutionary process can only be determined by knowing which features appeared first in the evolution leading to modern man. There are two methods of dating fossils; (1) relative dating, and (2) radioactive dating. A fossil is dated relatively when the position in which it is found is used as the reference point in time. In other words, what stratum the fossil was found in, along with the depth of the find and the other fossils occurring with it can be used to determine its relative age. Radioactive dating is more exact than relative dating. There are several different types of radioactive dating, for example, carbon-14, potassium-argon, and uranium-lead. They all work on the principle of half-lives. Fossils are dated by measuring the decrease in the level of radioactivity from the calculated original value. C\rule{1em}{1pt}14 is the only method that directly dates the fossils themselves. The others determine the approximate year the fossil was deposited by calculating the age of the rock the fossil is found in. After enough fossils have been obtained and dated, and structural differences determined, the story of man's evolution can be better understood. The story will never be complete, but some trends in structural changes leading to modern man can be observed. Some forms of pre-historic man are believed to have become extinct due to changing climate, competition with other animals (including hominoid forms), and by disease. In the end, the hominoids who were our ancestors, must have possessed distinctive selection factors that permitted their survival, per-petuation and evolution.

Question:

Give a brief summary of the events occurring in spermatogenesis and oogenesis. Compare the two processes.

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Solution:

Spermatogenesis is the production of sperm (male sex cells). It begins with spermatogonia, which are primitive, unspecialized germ cells lining the walls of the tubules in the testes. Throughout embryonic development and during childhood, the diploid spermato-gonia divide mitotically, giving rise to additional spermatogonia to provide for growth of the testes. After sexual maturity is reached, some of the spermatogonia grow and enlarge into cells known as primary spermato-cytes (see figure), which are still diploid. The primary spermatocytes then undergo the first meiotic division, each producing two equal-sized secondary spermatocytes. Each of these in turn undergoes the second meiotic division, forming two spermatids of equal size. The haploid spermatids then undergo a complicated process of maturation, including development of a tail, to become functional sperm. Oogenesis is the process of ovum (female gamete) formation. The immature germ cells of the female are called oogonia and are located in the ovary. Early in development, they undergo numerous mitotic divisions. Then some oogonia develop and enlarge into primary oocytes. At maturation, the primary oocyte undergoes the first meiotic division. The nuclear events are the same as in spermatogenesis, but the cytoplasm divides unequally to produce two cells of different sizes. The small cell is the first polar body and the large cell is the secondary oocyte. This haploid secondary oocyte undergoes the second meiotic division, which also involves unequal, cytoplasmic distribution, and forms a second polar body and a large ootid. The first polar body divides to form two additional second polar bodies. The ootid matures into the egg, or ovum. The polar bodies disintegrate and do not become functional gametes. Both spermatogenesis and oogenesis involve the production and maturation of the gametes. However, there are a few notable differences intrinsic in each process with the most important being the differences in the amount of cytoplasm and the number of viable gametes produced. The egg cell accumulates much more nutrient material during its development than does the sperm. The sperm actually loses cytoplasm during maturation. In addition, the unequal meiotic division of the oocyte results in a much larger gamete than would normal division. As a result, one viable gamete with a considerable amount of cytoplasm is produced in oogenesis. In spermatogenesis, four viable gametes with very little cytoplasm are produced. These differences are a reflection of the different roles of the egg and sperm in reproduction. It is the fertilized egg which will ultimately develop into the new individual. Large amounts of cytoplasm are necessary to provide adequate nutrition for the developing embryo. The un-equal cytoplasmic division solves the problem of reducing chromosome number without losing cytoplasm. Sperm, on the other hand, are specialized for their role in fertilization. Their small size is necessary for motility. The production of large numbers of sperm, in contrast to eggs, is to ensure the occurrence of fertilization against the odds resulting from physical barriers and environmental factors. Fertilization of the haploid ovum by the haploid sperm restores the diploid number in the resultant zygote.

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Question:

The recessive gene s produces shrunken endosperm in corn kernels and its dominant allele S produces full, plump kernels. The recessive gene c produces colorless endosperm and its dominant allele C produces colored endosperm. Two homozygous plants are crossed, producing an F_1 which are all phenotypically plump and colored. The F_1 plants are test-crossed with homozygous recessive plants and produce the following progeny: shrunken, colorless4035 plump, colored4032 shrunken, colored149 plump, colorless152 a) What were the phenotypes and genotypes of the original parents?b) How are the genes linked in the F_1? c) Cal-culate the map distance between the two gene loci.

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Solution:

In solving this problem we will apply principles that have been demonstrated in earlier problems. determined by looking at the F_1 and F_2. We are told that all the F_1 offspring are plump and colored. This indicates that the genotypes of these offspring are either SSCC or SsCC, because they show the dominant trait. Now we must decide which one of the two alternatives is correct. This is where the F_2 comes in. If the F_1 plants were SSCC, only offspring expressing the dominant trait could have been produced in the F_2, since the other parent in the test cross was a homozygous recessive. All offspring would have been heterozygous, and in such a state would have expressed only the plump and colored traits. However, four phenotypes were actually produced in the F_1, including the homozygous recessive (shrunken, colorless). Therefore, the F_1 parent had to have carried a copy of both types of recessive genes. The genotype of the F_1 is then SsCc. Since we are told that both parents of the F_1 off-spring were homozygous, their genotypes must be either SSCC and sscc, with phenotypes plump, colored and shrunken, colorless respectively or SScc and ssCC with phenotypes plump, colorless and shrunken, colored re- spectively. We illustrate this with a diagramatic cross: b)Now, looking at the ratios of offspring in the F_2 generation, we see that Now, looking at the ratios of offspring in the F_2 generation, we see that there is a significant departure from the expected 1 : 1 : 1 : 1 ratio when a test cross of a heterozygote with a recessive homozygote is performed. This is indicative of crossover. (See previous problems for more detailed explanation of this and other concepts of linkage). Looking at the ratios in the F_2 generation, we see that the larger groups, which represent the parental linkage combinations, are plump, colored and shrunken, colorless. Each of these separate combinations represents the combination of genes on the parental chromosomes. There-fore, the parental genes must have been linkedSCandsc. We have determined the F_1 genotype to be SsCc. Therefore, in the F_1, the genes are linked . In order to prove this, let us look at how this configu-ration could indeed give rise to the progency in F_2. In meiosis, can produce four types of gametes if crossover occurs: The homozygous recessive can only produce gametes of the sc type. In the cross, then: To calculate the map distances, we first determine the recombination frequency (RF) between the two genes: RF = [(number of recombinants) / (total number of progeny)] = (149 + 152) / (149 + 152 + 4035 + 4032)] = (301) / (8368) = 0.036 An RF of 0.036 would give a map distance of 0.036 × 100 or 3.6 map units. Hence the map distance between the locus for shape and the locus for color is 3.6 map units apart.

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Question:

H_2O_3, hydrogen trioxide, a close relative of hydrogen peroxide, has recently been synthesized. It is extremely unstable and can be isolated only in very small quantities. Write a Lewis electron dot structure for H_2O_3.

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Solution:

A Lewis dot structure of a compound shows the arrangement of valence electrons. Valence electrons are defined as an element's outer electrons which participate in chemical bonding. Thus, to write an electron dot structure for H_2O_3, calculate the total number of valence electrons. This can be done by considering the electronic con-figurations of H (hydrogen) and O (oxygen). Hydrogen possesses one valence electron while oxygen has 6 valence electrons. In H_2O_3, a total of (2) 1 + (3) 6 = 20 valence electrons are involved. With this in mind, the Lewis electron dot structure becomes Notice, you have represented the required 20 valence electrons.

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Question:

A 1000-gram metal block fell through a distance of 10 meters . If all the energy of the block went into heat energy, how many units of heat energy appeared?

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Solution:

Work is given by the force acting on the block times the distance travelled by the block, when force and distance are in the same direction. Hence, Work = (980 × 1000)dynes× 1000 cm = 98 × 10^7 erg. Since 1 joule = 10^7 ergs, 98 × 10^7 ergs equals 98 joules. There are 4.19 joules/cal. Therefore, Heat=(98 joules)/(4.19 joules/cal) = 23.5 cal approximately.

Question:

Let the force constant k of the spring in the figure be 24 Let the force constant k of the spring in the figure be 24 newtons /meter, and let the mass of the body be 4 kgm. The body is initially at rest, and the spring is initially unstretched. Suppose that a con-stant force p^\ding{217} of 10 newtons is exerted on the body, and that there is no friction. What will be the speed of the body when it has moved 0.50 meters?

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Solution:

The equations of motion with constant acceleration cannot be used, since the resultant force on the body varies as the spring is stretched. However, the speed can be found using the work-energy theorem. This states that the work done by all the forces acting on a body is equal to the change in kinetic energy of the object. Then, if v_-F is the final speed of the block, work = \intp\ding{217}\bullet dx\ding{217}+ \intF\ding{217}\bullet dx\ding{217}= (1/2) m(v2_F - 0^2) = ∆E_Ksince the initial kinetic energy of the body is zero. P is constant, the angle \ding{217} \ding{217} \ding{217} \ding{217} between P^\ding{217} and dx is 0\textdegree and the angle between F^\ding{217} and dx^\ding{217} is 180\textdegree. Also the magnitude of the restoring force is F = kx, by Hookes law. Then P\intdx - \intkx dx= (1/2) mV2_F Px - 1/2 kx^2= (1/2) mV2F 10n ×0.5m - (1/2) × 24(n/m) × 0.25m^2 = 1/2 ×4kg × V2_F orV_F =1(m/sec)

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Question:

Determine the quantity of heat required to convert 10 g of ice at 0\textdegreeC to vapor at 100\textdegreeC, For water, heat of fusion = 80 cal/g, heat of vaporization = 540 cal/g, and specific heat = 1 cal/g -\textdegreeC.

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Solution:

Heat is absorbed when ice is transformed from the solid to the liquid state. The heat absorbed when 1 g of solid melts is called the heat of fusion. The heat of fusion for water is 80 cal/g. In calculating the amount of heat needed to raise the temperature of the liquid water, the specific heat must be used. The specific heat is de-fined as the number of calories of heat necessary to raise the temperature of a gram of a substance 1\textdegreeC. The specific heat of water is 1\textdegree, thus 1 calorie is needed to raise the temperature of 1 gram of water 1\textdegreeC. The quantity of heat necessary to convert 1 g of a liquid into a vapor is termed the heat of vaporization. For water, 540 calories are ne-cessary to change 1 g of liquid water at 100\textdegreeC into vapor (steam) at 100\textdegreeC. In this problem three steps are involved. 1) The ice must be melted. 2) The water must be heated from 0\textdegreeC to 100\textdegreeC. 3) The water must be vaporized to steam. Heat is absorbed in each of these steps, so that each must be accounted for. Melting the ice: Because the heat of fusion is 80 cal/g for water, one knows that for every gram of ice melted, 80 cal of heat must be absorbed. Here, there are 10 grams of ice be-ing melted, so the specific heat must be multiplied by 10 to find the amount of heat absorbed. 10 g × 80 cal/g = 800 cal. Therefore, 800 cal of heat are absorbed when the ice is melted. Heating the liquid: To calculate the amount of heat absorbed when the water is heated, the specific heat should be taken into account. The specific heat of water is 1 cal/g\textdegreeC. This means that for every degree of temperature each gram of water is raised, 1 cal of heat is absorbed. In this case, 10 grams of water are raised 100\textdegree. Therefore, the specific heat is multiplied by 10 to take into account the weight of the water and then multiplied (again) by 100 to account for the fact that the water is being raised 100\textdegree. 1 cal/g-deg × 10 g × 100\textdegree = 1000 cal. Therefore, 1000 calories are absorbed when the liquid is heated to 100\textdegreeC. Vaporization: To find the amount of heat absorbed when the water is vaporized, the heat of vaporization must be used. For water, the heat of vaporization is 540 cal/g, which means that for every gram of water vaporized, 540 calories are absorbed. Here, 10 g of water is vaporized so that the heat of vaporization must be multiplied by 10 g to find the amount of heat absorbed. 10 g × 540 cal/g = 5400 cal. To calculate the total amount of heat absorbed, the heat absorbed in all three steps must be added together. melting800 cal heating1000 cal vaporization5400 cal 7200 cal or 7.20 Kcal. Therefore, 7200 calories are needed to melt 10 g of ice, heat it to 100\textdegreeC and vaporize it to steam.

Question:

A chemist wants to calculate Avogadro's number by the inspection of a solid cube of AgCl. The density of the cube is 5.56 g/cm^3. The spacing between the Ag^+ and Cl^- ions in the cube is 2.773 × 10^-8 cm from their centers. From these data, perform this calculation.

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Solution:

To solve this problem, the following quantities must be determined: the volume of one mole of AgCl, cubic edge length of one mole, the number of ions on an edge, and the total number of ions. The total number of ions divided by 2 will yield Avogadro's number. From the atomic weights of Ag and Cl (107.868 amu and 35.453 amu) one mole of AgCl weighs 143.321 grams. Because density = mass/volume, the volume of one mole = (143.321 g/mole) / (5.56 g / cm^3) =25.78 cm^3 / mole. The volume of a cube = (e)^3, where "e" is the length of the cube edge. Therefore, e = ^3\surd{25.78 cm^3/moles} = 2.954 cm / ^3\surdmole. Because the spacing between ions is 2.77 × 10^-8 cm, the number of ions along the edge is (2.954 cm / ^3\surdmole.)^3 / (2.773 × 10^-8 cm) = 1.065 × 10^8 ions / ^3\surdmole. The total number of ions in 1 mole of AgCl must be {(1.065 × 10^8/ ^3\surdmole)^3} = 1.209 × 10^24mole. This number is not Avogadro's number. Because you have two ions per formula unit, there is one Avogadro's number of Ag^+ and one Avogadro's number of Cl^-. It follows, therefore, that you must divide by two. As such, (1.209 × 10^24 / mole) / 2 = 6.04 × 10^23 / mole = Avogadro's number.

Question:

Nitrogen reacts with hydrogen to form ammonia (NH_3). The weight-percent of nitrogen in ammonia is 82.25. The atomic weight of hydrogen is 1.008. Calculate the atomic weight of nitrogen.

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Solution:

When elements combine to form a given compound, they do so in a fixed and invariable ratio by weight. In a given amount of NH_3, 82.25% of its weight is contributed by nitrogen. The weight-percent of hydrogen in ammonia is, then 100 - 82.25 = 17.75%. One mole of nitrogen and 3 moles of hydrogen combine to form one mole of ammonia. Therefore, one mole of nitrogen constitutes 82.25% and 3 moles of hydrogen constitute 17.75% by weight of one mole of ammonia. The atomic weight of an element is equal to the weight of one mole of that element. The following ratio can be set up to solve for the atomic weight (AW) of nitrogen: AW_N / (3 × AW_H) = weight-percent N / weight-percent H Solving for AW_N AW_N= [(3 × AW_H) (weight-percent N)] / (weight-percent H) = [3 (1.008) (18.25)] / 17.75 = 14.01.

Question:

The captain of a submarine fitted with a directional transmitter finds that he receives bad echoes from the sea bed when he is submerged if the transmission di-rection makes an angle greater than 45\textdegree with the vertical [see Figure (A)]. He wishes to transmit a message to shore using radiation which is completely horizontally polarized. How far from the coast should he surface if the receiving point is a house on top of a coastal cliff 500 ft high and he intends to polarize his radiation by reflection on the sea surface? Take the location of the transmitter as 12 ft above the water when the submarine has surfaced.

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Solution:

If the transmission direction makes a small angle with the vertical, most of the radiation passes through the water surface into the air and only a small fraction is reflected back to the ocean bed to produce echoes. The amount reflected increases with the angle until, when the critical angle is reached, reflection is total, and the echo becomes troublesome. The re-fractive index of water for the radiation used is found from Snell's Law, n sin \texttheta_i = n_air sin \texttheta_r where \texttheta_i and \texttheta_r are the angles of incidence and re-fraction for the transmitted signal. We are told that the critical angle is \texttheta_i = 45\textdegree. At the critical angle, \texttheta_r = 90\textdegree. Hence, n = (1/sin 45\textdegree) = \surd2 = 1.414. When the submarine has surfaced, it can produce a completely plane-polarized beam by reflection of the radiation at the Brewster angle from the surface of the sea. The angle required is tan \textphi = n = 1.414.That is, \textphi = 54.75\textdegree. [See figure (B)]. Thus, the distance of the submarine from the cliff is d = s_1 + s_2 From figure (B), however, (12ft/s_1) = tan (90 \rule{1em}{1pt} \textphi) = tan 35.25\textdegree and(500 ft/s_2) = tan (90 - \textphi) = tan 35.25\textdegree ord = (500 ft/tan 35.25\textdegree) + (12 ft/tan 35.25\textdegree) = (512/0.707) ft = 724 ft.

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Question:

If all the chemical components found in a living cell were to beput together in the right proportion, the mixture would not functionas living matter. What characteristics would enable usto consider something living?

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/Users/wenhuchen/Documents/Crawler/Biology/F02-0063.htm

Solution:

Living things are characterized by a specific, complex organization: thesimplest functional unit of organization is the cell. The cell possesses a membrane, called the plasma membrane, which delimits its living substancesfrom the surroundings. It also has a nucleus which is a specializedregion of the cell controlling many important life processes. Living things have to consume energy in order to stay alive. This energymust come from food; plants (with some exceptions) have the abilityto manufacture their own food from sunlight and inorganic materials inthe soil and air. Animals, lacking this ability, have to depend ultimately onplants for food. Complex food substances are taken in and degraded intosimpler ones; and in the process, energy is liberated to carry out life sustainingfunctions. Living things are characterized by their ability to move. The movementof most animals is quite obvious - they crawl, swim, climb, run orfly. The movements of plants are much slower and less apparent, but nonethelessdo occur. A few animals, for example sponges and corals, aresedentary - they do not move from place to place. To procure food and othernecessities of life, these animals rely on the beating of hair-like projections, called cilia, to move their surroundings (from which they obtaintheir food) past their body. The ability to respond to physical or chemical changes in their immediatesurrounding is another charac-teristic of living things. Higher animalshave special-ized cells or complex organs to respond to certain typesof stimuli. Plants also show irritability; the VenusFlytrapis a remarkableexample. The presence of an insect on the leaf of this plant stimulatesthe leaf to fold; the edges come together and ensnare the fly. The fly provides a source of nitrogen for the VenusFlytrap. Living things grow, that is, increase in cellular mass. This growth canbe brought about by an increase in the size of individual cells, by an increasein the number of cells, or both. Some organisms, such as most trees, will grow indefinitely, but most animals have a definite growth period whichterminates in an adult form of a characteristic size or age. Another characteristic of living things is the ability to reproduce. No organismcan live forever; hence there must be some way for it to leave descendantsthat can continue its kind on earth. There are a variety of reproductionmechanisms. Some simple organisms split into two after havingattained a certain size. Higher organisms have complex reproductivepatterns involving partners of the opposite sex. The ability of a plant or animal to adapt to its environment is anothercharacteristic of living things. It enables an organism to survive theexigencies of a changing world. Each kind of organism can adapt by seekingout a favorable habitat or mode of life or by undergoing changes tomake itself better fit to live in its present environment.

Question:

Two different compounds of elements A and B were found to have the following composition: first compound, 1.188 g of A combined with 0.711 g of B; second compound, 0.396 g of A combined with 0.474 g of B. (a) Show that these data are in accord with the law of multiple proportions, (b) If the formula for the first compound is AB_2, what is the formula for the second?

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Solution:

(a) The law of multiple proportions can be stated: When two elements combine to form more than one compound, the different weights of one that combine with a fixed weight of the other are in the ratio of small whole numbers. This means that if one solves for the expected amount of B that is used in forming the second compound from the ratio of A : B in experiment one, the experimental amount should be a multiple of the calculated value. This is seen more clearly after looking at the data. In experiment 1, A combines with B in a ratio of 1.188 g A : 0.711 g B or 1 : .598. In experiment 2, A combined with B in a ratio of 0.396 A : 0.479 B or 1 : 1.20. The law of multiple proportions states that .598 should be a small multiple of 1.20. 1.20/.598 = 2, thus the law is supported. (b) If the formula for the first compound is AB_2, one knows that the proportion of the number of moles of A to B is 1 : 2. Thus 1 unit volume of A weighs 1.188 g and 2 units of B weigh 0.711 g. Therefore, 1 unit of B weighs 0.711 g/2 = 0.356 g. Using this data, one solves for the number of units of A in the second compound by dividing the weight of A in compound 2 by 1.188 g. no. of units of A in compound 2 = (.396g) / (1.188g/unit)= .311 units no. of units of B in compound 2 = (.474) / (.356 g/unit) = 1.331 units The ratio of A to B in compound 2 can now be found. Let x = no. of B atoms A / B = .311 / 1.331 = 1 / x x = 1.331 / .311 = 4.0 Therefore, the second compound is AB_4. This result could also have been obtained by using the data from part (a) . It was determined that twice as much B is present in the second compound. Thus, if the first compound is AB_2, then the second compound must be AB_4.

Question:

Consider the decay of the positively chargedpion, \pi^+ . Pion \ding{217}muon+ neutrino. The energy equivalent of the mass of thepionis about 140 MeV, and themuonand neutrino produced in the decay have rest masses corresponding to energies of 106MeVand zero, respect-ively. How much kinetic energy is available to the particles created in the decay?

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Solution:

The difference of the energy equivalents of the masses of the pion and its decay products is 140 MeV - 106 MeV = 34 MeV. Let us consider the frame of reference in which the pion was at rest. The total energy of the system is given by the energy equivalent of the pion mass. After the decay, the total mass of the decay products is less than thepionmass. The principle of con-servation of energy requires this mass difference to be transformed into another form of energy, hence the two daughter particles (muonand neutrino) share it as kinetic energy.

Question:

Find the speed of acompressionalwave in an iron rod whose specific gravity is 7.7 and whose Young's modulus is 27.5 × 10^6 1b/in^2.

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Solution:

A wave in an iron rod will move with a speed v = \surd(E/\rho) where E is Young's modulus, and \rho is the specific weight of iron. E = 27.5 × 10^6 1b/in ^2 = 27.5 × 10^6 × (1b/in^2) × (144 in^2/ft^2) where we've used the fact that 1^2 1b/in ^2 = 144 1b/ft^2 \rho = (D/g) = {(7.7 × 62.4 1b/ft^3)/(32 ft/sec^2)} = 15.0 slugs/ft^3 Hence, v = \surd{(27.5 × 10^6 × 144 1b/ft^2)/(15.0 slugs/ft^3)} = 1.6 × 10^4 ft/sec.

Question:

Describe the three major structural distinctions between DNA and RNA.

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Solution:

DNA or deoxyribonucleic acid is in the form of a double helix having a deoxyribose sugar and phosphate backbone. The two helices are linked together by hydrogen bonds between nitrogenous bases bound to the sugar moiety of the backbone. In DNA, the bases are adenine, guanine, cytosine and thymine. RNA or ribonucleic acid differs from DNA in three important respects. First, the sugar in the sugar-phosphate backbone of RNA is ribose rather than deoxyribose. Ribose has hydroxyl groups on the number 2 and 3 carbons, whereas deoxyribose has a hydroxyl on the number 3 carbon only (see figure 1). Secondly, RNA has only a single sugar-phosphate backbone with attached single bases. Although hydrogen bonding may occur between the bases of a single RNA strand causing it to fold back on itself, it is not regular like that of DNA, where each base has its complement on the other strand. Thus, RNA is similar to a single strand of DNA. Finally, the pyrimidine base uracil (see figure 2) is found in RNA instead of the pyrimidine base thymine found in DNA. Therefore, the nitrogenous bases present in RNA are adenine, guanine, cytosine and uracil.

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Question:

A 20 g chunk of Dry Ice (CO_2) is placed in an "empty\textquotedblright 0.75 liter wine bottle and tightly corked. What would be the final pressure in the bottle after all the CO_2 has evaporated and the temperature has reached 25\textdegreeC?

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Solution:

The final pressure in the bottle will equal the original pressure plus the pressure contributed by the CO_2. Since the bottle was originally open in atmos-pheric pressure, the original pressure is 1 atm. The pressure contributed by the CO_2 is found by using the Ideal Gas Law. This law is stated PV =nRTorP =nRT/ V where P is the pressure of the gas, n is the number of moles, R is the gas constant, 0.082 liter-atm/mole -\textdegreeK, V is the volume, and T is the absolute temperature. One is given R and V in the problem. n is found by dividing the number of grams of CO_2 present by the molecular weight (MW of CO_2 = 44 ). no. of moles = 20 g / (44 g/mole) = .45 moles One can convert \textdegreeC to \textdegreeK by adding 273\textdegreeto the temperature in \textdegreeC. \textdegreeK = \textdegreeC + 273\textdegree T = 25 + 273\textdegree= 298\textdegreeK Solving for P_(CO)2 : P_(CO)2 = [(.45 moles × .082 liter -atm/ mole - \textdegreeK × 298\textdegreeK) / (.75 liter)] = 14.81 atm. P_total= P_(CO)2 +P_original P_total=14.81atm+ 1atm= 15.81 atm.

Question:

What is the structure arid function of the chloroplasts in green plants?

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Solution:

The chloroplasts have the ability to transform the energy of the sun into chemical energy stored in bonds that hold together the atom of such foods (fuel) as glucose. By the process of photosynthesis, carbon dioxide and water react to produce carbohydrates with the simultaneous re-lease of oxygen. Photosynthesis, which occurs in the chloro-plasts, is driven forward by energy obtained from the sun. Photosynthesis involves two major sets of reactions, each consisting of many steps. One set depends on light and cannot occur in the dark; hence, this set is known as the light reaction. It is responsible for the production of ATP from sunlight and the release of oxygen derived from water. This process is known as photophosphorylation. The other set, referred to as the dark reaction, is not dependent on light. In the dark reaction, carbon dioxide is reduced to carbo-hydrates using the energy of ATP from the light reactions. Chlorophyll, the pigment contained in chloroplasts which gives plants their characteristic green color, is the molecule responsible for the initial trapping of light energy. Chlorophyll transforms light energy into chemical energy; then it passes this energy to a chain of other compounds involved in the light and dark reactions. Plastids are organelles which contain pigments and/or function in nutrient storage. Chloroplasts are but one example os a plastid; they give the green color to plants. Each chloroplast is bounded by two layers of membranes; the outer one is not connected with the other membranes of the cell. The inner membrane gives rise to the complex internal system of the chloroplast. Surrounding the in-ternal membranes is a ground substance termed the stroma. The stroma contains the enzymes which carry out the dark reactions of photosynthesis. Granules containing starch are also found in the stroma. The internal membrane system of the chloroplast are usually in the form of flattened sacs called thylakoids (see accompanying diagram) . A stack of thylakoids is called a granum (plural - grana). The lamel-lae contain the chlorophyll molecules and other pigments involved in photosynthesis. The layered structure of the granum is extremely crucial for the efficient transfer of energy from one pigment molecule to another without the dissipation of a great deal of energy in the process.

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Question:

It is easy to understand how selfish behavior can be explained by natural selection. But how can one account for altruism?

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Solution:

Selfish behavior is exemplified in the formation of dominance hierarchies. The high-ranking animals eat before the other animals when food is scarce and take preference in the selection of mates and resting places. Consequently, these individuals are generally healthier, live longer and reproduce more frequently. Natural selection favors these strong animals, in that they produce more offspring having their superior traits. Selfish behavior is expected to be transferred to future generations since it leads to the more successful survival of the animal. Altruism is self-sacrificial behavior that benefits other members in a group. An altruistic animal reduces its own fitness, and therefore its chances of survival and reproduction, in order to increase the success of another animal or of the group as a whole. For example, a male baboon remains to fight a leopard while the troop escapes. Natural selection would not favor the continuation of this costly behavior since those who do show it are less likely to survive and reproduce. One would therefore expect the genes for altruism to be lost, but they are not. The reason for this lies in the fact that the male baboon - only risks its life for baboons in his own troop, not for baboons in other troops. His behavior therefore benefits only those baboons related to itself; these baboons probably share common genes, among which are those for altruistic behavior. The al-truistic behavior thus preserves the genes in the other members that are shared with the altruistic member by common descent. For the altruistic genes to be favored in natural selection, the benefit that the related members receive must outweigh the loss of the altruistic animal. Although the baboon might die and lose its ability to pass on its genes, the baboon's brother can compensate for the loss. To do so, it must have more than twice as many offspring as a result of its brother's altruistic behavior. (Since about 1/2 of the genes of a given animal are shared by its brother or sister.) Similarly, since first cousins share 1/4 of their genes, on the average, compensation would involve quadrupling the survivor's offspring. Altruism directed at a variety of relatives, from offspring to cousins, results in a summation of the benefits so that the altruistic gene(s) may be preserved and the related behavior may evolve. Thus, the existence of altruism can be accounted for by natural selection. To demonstrate the compensation by a relative for the loss of an altruistic relative, we will assume that altruism is determined by one allele, A. If one offspring of F_1 generation dies, its brother can compensate by doubling its offspring, (F_2). This will double the frequency of A in F_2 (2 × 1/4) to give 1/2, the frequency that A would have had in F_1 if the brother had not died. Similarly, if one altruistic offspring in the F_2 generation dies, its cousin (also in F_2) can compensate by quadrupling its offspring, F_3. Four times the frequency of A in F_3 (4 × 1/8) gives 1/2, which would be the frequency of A in F_1 if the cousin had not died. It is important to note that compensation must always be referred back to the F_1 generation, since that is the only way to prevent a loss of the altruistic gene.

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Question:

A proton is accelerated from rest for 1 nanosecond [= 10^\rule{1em}{1pt}9 s] by an electric field E = 3 x 10^4 volts/meter. What is the final velocity (see figure)?

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Solution:

The general equation of motion involving time, constant acceleration, and velocity is v_f = v_0 + at. The acceleration is given by the electric force, eE, divided by the mass of the proton. v_0, the initial velocity, is zero. Therefore, the expression for v_f is: vf= at = (eE/Mp) t e = 1.6 x 10^\rule{1em}{1pt}19 coulombs E = 3 × 10^4 volts/meter M_p = 1.67 × 10^\rule{1em}{1pt}27 kg t= 1 x 10^\rule{1em}{1pt}9 sec v_f = [(1.6 × 10^\rule{1em}{1pt}19)(3 × 10^\rule{1em}{1pt}4)(1 × 10^\rule{1em}{1pt}9)] / [1.67 × 10\rule{1em}{1pt}27] \approx 2.9 × 10^3 meters/sec

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Question:

A sample of gaseous krypton, maintained at constant press-ure , is found to have a volume of 10.5 l at 25\textdegreeC. If the system is heated to 250\textdegreeC, what is the resulting volume?

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Solution:

We need a relationship between volume and temperature. Such a relationship is provided by Charles' Law, which states that volume V and absolute temperature T are proportional, or, as an equality, V =kT, where k is a constant. We can determine k for our system from the initial volume and temperature . Thus, K = (V/T) = [(10.5 l)/(25\textdegreeC)] = [(10.5 l)/(298.15\textdegreeK)] = 0.0352 l-\textdegreeK^-1 The volume corresponding to a temperature of 250\textdegreeC, which is 523.15\textdegreeK, is V =kT= 0.0352l-\textdegreeK^-1 × 523.15\textdegreeK = 18.41 l.

Question:

Describe the data types available in Pascal.

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Solution:

Pascal has 4 standard data types. Used to represent signed and unsigned integers. ex: -25, 600, 37523 Used to represent real numbers. ex: -7.18, 3.1459, 2.718, 1.414, 17,0 Also in scientific notation: 3.14E02, 3.14E-02 3. Char. Used to represent character data. ex: 'a' , ' '(blank), '$' , ')' 4.Boolean. Has only two values - TRUE and FALSE. In addition to these four standard types, there are enumer-ated, or user defined types. ex: TYPE DAYOFWEEK = (sun, mon, tue, wed, thu, fri, sat); color = (red, blue, green, yellow); There are also subrange types, which are some ranges of in-teger, char, or enumerated types. ex: TYPE WORKDAY = (mon . . fri); Letters = ('A' . . 'Z'); DIGITS = (0 . . 9); All of the above types are scalar types. In addition to these, Pascal also has structured data types, composed of a number of scalar types. Examples are sets, arrays and records.

Question:

A steel shaft 12 ft long and 8 in. in diameter is part of a hydraulic press used to raise up cars in a garage. When it is supporting a car weighing 3200 lb, what is the decrease in length of the shaft? Young's modulus for steel is 29 × 10^6 lb\bulletin^-2.

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Solution:

The formula for Young's modulus is Y = (Stress)/(Strain) = (F_n/A)/(∆l/l_0) where F_n is the normal force at the end of the shaft compressing it, A is the cross sectional area of the shaft, l0is the length of the shaft when no stress is present and ∆l is the change in length due to the compression. A is given by \piR^2 = (\piD^2)/4, where R and D are the radius and diameter of the cross section of the shaft. Thus, the decrease in length is ∆l = (F_nl_0)/(YA) = (3200 lb × 144 in.)/(29 × 10^6 lb\bulletin^2 × 16\pi in^2) = 3.16 × 10^-4 in.

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Question:

Write the structural formulas for all chlorine derivatives having molecular formula C_5H_11CI.

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Solution:

In hydrocarbon derivatives, the hydrogen atoms are replaced by other atoms or groups of atoms, such as oxygen, chlorine, hydroxyl (-OH), or nitro (-NO_2), which produce chemically active centers in an otherwise less active molecule. To solve this problem, one first writes the structural isomers with five carbons in a continuous chain, in which each different hydrogen atom of the alkane has been replaced by a chlorine atom: Their names are (a) 1-chloropentane, (b) 2-chloropentane, (c) 3-chloropentane. Second, one writes the structural isomers with four carbons as the parent molecule, in which two hydrogen atoms have been replaced by a chlorine atom and a methyl group: Their names are (d) 1-chloro-2-methylbutane, (e) 1- chloro - 3-methylbutane, (f) 2-chloro -2-methylbutane, and (g) 2-chloro -3-methylbutane. Third, one writes the structure of the isomer with three carbons for a parent molecule, in which three hydrogen atoms of the three carbon chain have been re-placed by a chlorine atom and two methyl groups: This compound's name is 1-chloro -2, 2-dimethylpropane or 1-chloro neopentane. Hence, there are eight structural isomers with the molecular formula C_5H_11CI.

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Question:

Determine the phase velocity of the de Broglie waves associated with a neutron which has an energy of 25 eV.

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Solution:

If the neutron of mass m has energy 25 eV = 25 × 1.062 × 10^-19 J = 4.00 × 10^-1 8 J, then its speed is given by the relation (1/2) mv^2 = 4.00 × 10-1 8Jor v = \surd[(8.00 ×10-1 8J)/(1.67 ×10-2 7kg)] = 6.92 × 10^4 m \textbullet s^-1 . The phase velocity of the associated de Broglie waves is then v_p = (c^2/v) = [{(3.00 × 10^8)^2m^2 \textbullet s^-2}/(6.92 × 10^4 m \textbullet s^-1)]. = 1.30 × 10^12 m\bullets^-1 The difference between the phase and group velocities can be seen in the figure. When sinusoidal waves of different frequency (or wavelength) are combined, they appear as in the figure, with their amplitude modulated. Clearly, the frequency of variation of the amplitude of this resultant wave is less than the frequency of variation of the wave enclosed by the amplitude envelope. Similarly, the velocity of the envelope (v_group) is different from the velocity of the enclosed wave (v_phase). However, the 2 are related by v_phase = (c^2/v_group) where c is the speed of light in the case of de Broglie waves. Since the envelope locates the approximate po-sition (due to the uncertainty principle) of the particle, the group velocity of the wave must represent the actual velocity of the particle, or the particle's wave would not be able to keep up with the particle.

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Question:

Explain the buffering action of a liter of 0.10 M acetic acid containing 0.1 mole of sodium acetate. In the ex-planation, use ionic equations.

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Solution:

By buffering action, one means the ability of a substance to maintain relatively constant conditions of pH in the face of changes that might otherwise affect the acidity orbasicityin solution. For example, a weak acid, such as acetic acid, will dissociate according to the following equation: HOAc+ H_2O \rightleftarrows H_3O^+ +OAc^- . The sodium acetate particle will be completely ionized in solution: NaOAc\rightleftarrows Na^+ +OAc^-. These are the two processes that occur in this buffer. Suppose one increases the acetate concentration. By doing this the equilibrium is shifted to the left, i.e., to acetic acid. k_a = {[H_3O^+][OAc^-]} / [HOAc]or[H_3O^+] = ([HOAc] / [OAc^-]) k_a . The key to the buffering action is this [HOAc] to [OAc^-] ratio. If [HOAc] changes, [OAc^-] will change accordingly so that the same value of the ratio is obtained. Thus, the ratio is a constant. This means, therefore, that [H_3O^+] is maintained as a constant. pH = - log [H^+], it is also constant.

Question:

Write a trapezoidal integration routine in FORTRAN to evaluate^.4\int0 sin2 xdx+ ^.3\int_.1 cos2 xdx.

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Solution:

Let us structure the problem as follows: there shall be a main routinewhich calls a trapezoid function routine, which in turn references twofunction routines defining the functions sin2x and cos2x. First note that accordingto the trapezoid rule ^b\int_af(x)dx\approx (h/2)[f(a) + 2f(a+h) + 2f(a+2h) + ... + 2f(a+(n-1)h)+f(b)], whereh = (b-a)/n is the number of trapezoids. Thus we have FUNCTIONTRAP(A,B,N,FX) H = (B - A)/N SUM = 0 K = N - 1 DO 6 I = 1,K 6SUM = SUM +FX(A + I\textasteriskcenteredH) TRAP = (FX(A) + FX(B) + 2.\textasteriskcenteredSUM)\textasteriskcentered(H/2.) RETURN END The main calling program and function routines are given below. EXTERNAL FX1, FX2 APPROX =TRAP(0.,.4,5,FX1) + TRAP(.1,.3,5,FX2) PRINT, APPROX STOP END FUNCTION FX1(X) FX1 = (SIN X)\textasteriskcentered\textasteriskcentered2 RETURN END FUNCTION FX2(X) FX2 = (COS X)\textasteriskcentered\textasteriskcentered2 RETURN END Notice that the main program uses an EXTERNAL statement. This declara-tionmust be used in every calling program which passes the name ofa subprogram or built-in function to another subprogram. Also rememberthat the main program is entered first. After the END statement inthe main program, all subroutines may be entered. In this program, it is up to you to define the accuracy of the integra-tion. In otherwords, you choose the value of N, which determines the number of iterationsto be done.

Question:

If you administer too much insulin to a diabetic, it may cause a condition called "insulin shock". Explain this condition. Why does it come about?

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Solution:

To solve this problem, you must know how insulin, a hormone, (chemicals that stimulate or depress certain functions of the body) works in the body. The addition of insulin in the body stimulatesglycogenesis, a process which removed glucose from the blood and converts it to glycogen. Insulin also facilitates the transport of glucose across the membranes of muscle cells and adipose (fatty) tissue. If too much insulin is added, too much glucose is removed from the blood. The concentration of glucose in the blood becomes inadequate, a condition called hypoglycemiaresults.This"insulin shock" or hypoglycemia may be characterized by all or any of the following symptoms: weakness, trembling, profuse perspiration, rapidheart beat, delirium, and loss of consciousness.

Question:

Distinguish between viviparous, ovoviviparous, and oviparous reproduction.

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Solution:

The females of all birds, amphibians and bony fish, most insects, and many aquatic invertebrates lay eggs from which the young eventually hatch; such animals are said to be oviparous (egg-bearing). In such cases, the major part of embryonic development takes place outside the female body, even though fertilization may be internal. The eggs of oviparous animals therefore contain relatively large amounts of yolk, which serves as a nutrient source for the developing embryo. Mammals, on the other hand, have small eggs with comparatively little yolk. The mammalian embryo develops within the female's body, deriving nutrients via the maternal bloodstream, until its development has proceeded to the stage where it can survive independently. Such animals are termed viviparous (live-bearing) . The third type of reproduction is intermediate in character. This is ovoviviparous reproduction and it in-volves the production of large, yolk- filled eggs which remain in the female reproductive tract for considerable periods following fertilization. The yolk of the egg is the nutrient source for the developing embryo, which in this case, usually forms no close connection with the wall of the oviduct or uterus and does not receive nourishment from the maternal blood. A diverse group of animals, including snails, trichina worms, flesh flies, sharks, and rattle-snakes are in this category.

Question:

Give a list of the built-in single-precision mathematical functions that can be used in FORTRAN language.

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Solution:

NAME FUNCTION TYPE OF ARGUMENT TYPE OF RESULT ABS(X) IABS(I) SQRT(X) SIN(X) COS(X) ATAN(X) ATAN(X,Y) ASIN(X) ACOS(X) TAN(X) COT(X) ALOG(X) AL0G 10(X) EXP (X) GAMMA(X) INT(X) FLOAT(I) IFIX(X) AMOD(X,Y) MOD(I, J) SIGN(X,Y) ISIGN(I,J) DIM(X,Y) IDIM(I,J) ERF(X) ERFC(X) AMAX 1(U,V,X...) MAX 0(I,J,K...) AMIN 1(U,V,X...) MIN 0(I,J,K...) Absolute value of X Absolute value of I Square root of X Sine of X Cosine of X Principal value of arc Tangent of X Arc-tangent of X/Y Principal value of arc sine of X Principal value of arc cosine of X Tangent of X Cotangent of X Natural logarithm of X Common logarithm of X e^X Gamma function of X Integer part of X Converts integer to real Converts real to integer Remainder of X/Y Remainder of I/J Sign of Y times ABS(X) Sign of J times IABS(I) Difference between X and the smaller of X and Y Difference between I and the smaller of I and J Error function of X Complement error function of X Largest of real constants Largest of integer constants Smallest of real constants Smallest of integer constants real integer real real real real real real real real real real real real real real integer real real integer real integer real integer real real real integer real integer real integer real real real real real real real real real real real real real integer real integer real integer real integer real integer real real real integer real integer

Question:

Through several successive reactions, a chemist uses carbon,CaO,HCland H_2 O to produce C_6 H_4, Cl_2. Assuming an efficiency of 65 %, how much C_6 H_4 Cl_2 can be produced from 500 grams of carbon? Assume that 1/3 of the carbon is lost as 3 moles CO.

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Solution:

In solving this problem, you must account for all carbon atoms and employ the mole concept. You need not be concerned with the actual sequence of reactions nor the roles ofCaOand H_2 O. Dichlorobenzene, C_6 H_4 Cl_2, consists of 6 carbon atoms. You can determine that there were originally 9 moles of carbon present since 3 moles of CO are produced and the carbon present in the CO represents 1/3 of the original amount of carbon present. A mole is defined as weight in grams/molecular weight. The molecular weight of carbon is 12. You started with 500 grams. Therefore, the number of moles of carbon is 500/12. It is stated above, however, that for every 9molesof C, 1 mole of C_6 H_4 Cl_2 was produced. Therefore, the number of moles of C_6 H_4 Cl_2 is 1/9 of the moles of carbon that you started with. Namely, the number of moles of C_6 H_4 Cl_2 is 1/9 × 500/12. The problem calls for an efficiency of 65 %. Therefore, we must multiply this number of moles by (65/100)or,(1/9) × (500/12) × (65/100). The molecular weight of C_6 H_4 Cl_2 is 147. Recalling the definition of a mole, the weight of C_6 H_4 Cl_2 produced is 147 × (1/9) × (500/12) × (65/100) = 442 grams.

Question:

A sealed glass bulb contained helium at a pressure of 750 mm Hg and 27\textdegreeC. The bulb was packed in dry ice at - 73\textdegreeC. What was the resultant pressure of the helium?

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Solution:

The only parameters mentioned in this problem are pressure (P) and temperature (T) . One is given the initial pressure and temperature and the final temperature. One is asked to determine the pressure at - 73\textdegreeC. The Law of Gay-Lussac relates temperature and pressure. It can be stated mathematically (P_1 /T_1 ) = (P_2 /T_2 ) where P_1 is the initial pressure, T_2 the initial absolute temperature, P_2 the final pressure and T_2 the final absolute temperature. The temperature in \textdegreeC is converted to \textdegreeK by adding 273. T_1 = 27\textdegreeC + 273 = 300\textdegreeK T_2 = - 73\textdegreeC + 273 = 200\textdegreeK Solving for P2 [(750 mm Hg)/(300\textdegreeK)] = [(P_2 )/200\textdegreeK)] P_2 = [{(750 mm Hg) (200\textdegreeK)}/(300\textdegreeK)] = 500 mm Hg.

Question:

A brass rod of density 8.5 g\textbulletcm ^\rule{1em}{1pt}3 and length 100 cm is clamped at the center. When set into longitudinal vibration it emits a note two octaves above the fundamental note emitted by a wire also of 100-cm length weighing 0.295 g and under a tension of 20 kg weight which is vibrating transversely. What is Young's modulus for brass?

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Solution:

The rod vibrates with its center clamped (see figure). Its fundamental frequency of vibration is thus such that the center of the rod is a node and each of the ends an antinode. The length of the rod, L, is half a wavelength (\lambda). Thus \lambda = 2L. Further, the speed of sound in the rod is c = \surdY/\rho, where Y is Young's modulus for brass and \rho its density. Hence the frequency of vibration is f = (c/\lambda) = (1/2L) \surd(Y/\rho) For the vibrating wire the length is the same and, if \mu is the mass per unit length of the wire (\mu = m/L) the frequency of the fundamental vibration is f^1 = (1/2L) \surd(S/\mu) = (1/2L) \surd(SL/m), where m is the mass of the wire. But f = 4f_1, since one frequency is two octaves above the other. Hence (1/2L) \surd(Y/\rho) = (4/2L) \surd(SL/m) orY = (16\rhoSL)/m = [(16 × 8.5 g\textbulletcm ^\rule{1em}{1pt}3 × 2 × 10^4 kg × 981 cm\textbullets ^\rule{1em}{1pt}2 × 100 cm)/(0.295 g)] = 9.04 × 101 1dynes\textbulletcm^\rule{1em}{1pt}2 .

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Question:

A current of 30 amp Is maintained in a thin, tightly wound coil of 15 turns, with a radius of 20 cm. What is the magnetic induction at the center of the coil?

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Solution:

The magnetic induction at the center of a coil of radius R with one turn can be found by using the Biot-Savart Law. dB^\ding{217} = [(\mu_0 i)/(4\pi)] [(dl^\ding{217} × R^\ding{217})/(R^3)] Where the permeability constant \mu_0 = 4\pi × 10^[-7\omega/(a-m)] , and dl^\ding{217} is an element of the coil in the direction of the current i. From the figure we have dB = [(\mu_0 i)/(4\pi)] [(dlR sin 90 o )/(R^3 )] = [(\mu_0 i)/(4\pi)] where R^\ding{217} and dl^\ding{217} are 90 o apart. We have used the magnitude of dB^\ding{217} because all the infinitesimal contributions to the magnetic induction from the infinitesimal lengths dl^\ding{217} are in a direction perpendicular to the plane of the coil. Therefore the total magnetic field at the center is the sum of the infinitesimal contributions dB or B = \intdB = [(\mu_0i)/(4\pi)] (1/R^2) \intdl = [(\mu_0i)/(4\pi)] [(2\piR)/(R^2) = (\mu_0/2) (i/R) The magnetic induction at the center of a coil containing N coils will be equal to the sum of the contributions due to each of the coils, or B_T = NB = [(\mu_0N^i)/(2R )] = [(4\pi × 10^-7 weber)/(amp-m)] [(15 × 30 amp)/(2 × 0.02m)] = 1.4 × 10^-3 weber/m^2 The direction of B_T is perpendicular to the plane of the orbit.

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Question:

The hydrated K^+ and Cl^- ions are approximately spherical in shape, with a diameter of 6.0 \AA. What fraction of the total volume of the water phase of an E. coli cell is occupied by these ions?

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Solution:

A hydrated ion is an ion that is completely surrounded by water molecules. Water gathers around ions because of the strong electrostatic attraction between the water dipoles and the ions. The O atom of a water molecule is slightly negative and is attracted to the positive K^+ ion and the H atom of the water is slightly positive and is attracted to the negative Cl^- ion. The concentration of KCl in an E. coli cell is (150 × 10^-3) M. Because KCl dissociated, into K^+ and Cl^\rule{1em}{1pt} ions, the ion concentration in the cell is 2(150 × 10^-3 M) . If one knows the volume of the aqueous phase of an E. coli cell, one can determine the number of moles of the ions present. An E. coli is cylindrical; it is 2.0\mu long and has a diameter of 1.0\mu. The volume of a cylinder is equal to l\pir^2, where l is the length of the cylinder and r is the radius of its cross-section. 1\mu = 10^-4 cm. Volume of a single E. coli cell = (2.0 × 10^-4 cm) (\pi) × (0.5 × 10^-4 cm)^2 = 1.57 × 10^-12 cm^3 = 1.57 × 10^-15 l 80% of the E. coli cell is in the aqueous phase. and the K^+ and Cl^- ions will only exist in this aqueous phase. Volume of the aqueous phase = .80 × (1.57 × 10^-15 l) = 1.26 × 10-15l . The number of moles of ions can now be found. Moles of ions = (2) (150 × 10^-3 moles / liter) (1.26 × 10^-15 l) = 3.78 × 10^-16 moles There are 6.02 × 10^23 ions per mole. No. of ions = (3.78 × 10^-16 moles) (6.02 × 10^23 ions / mole) = 2.28 × 10^8 ions. Each ion is taken to be spherical with a diameter of \AA. The volume of a sphere is equal to 4/3 \pir^3 , where r is the radius. 1 \AA = 10^-8 cm. Volume of 1 ion = (4 / 3) \pi(3.0 × 10^-8 cm)^3 = 1.131 × 10^-22 cm^3 The total volume occupied by the ions is 2.28 × 10^8 ions times the volume of 1 ion. Total volume of ions = (2.28 × 10^8 ions) × (1.131 × 10^-22 cm^3 / ion) = 2.58 × 10^-14 cm^3 The fraction of the total volume of the cell occupied by the ions is equal to the volume of the ions divided by the volume of the cell. fraction of volume occupied by the ions = (2.58 × 10^-14 cm^3) / (1.57 × 10^-12 cm^3) = 1.65 × 10^-2 or about 2%.

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Question:

Calculate and print out the area of a rectangle whose width and length vary in integral steps from 1 to 4.

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Solution:

In this problem a "FOR - NEXT\textquotedblright loop is being introduced. It has a general form of: FOR variable = expr_1 TO expr_2 STEP expr_3 ..................... NEXT variable where expr_1 is the initial value of the variable, expr_2 - the final value, and expr_3 indicates the value by which the variable has to be incremented after each pass through the loop. If the STEP part is omitted, it is assumed to be 1. We use two nested "FOR ... NEXT" loops as shown below. 1\O REM CALCULATES RECTANGLE'S AREA 2\O FOR W = 1 TO 4 3\O FOR L = 1 TO 4 4\O PRINT "L ="; L; "W = "; W; "A ="; L\textasteriskcenteredW 5\O NEXT L 6\O NEXT W 7\O END

Question:

The molar absorptivity of phytochome (a light-sensitive pigment thought to control the process of flowering in plants) is 76,000. What will be the absorbance of a 5.0 × 10^-6 molar aqueous solution of phytochome if the pathlength of the cell is 1.0 cm?

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Solution:

The fraction of the incident light absorbed by solution at a given wavelength is related to the thickness of the absorbing layer and to the concentration of the absorbing species. These two relationships are combined into the Lambert-Beer Law, given in integrated form as log (I_0 / I) = acl, where I_0 is the intensity of the incident light. I is the intensity of the transmitted light, a is the molar absorb-ing index (the molar extinction coefficient), c is the concentration of the absorbing species in moles per liter, and l is the thickness of the light absorbing sample. Log (I_0 / I) is also known as the absorbance A. Solving for A, A = acl A = (76,000) (5.0 × 10^-6 moles / liter) (1 cm) = 0.38.

Question:

Find the pH of a 0.1 M solution of ammonia, NH_3.pK_b= 1.76 × 10^-5.

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Solution:

This problem involves the dissociation of a weak base, which proceeds by the following general equation: B + H_2O \rightleftarrows BH^+ + OH^-, where B is the base and BH^+ and OH^- are the dissociation products, x moles of B dissociate into x moles of BH^+ and x moles of OH^-. This is more clearly seen below: NH_3 + H_2O \rightleftarrows NH_4^+ + OH^- Before:0.1 MOO After:0.1 - xxx If x moles of NH_3 dissociate, then, at equilibrium, one has 0.1 - x left, since 0.1 is the initial concen-tration. NH_3 is a weak base and very littledissociation occurs, which means x is extremely small. This allows one to approximate 0.1 - x as 0.1. By now, it should be evident that [H^+] does not appear in the equilibrium equation. The strategy will be to first find [OH^-], then calculating thepOH, and then using the equationpK=pOH+ pH = 14 to findpH. In order to calculate the value of [OH^-], set up the equilibrium constant equation for which K_b = 1.76 × 10^-5. Thus,K_b = {[NH_4^+] [OH^-]} / [NH_3]. Substituting, K_b = [(x)(x)] / [0.1] = 1.76 × 10^-5, so that x = [OH^-] = 1.33 × 10^-3. Then, one can calculatepOHby: pOH = - log [OH^-] = - log [1.33 × 10^-3] = 3 - log 1.33 = 2.88. Therefore, pH = 14 -pOH = 14 - 2.88 = 11.12.

Question:

Compute the electrostatic force of repulsion between two \alpha\rule{1em}{1pt}particles at a separation of 10^\rule{1em}{1pt}11 cm, and compare with the force of gravitational attraction between them.

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Solution:

For this problem, we use Coulomb's Law which states that the electrostatic force between two charges is directly proportional to the product of the two charges and inversely proportional to the square of the distance by which the charges are separated. The constant of proportionality is written as 1/4\pi\epsilon_0,where \epsilon_0 has the value of 8.85 ×10^\rule{1em}{1pt}12coul^2 /nt \rule{1em}{1pt} m^2 . Each \alpha-particle has a charge of +2e, or 2 × 1.60 × 10^\rule{1em}{1pt}19 = 3.20 × 10^-19 coul. The force of repulsion at a separation of 10^\rule{1em}{1pt}11 cm or 10^\rule{1em}{1pt}13 m is F = [1 / (4\pi\epsilon_0)] (qq' ) / (r^2) = 9 × 10^9[(3.20 ×10^\rule{1em}{1pt}19)^2 / (10^\rule{1em}{1pt}13)^2 ] = 9.18 × 10^-2 newton and since 1 newton = 10^5 dynes, F = 9180 dynes. This is a sizable force, equal to the weight of nearly 10 grams ! To find the force of gravitational attraction we use Newton's Law of Universal Gravitation. This has the same form as Coulomb's Law(an inverse square law) but instead of charges we have masses and the constant of proportionality is called G. The mass of an \alpha-particle (2 protons + 2 neutrons) is 4 × 1.67 x 10^\rule{1em}{1pt}24 = 6.68 x 10^\rule{1em}{1pt}24 gm = 6.68 x 10^\rule{1em}{1pt}27 kgm. The gravitational constant G is G = 6.67 × 10^\rule{1em}{1pt}11 (newton\rule{1em}{1pt}m^2 / kgm^2). The force of gravitational attraction is F = G(mm' / r^2 ) = 6.67 × 10^\rule{1em}{1pt}11 [(6.68 × 10^\rule{1em}{1pt}27)^2 / (10^\rule{1em}{1pt}13)^2 ] = 2.97 × 10^\rule{1em}{1pt}37 newton. The gravitational force is evidently negligible in comparison with the electrostatic force.

Question:

An object is placed 12 centimeters from a convex mirror whose focal length is 6 centimeters. Where will the image be formed?

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Solution:

First, we construct a ray-diagram. Here we trace the path of two rays - one parallel to the axis, and one through the center of curvature. The image is virtual and smaller than the object (for the convex mirror this is true for all positions of the object). We may now solve mathematically. D_0 is 12 centimeters, f is - 6 centimeters. Sub-stituting in (1/D_0) + (1/D_I) = (1/f) (1/12 cm) + (1/D_I) = (1/- 6 cm) whence D_I = - 4 cm D_I is negative implies that the image is on the same side of the mirror as the object and is virtual.

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Question:

Describe, by both formal and ionic equations, what takes placewhen 0.10 mole of sodium sulfite is added to 1 liter of 1 M hydrochloric acid.

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Solution:

The formal equation for the reaction is Na_2SO_3 + 2HCI \rightarrow 2NaCl + H_2O + SO_2. The ionic equation gives more detailed information than this. The soluble sodiumsulfite salt will react with excess hydrochloric acid according to the ionicreaction (2Na^+, SO_3^2-)+ 2(H_3O^+, CI^-) \rightarrow 2(Na^+, CI^-) + H_2SO_3 (ag) + H_2O. Sulfurous acid, H_2SO_3, exists in equilibrium with both sulfur dioxide and hydroniumions: H_2SO_3 \rightleftarrowsSO_2 (g) +H_2O , SO_3^2- + 2H_3O^+ \rightleftarrowsH_2SO_3 . SO_2 gas will escape from solution, thus pulling the re-action towards completion, to give the products 2(Na^+ , CI^- ), SO_2, and 2H_2O.

Question:

How do transpiration and the theory of water cohesion explain the process of water transport in plants? When does root pressure become important in water transport?

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Solution:

Water transport in plants can be explained by transpiration and the theory of cohesion-tension. When a leaf transpires, water is lost from the leaf cells, molecule by molecule . As a consequence of the water loss, theosmolarityof the leaf cells increases and water molecules enter these cells by osmosis from adjacent cells with a lowerosmolarity. These adjacent leaf cells in turn receive water from neighboring leaf cells and so on, which all ultimately get their water from the xylem cells in the vascular tissue. Since water molecules are linked to each other by hydrogen bonds into continuous columns within the xylem vessels right down to the root tip , a molecule leaving a xylem cell to enter amesophyllcell of the leaf will necessarily tug this entire column of water along behind it. Because water has sufficient tensile strength, the column does not snap when being pulled . The upward movement of this column causes a negative pressure to develop within the xylem. This negative pressure is similar to the suction produced while sucking a drink through a straw. This draws water molecules from the root tip cells into the vessels. This theory of water movement in the xylem is known as the cohesion-tension theory. In the root tip, water molecules from the soil pass down a gradient through the root hairs to the cells that have just lost water to the xylem. In this way, the upward stream of water is continuous. In short, while transpiration provides a pull at the top of the plant, the strong tendency of water molecules to stick together attributed to hydrogen bonds (cohesive force of water molecules) transmits this force through the length of the stem and roots and results in the movement of the entire water column up the xylem . In the spring, before leaves are formed and trans-piration becomes important , root pressure is probably the major force bringing about the rise of sap. In addition, under conditions of high humidity, still air currents, or extremely low temperatures, root pressure may become important in raising water to the leaves of some plants. Thus root pressure may contribute to the upward movement of water under some conditions, but it probably is not the principal force causing water to rise in the xylem of most plants most of the time.

Question:

What are the distinguishing characteristics of cardiac and smoothmuscle.

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Solution:

Smooth muscle fibers are considerably smaller than skeletal musclefibers. Each smooth muscle fiber has a single nucleus located in thecentral portion of the cell. By contrast, each skeletal muscle fiber is multi-nucleatedwith the nuclei located peripherally. The most noticeable morphologicalfactordistingushingsmooth from either skeletal or cardiac muscleis the absence of striated banding patterns in the cytoplasm of smoothmuscle. Smooth muscle does not contain myofibrils. However, myosinthick filaments andactinthin filaments can be seen distributed throughoutthe cytoplasm, oriented parallel to the muscle fiber, but not organizedinto regular units of filaments as in skeletal and cardiac muscle. It is believed that the molecular events of force generation in Smooth muscle cells are similar to those in skeletal muscle. Smooth muscleexhibits wide varia-tions in tonus; it may remain almost entirely relaxedor tightly contracted. Also, it apparently can maintain the contractedcondition of tonus without the expenditure of energy, perhaps owingto a reorganization of the protein chains making up the fibers. Smooth muscle cells are not under voluntary control. Their activity is underthe regulation of the autonomic nervous system and they have an abilityto perform work for long periods of time, since their contraction is slowand sustained. Smooth muscle is often referred to as a visceral musclesince it regulates the internal environment of a great many systemsand organs. It is found in the walls of hollow organs such as the intestinaltract, the bronchioles, the urinary blad-der, and the uterus. Vascular smooth muscle lines the walls of blood vessels. It may also be foundas single cells dis-tributed throughout an organ such as the spleen or insmall groups of cells attached to the hairs in the skin. Cardiac muscle has properties similar to those of skeletal muscle. It is multinucleated and striated, with its thick and thinmyofilaments organizedinto myofibrils. The sliding-filament type of contraction is also foundin cardiac muscle. Unlike skeletal muscle but like smooth muscle, cardiacmuscle is involuntary. Each beat of the heart represents a single twitch. Cardiac muscle has a long refractory period. Consequently, it is unableto contracttetanically, since one twitch cannot follow another quicklyenough to maintain a contracted state. A unique feature of cardiac muscleis its inherentrhyth-micity; it contracts at a rate of about 72 beats perminute. The muscle is innervated by nerves, but these nerves only serveto speed up or slow down the inherent cardiac rhythm. In addition, cardiacmuscle is unique in having intercalated discs, or tight junctions, betweencells; these aid in the transmission of electrical impulses throughout the heart.

Question:

Calculate the length (in \AA) of a polypeptide chain con-taining 105 amino acid residues if (a) it exists entirely in \alpha-helical form, or (b) it is fully extended.

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Solution:

(a) Peptide chains may assume the \alpha-helix con-figuration spontaneously, because this form is stable and has the least free energy, providing there are no opposing inter-actions of the R groups or of the solvent. The \alpha-helix is pictured in Figure A. As shown, when the peptide chain assume this configuration each amino acid residue contributes 1.5 A^\textdegree to the length of the chain. If a chain in the \alpha-helical con-figuration contains 105 amino acid residues, its length is 1.5 \AA × 105 = 157.5 A. (b) If the chain is fully extended, each amino acid residue contributes its full length to the chain. Figure B shows that the length of an average amino acid is 3.6 \AA. The length of the chain in this conformation is 3.6 \AA × 105 = 378.0 \AA.

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Question:

Give an example of a distributed lag model for the national economy, using the national accounting scheme. Is the computer essential for dealing with distributed lag models?

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Solution:

In simulation studies, a problem in the real world is solved using a simple mathematical model in conjunction with the computer. Distributed lag models belong to the class of discrete simulation models. Such models change only at fixed intervals of time, and current values of a given variable are determined by previous values of the same variable. Using Keynesian macroeconomic analysis we can construct a model of the national economy. Let C be consumption, I be investment, T be taxes, G be government expenditure and Y be national income. Then, a general macroeconomic model would be: C = a + c(Y-T) I = b + rY(1) T = tY Y = c + I + G Here, "a" represents consumption that is independent of income while "c" is the marginal propensity to consume. "c" represents the amount per dollar that is spent on con-sumption. "b" is investment that is independent of income, and "r" is the marginal propensity to invest. Finally, "t" is the tax rate per dollar of income. Note that o \leq c, r, t \leq 1. Since simulation deals with concrete situations, we should give values to the parameters in (1). Thus, C = 30 + 0.7(Y - T) I= 3+ 0.1(Y)(2) T = . 2Y Y = C + I + G. The above static model can be made dynamic by picking a fixed time interval and then expressing the current values of the variables in terms of values at previous intervals, or lagging them. Thus, we can lag all the variables in (2) by a year to obtain C = 30 + 0.7(Y-1- T_-1) I = 3 + 0.1(Y_-1)(3) T = 0.2Y_-1 Y = C_-1 + I_-1 + G_-1 Model (3) is to be understood as follows: Current consump-tion depends on disposable income from the previous year. Investment is a function of the previous year's national income, as is taxes. In large-scale models, like (3), most calculations can be done without the aid of the computer. But at lower levels of aggregation (say where investment in different industries is considered separately), the computer is indispensable.

Question:

A copper kettle, the circular bottom of which is 6.0 in. in diameter and 0.062 in. thick, is placed over a gas flame. On assuming that the average temperature of the outer surface of the copper is 214\textdegreeF and that the water in the kettle is at its normal boiling point, how much heat is conducted through the bottom in 5.0 sec? The thermal conductivity may be taken as 2480 Btu/(ft^2 hr F\textdegree/in).

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Solution:

The heat Q conducted through the bottom of the kettle in time t is given by Q = K At (∆T/∆x) where ∆T/∆x is the temperature gradient in F\textdegree/in, K is the thermal conductivity and A is the area of the bottom in ft^2. We have A = \pir^2 = \pi[(30/12) ft]^2 = 0.20 ft^2 t = 5.0 sec = (5.0/3600) hr = 0.0014 hr The temperature on the inside of the bottom of the kettle is the same as that of boiling water (212\textdegreeF). Since the temperature on the outside of the bottom is 214\textdegreeF and the thickness of the bottom of the kettle is 0.062 in, the temperature gradient across the bottom is ∆T/∆x = (214\textdegreeF - 212\textdegreeF)/(0.62 in) = 32\textdegreeF/in The heat conducted through the bottom is then Q = [2480 {(Btu)/(ft^2 hr \textdegreeF/in)}] (0.20 ft^2)(0.0014 hr)(32\textdegreeF/in) = 22 Btu

Question:

Examine the DATA DIVISION of Fig. 1 and explain the usage and effects of File Description entries.

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Solution:

The DATA DIVISION of a COBOL program is divided into sections such as WORKING-STORAGE section, and the FILE Section. The File Section always precedes the Working-Storage Section and includes descriptions of all the files and their associated records. File Section header begins with margin A. In the given example program FD means File Description and it describes 2 files, PARTS-FILE and OUTFILE. Both PARTS-FILE and OUTFILE must have been specified in an ENVIRONMENT DIVISION SELECT Statement. Following this, LABEL and DATA records are also included and are very impor-tant in the FD section. Every FD entry must include a LABEL RECORDS clause. Many files use labels to tell the computer where the particular file begins and ends. These records may by unique to that file, they may be standard, or in some files they may be omitted. For unit record files, such as card files or printer files, the label clause reads LABEL RECORDS ARE OMITTED. The DATA RECORD clause specifies the record name associated with the file being described. The DATA RECORD clause is not required in most computer systems since the description of the associated data record must directly follow each FD entry. However, the LABEL RECORDS clause is always required for every FD entry. The data records P_RECORD and SUMMARY are used in the PROCEDURE DIVISION to store the data read into or out of their asso-ciated files.

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Question:

Discuss the structural and hormonal aspects of insect metamorphosis.

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Solution:

Basically, there are two kinds of metamorphosis: incomplete metamorphosis which results in a similiar but larger form, and complete metamorphosis, which gives rise to strikingly different forms. Many insects, such as moths, butterflies and flies undergo metamorphosis. Generally, a wormlike larva \rule{1em}{1pt} called a caterpillar in moths, maggot in flies, and grub in bees - hatches from the egg. The larva is a relatively active form; it crawls about, eats voraciously, and molts several times, each time becoming larger in size. The last larval molt gives rise to a pupa, an inactive form which typically does not move or eat. Moth and butterfly larvae spin a cocoon around their bodies and molt within the cocoon to form the pupae. During this process all the structures of the larva are broken down and used as raw materials for the development of the adult. Each part of the adult body develops from a group of cells called a disc. The discs are embryonic cells derived directly from the egg and remain quiescent during larval stages. During the pupal stage they grow and differentiate into adult structures such as wings, legs, and eyes. These structures remain collapsed, folded and thus nonfunctional at first; when the pupa molts into the adult form, blood is pumped in to inflate them, and chitin is deposited to make them hard. Metamorphosis in insects is characterized not only by sharp changes in appearance from the larval to the adult form, but also by striking changes in their modes of life. The butterfly larva eats leaves whereas the adult feeds on nectar from flowers. The mosquito larva lives in ponds and eats algae and protozoa, while the adult sucks the blood of humans and other mammals. The process of insect metamorphosis involves a brain hormone, a molting hormone called ecdysone, and a regulatory hormone called the juvenile hormone. The brain hormone stimulates glands in the insect's prothorax (the part of the body immediately behind the head, to which the first pair of legs is attached). The prothoracic glands respond by secreting ecdysone which induces molting. Ecdysone is believed to be involved also in many growth and developmental processes in insects. Juvenile hormone is produced by a pair of glands, called corpora allata, which is located just behind the brain and closely associated with it. High concentrations of juvenile hormone at the time of molting result in another immature stage following the molt, while low concentrations lead to a more advanced stage after molting. This is supported by the demonstration that removal of the corpora allata from insects in a highly immature stage resulted in pupation at the next molt, followed by a molt that resulted in a midget adult. Conversely, implantation of an active corpora allata into an advanced pupa resulted in another immature stage after molting rather than an adult. In the larva, juvenile hormone is present in a high concentration, hence the larva molts to form a pupa. In the pupa, juvenile hormone is absent and thus when the pupa molts an adult results. Notice that insect metamorphosis involves the nervous system as well as endocrine glands. Of the three hormones discussed, one is released directly by the brain (the brain hormone), one is secreted under the influence of the brain hormone (ecdysone),and the third is secreted by glands closely associated with the brain (juvenile hormone). The interactions between these three hormones are shown in the accompanying figure.

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Question:

Briefly discuss the major functional unit of the Central Processing Unit (CPU).

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Solution:

The CPU (Central Processing Units), or simply PU or PE (Processing Element) since there may be multiple cooperating units, is the engine that executes the instructions (software). The conventional CPU, also referred to as the Von Neumann type processor is sequential. It fetches , decodes, and executes in-structions one at a time. Generally, the CPU (or simply the processor) consists of three major functional units: Control Unit (CU): Controls the fetch, decode, execute cycles for instructions stored in memory. Arithmetic Control Unit (ALU): Accomplishes arithmetic operations and logical comparisons of words of data (a word of data is as long as the length of register in bits). Internal Storage Unit (ISU): This is the fast internal memory, not to be confused with primary memory that is external to a processor chip, that temporarily stores and manipulates data. Typically, it consists of a well defined set of regis-ters. The registers hold small chunks of data (8 bits, 16 bits , 32 bits, 64 bits) to be manipulated. In addition, this unit contains the address and data busses, in a way analogous to highways, for data to move data to and from memory. Finally, this unit contains busses (a bunch of wires) for I/O, as when a signal needs to be sent to CPU (e.g. and I/O interrupt ) regard-ing the status of an attached I/O device (also referred to as peripheral).

Question:

Let (X_i Y_i) be a set of points in the X,Y plane. Devise a flow-chart for a program that will find the regression line \^{y} =\alpha˄+ \beta˄ x_i that minimizes the sum of the squared \alpha˄ deviations from the line.

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Solution:

The simple linear regression model is as shown in Fig.1. It is required to find the equation of the straight line, \^{y} =\alpha˄+ \beta˄ x_i. The \alpha˄ unknowns are\alpha˄, the y- intercept and \beta˄- the slop of the line or the \alpha˄ regression coefficient. Assuming that such a line exists, it is required to minimize ^n\sum_i=1 (y_i - \^{y})^2 , the sum of squared y_i deviations. But (y_i - \^{y}) = y_i - (\alpha˄+ \beta˄ x_i). \alpha˄ Thus, the problem is to Minimize^n\sum_i=1 (y_i -\alpha˄+ \beta˄ x_i)^2(1) \alpha˄ where x_i ,y_i are known data points and\alpha˄, \beta˄ are the variables. \alpha˄ Taking the partial derivatives of (1) with respect to\alpha˄, \beta˄, setting them \alpha˄ equal to zero to find the minimum of (1) and solving the result-ing set of linear equations yields \beta˄ = [^n\sum_i=1 (y_i -y) (x_i -x)] / [^n\sum_i=1 (x_i -x)^2] \alpha˄=y- \beta˄x \alpha˄ \alpha˄ where y= (1/n)^n\sum_i=1 y_iandx= ^n\sum_i=1 x_i. In order to convert the above program from algebraic to computer form, the following steps must be performed: 1.The points xi,y_i are read and stored in arrays X and Y. 2.X,Yare calculated. 3.\sum(x_i -x)^2 , \sum(y_i -y)^2, \sum(y_i -y) (x_i -x) are calculated. 4.\alpha˄, \beta˄ and \sum(y_i - (\alpha˄+ \beta˄ x_i)^2 are computed. \alpha˄ \alpha˄ \alpha˄ The flow-chart of the procedure is as shown in Fig. 2

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Question:

A 1200-lb sled is pulled along a horizontal surface at uniform speed by means of a rope that makes an angle of 30\textdegree above the horizontal (see figure (a)). If the tension in the rope is 100 lb, what is the coefficient of friction?

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Solution:

Since the sled is being pulled at constant velo-city, there are no unbalanced forces. We break up the ten-sion in the rope into components parallel and perpendicular to the horizontal (see figure (a)). By Newton's second law \sumF_x = 0 therefore F_friction = \muN = T cos 30\textdegree(1) \sumF_y = 0 therefore N + T sin 30\textdegree = 1200 N = 1200 lb - T sin 30\textdegree(2) From (1), \mu = (T cos 30\textdegree)/N Substituting (2) into this expression, \mu = (T cos 30\textdegree)/(1200 lb - T sin 30\textdegree) = [(100 lb)(.866)]/[1200 lb - (100)(1/2)lb] = (86.6)/(1150) = 0.0753.

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Question:

Define the term mutation. What types of mutations can be distinguished and how may each be produced?

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Solution:

A mutation can be defined as any inheritable change in a gene not due to segregation or to the normal recombination of unchanged genetic material. There are two major types of mutation: chromosomal mutation, which can involve an extensive chemical change in the structure of a chromosome, and point mutation (or gene mutation) which involves a single change in molecular structure at a given locus. There are a variety of types of chromosomal muta-tions (see fig. 1). A deletion is a mutation in which a segment of the chromosome is missing. In duplication, a portion of the chromosome is represented twice. When a segment of one chromosome is transferred to another non- homologous chromosome, the mutation is known as a trans-location. An inversion results when a segment is removed and reinserted in the same location, but in the opposite direction. (See Figure 1). Chromosomal mutations can also involve only one nucleotide. The insertion or deletion of a single nucleotide can have extensive effects. Such an occurrence results in a frame-shift by throwing the entire message out of register. (Recall that the gene message is read continuously from triplet to triplet, without "punctuation" between the codons.) Frame-shift mutations usually lead to the production of completely non-functional gene products. Point mutations involve some change in a nucleotide of the DNA molecule, usually the substitution of one nucleotide for another. Point mutations can result from exposure to x-rays, gamma rays, ultraviolet rays and other types of radiation, from errors in base pairing during replication, and from interaction with chemical mutagens. How radiation leads to changes in base pairs is not clear, but the radiant energy may react with water molecules to release short-lived, highly reactive radicals that attack and react with specific bases to cause chemical changes. These changes can block normal replication or cause errors is base pairing. Errors arising from DNA replication are better understood. There exist analogs of nitrogenous bases, such as azoguanine and bromouracil. These analogs possess structures similar to purines or pyrimidines, but have different chemical properties. For example, bromouracil, an analog of thymine, can pair with adenine during repli-cation. However, bromouracil can also pair with guanine and if it does that, subsequent replication results in a strand having a G-C base pair, replacing the A-T base pair of the original strand. (See Fig. 2) Mutagenic chemicals - that is, chemicals which induce mutations - include nitrogen mustards, epoxides, nitrous acids and alkylating agents. These are chemicals that react with specific nucleotide bases in DNA and can convert one base to another or cause mispairing of bases during replication. If a mutated gene is transcribed, and results in changes at or near the active site of an enzyme, the altered enzyme may have markedly decreased or altered enzymatic properties. However, if the mutation occurs elsewhere in the enzyme molecule, it may have little or no effect on the catalytic activities of the enzyme and may go undetected. Therefore, the true number of mutations may indeed be much greater than the number observed.

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Question:

Write the micro-instructions for the interrupt cycle of the computer. What changes must be made in the execute cycle of the previous instruction to accommodate the interrupt cycle?

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Solution:

Previously, after executing the instruction the F register would be cleared and control would then go to the fetch cycle. To facilitate the interrupt capability, however, the IEN flag must be checked at the end of each execute instruction. If IEN = 0 the interrupt capability is not used and control then goes to the fetch cycle. If IEN = 1 the FGO and FGI flags must be checked to see if the teletype needs service. If FGO and FGI are both clear (zero), then control goes to the fetch cycle, but if FGO or FGI are set, then con-trol must go to the interrupt cycle. Hence, the end of the execute cycle must be changed from this: c_2t_3 :F\leftarrow0 to this:c_2t_3 { [IEN] \bullet [ [FGI] + [FGO] ] } :R\leftarrow1 c_2t_3 { [IEN] \textbullet [ [FGI] + [FGO] ] } :F\leftarrow0 When the computer enters the interrupt cycle it goes to a subroutine specifically designed for interrupts. The return address is stored in a special location in memory (location 0), that can be recalled at the end of the sub-routine. The IEN is cleared so another interrupt will not occur until the present one is dealt with. c_3t_0 :MBR_4-15 \leftarrow[PC]\mid PC\leftarrow0\midGo to Interrupt Subroutine c_3t_1 :MAR\leftarrow [PC]\midlocation1, Save return address PC \leftarrow [PC] + 1\midin location O, Clear IEN. c_3t_2 :M\leftarrow[MBR]\mid IEN \leftarrow 0\mid c_3t_3 :F\leftarrow0\midGo to fetch cycle R\leftarrow0\mid The FGO and FGI flags will be checked in the interrupt subroutine to determine if data must enter or leave the AC. After the teletype is serviced the IEN is set to 1 and control leaves the subroutine at the BUN instruction. The computer then fetches the instruction addressed by the PC contents saved in location O.

Question:

Use the quantum aspect of light to derive the formula for the Doppler effect. Assume that the light source moves with a nonrelativistic velocity with respect to the observer.

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/Users/wenhuchen/Documents/Crawler/Physics/D33-1012.htm

Solution:

Let M be the mass of the source and v^\ding{217} its velocity. In order to emit photons, the atoms of the source must initially be in excited states. If the inter-nal energy of the excited source is E, the total energy of the source before emission is E_t = (1/2) Mv^2 + E. When a photon is emitted, the internal energy changes by an amount hv_0 where v_0 is the frequency of the photon observed by the source. The source suffers a recoil as a result of emission and acquires a new velocity v'. By the law of conservation of energy, (1/2) Mv^2 + E = (1/2) Mv'^2 + E' + hv(1) where v is the frequency of the photon as measured by the observer. From the figure, we write the conservation of the momentum in directions parallel and perpendicular tovas Mv = Mv' cos\alpha + (hv/c) cos\texttheta.(2) 0 = mv' sin\alpha + (hv/c) sin\texttheta.(3) Since E' = E - hv_0' (1) can be written as Mv^2 = Mv'^2 + 2h(v - v_0).(4) Rewriting (2) and (3) as Mv' cos \alpha = Mv - (hv/c) cos \texttheta Mv' sin \alpha = (hv/c) sin \texttheta and squaring both sides M^2v'^2 cos^2 \alpha = M^2v^2 + (h^2v^2/c^2) cos^2 \texttheta - 2Mv (hv/c) cos \texttheta M^2v'^2 sin^2 \alpha = (h^2v^2/c^2) sin^2 \texttheta and adding the two, we get M^2v'^2 = M^2v^2 + [(h^2v^2)/c^2] - 2Mv (hv/c) cos \texttheta.(5) Here we made use of the trigonometric identity cos^2 \texttheta + sin^2 \texttheta =1. Combining (4) and (5), we have M^2v'^2 = M^2v2- 2hM(v - v_0) = M^2v^2 + [(h^2v^2)/c^2] - 2Mv (hv/c) cos \texttheta or2hM(v - v_0) - 2Mv (hv/c) cos \texttheta + [(h^2v^2)/c^2] = 0. If the mass M is sufficiently large, we may neglect the last term in the preceding equation, 2hM(v - v_0) = 2Mv (hv/c) cos \texttheta givingv = [v_0/{1 - (v/c) cos \texttheta}]. Therefore, the frequency of the emitted light gets smaller as the source moves faster with respect to the observer.

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Question:

A microbiologist takes a stool specimen from you and a few days later tells you that Escherichia coli andSalmonella typhi are growing in your intestinal tract. Which typeshould you be concerned about?

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Solution:

Escherichiacoli, orE.coli, is a species of bacteria which normally inhabits the intestine of man and other animals. Only rarely canE.colibe pathogenic . For example, they may cause diarrheal disease in infants, and are occasionally found in infections of theurogenitaltract. The presence of E .coliin the stool should not cause any alarm, since it is one of the many types of bacteria that normally inhabit our intestine. The genusSalmonella, however, includes several species which are pathogenic to man and other animals.Salmonellatyphi, for example, causes the acute infectious disease known as typhoid fever. Typhoid fever is character-ized by a fever, inflammation of the intestine, intestinal ulcers, and a toxemia (presence of toxins in the blood). The presence of Salmonella is therefore ample reason for concern. Typhoid fever, like most intestinal infections, is transmitted from one person to another through food and water. Transmission may be indirect . For example, wastes from an infected person can pollute drinking water or food, or the infected person may handle food at some point in its processing or distribution, and contaminate it, af-fecting the consumer. The common housefly can also trans-mitSalmonellafrom wastes to food. Typhoid fever occurs in all parts of the world. In locations where good sanita-tion (proper disposal and treatment of biological waste, and water purification ) is practiced, typhoid fever inci-dence is very low. Carriers are people who are infected withSalmonellatyphi, but who have had only a slight intestinal infection, and hence do not knowthayharbor the pathogenic organism. Carriers should not be allowed to handle or prepare food .

Question:

Consider a satellite in a circular orbit concentric and coplanar with the equator of the earth. At what radius r of the orbit will the satellite appear to remain stationary when viewed by observers fixed on the earth? We suppose the sense of rotation of the orbit is the same as that of the earth.

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Solution:

The satellite is being pulled towards the earth by a gravitational force F\ding{217}_g. By Newton's second law, this force equals ma where m is the \ding{217} mass of the satellite and a is the linear acceleration along the direction parallel to the force F\ding{217}_g (which acts as the centripetal force in this case \ding{217} since the motion is circular). Therefore F_g = ma Furthermore, we know that in circular motion, linear acceleration is given by a = \omega^2r where \omega is the angular velocity of m. The force of gravity (F_g) is given by GmM/r^2 where G is the gravitational constant and M is the mass of the earth. Substituting in the equation above we have GmM/r^2 = m\omega^2r Solving for r we have r^3 = GM/\omega^2 In this equation we note that G is a constant as is M (the mass of the earth). Therefore, r varies as a function of \omega (or vice versa). If we fix \omega, we necess-arily fix r. For the satellite to appear to remain stationary when viewed by observers fixed on the earth, the satellite must have the same angular velocity as the earth. The angular velocity of the earth (\omega_e = ∆\texttheta/∆t, or change of angle per unit time) is 2\pi/1 day= 2\pi/[(60 sec/min)(60 min/hr)(24 hr/day)] = 2\pi/(8.64 × 10^4 sec^-1) = 7.3 × 10^-5 sec^-1 Substituting in the above equation we have: r^3= [(6.67×10^-11N\bulletm^2/kg^2)(5.98×10^24kg)]/(7.3×10^-5 sec^-1)^2 r^3\approx 7.49×10^22 m^3 r\approx 4.2×10^7 m r\approx 4.2×10^9 cm The radius of the earth is 6.38×10^8 cm. The distance calculated is roughly one-tenth of the distance to the moon.

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Question:

Write a FORTRAN program which uses the modified Euler method to simulate the competition of two species of population, N_1(t) and N_2(t), isolated from the environment, from time t = 0 to t =t_f, if it is observed that: Ṅ_1 = (A_1 - K_11N_1 - K_12N_2) N_1(1) Ṅ_2 = (A_2 - K_21N_1 - K_22N_2) N_1(2) where Ṅ_1Ṅ_2 are the time rates of change of N_1(t) and N_2(t), respectively. A_1 and A_2 are positive constants involving natural birth/death rates for each species when isolated. The K_ij-are positive constants involving cross-effects between species. Initially, N_1(0) = N_10, and N_2(0) = N_20 .

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G24-0590.htm

Solution:

First, note that equations (1) and (2) are coupled, non-linear, first-order differential equations. Since exact solutions of such equations are rare, a numerical method, such as the modified Euler's method, is essential. The program first applies the modified method to equation (1), then to equation (2). By putting N1 and N2 in a COMMON block, the new computed value of N1 will be used in determining N2. A logical flag, TWO, is used to determine which equation is being integrated: REAL T/0.0/,TFIN,DT,REALN,ACCUR,N1, 1N2,N10,N20,A1,A2,K11,K12, 1K21,K22 INTEGER I,N COMMON TW0,N1 ,N2 ,A1 ,A2, K11, K12, K21, K22 LOGICAL TWO/.FALSE./ READ, N,TFIN,ACCUR, N10,N20,A1,A2, K11, K12, K21,K22 PRINT, T,N10,N20 REALN = N DT = TFIN/REALN N1 = N10 N2 = N20 DO 10 I = 1,N T = T + DT CALL MEULER(T,N1,ACCUR,DT) TWO = .TRUE. CALL MEULER(T,N2,ACCUR,DT) TWO = .FALSE. PRINT,T,N1,N2 10CONTINUE STOP END FUNCTION G(W) COMMON TWO,N1, N2,A1,A2, K11, KL2, K21, K22 REAL G, W,N1 ,N2 ,A1 ,A2, K11, K12, K21, K22 LOGICAL TWO G = (A1 - K11\textasteriskcenteredW - K12\textasteriskcenteredN2)\textasteriskcenteredW IF (TWO) G = (A2 - K21\textasteriskcenteredN1 - K22\textasteriskcenteredW)\textasteriskcenteredW RETURN END

Question:

When heated to 600\textdegreeC, acetone (CH_3COCH_3) decomposes to give CO and various hydrocarbons. The reaction is found to be first order in acetone concentration with a half-life of 81 sec. Given at 600\textdegreeCa 1-liter container into which acetone is injected at 0.48 atm, approximate how long would it take for the acetone pressure to drop to 0.45 atm?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E13-0466.htm

Solution:

When a reaction is said to be first order in a particular reactant it means that the rate of the red-action is proportional to the concentration of that re-actant. Thus the rate law for this reaction can be written : Rate = k [CH_3COCH_3] where k is the rate constant and[ ] indicate concentration. The rate constant is related to the half-life (t_1/2) by the following equation: t_1/2 = 0.693 / k Solving for k:t_1/2 = 81 sec 81 sec = 0.693 / k k = 8.56 × 10^-3 sec^-1 To solve for the time it takes for a reaction to proceed to a certain degree one uses the equation: dln [A] = - k dt where dln[A] is the natural logarithm of the original con-centration subtracted from the natural log of the final concentration of acetone, k is the rate constant and dt is the time elapsed. Because acetone is a gas in this ex-periment the pressure is proportional to the concentration. Therefore one can let [A]_ original = 0.48 and[A]_final =0.45. Solving for dt: ln 0.48- ln 0.45 = (- 8.56 × 10^-3 sec^-1)(dt) (-.734) - (-.799) = (- 8.56 × 10^-3 sec^-1)(dt) (-.734) - (-.799) =dt -8.56 × 10^-3 sec^-1 -7.5 sec = dt Therefore 7.5 sec have elapsed since the beginning of the reaction.

Question:

Develop a program to find rare prime decades. Consider a decade to be a series of numbers between and including 10a + 1 and 10a + 10, where a = 0,1,2,3... Rare decades are those which contain four primes. Output such decades less than 5000.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G22-0548.htm

Solution:

The rarity of this occurrence is due to the fact that in any decade, there can be at most four primes. If you consider that in each decade there are five even numbers and one odd number that is a multiple of five, you should recognize that the remaining four possible primes must end with 1,3,7, or 9. The only exception to this rule is the first decade, the primes of which are 2,3,5 and 7. The needed algorithm must check the odd integers between 3 and 5000 to establish the existence of primes. To do this task, a FUNCTION subprogram IPRIME is used. This subprogram may use the Sieve of Eratosthenes (developed earlier) or some other method. It is assumed that the subprogram returns either of the two values: a 1 if the integer is a prime, a 0 if it is composite. The main program makes use of a four element queue. Each time a prime is found, it is stored in the array K. When four consecutive primes are in the queue, a check is made to see if they are included in one decade. If so, the entire queue is printed out. If not, the first prime in the queue is removed, the remaining three elements move up one position, and the next prime is inserted at the end of the queue. This process continues until I reaches 5000. The program looks as follows: DIMENSION K(4) K(1) = 2 ISUM = 1 J = 2 I = 3 II = 1 CDO WHILE I IS LESS THAN 5000 20IF (I.GE. 5000) GO TO 60 CASSIGN A NEW DECADE TO ISUM IF II = 5 IF (II.LT. 5) GO TO 5 ISUM = ISUM + 10 II = 0 CIF I IS A PRIME, ASSIGN IT TO K(J) 5CALL IPRIME (I) IF (IPRIME.EQ. 0) GO TO 57 K(J) = I J = J + 1 I = 1+2 II = II + 1 IF (J.LE.4) GO TO 20 CCHECK IF ALL 4 PRIMES IN ARRAY K ARE WITHIN CTHE SAME DECADE IF (K(1).LT.ISUM) GO TO 30 IF (K(4). GE. ISUM + 10) GO TO 30 CIF, IN FACT, THEY ARE - OUTPUT THE ARRAY K WRITE (5,100)(K(J), J = 1,4) 100FORMAT (4(2X,I4)/) J = 1 GO TO 20 CIF THEY ARE NOT WITHIN ONE DECADE 30DO 35 N = 2,4 35K(N-1) = K(N) J = 4 GO TO 20 57I = I + 2 II = II + 1 GO TO 20 60STOP END

Question:

How does asexual reproduction in plants differ from sexual reproduction? What are the evolutionary advantages of the latter?

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/Users/wenhuchen/Documents/Crawler/Biology/F09-0209.htm

Solution:

Asexual reproduction in plants takes place by the separation of any portion, or specialized portion, of a parent plant to form new individuals, or by the for-mation and germination of spores or single cells specialized in the replication of new individuals. Both these methods preserve the genetic makeup of the parent plants. Sexual reproduction involves the union of two gametes. The combination of two different sets of chromosomes from the gametes results in a new individual with a new genetic composition. Sexual reproduction therefore gives rise to new genotypes while asexual reproduction perpetuates the same ones. Sexual reproduction is an important process in the evolution of a species, and is considered much more ad-vantageous than asexual reproduction. The fusion of gametes from two genetically different parents makes possible new combinations of genetic material and thus increases the variation among individuals in the population. Genetic variation provides a pool of diversified phenotypes upon which natural selection can act. By this process, natural forces determine which individuals of the species will survive and which will become extinct, ultimately improving the species by sorting out the advantageous gene in the pool. An asexually reproducing species offers little genetic variation for selection and evolution is minimized. In addition, the natural forces acting upon any species are always changing. The more a species is able to respond to these forces with evolutionary changes that maintain or increase its fitness, the more likely it is to survive and multiply. Such evolutionary changes can occur through genetic mutations, expressed in phenotypes which are then acted upon by natural selection. Because a mutated gene is usually recessive, sexual reproduction increases its chance of expression by bringing two mutated alleles together in the process ofgameticfusion. Asexual reproduction will need a double mutation, which occurs in the order of 10^-12 per generation, to result in an altered phenotype. Therefore, without sexual reproduction there could be little genotypic and thus phenotypic changes. Evolution could not progress, as there could be no response to inevitable environmental changes, and extinction of the species would soon follow.

Question:

In an earlier question, we learned about the various factors which are involved in the clotting mechanism. Show how these factors are affected in conditions that cause excessive bleeding in humans, and in conditions that cause intravascular clotting.

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/Users/wenhuchen/Documents/Crawler/Biology/F14-0348.htm

Solution:

Excessive bleeding can result from a deficiency of any one of the blood clotting factors. An insufficiency ofprothrombin, one of the intermediates in the clotting mechanism, can cause a patient to develop a severe tendency to bleed. Both hepatitis and cirrhosis (diseases of the liver) can depress the formation ofprothrombinin the liver. Vitamin K deficiency also depresses the levels ofprothrombin. Vitamin K deficiency does not result from the absence of the vitamin from the diet, since it is continually synthesized in the gastrointestinal tract by bacteria. The deficiency results from poor absorption of fats (vitamin K is fat soluble) from the tract due to a lack of bile, which is secreted by the liver. Hemophilia, or "bleeder's disease," is a term loosely applied to several different hereditary deficiencies in coagulation, resulting in bleeding tendencies. Mosthemo-philiasresult from a deficiency in one of the protein factors called Factor VIII or theantihemophilicfactor necessary for production ofprothrombinaseby the platelets. Thrombocytopenia, the presence of a very low quantity of platelets in the blood, also causes excessive bleeding. With adefinciencyof platelets, not enoughprothrombinasecan be synthesized. One major type of thrombocytopenia results from the development of immunity to one's own plate-lets. [Normally, the immune system does not develop immunity against the body's own proteins, but sometimes one can deve-lop this "autoiInmunity".] The antibodies (specific proteins in the blood) attack and destroy the platelets in the person's own blood. Thrombocytopenia also results from pernicious anemia and certain drug therapies. These - abnormalities cause excessive bleeding; but there are also pathological conditions caused by clotting when it should not normally occur. An abnormal clot that develops in a blood vessel is called a thrombus. If the thrombus breaks away from its attachment and flows freely in the bloodstream, it is termed an embolus. Should an embolus block an important blood vessel (to the heart, lungs, or brain), death could occur. What causes intravascular (within a blood vessel) clotting? Any roughened inner surface of a blood vessel, which may result from arteriosclerosis, bac-terial infection, or physical injury, can initiate the clotting mechanism by releasingthromboplastinfrom the platelets. Blood which flows too slowly may also cause clotting. Since small amounts of thrombin are always being produced, a hampered flow can increase the concentration of thrombin in a specific area, so that a thrombus results. The immobility of bed patients presents this problem. Heparin is a strong anticoagulant normally produced in small amounts by special cells in the body. Heparin can be used as an intravenous anticoagulant in the prevention of thrombi formation.

Question:

For the reaction PbSO_4 (s) \rightarrow Pb^2+ + S0^2-_4, ∆H = + 2990 cal/mole. Will the solubility of PbSO_4 increase or decrease with increasing temperature?K_sp=1.8 × 10^-8 at 25\textdegreeC. Find itsk_spat 55\textdegreeC.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E15-0544.htm

Solution:

Whether the solubility of a salt increases or decreases with an increase in temperature can be determined by an investigation of its ∆H, the enthalpy change. The formula for the temperature dependence on k is log k = [(∆H\textdegree)/(2.303 RT)] + [(∆S\textdegree)/(2.303 R)]. A positive ∆H\textdegree suggests that solubility increases with increasing temperature. When ∆H\textdegree is positive, the first term remains a negative value. When T increases, the first term becomes a smaller negative number. Thus, k increases. Since k measures solubility, the solubility increases. To find thek_spof PbSO_4 at 55\textdegreeC, use the fact that log (k_2 /k_1) = [(- ∆H\textdegree) / (2.303 R)] [(1/T_2) - (1/T_1)], where k_2 = solubility constant at the temperature in Kelvin (Celsius plus 273\textdegree) of T_2; k_1 = solubility constant at a temperature of T_1. Thus, if thek_sp at one temperature is known, thek_spat another temperature can be found given ∆H\textdegree. Let T_2 = 55\textdegreeC or 328\textdegreeK and T_1 = 298\textdegreeK. (25\textdegreeC).Thus log [(k_328)/(k_298)] = [(- ∆H\textdegree)/(2.303 R)] [(1/328) - (1/298)] log [(k_328 )/(1.8 × 10^-8)] = [{- (2990 cal/mole) (- 3.07 × 10^-4/k)} /{(2.303) 1.987 cal/mole K)}] = 0.200 = 0.200 k_328 = (1.8 × 10^-8) (10^0.200) = (1.8 × 10^-8) (1.58) = 2.8 × 10^-8.

Question:

Write a program to solve for integral zero of a third-degree polynomial

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G21-0519.htm

Solution:

The flowchart illustrates the basic program. Now we give an explanation of the theory behind our search. Which in-tegers do we try for Z to test P(z) for 0? Let us assume that there are N complex zeros denoted by. Z_N, Z_N-1 ,..., Z_2, Z_1. The fac-tor theorem for polynomials states that if f(z1) = 0, i.e., z1 is a root of f(z), then f(z) = (z - z1)g(z) where g(z) is a polynomial one degree lower than f(z). Then, by repeated applications of the factor theorem, we obtain (X - Z_N) (X - Z_N-1) ...(X - Z_2) (X - Z_1) = A_NX^N + A_N-1X^N-1 +... +A_1X + A_0.(1) Multiplying out the left hand side of (1), we see that the only constant term in the product is (-1)^n (Z1)\bullet(Z2)...(Z_N) = A_0. Hence, we can conclude that if a polynomial has any integral zeros, they must be factors of the constant term A_0 . Note that the converse need not be true, i.e., a factor of A_0 need not be a root. Note, however, that any factor of P(1) = A_0 will lie in the interval -\vertP(1)\vert to \vertP(1) \vert. The actual program makes use of the sign function SGN to assure stopping in the right direction. 10DEF FNP(X) = P(4)\textasteriskcenteredX\uparrow3 + P(3)\textasteriskcenteredX\uparrow2 + P(2)\textasteriskcenteredX + P(1) 20PRINT 25FOR S = 4 TO 1 STEP - 1 30READ P(S) 40PRINT P(S); 50NEXT S 60PRINT "INTEGRAL ZERO(S): " ; 68REM TURN SWITCH OFF 70LET K = 0 78REM STUDY LINE 80 CAREFULLY 80FOR X = -P(1) TO P(1) STEP SGN (P(1)) 88REM LINE 90 PREVENTS ERROR MSG CAUSED BY 89REM DIVIDING BY ZERO 90IF X = 0 THEN 140 98REM IS X A FACTOR, OF P(1)? 100IF P(1)/X = INT(P(1)/X) THEN 140 108REM IS THE REMAINDER ZERO? 110IF FNP(X) < > 0 THEN 140 118REM IF THE COMPUTER GETS THROUGH HERE, THE 119REM VALUE OF X IS A ZERO OF THE FUNCTION 120PRINT X; 128REM TURN THE SWITCH ON - WE HAVE A ZERO 130LET K = 1 140NEXT X 150IF K = 1 THEN 20 160PRINT "NONE FOUND"; 165GO TO 20 170DATA 1, -2, -11, 12 180DATA. 1, 1, -5, -2 190DATA 1, -2, 3, -4 200DATA 2, -3, -10, 3 210END

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Question:

What is the resultant force on a body of mass 48 kg when its ac-celeration is 6 m/sec^2?

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Solution:

The relationship between a body's acceleration and the net force on it is given by Newton's Second Law. The mass of the body is given, hence the net force on the body is \sumF = ma = 48 kg × 6(m/sec^2) = 288newtons.

Question:

Write a program in BASIC to find the sum and number of factors for the integers between 1000 and 1050, not including 1 and the integer itself.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G22-0550.htm

Solution:

As an example, consider the integer 36, It has 7 factors: 2,3,4,6,9,12,18, the sum of which is 54. The presented program outputs the considered integer, sum of its factors, and the number of factors. The program terminates when the considered number is greater than 1050. The program looks as follows: 15LET F = 1133 20LET F = F + 1 25IF F < 1175 THEN 30 28STOP 30LET M = 0 32LET N = 0 34LET D = 2 35LET Y = F/D 40LET W = INT(Y) 45IF W < > Y THEN 70 50LET M = M + D 55LET N = N + 1 60IF D = Y THEN 70 65LET M = M + Y 68LET N = N + 1 70LET D = D + 1 75IF D\textasteriskcenteredD < = F THEN 35 80PRINT F,M,N 90GO TO 20 95END

Question:

A uniform rod of mass m and length 2a stands vertically on a rough horizontal floor and is allowed to fall. Assuming that slipping has not occurred, show that, when the rod makes an angle \texttheta with the vertical, \omega^2 = (3g/2a)(1 - cos \texttheta) where \omega is the rod's angular velocity. Also find the normal force exerted by the floor on the rod in this position, and the coefficient of static friction involved if slipping occurs when \texttheta = 30\textdegree.

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Solution:

The forces acting on the rod are the weight mg^\ding{217} acting downward and the normal force N^\ding{217} and the frictional force F^\ding{217} of magnitude \muN exerted by the floor at the end 0 in contact with the floor. In order to find \omega, we re-late the net torque \cyrchar\cyrt on the rod to the rod's angular acceleration a by using \cyrchar\cyrt = I\alpha Here, I is the rod's moment of inertia. We will then be able to solve for \omega. When one takes moments about 0, the only force producing rotation about 0 is the weight of the rod. Hence \cyrchar\CYRT = mga sin \texttheta = I_0 \alpha = (4/3)ma^2 \alpha Here, I_0 is the rod's moment of inertia about 0. Now, \alpha = (d\omega/dt) = (d\omega/d\texttheta) × (d\texttheta/dt) = \omega(d\omega/d\texttheta) = (3/4)(g/a) sin \texttheta. ^\omega\int_0 \omega d\omega = ^\texttheta\int_0(3/4)(g/a) sin \texttheta d\texttheta. [(1/2) \omega^2]^\omega_0 = [-(3/4)(g/a) cos \texttheta]^\texttheta_0 or \omega^2 = [(3g)/(2a)] ( 1 - cos \texttheta). The center of gravity G has an angular acceleration \alpha about 0, and thus a linear acceleration a\alpha at right angles to the direction of the rod. This linear acceler-ation can be split into two components, a\alpha cos \texttheta horiz-ontally and a\alpha sin \texttheta vertically downward. The horizontal acceleration of the center of gravity is due to the force \muN and the vertical acceleration is due to the net effect of the forces mg and N. Thus, using Newton's Second Law, and taking the positive direction downward, mg - N = ma\alpha sin \texttheta = (3/4) mg sin^2 \texttheta(1) andF_max = \muN = ma\alpha cos \texttheta = (3/4)mg sin \texttheta cos \texttheta.(2) From (1) N = mg - (3/4) mg sin^2 \texttheta= [(mg)/(4)] (4 - 3 sin^2 \texttheta) . But when \texttheta = 30\textdegree, slipping just commences. At this angle F has its limiting, maximum value of F_max. We have \mu_s= (F_max)/N = [(3/4)mg sin \texttheta cos \texttheta]/[(mg)/(4) (4 - 3 sin^2 \texttheta)] = (3 sin \texttheta cos \texttheta)/(4 - 3 sin^2 \texttheta) = [3 × (1/2)× (\surd3/2)]/[4 - (3/4)] = (3\surd3)/(13) = 0.400.

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Question:

What is the least radiation frequency v, capable ofpositron- electronpair production?

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Solution:

The process in which a photon disappears and an electron-positron pairappears, is known as pair production. The energy of the photon shouldat least be large enough to provide for the rest masses of the electronand proton. Therefore E_photon= m_electronc^2 + m_positronc^2 hv= 2m_ec^2 wherev is the photon frequency. Hence v = (2m_ec^2/h) = [{2 x (9.11 x 10^-31 kg) x (3 x 10^8m/s)^2} / (6.63 x 10^-34 J.S)] = 2.47 x 10^20 (1/sec).

Question:

Given a one-dimensional array Y with 50 elements, and numbersU and I, write statements to compute the value of S fromthe following equation: S =y_i+ U {(y_i+ 1-y_i- 1) / 2} + (U^2 / 2) (y_i+ 1- 2y_i +y_i- 1) S =y_i+ U {(y_i+ 1-y_i- 1) / 2} + (U^2 / 2) (y_i+ 1- 2y_i +y_i- 1)

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Solution:

Note that since the subscripts on Y can be i+1,i, or i-1 where the Y arrayis dimensioned 50, then we must restrict I to being an integer from 2 to49, i.e., there are 48 possible values of I. With this restriction in mind we cansimply write: DIMENSIONY(50) DO 100 I = 2,49 S =Y(I) + U \textasteriskcentered (Y(I + 1) - Y(I - 1)/2.0) 1+ U \textasteriskcentered U \textasteriskcentered (Y(I + 1) - 2\textasteriskcenteredY (I) 1+Y(I - 1))/2.0 100CONTINUE STOP END Further remarks on interpretation: This formula is calledStirling's interpolationformula [S stands forStirling!] through second differences andis described below. We have three points on a curve: (Xi -1,Yi - 1), (X_i, Y_i), and (Xi + 1, Yi + 1) such that Xi + 1- X_i = X_i - Xi - 1= h , (i.e., the x coordinates are equally spaced) and have a value of x for which wewant to guess Y(x). We write U = (x -x_i )/h. Then the formula stated givesthe interpolated value of Y corresponding to x, found by passing a quadraticthrough the three given points.

Question:

If a 1,000\rule{1em}{1pt}V battery is connected to two parallel plates separated by 1mm (10^\rule{1em}{1pt}3m), what is the electric field?

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/Users/wenhuchen/Documents/Crawler/Physics/D18-0605.htm

Solution:

The potential of the battery, V = 1,000 V, and the distance between the plates, d = 10^\rule{1em}{1pt}3 m, are given. By definition, the difference in potential experienced by moving a charge from point A to point B is V_B \rule{1em}{1pt} V_A = \rule{1em}{1pt} ^B\int_A E \bullet dl(1) where E is the electric field and d is an element of the path traversed in moving the charge. Now, looking at the figure, we see that, for the plates of a battery, E is perpendicular to the plates. If we evaluate (1) over a straight line path parallel to E, we find V_B \rule{1em}{1pt} V_A = \rule{1em}{1pt}B\int_A Edl = \rule{1em}{1pt}d\int_0 Edl = \rule{1em}{1pt} E d where d is the plate separation. Then │E│ = │(v_b \rule{1em}{1pt} v_a)│ / d = (10^3 v) / (10^\rule{1em}{1pt}3m) = 10^6 v / m.

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Question:

Artificial kidneys have been devised for patients with kidney disease. How do these artificial kidneys work? What is the basic principle involved?

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/Users/wenhuchen/Documents/Crawler/Biology/F18-0459.htm

Solution:

The artificial kidney is used to replace the diseased kidney in eliminating the excess ions and wastes which would accumulate in the blood as a result of failure of the latter. In an artificial kidney, the patient's blood is passed through a system of very fine tubes bounded by thin membranes. The other side of the membrane is bathed by a dialysis fluid into which waste products can pass from the blood. The fine tubes con-verge into a tubing which then conducts the blood back into the patient's body through a vein. The fine tubes of the artificial kidney are made of cellophane. This material is used because it has much the same characteristics as the endothelium of blood capillaries; it is highly permeable to most small solutes but relatively impermeable to protein. The dialysis fluid which bathes the cellophane tubes is a salt solution with ionic concentrations similar to those of blood plasma. The function of the artificial kidney can be ex-plained by the principle of diffusion. Since the cel-lophane membrane is permeable to most small solutes, the concentrations of solutes in the blood, as it flows through the tubes, tend to equal those in the dialysis fluid. However, if there is an above normal level of a certain solute in the blood, this solute will diffuse out of the blood into the surrounding fluid, which has a lower concentration of the solute. In this way, waste products and other fluid substances in excess will leave the blood and pass into the dialysis fluid since their concentrations in the fluid are very low or non-existent. To prevent waste solutes from building up in the fluid and inter-fering with diffusion, the fluid is continually being re-placed by fresh supply. Great care must be taken to maintain the sterility of the artificial kidney. The total amount of blood in an artificial kidney at any one moment is roughly 400-500 ml. Usually heparin, an anticoagulant, is added to the blood as it enters the artificial kidney to prevent coagulation; an antiheparinis added to the blood before returning it to the patient's body in order to allow normal blood clotting to take place. An artificial kidney can clear urea from the blood at a rate of about 200 ml. of plasma per minute, - that is, 200 ml of plasma can be completely cleared of urea in one minute - this being almost three times as fast as the clearance rate of both normal kidneys working together. Thus we can see that the artificial kidney can be very effective, but it can be used for only 12 hours every 3 or 4 days, because of the danger to the blood's clotting mechanisms due to the addition of heparin.

Question:

Show that the area of a parallelogram, whose sides are formed by the vectors A\ding{217} and B\ding{217} (see figure) is given by Area = \vertA\ding{217} × B\ding{217}\vert .

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/Users/wenhuchen/Documents/Crawler/Physics/D01-0004.htm

Solution:

The area of the parallelogram shown in the figure is Area = bh But h = \vertA\ding{217}\vert sin \texttheta and b = \vertB\ding{217}\vert Area = \midA\ding{217}\mid \midB\ding{217}\mid sin \texttheta(1) The left side of (1) is the magnitude of A\ding{217} × B\ding{217}, hence Area = \midA\ding{217} × B\ding{217}\mid = \midA\ding{217}\mid \midB\ding{217}\mid sin \texttheta If we are interested in obtaining a vector area, we may write Area = A\ding{217} × B\ding{217} where the direction of the area is the direction of A\ding{217} × B\ding{217} . Such vector areas are useful in defining certain surface integrals used in physics.

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Question:

How is the human eye regulated for far and near vision? In what respects does the frog eye differ in these regulatory mechanisms ?

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/Users/wenhuchen/Documents/Crawler/Biology/F20-0495.htm

Solution:

The human eye can focus near or distant images by changing the curvature of the lens. The lens is bound tociliarymuscles via the suspensory ligaments. When inverted to focus on a distant object, the ciliary muscles contract, stretching thesuspensoryligaments as well as the flexible lens. A flat lens correctly focuses the image of a distant object on the retina.Ciliarymuscles relax when one focuses on a near object. This allows the lens to contract into a shape that will correctly focus the closer object. Lens regulation in frogs differs from man in one impor-tant aspect. The frog focuses objects by moving the eye lens forward or backward, whereas inmanaccommodation is achieved by changing the shape of the lens , without any change in its position.

Question:

Calculate the pH of a 0.10 M solution of sodium acetate, NaOAc. The equilibrium constant for the dissociation of acetic acid is 1.8 × 10^-5.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E12-0438.htm

Solution:

The pH will be governed by two processes: the ionization of NaOAc, and the hydrolysis of acetate ion,OAc^-. NaOAcdissociates according to the equation NaOAc\rightleftarrows Na^+ +OAc^-. If we assume that dissociation is complete (a reasonable assumption for dilute salt solutions), the concentrations of Na^+ andOAc^- are equal to the initial concentration ofNaOAc: [Na^+] = [OAc^-] = 0.10 M. Hydrolysis of acetate ion proceeds according to the equation OAc^- + H_2O \rightleftarrowsHOAc+ OH^-. Let x denote the concentration ofundissociatedHOAcformed by hydrolysis. Then, since one mole of OH^- is formed and one mole ofOAc^- is consumed per mole ofHOAcformed, the concentration of hydroxide ion formed is x, and the concentration of acetate ion remaining is 0.10 - x. That is, [OH^-] = x,[OAc^-] = 0.10 - x. These concentrations can be substituted into the expression for the hydrolysis constant, K_h= {[HOAc] [OH^-]} / [OAc^-] to find x = [OH^-].K_hcan be found by using the water constant,K_w = 1.0 × 10^-14 = [H_3O+] [OH^-] and the equilibrium constant for dissociation of acetic acid (HOAc), K_a = 1.8 × 10^-5 = {[H_3O^+] [OAc^-]} / [HOAc]. Then, K_h= {[HOAc] [OH^-]} / [OAc^-] = [H_3O+] [0H^-] × {[HOAc] / ([H_3O^+] [OAc^-])} =K_w/ K_a = (1.0 × 10^-14) / (1.8 × 10^-5) = 5.6 × 10^-10. Substituting the unknown concentrations into the expression gives K_h= 5.6 × 10^-10 = {[HOAc] [OH^-]} / [OAc^-] = {(x) (x)} / (0.10 - x) \allequal x^2 / 0.10 = 10 x^2, where we have assumed that x is much smaller than 0.10 (hence 0.10 - x \allequal 0.10). Then 10 x^2 = 5.6 × 10^-10 x = (5.6 × 10^-11)^1/2 = 7.5 × 10^-6 Therefore, [OH^-] = x = 7.5 × 10^-6. Using the water constant, [H_3O+] [OH^-] = 1.0 × 10^-14 [H_3O^+] = [1.0 × 10^-14] / [OH^-] = [1.0 × 10^-14] / [7.5 × 10^-6] = 1.3 × 10^-9 and the pH of the solution is pH = - log [H_3O^+] = - log (1.3 × 10^-9) = 8.9.

Question:

What is the average velocity of the molecules of the air at 27\textdegree C?

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/Users/wenhuchen/Documents/Crawler/Physics/D14-0490.htm

Solution:

From a simple atomic model in which we consider the atom as a hard spherical body subject to completely elastic collisions we can develop an equation for the average kinetic energy of a molecule which is: (1/2)mv^2_a = (3kT)/2 where m is the mass of one molecule,v_ais its average velocity, k is Boltzmann's constant, and T is the absolute temperature of the environment of the molecule. Multiply both sides by Avogadro's number N_A and 2 N_Amv^2_a = 3N_AkT ButN_Am= M_m andN_Ak= R since M_m is the mass of one molecules, and R is the gas constant which is Boltzmann's constant times Avogadro's number. Substitut-ing and rearranging we have: v^2_a = (3RT)/(M_m). Air consists mainly of nitrogen, which is diatomic, and its effective molecular mass is approximately twice the atomic mass of nitrogen. So M_m \approx 2 × 14 = 28 gm/mole v^2_a = [(3)(8.32 Joule/mole \textdegreeK)(273 + 27)\textdegreeK]/[28 gm/mole] where we have used the fact that 27\textdegree C = (273 + 27)\textdegreeK. Hence v^2a= 267.43 Joule/gm But 1 Joule = 1nt\textbullet m = 1 kg \textbullet m^2/s^2 =10^7 gm \textbullet cm^2/s^2 and v^2_a^ = 267.43 × 10^7 [(gm \bullet cm^2)/(gm \bullet s^2)]^ = 267.43 × 10^7 cm^2/s2 Thereforev_a= 5.17 × 10^4 cm/s. This is equivalent to 1,160 miles per hour!

Question:

The freezing point of silver is 960.8\textdegreeC and the freezing point of gold is 1063.0\textdegreeC. Convert these two readings to Kelvin (\textdegreeK), Fahrenheit (\textdegreeF), and Rankine (\textdegreeR).

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/Users/wenhuchen/Documents/Crawler/Chemistry/E01-0021.htm

Solution:

Kelvin: Temperatures measured in Celsius (\textdegreeC) are converted to \textdegreeK by adding 273.15 to the original measurement. freezing point of silver = 960.8\textdegreeC + 273.15 = 1234\textdegreeK freezing point of gold = 1063.0\textdegreeC + 273.15 = 1336.2\textdegreeK Fahrenheit: \textdegreeC are converted to \textdegreeF by using the equation \textdegreeF = 9/5 (\textdegreeC) + 32. freezing point of silver = 9/5 (960.8\textdegreeC) + 32 = 1761\textdegreeF freezing point of gold = 9/5 (1063.0\textdegreeC) + 32 = 1945\textdegreeF Rankine: The Rankine scale is an absolute scale used by engineers. Its unit is the Fahrenheit degree. Absolute zero is equal to zero degrees Rankine. Convert \textdegreeK to \textdegreeR by using the equation \textdegreeR = 9/5 (\textdegreeK) freezing point of silver = 9/5 (1234\textdegreeK) = 2221\textdegreeR freezing point of gold = 9/5 (1336.2\textdegreeK) = 2405\textdegreeR.

Question:

If you mix 50ml of 5.0 × 10-^4 M Ca(NO_3)_2 and 50ml of 2.0 × 10-^4 M NaF to give 100ml of solution, will precipitation occur? The K_sp of CaF_2 is 1.7 × 10-^10 .

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/Users/wenhuchen/Documents/Crawler/Chemistry/E11-0401.htm

Solution:

Whether or not precipitation will occur, when these two solutions are mixed, depends upon the ion product. If two solutions containing the ions of a salt are mixed, and if the ion product ex-ceeds the K_sp , then precipitation will occur. You need to determine the concentrations of Ca^2+ and F- and to see if the product exceeds the given K_sp of 1.7 × 10-^10 for CaF_2 . The equation for this reaction is CaF_2 (s) \rightleftarrows Ca^2+ + 2F- . Thus, the K_sp = [Ca^2+] [F-]^2 . You are told that a 50ml solution of Ca(NO_3 )_2 has a molarity of 5 × 10-^4 M. Each Ca(NO_3 )_2 dissociates to yield one Ca++ ion which means the molarity of Ca++ is also 5 × 10-^4 M. However , it is 5 × 10-^4 M in 50ml , and the mixture is 100ml . Therefore , the molarity must be divided by two . The molarity of NaF, 2.0 × 10-^4 , and therefore Na+ , because of its dissociation, must also be divided by two for the same reasons . As such , in the mixture, [Ca^2+ ] = 2.5 × 10-^4 M and [F-] = 1.0 × 10-^4 M. Recalling that K_sp = [Ca^2+ ] [F-]^2 , you have K_sp = (2.5 × 10-^4 ) (1.0 × 10-^4 )^2 = 2.5 × 10-^12 . You were given that the K_sp for a saturated solution of CaF_2 is 1.7 × 10-^10 . Since 2.5 × 10-^12 is less than 1.7 × 10-^10 , the mixture is unsaturated, and precipitation should NOT occur .

Question:

A plane can travel 100 miles per hour (mph) without any wind. Its fuel supply is 3 hours. (This means that it can fly 300 miles without a wind). The pilot now proposes to fly east with a tail wind of 10 mph and to return with, of course, a head wind of 10 mph. How far out can he fly and return without running out of fuel? (The student should first reflect on whether or not the answer is the same as it is without a wind, namely, 150 miles.)

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/Users/wenhuchen/Documents/Crawler/Physics/D03-0086.htm

Solution:

Our basic equation is (time out) + (time back) = 3 hours We now use the fact that time = (distance)/(speed) so that if d represents the distance out (= distance back) in miles we have [d/(110 miles/hr)] + [d/(90 miles/hr)] = 3 hours since the speed out is (100 + 10)and the speed back is (100 - 10)mph. Thus: (90 d + 110 d)/(9900 miles/hr) = (200 d)/(9900 miles/hr) =3 hours d = 148.5 miles.

Question:

An electric motor, rated at 1 hp, operates from a 100-V line for 1 hr. How much current does the motor require and how much electrical energy was expended?

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/Users/wenhuchen/Documents/Crawler/Physics/D24-0785.htm

Solution:

The power expended in an electric circuit is given by the product of the voltage across its terminals and the current passing through the terminals. Hence, the current in the motor is I = (P_E/V) = (1 hp/100 V) = (746 W/100 V) = 7.46 A The power of the motor indicates the amount of electrical energy it draws from the electrical source. per unit time and converts to mechani-cal energy. Then, the electrical energy expended in one hour is E_E = P_E × t = 1 hp × 1 hr = 746 W-hr = 0.746 kW-hr or, in CGS units, E_E = (746 W-hr) × {(10^7 ergs/sec)/1W} × (3600 sec/1 hr) = 2.68 × 10^13 ergs

Question:

Explain how the sense organ of equilibrium functions in the crayfish.

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/Users/wenhuchen/Documents/Crawler/Biology/F12-0309.htm

Solution:

The sense organ of equilibrium in the crayfish is the antenna. All crustaceans have two pairs of antennae. The second pair in lobsters and crayfish is extremely long; this pair functions in touch and taste. At each base of the first pair, astatocystis present, which functions in the maintenance of equilibrium. Thestatocystsare sacs that open to the exterior. Inside the sac is astatolith, com-posed of fine sand grains cemented together by secretions from thestatocystwall. Along the floor of the sac are a number of rows of sensory hairs. These hairs arise from sensory receptor cells and are innervated by theantennularnerve. When the crayfish is in an upright po-sition, continual impulses from both right and leftstatocystscounterbalance each other and the crayfish does not lean to either side. If the animal is rotated onto its right side, thestatolith of the left side exerts a gravi-tational pull on the sensory hairs, which send impulses to the brain. Thestatolithon the right side does not exert a pull on the sensory hairs, and the crayfish as a result rotates to the left, thus balancing itself. If the animal is turned so that bothstatolithsare exerting forces, and impulses are sent from bothstatocysts, the crayfish rotates in the direction of whicheverstatocystis most stimulated. Thestatocystis aninvaginationof the ectoderm, which forms the exoskeleton, and is shed with each molt. The animal must gather new sand grains to form thestatolith. This is done by inserting sand into the statocyst sac or by burying the head in sand. Studies of thestatocyst's function were done by inserting' iron filings into sand when crayfish molted, and using a magnet to directstatolithforces on the sensory hairs. Thestatocystmay also function to maintain position during movement. For this purpose there are free sensory hairs, which are not in contact with thestatolith. These hairs are stimulated by the motion of fluid in the chamber when the animal moves.

Question:

Each liter of human blood serum contains about 3.4 g of sodium ions (Na^+, ionic weight = 23 g/mole) . What is the molarityof Na^+ in human blood serum?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E08-0282.htm

Solution:

We are given that the concentration of Na^+ is 3.4 g/liter. To convert this concentration tomolarity, we need to convert 3.4 g to the corresponding number of moles. This is accomplished by dividing 3.4 g by the ionic weight of Na^+ (which is essentially the atomic weight, since the missing electron does not detract much from the weight) to obtain 3.4 g/23 g/mole = 0.15 mole. The con-centration is thus 0.15 mole/liter = 0.15 molar = 0.15 M.

Question:

When 100 g of water at 0\textdegreeC are mixed with 50 g of water at 50\textdegreeC, what is the change of entropy on mixing?

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Solution:

The 100 g of water at 0\textdegreeC are arbitrarily said to have zero entropy. The 50 g of water at 50\textdegreeC have a greater entropy than the same quantity of water at 0\textdegreeC, since it contains more heat energy. Entropy S is defined as ds ds =dQ/T(1) =dQ/T(1) wheredsis the infinitesimal change in the entropy due to an infinitesimal quantity of heatdQin the system. T is the instantaneous temperature in Kelvin degrees. Integrating both sides of equation (1) gives S_2 - S_1 = ^Q2\int_Q1dQ/T.(2) In raising its temperature from a temperature t_1 to t_2 a substance absorbs heatdQgiven by dQ=mcdT(3) where m is the mass of the substance, c is its specific heat, anddTis its change in temperature in Kelvin degrees (same as change in Celsius degrees). Substitution of equa-tion (3) into equation (2) yields S_2 - S_1 = ^T2\int_T1 mc (dT/T)(4) In degree Kelvin, 0\textdegreeC = (0 + 273)\textdegreeK = 273\textdegreeK and 50\textdegreeC = (50 + 273) \textdegreeK = 323\textdegreeK. Let m_2 = 50 gm and m_1 = 100 g. The specific heat of water is given by c = 1 cal \textbullet g^-1 \textbullet K deg^-1 . S_2 - S_1 = ^323\textdegreeK\int_273\textdegreeK m_2c (dT/T) = m_2c 1n(323/273) = 50 g × 1 cal \textbullet g^-1 \textbullet K deg^-1 × 2.303 × 0.0730 = 8.4 cal \textbullet K deg^-1. Since S_1 = 0, it follows that S_2 = 8.4 cal \textbullet K deg^-1. When the water is mixed, the heat gained by the cold water is equal to the heat lost by the hot water. There-fore, m_1c(t_3 - t_1) = m^2c(t_2 - t_3), where t_1 is the original temperature (0\textdegreeC) of the 100 g of water, t_2 is the original temperature (50\textdegreeC) of the 50 g of water and t_3 is the final, intermediate temperature of the system. 100 g × (t^3 - 0\textdegreec) = 50 g × (50\textdegreeC - t_3).\thereforet_3 = (2500\textdegreeC)/150 = 16.67\textdegreeC. Converting to Kelvin degrees in order to be able to use equation (4), we have t_3 = 16.67\textdegreeC = (16.67 + 273)\textdegreeK = 289.67\textdegreeK. The entropy of the final mixture is S_3 = ^t3\int_t1 dQ dQ /T /T = ^289.67\textdegreeK\int273\textdegreeK(m_1 + m_2)c (dT/T) = ^289.67\textdegreeK\int273\textdegreeK(m_1 + m_2)c (dT/T) = = (m_1 + m_2)cln(289.67/273) = 150 g × 1 cal \textbullet g^-1 \textbullet K deg^-1 × 2.303 × 0.0257 = 8.9 cal \textbullet K deg^-1. The increase in entropy is thus 0.5 cal \textbullet K deg^-1.

Question:

In an inertial frame a body moves freely with a trajectory given by x_I = v_Ot;y_I = 0;z_I = 0. What is the trajectory in a frame rotating with constant angular velocity \omega counterclockwise about the z_I axis?

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Solution:

In the figure the particle is travelling at velocity v_0 in the inertial coordinate system. After time t, the body has traveled a distance v_0t along the x-axis. Now, we consider a coordinate system which rotates with angular velocity \omega about the z_I axis. Therefore the z- axis of the new system coincides with the inertial z_I axis and the angle between the rotating (x_R, y_R) axes and the inertial (x_I, y_I) axes is by definition of \omega, \texttheta = \omegat. The coordinates of the position of the particle in the rotating frame in terms of the coordinates of the particle in the inertial frame can be read off the figure as x_R = v_0t cos \omegat y_R = -v_0t sin \omegat z_R = 0.

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Question:

Wave number (ѵ̅) are the reciprocals of wavelengths, \lambda, and are given by the expression ѵ̅ = 1/\lambda. For the hydrogen atom, the Bohr theory predicts that the wave number for the emission line associated with an electronic transition from the energy level having principal quantum number n_2 to that with principal quantum number n_1 is ѵ̅ = R_H [(1/n_1 ^2) - (1/n_2 ^2)], where R_H is theRydbergconstant. In what region of the electromagnetic spectrum would there appear a spectral line resulting from the transition from the tenth to the fifth electronic level in hydrogen?

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Solution:

This problem can be solved by applying theRydbergformula used for determining thewavenumbersof the spectral lines. TheRydbergconstant is 1.10 × 10^5 cm^-1. n_2 = 10 and n_1 = 5. Substituting: ѵ̅ = R_H [(1/n_1 ^2) - (1/n_2 ^2)] = 1.10 × 10^5 cm^-1 [(1/5^2) - (1/10^2)] = 1.10 × 10^5 cm^-1 [(1/25) - (1/100)] = 1.10 × 10^5 cm^-1 (0.04 - 0.01) = 1.10 × 10^5 cm^-1 × 0.03 = 3.3 × 10^3 cm^-1. This line appears in the infrared region of the spectrum (approximately 20 cm^-1 to 10^4 cm^-1).

Question:

Write a FORTRAN program which uses the modified Euler method to simulate the growth of an isolated species from time t = 0 to t = t_f, if the number of individuals at time t is N(t), N(0) = N_0 and the birth rate B and the mortality rate M (per unit of population per unit of time) are given by: Case (1): B = b_1; M = m_1 where b_1 and m_2 are positive constants. Case (2): B = b_2 > 0. Competition for food causes the death rate to increase in direct proportion to the ratio N(t)/2N_0 with the con-stant of proportionality equal to m_2 where 0 < m_2 < 1, and N_0 > 0. Compare the approximations with the exact values.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G24-0583.htm

Solution:

An initial schematic diagram of the system looks as shown in figure #1. The rate of flow of individuals into and out of the system (per unit of time) is given by BN and MN respectively. Thus, B and M act as valves (see Fig. 02). Thus, the net rate of change in population, N, is the difference between the entrance and exit rates: Ṅ = BN - MN = (B - M)N, and the block diagram of the process is shown in figure #3. Case (1):If B = b_1 and M = m_1 are positive constants, B-M = b_1 - m_1 = r Case (1):If B = b_1 and M = m_1 are positive constants, B-M = b_1 - m_1 = r will also be a constant. If b_1 > m_1, then r > 0 and the population will rise indefinitely. If b_1 < m_1, then r < 0 and the population will continuously decrease to zero. If b_1 = m_1, r = 0 and there will be no change in population. In any case, one can write: Ṅ = (B - M)N = rN(2) The exact solution to equation (2)is N(t) = N_0e^rt Note that when r > 0, t \ding{217} \infty \Rightarrow N(t) \Rightarrow \infty . When r < 0, t\ding{217} \infty \Rightarrow N(t) \Rightarrow 0 When r = 0, N(t) = N_0e^0 = N_0(1) = N_0. The program uses POP for N(t). REAL T/0.0/, Ml COMMON R READ, N,TFIN,ACCUR,B1,Ml, POPO PRINT, T,POPO REALN = N DT = TFIN/REALN POP = POPO R = B1 - Ml DO 10 I = 1,N T = T + DT CALL MEULER(T,P0P,ACCUR,DT) EXACT = POPO\textasteriskcenteredEXP(R\textasteriskcenteredT) ERROR - ABS((EXACT-POP)/EXACT)\textasteriskcentered100 PRINT,T,POP,EXACT,ERROR 10CONTINUE STOP END FUNCTION G(W) COMMON R G = R\textasteriskcenteredW RETURN END Case (2): In this case, equation (1) becomes: Ṅ = (b_2 - m_2N/2N_0)N = b_2N - (m_2N^2/2N_0)(3) Equation (3) is a nonlinear differential equation (because of the N^2 term) . Though, generally, solutions for nonlinear equations are not easy to find, equation (3) is known as the Bernoulli equation with the solution: N(t) = (2b_2N_0) / (m2+ (2b_2 - m_2)e^-b2t) There are three possible cases: (i)m_2 = 2b_2 . Substituting for m_2 in equation (3) at time t = 0 when N = N_0 gives: Ṅ = b_2N_2 - (2b_2N_0^2 / 2N_0) = b_2N_2 - b_2N_0 = 0 which means population remains at initial level. This can also be seen by substituting for m_2 in equation (4): N(t) = 2b_2N_0/ {2b_2 (2b_2 - 2b_2 )e^-(b)2t } = 2b_2N_2 / 2b_2 = N_0 . (5) Since t does not appear in equation (5), in this case, N(t) is the constant function, N(t) = N_0. What this means to the programmer is that for the pair of parameters (b2 ,m_2), where m_2 = 2b_2 any initial population will be an equilibrium point; and the system is stable at this point. (ii)m_2 < 2b_2 . m_2 < 2b_2 \Rightarrow m_2 > 0 \Rightarrow as t \ding{217} \infty : (2b_2 - m_2)e^-bt \ding{217} 0^+ m_2 + (2b2- m_2)e^-bt 2b_2 / {m_2 + (2b_2 - m_2)e^-bt} \ding{217} [2b_2 / m_2] > 1 (since m_2 < 2b_2) N(t) \ding{217} (2b_2 / m_2)N_0 > N_0(6) From equation (6), we see that population will increase to the equilibrium value 2b_2N_0/m_2 . (iii)m_2> 2b_2 . An analysis similar to that used in case (ii) yields: N(t) \ding{217} (2b_2 / m_2 )N_0 < N0= (since m_2 > 2b_2 \Rightarrow {2b_2 / m_2} < 1)(7) as t \ding{217} \infty. From equation (7), the population will decrease to the equilibrium level . It follows that the system is asymptotically stable, since it converges back to its equilibrium point as a result of any arbitrary disturbance: For this first-order case, the equilibrium point could also have been found by the graphical method, plotting B and M as functions of N: The equilibrium point is the point of intersection, where the birth rate just equals the death rate, since the net rate of change In population, N = (B - M)N, will be zero. But B - M = 0 \Rightarrow b_2N - (m_2 / 2N_0)N^2 = 0 N{b_2 - (m_2/2N_0)N} = 0 \Rightarrow N = 0 or If N > 0, b_2 - (m_2/ 2N_0) N = 0 (m_2/2N_0)N = b_2 \Rightarrow N = (2b_2/m_2) N_0 which is the same result obtained in equations (6) and (7), using the exact solution. Some sort of analysis of the system under study is an invaluable aid to the programmer in distinguishing between peculiar system behavior due to system stability or instability and peculiar results due to programming or logical errors. The program uses POP for N(t), POPO for N_0 . REAL T/0.0/, NUMBER,M2 COMMON B2,M2,POPO READ,N,TFIN,ACCUR,B2,M2,POPO PRINT ,T, POPO REALN = N DT = TFIN/REALN POP = POPO CCONSTANT TERMS IN EXPRESSION FOR EXACT NUMER = 2.\textasteriskcenteredB2\textasteriskcenteredPOPO DENOM = 2.\textasteriskcenteredB2 - M2 DO 10 I = 1,N T = T + DT CALL MEULER(T,POP,ACCUR,DT) EXACT = NUMER/ (M2+(DEN0M\textasteriskcenteredEXP(-B2\textasteriskcenteredT))) ERROR = ABS((EXACT-POP)/EXACT)\textasteriskcentered100. PRINT,T,POP,EXACT,ERROR 10CONTINUE STOP END FUNCTION G(W) COMMON B2,M2,POPO REAL M2 G = (B2-(M2\textasteriskcenteredW)/(2.\textasteriskcenteredPOPO))\textasteriskcenteredW RETURN END

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Question:

The angular velocity of a body is 4 rad/sec at time t = 0, and its angular acceleration is constant and equal to 2 rad/sec^2. A line OP in the body is horizontal ( \texttheta_0 = 0) at time t = 0. (a) What angle does this line make with the horizontal at time t = 3 sec? (b) What is the angular velocity at this time?

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/Users/wenhuchen/Documents/Crawler/Physics/D04-0172.htm

Solution:

The angular kinematics equations for constant angular acceleration are identical in form to the linear kinematic equations with a corresponding to a, \omega to v, and \texttheta to x. (a) Comparable to x = x_0 + \omega_0t + (1/2)at^2, we have \texttheta = \texttheta_0 + \omega_0t + (1/2)\alphat^2 where \texttheta_0, \omega_0 are the initial angular position and velocity of the body. Since \texttheta_0 = 0, we have \texttheta_-= \omega_0t + (1/2)\alphat^2 = 4 (rad/sec) × 3 sec + (1/2) × 2 (rad/sec^2) × (3 sec)^2 =21 radians = 21 radians [(1 revolution)/(2\pi radians)] = 3.34 revolutions. The angle \texttheta is then \texttheta= 0.34 × one revolution = 0.34 × 360\textdegree \approx 122\textdegree. (b)\omega= \omega_0 + \alphat = 4(rad/sec) + 2(rad/sec) × 3 sec = 10(rad/sec). Alternatively, \omega^2= \omega^2_0 + 2\alpha\texttheta = 4(rad/sec)^2 + 2 × 2(rad/sec^2) × 21 rad = 100(rad^2/sec^2)\omega = 10(rad/sec).

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Question:

The mass of an electron is 9.11 × 10^-31 and Planck's constant = 6.63 × 10^-34 J sec. How much energy does an electron possess in its five lowest states, if it is placed in a cube of dimension 50 nm?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E18-0680.htm

Solution:

Wave function theory states that the permitted values of the energy of a particle in a box is given as E = [(N^2 h^2)/(8m L^2)], where E = energy, N = principal quantum number, h = Planck's constant, m = mass of particle, and L = length of box. Substitute in the known values and solve for E. The principal quantum number N corresponds to the energy levels of the electrons. The five lowest energy states are N = 1, N = 2, N = 3, N = 4, and N = 5, where N = 1 is the lowest and N = 5 the highest energy level. Substituting: E = [(N^2 h^2)/(8m L^2)] = [{N^2 (6.63 × 10^-34 J sec)^2} / {8 (9.11 × 10^-31 kg) (5 × 10^-8 M)^2}] = N^2 (2.41 × 10^-23) (Note: 50 nm = 5.0 × 10^-8 M) 1st energy level (lowest): N^2 = 1, E = 2.41 × 10^-23 J 2nd energy level: N^2 = 4, E = 9.64 × 10^-23 J 3rd energy level: N^2 = 9, E = 2.16 × 10^-22 J 4th energy level: N^2 = 16, E = 3.85 × 10^-22 J 5th energy level: N^2 = 25, E = 6.025 × 10^-22 J.

Question:

Construct a galvanic cell based on the reaction3Fe(s) Construct a galvanic cell based on the reaction3Fe(s) + 2Au^+3 \rightarrow 3Fe^+2 + 2Au(s) with ∆H negative.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E16-0591.htm

Solution:

A galvanic cell may be defined as one in which spontaneous chemical reactions occur at each of the electrodes of the cell to produce an electric current. Thus, to construct this cell, you need to know those reactions that occur at the electrodes. The two electrodes are the anode and cathode. At the anode oxidation occurs, which means you have the loss of electrons. At the cathode, reduction occurs, which means you have the gain of electrons. The oxida-tion and reduction reactions are each called half-reactions. These are the reactions that occur at the electrodes. From the given reaction, it becomes apparent that oxidation is given by Fe(s) \rightarrow Fe^+2 + 2e^- and re-duction by Au^+3 + 3e^- \rightarrow Au(s). Thus, the anode half-cell would consist of an iron electrode in a solution containing Fe^+2 ions, while the cathode half-cell would consist of a gold electrode in a solution con-taining Au^+3 ions. A compound such as FeSO_4 could serve as a source of Fe^2+ ions, and AuCl_3 as a source of Au^+3 ions. The cell might appear as in the figure shown.

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Question:

Both myoglobin and hemoglobin are oxygen-carrying molecules. Why is hemoglobin the molecule of choice to carry oxygen in the blood?

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/Users/wenhuchen/Documents/Crawler/Biology/F14-0351.htm

Solution:

In vertebrates, both myoglobin and hemoglobin act as oxygen carriers. Hemoglobin, however, is the only one that acts in the blood, carrying oxygen from the lungs to the tissues where it is needed for metabolic processes. Myoglobin is located in muscle, and serves as a reserve supply of oxygen. What accounts for the functional differ-ences between these two proteins? Myoglobin is a single-chain protein containing only one heme group, which serves to bind an oxygen molecule. Hemoglobin is composed of four polypeptide chains (each chain similar to the one of myoglobin) and thus has four binding sites for oxygen. It is the interaction of these four polypeptide chains that confers special properties to the hemoglobin molecule, making it a better oxygen-carrier in the blood. One of these important properties is the ability to transport CO_2 and H^+ in addition to O_2. Myoglobin does not have this ability. The binding of oxygen by hemoglobin is regulated by specific substances in its environment such as H^+, CO_2, and organic phosphate compounds. These regu-latory substances bind to sites on hemoglobin that are far from the heme groups. The binding of these substances affects the binding of oxygen by producing conformational changes in the protein, that lower hemoglobin's affinity for oxygen. A change at the regulatory site (where these regulatory substances bind) is translated to the heme site by changes in the way the four polypeptide chains are spatially arranged. This results in a structural change in the protein. Such interactions between spatially distinct sites are termed allosteric interactions. Hemoglobin is an allosteric protein (whereas myoglobin is not) and there fore exhibits certain allosteric effects which account for the functional differences between the molecules. This essential difference between myoglobin and hemoglo-bin is reflected in their respective oxygen dissociation curves. The curve is a plot of the amount of saturation of the oxygen-binding sites as a function of the partial pressure of oxygen: The shape of the oxygen dissociation curve reflects the crucial difference: the shape of the hemoglobin curve is sigmoidal (S-shaped) whereas that of myoglobin is hyper-bolic. The sigmoidal shape is ideally suited to hemo globin's role as an oxygen carrier in the blood. When the partial pressure of the oxygen is high, hemoglobin tends to bind oxygen; that is, many of the binding sites are filled with oxygen. There is a high partial pres-sure of oxygen (\sim 100 torrs) in the alveoli of the lungs, so that hemoglobin tends to pick up oxygen there. When the partial pressure of oxygen is low, hemoglobin tends to release oxygen. There is a low partial pres-sure of oxygen (\sim 20 torrs) in the capillaries in active muscle, where oxygen is being rapidly consumed by cellular respiration. CO_2 is being rapidly produced, and its concentration is correspondingly high. Hemo-globin tends to release oxygen. Hemoglobin thus acts to pick up oxygen where it is available and release it where it is needed. Myoglobin, on the other hand, has a higher affinity for oxygen than does hemoglobin. Thus, even at relatively low partial pressures, myoglobin still tends to bind oxygen. Only when the pressure is very low does myoglobin release oxygen. This behavior makes it unsuitable to carry oxygen in the blood. An oxygen carrier in the blood must be able to readily release oxygen at places where oxygen is needed. On a molecular level , the sigmoidal shape means that the binding of oxygen to hemoglobin is cooperative- that is, the binding of oxygen at one heme facilitates the binding of oxygen at another heme on the same hemoglobin molecule. The binding of the first oxygen molecule is thought to be most thermodynamically unfavorable. Sub-sequent oxygen molecules can bind more easily because the first molecule alters the structure (disrupt certain electostatic interactions) upon binding. The increasing affinity of the hemoglobin molecule for oxygen thus accounts for its sigmoidal dissociation curve.

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Question:

An object, starting from rest, is given an acceleration of 16 feet per second^2 for 3 seconds. What is its speed at the end of 3 seconds?

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/Users/wenhuchen/Documents/Crawler/Physics/D03-0060.htm

Solution:

Since the acceleration is constant, we have a = (v_final-v_initial)/t orv_final= v_initial + at Butv_initial= 0 for the object started from rest. Therefore v_final= a × t = [(16 ft)/(sec^2)] × 3sec = 48 ft per sec.

Question:

Summarize the complete oxidation of glucose to CO2 and H2O.

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/Users/wenhuchen/Documents/Crawler/Biology/F03-0100.htm

Solution:

See accompanying figure. The complete oxidation is written as: Glucose + 6 O2 \textemdash\textemdash> 6 CO2 + 6 H2O + 36 ATP Glucose enters the cell. In the cell cytoplasm,glycolysisbreaks glucose downtopyruvate.Pyruvateenters the mitochon-drion. In the matrix, pyruvateis converted to acetyl-CoA. Acetyl-CoAenters the Krebs cycle by combiningwithoxaloa-ceticacid (OAA), a four carbon molecule. Acetyl- CoAis 2 carbons, so when it joins with OAA it forms a 6-c citrate molecule. By a series of oxidation-reduction steps, OAA is regenerated to keepthe cycle spinning,2 CO2s are produced per turn of the cycle, and the reducingequivalents NADH and FADH2 are produced. The reducing equivalentsgo to the in-ner mitochondrial membrane and pass their electronsalong. At three sites, a pair of protons are extruded into the in- termembranespace (only two pairs for FADH2). The last en-zyme of the respiratorychain iscytochromeoxidase: oxygen accepts the electrons, picksup a pair of H+s, and forms wat-er. Concomitant with the flow of electronsis the synthesis of ATP: The extruded H+s flow into the matrix viachannels on the "lollypop" enzymes. One ATP is produced per pair of H+s. Note on the figure that the only substances entering the cell are glucose andoxygen (circled) and the only sub-stances leaving the cell are water, carbondioxide and ATP (circled). This is in agreement with the equation above. When glucose is oxidized completely, 36 ATPs are produced. Fatty acids andamino acids also feed into this scheme at acetyl-CoA, but different amountsof ATP are produced de-pending on the molecule.

Question:

How much silver is deposited on a spoon which is plated for 20 minutes at 0.50 ampere?

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/Users/wenhuchen/Documents/Crawler/Physics/D19-0634.htm

Solution:

In order to check this problem, we must under-stand what happens during the electroplating process. As each silver ion migrates to the spoon, it is reduced to a metallic silver atom. All the silver ions are in a plus 1 ionized state. This means that for every electron pass-ing through the circuit, one silver ion will be turned into metallic silver and deposited. The problem then reduces to one of charge transport - as many atoms (or moles of atoms) of silver will be deposited as electrons (or moles of electrons) that pass through the circuit. We merely have to equate the two. The number of moles of deposited silver is given by the mass of the liberated silver (an unknown quantity) divided by the atomic mass of silver (107.9): Mass liberated in grams / (107.9gm / mole) To discover the amount of charge transported, and hence the number of moles transported, we use the relationship I = q / torq = It where I is current, q is charge, and t is time. Since we know that there are 96,500 coulombs of charge per mole of electrons, we can calculate the number of moles of electrons by using q / (96,500 coulombs /mole)orIt / (96,500 coulombs/mole) Equating the number of moles of silver plated, and the number of electrons which have passed through the circuit during the given time period, we have [Mass liberated in grams / (107.9gm / mole)] = [It / (96,500 coulombs / mole)] Since I = 0.50 amp and t = 20 min = 1200 sec. we have Mass / (107.9gm / mole) = (0.50 amp × 1200 sec) / (69,500 coulombs / mole) Solving, we obtain Mass = 0.671 gm, approximately

Question:

The animals having radial symmetry are considered to be at a lower evolutionary stage than those having bilateral symmetry. Why is this true? What is meant by the term secondarily radial?

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/Users/wenhuchen/Documents/Crawler/Biology/F11-0290.htm

Solution:

The bodies of most animals are symmetrical, that is, the body can be cut into two equivalent halves. In radial symmetry any plane which runs through the central axis from top to bottom divides the body into two equal halves. In bilateral symmetry, only one plane passes through the central body axis, that can divide the body into equal halves. In a bilateral body plant, six sides are distinguished: front (ventral), back (dorsal), head-end (anterior),tailend(posterior), left, and right. (Most animals are not perfectly symmetrical; for example, in man, the heart is located more to the left, the right lung is larger than the left lung, and the liver is found on the right side of the body.) Coelenterata(including the hydras, the true jelly-fishes, the sea anemones, sea fans and corals) andCtenophora(including the comb jellies and sea walnuts) are two radiate phyla. Their members have radially sym-metrical bodies that are at a relatively simple level of construction. They have no distinct internal organs, no head, and no central nervous systems though they possess nerve nets. There is a digestive tract with only one opening serving as both mouth and anus, and there is no internal space orcoelombetween the wall of the digestive cavity and the outer body wall. These animals have two distinct tissue layers: an outer epidermis (derived from embryonic ectoderm) and an innergastrodermis(derived from em-bryonic endoderm). A third tissue layer,mesoglea(mesoderm) is usually also present between the epidermis andgastrodermis. Often gelatinous in nature, it is not a well- developed layer, and has only a few scattered cells, which may be amoeboid or fibrous. Coelenterates and ctenophores are mostly sedentary organisms. Some are sessile at some stage of their life cycles, while others are completely sessile throughout their lives. The higher animals are usually bilateral in symmetry. The flatworms and the proboscis worms are regarded as the most primitive bilaterally symmetrical animals. They are however far more advanced than the coelenterates and ctenophores. Both have bodies composed of three well-developed tissue layers - ectoderm, mesoderm, and endoderm. Their body structures show a greater degree of organization than theradially symmetrical animals. Although there are no respiratory and circulatory systems, there is a flame- cell excretory system, and well-developed reproductive organs (usually both male and female in each individual). Mesodermal muscles show an advance in construction; cir-cular and longitudinal muscle layers are developed for purposes of locomotion and/or alteration of body shape. Several longitudinal nerve cords running the length of the body and a tiny "brain" ganglion located in the head are present which together constitute a central nervous system. In higher bilaterally symmetrical animals, there is observed a trend toward more complicated construction of the body. Separate organ systems are developed, specializing in dif-ferent functions. There is also a separation of sexes in individuals so that each individual produces only one kind of gametes (male or female but not both). The echinoderms areradiallysymmetrical animals. The phylum Echinodermata includes sea stars (starfish), sea urchins, sea cucumbers, sand dollars, brittle stars, and sealillies. These animals are fairly complex; they have a digestive organ system, a nervous system, and a reproductive system. The echinoderms are believed to have evolved from a bilaterally symmetrical ancestor. Also, whereas the adults exhibit radial symmetry, echinoderm larvae are bilaterally symmetrical. For these reasons, the echinoderms are considered to be secondarilyradiallysymmetrical. (The echinoderms will be discussed in greater detail in Chapter 12.)

Question:

Design an experiment to demonstrate that florigen, the flowering hormone, is indeed produced in the leaves and travels to the flower bud. 12 hrs per day - change to flower - inducing conditions 18 hrs per day - change to flower - inhibiting conditions A \rightarrow a B \rightarrow b C \rightarrow c D \rightarrow d

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/Users/wenhuchen/Documents/Crawler/Biology/F09-0239.htm

Solution:

To demonstrate that florigen is indeed produced in the leaves, a flowering plant is defoliated and then exposed to appropriate conditions under which flowering normally occurs. The observation that the plant fails to flower indicates that florigen must have been produced in the leaves, and that florigen has been removed as a result of defoliation. A control experiment must of course be set up to show that defoliation of the plant produces no adverse effect on the plant. It is postulated that in order to exert its effect on the buds, florigen must be transported from the leaves to the flower buds. This can be tested by defoliating the lower part of a plant and exposing it to flower inhibiting factors, while maintaning the upper, leafy part under flower-promoting conditions. The leafy part is found to flower as expected. Flowering in the lower part, however, can best be explained by the diffusion of florigen from the leaves of the upper part to the buds of the lower part of the plant. The defoliated lower part of the plant can never flower on its own because of the absence of florigen. Because flowers do appear on the defoliated part of the plant, it is concluded that florigen is a diffusible sub-stance made in leaves. Even more compelling evidence for this conclusion comes from grafting experiments. One can grow two plants, one plant exposed to flowering conditions, the other not. The plant exposed to flowering conditions will flower (fig. 1a). The flowering plant is now grafted onto the non-flowering plant, and a partition is used so that the grafted portion continues to be exposed to flowering conditions and the non-flowering plant to flower-inhibiting conditions (fig. 1b). Gradually, the non-flowering plant will begin to produce flowers, usually at the point nearest the graft (fig. 1c) and eventually throughout the entire plant (fig. 1d). If one were to graft a short-day plant onto a long-day plant, and expose both to a short-day period, the short-day graft induces flowering in the long-day plant (fig. 2a). Likewise, one can expose both to a long- day period, and both portions will flower, the short-day portion having been induced by the long-day portions (fig. 2b) . Similar results can be obtained using day- neutral plants. Thus florigen appears to be physiologically equivalent in the three groups of flowering plants. Graft-induced flowering is successful only if there is a living tissue connection between the two plants. In addition, flower-induction is inhibited if one removes the phloem containing tissue in the graft. This supports the conclusion that florigen is transported via the phloem system of the plant. Some plants cannot be induced to flower by the above methods unless they are first defoliated. This suggests that, under flower- inhibiting conditions, some inhibiting substance may be produced by the leaves. Flowering may thus involve the interaction of both inducing and inhibiting factors.

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Question:

What is an antipyretic ? Explain how it functions in humans.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E32-0938.htm

Solution:

Antipyretics are fever-reducing substances that relieve headaches and minor pains. They are also important in the treatment of arthritis and rheumatic fever. They belong to the class of medicinal compounds called sali-cylates, which, in turn, are termed analgesics or pain-killers. The antipyretic action of these drugs appears to be accomplished by the influence of the drugs on a center in the hypothalamus of the brain. The hypothalamus causes small blood vessels in the skin to dilate (to swell or ex-pand) thus enabling the body to lose heat.

Question:

The following table gives the number of items sold daily at a super-market. Week 1 Week 2 Sunday 900 800 Monday 400 500 Tuesday 500 300 Wednesday 600 300 Thursday 300 400 Friday 700 600 Saturday 1100 900 Write a FORTRAN program to compute a seven day moving average.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G19-0486.htm

Solution:

Observations taken over periods of time typically show seasonal variability. The purpose of moving averages is to reduce these variations to enable us to spot the trend. First, find the average daily sales from Sunday to Saturday of the first week. This figure is placed in the table next to Wednesday - the midpoint of the examined period. Then find the average from Monday of the first week to Sunday of the second week, etc. Proceed-ing in this way we obtain the seven-day moving average for the given data. 7 day M.A. 7 day M.A. Sunday 571 Monday 586 Tuesday 571 Wednesday 643 543 Thursday 629 Friday 643 Saturday 614 From the table, it can be seen that the sales are falling. Perhaps the supermarket should start an advertising campaign; The program looks as follows: DIMENSION X(14), AV(7) L = 1 M = 7 SUM = 0 DO 2 I = 1,14 READ X(I) 2CONTINUE 10DO L J = L,M SUM = X(J) + SUM 1CONTINUE AV(L) = SUM/7 L = L + 1 M = M + 1 IF (M.EQ. 15) GO TO 11 GO TO 10 11DO 12 I = 1,7 PRINT AV(I) 12CONTINUE STOP END

Question:

How many cubic feet of life preserver of specific gravity .3, when worn by a boy of weight 125 lb and having a specific gravity .9, will just support him 8/10 submerged in fresh water of which 1 cu ft weighs 62.4 lb?

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/Users/wenhuchen/Documents/Crawler/Physics/D10-0401.htm

Solution:

In this problem the boy b is 8/10 submerged while the life preserver p must be completely sub-merged to give the maximum buoyancy. p must be completely sub-merged to give the maximum buoyancy. The weight of the boy W_b and the weight of the preserver W_p acting downward are just balanced by the buoyant force of the preserver B_p and the buoyant force of the boy B_b. B_b + B_p = W_b + W_p(1) B_b and B_p are equal to the weight of fluid displaced by the boy and the preserver respectively. Hence B_p = Mg where M is the mass of the water displaced by the boy. Since density d = M/V and the volume of the displaced water V = 8/10 V_b where V_b is the volume of the boy, then B_b = [(8/10) V_b] d_w g where d_w is the density of water. The weight of the boy is W_b = (V_bd_b)g =125 lb W_b = (V_bd_b)g =125 lb where V_bd_b is the mass of the boy and d_b is the density of the boy. Similarly, B_p = V_pd_w gandW_p = V_pd_pg where V_p is the volume of fluid displaced by the preserver, and d_p is the density of the preserver. V_b= W_b/gd_b Now,dw = (mass of 1 cu ft of water) / (1 cu ft of water) = [(weight of 1 cu ft of water) / g] / (1 cu ft of water) = [(62.4 lb) / (1 cu ft of water)] × (1/g) Also, the specific gravity (or relative density) of the preserver with respect to water S_p = d_p/d_w = 3/10 ; and s_boy = (d_b/d_w) = 9/10 S_p = d_p/d_w = 3/10 ; and s_boy = (d_b/d_w) = 9/10 Therefore, equation (1) becomes \therefore(8/10) V_bd_wg_- + V_pd_wg = W_b + V_pd_pg (8/10)(125/gd_b) d_wg_- + V_pd_wg = 125 + V_p (3/10) d_wg V_p = [(8/10)(125)(d_w/d_b) - 125] / [(3/10) d_wg - d_wg] = [(8/10)(125)(1/s_b) - 125] / [-(7/10) d_wg] = [(8/10){(125)/(9/10)} - 125] / [-(7/10)(62.4/g)g] = [125 - (8/9)(125)] / [(7/10)(62.4)] = [(1/9)(125)] / [(7/10)(62.4)] = 13.9/43.7 = .32 cu ft (approx).

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Question:

What is the force due to the liquid only acting on a circular plate 2 in. in diameter which covers a hole in the bottom of a tank of oil 4 ft high, if the specific gravity of oil is .5?

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/Users/wenhuchen/Documents/Crawler/Physics/D10-0414.htm

Solution:

Since the liquid in the tank is in equilibrium, it can exert only a vertical force on the bottom of the tank, as shown in the figure. Furthermore, the net force on the bottom surface of the tank is due only to the weight of the water above this surface. Hence, F_net = Mg(1) where M is the mass of fluid in the tank. Since the density of the fluid is \rho = M/V where V is the tank volume, we obtain M = \rhoV andF_net = \rhogV(2) The force on the circular plate of radius r is, using (2) F_plate = F_net (\pir^2/A) = \rhogV(\pir^2/A)(3) where A is the area of the tank bottom. We may write the volume of the tank asV = hA(4) Here, h is the depth of fluid in the tank. Combining (4) and (3) F_plate = \rhoghA(\pir^2/A) = \pi\rhoghr^2 F_plate = \pi\rhoghr^2 Now, the specific gravity of a substance, \sigma, is defined as \sigma = \rho/\rho_w where \rho' is the density of the substance and \rho_w is water's density. Hence F_plate = \pi\rho_w\sigmaghr^2 F_plate = (3.14) (1.94 sl/ft^3) (.5) (32 ft/s^2) (4 ft) (1 in^2) In order to keep our units consistent, 1 in = 1/12 ft 1 in^2 = 1/144 ft^2 whenceF_plate = (3.14) (1.94 sl/ft^3) (.5) (32 ft/s^2) (4 ft) (1/144 ft^2) = 2.71 lb.

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Question:

How is DNA replicated ?

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/Users/wenhuchen/Documents/Crawler/Biology/F24-0615.htm

Solution:

By studying the Watson-Crick model of DNA, Kornberg and his colleagues determined the mechanism of DNA replication. Since DNA is the genetic material of the cell, it must have the information to replicate itself built into it. The Watson-Crick model of DNA seems to offer a good explanation of how DNA is replicated. There are only four different nucleotides found in DNA. Chargaff's rule of. specific base pairing states that these nucleotides (adenine, guanine, thymine and cytosine) are ordered on both strands of the DNA helix so that adenine is always paired to thymine and cytosine to guanine. Thus, in order for DNA to be replicated, the helix need only be unwound to form a template. The nucleotide building blocks can line up in a sequence complementary to the order presented. In this way, two double stranded helices identical to the original molecule are synthesized in the nucleus. This is a general overview of the mechanism of DNA replication. Actually, the first step of DNA replication occurs when unwinding proteins uncoil a part of the DNA helix (see figure - step A). At this time, RNA polymerase combines with the sigma factor to initiate RNA synthesis. This is an essential preliminary step because the RNA that is synthesized serves as a primer which is necessary before DNA replication can occur (Figure-step B). The RNA that is synthesized is not a final product because before DNA replication is completed, it is destroyed by endonucleases and exonucleases. After the primer is made, DNA polymerase initiates DNA replication. There are free nucleotides in the nucleus which complementarily join to form a new strand along the 5' to 3' direction of the unwound portion, however, the DNA strand itself is synthesized in a 3' to 5' manner (Figure-step C). DNA replication is thought to occur in short, uncontinuous bursts allowing the DNA helix to unwind farther (Figure-step D). DNA polymerase then fills in the gaps between the fragments synthesized. DNA ligase joins the 5' end of one fragment to the 3' end of the adjacent fragment. In this way, a strand of DNA is replicated accor-ding to the semi-conservative model, a parental DNA replica-tes to give two first generation hybrids containing one strand of the parental DNA and one newly synthesized strand. (Figure-step E).

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Question:

Given that theelectronegativitiesof F,Cl, Br, and I are, respectively, 4.0, 3.0, 2.8, and 2.5, account for the fact that the dipole moment decreases in the sequence HF,HCl,HBr, and HI, even though bond length and the number of electrons increase.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E17-0645.htm

Solution:

A dipole is defined as a molecule which is electrically unsymmetrical - that is, the centers of positive and negative charges are not located at the same point within the molecule. Quantitatively, a dipole is described by its dipole moment, which is equal to the charge times the distance between the positive and negative centers. The decrease in dipole moment will reflect a decrease in either the value of the charges or the distance between them. It is given that the bond length increases in this sequence, which means that distance increases. This suggests an in-crease in dipole moment within the sequence. But the dipole moments decrease, this means that the charge must be de-creasing. This can be explained by noting that the electro-negativity values are decreasing within the sequence. Electronegativity measures the tendency of an atom to attract shared electrons in a molecule. By attracting these electrons, it develops a negative charge. If theelectronegativitydecreases, so does the charge, which would then account for the decrease in dipole moment.

Question:

An electron travels from one to the other of two plates, between which is maintained a potential difference of 1000V. With what speed and with what energy does the electron arrive at the positive plate? A positively charged particle with equal and oppos-ite charge but 3680 times the mass is then released at the positive plate. With what velocity and what energy does it reach the negative plate?

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/Users/wenhuchen/Documents/Crawler/Physics/D18-0616.htm

Solution:

The electron travels through a potential difference of 1000 V. The energy acquired is thus E = eV, where e is the charge of the electron ( = 1.60 × 10^\rule{1em}{1pt}19 C) and V is the potential difference between the plates. Thus E (1.60 × 10^\rule{1em}{1pt}19 c) (1000 V)= 1.60 × 10^\rule{1em}{1pt}16 J. However, the energy required to move an electron through a potential difference of 1 volt is an electron volt (eV). Hence, the energy required to move the electron through a potential difference of 1000 volt is 1000 eV, whence 1000 eV = 1.60 × 10^\rule{1em}{1pt}16 J. As required by the conservation of energy, all this potential energy is converted to kinetic energy on arrival of the electron at the positive plate. Hence (1 /2) mv^2 = 1.60 × 10^\rule{1em}{1pt}16 Jor v^2 = (3.20 × 10^\rule{1em}{1pt}16 J .) / (9.1 × 10^\rule{1em}{1pt}31 kg) \therefore v = 1.88 × 10^7 m \textbullet s^\rule{1em}{1pt}1. For the positively charged particle, the charge it possesses is the same in magnitude as that of the electron and it moves through the same potential differ-ence in the opposite direction. Hence, it acquires the same energy of 1.60 × 10^\rule{1em}{1pt}16 J. The velocity, however, is smaller because of the much larger mass. Thus (1 /2) m_1v^2_1 = 1.60 × 10^\rule{1em}{1pt}16 Jor V^2_1 = (3.20 × 10^\rule{1em}{1pt}16 J.) / (9.1 × 10^\rule{1em}{1pt}31 × 3680 kg) \thereforev_1 = 3.10 × 10^5 m \bullet s^\rule{1em}{1pt}1.

Question:

In Figure 1, a small object of mass m is attached to a light string which passes through a hollow tube. The tube is held by one hand and the string by the other. The object is set into rotation in a circle of radius r_1 with a speed v_1. The string is then pulled down, shortening the radius of the path to r_2. Find the new linear speed v_2 and the new angular speed \omega_2 of the object in terms of the initial values v_1 and \omega_1 and the two radii.

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/Users/wenhuchen/Documents/Crawler/Physics/D37-1069.htm

Solution:

The downward pull on the string is transmitted as a radial force on the object. Such a force exerts a zero torque on the object about the center of rotation. Since no torque acts on the object about its axis of rotation, its angular momentum in that direction is constant. Hence initial angular momentum = final angular momentum, mv_1r_1 = mv_2r_2, and v_2 = v_1(r_1 / r_2). Since r_1 > r_2, the object speeds up on being pulled in. In terms of angular speed, since v_1 equals \omega_1r_1 and v_2 equals \omega_2 r_2, mr_1 ^2\omega_1 = mr 2 2 \omega_2 and \omega_2 = (r_1 / r_2) ^2\omega_1, so that there is an even greater increase in angular speed over the initial value.

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Question:

In what ways are slime molds like true fungi? In what ways do they resemble animals?

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/Users/wenhuchen/Documents/Crawler/Biology/F06-0168.htm

Solution:

Both cellular slime molds (acrasiomycota) and true slime molds (myxomycota) have unusual life cycles con-taining fungus-like and animal like stages. Slime molds have membrane-bound nuclei, are heterotrophic, ingest food, lack cell walls and produce fruiting bodies. They belong to the kingdom Protista in the phylum Gymnomycota. The true slime mold's adult vegetative stage is decid-edly animal like. At this stage, the slime mold is a large, diploid, multinucleated amoeboid mass called a plasmodium. It moves about slowly and feeds on organic material by phagocytosis. Plasmodium growth continues as long as an adequate-food supply and moisture are available. When these run short, the plasmodium becomes stationary and de-velops organs specialized to produce haploid spores, known as the fruiting bodies. At this stage, the true slime mold is similar to the fungi. Meiosis occurs in the fruiting body, and spores with cellulose cell walls are released. Spores are a resistant and dormant form of a slime mold. When thw spores germinate, they lose their cell walls and become flagellated gametes. Gametes fuse to become zygotes. The zygotes lose their flagella, become amoeboid, and grow into multinucleated plasmodial slime molds. Cellular slim molds differ from the true (acellular) slime molds in being haploid and in that the amoeboid cells, on swarming together, retain their identity as individual cells. The cellular slime molds resemble amoebas through-out most of their life cycle: they lack cell walls, move about and ingest particulate matter. Under certain con-ditions, the amoebas aggregate to form a multicellular pseudoplasmodium, called a slug. The pseudoplasmodium becomes stationary and fruiting bodies are formed. Spores are not produced by cellular division, but by the formation of cell walls around the individual amoeboid cells. Each spore becomes a new amoeboid cell when it germinates. Cellular slime molds are haploid throughout their life cycles. The formation of a fruiting, spore-forming body is characteristic of fungi.

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Question:

Near the absolute zero of temperature, the specific heats of solids obey the Debye equation, c = kT^3, where T is measured in \textdegreeK. For a particular solid k has the value 2.85 × 10^-2 cal/g\bulletK deg^4. Calculate the heat that must be supplied to raise 50 g of the solid from 10\textdegreeK to 20\textdegreeK and the mean specific heat capacity of the solid in this interval.

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/Users/wenhuchen/Documents/Crawler/Physics/D13-0469.htm

Solution:

The specific heat capacity varies markedly with temperature. The mean value of the specific heat is defined by c= [1/(T_2 - T_1)] ^(T)2\int_(T)1 cdT Over the range 10\textdegreeK to 20\textdegreeK its mean value will be c= [1/(20 - 10)0K] ^200K\int_100K cdT = 1/10\textdegreeK ^200K\int_100K kT^3 dT = [1/100K] [(1/4) kT^4]200K100K= [1/400K] × 2.85 × 10^-2 cal/g\bulletK deg^4 [20^4 - 10^4] K deg^4 = 106.9 cal/g\bulletK deg. This compares with a magnitude for c of 28.5 cal/g\bulletK deg at 10\textdegreeC and 228 cal/g\bulletK deg at 20\textdegreeC. The heat that must be applied to raise the temper-ature of 50 g through the range of temperature is Q = mc(T_2 - T_1) = 50 g × 106.9 cal/g\bulletK deg × 10 K deg = 53,450 cal.

Question:

What is meant by "genetic drift"? What role may this playin evolution?

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/Users/wenhuchen/Documents/Crawler/Biology/F27-0706.htm

Solution:

The random fluctuation of gene frequencies due to chance processesin a finite population is known as genetic drift. Genetic drift may occurby what is commonly referred to as the Founder Principle. This principleoperates by chance factors alone. A group of individuals may leavetheir dwelling and establish a new feeding or breeding ground in an unexploitedarea. The individuals that find the new area may not represent thesame genetic makeup as the original population, since the individuals thatleft were a random group. This exhibits how one population may establishitself with a new genetic composition by chance alone. Genetic driftmay also affect a population without the migration of individuals. Catastrophes in one area of a certain popu-lation, such as a flood, may randomlykill individuals of a population. Those individuals which survive, doso by chance alone and not by possessing any special selective advantage. Thus genetic drift may result from natural catastrophes where arandom group of individuals may survive by chance alone and change theoriginal population's allelic frequencies. The alteration of allelic frequenciesby chance due to genetic drift is unlike the directional movementscaused by the systematic pressures of mutation, selection, anddifferential migration. The situation is quite different in a large population. In large populations, the relatively small numbers of chance variationsin gene frequencies are absorbed into the population in succeedinggenerations and the overall effect is negligible. In small interbreedingpo-pulations, however, where there are limited numbers of progenyin a given generation, random variation has a significant effect andgenetic drift can become an important factor. Random fluctuation can mostoften lead to fixation of a given allele in a small population. That is, one allele is slowly replacing the other as a result of chance, and the former becomes fixed in the population as the latter is lost. This process of reduction ofheterozygositythrough loss and fixation atvarious loci is also known as the decay of variability. The fixation of certain genes by means of genetic drift may explain theappearance of seemingly unimportant and useless structures in some organisms. These structures may be the expression of homozygous gene pairswhich have accumulated in the population by chance alone. Such a lossof variability can also inactivate natural selection, which can act only whena certain degree of phenotypic variation is present. If there is a lack ofvariation, any adverse situation, such as a period of extreme cold or the arrivalof a group of predators, could terminate an entire population. In sucha homogeneous population, only genetic mutation could hopefully preservethe population. This happens as some mutants arise which are resistantto the unfavorable force. However, since the frequency of an allelicmutation is in the area of 10-6, a homogeneous population could find itselfin great danger of extinction.

Question:

Natural chlorine is a mixture of isotopes. Determine its atomic weight if 75.53% of the naturally occurring element is chlorine 35, which has a mass of 34.968, and 24.47% is chlorine 37, which has a mass of 36.956.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E19-0708.htm

Solution:

If naturally occurring chlorine contains 75.53% chlorine 35 and 24.47% chlorine 37, one mole of chlorine contains .7553 mole chlorine 35 and .2447 mole chlorine 37. The weight of one mole of chlorine is then, equal to the sum of the weight of .7553 mole ^35CL and .2447 mole ^37CL. atomic weight of CL = (.7553) (34.968) + (.2447) (36 .956) = 26.34 + 9.11 = 35.45.

Question:

Most bacteria are heterotrophic, requiring an organic form of carbon, such as glucose, which they oxidize to obtain energy. How do chemotrophic bacteria which can utilize carbon dioxide as a sole carbon source acquire energy?

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/Users/wenhuchen/Documents/Crawler/Biology/F05-0129.htm

Solution:

Autotrophs are organisms which require only carbon dioxide as a carbon source, from which they can construct the carbon skeletons of all their organic biomole-cules. The carbon dioxide is reduced to glyceraldehyde-3-4 phosphate in order to synthesize carbohydrates. This reductive process requires much energy. Since glucose is not available to be oxidized for energy, these autotrophic bacteria must oxidize inorganic compounds. They are called chemoautotrophs, since they obtain their energy by oxidizing chemical compounds (as opposed to photoautotrophs, which obtain their energy from light). The inorganic compounds oxidized by the various bacteria include molecular hydrogen, ammonia, nitrite, sulfur (sulfide ions), and iron (ferrous ions). These oxidations result in electrons which enter the respiratory chain, with the concomitant production of ATP. ATP is then used as a source of energy in the reduction of CO_2. Bacteria of theHydrogenomonasgenus obtain energy through the oxidation of hydrogen gas. They possess an enzyme, hydrogenase, which catalyzes the following reaction: H_2 + (1/2) O_2\rightarrow H_2O + 2e^\rule{1em}{1pt} The electrons are transferred to NAD to form NADH_2, which is then oxidized in the respiratory chain, yielding ATP. The bacteriumNitrosomonasobtains energy by the oxidation of the ammonium ion: (2NH_4^+)^ + + 3O_2 \rightarrow 2NO_2^\rule{1em}{1pt} + 2H_2O + 4H^+ + 2e^\rule{1em}{1pt} The electrons produced enter the respiratory chain where they are passed down to O_2, forming ATP along the chain. Bacteria of theNitrobactergenus obtain their energy through the oxidation of nitrite ions into nitrate ions: 2NO_2^\rule{1em}{1pt} + O_2 \rightarrow 2NO_3^\rule{1em}{1pt} + 2e^\rule{1em}{1pt} Again the electrons produced are used in the formation of ATP. The oxidation of ammonia into nitrate is called nitrification. It is one of the most important activities of autotrophic bacteria since it provides the form of nitrogen most available to plants. Nitrification is carried out in two stages, with the first stage involvingNitrosomonasand the second stage featuring the oxidation of nitrites byNitrobacter. The opposite process, called nitrate respiration, is the reduction of nitrate to ammonia. This process is carried out by several heterotrophic bacteria under anaerobic conditions. The oxygen of the nitrate serves as the hydrogen acceptor (under aerobic conditions, molecular oxygen would normally serve as the final electron or hydrogen acceptor). The overall reaction is as follows: HNO_3 + 4H_2 \rightarrow NH_3 + 3H_2O Nitrification should not be confused with nitrogen fixation or denitrification. Nitrogen-fixing microorganisms in the soil use molecular nitrogen in the atmosphere as their source of nitrogen and convert it into ammonia. The ammonia is then used in the synthesis of proteins and other nitro-genous substances. Denitrification is the reduction of nitrates to molecular nitrogen carried out by certain bacteria, such asPseudomonas. The different kinds of reactions are outlined as follows:

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Question:

A and B are the players. A tells B to pick a number between 1 and 100. He then asks B to tell him the remainders when the secret chosen number is divided successively by 3, 5 and 7. A then tells B the number he originally chose.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G23-0560.htm

Solution:

The game is centered on the notion of congruence. Let x be the unknown number chosen by B, 1 \leq x \leq 100. Now x divided by 3 can give a remainder of 0, 1 or 2. The numbers between 1 and 100 can be arranged in three sets S_0 = {3,6,9,...,90,93,96,99} S_1 = {1,4,7,...,94, 97,100} S_2 = {2,5,8,...,92,95,98} Note that {S_0,S_1,S_2} forms a partition of 1 - 100. Similarly, on division by 5, the possible remainders 0,1,2,3,4 give rise to the disjoint sets S_0 = { ,5,10,...,90,95,100} S_1 = {1, 6,11,...,91,96} S_2 = {2,7,12,...,92,97} S_3 = {3,8,13,...93,98} S_4 = {4,9,14,...,94,99} Finally division by 7 yields the following Sets:S_0 = {7,14,...,91,98} S_1 = {1,8,15,...,92,99}S_2 = {2,9,16,...,93,100} S_3 = {3,10,17,...,87,94}S_4 = {4,11,...,88,95} S_5 = {5,12,...,89,96}S_6 = {6,13,...,90,97}. To be concrete, suppose x = 41. Then the remainders are 2,1, 6 when 41 is divided by 3,5 and 7 respectively. Thus, x = 41 lies in S_2 = {2,5,8,...,95,98} S_1 = {1,6,11,...,91,96} and S_6 = {6,13,...,90,97}. In fact, 41 is the only number that simultaneously lies in these three sets ! Now memorizing the above sets is one way of mastering this puzzle but fortunately, a much simpler algorithm exists. It uses the fact that 3 × 5 × 7 = 105. Let the secret number be x and let A, B, C be the remainders. Then x \equiv D = 70A + 21B + 15C (mod 105)(1) Thus, for x = 41, 70(2) + 21(1) + 15(6) = 140 + 21 + 90 = 251. But we need x \leq 100. Note that 251 - 105 = 146 and 146 - 105 = 41. Thus, 251 when divided by an integral multiple of 105 leaves a remainder of 41. We say 251 is congruent to 41 modu-lo 105 and write 41 \equiv 251 (mod 105). The program follows: 10PRINT \textquotedblleftTHINK OF A NUMBER BETWEEN 1 and 100." 20PRINT "YOUR NUMBER DIVIDED BY 3 HAS A REMAINDER OF\textquotedblright; 30INPUT A 40PRINT "YOUR NUMBER DIVIDED BY 5 HAS A REMAINDER OF\textquotedblright; 50INPUT B 60PRINT "YOUR NUMBER DIVIDED BY 7 HAS A REMAINDER OF\textquotedblright; 70INPUT C 80D = 70\textasteriskcenteredA + 21\textasteriskcenteredB + 15\textasteriskcenteredC 90IF D < = 105 THEN 120 100D = D - 105 110GOTO 90 120PRINT "YOUR NUMBER WAS\textquotedblright; D;", RIGHT?"; 130INPUT A$ 140IF A$ = "YES" THEN 170 150REM SOMETIMES HUMANS MAKE ERRORS 160IF A$ = "NO" THEN 190 165PRINT "VERY FUNNY! NOW TRY 'YES' OR 'NO'." 166GO TO 120 170PRINT "HAL HAS SPOKEN TRULY" 180GOTO 200 190PRINT "CHECK YOUR ARITHMETIC" 200PRINT "LET'S TRY AGAIN." 210GOTO 10 220END

Question:

(1)Obtain the SUM and CARRY functions for a FULL ADDER and give its circuit implementation. (2)Obtain the Difference and Borrow functions for the FULL SUBTRACTOR and give its circuit implementation.

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Solution:

In a FULL-ADDER we have to add three bits, the Augend bit, Addend bit and previous carry bit. Using the rules for addition in binary, create the truth table shown in Figure 1. A B C_i-1 SUM CARRY 0 0 0 0 0 0 0 1 1 0 0 1 0 1 0 0 1 1 0 1 1 0 0 1 0 1 0 1 0 1 1 1 0 0 1 1 1 1 1 1 Fig. 1 Now use the Karnaugh Maps of Figure 2 to find the logic expressions. SUM = AB'C' +A'BC' +ABC + A'B'CCARRY = AC+AB+BC = (AB' +A'B) C' + (AB+A'B') C= AB + (A 'B+AB') C (Distributive Law) = (A+B)C' + (A+B)' C= AB + (A+B) C (NOR Definition) = A+B+C(NOR definition) The Circuit Implementation of a full adder is shown in Figure 3. Figure 3. FULL SUBTRACTOR:A full subtractor is a combinational circuit that performs a subtraction between two bits while taking into account the fact that a 1 may have been borrowed by a lower significant position. The three inputs A,B,C denote the minuend, subtrahend, and previous borrow, respectively. The two outputs D and K represent the difference and next borrow, respectively. The truth table for the circuit is shown in Figure 4. A B C Diff Borrow 0 0 0 0 0 0 0 1 1 1 0 1 0 1 1 0 1 1 0 1 1 0 0 1 0 1 0 1 0 0 1 1 0 0 0 1 1 1 1 1 Fig. 4 Now use the Karnaugh's Maps of Figure 5 to find the logic equations. Difference = AB'C' + A 'BC'Borrow = A 'B'C + A 'BC' + ABC + A'B'C+ BC = A + B + C= A' [B'C + BC'] (same as full-adder)+ BC = A'(B+C) + BC The Circuit Implementation of a full subtractor is shown in Figure 6. in Figure 6.

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Question:

An automobile of mass 50 slugs accelerates from rest. During the first 10 sec, the resultant force acting on it is given by \sumF = F_0 -kt, where F_0 = 200 lb, k = 10 lb/sec, and t is the time in seconds after the start. Find the velocity at the end of 10 sec, and the distance covered in this time.

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/Users/wenhuchen/Documents/Crawler/Physics/D04-0131.htm

Solution:

The resultant force decreases with time, since F(t) = F_0 -ktwhere t is the elapsed time in seconds. Newton's Second Law tells us that at any time F = ma, where F is the net force on a mass m, and a is its acceleration. Hence F(t) = m a(t);a(t) = [{F(t)} /m] = (F_0 -kt)/m We integrate to find v(t), the instantaneous velocity at time t. v(t) = \inta(t)dt= \int{(F_0/m) - (kt/m)}dt= (F_0/m)t - (k/m) (t^2/2) + c_1 To evaluate the constant of integration, C_1 we use our knowledge that the car accelerates from rest. Hence, v(0) = 0 v(0) = (F_0/m)(0) - (k/2m)(0)^2 + c_1 = 0 \therefore c_1 = 0. The velocity function is therefore v(t) = (F_0/m)(t) - (k/2m)(t)^2 and, at t = 10s, v = (200 lb/50sl)(10 s) - {(10 lb/s)/100sl} 100 s^2 = Again, we integrate v(t) to find s(t), the instantaneous position at time t. s(t) = \intv(t)dt= \int(F_0/m)t - (k/2m)t^2 = (F_0/2m)t^2 - (k/6m)t^3 + c_2 We assume that the car starts at position s(0) = 0. Evaluating c_2, s(0) = (F_0/2m)(0)^2 - (k/6m)(0)^3 + c_2 = 0 \therefore c_2 = 0 The position (and displacement) function is given by s(t) = (F_0/2m)^2 - (k/6m)t^3 Att = 10, the displacement from the origin is s(10 s) = (200 lb/100sl)(100 s^2) - {(10 lb/s)/300sl}(1000 s^3) s(10 s) = 200 f - 33 (1/3) f = 166 (2/3) f.

Question:

There is a disease of tobacco plants called tobacco mosaic. Explain how one can demonstrate the causative agent.

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/Users/wenhuchen/Documents/Crawler/Biology/F05-0148.htm

Solution:

Tobacco mosaic disease causes the leaves of the tobacco plant to become wrinkled and mottled. By grinding these leaves, one can extract the juice from the infected plant. If this juice is rubbed onto the leaves of a healthy plant, it becomes infected. The agent can therefore be transmitted in the juice. If the juice is boiled before being rubbed on a healthy plant, no disease develops. The agent might therefore be a bacterium, since most bacteria cannot sur-vive temperatures above 70\textdegreeC. To isolate the bacteria, we could use filters with known pore diameters. Since most bacteria are larger than .5 \mu (a micron, \mu, is equal to 10^-6 meter), we could use a filter with a pore size slightly less than this. The liquid is passed through this very fine filter in order to remove the bacteria, and the filtrate is checked for the absence of bacteria. When the filtrate is rubbed on the leaves of a healthy plant, it still causes infection. The causative agent could therefore be a toxin produced by some bacteria or a bacterium smaller than any known . It can be demonstrated that the agent is not a toxin by showing that the agent can re- produce. If the filtrate is used to infect a healthy plant , and this plant is then ground up to obtain a new filtrate, this new filtrate could be used to infect another plant. If this procedure is repeated, and if the extent of mottling becomes decreased with subsequent infections , the agent is most probably a toxin. The attenuation of the disease can be attributed to the dilution of the toxin. However, in tobacco mosaic disease the extent of the mottling does not decrease with repetitions of filtration. One can assume that the causative agent does not become diluted. Another process must be occurring, since the agent is reproducing . Further experiments can show that the causative agent reproduces itself only inside the living plant; it cannot grow on artificial media . It is therefore not a bacterium, since bacteria do not require living host cells to reproduce. The microbial agent of this disease is termed a virus: the tobacco mosaic virus, or TMV. It can be iso-lated and crystallized and observed using the electron microscope. It is a rod-shaped virus, much smaller than any known group of bacteria: .28 \mu in length, .015 \mu in diameter.

Question:

Give an example, in pseudocode, of the REPEAT-UNTIL construct. Compare it to the DO-WHILE construct. Are there any similarities? differences?

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Solution:

In pseudocode, the REPEAT-UNTIL construct is rendered as repeat tasks A - Z until(X) The condition X can be either logical or numerical. When condition X is satisfied, program control passes out of the loop. Notice also that no matter what the status of the condition X, the enclosed statements (tasks A - Z) are executed at least once. Let us write out the REPEAT-UNTIL construct again, but this time we will include the DO-WHILE construct alongside for comparison. repeatdo while (X) tasks A - Ztasks A - Z until (X)end do while When program control passes into a DO-WHILE construct, the condition, or predicate, is tested before the enclosed statements are executed. The REPEAT-UNTIL construct, as previously mentioned, allows the enclosed statements to be executed at least once. A flowchart representation classifies the structure: The DO-WHILE construct, on the other hand, has this flowchart The DO-WHILE construct, on the other hand, has this flowchart rep-resentation: A general rule can be derived from the similarity of the two constructs: any REPEAT-UNTIL construct can be coded using the DO-WHILE construct plus a Boolean variable. For example, suppose that condition X in the earlier pseudocode is a logical variable TEST. Before entering the loop, TEST = FALSE. To exit, the loop TEST must become TRUE. However, tasks A through Z must be executed at least once, a requirement of the REPEAT- UNTIL construct. To render these notions into a pseudocoded DO-WHILE, we can write: TEST = FALSE do while \textasciitilde (TEST) tasks A - Z TEST = Q end do while The variable Q is dependent upon the execution of tasks A through Z. After evaluating tasks A through Z, Q can become either TRUE or FALSE. The loop is terminated when Q is true; the condition \textasciitilde (TEST) is not satisfied.

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Question:

The combined effects of pollution have been known to bring about the phenomena of synergism and antagonism. Explain each of these terms. Provide specific examples of each.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E24-0857.htm

Solution:

The effects of individual pollutants can be disastrous to health, while the combined action of two or more pollutants can be completely different on health and/or the environment. When the combined effects of two or more pollutants are more severe or qualitatively different from the indi-vidual effects, a condition called synergism exists. For example, in 1966 and 1967, there was injury to peanut crops due to the synergistic action of ozone and sulfur dioxide. A condition termed antagonism exists when the combined effects are less severe. For example, cyanides in industrial wastes are quite poisonous to aquatic life, however, in the presence of nickel, a nickel cyanide complex is formed. This nickel-cyanide complex is not as toxic as the cyanides originally produced.

Question:

What is the molarity of ethanol in 90-proof whiskey? Density of alcohol = 0.8 g/ml.

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Solution:

Molarity is defined as the number of moles of solute per liter of solution. By definition, 90-proof whiskey is 45 % ethanol by volume. Thus in 1 l (= 1000 ml) of whiskey there is (45 %) (1000 ml) = 450 ml of ethanol. To determine the number of moles of ethanol in 1 l of whiskey first calculate the weight of ethanol in 1 l by multiplying the density of ethanol by the volume of ethanol in 1 l. Then, divide the weight of ethanol in 1 l by its molecular weight (MW ethanol = 46 g/mole). Therefore, (450 ml ethanol) (0.8 g/ml) = 360 g ethanol molarity = [(number moles ethanol) / (1 l whiskey)] = [(360 g/46 g/ mole) / (1 l)] = 7.83 M.

Question:

The engine of a jet aircraft develops a thrust of 3000 lb. What horsepower does it develop at a velocity of 600 mi/hr = 800 ft/sec?

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Solution:

Power = Work/Time = Fs/T(1) where F is the force acting on a body and s is the displacement of the object in the direction of F in time t. By definition of velo-city v = s/t(2) v = s/t(2) Therefore, combining equations (1) and (2) P = Fv In this example, P = Fv = 3000 lb × 880 ft/sec = 2,640,000 ft\bulletlb/sec Since 1 hp = 550 ft\bulletlb/sec = [2.64 × 10^6 ft\bulletlb/sec]/[550 (ft\textbulletlb/sec)/hp] = 4800 hp.

Question:

The density, \rho of air at STP is 0.00129 g/cm^3 , the specific heat capacity at constant pressure is 0.238 cal/g - 0K, and the ratio of the principal specific heats is 1.40. what is the mechanical equivalent of heat?

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/Users/wenhuchen/Documents/Crawler/Physics/D13-0462.htm

Solution:

The equation of state of an ideal gas for one mole is PV = RT, where R is the gas constant, P is the pressure of the gas, T is its temperature in 0K and V is the molar volume. This equation may be written as, P(M/ \rho) = RT where M is the molecular weight, and \rho is the density of air. Thus, [R/M] = [P/\rhoT] = [(1.013 x 10^6 dynes cm^-2)/(0.00129 g\bulletcm^-3 × 237\textdegreeK)] = [(1.013 × 10^9)/(1.29 × 273)] ergs/g -\textdegreeK Also C_P - C_V = R, where C_P and C_V are the molar heat capacities at constant pressure and constant volume respective-ly. Thus C_p -C_v= R/M, where C_p andC_vare the corres-ponding specific heat capacities per unit mass. Further, C_P /C_V = \Upsilon. Therefore, R/M = (C_p - C_v C_v ) = C_p[1 - (1/\Upsilon)] = 0.238 cal/g - \textdegreeK × [1 -(1/1.40) ) = C_p[1 - (1/\Upsilon)] = 0.238 cal/g - \textdegreeK × [1 -(1/1.40) ] =[(0.2380.40)/ 1.40] cal/g - \textdegreeK. The value of R/M is thus given in two systems of units, one mechanical and the other thermal. The mechanical equivalent of heat is thus obtained by dividing one by the other. Hence J = [(1.013 × 10^9)/(1.29 × 273)]ergs/g -0K × [1.40/(0.238 × 0.40 cal/g -0K)] = 4.23 × 10^7 ergs/cal.

Question:

A lump of clay of mass 30 gm traveling with a velocity of 25 cm/sec collides head on with another lump of clay of mass 50 gm traveling with a velocity of 40 cm/sec in exactly the opposite direction. If the two lumps coalesce (a) what is the velocity of the combined lump after the collision, assuming that no external forces act on the system? (See figure.) (b) What is the energy lost due to collision?

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Solution:

(a) Although all the velocities are in the same straight line, it is important to remember that momentum is really a vector and to distinguish care-fully between positive and negative directions. Choose the positive direction to be to the right, in the same direction as the initial velocity of the 30 gm lump of clay. Then, Initial momentum of 30 gm lump = (30 gm) (25 cm/sec) = 750 gm - cm/sec Initial momentum of 50 gm lump = (50 gm) (- 40 cm/sec) = - 2,000 gm-cm/sec Therefore, total initial momentum =750 gm-cm/sec - 2,000 gm-cm/sec = - 1250 gm-cm/sec. If V is the velocity of the combined 80 gm lump after the collision, Momentum after collision = (80 gm)V The law of conservation of linear momentum assures that the total momentum of the system before collision is equal to the system's momentum after collision. Equating the total momentum before to the total momentum after the collision, - 1250 gm-cm/sec = (80 gm)V V = -15.6 cm/sec. The negative sign indicates that the combined lump is really traveling to the left in the diagram. There-fore, after the collision the combined lump has a velocity of 15.6 cm/sec in the same direction as the initial velocity of the 50 gm lump. The energy lost in the collision is just the difference between the kinetic energy before collision and the kinetic energy after collision: \DeltaE = Ei- E_f = [(1/2) (30 gm) (25 cm/sec)^2 + (1/2) (50 gm) (- 40 cm/sec)^2] - (1/2) (80 gm) (- 15.6 cm/sec)^2 = 49,375 ergs - 9,734 ergs = 39,641 ergs

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Question:

Find the total resistance in the circuit in the figure.

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/Users/wenhuchen/Documents/Crawler/Physics/D20-0667.htm

Solution:

This problem is simplified by observing that the circuit can be resolved into its component parts. Since resistors R_1 and R_2 are in series, we may take their sum and consider it as a single resistor (see diagram) with resistance R. Combining the first two resistances, R = R_1 + R_2 = 4 ohms + 2 ohms = 6 ohms Now we have a simple parallel circuit and can easily solve for the total resistance of the circuit. Letting R' be the total resistance of the circuit we obtain: (1/R') = (1/R) + (1/R_3) = (1/6 ohms) + (1/6 ohms) = (1/3 ohms) whence R' = 3 ohms.

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Question:

When files are used in a COBOL program, the physical envi-ronment of the program changes. Write a complete ENVIRON-MENT DIVISION for a COBOL program which will make the usage of the file called MEDICAL-DATA possible, using a card reader.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G11-0253.htm

Solution:

When files are used in COBOL programming the INPUT-OUTPUT SECTION must be included in the ENVIRONMENT DIVISION. INPUT-OUTPUT SECTION follows the CONFIGURATION SECTION, it also begins with Margin A. FILE-CONTROL, which is the first statement and paragraph header in INPUT-OUTPUT SECTION begins with margin A. The two clauses that make the FILE-CONTROL paragraph are the words SELECT and ASSIGN, which must begin on margin B. The file-name specified in the SELECT clause must be the one the programmer will call in the program. The ASSIGN clause describes to the computer exactly what physical equipment is required for that file. The general formats for ASSIGN, and SELECT are, ASSIGN system-name SELECT file-name Fig. 1 shows the complete ENVIRONMENT DIVISION with INPUT- OUTPUT SECTION entries included Variables UR-S-CARDX. following the ASSIGN clause indicate that the file is used for card input (which is a reading device) with external name CARDX. These variables change from system to system and each device has its own code name.

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Question:

On a hot day, the pressure in an automobile tire increases. Assuming that the air in a tire at 59\textdegreeF increases in pressure from 28.0 lbs/in^2 to 30.0 lbs/in^2, (a)what is the temperature of the air in the tire, assuming no change in volume? (b) what will the pressure be if the temperature rises to 106\textdegreeF?

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Solution:

(a) Because pressure, volume, and temperature are involved, one thinks of the combined gas law. This law can be stated: For a given mass of gas, the volume is inversely proportional to the pressure and directly proportional to the absolute temperature. This law can be stated algebraically as: (PV/T) = K, where P is the pressure, V is the volume, T is the absolute temperature, and K is a constant. This means that if two of the parameters are changed, the third will adjust itself so that PV/T = K is still true. The following equation is thus true: (P_1 V_1 )/T_1= (P_2 V_2 )/T_2 where P_1 is the original temperature, V_1 is the original volume, T_1 is the original absolute temperature, P_2 is the new pressure, V_2 is the new volume, and T_2 is the new absolute temperature, when V_1 = V_2, as in this problem, it follows from the formula that pressure is directly proportional to absolute temperature. In this case, the law can be written (P_1 /T_2 ) = (P_2 /T_2 ) In this problem, the temperature is given in \textdegreeF, and must be converted to \textdegreeK before it is used in the combined gas law. This is done by first converting it to \textdegreeC and then adding 273.\textdegreeP are converted to \textdegreeC by use of the following equation: \textdegreeC = 5/9 (\textdegreeF - 32) T_1 in \textdegreeC = 5/9 (59\textdegree - 32\textdegree) = 15\textdegreeC T_1 = 15 + 273 = 288\textdegreeK. One is given P_1, P_2, and T_1 and asked to find T_2. (P_1 /T_1 ) = (P_2 /T_2 )P_1 = 28.0 1bs/in^2 T_1 = 288\textdegreeK P_2 = 30.0 1bs/in^2 T^2 = ? [(28.0 1bs/in^2 )/( 288\textdegreeK )] = [(30.0 1bs/in^2 /(T^2 )] T_2 = [(30.0 1bs/in^2 × 288\textdegreeK)/(28.0 1bs/in^2 )] = 309\textdegreeK The absolute temperature can now be converted to \textdegreeF by first subtracting 273\textdegree, and then using the equation for conversion from \textdegreeC to \textdegreeF. \textdegreeF = 9/5 \textdegreeC + 32 T_2 in \textdegreeC = 309\textdegree - 273 = 36\textdegreeC T_2 in F = 9/5 (36\textdegree) + 32 = 97\textdegreeF. (b)One can again use the shortened form of the combined gas law, her (P_1 /T_1 ) = (P_2 /T_2 ) Here, one is given T_1, T_2, and P_1 and asked to find P_2. The temperatures must be converted to the absolute scale before use. The same method as used in part (a) will be applied. T_1 in \textdegreeC = 5/9 (59 - 32) = 15\textdegreeC T_1 in \textdegreeK = 15 + 273 = 288\textdegreeK T_2 in \textdegreeC = 5/9 (106 - 32) = 41\textdegreeC T_2 = 41 + 273 = 314\textdegreeK The P_2 can now be found (P_1 /T_1 ) = (P_2 /T_2 )P_1 = 28.0 1bs/in^2 T_1 = 288\textdegreeK P^2 = ? T_2 = 314\textdegreeK [(28.0 1bs/in^2 )/(288\textdegreeK)] = (P_2 /314\textdegreeK) P_2 = [(314\textdegreeK × 28.0 1bs/in^2 )/(288\textdegreeK)] = 30.5 1bs/in^2 .

Question:

A string 100 centimeters long has a mass of 4 grams and emits a frequency of 250 vibrations per second. Another string 100 centimeter long has a mass of 1 gram. What is the frequency of the second string if it is put under the same tension as the first?

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Solution:

The second string has one-fourth the mass of the first. The square root of 1/4 is 1/2 . Since the frequency varies inversely as the square root of the mass per unit length, the frequency of the second string is 2 × 250, or 500vps.

Question:

An army captain wants to fire an artillery shell deep into the enemy's flank. However, he knows that there are strong winds blowing above at height H that would blow his shells off course. If his artillery fires shells with muzzle velocity v_0, what is the farthest that he can fire them without their going off course?

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/Users/wenhuchen/Documents/Crawler/Physics/D03-0092.htm

Solution:

The equations of motion for such a projectile are the kinematical equations for constant acceleration, with v_0x = v_0 cos \texttheta and v_0y = v_0 sin \texttheta. (Here, v_0 is the initial velocity). x = (v_0 cos \texttheta)t y = v_0 sin \texttheta)t - (1/2) gt^2 where we take the +x-direction as pointing toward the enemy, the +y direction as going straight up, -g is the acceleration of gravity and \texttheta is the angle the shell makes with the horizontal axis as it is fired. We can solve for the time t' at which the shell is at its maximum height since at that point the shell's velocity in the y-direction is zero: v_y = (dy/dt) = v_0 sin \texttheta - gt = 0; t' = (v_0 sin \texttheta)/g The maximum height that the shell reaches is: y_max = (v_0 sin \texttheta)t' - (1/2) gt'^2 = (v_0 sin \texttheta) [(v_0 sin \texttheta)/g] - (1/2) g [(v_0 sin \texttheta)/g]^2 = [(v^2_0 sin^2 \texttheta)/g] - [(v^2_0 sin^2 \texttheta)/2g] = [(v^2_0 sin^2 \texttheta)/2g] But we know that the captain must aim his artillery at an angle \texttheta such that y_max = H, so that the shells just pass under the wind that would blow them off course. Thus we can solve for \texttheta: y_max =[(v^2_0 sin^2 \texttheta)/2g] = H sin \texttheta = \surd[(2gH)/(v_0)] Since the 'shell follows a parabolic path through the air, at time t', when it reaches its maximum height, it has traveled half of its maximum horizontal distance or range. Thus we can solve for its range R: (1/2)R = (v_0 cos \texttheta)t' R = 2(v_0 cos \texttheta)[(v_0 cos \texttheta)/g] = [(2v^2_0 sin \texttheta cos \texttheta)/(g)] Before we can calculate R we must determine cos \texttheta (see figure (b)). To find cos \texttheta, we must find the value of side x in the right triangle, since cos \texttheta = x/v_0. From the Pythagorean theorem: x^2 + 2gH = v^2_0 x = \surd(v^2_0 - 2gH) cos \texttheta = [\surd(v^2_0 - 2gH)]/v_0 Finally, R = [2 v^2_0 {\surd(2gH)/v_0} {\surd(v^2_0 - 2gH)/(v_0)}] / g = [2 \surd{2gH(v^2_0 - 2gH)} ]/g

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Question:

At a temperature of 18\textdegree C a beam of diffracted mono- chromatic x-rays is observed at an angle of 150.8\textdegree to the incident beam after being diffracted by a crystal with cubic structure. At a temperature of 318\textdegreeC the corresponding beam makes an angle of 141.6\textdegree with the incident beam. What is the mean co-efficient of linear expansion of the crystal in the given temperature range?

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/Users/wenhuchen/Documents/Crawler/Physics/D35-1039.htm

Solution:

If a beam of monochromatic x-rays is deviated by an angle \varphi after diffraction from a set of crystal planes, it is seen from the diagram that, if \texttheta is the glancing angle of the x-rays on this set of planes, 2\texttheta + (180 - \varphi) = 180\textdegreeor\texttheta = \varphi/2 . When the temperature is 18\textdegreeC, the appropriate interplanar spacing is d, and the glancing angle for diffraction from this set of planes is 150.8\textdegree/2 = 75.4\textdegree. Bragg's Law is m\lambda = 2d sin \texttheta where m is the order number, d is the interplanar spacing, \lambda is the x-ray wavelength and \texttheta is the angle the incident x-ray beam makes with the atomic plane from which it is reflected (see figure). Thus 2d sin 75.4\textdegree = m\lambda When the temperature is 318\textdegreeC, the interplanar spacing has increased to d', and the glancing angle has dropped to 141.6\textdegree/2 = 70.8\textdegree. Thus 2d' sin 70.8\textdegree = m\lambda, the wavelength and order of diffraction being unchanged. Hence, d'/d =sin 75.4\textdegree / sin 70.8\textdegree = 1.024. But d' = d(1 + \alpha × 300 C deg), where \alpha is the mean coefficient of linear expansion at right angles to the set of planes considered. Since the crystal is cubic, the coefficient of linear expansion is the same in all directions. \therefore1 + 300 C deg × \alpha = 1.024. \therefore\alpha = 0.024 / 300 C deg = 8.3 × 10^-5 per C deg .

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Question:

Describe the operations and built-in functions used for char-acterstring manipulations.

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Solution:

One of the most useful operations on character strings is called concatenation. Concatenation is used to join two characterstring operands, giving a single character string as a result. For example, 'JANEb' \vert\vert 'FONDA' = 'JANEbFONDA' The concatenation operator must connect the character string JANEbwith the character string FONDA. The stringJANEbFONDAwill result. The concatenation operator is represented by \vert\vert. The string resulting from the execution of a concatenation oper-ation has a length equal to sum of the lengths of the two operands (character strings). To understand the processing of variable length and fixed length string data, consider the following example. DECLARE[(FIRST_NAME, LAST_NAME) CHAR(10), STUDENT_NAME CHAR(15)] VARYING; GET LIST(FIRST_NAME, LAST.NAME); STUDENT_NAME = FIRST_NAME \vert\vertb\vert\vert LAST_NAME; If on the data card we have 'JANE', 'FONDA' thentheGET statement creates the followings VariableValueCurrent length FIRST_NAMEJANE4 LAST_NAMEFONDA5 but due to the varying attribute, no blanks are added, and STUDENT_NAME =JANEbFONDA butif the varying attribute is not specified thentheGET statement creates thefollowing: VariableValueCurrent length FIRST_NAMEJANEbbbbbb10 LAST_NAMEFONDAbbbbbb10 Hence STUDENT_NAME =JANEbbbbbbFONDAbbbbbb. It has a length of21 characters. But the declared length is 15 so the last 6 characters will betruncated, and hence the following is obtained: STUDENT_NAME =JANEbbbbbbFOND Condition Testing and comparison Operations: All eight comparison operators (>,>=,= ,<=,<,\lnot>,\textquotedblleft\lnot= ,\lnot<) , together withthe condition-testing IF statement can be used with character type operands. Comparison can only be made with character strings of the same length. If the operands are of different lengths, the shorter character string valueis expanded at the right with blank characters. If two character stringsof length zero (i.e., null strings) are compared, the result of the '=' comparisonis true. All other comparison operators re-turn a false when appliedto two null-string operands. Character strings of more than one characterare compared, character by character, in a left-to-right order. STRING BUILT-IN FUNCTIONS: For string manipulations, three built-in functions are provided in PL/I. 1) LENGTH built-in function 2) SUBSTR built-in function 3) INDEX built-in function LENGTH Built-in function:The length of character string variables declaredwith varying attributes may change during the execution of the PL/Iprogrma.It may be necessary to utilize the current length of a characterstring during cal-culations . The built-in function 'LENGTH' returns an integer value which is the currentlength of the character string. e.g.STUDENT_NAME = 'JERRY' N = LENGTH (STUDENT_NAME); LENGTH returns a value of 5 as the length of the character string. The SUBSTR Built-in function:The substring function extracts a segmentof a string (i.e. a substring) from a given character string. The functionreference uses three arguments. e.g.TAR = SUBSTR (CHAR-STRING,i, j) 1. The first argument CHAR-STRING identifies thechar-acterstring fromwhich the selection process is to take place. 2. The second argument 'i' identifies the starting position in the characterstring from where extraction of the substring is to be done. It isalways an integer value. If k is the current length of a character string, then, 0

Question:

What is the wavelength of a 10-eV electron?

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Solution:

For an electron of this low energy we can neglect the change of electron mass [^=^mrest/\surd{1 - (v^2/c^2)}] with velocity. Since kinetic energy KE = (1/2)mv^2 and momentum p =mv, we have, because v = \surd[{2(KE)}/m] p = \surd(2mKE). The de Broglie wave theory of matter tells us that every particle has wave-like characteristics. The electron, then, has a wave- length \lambda associated with it. It is given by the de Broglie relation, \lambda = (h/p) = [h/\surd(2m KE)] = (6.62 × 10^-27 erg-sec)/{\surd[2 ×(9.11 × 10^-28g) × (10eV) × (1.60 × 10^-12erg/eV)]} = 3.86 × 10^-8 cm = 3.86 \AA where 1\AA = 10^-8 cm. (The unit of energy was converted from electron volts to ergs, because the CGS System is being used.) A photon with this same wavelength would have an energy \epsilon =hvbut v = (c/\lambda) for a photon. Therefore \epsilon =(hc/\lambda) = [{(6.62 × 10^-27 erg-sec) × (3 × 10^10cm/sec)}/{(3.86 × 10^-8cm) × (1.60 × 10^-12erg/eV)}] = 3.2keV Such an energetic photon is an X-ray.

Question:

There exists a 0.5 M HF solution for whichK_diss= 6.71 × 10^-4. Determine how much dilution is necessary to double the percent dissociation of HF.

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Solution:

Percent dissociation means the ratio of [H^+] to original [HF] concentration times 100. To find the amount of dilution necessary to double the percent dis-sociation of an acid, first, establish what the percent dissociation is before dilution. This can be determined fromK_diss_' the equilibrium dissociation constants, which measures the ratio of products to reactants (i.e. their concentrations), each raised to the power of their co-efficients in the balanced chemical equation. The general reaction for the dissociation of an acid, e.g., HA, is HA + H_2O \rightleftarrows H_3O^+ + A^-. Thus, for HF;HF + H_2O \rightleftarrows H_3O^+ + F- for the dissociation reaction. This means that K_diss= 6.71 × 10^-4 = {[H_3O^+][F^-]}/[HF] . It is given that the initial concentration of HF is 0.5. Thus, to find percent dissociation, one needs to know [H_3O^+]. To find this, perform the following operations: Let x = [H_3O^+]. Since H_3O^+ and F^- are formed inequimolar amounts, as can be seen in the chemical re-action, [H_3O^+] = [F^-] = x. If the initial [HF] = 0.5, and x moles/liter dissociate to give [H_3O^+] and [F^-], then, at equilibrium, [HF] = 0.5 - x. Substituting these values: [(x \textbullet x)/(0.5 - x)] = 6.71 × 10^-4 . Solving for x, using the quadratic formula, x = 0.018 = [H_3O^+]. Thus, percentage dissociation becomes (0.018/0.5) × 100 = 3.6 % HF dissociated. Then, to determine the final answer, 7.2 % HF dissociation is needed when one dilutes. Recall that [H^+] = [F^-]. At 7.2 % dissociation, [F^-]/[HF] = [(0.072) / (1 - .072)] = (0.072/0.928). From theK_dissexpression, one has K_diss= 6.71 × 10^-4 = {[H^+][F^-]} / [HF]or (6.71 × 10^-4) / [H^+] = [F^-] / [HF] . [F^-] / [HF] = (0.072/0928) , when percent dissociation is doubled to 7.2 %. After substitution, (6.71 × 10^-4) / [H^+] = (0.072/0928) . Solving for [H^+], [H^+] = 8.65 × 10^-3 = [F^-], Substituting these actual molar concentrations into the equilibrium constant expression, one obtains [(6.71 × 10^-4) / ( 8.65 × 10^-3)] = (8.65 × 10^-3) / [HF] Solving:[HF] = 0.112 at equilibrium [H^+] = 8.65 × 10^-3. Before dissociation [HF] = amount at equilibrium plus amount dissociated = 8.65 × 10^-3 + 0.112 = 0.121 M, Thus, when the percent dissociation is doubled, the initial amount equals 0.121 M for [HF]. The [HF], when the percent dissociation was unchanged, was 0.5 M (given). Therefore, dilute by factor 0.500/0.121 = 4.13. Remember that con-centration ormolarityis a parameter of volume, actually M = moles/liter. One is going from 0.5 M to 0.121 M, which means volume must be increased by a certain factor. When one dilutes a solution, one is adding volume to it and, dividing the solution. This account for using division to obtain the factor of dilution, once initial concentrations are known.

Question:

Spherical particles of pollen are shakenup in water and allowed to stand. The depth of water is 2 cm. What is the diameter of the largest particles remaining in suspension 1 hr later? Density of pollen =1.8 g/cm^3.

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Solution:

The terminal velocity of the particles after they are allowed to settle will very quickly be reached. After 1 hr the only particles left in suspension are those which take longer than 1 hr to fall 2 cm. The larger, heavier particles have already settled. The particles which have just not settled are those which take exactly 1 hr to fall 2 cm. That is, V_T = [2 cm]/[(1 hr)(3600 s/hr)] = 1/1800cm/sec We need another expression for the terminal velocity.Stoke'slaw states that when a sphere moves through a viscous fluid at rest, the resisting force f exerted by the fluid on the sphere is given by f = 6\pi\etarv where \eta is the viscosity of the fluid, r is the radius of the sphere, and v is its velocity with respect to the fluid. The other forces which act on the sphere are its weight mg and the upward buoyant force B of the fluid. Let \rho be the density of the sphere and \rho' the density of the fluid. Then mg = (4/3) \pir^3\rhog B=(4/3) \pir^3 \rho' g(Archimedes' principle) The net force on the sphere equals the product of its mass and acceleration. Taking the downward direction as positive, mg - B - R = ma a = g - [(B + R)/m] Assuming the initial velocity is zero, this net acceler-ation imparts a downward velocity to the sphere. As this velocity increases, so does the retarding force. At some terminalv_T, the retarding force has increased an amount such that the downward acceleration equals zero. At this point, the velocity of the sphere stays constant and is found by setting the acceleration equal to zero. Then mg = B + R (4/3) \pir^3\rhog= (4/3) \pir^3 \rho' g + 6\pi\etarv_T v_T= (2/9) (r^2g/\eta) (\rho - \rho') The radius of the largest particles still just in suspension is thus given by r^2 = (9/2) [(\etav_T)/{g(\rho - \rho')}] = (9/2) [(1 × 10^-2 poise × 1/1800 cm/s)/{980 cm/s^2(1.8 - 1 )g/cm^3}] = [(10^-4)/(64 × 49)] cm^2 d = 2r = [(2 × 10^-2)/(8 × 7)] cm = 3.57 × 10^-4 cm.

Question:

A 2.4 kilowatt generator delivers 10 amperes. At what potential difference does the generator operate?

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Solution:

In most situations we cure given the voltage and amperage of a device, and are asked to calculate its power. Here we must work in reverse, calculating the potential difference (voltage) from the power and the current. Nevertheless, we resort to the same formula, namely P = El. P = 2.4 kilowatts = 2400 watts, I = 10 amp P = E × I Then E = (P/I) = (2400 watts/10 amp) = 240 volts.

Question:

An electric heater takes 4 amperes from a 120-volt line. What is its power rating?

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Solution:

E = 120 volts,I = 4 amp P = E × I = 120 volts × 4 amp = 480 watts.

Question:

Compare the bond order of He_2 and He_2^+.

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Solution:

The bond order, or number of bonds in a molecule, is equal to the difference in the sum of the number of bonding electrons and the number ofantibondingelectrons divided by two. Bond order = [(no. of bonding electrons-no. ofantibondingelectrons)/(2)] This means that the number of bonding andantibondingelectrons must be determined. There are 2 electrons in He, thus in He_2 there are 4. These electrons are all in the 1s level. For each level, there exists bonding andantibondingorbitals, each of which holds 2 electrons. Thus, in He_2, 2 electrons are bonding and 2 areantibonding. From this, the Bond order = [(2 - 2)/(2)] = 0. Thus, there are no bonds in He_2; and two He atoms will not bond together to form a molecule of He_2. In He_2 +, one electron is removed from He_2, which means that there are now three electrons present. They are all in the 1s level. This is the lowest energy level that an electron can assume. The three electrons are distributed so that two are in bondingorbitalsand one is in anantibonding orbital. Thus, no. of bonds = [(2 - 1)/(2)] = 0.5 = bond order. Because the bond order is not zero, this molecule can form.

Question:

Determine the number of cubic centimeter in one cubic inch.

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Solution:

One meter equals 39.37 inches and, since there are 100 centimetersin 1 meter, there are 39.37 inches in 100 cm. Thus, 1 inch is equalto 100/39.37 cm. 1 inch = (100/39.37)cm= 2.54 cm. By cubing both sides of this equation, one can solve for the number of cubiccentimeters in 1 cubic inch. (1 inch)^3= (2.54 cm)^3 1 inch^3 = 16.4 cc.

Question:

Why are magnolias found both in the eastern United States andeastern China, which belong to different realms, and hardlyanywhere else?

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Solution:

Early in the Cenozoic Era, the northern hemisphere was much flatterthan it is now and North America was geographically continuous withAsia by means of a land bridge at the Bering Strait. The climate then wasmuch warmer than at present, and magnolias spread over this entire connectedregion. In the later Cenozoic, the Rockies increased in height, andthe western part of North America grew colder and drier. Magnolias, whichwere adapted to a warm moist climate, disappeared from western North America. Subsequentglaciationmoving from the north wiped out anysurviving temperate plants in western North America, and many in Europe,and the western part of China. Because only regions in the southeasternUnited States and eastern China were untouched by the glaciation, magnolias were left growing in these two regions. North America, however, soon broke away from the land mass of Asia, and two separatecontinents were formed. Because of the resulting geographic barrier(ocean) between the magnolias of the two countries, the plants for severalmillion years followed separate evolutionary pathways, and have consequently become slightly different from each other.

Question:

Consider an octahedral complex having a univalent negative ion at each vertex of the octagon. Explain why the d(x)2 - (y)2 orbital of the central atom is less stable relative to the d_xy orbital.

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Solution:

Both the d(x)2 - (y)2and the d_xy orbitals are centered about the x - y plane; hence, the difference in the stability of the two orbitals must be related to their different orientations along the x- and y - axes. In the d(x)2 - (y)2orbital, the lobes are directed along the axes; whereas in the d_xy orbital, the lobes are di-rected along the bisectors of the angles between the axes (see Figure A). In the octahedral configuration the ligands (the negatively charged ions electrostatically bound to the central atom) lie on the axes (see Figure B) . Hence, the d(x)2 - (y)2orbital brings the negatively charged electrons in the lobes closer to the negatively charged ligands than does the d_xy orbital. Since the negative charge of the electrons repels that of the ligands, and this electrostatic repulsion falls off as the reciprocal of the square of the distance between the charges, the greater degree of repulsion between the d(x)2 - (y)2orbital and the ligands, relative to the repulsion between the d_xy orbital and the ligands, decreases the stability of the d(x)2 - (y)2orbital.

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Question:

Consider the following structure struct element { char keyfield; struct element \textasteriskcenteredback; struct element \textasteriskcenterednext; } } ; ; Now 1) create a doubly linked list of 5 elements. 2) Explain the logic of doubly linked lists.

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Solution:

In a doubly linked list the structure contains a) pointer to the next node in the list. b) pointer to the previous node in the list. The logic followed is that the given structure has two pointers i.e. one for the next node and one for the back or previous node. Connecting the pointers in such a way that one single node can refer its previous and its next node is the scheme followed in the program. The first element will have in back NULL value while the next will point to the next node in the list. Similarly, the last element will have a null value in its 'next' and the 'back' will point to the previous node in the list. #define NULL '\textbackslash0' structelement { char keyfieId; struct element \textasteriskcenteredback; struct element \textasteriskcenterednext; }; }; main ( ) { int i; char \textasteriskcenteredcalloc ( ); struct element \textasteriskcenteredstart, \textasteriskcenteredptr, \textasteriskcenterednew _ ptr; ptr = (structelement \textasteriskcentered) calloc(1, sizeof (struct ele-ment)); \textasteriskcentered/ /\textasteriskcenteredfollowing segment creates the 1st element ptr \rightarrow next = NULL; ptr \rightarrow back = NULL; start = pitr; /\textasteriskcenteredfollowing segment creates 4 more elements \textasteriskcentered/ for (i = 1; i < 5; i++) { new _ ptr = (structelement \textasteriskcentered) calloc(1, sizeof (structelement)); new _ ptr \rightarrow back = ptr; new _ ptr \rightarrow next = ptr \rightarrow next; ptr \rightarrow next = new _ ptr ptr = new _ ptr; } } } Diagram Representation:

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Question:

A student made three successiveweighingsof an object as follows: 9.17 g, 9.15 g, and 9.20 g. What is the average weight of the object in milligrams?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E01-0011.htm

Solution:

The average of a set of weights is found by adding together all of the weights and then dividing by the number of weighings used. avg. weight = [(9.17 g + 9.15 g + 9.20 g) / 3] = (27.52 g / 3) = 9.17 g. Now that the average weight in grams has been determined, convert it to milligrams using the conversion factor of 1,000 mg/g. 9.17 g × (1000 mg / g) = 9170 mg.

Question:

Design a 4 × 3 random access memory (RAM) using one-bit memory cells, OR gates, and a decoder.

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Solution:

Figure 1 is the configuration of a 4 × 3 RAM. Each box labeled M.C. is a one-bit memory cell whose inputs and output are designated in Figure 2. The enable input of the decoder is used as a Memory enable line. When the line is 0 the decoder is disabled and none of the memory words are selected. When the Memory enable line is 1 the decoder is enabled and one of the four words is selected, depending upon the address. The Read/Write control is connected to every M.C. When it is 1, the outputs of the selected M.C.s pass through the three OR gates to the Data-output terminals. The outputs of the non-selected M.C.s are 0, having no effect on the OR gates. When the Read/Write control is 0 the information on the data-input lines is transferred into the flip-flops of the selected word. The M.C.s in the non-selected words are disabled by the selection line so their values remain unchanged.

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Question:

C0_2(g) + H_2(g) \rightleftarrows C0(g) + H_20(g) . Initially, 49.3 mole percent CO_2 is mixed with 50.7 mole per cent H_2 . At equilibrium, you find 21.4 mole percent CO_2 , 22.8 mole percent H_2 , and 27.9 mole percent of CO and H_20 . Find K. If you start with a mole percent ratio of 60:40, C0_2 to H_2 , find the equilibrium concentrations of both reactants and products.

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Solution:

An equilibrium constant K_eg measures the ratio of the concentrations of products to reactants, each raised to the power of their respective coefficients in the chemical equation. Thus, K_eg for this reaction = { [CO] [H_2O] } / { [CO_2] [H_2] } . If you assume each substance occupies the same volume in liters, the concentration can be expressed in moles because concentration = moles/liter , i.e., liters cancel out of the equilibrium constant expression. Thus, to find K , you need to find the number of moles of each of the products and reactants and then to substitute into the equilibrium expression. You are told the final product mole per-cents. The reactants, then, at equilibrium, have mole percents that equal their initial amounts minus the amount that decomposed to pro-duce the products. Thus, at equilibrium, [CO_2] = 49.3\Elzbar 21.4 = 27.9, which was given. Similarly [H_2] = 50.7\Elzbar 22.8 = 27.9, which was given. Thus, by substitution into K = { [CO] [H_2O] } / {[CO_2] [H_2]} ,you obtain K = { (.279) (.279) } / { (.214) (.228) } = 1.60, which is the equilibrium constant for this reaction. The second part follows. You begin with a 60:40 ratio of CO_2 : H_2 , which means that initially you have .600 moles CO_2 and .400 moles H_2 . To find the equilibrium concentrations, let x = moles of CO formed. Thus, x = moles H_2 O formed since coefficients tell us they are formed in equimolar amounts. The fact that moles of a product form, means that x moles of a reactant must have de-composed. Thus, at equilibrium, you have 0.600-x moles of CO_2 and 0.400-x moles of H_2. Recalling, K = {[CO] [H_2O]} / {[CO_2] [H_2]} , you can now substitute these values to give K = [x^2 / {(0.6\Elzbar x)(0.4\Elzbar x)}]. From previous part, K = 1.60, therefore, 1.60 = x^2 / {(0.6\Elzbar x)(0.4\Elzbar x)} . Solving, x = 0.267 . Thus, [CO] = [H_2 O] = x = 0.267M, [CO_2 ] = 0.6\Elzbar x = 0.333M and [H_2 ] = 0.4\Elzbar x = 0.133M.

Question:

(1) Cu(NH_3)^2_4^+ \rightleftarrows Cu^2+ + 4NH_3k = 1.0 × 10^-12 (2) Co(NH_3)^2_6^+ \rightleftarrows Co^2+ + 6NH_3k = 4.0 × 10^-5 (3) Co(NH_3)^3_6^+ \rightleftarrows Co^3+ + 6NH_3k = 6.3 × 10^-36 (4) Ag(NH_3)^+_2 \rightleftarrows Ag^+ + 2NH_3k = 6.0 × 10^-8. Arrange them in order of increasing NH_3 concentration that would be in equilibrium with complex ion, if you start with 1 M of each.

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Solution:

To solve this problem determine the [NH_3] for each complex ion equilibria. This necessitates writing the equilibrium constant expression, which measures the ratio of concentrations of products to reactants, each raised to the power of their coefficients in the chemical reaction. Proceed as follows: k = 1.0 × 10^-12 = {[Cu^2+][NH_3]^4} / [Cu(NH_3)^2_4^+] Start with 1 M of Cu(NH_3)^2_4^+. Let x = [NH_3] at equi-librium. From the chemical reaction one can see that 1 mole of Cu^2+ is formed as 4 moles of NH_3 are formed. Thus, [Cu^2+] = (1/4)[NH_3] = 0.25 x. If x moles/liter of NH_3 form, and the only source is Cu(NH_3)^2_4^+, then, at equi-librium, [Cu(NH_3)^2_4^+] = 1 - 0.25 x. It is initial con-centration minus 1/4 of [NH_3], since they have a mole ratio of 1 : 4, as is seen in the chemical reaction. Substituting into the equilibrium equation: [{(0.25x) (x)^4} / (1 - 0.25 x)] = 1.0 × 10^-12 . Solving for x, x = [NH_3] = 5.3 x 10^-3. To obtain this answer the denominator (1 - 0.025x) is approximated as 1, since the value for x is relatively small compared to 1. To find [NH_3] at equilibrium in (2), (3), and (4), employ the same type of logic and reasoning. The only differences are the values of K and the mole ratios between the substances. For Co^+2 : 1/6 x^7 = 4.0 × 10^-5 [NH_3] = x = 0.3 For Co^+3 : 1/6 x^7 = 6.3 × 10^-36 [NH_3] = x = 9.4 × 10^-6 For Ag^+:1/2 x^3 = 6.0 × 10^-8 [NH_3] = x = 5.0 ×10^-3 In increasing order, rank them as Co^3+, Ag^+, Cu^2+, Co^2+ / increasing [NH_3] .

Question:

If you place HClO_4, HNO_3 orHClin water, you find that they are strong acids. However, they show distinct differences in acidities when dissolved in acetic acid. Such an occurrence is referred to as the leveling effect of the solvent, water: a) Explain the basis for this leveling effect by comparing acid reactions in the water solvent system to the acetic acid solvent system. b) Discuss the leveling effect in terms of basicitiesinstead of acidities.

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Solution:

(a) To explain the leveling effect, you must consider the relative acidic or basic properties of the species involved. In water, the general reaction for the acid is HA + H_2O \rightleftarrows H_3O+ + A^-. in acetic acid, however, the reaction is (assuming HA is a stronger acid than acetic acid (CH_3COOH), and the three acids given are) HA + CH_3COOH \rightleftarrows CH_3COOH_2^+ + A^-. Let us consider the strengths of the acids in these two solvent systems. Water is less acidic, thus more basic and more strongly proton-attracting, than acetic acid. This means that when strong acids are dissolved the equilibrium will be shifted far to the right in water, but not as far in acetic acid. Thus, more products will be produced in the water solution than in the acetic acid. The acidities of the 3 given acids depend upon how much H^+ ion they produce. If the equilibrium is not shifted to the right, less H^+ ion is being produced. Thus, theacidicitiesof the three given acids will be less in acetic acid, since the equi-librium is shifted less to the right, and thus, not much H^+ ion is generated. (b) The leveling effect can also be thought of in terms ofbasicity. For the two types of bases, you have B + H_2O \rightleftarrows BH^+ + OH^- in water and B + CH_3COOH \rightleftarrows BH^+ +CH_3COO^- in acetic acid. Now OH^- is a stronger base than CH_3COO^-, and, therefore, the equilibrium is further to the left in water than in acetic acid. Thus, when a base is added to water it will ionize less than in acetic acid.

Question:

A stream of gas is escaping through a small opening at one end of a large cylinder under the action of an excess pressure (relative to air pressure) ∆P = 10^4 dynes/cm^2. If the density of gas in the cylinder is \varphi = 8 × 10^-4 gm/cm^3, find the escape velocity v.

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Solution:

The excess pressure is about 10^4 dynes/cm^2 \approx 10.2 cm of water, whereas the normal air pressure P_air is about 98 3 cm of water. Since the excess pressure is much smaller than the outside pressure, we can assume that there is no appreciable compression of the gas. Then, we can treat the gas escaping through the hole as incompressible and apply Bernoulli's equation. The pressures inside and outside the cylinder are related by P_inside = P_outside + (1/2) \rhov^2 orv^2 = (2/\rho) (P_a + ∆P - P_a) orv^2 = (2/\rho) (P_a + ∆P - P_a) v= \surd(2∆P/\rho) v= \surd(2∆P/\rho) = \surd[(2 × 10^4 = \surd[(2 × 10^4 dyne/cm^2)/(8 × 10^-4 gm/cm^3)] = 5 × 10^3 cm/sec.

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Question:

The trait represented in the pedigree below is inherited through a single dominant gene. Calculate the probability of the trait appearing in the offspring if the following cousins should marry.

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Solution:

Let R represent the dominant allele for the trait, and r the recessive allele. First, we can determine the possible genotypes of the individuals involved. We know that F_1,2 ,F_1,3 , F_1,4 ,F_2,1 ,F_2,3 and F_2,4 are homozygous recessive (rr), since they do not express the trait. P_1 and P_2 must be heterozygous dominant (Rr) since they produced a child F_1,4 who is homozygous recessive. F_1 , could be either RR or Rr. However, F_2 , is homozygous recessive. Therefore F_1,1 must carry the recessive allele and be Rr. F_2,2 is also a heterozygote. We know this because one of his parents (F_1,2) is a homozygous recessive, and the cross (Rr × rr) is incapable of producing a homozygous dominant. In this cross, half the offspring are heterozygous dominant and half are monozygous recessive. Thus we can say that there is a .5 probability that any offspring of the mating will express the dominant trait. Since neither parent carries the dominant allele, it should be obvious even without doing the cross that no offspring expressing the dominant trait will be produced. Thus we can say that there is O probability that any offspring will express the trait.

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Question:

What are the basic differences in themechanical arrangement of vertebrate and arthropod joints?

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Solution:

The vertebrates are characterized by having an endoskeleton - a bony or cartilaginous framework lying within the body - surrounded by muscles . The arthropods, on the other hand, have an exoskeleton. The arthropod exoskeleton is achitinousframework on the outside of the body surrounding the muscles. The contraction of muscles in the vertebrates move one bone with respect to another. One end of the muscle is attached to one bone and the other end is attached to another bone. When the muscle contracts, one end - called the origin - remains relatively fixed while the other - called the insertion - moves. The insertion pulls along the bone to which it is attached, and bone movement results. The muscles of the arthropod lie within the skeleton and are attached to the skeleton's inner surface. Regions in which the exoskeleton is thin and flexible allow bending to occur and serve as the joints in the arthropod exoskeleton . A muscle may stretch across the joint so that its contraction will move one part on the next; or a muscle may be located entirely within one section of the body and may be attached at one end to a tough apodeme - a long, thin, firm part of the exoskeleton extending into that section from the adjoining one. (See Figure)

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Question:

Using the data from the accompanying figure, calculate the heat of reaction for the following process at 25\textdegreeC: a) CaCl_2 (s) + Na_2CO_3(s) \rightarrow 2NaCl(s) + CaCO_3(s) b) H_2 SO_4 (I) + 2NaCl(s) \rightarrow Na_2SO_4 (s) + 2HCl(g) Standard Heats of Formation, ∆H\textdegree, in kcal/mole at 25\textdegreeC. Standard Heats of Formation, ∆H\textdegree, in kcal/mole at 25\textdegreeC. Substance ∆H\textdegree CaCl_2 (s) - 190.0 Na_2CO_3(s) - 270.3 NaCl(s) - 98.2 CaCO_3(s) - 288.4 H_2 SO_4 (I) - 193.9 HCl (g) - 22.1 Na_2SO_4 (s) - 330.9

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Solution:

Calculate the amount of heat released (or absorbed) from the heats of formation. In general, ∆H for a reaction will be the sum of the heats of formation of products minus the heats of formation of reactants, each of which is multiplied by its coefficient in the equation. Once ∆H is determined, ∆E can be found from ∆E = ∆H - (∆n)RT, where ∆n = change of moles, R = universal gas constant and T = temperature in Kelvin (Celsius plus 273\textdegree) Proceed as follows: (a) CaCl_2 (s) + Na_2CO_3(s) \rightarrow 2NaCl(s) + CaCO_3(s) Thus, ∆H= 2∆H\textdegree_NaCl(s) + ∆H\textdegreeCaCO3 (s)-∆H\textdegree_CaCl2 (s) -∆H\textdegreeNa2CO3 (s) = 2(-98.2) + (-288.4) - (-190.) - (270.3) = - 24.5 Kcal/mole. ∆E= ∆H - ∆nRT. But, in this reaction, no gases appear. This means ∆nRT becomes zero. As such, ∆E = ∆H = -24.5 Kcal/mole. (b) H_2 SO_4 (I) + 2NaCl(s) \rightarrow Na_2SO_4 (s) + 2HCl(g) ∆H= ∆H\textdegreeNa2 SO4_(s) + 2∆H\textdegree_HCl _(g) - ∆H\textdegreeH2 SO4_(l) - 2∆H\textdegree_NaCl(s) = (-330.9) + 2(-22.1) - (-193.9) - 2(-98.2) = +15.2 kcal/mole. ∆E= heat of reaction = ∆H - (∆n)RT. A gas, HCI, is involved in this reaction. ∆n= moles products - moles reactants =2 - 0 = 2 T = 25 + 273 = 298\textdegreeK; R is in terms of cal. and ∆H in terms of Kcal, so that you must use the conversion factor of 1Kcal/1000 cal. Thus, ∆E= -15.2 - (2)(1.987)(298)(1/1000) = - 15.2 - 1.2 = - 16.4 Kcal/mole.

Question:

Genetic variation occurs in bacteria as a result of mutations. This would seem to be the only process giving rise to variation , for bacteria reproduce through binary fission. Design an experiment to support the existence of another process that results in variation.

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Solution:

It was originally thought that all bacterial cells arose from other cells by binary fission, which is the simple division of a parent bacterium. The two daughter cells are genetically identical because the paren-tal chromosome is simply replicated, with each cell getting a copy. Any genetic variation was thought to occur solely from mutations. However, it can be shown that genetic variation also results from a mating process in which genetic information is exchanged. One can show this recombination of genetic traits by using two mutant strains ofE.coli, which lack the ability to synthesize two amino acids . One mutant strain is unable to synthesize amino acids A and B, while the other strain is unable to synthesize amino acids C and D. Both these mutant strains can be grown only on nutrient media, which contain all essential amino acids. When both strains are plated on a selective medium lacking all four amino acids in question, some colonies of prototrophic cells appear which can synthesize all four amino acids (A, B, C, and D) .When the two strains are plated on separate minimal medium plates., no recombination occurs, and noprototrophiccolonies appear. These results may be proposed to be actually a spon-taneous reversion of the mutations back to wild type (normal prototroph) rather than recombination. If a single mutation reverts to wild-type at a frequency of 10^-6 muta-tions per cell per generation, two mutations would simul- taneously revert at a frequency of 10^-6 × 10^-6 = 10^-12 mutations per cell per generation. If one plates about 10^9 bacteria, no mutationalrevertants for both deficiencies should occur. Recombination occurs at a frequency of 10^-7, and thus if 10^9 bacteria are plated,prototrophiccolonies should be found. This recombination of traits from the two parent mutant strains is brought about through a process called conjugation. During conjugation, two bacterial cells lie close to one another and acytoplasmicbridge forms be-tween them. Parts of one bacterium's chromosome are transferred through this tube to the recipient bacterium. The transferred chromosomal piece may or may not get in-corporated into the recipient bacterium's chromosome .

Question:

The easiest method of measuring the refracting angle of a prism is to direct a parallel beam of light on to the angle (vertex A in figure) and measure the angular separation of the beams reflected from the two sides of the prism containing the refracting angle. Show that this angular separation is twice the angle of the prism.

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Solution:

Consider three incoming rays, all parallel and striking the prism at points A, B and C. Erect normals to AB at A and B and to AC at A and C. Designate the angles as in the diagram. The rays striking at B and C are reflected according to the laws of optics, as shown and the angle between the reflected rays is \beta. E is the point at which the normals at B and C meet. The sum of the angles of quadrilateral ABEC is 360\textdegree. Since \angle ABE and \angle ACE are each 90\textdegree, \alpha + \Upsilon = 180\textdegree.(1) In the quadrilateral BDCE, \beta + \Upsilon + \texttheta_1 + \texttheta_2 = 360\textdegree.(2) Since two of the angles surrounding A are right angles, \alpha + \texttheta_1 + \texttheta_2 = 180\textdegree.(3) Add Eqs.(1) and (3) 2\alpha + \Upsilon + \texttheta_1 + \texttheta_2 = 360\textdegree.(4) Subtract (2) from (4) 2\alpha - \beta = 0 or2\alpha = \beta Thus, the angle between the reflected beams, \beta, is twice the refracting angle of the prism, \alpha.

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Question:

Describe the growth zones of a typical root.

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Solution:

The tip of each root is covered by a protective root cap, a thimble-shaped group of cells. This root cap protects the rapidly growing meristemic region of the root tip. The outer part of the root cap is rough and uneven because its cells are constantly being worn away as the root pushes its way through the soil. The meristem, protected by the root cap, consists of actively dividing cells from which all the other tissues of the root are formed. It also gives rise to new root cap cells to replace the ones that are sloughed off. Immediately behind the meristem is the zone of elongation. The cells of this zone remain undifferen-tiated but grow rapidly in length by taking in large amounts of water. The meristematic region and the zone of elongation together account for the increase in length In the zone of maturation (or differentiation) of the root cells differentiate into the permanent tissues of the root, for example, the xylem and the phloem tissues. Present in this region too, are slender, elongated, hairlike projections called root hairs that arise laterally. There is one root hair to each epidermal cell. Root hairs greatly increase the surface area of the root and enhance the absorption of the water and minerals into the root. As the root elongates, the delicate, short--lived root hairs wither and die, and are replaced by the ones newly formed by the meristematic region. Root hairs occur only in a short lower segment of the zone of maturation, usually only 1 to 6 cm long.

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Question:

An airplane is traveling horizontally at 480 mph at a height of 6400 ft. The airplane drops a bomb aimed at a stationary target on the ground. To an observer on the aircraft, what angle must the target make with the vertical, when the bomb is dropped, if the bomb is to hit the target? (See the figure.) Suppose that the target is a ship which is steaming at 20 mph away from the aircraft along its line of flight. What alterations would need to be made to the previous calculations?

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Solution:

At the moment of release of the bomb, time t = 0, the airplane is at the point which is taken as the origin of the coordinate system, traveling in the positive x-direction with a speed u of 480 mph. 480 mph = 480 (mile/hr) × ( 5280 ft/mile) × (1hr/60 min) × (1min/ 60 sec) = 704 ft/sec. The bomb has the same initial speed. There is no acceleration in the x-direction, for no horizontal force acts on the system. Hence, after time t, when the bomb strikes the target, the distance traveled by the bomb in this direction is given by the kinematic equation for constant velocity, x_0 = ut. The airplane and bomb have no initial speed in the y-direction, but the acceleration g acts in this direction. After time t, the downward distance travel-ed by the released bomb will be, using the kinematic equation for constant acceleration Since ^vo_y = o, this becomes y_0 = 1/2gt^2 \bullet y_o = ^vo_y t + 1/2 gt^2. But y_0 = 6400 ft in this problem, and the time it takes the object to fall this distance is therefore t = \surd[(2y_0)/(g)] = \surd[(2 × 6400 ft)/(32 ft\bullets^-2)] = 20 s. Thus, in the same time, the object moves a horizontal distance x_0 = ut = 704 ft\bullets^-1 × 20 s = 14,080 ft andtan \texttheta = x_0/y_0 = (14080 ft)/(6400 ft) = 2.2 or \texttheta = 65.5\textdegree. The bomb should be released when the target is seen at an angle of 65.5\textdegree to the vertical. If the target is moving, the relative velocity, between plane and ship is the important velocity. For, relative to the ship, the bomb has an initial velocity v_BS\ding{217} = v_BW\ding{217} + v_WS\ding{217}, where v_BW\ding{217} is the initial velocity of the bomb relative to the water, and v_WS\ding{217} the velocity of the water relative to the ship. Since the velocity of the ship relative to water (v_SW\ding{217}) is given as 20 mph, the v_WS\ding{217} = - 20 mph. Thus v_BS = (480 - 20)mph = 460 mph = 450 (mile/hr) × (5280 ft/mile) × (1 hr/60min) × (1 min/60 sec) = 674.6 ft/sec The foregoing analysis can thus be carried out once more, with v_BS in place of u. Thus x_0 = v_BSt' = 674.6 ft/sec × 20 sec = 13,492 ft tan \texttheta' = (13,492 ft)/(6400 ft) = 2.1 and the bomb should now be released when the target is seen at an angle \texttheta' = 64.5\textdegree to the vertical.

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Question:

Given the rest masses of the various particles and nuclei involved in the fission _0n^1 +_92U235\ding{217} _56Ba^141 + _36Kr^92 + 3_0n^1 calculate the energy released in ergs and electron volts.

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Solution:

The nuclear masses will be quoted in atomic mass units or u. The atomic mass unit is 1/12 of the mass of a _6C^12 atom. The reaction is _0n^1 +_92U235\ding{217} _56Ba^141 + _36Kr^92 + 3_0n^1 The masses of the reacting particles are Mass of neutron , _0n^1= 1.0087 u Mass of _92U^235=235.0439 u Combined mass of reactants = 236.0526 u The masses of the products are Mass of _56Ba^141= 140.9139 u Mass of_36Kr^92=91.8973 u Mass of three neutrons,3_0n^1 =3.0261 u Combined mass of products = 235.8373 u The mass of the reactants is greater than the mass of the products by 0.2153 u. Hence, If all this mass is converted to energy, we have, by Einstein's mass energy relation, an equivalent amount of energy of E= mc^2 - (0.2153 u)c^2 But,lu. = 1.66 × 10^-24gm E= (0.2153 u) (1.66 × 10^-24gm/u) (9 × 10^20cm^2/s^2) = 3.27 × 10^-4erg But ,1 erg = {1 / (1.6 × 10^-12)} eVand E = (3.21 × 10^-4erg) / (1.6 × 10-12erg / eV) = 2.01 × 10^8eV(1) The energy released in the fission of one uranium 235 nucleus is 201 MeV. Conventional fuels and conventional explosives release their energy by means of chemical reactions. The energy liberated in a chemical reaction is due to a readjustment of the behavior of electrons in atoms and molecules. A good idea of its order of magnitude may be obtained by comparing the E calculated above with the energy needed to re-move an electron from a hydrogen atom(the ionization energy of hydrogen) . Ionization energy of hydrogen = 2.18 × 10^-11erg ={(2.18 × 10^-11) / (1.60 × 10^-12)}eV = 13.6 eV(2) Comparing equations (1) and (2) we see that the fission of one nucleus of uranium 235 releases about ten million times as much energy as a chemical reaction between one or two atoms. One pound of uranium releases as much energy as several thousand tons of a conventional fuel.

Question:

Write the equations for the stepwise dissociation of pyrophosphoricacid, H_4P_2O_7. Identify all conjugate acid- basepairs.

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Solution:

Pyrophosphoricacid is an example of apolyproticacid.Polyprotic acidsfurnish more than one proton per molecule. From its molecular formula, H_4P_2O_7, one can see there exist four hydrogen atoms. This might leadone to suspect that it istetraprotic, i.e. having 4 protons that can be donatedper molecule. This is in fact the case, which means there exist fourdissociation reactions. In general, the equation for a dissociation reactionis, HA + H_2O \rightarrow H_3O^+ + A^- . Polyproticacids follow this pattern. Thus, one can write the followingequations for the step-wise dissociation of H_4P_2O_7. (1) H_4P_2O_7 + H_2O\rightarrowH_3O^+ + H_3P_2O_7^- (2) H_3P_2O_7^- + H_2O\rightarrowH_3O^+ + H_2P_2O_7^-2 (3) H_2P_2O_7^-2 + H_2O\rightarrowH_3O^+ + HP_2O_7^-3 (4) HP_2O_7^-3 + H_2O\rightarrowH_3O^+ + P_2O_7^-4- To identify all conjugate acid-base pairs, note the definition of the term. The base that results when an acid donates its proton is called the conjugatebase. The acid that results when a base accepts a proton is calledthe conjugate acid. From these definitions, one sees that in all casesH_3O^+ is the conjugate acid of H_2O (the base in these reactions) and H_3P_2O_7^-,H_2P_2O_7^-2, HP_2O_7^-3 and P_2O_7^-4- are the conjugate bases of H_4P_2O_7 , H_3P_2O_7^-,H_2P_2O_7^-2 and HP_2O_7^-3 , respectively.

Question:

It is known that, when exposed to air, beryllium does not corrode but barium does. One explanation is that beryllium (Be) forms a tightly protective oxide coat whereas barium (Ba) does not. The density of BeO = 3.01 g/cc and BaO = 5.72 g/cc, find what happens to the volume per atom when the metals become oxides. The density of Be = 1.86 g/cc and of Ba = 3.598 g/cc.

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Solution:

To solve this problem, first calculate the volume per atom of Be and Ba, and then compare it with the volume per atom of their oxides. From this comparison, expansion or shrinkage can be determined. The volume per atom of any element can be found by knowing the atomic weight, the number of atoms per mole and the density of the element. Thus, there are Avogadro's number or 6.02 × 10^23 atoms per mole. You are given the densities of the sub-stances involved in this problem. Thus, by substitution you find volume of Be = MW / {(6.02 × 10^23) (density)} = (9.01 g /mole) / (6.02 × 1023a/molex1.86 g/cc) = 8.05 × 10^-24 cc/atom From similar computations, you find that BeO = 1.38 × 10^-23 cc/atom, Ba = 6.34 × 10^-23 cc/atom, and BaO = 4.45 × 10^-23. One can find the comparative size of BeO to Be by subtracting the volumeper atom of Be from that of BeO. This difference is then divided by the volume of Be and multiplied by 100. This product gives the percentage in-crease in size of the initial and final atoms when Be is oxidized. A similar process is used for Ba. For Be: {(1.38 × 10^-23 cc - 8.05 × 10^-24 cc) / (8.05 × 10^-24 )} × 100 = 71% For Ba: {(4.45 × 10^-23 cc - 6.34 × 10^-23 cc) / (6.34 × 10^-24 )} × 100 = -30% From a comparison of the atomic volumes of Be andBeO, you see that Be expands by 71 % when it forms the oxide. From a comparison of BaandBaO, you see that Ba shrinks by 30 %, when it forms the oxide.

Question:

The mass absorption coefficients for theK_\alphaandK_\betaradiations ofsilver are 13.6 cm^2 \textbulletg\Elzbar1and 58.0cm^2 \textbulletg\Elzbar1when palladium isused as an absorber. What thickness of palladium foil of density11.4g\textbulletcm\Elzbar3reduces the intensity of theK_\alpharadiation toone-tenth of its incid-ent value? What is then the percentagereduction in the intensity of theK_\betaradiation?

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Solution:

The relationwhichdescribes the absorption of x-radiation is I = I_0e\Elzbar^\muz. Solving for z (which locates a point inside the absorber) ln(I / I_0) =\Elzbar\muz orz =\Elzbar (1/\mu)ln(I / I_0) z = (1/\mu) In (I / I_0) = [1/ (\mu_mp)] In (I / I_0), where\mu is the linear absorption coefficient,\mu_mthe mass absorption coefficient, p the density of the absorber and I_0 the intensity of the incident radiation. If theK_\alpharadiation is to be reduced to one-tenth of its incident value, the thickness required is z =[ 1/ (13.6 cm^2 \textbulletg\Elzbar1× 11.4g\textbulletcm\Elzbar3)]ln10 = 1.49 ×10\Elzbar^2 cm. For theK_\betaradiation with this thickness of absorber, (I / I_0) = exp(\Elzbar58.0cm^2 \textbulletg\Elzbar1× 11.4g\textbulletcm\Elzbar3×1.49 × 10\Elzbar2cm ) = exp(\Elzbar9.85)= 5.28 ×10\Elzbar5. Hence, [(I_0 / I) / (I_0)] × 100 = [1\Elzbar(I / I_0) ] × 100 = (1\Elzbar 5.28 × 10\Elzbar5) × 100 [(I_0 / I) / (I_0)] × 100 = 99.995 % whichis the percentage reduction in the intensity of theK_\betaradiation.

Question:

A string 4.0 m long has a mass of 3.0 gm. One end of the string is fastened to a stop, and the other end hangs over a pulley with a 2.0-kg mass attached. What is the speed of a transverse wave in this string?

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Solution:

The string is stretched by a force (the tension), which is in equilibrium with the weight of the mass: T = Mg = 2.0 kg × 9.8 m/sec^2 = 19.6 nt. The linear density \mu (mass per unit length) of the string is \mu = (m/1) = 0.0030 kg/4.0m = 7.5 × 10^\rule{1em}{1pt}4 kg/m A transverse wave can be set up in the string by moving the weight at one end up and down. The speed v of this wave is given by v = \surd(T/\mu) = \surd{(19.6 kg\rule{1em}{1pt}m/sec^2)/(7.5 × 10^\rule{1em}{1pt}4 kg/m)} = \surd(2.6 × 10^4 m^2/sec^2) = 160 m/sec

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Question:

A "point-source" unshaded electric lamp of luminous in-tensity 100 candles is 4.0 ft above the top of a table. Find the illuminance of the table (a) at a point directly below the lamp and (b) at a point 3.0 ft from the point directly below the lamp.

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Solution:

a) Illuminance E is the luminousflux F incident per unit area A. Then E = F/A AlsoF = I\cyrchar\cyromega and\cyrchar\cyromega = A / R^2 If the area A is perpendicular to the path of radia-tion from the point source, the illuminance can also be expressed as E = F / A = I\cyrchar\cyromega/A = [{I(A/R^2)}/A] = I / R^2 which is the equation needed since the given quantities are the luminous intensity I and the distance R from the point source. Furthermore, we know that the area is per-pendicular to the radiation being emitted by the lamp. Therefore, E = I / R^2 = 100 candles / (4.0 ft)^2 = 6.25 lu/ft^2 b) For this second point, the area illuminated is a distance R = \surd[(4.0 ft)^2 + (3.0 ft)^2] = 5.0 ft from the lamp, which was found by using the trigono-metric relation for a right triangle. The illuminance is E = F/A whereF = I\cyrchar\cyromega. The solid angle \cyrchar\cyromega is defined as the ratio of the area upon which the source radiates at a radius R to this distance R. The area upon which the source radiates for this case can be seen from the figure to be A cos \texttheta. Therefore, \cyrchar\cyromega = (A cos \texttheta)/R^2 and E = F/A = I\cyrchar\cyromega/A = {I(A cos \texttheta/R^2)} / A = (I cos \texttheta) / R^2 Using trigonometry, cos \texttheta is found to be cos \texttheta = 4.0ft / 5.0ft = 0.80 Then the illuminance on the area A is E = (I cos \texttheta)/R2= {(100 candles)(0.80)} / (5.0ft)^2 = 3.2 lu/ft2

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Question:

A current of 10 A is flowing in a long straight wire situated near a rectangular loop, as indicated in the diagram. The current is switched off and falls to zero in 0.02 s. Find the emf induced in the loop and indicate the direction in which the induced current flows.

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Solution:

Consider the diagram, and in particular the shaded portion of width dy situated distance y from the straight current-carrying wire. The magnetic induction at all points of the shaded portion due to the current- carrying wire has the value dB = (\mu_0 /2\pi) (I/y) . where the permeability\mu_0 = 4\pi × 10^-7 N \bullet A^-2 . Therefore the flux through the shaded area is d\varphi = (dB^\ding{217}) \textbullet dA^\ding{217}. Since dB^\ding{217} is constant throughout the shaded portion and dB^\ding{217} is parallel to dA^\ding{217} where dA^\ding{217} is an element of area within the shaded portion, then d\varphi = (dB) \int dA = (dB) (bdy) = (\mu_0 /2\pi) (I/y) b dy. The flux through the whole loop is thus \varphi = \int d\varphi = (\mu_0 /2\pi) Ib0.15 m\int0.05 m(dy/y) = (\mu0 /2\pi) Ib 1n 3.(1) The limits of integration are taken from y = 5cm = 0.05 m to y = 15 cm = 0.15 m to include the flux within the whole loop. The emf induced in the loop is by Faraday's law, \epsilon = - [(∆\varphi)/∆t)]. Thus, the magnitude of the induced emf is (since the flux goes from the value in (1) to 0 in .02 s) \epsilon = [(\mu_0 Ib)/(2\pi) 1n 3 × \epsilon = [(\mu_0 Ib)/(2\pi) 1n 3 × (1/t∆) = [(4\pi × 10^-7 N \bullet A^-2 × 10 A × 0.2m × 1n 3)/(2\pi × 0.02 s)] = 22.0 × 10-6 V. The flux through the loop is decreasing. The induced current must produce effects tending to oppose the change causing it. The current must there-fore be in such a direction as to produce a flux through the loop in the same direction as that produc-ed by the current in the straight wire, i.e., into the plane of the paper. Thus the induced current traverses the loop in a clockwise direction. The negative sign in Faraday's law signifies this opposi-tion of the induced emf to the charge causing it (Lenz's law).

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Question:

A ball bearing is released from rest and drops through a viscous medium. The retarding force acting on the ball bearing has magnitude kv, where k is a constant depending on the radius of the ball and the viscosity of the medium, and v is the bearing's velocity. Find the terminal velocity acquired by the ball bearing and the time it takes to reach a speed of half the terminal velocity.

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Solution:

As the ball falls through the medium, it is accelerated by gravity and the viscous force. To find the acceleration of the bear-ing, we use Newton's Second Law to relate the net force on the ball to its acceleration. Taking the positive direction downward (see figure). mg - kv = ma, where a is the acceleration produced at any time. The initial val-ue of a is g, since at the moment of release v = 0. As the value of v increases, the acceleration decreases until, when v = v_0, the terminal velocity, a = 0. Thus mg - kv_0 = 0. Therefore v_0 = (m/k)g. In order to find out at what time v = v_0, we must calculate v as a function of t (or vice versa). At any time t, it will be found that mg - kv = ma = m(dv/dt). or(mg - kv)dt = mdv dt = (mdv) / (mg - kv) = mdv/{m(g - (k/m)v)} dt = dv/{g - (k/m)v};and^t\int_0 dt = ^v\int_0 [{dv} / {g - (k/m)v}]. where, in the integration limits, v = 0 at t = 0 and v = v at t = t. Hence t = ^v\int_0 [{dv} / {g - (k/m)v}] Lettingu = {g - (k/m)v} du = -(k/m) dv To find the new integration limits, we realize that when v = 0, u = g and when v = v, u = g - (k/m)v, whence t = ^g-(k/m)v\int_g [{-(m/k) du} / {u}] = (-m/k) ln (\vertu\vert) ^g-(k/m)v\vert_g t = (-m/k) {ln [\vert{(g - (k/m)v}\vert] - ln(g)} t = -(m/k) ln [\vert{(g - (k/m)v}/g\vert] t = -(m/k) ln \vert1 - (k/m)v\vert(1) The time to acquire half the terminal velocity, T, is thus found by inserting v = v_0/2 in (1) T = -(m/k) ln \vert1 - (k/mg) \bullet (mg/2k)\vert = -(m/k) ln \vert1/2\vert = +m/k ln \vert2\vert = 0.69 (m/k).

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Question:

Define the terms haploid, dipolid, gamete, zygote, sporophyte and gametophyte.

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Solution:

These terms refer to the life cycle and reproductive processes of algae, fungi, and higher plants. The chromosome number is termed haploid or diploid. A haploid organism is one which possesses a single complete set of chromosomes while a diploid organism has two sets of chromosomes. A gamete is a cell specialized for sexual repro-duction, and is always haploid. Two gametes fuse to form a diploid cell, which may give rise to a new multi-cellular individual via mitotic division, or which may itself undergo meiosis to form new haploid cells. The diploid cell that results when two gametes fuse is termed a zygote. The zygote is always diploid and is always a single cell. Life cycles of algae and fungi show much more variability than life cycles of higher plants and animals. The different species within a phylum often have very different life cycles, and most algae and fungi are capable of both sexual and asexual reproduction. Sexual reproduction occurs when two haploid gametes fuse to form a zygote. Asexual reproduction occurs by frag-mentation, with each fragment growing into a new plant or thallus. Asexual reproduction also involves mitotic division of a single cell of a thallus to produce daughter cells which are released from the parent thallus to grow into new algae or fungi. Daughter cells which are motile and specialized for asexual reproduction are termed zoospores. In certain species, alternation of generations occurs. This type of life cycle includes both a multi-cellular haploid plant and a multicellular diploid plant, and both sexual and asexual reproduction. In algae, fungi, and most other lower plant forms, the dominant generation is the haploid one, and the reverse is true in the higher plants. Ulva (sea lettuce) , a green alga, and Ectocarpus (a brown alga) are representative of organisms which show alternation of generations. The entire life cycle can be summarized as follows: Haploid zoospores (stage 1) divide mitotically to produce a haploid multicellular thallus (stage 2). These haploid thalli may reproduce asexually by means of zoospores or sexually by means of gametes (stage 3). Gametes fuse to produce a diploid zygote (stage 4). The zygote divides mitotically to produce a diploid multi-cellular thallus (stage 5). Certain reproductive cells of the diploid thallus divide by meiosis to produce zoospores (stage 1). The zoospores are released, and begin the cycle anew. The haploid multicellular stage is called a gametophyte because it is a plant that produces gametes. The diploid multicellular stage is called a sporophyte because it is a plant that produces spores.

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Question:

Fibbonacci.numbersare a sequence of numbers such that the first number is 0, the second number is 1, and each subsequent number is the sum of two preceding numbers, e.g. F_1 = 0 Fn- 1 + Fn - 2, n \geq 2 F_2 = 1 Write a Pascal program to print out the first 40Fibbonacci numbers and their sum.

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Solution:

We will need five variables for this program: I - to count the number ofiterations .We want only 40 iterations. SUM - to keep the current sum ofnumbers . A, B - the initial two numbers (0 and 1), they will also re-present the two predecessorsof the current number. X - stores the current number, the sum of the above two. PROGRAMFibbonacci(output); VAR A, B, X, I,SUM :integer; BEGIN A : = 0; B : =1; SUM: = 0; {initialize the sum and the 2 initial numbers} FOR I:=1 to 40 DO {Repeat 40 times} BEGIN WRITELN (A); {A is the current number} SUM: = SUM + A; {Update SUM} X: = A + B; {get the next number} A: = B; {A becomes the next predecessor} B: = X {transfer the current number to B} END;{all 40 numbers printed out} WRITEIN ('Thesum of 40Fibbonaccinumbers is', SUM) END.

Question:

Describe and compare the different forms of currently used methods of contraception. Discuss the safety and efficiency of each.

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Solution:

Reproduction in man is an extremely complex process involving many different events. Contraception or birth control acts on the principle that these events are sequential; preventing any given event from occurring will prevent the end result, which is pregnancy. The currently used methods of birth control interfere with the normal reproductive process by (1) suppressing the formation and/or release of gametes; (2) preventing the union of gametes in fertilization; or (3) preventing implantation of the fertilized egg. The oral contraceptive or "pill" acts to prevent ovulation in the female. The pill contains combinations of estrogen and progesterone, which inhibit the release of FSH (follicle stimulating hormone) and LH (luteinizing hormone) by the pituitary via a feedback mechanism; these hormones respectively stimulate follicular growth and trigger ovulation. A woman who is taking the pill in-hibits their release and does not ovulate. The pill is usually taken daily from the fifth to the twenty-fifth day of the female cycle. Withdrawal of the first five days triggers menstruation. Some synthetic hormones used in oral contraception have additional antifertility effects in that they cause a thickening of the cervical mucus and as a result, sperm have more difficulty entering the uterus. The effects of long-term use of the pill are not yet known. There does exist a relationship between increased amounts of estrogen and a higher risk of thromboembolism (the formation of blood clots). Presently, the efficiency of the oral contraceptive is close to 100%. Women using the pill have between zero and one pregnancy per hundred woman-years (a group of one hundred women using that method of contraception for one year). Fifty to eighty pregnancies per hungred woman-years will occur if no attempt is made to restrict conception. Condoms and diaphragms interpose a barrier between the male and female gametes. A condom prevents semen from entering the vagina. A diaphragm blocks the cervix and prevents sperm entry into the uterus. Spermicidal jellies and foams are used with the diaphragm to kill sperm. Condoms and diaphragms are relatively safe, and their effi-ciency results in about 12 to 14 pregnancies per hundred woman-years. The intrauterine device or IUD prevents the implan-tation of the fertilized egg. The IUD is a small object, such as a coil, placed within the uterus, which causes a slight local inflammatory response in the endometrium. This interferes with the endometrial preparation for proper implantation of the blastocyst. In some women, the IUD causes severe cramping and is sometimes spontaneously expelled. Its efficiency is about two pregnancies per hundred woman-years. Restricting intercourse to a "safe" period is known as the rhythm method. Normally, a woman ovulates once a month and abstaining from intercourse during this time can be used as a form of contraception. However, it is extremely diff-icult to precisely determine when ovulation occurs, even when taking body temperature daily (the body temperature rises slighlty following ovulation). Therefore this method is not a very reliable method. About twenty-four pregnancies per hundred woman-years occur using the rhythm method. A more permanent form of birth control is sterilization. In females, this is accomplished by tubal ligation, an operation in which the Fallopian tubes are cut and tied, preventing passage of the ovum from the ovaries to the uterus. This method is 100% effective if performed properly, but involves obvious surgical risks. Male sterilization involves a vasectomy, the cutting and tying off of the vas deferens. Again, if done properly, it is 100% effective and does not involve complicated surgery or anesthesia. The development of the vasectomy marks the beginning of the search for effective male contraceptive techniques.

Question:

The transformation of insect larvae into mature adults is known to be under hormonal control. What hormones are involved and what are their effects?

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Solution:

A hormonal secretion from the brain stimulates the release of the hormone ecdysone by the prothoracic gland. Ecdysone is the hormone that stimulates growth and molting in insects. Another hormone, present in insects prior to adulthood, is secreted by a portion of the brain known as the corpora allata. This hormone, referred to as juvenile hormone, is responsible for the maintenance of larval structures and inhibits metamorphosis. The juvenile hormone can exert its effect only after the molting process has been initiated. It thus must act in conjunction with the prothoracis hormone. When a relatively high level of juvenile hormone is present in the blood, the result is a larva-to-larva molt. When the level of the juvenile hormone is low, the molt is larva-to-pupa. Finally, in the absence of the juvenile hormone, molting gives rise to adults. Ecdysone has been chemically identified as a steroid and it has been obtained in pure form. The pure extract of ecdysone has been used to investigate the functional mechanism of the hormone. In one recent experiment, small amounts of ecdysone were injected into a gnat. Within fifteen minutes, chromosome puffs (enlargements of certain regions of the chromosome) were observed to form at certain sites on the salivary gland. Chromosomes puff formation is taken as an indication of the activation of a certain gene or set of genes. This is one piece of evidence that certain genes may be activated by hormones.

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Question:

Find the wavelength of light required for the ionization of sodium atoms to occur. The ionization potential of sodium is 8.17 × 10^-19 J.

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Solution:

The energy necessary to ionize a sodium atom is equal to the ionization potential. Here, one is looking for the wavelength of light, which, when flashed upon sodium, will possess an energy equal to the ionization potential. Wavelength is related to energy by the equation E = hc/\lambda, where E is the energy, h Planck's constant (6.626 × 10^-34 J sec), c the speed of light (3.0 × 10^8 m/sec) and \lambda the wavelength. One can rewrite this equation as \lambda = hc/E and then substitute the given to solve for \lambda. \lambda = [{(6.626 × 10^-34 j sec) (3.0 × 10^8 m/s)} / (8.17 × 10^-19 J)] = 2.43 × 10^-7 m = 2.43 × 10^-7 m × [(1 \AA) / (10^-10 m)] = 2430 \AA.

Question:

Explain bus transfer of data. Show the circuit elements Explain bus transfer of data. Show the circuit elements involved in parallel bus transfer and the symbolic representation of data transfer.

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Solution:

In a system with many registers, the transfer from one register to another requires that lines be con-nected from the output of each flip-flop in one register to input of each flip-flop in another register. Consider for example, the requirement for transfer among three registers as shown in Figure 1. There are six data paths between registers. If each register consists of F flip-flops, there is need of 6F lines for parallel transfer from one register to another. As the number of registers goes to n, there will be (2^n - 2) × F lines (where F is number of flip-flops in each register). This greatly enhances the complexity of connections and cost per bit transfer. The idea of bus transfer is analogous to a central transportation system. Instead of each commuter using his own private car to go from one location to another, a bus system is used with each commuter waiting in line until transportation is available. A bus system is formed with multiplexer and decoder circuit elements. Consider Figure 2 where there are four registers (each register has four flip-flops) and four 4 × 1 multi-plexers. The four inputs to Ml are A_1, B_1, C_1 and D_1. Similarly, for M_2, M_3 and M_4, each multiplexer is con-nected to the same two selection lines x and y. When xy = 00, I_O line of all four multiplexers gets selected and at output we have A_1A_2A_3A_4. Similarly xy = 01 selects B_1B_2B_3B_4. When xy = 10 the selection is C_1C_2C_3C_4, and xy = 11 selects D_1D_2D_3D_4. Figure 2 is symbolically represented as Figure 3. Thus, by proper selection of x and y select lines, we have the desired data on Bus. Now, the question is: How do we transfer the data on bus to the desired register? The technique is to use a Decoder whose output is used to enable the desired register in which the data is to be transferred. As shown in Figure 4, we have a 2 × 4 Decoder whose output is connected to the enable input of the registers. The Bus line carries four output wires from multiplexer. These wires, in the form of a bus, are connected to all four registers R_1, R_2, R_3 and R_4. On selection of xy = 00 we have A_1A_2A_3A_4 on the bus line; all these are going as input to the four registers of Figure 4. Yet the data is not transferred to them until they are enabled. En-abling the register is done by the 2 × 4 decoder, depending on the input v and w. If vw = 00, I_O is on and register R_1 gets enabled. Similarly, if vw = 01, I_1 is on, vw = 10, I_2 is on and if vw = 11, I_3 is on, enabling registers R_2, R_3 and R_4 respectively. Hence the data from register A is transferred to register R_2 by making xy = 00 and vw = 01. Symbolically, this is represented as x̅ y̅:BUS\leftarrowA v̅w:R2\leftarrowBUS where x̅ y̅ or v̅w before the colon (:) represent conditions and the transfer occurs only if the conditions are true.

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Question:

A rectangular post 4 in. thick is floating in a pond with three- quarters of its volume immersed. An oil tanker skids off the road and ends up overturned at the edge of the pond with oil of 1.26 slugs \textbullet ft^-3 density leaking from it into the water. When the upper face of the post is just level with the surface of the liquid, what is the depth of the oil layer? What happens if more oil keeps pouring into the pond?

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Solution:

Before the oil is spilled, the post is floating in the water symmetrically. Let its cross- sectional area be A and its density \rho. Then, if three- quarters of the volume is immersed, only 1 in. is above the surface and 3 in. below the surface. The weight downward, F_G and the buoyant force F_B balance, since the post is in equilibrium vertically. Hence if \rho_0 is the density of water, we may write F_G - F_B = 0 \rhoAg × 4" - \rho_0Ag × 3" = 0(1) or\rho/\rho_0 = 3/4 Note that the buoyant force is equal to the weight of water displaced by the post. (We have taken the positive direction downward). When the oil density \rho' pours on, it stays above the water. The water extends up to a height y and the oil fills the other (4 in. - y). Since equilibrium is still achieved, the weight downward must equal the sum of the two buoyant forces due to the water and the oil. Hence \rhoAg × 4" = \rho_0Agy + \rho'Ag(4" - y). Using (1) \rho× 4" = \rho_0 × 3" = \rho_0y + p'(4" - y). \thereforey = [\rho_0 × 3" - p" × 4"]/[\rho_0 - \rho'] = [1.94 slugs \bullet ft^-3 × 3" - 1.26 slugs \bullet ft^-3 × 4"]/[(1.94 - 1.26) slugs \textbullet ft^-3] = (0.78)/(0.68) "= 1.15 " Thus the depth of the oil layer is 2.85". If oil keeps pouring onto the pond, the post must stay as it is with respect to the water-oil interface. While the oil poured on initially, the post rose from the water to compensate for the extra up thrust from the oil by diminishing the up thrust from the water. Once the post is totally immersed, it is at the correct po-sition with respect to the water-oil interface for the sum of the two up thrusts to equal the weight. Adding further oil cannot alter this.

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Question:

Distinguish betweenapoenzymesandcofactors .

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Solution:

Some enzymes consist of two parts: a protein constituent called an apoenzymeand a smaller non-protein portion called a cofactor. The apoenzymecannot perform enzymatic functions without its respective cofactor. Some cofactors, however, are able to perform enzymatic reactionswithout anapoenzyme, although the reactions proceed at a muchslower rate than they would if the cofactor andapoenzymewere joinedtogether to form what is called aholoenzyme. The cofactor may be either a metal ion or an organic molecule calleda coenzyme. A coenzyme that is very tightly bound to the apoenzymeis called a prosthetic group. Often, both a metal ion and a coenzymeare required in aholoenzyme. Metalloenzymesare enzymes that require a metal ion;catalase, for example, which rapidly catalyzes the degradation of H_2O_2 to H_2O and O_2, needsFe+^2 or Fe+^3. Iron salts can catalyze the same reaction on their own, butthe reaction proceeds much more quickly when the iron combines with theapoenzyme. Coenzymes usually function as intermediate carriers of thefunctional groups, atoms, or electrons that are transferred in an overall enzymatictransfer reaction.

Question:

How fast does light travel in glass of refractive index 1.5?

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Solution:

By definition, refractive index n is the ratio of the velocity of light in vacuum (3.00 × 10^10 cm/sec) to the velocity of light in the medium in question. Thereforen = [(3 × 10^10 cm/s)/v] = 1.5 Thereforev = [(3 × 10^10 cm/s)/1.5] = 2.00 × 10^10 cm/sec.

Question:

Compare the processes of oxidativephosphorylationand substratelevelphosphorylation. How can certain evolu-tionarychanges be explained by these differences?

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Solution:

In the complete oxidation of glucose to carbon dioxide and water, NADH (reducednicotinamideadeninedinucleotide) is produced. High energyelectrons are transferred from NADH to oxygen via the electron transportsystem. This transfer of electrons involves a series of enzyme catalyzedoxidation - reduction reactions in which the energy released is usedin the production of high energy phosphate bonds. Three high energyphosphate bonds are made for each pair of electrons transferred tooxygen from NADH. This process of harnessing the energy released in theelectron transport chain to synthesize ATP from ADP and inorganic phosphateis termed oxidativephosphorylation. The electron transport chainis not the only system that produces high energy bonds.Atpis pro-ducedin bothglycolysisand the Krebs cycle. Any reaction which synthesizes high energy phosphate bonds but isnot involved in the electron transport system, is called a substrate level phosphorylation. Inglycolysis, a net total of 2 \simP s (high energy phosphate bonds) are produced at the substrate level; in the TCA cycle, one \simP is synthesizedat the substrate level. Most of the ATP produced during aerobicrespiration, however, is synthesized by oxidativephosphorylation. Indeed, therein lies the source of the differences between aerobic and anaerobicorganisms, and the reason that anaerobic organisms must be verysmall, and almost always unicellular. The high energy requirements ofa larger organism could not beaccomodatedthrough the 3 ATP producedvia substrate levelphosphorylation.

Question:

An important application of computers to market research is in computer simulation of continuous models. As an example, there appears to be a positive relationship be-tween the number of new houses built and the sale of wash-ing machines. Give a differential equation connecting the growth of houses and washing machines. Illustrate with graphs.

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Solution:

A mathematical model is said to be continuous when the variables of the system change smoothly over time. Quantities that change continuously over time are best studied with the use of differential equations. Let H be the potential number of families and y be the number of families with houses. H is a constant upper limit on the number of houses and, thus, the number of washing machines that can be sold. Furthermore, assuming that every new house requires a washing machine, the un-filled market is the difference between houses sold and washing machines sold. Finally, as the number of new houses approaches H, the market approaches saturation and the rate of selling houses and, by extension, washing machines drops. The given figure applies: From the figure, the slope of the function between number of houses sold and time is seen to decrease with time, reaching zero at the Time when y(t) = H. Let y = y(t).Then y(0) = 0. If we assume that the rate of new houses sold is a constant function of the remaining potential market, the differen-tial equation is ẏ = k_1 (H - y)y(0) = 0(1)

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Question:

An excellent method of making alcohols is the reaction of an aldehydeon a Grignard reagent. If you perform this reaction, you must scrupulously dry thealdehydebefore adding it to the Grignard reagent. Why?

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Solution:

To answer this question, you must know what a Grignard reagent is and some of its properties. The general formula of agrignardreagent is RMgX, where R is an alkyl or aryl group, Mg is magnesium, and X is a halogen atom (usually bromine or chlorine). Grignard reagents are extremely reactive. You can consider it to be the magnesium salt of the weak acid R\ElzbarH. Thus, if there exists in solution a stronger acid, R\ElzbarH will be displaced from its salt. Water is a stronger acid than R\ElzbarH. Thus, if H_2O (water) is present, the following re-action occurs: RMgX+ H_2O \rightarrow RH + Mg(OH)X. Thus, if thealdehydeis not dried, water will be present, and it will then react with thegrignardreagent producing RH. The presence of water reduces or eliminates thegrignardreagent from solution, so that little or none is present to react with thealdehyde.

Question:

Obtain the 1's and 2's complements of the following binary numbers: a)1010101 b)0111000 c)0000001 d)10000 e)0000

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Solution:

Forming complements is useful when one wants to perform subtractionwithout borrowing from another column. This is because subtractionis the same as adding the two's complement. a) The 1's complement of 1010101_2 is found by subtracting it from 1111111 .Thus 1111111 -1010101 0101010 The two's complement is formed by subtracting1010101_2 .from 10000000. Thus, 10000000 - 1010101 00101011 Forming the two's complement by binary subtraction is difficult. b) Since the base, or radix, of the binary systemis2, the one's complementis referred to as the radix minus one complement. Similarly, thetwo's complement is called the radix complement. The radix minus one complementof 0111000_2 is 1111111 - 0111000 1000111 The radix complement of 0111000_2 is 10000000 - 0111000 1001000 While examining a) and b), note that finding the radix minus one complementsof the two given numbers is equivalent to changing 0's to 1's and1's to 0's in the given numbers. c) The radix minus one complement of 0000001 is 1111110. The radix complementis 10000000 - 0000001 1111111 Examining the radix complements of a), b) and c) observe that Radix complement(X)_2 = Radix complement minus one(X)_2 + 1(1) whereX is a binary number. d) The one's complement of 10000 is 01111 or 1111. The two's complement, using (1) is 1111 +1 10000 or, the number itself. e) The radix minus one complement of 0000 is 1111 and the radix complementis 1111 +1 10000 To illustrate the advantage of representing numbers by their radix complements, suppose we wish to store the negative of d), i.e., -10000, in a12 - bit word. The left-most bit has value 1 to represent the minus sign: 1000 0001 0000 Now suppose word size is increased to 16 bits. This is most easily done byadding 4 bits with zero value on the left. But the left-most bit must be changedto 1 and the 1 in the third quartet of bits (third from the left) must bechanged to zero: 1000 0000 0001 0000 Suppose ,instead, that the radix complement of -10000 is in a 12-bit storage. The radix complement of -10000 in a 12-bit storage is 111111111111 -10000 111111101111 +1 111111110000 Now, if word length is to be extended to 16 bits, simply add a quartet of 1's. No special provisions need be made for the sign bit. In other words, whenword size is increased, simply copy the sign bit into all the bits to its left. Finally, the two's complement representation is preferred to the one'scomplement because + 0 \not = - 0 in the one's complement, but + 0 = - 0 in the 2's complement.

Question:

A sample of gold is exposed to a beam of neutrons, and the reaction Au^197 + n = Au^198 + \Upsilon absorbs 10^6 neutrons per second. Au^198 emits \beta-particles and has a half-life of 2.70 days. How many atoms of Au^198 are present after 2 days of continuous irradiation?

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Solution:

At any instant during the irradiation, the increase in the number of Au^198 atoms in unit time will be the number being produced less the number disintegrat-ing. Thus, if \lambda is the decay constant for the \beta decay, dN/dt= 10^6 s\Elzbar1\Elzbar\lambdaN. This follows because the reaction occurs once for every neutron (n) absorbed. If 10^6 neutrons are absorbed per second, 10^6 Au^198 atoms (as well as 10^6 \Upsilon ray photons) are produced per second. dN/(10^6 s\Elzbar1\Elzbar\lambdaN) =dt^N\int_0dN/(10^6 s\Elzbar1\Elzbar\lambdaN) =^t\int_0dt(1) there being no atoms of Au^198 present at time t = 0 when the irradiation starts. Letting u = 10^6 s\Elzbar1\Elzbar\lambdaN du =\Elzbar\lambdadN Since u = 10^6 s\Elzbar1when N = 0, and u = 10^6 s\Elzbar1\Elzbar\lambdaNwhen N = N, (1) becomes \Elzbar(1/\lambda)(10)6(s)\Elzbar1\Elzbar(\lambdaN)\int10^6_s\Elzbar1du/u = ^t\int_0dt or\Elzbar(1/\lambda)In (u)(10)6(s)\Elzbar1\Elzbar(\lambdaN)▏_10^6_s\Elzbar1= t \Elzbar(1/\lambda)In [(10^6s\Elzbar1\Elzbar\lambdaN) / (10^6s\Elzbar1)] = t In[1\Elzbar {(\lambdaN) / (10^6s\Elzbar1)}] =\Elzbar\lambdat and1\Elzbar {(\lambdaN) / (10^6s\Elzbar1)} = exp (\Elzbar\lambdat) N = [(10^6s\Elzbar1) / \lambda] [1\Elzbarexp(\Elzbar\lambdat)](2) Note that in general, the equation of radioactive decay is N = N_0e\Elzbar\lambdat if N_0 is the original number of atoms which are de-caying. After one half life (\tau) , N0/2 atoms are left and N(\tau) = N0/2 = N_0e\Elzbar\lambda\tau or2 = e\lambda\tau Hence\lambda\tau= In 2 and \lambda = 1/\tauIn 2. Using this fact in (2) N = [{(10^6 s\Elzbar1)(\tau)} / (In 2) ] [ 1\Elzbarexp{\Elzbar[(In 2) t} / \tau] } ] where \tau is the half-life period of the \beta-activity. After 2 days, N = [{(10^6 s\Elzbar1) (2.7 days)} / (.6932) ] [ 1\Elzbarexp[\Elzbar[{(In 2)(2 days)} / (2.7days) ] ] ] Since 1 day = 86,400secs N = [{(10^6 ) (86,400)} / (.6932) ] [ 1\Elzbarexp{\Elzbar (2 / 2.7) In 2} ] N = 1.35 x 10^11 atoms.

Question:

A damped harmonic oscillator (DHO) is a system whose behavior can be described by a second-order, linear differential equation of the form: ӱ =-A_1ẏ - A_2y(t)(1) where A_1, A_2 are positive constants. One DHO system is a spring- mass-damper system shown in Figure #1. This could be a simple model of an automobile wheel suspension system (assuming the automobile's body is immobile in a vertical direction). Then the damper acts as a shock absorber. Another DHO system is an RLC electric circuit: The analogy between these two DHO systems is given in Table #1. DHO Y(t) Ẏ = dy/dt A1 A_2 Spring- mass- damper x-displacement V = velocity D/M K/M RLC circuit q-charge in coulombs i = current in amperes R/L 1/LC Given the initial conditions y(0) = y_0 and ẏ(0) = z_0 , write a Given the initial conditions y(0) = y_0 and ẏ(0) = z_0 , write a FORTRAN program that uses the modified Euler method to simulate this system from t = 0 to t = t_f- .

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Solution:

Note that equation (1) is of the form d^2y/dx^2 = g(y,y'). The modified Euler subroutine will be written for this form. First, rewrite equation (1) as a system of simultaneous first-order equations: y = dy/dt = z, z(0) = y' (0) = z_0(2) z = dz/dt = -A_1Z - A_2y , y(0) = y_0(3) Now, use the modified Euler method to find approximate solutions to both (3) and (4). REAL T/0.0/,TFIN,N,ACCUR,Y0,Z0,A1,A2 COMMON A1,A2 READ, N, TFIN, ACCUR, YO, ZO, A1, A2 CALL EULER2(N,TFIN,T,Y0,Z0,ACCUR) STOP END FUNCTION G2(W,X) REAL G2,W,A1,A2 COMMON A1,A2 G2 = -A1\textasteriskcenteredX-A2\textasteriskcenteredW RETURN END SUBROUTINE EULER2 (N,XN,X,YCOR, ZCOR,ACCUR) REAL N DELTAX = (XN - X)/N 50Y = YCOR Z = ZCOR G2OLD = G2(Y,Z) PRINT,X,Y,Z X = X + DELTAX IF (X.GT.XN) STOP YPRED = Y + Z\textasteriskcenteredDELTAX ZPRED = Z + G2(YPRED,Z)\textasteriskcenteredDELTAX 100YCOR = Y + 0.5\textasteriskcentered(Z + ZPRED)\textasteriskcenteredDELTAX ZCOR = Z + 0.5\textasteriskcentered(G2QLD + G2(YPRED,ZPRED))\textasteriskcenteredDELTAX YDIF = ABS(YCOR - YPRED) ZDIF = ABS(ZCOR - ZPRED) IF(YDIF.LE.ACCUR.AND.ZDIF.LE.ACCUR) GO TO 50 YPRED = YCOR ZPRED = ZCOR GO TO 100 RETURN END

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Question:

A chemist wants the percent dissociation of HC_2H_3O_2 to be 1 %. If theK_dissof HC_2H_3O_2 is 1.8 × 10^-5, what con-centration of HC_2H_3O_2 is required?

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Solution:

To answer this question, consider the dissociation of an acid in H_2O. Then write anequlibriumconstant ex-pression which relates the equlibriumconstant for dis-sociation,K_dissto the ratio of the concentrations of products to reactants, each raised to the power of its coefficients in the reaction. In general, for an acid, there exists the following reaction for dissociation: HA + H_2O \rightarrow H_3O^+ + A^-, where HA is the acid. For this reaction, HC_2H_3O_2 + H_2O \rightarrow H_3O^+ + C_2H_3O_2^- . Theequlibriumconstant expression is K_diss= 1.8 × 10^-5 = {[H_3O^+][C_2H_3O_2^-]} / [HC_2H_3O_2] One wants to find [HC_2H_3O_2] with the knowledge that the percent dissociation is 1 % (note: exclude water from the expression, as it is assumed to be constant). If the percent dissociation is 1 %, then [H_3O^+] / [ HC_2H_3O_2] = 1/99. The H^+ can only come from HC_2H_3O_2. If 1 % of H^+ exists, then 100% (initial percentage) - 1% = 99% of HC_2H_3O_2 must be left at equilibrium. Thus, 1.8 × 10^-5 = (1/99) [C_2H_3O_2^-]. Solving for [C_2H_3O_2^-]; [C_2H_3O_2^-] = 1.78 × 10^-3 M. This concentration of C_2H_3O_2^- must equal the con-centration of H_3O^+, since they are formed inequimolaramounts. One can substitute into the equation 1.8 × 10^-5 = {[H_3O^+][C_2H_3O_2^-]} / [HC_2H_3O_2] to obtain{(1.78 × 10^-3)( 1.78 × 10^-3)} / [HC_2H_3O_2] = 1.8 × 10^-5 . Solving for [HC_2H_3O_2)], 1.76 × 10^-1 M.

Question:

For the following PL/I program, a) Explain how a condition called ENDFILE CONDITION' arises. b) Show what will be printed out if the program is run. c) Write out a corrected program to avoid errors due to this condition . d) Show what the printout from the corrected program will be . EXAMPLE:PROCOPTIONS(MAIN); DCL(X,Y,Z) FIXED(3) INIT(0); GETLIST(Z); DO I = 1 TO Z; GET LIST(X); Y = Y + X; PUTLIST(I,Y); END; PUTLIST(21,Y); END EXAMPLE; The DATA CARD contains the following data: 8, 0, 1, 2, 3, 4, 5, 6

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Solution:

a) The purpose of the above program is to illus-trate a condition knownas ENDFILE CONDITION. In this program, the GETLIST(Z) statement reads a value for Z fromthe DATA CARD. Hence, Z = 8. This value of Z be-comes the upper limitfor the loop of the next statement. Hence, the loop statement becomes: DO I = 1 TO 8. During each loop, a value of X is read from the card and the sum Y = Y + x is calculated. Now, observe that there are only 7 values of X (0 to 6) punched on theDATA CARD. Hence, when the program enters the eighth loop, the computerfinds that the data have run out and terminates the program. This condition of data running out is known as an ENDFILE CONDITION, b) The printout of the computer is as follows: 10 21 33 46 510 615 721 After the seventh loop, Igetsa value of 8. Now the program tries to executeGET LIST(X). But since there is no 8th value ofX,the program haltsgiving out an error message. So there is no chance for executing the otherPUT LIST statement which follows outside the loop. c) The program can be corrected by inserting an ON statement of the generalform ON ENDPILE (FILE NAME) ACTION; The above statement prevents termination of the program abrupt-ly on the occurrenceof the ENDFILE CONDITION. Inside the parentheses, the name of the file,which is to be used, shouldbe specified. For example, if punched cards are being used for input, the name of the input file is (SYSIN). 'ACTION' in the above statement means some sort of in-struction to thecomputer telling it what to do if the data on the file run out. The correctedprogram can therefore be written as follows: EXAMPLE:PROCOPTIONS(MAIN); DCL (X,Y,Z) FIXED(3), INIT(O) ; ONENDFILE(SYSIN) GOTO A ; GETLIST(Z); DO I = 1 TO Z; GET LIST(X) ; Y = Y +X ; PUTLIST(I,Y) ; END ; A:PUTLIST(21,Y); END EXAMPLE; d) The printout from the corrected program will be: 10 21 33 46 510 615 721 1621 Note that after the seventh loop is completed, the computer goes back to theloop head, viz., DO I = 1 TO Z. As Z = 8, therefore I is incremented by 1 from 7 to 8. Now, on trying to executethe next statement, viz., GET LIST(X), the computer finds that an ENDFILE CONDITION has occurred on the SYSIN file. Hence, the programgoes to the statementlabelled'A'. This statement is PUT LIST(2I,Y);. The lat-est values of I and Y lying in the program are 7 and 22 re-spectively; therefore 16 and 22 are printed out. The pro-gram now goes tothe next statement END EXAMPLE which is a normal end of the program.

Question:

Write a BASIC program to reduce a fraction to lowest terms. Test your program with sample data.

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Solution:

We make use of the INT function to find the largest factor of an integer. What we need to do is find the largest common factor of the numerator and denominator. A flowchart illustrating the program logic is given below: 1\O READ N,D 2\O FOR P = N TO 2 STEP - 1 3\O IF N/P = INT(N/P) THEN 7\O 4\O NEXT P 5\O PRINT N'V'D 6\O GO TO 1\O 7\O IF D/P = INT(D/P) THEN 9\O 8\O GO TO 4\O 9\O PRINT N"/"D" ="N/P"/"D/P 1\O\O GO TO 1\O 11\O DATA 5,6 12\O DATA 82, 48 13\O END Note that the absence of commas in both PRINT statements tells the computer to write the numerical values of the variables and the arith-metic operators right after each other, not leaving any blank space in between.

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Question:

Calculate, approximately, the relative intensities of the secondary maxima in the single-slitFraunhoferdiffraction pattern.

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Solution:

Theseconderymaxima lie approximately halfway between the minima and are found from \alpha \cong (m + 1/ 2)\pim = 1,2,3,.... Substituting into the formula for intensity in single-slit diffraction yields I_\texttheta=I_m[{sin(m + 1 / 2)\pi} / {(m + 1 / 2)\pi}]^2 , which reduces to (I_\texttheta/I_m) = [1 / {(m + 1 / 2)^2\pi^2}] . This yields, for m = 1,2,3,...,I_\texttheta/I_m= 0.045, 0.016, 0.0083, etc. The successive maxima decrease rapidly in intensity.

Question:

0.1 moles of Cu(NO_3)_2 are added to 1.0molesof NH_3 and then diluted to 1000ml. Find the concentration of Cu^2+ formed. K_eq= 8.5 × 10^-13 .

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Solution:

The first thing to realize is that the complex Cu(NO_3)_2 dissociates to Cu^2+ + 2NO_3^+ in H_2O Cu(NO_3)_2 \rightleftarrows Cu^2+ + 2NO_3^+ when NH is added to the solution the following reaction occurs. Cu^2+ + 4NH_3 \rightleftarrows Cu(NH_3)_4^2+ Cu(NH_3)_4^2+, is called a complex ion. Cu(NH_3)_4^2+, will be in equilibrium with Cu^2+ 4NH_3 . You are given theK_eqfor this reaction which is defined as K = {[Cu^2+][NH_3]^4} / [Cu(NH_3)_4^2+] . In general, if you have the equilibrium: H(A)_X \rightarrow H^+ + XA^- , then K ={[H^+][A^-]^X} / {H(A)_X}]. That is, the coefficients of the balanced equation serve as the ex-ponents in the K expression. The problem stated K = 8.5 × 10^-13 . Therefore, {[Cu^2+][NH_3]^4)} / [Cu(NH_3)_4^2+] = 8.5 × 10^-13 . You are solving for [Cu^2+]. Let x = [Cu^2+], then [Cu(NH_3)_4^2+] = .1-x, since it must be the original amount of Cu(NO_3)_2. present minus the Cu^2+ formed. There is originally 1.0 moles NH_3, at equilibrium 4 moles of NH_3 will have reacted with each mole of Cu^2+ to form Cu(NH_3)_4^2+, thus, [NH_3] may be represented as 1.0-4 × [Cu(NH_3)_4^2+] = 1.0-4(.10-x). Since most of the Cu^2+ present will be in the form of Cu(NH_3)_4^2+ and that there was only .1M of it in the first place, one can make the assumption that x << .1, therefore, .10-x = .1 and 1.0 - 4(.10-x) = 1.0 - 4(.1) = 0.6 = [NH_3] . Using this same assump-tion, [Cu(NH_3)_4^2+] = .1-x = .1 . You now substitute these values into the previous expression. That is, [{x(0.6)^4}/(.1)] = 8.5 × 10^-13 , x = 6.5 × 10^-13 . Therefore, [Cu^2+] = 6.5 × 10^-13 M.

Question:

A shower of protons from outer space deposits equal charges +q on the earth and the moon, and the electrostatic repulsion then exactly counterbalances the gravitational attraction. How large is q?

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Solution:

If R is the distance between the earth and the moon, the electrostatic force in the CGS system is F_e = q^2 / R^2 If M_e and M_i are the masses of the earth and the moon respectively, the gravitationalforce is F_G = (GM_eM_i) / (R^2) Since the two forces are equal, (q^2) / (R^2) = (GM_eM_i) / (R^2) q = \surd(GM_eM_i)

Question:

A small object of weight w^\ding{217} hangs from a string of length l. as shown in the figure. A variable horiz-ontal force P^\ding{217}, which starts at zero and gradually increases, is used to pull the object very slowly (so that equilibrium exists at all times) until the string makes an angle \texttheta with the vertical. Calculate the work of the force P^\ding{217}.

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Solution:

The object is in equilibrium, meaning that its acceleration is zero and the net force acting on the weight is zero. Consider the forces acting on the object, as shown in the diagram. We can say \sumF_x = 0 = P - T sin \texttheta(1) P = T sin \texttheta(2) and\sumF_y- = 0 = T cos \texttheta - W(3) W = T cos \texttheta(4) Dividing eq. (2) by eq. (4), we get P = W tan \texttheta Since P is variable, the work done by it must be found through integration. Recall that work is the integral of the dot product of the force F^\ding{217} and the displacement vector dx^\ding{217}: W = ^x2\int_x1 F^\ding{217} \bullet dx^\ding{217} = ^x2\int_x1 F cos \Upsilon dx where \Upsilon is the angle between F^\ding{217} and dx^\ding{217}. In this case, the force is P, the differential displacement is ld\texttheta and the angle between the two is \texttheta. Substituting these expressions, we have W = ^\texttheta\int_0 P^\ding{217} \bullet ld\texttheta^\ding{217} = ^\texttheta\int_0 (\omega tan \texttheta)(cos \texttheta)(l) d\texttheta = \omegal^\texttheta\int_0 sin \texttheta d\texttheta = - \omegal cos \texttheta\vert^\texttheta_0 = \omegal(1 - cos \texttheta) This result can also be derived using conservation of energy. Since the object's initial and final velocity is zero, kinetic energy is not involved. The change in the object's potential energy must be due completely to the work done on the weight by the force P^\ding{217}. This change in potential energy, ∆PE, is ∆PE = Wh = W(l - x) Butx = l cos \texttheta Therefore, we have ∆PE = \omega(l - l cos \texttheta) = \omegal(1 - cos \texttheta) This is equal to the work: W = \omegal (1 - cos \texttheta)

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Question:

A particle of rest mass M moving with speedv_ocollides with and sticks to a stationary particle of rest massm. What is the speed of the composite system after collision?

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Solution:

In fact, we are asked to find the speed ofthe center of mass of the system, since it coin-cides with the composite particle after the collision. The velocity, momentum and total energy of a relativ-istic particle are related to each other as v^\ding{217} / c = (p^\ding{217} c) / E Therefore, we can define the center of mass of a composite system as v^\ding{217}_cm/ c= (\sum_ip_i^\ding{217} c) / (\sum_iE_i) Since there are no forces acting on the system, the velocity of the center of mass does not change during the collision. Before the collision, we have p^\ding{217}_m= 0 ,E_m= mc2 p^\ding{217}_M = M \Upsilon_0 v^\ding{217}_0,E_M = MC^2 \Upsilon0 , where \Upsilon_0 = 1 /\surd{1 ─ (v_0^2 / c^2)} . Hence v^\ding{217}_cm/ c = (p^\ding{217}_Mc) / (mc2+ MC^2 \Upsilon_0 ) v_cm_ / c= (M\Upsilon_0v_0)/ [(m + M\Upsilon_0)c՛] v_cm_ = (M\Upsilon_0)/ [(m + M\Upsilon_0) v_0].

Question:

The heart is removed from the body and placed in an isosmotic solution. Although it is completely separated from nerves, it continues to beat. Explain.

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Solution:

The initiation of the heartbeat and the beat itself are intrinsic properties of the heart and are not dependent upon stimulation from the central nervous system. The heart is stimulated by two sets of nerves (the sympathetic and vagus nerves), but these only partly regulate the rate of the beat, and are not responsible for the beat itself. The initiation of the heartbeat originates from a small strip of specialized muscle in the wall of the right atrium called the sino-atrial (S-A) node, which is also called the pacemaker of the heart. It is the S-A node which generates the rhythmic self-excitatory impulse, causing a wave of con-traction across the walls of the atria. (See figure). This wave of contraction reaches a second mass of nodal muscle called the atrio-ventricular node, or A-V node. The atrio-ventricular node is found in the lower part of the interatrial septum. As the wave of contraction reaches the A-V node, the wave stimulates the node, which produces an excitatory impulse. This impulse is rapidly transmitted to a bundle of nodal fibers called the A-V bundle or bundle of His. This bundle divides into right and left bundle bunches which ramify to the ventricular myocarium (the muscular layer of the heart wall) via the Purkinje fibers. The impulse is then transmitted to all parts of the ventricles causing them to contract as a unit. As the cardiac impulse travels from the S-A node through the atria, the wave of contraction forces blood through the valves and into the ventricles. (In the atria, the impulse is transmit-ted from cell to cell across the intercalated discs) . It is a "wave" of contraction because parts of the atria farthest from the S-A node contract later than do the parts closest to the node. The parts of the atria that contract first are also the first to relax, thus causing a wavelike motion upon contraction. Once the impulse reaches the A-V node, it is delayed (by a fraction of a second) before it is passed on to the ventricles. This transmission delay allows time for the atria to pump the blood into the ventricles before ventricular contraction begins. Unlike the atria, the ventricles contract as a unit. This is due to the rapid trans-mission of the impulse by the Purkinje fibers. For example, in the atria, the impulse travels at a velocity of .3 meter per second while in the Purkinje fibers, the impulse travels at about 2 meters per second. This allows almost immediate transmission of the impulse throughout the ventricular system. The adaptive significance of the contraction of the ventricles as a unit rather than as a wave is clear. The ventricles must exert as great a pressure as possible in order to force the blood throughout a long system of arteries, capillaries and veins. A greater pressure is more easily obtained if the ventricles con-tract as a unit, rather than in waves.

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Question:

A resistor R = 10 megohms is connected in series with a capacitor 1 \muf. What is the time constant and half-life of this circuit?

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Solution:

Suppose that we introduce an e.m.f. in the circuit above, for example by connecting a battery to the circuit. (See the figure). In this case a current I will flow in the loop until C is charged up to the battery voltage V. Therefore, by Kirchoff's Voltage Law, V_R and I approach zero as V_C increases. The same current goes through R and C in the loop as we close the switch S; I = (V_R /R) = [(dQ)/(dt)] = (d/dt) (C V_C ) = C [(dV_C )/(dt)] where Q is the charge on the capacitor. As a result of Kirchoff's second rule, V_C + V_R = V and we obtain RC [(dV_C )/(dt)] + (V_C - V) = 0 or[(dV_C )/(dt)] + (1/RC) (V_C - V) = 0 The general solution of this differential equation is V_C (t) = - A e^-t/\cyrchar\cyrt + V(1) where \cyrchar\cyrt = RC is the time constant of the circuit and A is the integration constant. At t = 0, V_C = 0 because the voltage is initially only across R. Thus, substituting (1), A = V, and we get V_C (t) = V (1 - e^-t/\cyrchar\cyrt ). Note that as t becomes very large V_C (t = \infty) = V. The current I is I (t) = C [(dV_C )/(dt)] = C [(d)/dt)] [V (1 - e^-t/\cyrchar\cyrt )] I (t) = [(CV)/(\cyrchar\cyrt)] e^-t/\cyrchar\cyrt = (V/R) e^-t/\cyrchar\cyrt For the circuit under consideration \cyrchar\cyrt = RC = 10 × 10^6 ohms × 10^-6 f = 10 sec. The half-life is the time required for the decay factor (e^-t/\cyrchar\cyrt ) to be (1/2); or e-[{(t)h}/\cyrchar\cyrt] = (1/2) t_h = \cyrchar\cyrtln 2 t_h = 10 sec × ln 2 = 6.9 sec. On the other hand, if R = 10 ohms the time constant is only 10 × 10^-6 sec, or 10 \musec.

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Question:

A tomato weighing 60.0 g is analyzed for sulfur content by digestion in concentrated HNO_3 followed by precipitation with Ba to form BaSO_4. 0.466 g of BaSO_4 is obtained. If all of the sulfur is converted to BaSO_4, what is the percentage of sulfur in the tomato?

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Solution:

The percentage of sulfur in the tomato is the weight of sulfur in the tomato divided by the total weight of the tomato multiplied by a factor of 100. To find the percentage, the weight of sulfur must be calculated. This is done by calculating the amount of sulfur precipitated out or, in other words, the amount of sulfur present in 0.466 g of BaSO_4. Using the Law of Definite Proportions, 32 g (1mole) of sulfur are consumed to form 233 g (1mole) of BaSO_4. Thus, once the percentage by weight of S in BaSO_4, is determined, then the weight of sulfur can be calculated. The fraction of S in BaSO_4 is [MW of S / MW of BaSO_4]= [(32 g / mole) / (233 g / mole)] = 0.14 The weight of sulfur in 0.466 g of BaSO_4 is thus, (0.14) (0.466 g)= 0.065 a. Now, that the weight of sulfur is known, the percentage of sulfur in the tomato is, [0.065 g of S / 60.0 g of tomato] [0.065 g of S / 60.0 g of tomato] × 100 = 0.11%.

Question:

An automobile tire whose volume is 1500 in^3. is found to have a pressure of 20.0 lb/in^2. When read on the tire gage. How much air (at standard pressure) must be forced in to bring the pressure to 35.0 lb/in^2. ?

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Solution:

The 1500 in^3. of the air at 20 lb/in^2. is compressed into a smaller volume at 35.0 lb/in.^2 P_1V_1 = P_2V2 We must remember to add, 14.7 lb/in^2. , the atmospheric pres-sure to the value read on the tire gauge. P_1 = 20.0 lb/in.^2 + 14.7 lb/in.^2 = 34.7 lb/in.^2 P_2 = 35.0 lb/in.^2 + 14.7 lb/in.^2 = 49.7 lb/in.^2 (34.7 lb/in.^2)(l500 in.^3) = (49.7 lb/in.^2)v_2 V_2 = 1050 in.^3 The volume of air added to the tire is 1500 in.^3 - 1050 in.^3 = 450 in.^3 when its gage pressure is 35.0 lb/in^2 . The volume at atmospheric pressure will be found from Boyle's law, 14.7 lb/in.^2 × V = 49.7 lb/in.^2 × 450 in.^3 V= (49.7/14.7) 450 in.^3 = 1500 in.^3.

Question:

Assuming theK_spfor radium sulfate is 4 × 10-^11 , what is its solubility in (a) pure water, and (b) ,1M Na_2 SO_4 ?

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Solution:

Whenever an ionic solid is placed in water, an equilibrium is established between its ions and the excess solid phase. A solu-bility constant,K_sp, measures this equilibrium. The concentration of solid is always constant, no matter how much it is in contact with the ions. This means the solid phase will not appear in the equili-brium constant expression. The dissociation is RaSO_4 (s) \rightleftarrows Ra^2+ + SO^2-_4 . TheK_spis, thus, [Ra^2+] [SO^2-_4 ] = 4 × 10-^11 . From the chemical reaction, it becomes evident that the concentration of Ra^2+ must equal that of SO^2- _4 , since they are generated inequimolaramounts. Therefore, let x = the solubility of each. Substituting into [Ra^2+] [SO^2-_4 ] = 4 × 10-^11 , you have x \bullet x = 4 × 10-^11 . Solving for x, you obtain x = 6 × 10-^6 mol/liter. Thus, the solubility of RaSO_4is 6 × 10-^6 mol/liter, the solution is 6 × 10-^6 M Ra^2+ and 6 × 10-^6 SO^2-_4 in water. b) The equation for this reaction in water is RaSO_4 \rightleftarrows Ra^2+ + SO^2-_4 . Na_2 SO_4 is a salt and therefore will ionize completely in water. The concentration of the Na_2 SO_4 solution is .1M, which is equivalent to a solution of .2M of Na+ and .1M SO-_4 . If one adds more SO-_4 to the system, the equilibrium is forced to the left. Therefore, there will be even less Ra^2+ present than in pure water. Thus, [Ra^2+ ] << 6.0 × 10-^6 M when additional SO-_4 is present. Let y = concentration of Ra+. The solubility of RaSO_4 is equal to the product of [Ra+ ] and [SO-_4 ] in this solution. Solving: [Ra+] is equal to y. Because RaSO_4 dissociates into Ra++ and SO-_4 , the RaSO_4 in this con-tributes y to [SO-_4] . There is also SO-_4 already present in this solution, its concentration is .1M, therefore [SO-_4 ] = (y + .1). Writing the equation for the solubility solubility of RaSO_4 = [Ra++] [SO-_4] = y(y + .1). It has been shown that y is much smaller than .1 and the following approximation can be made, y + .10 \cong .1 . Solving for . y: y(y + .1) = 4 × 10-^11 .1y = 4 × 10-^11 y = 4 × 10-^10 .

Question:

Allow exactly 100 ml of 1.5 NNaOHsolution to be mixed with 100 ml 3.0 N H_3PO_4 solution and allow them to reach equilibrium. (1) Determine what species will be present at equilibrium. (2) Find the pH of the solution. k_2 = 6.2 × 10^-8.

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Solution:

To solve this problem, first consider the type of reaction, if any, that occurs and to what extent it has proceeded. Once this is done, the answers to (1) and (2) follow directly. One is adding a base of a given amount to an acid of a given amount. This is a neutralization; i.e., the reaction of an acid and base to produce a salt and water. One is told there are 100 ml of 1.5 NNaOHand 100 ml of 3.0 N H_3PO_4 solution. N = normality, which is defined as equivalents/liter. An equivalent = the number ofhydroniumions that an acid can produce or hydroxide ions that a base can produce per mole of that substance. In a neutral solution the number of equivalents of base is equal to the number of equivalents of acid. Thus, to see to what extent the reaction has proceeded, calculate the equivalents of base and acid. If they are equal, the re-action went to completion. Recall, N = equivalents/liter. Thus, equivalentsNaOH = (1.5 equiv/l)100ml/1000 m/l = 0.15 equiv. Equivalents H_3PO_4 = (3.0 equiv/l) 100ml/1000l = 0.30 equiv. The equivalents of acid do NOT equal those of the base. The H_3PO_4 has 0.30 - 0.15 = 0.15 equivalents that do not react. The rest has been neutralized by the base to form a salt and water. H_3PO_4 is a polyproticacid, which means it can donate 3 protons when a mole of it exists. Since only one-half of the equivalents are neutralized, only one-half of the protons were used. This means that the reaction H_3PO_4 + OH^- \rightleftarrows H_2PO_4^- + H_2O goes to completion, but H_2PO_4- + OH^- \rightleftarrows HPO^2_4^- + H_2O will only be half complete. With this information, (1) can be answered. One started with 1.5 N of 100 mlNaOHsolution, which indicates that 0.15 moles of Na^+ is present.NaOHcan react with only 1 mole ofhydronium ion. Thus, its equi-valents correspond to its moles. 0.30 equivalents of H_3PO_4 correspond to 0.10 moles, since each mole generates 3hydronium ions. In (0.1) H_3PO_4 + (0.15)OH^- \rightleftarrows (0.1)H_2PO_4^- + (0.1)H_2O + (.05)OH^- there is a total 0.1 moles of H_2PO_4^- generated, since H_3PO_4 is the limiting reagent with only 0.1 moles compared with OH^-'s 0.15 moles. This means that for the reaction; (0.1) H_2PO_4^- + (0.05)OH^- \rightleftarrows (0.05) HPO^2_4^- + (0.05)H_2O + (0.05)H_2PO_4^-, 0.05 moles of OH^- is left to react with the 0.1 moles of H_2PO_4^- generated previously. Thus, only 0.05molesof H_2PO_4^- is consumed in producing 0.05 moles of HPO^2_4^-. Thus, anunreactedamount of 0.05 moles of HPO_4^- is left. In summary, the species in solution are 0.05 moles each of H_2PO^2_4^- and HPO^2_4^- with small amounts of H_3O^+, H_3PO_4, PO^3_4^-, OH^- and 0.15 moles Na+. One can now calculate the pH of this. There exists some H_2PO_4^- in solution. This is the source ofhydroniumions, whose pH is measured by pH = - log [H_3O^+]. The reaction is H_2PO_4^- + H_2O \rightleftarrows H_3O^+ + HPO^2_4^-. The equilibrium constant, k_2 = {[H_3O^+] [HPO^2_4^-]} / [H_2PO_4^-] = 6.2 × 10^-8. One can calculate [HPO^2_4^-] and [H_2PO_4^-], since one calculated the mole amounts. The volume of the solution is 0.2 liters, and the concentration in the brackets, are in moles/liter. Substituting these values, 6.2 × 10^-8 = {[H_3O^+] (0.05/.2)} / (0.05/.2). Solving for [H_3O^+], one obtains [H_3O^+] = 6.2 × 10^-8 M. pH = - log [H_3O^+], Thus, pH = - log [6.2 × 10^-8] = 7.21.

Question:

The spherical shell of a Van de Graaff generator is to be charged to a potential of 10^6 V. Calculate the minimum radius the shell can have if the dielectric strength of air is 3 × 10^6 V\bulletm^-1.

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Solution:

The potential and electric intensity at the surface of a sphere of radius R are V = Q/4\pi\epsilon_0RandE = Q/4\pi\epsilon_0R^2 = V/R Note that these relations are the same as the formulae for the potential and electric field of a point charge at a distant R from the charge. Thus, the field due to a sphere of charge Q is the same as that due to a point charge of charge Q. Therefore, R = V/E. But the maximum acceptable value of E is 3 × 10^6 V\bulletm^-1 for, at any higher value of E, the air will break down, and arc discharges through the air will result. Hence, the maximum radius for the spherical shell is R = (10^6 V)/(3 × 10^6 V\bulletm^-1) = 1/3m.

Question:

Design a 5 × 32 decoder using four 3 × 8 decoders (with enable inputs) and one 2 × 4 decoder.

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Solution:

a decoder is a digital device with n input lines and 2^n output lines. A 2 × 4 decoder with inputs X and Y and outputs Z_0 Z_1 Z_2 Z_3 has the following truth table, shown in figure 1. Inputs XY Outputs Z_0Z_1Z_2Z_3 00 1000 01 0100 10 0010 11 0001 Fig. 1 Hence, for each of the four combinations of 1's and 0's there is one and only one output line that assumes the value of 1. Similarly, the truth table for a 3 × 8 de-coder with an enable input is shown in figure 2. Inputs enableWXY Outputs Z_0Z_1Z_2Z_3Z_4Z_5Z_6Z_7 0XXX 00000000 1000 10000000 1001 01000000 1010 00100000 1 011 00010000 1100 00001000 1101 00000100 1110 00000010 1111 00000001 Fig. 2 Note: X denotes "don't care" The 5 × 32 decoder is shown in fig. 3. The 2 × 4 decoder decodes bits A_3 and A_4 and ensures that only one of the 3 × 8 decoders is enabled. Hence bits A_3 and A_4 determine which of the 3 × 8 decoders to enable and bits A_0, A_1, and A_2 determine which output of the enabled decoder will become 1.

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Question:

Explain why aldehydes but not ketones can be oxidized to carboxylic acids.

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Solution:

An examination of the structures of aldehydes, ketones, and carboxylic acids will reveal the solution to this problem. In answering this problem, keep in mind that a carbon atom can participate in only four bonds. Aldehydes and ketones have the carbonyl functional group; i.e., The aldehyde has the general formula where R is any carbon group. The ketone's formula is where R and R' are different or the same carbon groups. A carboxylic acid has the formula where R is any carbon group. Now, let us look at the oxidation: But, notice, in the oxidation from the ketone, the resulting compound has a carbon atom with 5 bonds, which is not permissible. Such a situation does not occur with oxidation in aldehydes. This explains why aldehydes, but not ketones,can be oxidized to carboxylic acid.

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Question:

Let us estimate the gravitational energy of the Galaxy. We omit from the calculation the gravitational self-energy of the individual stars.

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Solution:

The gravitational energy of an arbitrary system of N parti-cles consists of the sum of the mutual potential energies of each pair of particles. Hence, U = (1/2) [(U_12 + U_13 + U_14 + ... U_1N) + (U_21 + U_23 + U_24 + ... + U_2N) + (U_31 + U_32 + U_34 +...+ U_3N) +...+ ( U_NI + U_N2 + U_N3 +...+ U_NN-1 )](1) The terms U_12, etc., represent the mutual potential energy of particles 1 and 2. By Including sets of terms such as U_12 and U_21 we have double counted, because these represent the mutual potential energies of particles 1 and 2, and particles 2 and 1, respectively. However, these 2 terms are the same. Hence, the factor (1/2) must be included in (1) to negate the process of double counting. Furthermore, terms such as U_11 are omitted because they represent the mutual potential energy of particle 1 with particle 1, i.e., they are self energies. We approximate the gross composition of the Galaxy by N stars, each of mass M, and with each pair of stars at a mutual separation of the order of R. From the definition of potential energy U_ij= (-Gm_im_i)/(r_ij) where m_i andm_jare the masses of particlesiand J, respectively, G = 6.67 × 10^-11 N\bulletm^2/kg^2, is their mutual separation. For our case r_ij= R andm_i =m_j= M for all pairs of particles. Then U_ij= (-GM^2)/R(2) for any 2 particles. Notice that in equation (1), the first parenthesis has N-1 terms of the type (2), the second parenthesis has N-1 terms of this type, and similarly for all Nparenthesises. Altogether, there are N(N-1) terms of type (2) in (1). Therefore, U = (1/2)(N)(N - 1)[(-GM^2)/R] U = -(1/2)[{N(N - 1)GM^2} / (R)] If N \approx 1.6 × 10^11, R \approx 10^21m, and M \approx 2 × 10^30 kg, then U \approx [-(1/2)(1.6 × 10^11)(1.6 × 10^11 - 1)(6.67 × 10^-11N\bulletm^2/kg^2) (2 × 10^30 kg)^2] / [10^21m] U \approx[ (-34.15 × 10^71)/(10^21)] J = -34.15 × 10^50J U \approx -3.42 × 10^51J

Question:

How many moles of hydrogen gas are present in a 50 liter steel cylinder if the pressure is 10 atmospheres and the temperature is 27\textdegreeC? R = .082 liter-atm/mole \textdegreeK.

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Solution:

In this problem, one is asked to find the number of moles of hydrogen gas present where the volume, pressure and temperature are given. This would indicate that the Ideal Gas Law should be used because this law relates these quantities to each other. The Ideal Gas Law can be stated: PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant (.082 liter-atm/mole \textdegreeK), and T is the absolute temperature. Here, one is given the temperature in \textdegreeC, which means it must be converted to the absolute scale. P, V, and R are also known. To convert a temperature in \textdegreeC to the absolute scale, add 273 to the temperature in \textdegreeC, T = 27 + 273 = 300\textdegreeK Using the Ideal Gas Law: PV = nRTorn = (PV/RT) P = 10 atm V = 50 liters R = .082 liter-atm/mole \textdegreeK T = 300\textdegreeK n = number of moles of H_2 present n = [{(10 atm) (50 liters)/{(.082 liter-atm/mole \textdegreeK) (300\textdegreeK)}] = 20 moles.

Question:

Show that the optical length of a light path, defined as the geometricallength times the refractive index of the medium inwhich the light is moving, is the equivalent distance which thelight would have traveled in a vacuum.

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Solution:

Suppose that light travels a distance l in a medium of refractive indexn. The optical length is then opticallength =nl and, since n = c/v opticallength =cl/v Here c and v are the speeds of light in vacuum and the medium, respectively. But light travels with constant velocity in the medium, and hence l/v = t, where t is the time taken to traverse the light path. nl= (cl/v) = ct = l_0 wherel_0 is the distance the light would have traveled at velocity c, that is, ina vacuum. Thus the optical length is the equivalent distance which the light would have traveled in the same time in a vacuum.

Question:

Explain structures in C. Give an example and show how they can be referenced.

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Solution:

Structures in C are created by astructstatement. It is similar to the typedef statement in the sense that it does not create a variable we can use but provides a format for future variable declarations. Suppose we want to create a structure of students in which we want to store their name, id and theirgpa. The following format is used. #define MAX_NAME 30 #define MAX_ID 10 #define NUMBER _ of _ STUDENTS 100 structstudents { char name [MAX_NAME]; char id [MAX_ID]; float gpa; } stud[NUMBER _ of _ STUDENTS]; In the above declaration stud is the variable created that will use the structure students. Students in this case are called a tag and each subfield within the structure i.e. name, id,gpa, etc., are called a member. We can also define the variableNYCstudof type students for example, structstudentsNYCstud; We can reference individual students in the following manner. stud [O].name; stud [0].id; stud [0].gpa; Thus if we want to printgpaof all students along with their names then for(i= 0;i< NUMBER _ of _ STUDENTS; ++i) printf("%s%1.1f\textbackslashn", stud [i].name, stud [i].gpa); prints the name andgpaof each student.

Question:

A chemist decides to prepare some chlorine gas by the following reaction: MnO_2 + 4HCl \rightarrow MnCl_2 + 2H_2 O + Cl_2 \uparrow If he uses 100 g of MnO_2, what is the maximum volume of chlorine gas that can be obtained at standard temperature and pressure (STP)?

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Solution:

The solution of this problem entails a combination of stoichiometry and gas law theory. The first thing to determine is how many moles of chlorine will be produced. The volume can be determined using the fact that each mole, at STP, occupies 22.4 l. Since the equation is balanced, you must look at the coefficients. The equation indicates that for every mole of MnO_2, 1 mole of chlorine gas is produced. A mole is defined as being equal to weights in grams of material/molecular weight of material. The molecular weight of MnO_2 is 86.9. It follows, then, that the number of moles of MnO_2 is 100/86.9. This will be equated to the number of moles of chlorine gas produced. At STP one mole of gas occupies 22.4 liters. However, we do not have 1 mole, but 100/86.9 moles. To calculate the new volume, you multiply the STP volume by the number of moles of chlorine gas produced. In other words, the volume of the chlorine gas is 22.4 × = [(100) / (86.9)] = 25.8 liters.

Question:

Water flow into a tank is regulated by the position of a valve. The position of the valve is controlled by the height of a float: as the float drops, the valve is turned on to admit the water, while the rising water level causes the float to rise, which gradually shuts off the valve, stopping the flow of water. Write a FORTRAN program to simulate this system, using the modified Euler's method, from time t = 0 to t = tf, if the volume at time t is V(t), V(0) = 0, and the capacity of the tank is C units of volume.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G24-0579.htm

Solution:

If V̇ is the flow rate, the behavior of the system can be diagrammed as shown in Fig. 2. The present water level is an accumulation of all past flow-rates. The next flow rate is determined by the present water level. Mathe-matically expressed: V̇ = f(V). But note that the flow rate is the greatest when the tank is empty and decreases to zero as the tank fills up. Stating this mathematically: V̇ = K(C - V(t))(1) where 0

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Question:

A rifle weighing 7 pounds shoots a bullet weighing 1 ounce, giving the bullet a speed of 1120 feet per second. (a) If the rifle is free to move, what is its recoil speed? (b) If the rifle is held tight against the shoulder of a man weighing 133 pounds and if he were free to move, what would be the recoil speed of the rifle and man? (c) If the bullet imbeds itself in a block of wood weighing 3 pounds and 7 ounces and if the block were free to move, what would be the speed of the block plus bullet?

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Solution:

The law of conservation of momentum may be applied in an isolated system where no external forces are applied. (a) The momentum of the gun plus bullet before firing is zero, and it is therefore also zero after firing. The momentum after firing is M_bullet × vbullet+ M_gun × v_gun = 0(1) Since the law of conservation of momentum involves mass and not weight, we must convert weight into mass by dividing by the acceleration of gravity [M = {(Wt)/g} ].Therefore M_bullet = [(1/16) lb] / [32 ft/s^2] = .001953 slugs and M_gun = (7 lb) / (32 ft/s^2) = 7/32 slugs Equation (1) then becomes .001953 slugs × 1120 ft/sec + 7/32 slugs × v_gun = 0 Whence v_gun = -10 ft/sec, or 10 ft/sec backwards (b) The momentum after firing is M_bullet × v_bullet + M_gun + _man × v_gun _+ _man = 0or .001953 slugs × 1120 ft /sec + [(132/32) slugs + (7/32) slugs] × vgun + man= 0 whence vgun + man= - 0.5 ft/sec,or0.5 ft/sec backwards (c) The momentum of the bullet before the collision with the block is M_bullet × v_bullet = .001953 slugs × 1120 ft/sec × 32 ft/s^2 = 70 lb-ft/sec The momentum after collision is the same, 70 lb-ft/sec. Then 70 lb-ft/sec =(Mbullet + block) × (vbullet + block) = [ (3/32) slugs + (7/2) slugs + 2 slugs] × (v_bullet+block) whence (vbullet + block) = 20 ft/sec forwards.

Question:

An unbanked curve has a radius of 80.0 m. What is the maximum speed at which a car can make the turn if the coefficient of friction \mu_s is 0.81?

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/Users/wenhuchen/Documents/Crawler/Physics/D04-0167.htm

Solution:

We assume that the car is travelling in a circular path (see fig.(a)) at a constant speed. However, its velocity is constantly changing in direction. Hence, the car is accelerating, and, therefore, a force must be acting on the car. This force accelerates the car towards the center of the circular path and is therefore centri-petal. Applying Newton's Second Law, F = ma, to the car, we obtain f = ma(1) where f is the frictional force and a is the acceleration of the car. But, in uniform circular motion a = v^2/R(2) where v is the speed of the car and R is the radius of the circle. Furthermore, the frictional force f is f \leq \musN(3) where N is the normal force exerted by the road on the car. (Note that if f = \mu_s N, the car will begin to slip relative to the road.) Combining (1), (2) and (3) mv^2/R = f \leq \musN mv^2/R \leq \mu_s N(4) Applying Newton's Second Law to the vertical direction of motion (see fig. (b)), we obtain N - mg = 0 because there is no acceleration of the car in this direction. Using this in (4) mv^2/R \leq \mu_s mg Solving for v v \leq \surd(\mu_s gR) v_max = \surd(\mu_s gR) v_max = \surd[(.81) (9.8 m/s^2) (80 m)] v_max = 25 m/s

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Question:

(a) Explain what is meant by a paritycheck; (b) Encode the decimal digits 0-9 into their (i)7421code(ii) two-out-of-five code (iii)biquinarycode representations.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G06-0128.htm

Solution:

(a) A parity check is simply the technique of counting the number of0's or 1' s in a data word. In some systems, the result then determines thevalue assigned to an extra "parity bit", which is carried along as a check. (b) (i) Codes are sometimes identified by their positional values. The binarycode may be referred to as the 8421 code (since a 1 in the leftmost positionhas a value of 8, a 1 in the rightmost position has a value of 1, etc.) The 7421 code is identical to the 8421 code for the first seven digits, butthe high-order bit is used to represent decimal 7 rather than decimal 8 asin 8421 code. The codes are shown in fig. 1. Decimal 8421 7421 0 1 2 3 4 5 6 0000 0001 0010 0011 0100 0101 0110 0000 0001 0010 0011 0100 0101 0110 7 8 9 0111 1000 1001 1000 1001 1010 Note that no legal 7421 word has more than two1' s. Therefore, if a parity checkindicated that a received word had three or four 1' s, an error conditionwould besignalled. (ii) The two-out-of-five code can be obtained from the 7421 code by addinga fifth bit. This new bit, added in the least significant position, is calledthe paritybyt. If the 7421 code representation of a particular decimalinteger has a single 1 bit, the parity bit is 1. If the 7421 code for a decimalinteger has zero or two 1 bits, the parity bit is 0, The two-out-of- fivecode is shown in fig. 2. Decimal 7421 Two-out-of-five 0 0000 00000 1 0001 00011 2 0010 00101 3 0011 00110 4 0100 01001 5 0101 01010 6 0110 01100 7 1000 10001 8 1001 10010 9 1010 10100 Note that every legal two-out-of-five code word has a pair of 1's, except 00000. Thus, the presence of a single 1 bit or more than two 1 bits would signalan error. (iii) The problem with the two-out-of-five code is that there is no parity checkfor zero, since zero is represented by a code consisting only of zeros. If a digit is being transmitted and for some reason the 1's are suppressed, the result will be 00000, and the computer has no way of knowingif this is an error or the number is supposed to be zero. In the biquinarycode, zero is indicated as 01 00001. The value of each succeedinginteger is increased by 1 by moving the 1 left one column until decimal5 is reached. Decimal 5isrepresented as 1000001. The values of decimals6-9 are obtained by moving the 1 to the left, again. Thebiquinary codeis shown in fig. 3. Decimal Biquinary 0 0100001 1 0100010 2 0100100 3 0101000 4 0110000 5 1000001 6 1000010 7 1000100 8 1001000 9 1010000 Every legalbiquinarycode word must contain a pair of 1's. Thus, the presenceof a single 1 bit or more than two 1 bits would signal an error.

Question:

In an experiment, x-rays of frequency f are scattered by the electrons in a block of paraffin. It is found that the x-rays scat-tered at an angle of less than 900, have a wavelength f ', greater than f. Interpret this result in terms of a collision of a photon with an electron.

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/Users/wenhuchen/Documents/Crawler/Physics/D35-1037.htm

Solution:

The collision is shown in the diagram. The photon is scat-tered through an angle \texttheta, and the electron through an angle \varphi. The momentum of the photon is given by the de Broglie relationship. p = h/\lambda = hf/cusing \lambdaf = c As in all collision problems, the law of conservation of momentum may be applied to this collision. The momentum of the photon-electron system, then must remain constant throughout the collision. Since momentum is a vector quantity, (having both direction and mag-nitude) we must resolve the momenta of the electron and photon into x and y components. Conservation of momentum can then be applied as follows: hf/c = (hf '/c)cos \texttheta + mv cos \varphi (x component)(1) (hf '/c)sin \texttheta = mv sin \varphi (y component)(2) The law of conservation of energy can also be applied to the collision analysis because the collision is elastic. The total energy of the electron- photon system must then remain constant throughout the col-lision. Or hf + m_0c^2 = hf ' + mc^2(3) where m_0 is the rest mass of the electron. And m is the new relativistic mass [= m_0 / \surd{1 - (v^2/c^2)}] It incorporates the kinetic energy the electron has, after the col-lision. The angle \varphi can be eliminated by squaring (1) and (2) and adding. Then [(hf/c) - (hf '/c) cos \texttheta]^2 = m^2v^2 cos^2 \varphi [(h^2f '^2)/c^2] sin^2 \texttheta = m^2v^2 sin^2 \varphi h^2/c^2(f^2 + f '^2 - 2ff ' cos \texttheta) = m^2v^2(cos^2\varphi + sin^2\varphi) orh^2(f^2 + f '^2 - 2ff ' cos \texttheta) = m^2v^2c^2(4) forcos^2 \varphi + sin^2 \varphi = 1 Next the energy equation is squared to obtain [h(f - f ') + m_0c^2]^2 = m^2c^4(5) Consider now the quantity m^2c^4 - m^2v^2c^2 = c^2m^2(c^2 - v^2)(6) We know that m^2 = m2_0 / {1 - (v^2/c^2)} Thus m^2(c^2 - v^2) = m2_0c^2 and (6) simplifies to m^2 _0c^4. Therefore by subtracting (4) from (5) we obtain [h(f - f') + m_0c^2]^2 - h^2(f^2 + f'^2 - 2ff' cos \texttheta) = m^2c^2(c^2 - v^2) = m2_0c^4. Simplifying this equation gives the result 2hm_0c^2(f - f') - 2h^2ff'(1 - cos \texttheta) = 0 or 1 - cos \texttheta = [(m_0c^2)/h] [(f - f')/ff'](7) Using the relationship \lambdaf = c, equation (7) is reduced to 1 - cos \texttheta = (m_0c / h) (\lambda' - \lambda)(8) This process of scattering of photons is known as the Compton effect, and is a clear indication of an instance in which a wave motion be-haves like a particle. Notice that the change in wavelength depends only on the angle of scattering \texttheta and is independent of the wave-length itself. The quantity \lambda_0 = h / m_0c =0.0242\AA is known as the Compton wavelength.

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Question:

10 ml of 0.200 M HC_2H_3O_2 is added gradually to 25 ml of 0.200 M NaOH. Calculate the pH of the initial 0.200 M NaOH solution and after each successive addition of 5 ml of acid. K_w = 10^-14.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E12-0434.htm

Solution:

This is a titration problem. One is asked to find the pH of the 0.200 M NaOH at different concentration levels. First one wants to calculate the pH before any acid is added. To do this, consider the fact that the equilibrium constant for the autodissociation of water, K_w ,is defined as K_w = [H_3O^+] [OH^-]. Since, pH = - log [H_3O^+], one needs to calculate [H_3O^+], which can be determined from K_w,once [OH^-] is known. One knows [OH^-] from the fact that one has a 0.2 M NaOH solution. M = molarity = moles/liter. In solution, NaOH dissociates according to NaOH \rightleftarrows Na^+ + OH^-. Thus, [OH^-] = [NaOH]. Since [NaOH] = 0.2, [OH^-] = 0.2, which allows for substitution into the equilibrium expression, K_w = 1.0 × 10^-14 = [H_3O^+] [0.2], or [H_3O+] = [(1.0 × 10^-14) / 0.2] = 5.0 × 10^-14, which means pH = - log [5.0 × 10^-14] = 13.3. What is the pH when 5 ml of acid (HC_2H_3O_2) is added? To answer this, note that a neutralizing reaction occurs, (i.e. a reaction between an acid and base to produce a salt and water) when the acid is added to the NaOH solution. HC_2H_3O_2 + NaOH \rightarrow HOH + NaC_2H_3O_2 . It is necessary to find how much base is consumed, since this will reflect the remaining [OH^-]. From this, return to the expression for K_w and substitute this new value for [OH^-] to determine [H_3O^+] from which the pH can be calculated. To find the amount of NaOH that was consumed, use Molarity = moles/liter. If the acid is 0.200 M and one adds in 5 ml or 0.005 l (1000 ml =1 l), one has (0.2) (0.005) = 0.001 moles of acid. Amount of base present in moles equals: molarity × volume = (0.2) (0.025) = (0.005) moles. According to the reaction, they will react in equi-molar amounts. Thus, since only 0.001 moles of acid is present, only 0.001 moles of base react. This leaves 0.004 moles of base (NaOH), and as such 0.004 moles of OH^-. Since volume = base (l) + acid (l) = 0.03, M = 0.004 / 0.03 = 0.13 M. Recalling, K_w = 1.0 × 10^-14 = [H_3O^+] [OH^-] and that [OH^-] = 0.13, such that [H_3O^+] = (1.0 × 10^-14) / 0.13 = 7.5 × 10^-14. Thus, pH = - log [7.5 × 10^-14] = 13.13. If one adds another 5 ml of acid, the total acid added is 10 ml. The same procedure is followed, except that volume of acid = 10 ml or 0.01 l. Molarity of acid = 0.2, so that (0.2) (0.01) = 0.002 moles of acid are present. From the previous calculation, it can be seen that 0.005 - 0.002 = 0.003 moles of OH^- are now left in a volume of 0.025 l from base + 0.010 l from acid or 0.035 liters. Thus, its molarity is 0.003 / 0.035 = 0.086 M. K_w = 1.0 × 10^-14 = [H_3O^+] [OH^-] = [H_3O^+] [0.086]. Solving, [H_3O^+] = 1.17 × 10^-13.This means, pH = - log [H_3O^+] = - log [1.17 × 10^-13] = 12.9.

Question:

What is meant by the term "renal threshold"?"renal clearance"?

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/Users/wenhuchen/Documents/Crawler/Biology/F18-0447.htm

Solution:

Although glucose is present in theglomerularfiltrate, there is normally little or no glucose in the urine, due to itsreabsorptionby the cells of the renal tubules. If the plasma level of glucose were increased drastically, the level of glucose in theglomerularfiltrate would likewise increase. In this situation, not all the glucose could be reabsorbed as the filtrate passed through the kidney tubules, and some glucose would appear in the urine. The concentration in the plasma of a sub-stance such as glucose at the point where it just begins to appear in the urine is referred to as the "renal threshold" of the substance. The renal threshold for glucose is about 150 mgs. of glucose per 100mls. of blood. When the concentration of glucose in the blood exceeds this level, glucose will begin to appear in the urine. There are comparable thresholds for many other substances. Physiologists have adopted the concept of "renal clearance" to express quantitatively the kidneys' ability to eliminate a given substance from the blood. The elimination of a substance is dependent on its concentration in the blood, the filtration rate at theglomerulus(or volume of blood filtered per minute), and the rate of secretion orreabsorptionof that substance by the renal tubules. The renal clearance relates the rate of appearance of a substance in the urine to its concen-tration in the blood, and is expressed as the volume of plasma cleared of that substance per minute. One can also think of renal clearance as the volume of plasma that would contain the amount of a substance excreted in the urine per minute. The renal clearance of a substance x is defined as C_X=[(mgs. of substance × secreted in urine per minute) / (mgs. of substance × per ml. of plasma)] Thus, if x is present in the plasma in a concentration of .02 mg/ml, but secreted at a rate of .2 mg/min, its clearance would be CX= (.2 mg x/min.) / (.02 mg x/ml plasma) = 10mlsplasma/min. In other words, 10mlsof plasma are completely cleared of substance x per minute, thus .2 mgs. of that substance appears in the urine per minute.

Question:

You are in a room with 29 other people. You are asked to write a program which computes the probability of at least two people having the samebirthdate.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G19-0472.htm

Solution:

Denote the requested probability by Q. Then Q = 1 - P where P is the probability of no two identical birthdates. From elementary probability theory P = (365/365) × [(365-1)/365] ×...× [(365-29)/365] = (365 × 364 × 363 × ... × 336)/365^30 Here, the probability is 365/365 that one person will not have two identical birthdates. Given two people, the probability is 364/365 since the first person has 365 choices and the second only has 364. Continuing, we obtain P. The program appears as follows : 1\O LET P = 1 2\O FOR D = 365 TO 336 STEP -1 3\O LET P = P \textasteriskcentered D/365 4\O NEXT D 5\O LET Q = 1 - P 6\O PRINT "THE PROBABILITY OF TWO OR MORE" 7\O PRINT "IDENTICAL BIRTHDATES AMONG 30 PEOPLE IS"; Q 8\O END

Question:

The double-slit experiment can be simulated using two small blocks of wood oscillating up and down together in a pool of water as shown in the figure. Suppose that the blocks are 3 cm apart and the wavelength of the water waves is 1 cm. Locate the points 10 cm from one of the blocks where constructive interference occurs.

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Solution:

The motion of the blocks produce circular waves which are emanating from two sources S_1 and S_2 separated by d = 3 cm. In order to observe the interference pattern that forms at 10 cm from one of the blocks, draw a circle of radius 10 cm around source S_2 as shown in the figure. Constructive interference occurs at points where the dif-ference in the distances to the two blocks is either zero or some whole number of wavelengths. Let A be a point on the circle at a distanceL_1 from S_1. All the points on the circle are L_2 = 10 cm from S_2. Constructive interference occurs at points for which the length \vert L_1 - L_2 \vert is an integer multiple of the wavelength \lambda = 1 cm. Therefore, we have constructive interference at points on the circle that are L_1 = 10 cm - 3 cm = 7 cm = 10 cm - 2 cm = 8 cm = 10 cm - 1 cm = 9 cm = 10cm = 10 cm + 1 cm = 11 cm = 10 cm + 2 cm = 12 cm = 10 cm + 3 cm = 13 cm from S_1.

Question:

Describe the structure of an insect wing. How are wings used in flight?

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Solution:

The wings of an insect are evaginations or folds of the integument, and are composed of two sheets of cuticle. A vein runs through the wing at a point where the two cuticular membranes are thickened and separated, forming an effective supporting skeletal rod for the wing. The wings of the more primitive insects are netlike, but there has been a general tendency in the evolution of wings toward reduction of this netlike appearance. Each wing articulates with the edge of the dorsal cuticle, called the tergum, but its inner end rests on a dorsal pleural process (the pleura is the cuticle covering the side of the body), which acts as a fulcrum. The wing is thus somewhat analogous to an off-centered seesaw. Upward movement of the wings results indirectly from the contraction of vertical muscles within the thorax, depressing the tergum (see Figure A). Downward movement of the wings is produced either directly, by contraction of muscles attached to the wingbase, or indirectly, by the contraction of transverse horizontal muscles raising the tergum. Insects such as dragonflies and roaches exhi-bit direct contraction, while bees, wasps, and flies show indirect contraction. Downward movement can come about by both direct and indirect muscles in insects such as grasshoppers and beetles. The raising or lowering of the wings involves the alternate contraction of antago-nistic muscles. Up and down movement alone is not sufficient for flight; the wings must also move forward and backward. A complete cycle of a single wing beat describes an ellipse in grasshoppers, and a figure eight in bees and flies; the wings are held at different angles to provide both lift and forward thrust. The wing beat frequency varies from 4 beats per second in certain gnats. The fastest flying insects are the moths and horse flies, which can fly over 33 miles per hour. Gliding, an important form of flight in birds, occurs in only a few large insects. Insect flight muscles are very powerful. The fibrils in the muscle cells are relatively large and the mitochond-ria are huge. Insects are the only poikilothermic (cold-blooded) fliers; their low body temperature and corres-pondingly low metabolic rate impose limitations on mobility. On a cold day some butterflies are known to literally "warm up" before flight. They remain stationary on a tree trunk or some other location, and move their wings up and down until sufficient internal heat is generated to permit the stroke rate necessary for flight.

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Question:

Using APL, simulate the game of craps.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G23-0555.htm

Solution:

First of all, let us describe how the game of craps is played: A player rolls two dice. If the sum of the two faces on the first throw is 7 or 11, he wins the amount bet. If the sum turns up to be a 4, 5, 6, 8, 9 or 10, the player continues to roll until he obtains (1) a 7 in which case he loses, or (2) the sum he obtained on the first throw; in this case he wins. Our APL program for craps is: \nablaCRAPS [1]A \leftarrow 10 [2]AGAIN:A \leftarrow A, B \leftarrow +/?2\rho6 [3]\rightarrow( ( (1 =\rhoA)\wedge\simB \epsilon 7 11 2 3 12) \vee (1 \not =\rhoA) \sim \wedge\simB \epsilon (A [1] , 7) ) /AGAIN \sim \sim [4]\rightarrow( ( (1 =\rhoA) \wedge B \epsilon 7 11) \vee (1 \not =\rhoA) \wedge B =A [1] ) /WIN [5]`YOU LOSE WITH' ; A [6]\rightarrow0 [7]WIN:`YOU WIN WITH' ; A [8]\nabla In step [1] weintializethe vector A to be the null vec-tor. In [2] we find the sum of two random numbers. The random numbers are positive integers between 1 and 6. Wecatenateall sums in the vector A. Step [3] checks two things: (1) whether it was the first roll and if the roll automatically gave you a win (7 or 11) or a loss (2, 3, 12), (2) whether you rolled more than once, and if the last roll gave you a 7 (loss) or the same sum as the one obtained on the first throw (win). If you do not win or lose in step [3] you must roll again (go to step 2). In step 4: If you won, you are sent to step 7, where it is printed out that you win and vector A is also printed out. If you do not win, step [5] will indicate that you are a loser.

Question:

At a counting rate of 5 atoms per second, how long would it take to count the atoms in a spherical droplet of water 0.1 take to count the atoms in a spherical droplet of water 0.1 cm in diameter? cm in diameter?

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Solution:

We can calculate the number of atoms in the water droplet from the density of water and the gram molecular weight of a water molecule. The density of water is defined as its mass per unit volume, or \rho = m/v Hence,m =\rhov.(1) Now, the volume of water in the spherical droplet is the volume of a sphere of radius .05 cm. The volume of a sphere of radius r = 4/3 \pir^3. Volume of droplet = 4/3 \pi(0.05)^3 =5.24 × 10^-4 cm^3 The density of water is 1 gm per cubic centimeter, and from (1), the mass of water in the droplet is m = (1 gm/cm^3) (5.24 × 10^-4 cm^3) m = 5.24 × 10^-4 gm We may now find the number of molecules in this mass by dividing it by the mass of one molecule. A molecule of water contains two atoms of _1H^1 and one atom of _8o^16 and its molecular mass is therefore (2 × 1.0078amu) + 15.995amu= 18amu But 1amu1.66 × 10^-24gms. Therefore, a water molecule has a mass, in grams, of (18) (1.66 × 10^-24gms) = 29.88 × 10^-24gms The total number of molecules in 5.24 × 10^-24gmsis then n = [{5.24 × 10^-4 gm} / {2.99 × 10^-23 (gm/molec)]. = 1.75 × 10^19 molecules to sufficient accuracy. Since each molecule contains 3 atoms, the number of atoms in the water droplet is 3n = [(1.75 × 10^-19 molecules) (3 atoms/molecules)] 3n = 5.25 × 10^19 atoms At a rate of 5 per second the time taken to count these atoms would be [{5.25 × 10^19 atoms} / {5 (atoms/sec)}] = 1.05 × 10^19 sec There are approximately 3.16 x 10^7 sec in a year, so the time taken would be [{1.05 × 10^19 sec} / {3.16 × 10^7(sec/yr)}] = 3.3 × 10^11 years.

Question:

Given 0.207 nm for the S-S distance and 105\textdegree for the S-S-S bond angle, calculate the S-to-S distance across the S_8 ring.

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Solution:

One is asked to find the distance from S_1 to S_2 as seen in figure A. This can be done by using trigo-nometry. Draw a line perpendicular to S_1S2 to A from S_0. One can solve for the length of S_2A with the knowledge that the angle S_2- S_0-A is equal to 1/2 of the S-S-S angle or 52.5\textdegree. This is true because the triangle S_0S_1S_2 is isosceles. sin 52.5o= (S_2A / S_2S_0) = 0.793 S_2A= 0.793 × S_2S_0 = (0.207) × (0.207 nm) = 0.164 nm. S_2S_1= 2 × S_2A = 2 × 0.164 nm = 0.328 nm.

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Question:

Find the theoretical speed of sound in hydrogen at 0\textdegreeC. For a diatomic gas \Elzpbgam = 1.40, and for hydrogen M = 2.016 gm/mole.

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Solution:

Sound waves are longitudinal mechanical waves. That is, they are propagated in matter and the particles transmitting the wave oscillate in the di-rection of propagation of the wave. The speed of the wave is determined by the elastic and inertial pro-perties of the medium. For a gaseous medium, the elastic property depends on the undisturbed gas pressure p_0 and the inertial property onl_o, the density/mole of the gas. With \Elzpbgam a constant called the ratio of specific heats for the gas, the velocity is v = \surd[(\Elzpbgamp_o)/(p_o)] Since for one mole of gas, p_0 = RT/V = (RTp_0)/Mwe have v = \surd[(\ElzpbgamRT)/M] = \surd([1.40{8.317 joules/(mole K0)} (2730K)] / {2.016 × 10^-3 kg/mole}) = \surd[{1.40 × 8.317 × 273 joules} / {2.016 × 10^-3 kg}] = 1.25 × 10^3 m/sec

Question:

A capillary tube of internal radius 0.25 mm is dipped into water of surface tension 72 dynes \textbullet cm^-1. How high does the water rise in the tube? The capillary tube is gradually lowered into the water until only 1 cm is left above the surface. Explain what happens to the water in the tube.

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/Users/wenhuchen/Documents/Crawler/Physics/D10-0422.htm

Solution:

Consider a capillary tube of radius r. The liquid in it makes contact with the tube along a line of length 2\pir. Let the liquid in the tube be a height y above the surface of the liquid in which it is dipped. The up-ward force, T, is defined as the product of the surface tension, \Upsilon, and the length perpendicular to the force, along which it acts. Then T = (\Upsilon) (2\pir) (cos \texttheta) where \texttheta is the contact angle. The downward force on the liquid in the tube is equal to its weight w. Then w equals the liquid's weight density, \rhog, multiplied by its volume \pir^2y. w = \rhog\pir^2y For the liquid in the tube to be in equilibrium, these forces must be equal. w = T \rhog\pir^2y = \Upsilon^2\pir cos \texttheta Theny =2\Upsilon cos \texttheta/ \rhogr. Since the liquid in this case is water, the angle of contact is 0\textdegree, and the density is 1 g\bulletm^-3. Hence y = (2 × 72 dynes \bullet cm^-1)/(1 g \textbullet cm^-1 × 980 cm \textbullet s^-2 × 0.025 cm) = 5.88 cm. As long as more than 5.88 cm of tube shows above the liquid surface, there is no problem. The liquid rises to that height. But, as the tube is lowered, a stage will be reached when less than 5.88 cm are above the surface. What cannot happen is that liquid pour out over the top. If it did, the liquid pouring over the edge could be used to drive a water wheel to provide energy; and the process would continue as liquid would always rise up the tube to take the place of that pouring from the end. In other words, a perpetual motion machine would be establish-ed, which is in direct contradiction with the principle of conservation of energy. What does happen is that the angle of contact at the top of the tube increases. Only the vertical component of the surface tension is used to balance the weight of the water column held up. As the height of the projecting tube gets smaller and smaller, the angle of contact gets larger and larger until, with y = 0, \texttheta = 90\textdegree, and the surface at the top of the tube is flat. In particular, when y = 1 cm, 1 cm = (2\Upsilon cos \texttheta)/(\rhogr) = 5.88 cos \texttheta cm. \thereforecos \texttheta = 1/5.88 = 0.17\therefore \texttheta = 80.2\textdegree.

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Question:

Natural boron is made up of 20 per cent^10B and 80 per cent 11 B. What is the approximate atomic weight of natural boron on the scale of 16 units for ^16o (chemical scale)?

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Solution:

The scale of 16 units for ^16o means that the superscript represents the atomic weight of the atom. There-fore: atomic weight of ^10B = 10 atomic weight of ^11B = 11 average atomic weight of boron (natural) = [(20 × 10 + 80 × 11) / (100)] = 10.8 .

Question:

Write an APL program to evaluate the roots of the quadratic equation of the form ax^2 +bx+ c = 0.

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Solution:

To do this, we use the quadratic formula: ROOT = [-B \pm \surd(B^2 - 4AC)] / [2A] The quantity given by B^2 - 4AC is named thediscriminant. If it is negative, then the roots are complex numbers. In that case the program will be terminated. (Complex numbers can be evaluated by a built-in library function, but we will omit this for simplicity sake.). If thediscriminantis positive, then we want to calculate the roots. The program is written below: \nablaQUADRATIC [1]DISCRIM \leftarrow (B\textasteriskcentered2) - 4 × A × C [2]\rightarrow (0 > DISCRIM) / 8 [3]'THE ROOTS ARE' [4]ROOT1 \leftarrow (-B-DISCRIM\textasteriskcentered0.5) \div 2 × A [5]ROOT2 \leftarrow (-B + DISCRIM\textasteriskcentered0.5) \div 2 × A [6]ROOT1, ROOT2 [7]\rightarrow0 [8]'THE ROOTS ARE COMPLEX' \nabla Line [2] may be translated as follows: "If zero is greater than the discriminant , then go to line [8]. If zero is less than or equal to the discriminant , proceed with the next line." This is called a branch statement. Line [7] is also a branch: after the roots have been cal-culated, the program goes to line [0], which actually means the end of the program. Literal text has been added to make clear the meaning of the values being computed. Now, for the polynomial 3x^2 - 19x - 14 = 0, we can give a sample output, with A = 3, B = -19, and C = -14. This polynomial has real roots. A \leftarrow 3 B \leftarrow -19 C \leftarrow -14 QUADRATIC The roots are -0.6666666667 +7.000000000.

Question:

A physicist, Mavis, walks at a speed of 1 m/s past a stationary physicist, Stan. They each have measuring sticks and clocks and they observe a bird hop from a tree branch to the lawn. Each measures the position and the time when this event occurred. Show in the special case of small velocities (v very much smaller than c) that the Lorentz equations may be replaced by the following equations: x' = x - vt y' = y t' = t where the primed and unprimed frames represent the frames of Mavis and Stan respectively.

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Solution:

When the velocities with which we deal approach the speed of light, we can no longer use classical mechanics, and must replace this theory by relativistic mechanics. The purpose of this problem is to show us how the Lorentz transformation (which is part of the relativis-tic mechanics) reduces to the Galilean transformation (which is part of classical mechanics) when the velocities we are concerned with are small when compared with c. The Lorentz transformation, relating the space and time coordinates of an event as observed in S' to the space and time coordinates of the same event as observed in S is x' = (x - vt) / \surd{1 -(v^2c^2)}(1) y' = y z' = z t' = {t - (xv/c^2)} / \surd{1 -(v^2c^2)} where v Is the relative velocity of S and S'. The event which we wish to locate in S and S' landing of the bird. Since S' moves relative to S at a velocity of v = 1 m/s, and the speed of light is 3 × 10^8 m/s, we find v/c = 1/(3 × 10^8) = 0.333 × 10^-8 \ll 1 Hence, we may neglect v^2 / c^2 in the equation of (1) because it is negligible when compared with 1. Therefore x' = (x - vt) / \surd1 = x - vt y' = y(2) z' = z t' = {t - (xv/c^2)} / \surd1 = {t - (xv / c^2)} Furthermore , v/c^2 = 1 / ( 9 × 10^16) = 0.111 × 10^-16 << 1 Now x would have to be enormous if xv/c^2 were to be comparable to t. Whence, we may neglect xv/c^2 in the last equation of (2) and t'= t

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Question:

Define the term catalyst. Give an example of an inorganic catalyst and of an organic catalyst.

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Solution:

A catalyst is a substance which affects the speed of a chemical reaction without affecting its final equilibrium point. The addition of a catalyst enables the reaction to reach equilibrium at a faster rate than it otherwise would. A catalyst is not incorporated into the products of the reaction and hence is capable of being regenerated. To understand how a catalyst speeds up a chemical reaction, it is first necessary to define the activation energy of a reaction. There is usually an energy barrier to every chemical reaction which prevents the reaction from spontaneously occurring, even if the reaction is exergonic, that is, one with a negative ∆G (a decrease in free energy). This energy barrier is called the activation energy (Ea). One way to overcome the energy barrier, is through the addition of heat. This increases the internal energy content and therefore the frequency of collisions and reactions. A catalyst can also overcome the energy barrier by lowering the activation energy, and thereby increasing the number of molecules with a suffi-cient energy content to react and form a product. The catalyst acts by forming an unstable intermediate complex with the reactants. This complex is subsequently trans-formed into the product with the release of the catalyst. (C + R)\rightleftharpoonsC \textbullet R\rightleftharpoons(C + P) (catalyst + reactant)(complex)(catalyst + product) The catalyst is therefore free to complex with a second molecule of the reactant. The intermediate complex formed provides an alternative pathway of reaction with a corres-pondingly lower activation energy, thus enabling the reaction to proceed at a faster rate. Figure showing a higher activation energy for an uncata-lyzed reaction than for a catalyzed reaction, note that ∆G remains the same for both types of reactions. There are both inorganic catalysts and organic catalysts. Most inorganic catalysts are non-specific, i.e., they will catalyze various similar reactions. For example, platinum black will catalyze any reaction involving H_2, since it serves to weaken, the bond between the hydrogen atoms. Other inorganic catalysts include water, iron, nickel, and palladium. On the other hand, organic catalysts such as enzymes are highly selective as to the reactions they will catalyze. The specific molecules which an enzyme will bind to and react with are called substrates. There are many enzymes synthesized by living cells, all with specific functions and names. An example of an organic catalyst is sucrase, which is the enzyme that hydrolyzes sucrose into glucose and fructose. Other important enzymes include lysozyme, chymotrypsin, trypsin, and elastase.

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Question:

Given that the surface tension of water is 72.62 (dyn / cm) at 20\textdegreeC, how high should water rise in a capillary that is 1.0 mm in diameter?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E07-0272.htm

Solution:

The surface tension is related to the radius of the capillary tube by the equation: \Upsilon = (1/2) rh\rhog where \Upsilon is the surface tension, r is the radius of the capillary tube, h is the height the water rises, \rho is the density and g is the downward force of gravity. The density of H_2O is (1 g / cm^3), the force of gravity is 980 (cm / s^2) and the surface tension is given as 72.62 (dyn / cm). The radius is found by dividing the diameter of the tube by 2. (1 mm =10^-1 cm). Solving for h: 72.62 (dyn / cm) = 72.62 (g / sec^2) 72.62 (g/sec^2) = (1/2) (10^-1 / 2 cm) (h) (1 g / cm^3) (980 cm / sec^2) h = [(72.62 g / sec^2) (2)(2)] / [(10^-1 cm) (1 g / cm^3) (980 cm / sec^2)] h = 2.96 cm.

Question:

What are the distinguishing features of the several orders of insects?

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/Users/wenhuchen/Documents/Crawler/Biology/F12-0303.htm

Solution:

Insects are divided into different orders based primarily upon wing structure, type of mouthparts, and the type of metamorphosis they undergo, i.e., complete or incomplete. There are approximately twenty-five orders of in-sects. The class Insecta (Hexapoda) is divided into two subclasses. SubclassApterygota contains the wingless insects, which are believed to be the most primitive living insects. SubclassPterygotacontains most of the insect orders. These are winged insects, or if wingless, the loss of wings is secondary. Some of the better known orders will be further discussed. OrderThysanuracontains primitive wingless insects with chewing mouthparts and long tail-like appendages. Silverfish and bristletails are members of this order. Some species are commonly found in houses and eat books and clothing. OrderOdonatacontains dragonflies and damsel flies. The two pairs of long membranous wings are held permanent-ly at right angles to the body. They have chewing mouth-parts and large compound eyes. Immature forms of these insects (nymphs) are aquatic (fresh water). Other winged insects can fold their wings back over the body when they are not flying. OrderOrthopteracontains grasshoppers, crickets and cockroaches. The forewings are usually leather-like. They do not function in flying, but function as covers for the folded hind-wings. The chewing mouthparts are strong. Termites are social insects that belong to the orderIsoptera. Both winged and wingless varieties comprise the termite colony. OrderHemipteracontains the true bugs. They have piercing sucking mouthparts. The forewing has a distal membranous half and a basal, leathery, thick half. OrderAnopluracontains the sucking lice. These insects are wingless and have piercing-sucking mouthparts. The legs are adapted for attachment to the host. These lice are external parasites on birds and mammals- theheadlouseand crablouse are parasites on man. They are often vectors of disease, such as typhus. All of the orders just discussed have incomplete metamorphosis. Order Lepidoptera contains butterflies and moths. They undergo complete metamorphosis, as do the rest of the orders to be discussed. Lepidoptera have two large pairs of scale covered wings, and sucking mouthparts (in adults). OrderDipteracontains true flies, mosquitoes, gnats, and horseflies. They have one pair of flying wings, with thehindwingsmodified as balancing organs. Mouthparts are piercing-sucking, or licking. Adults are often disease vectors. OrderColeoptera, containing beetles and weevils, is the largest order. They have hard forewings which cover membranoushindwings. They have chewing mouth-parts and undergo complete metamorphosis. OrderSiphonapteracontains the fleas. They are small, wingless parasites. They have piercing-sucking mouthparts, and long legs adapted for jumping. Order Hymenoptera contains ants, wasps, bees, and sawflies. There are winged and wingless species. Wings, when present, are two membranous pairs which interlock in flight. Mouthparts are for chewing, chewing- sucking or chewing-lapping.

Question:

Zn, Zn^+2(.50m) \vert\vert Cu^2+(0.20m), Cu . The concentrations of the ions are given in parentheses. Find the E for this cell at 25\textdegreec. Assume the following standard oxidation potentials: Zn \rightarrow Zn^+2 + 2e^-E\textdegree = .763 Cu \rightarrow Cu^+2 + 2e^-E\textdegree = -.337 .

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Solution:

First of all, it is important to note that a Daniell cell is a typical galvanic cell, which means that spontaneous chemical reactions occur at each electrode of the cell to produce an electric current. To calculate E, the potential at other than standard conditions (25\textdegreec and 1 molar concentrations are the standard conditions), you can employ the Nernst equation. It is stated E = E\textdegree -(.0592 / n) log K, where E\textdegree = standard electrode potential, n = number of electrons gained or lost as shown in the overall reaction and K is the ratio of con-centration of ions on right side of equation to those on left side. The overall equation for this cell is given by Zn + Cu^+2 \rightarrow Zn^+2 + Cu . Thus, for this problem, K = [Zn^+2] / [Cu^+2] First, let us try to find the values of these parameters of the Nernst equation, so that E can be found. E\textdegree = E\textdegree_oxidation + E\textdegree_reduction . For the reaction to be spontaneous, E\textdegree must equal a positive value. Thus, you want to choose the oxidation half-reaction (loss of electrons) and reduction half-reaction (gain of electrons) carefully, so that their sum, E\textdegree, is a positive value. You are given the standard oxidation potentials, E\textdegree_ox , of all species in question. The reduction will be the reverse of this reaction, with a reduction potential, E\textdegree_red to negative of the E\textdegree_ox given. The electrode potential of the overall reaction is the sum of these two half-reactions.. For the E\textdegree (overall) to be positive, it follows that the oxidation is Zn \rightarrow Zn^+2 + 2e^-E\textdegree = .763 and the reduction is Cu^2+ + 2e^- \rightarrow CuE\textdegree = -(-.337) overall reaction Zn + Cu^2+ \rightarrow Zn^+2 +Cuwith E\textdegree = .763 + [-(-.337)] =1.1v. From the overall reaction, recall K= [Zn+^2] / (Cu^+2]. (The concentrations of solids are not included in a dissociation constant). You are told Zn^+2 = .5m and Cu^+2 = .2m, so that K = (.50/.20). In the overall reaction, 2 electrons are lost and gained, which means n = 2. If you substitute these values in the Nernst equation, you obtain E = 1.1 - (.-592 / 2) log (.50 / .20). Solving, E = 1.088v for this Daniell cell.

Question:

A list of numbers (1 through 9) is given. The object is to arrange the numbers in ascending order. A move is made by specifying how many numbers (starting from the left) are to be placed in reverse order. For example, suppose 214356789 is the starting posi-tion. Then "reverse 2" gives 124356789. "Reverse 5" gives 534126789. Write a Basic program to play Reverse with the computer.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G23-0559.htm

Solution:

There are two approaches to the game - one algorithmic, and the other heuristic. The algorithmic solution guarantees a win in a certain number of moves (if there are N numbers, it takes 2N - 3 moves). The heuristic method takes advantage of the information at any moment to improve the strategy for minimizing the number of moves. The player decides which method to pursue. Here is the program for this game: 5PRINT "REVERSE - A GAME OF SKILL" 10DIM A (20) 20REMN = NUMBER OF NUMBERS 30N = 9 40REMMAKE A RANDOM LIST A (1) TO A (N) 50A (1) = INT ((N - 1) \textasteriskcenteredRND (1) + 2) 60FOR K = 2 TO N 70A (K) = INT (N\textasteriskcenteredRND (1) + 1) 80FOR J = 1 TO K - 1 90IF A (K) = A (J) THEN 70 100NEXT J 110NEXT K 120REM PRINT ORIGINAL LIST AND START GAME 130PRINT "THE LIST IS:" 140T = 0 150GOSUB 430 160PRINT "HOW MANY TO BE REVERSED?\textquotedblright 170INPUT R 180IF R = 0 THEN 370 190IF R < = N THEN 240 200REM THE COMPUTER CANNOT REVERSE MORE THAN 210REM N NUMBERS 220PRINT "TOO MANY. I CAN REVERSE AT MOST" ;N 230GOTO 160 240T = T + 1 250REM REVERSE R NUMBERS AND PRINT NEW LIST 260FOR K = 1 TO INT(R/2) 270Z = A (K) 280A (K) = A (R - K + 1) 290A (R - K + 1) = Z 300NEXT K 310GOSUB 430 320REM CHECK FOR A WIN 330FOR K = 1 TO N 340IF A(K) <> K THEN 160 350NEXT K 360PRINT "YOU WON IT IN"; T; "MOVES!" 370PRINT "TRY AGAIN (YES OR NO)"; 380INPUT A$ 390IF A$ = "YES" THEN 50 400PRINT "GOOD GAME, THANKS." 410GOTO 470 420REM SUBROUTINE TO PRINT LIST 430FOR K = 1 TO N 440PRINT A (K) 450NEXT K 460RETURN 470END

Question:

In an experiment, a block of beryllium is bombarded with \alpha- particles. A nearby block of paraffin, shielded from the \alpha- particles, is observed to be emitting protons and nitrogen nuclei in separate events. The ratio of the velocity of the proton to that of the nitrogen nucleus is measured to be 7.5. It is suspected that this is a result of a chargeless particle being emitted by the beryllium and absorbed by the paraffin. What is the mass of this particle?

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Solution:

Let the mass of the unknown particle be m and its velocity \upsilon. Paraffin contains many hydrogen atoms (for it is a hydrocarbon). Hence we assume that protons are emitted after a head-on collision with one of the hydrogen atoms, (initially at rest). The momentum of the unknown "particle-hydrogen" system, before the collision, must equal the total momentum of the system after the collision, by the law conservation of momentum. m\upsilon = m_p\upsilon_p' + m\upsilon'(1) where \upsilon_p' and \upsilon' are the new velocities of the emitted proton and of the unknown particle, respectively. By con-servation of energy, the sum of the kinetic energies of the particles before the collision must be equal to the total kinetic energy after the collision or (1/2) m\upsilon^2 = (1/2) (m_p\upsilon'_p)^2 + (1/2) m (1/2) m\upsilon^2 = (1/2) (m_p\upsilon'_p)^2 + (1/2) m \upsilon'^2(2) Equation (1) is rewritten as \upsilon - \upsilon' = [(m_p)/(m)] \upsilon' \upsilon - \upsilon' = [(m_p)/(m)] \upsilon' p (3) and (2) as \upsilon^2 - \upsilon'^2 = \upsilon^2 - \upsilon'^2 = [(m_p)/(m)] \upsilon'_p^2(4) Dividing (4) by (3) gives \upsilon + \upsilon' = \upsilon'_p(5) Then combining (3) and (5) we obtain \upsilon = [(m_p + m)/(2m)]\upsilon'_p(6) Similarly for the collision with the nitrogen atom we obtain \upsilon = [(m_N + m)/(2m)] \upsilon'_N(7) Thus from (6) and (7) [(\upsilon'_p)/(\upsilon ׳ N )] = (m_N +m)/(m_p + m) This ratio equals 7.5 so that 7.5 = (m_N + m)/(m_p + m) 7.5 m_p + 7.5 m = m_N + m 7.5 m_p - m_N = - 6.5 m (m_N - 7.5 m_p)/(6.5) = m Butm_N = 14.008 am\upsilon m_p = 1.008 am\upsilon Hence,m = [(6.458)/(6.5)] am\upsilon= .99 am\upsilon m = (m_N - 7.5 m_p)/(6.5) In other words the unknown particle has a mass almost identical to that of the proton. The particle is called a neutron, and actually has a mass larger than the proton by about 0.18 per cent.

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Question:

40.0 kJ of energy is added to 1.00 gram of magnesium atoms in the vapor state. What is the composition of the final mixture? The first ionization potential of Mg is 735 kJ/mole and the second ionization potential is 1447 kJ/mole.

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Solution:

Ionization potential may be defined as the energy required to pull an electron away from an isolated atom. The second ionization potential is the amount of energy re-quired to pull off a second electron after the first has been removed. The composition of the final mixture is determined by calculating the number of electrons that will be removed from the magnesium ions. To do this one must determine the number of moles of Mg present in 1 g. From this one can determine the number of electrons that will be liberated by using the values for the first and second ionization po-tentials of Mg. The atomic weight of Mg is 24.3. Since moles = grams/atomic weight, there are in 1 gram of Mg, 1 / 24.30 or 4.11 × 10^-2 moles present. The first ionization potential of Mg is 735 kJ/mole. Therefore, 4.11 × 10-2 moles of Mg re-quires 4.11 × 10^-2 moles × 735 kJ/mole or 30.2 kJ to ionize all of the atoms once. 40 kJ was added to the system leaving 40 kJ - 30.2 kJ or 9.8 kJ to remove the second electron. If one has 9.8 kJ and 1447 kJ/mole is required to remove the second electrons, then [9.8 KJ / (1447 KJ/mole)] = 6.77 × 10^-3 moles of atoms can have their second electron removed. 4.11 × 10^-2 moles of Mg are present. [(6.77 × 10^-3) / (4.11 × 10^-2)] × 100 = 16.5%. This means that 16.5% of the atoms can have a second electron removed. Therefore, the composition of the mixture is: Mg^++ 16.5%, Mg^+ 100 - 16.5 or 83.5%.

Question:

What is meant by a hydrostatic skeleton? In which organisms is such a structure found?

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Solution:

Cnidarians (coelenterata) such as hydra, flatworms such as planaria , and annelids like the earth worms all move by the same basic principle of antagonistic muscles. They have no hardexo- or endo - skeleton to anchor the ends of their muscles. Instead the noncompressible fluid contents of the body cavity serve as a hydrostatic skeleton. Such animals typically have a set of circular muscles, the contraction of which decreases the diameter and increases the length of the animal, and a set of antagonistic longitudinal muscles, which, when contracted, decrease the length and increase the diameter of the animal. As an example, the leech is often used. The leech attaches its posterior end to the substrate by means of a sucker, extends its body forward by contraction of the circular muscles, attaches its front end by a sucker, and then detaches the posterior end and draws it forward by contraction of the longitudinal muscles. Note that circular and longitudinal muscle are found in man: the smooth muscle lining his hollow visceral organs, for example. By alternate contractions of each muscle layer, peristaltic waves propel the contents along the hollow organ. For instance, this occurs in the esopha-gus, intestines, oviducts,ureters, etc. The relatively soft internal tissues of solid-bodied animals like the leech can function as a hydrostatic skeleton only to some extent. More active wormlike animals, like the annelids, have evolved segmented bodies in which fluid is contained in a partitioned series of cavities. In addition to segmentation of the body cavity, there is a segmentation of the musculature; the fact that each segment of the body has its circular and longi-tudinal muscles makes possible effective use of the compartmented hydrostatic skeleton. The arrangement of the hyd-rostatic skeleton allows for peristaltic movements down the length of the organism. This aids the organism in burrowing. Some marine worms have additional diagonally arranged muscles that permit more complex movements- of the body and thepaddlelike parapodia that extend laterally from the body wall. Many marine worms live in tubes and the movements of theparapodiaare important not only in locomotion, but in moving currents of water laden with oxygen and nutrients through these tubes.

Question:

A piece of uniform wire is made up into two squares with a common side of length 4 in. A current enters the rectangular system at one of the corners and leaves at the diagonally opposite corner. Show that the current in the common side is one-fifth of the entering current. What length of wire connected between input and output terminals would have an equivalent resistive effect?

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Solution:

Let each side of the double square have resistance R, and let the lettering of the diagram and the currents flowing be as shown in Figure A. Applying the Kirchoff current rule, \sumI = 0, to the junctions A, B and E in turn (with the convention that current entering a junction is positive, and current leaving a junction is negative), gives I_1 - I_2 - I_3 = 0,(1) I_2 - I_4 - I_5 = 0,(2) I_3 + I_4 - I_6 = 0,(3) Applying the Kirchoff voltage rule to the loops ABED and BCFE gives I_2 R + I_4 R - I_3 × 2R = 0,(4) I_5 × 2R - I_6 R - I_4 R = 0.(5) Eliminating I_5 and I_6 from Eqs. (2), (3) and (5), we obtain I_1 - I_2 - I_3 = 0(6) I_2 - 2I_3 + I_4 = 0(7) 2I_2 - I_3 - 4I_4 = 0(8) Eliminating I_2 from these three equations gives I_1 - 3I_3 + I_4 = 0,(9) and2I_1 - 3I_3 - 4I_4 = 0,(10) orI_4 = (1/5) I_1 .(11) Further, the potential drop from A to F by route ADEF is V_AF = I_3 × 2R + I_6 × R = R(2I_3 + I_3 + I_4) using equation (3). By (9) this becomes V_AF = R (I_1 + 2I_4) and using (11),V_AF = I_1 R [1 + (2/5)] = (7/5) RI_1 The equivalent effect is therefore obtained if a wire 7/5 times the length of any side of the square is connected between A and F, because it produces the same potential drop as the double square between these points (see fig. B).

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Question:

Calculate the mean free path for oxygen at 25\textdegreeC at (a) 1atm pressure and (b) 10^-3torr.

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Solution:

The mean free path is the average distance traversed by a molecule between collisions. In a second, a molecule will, on the average, traverse v meters and collide Z times (where Z = \surd2 \pi\sigma^2vn) . Thus, the mean free path is l = v/Z = 1/(\surd2 \pi\sigma^2n) Substituting n = P/kTfrom the ideal gas law, we have l =kT/(\surd2 \pi\sigma^2P) To solve this problem, we must 1) find n, the con-centration, 2) substitute the known values into the equation for the mean free path a) P = 1atm\sigma. = 3.61 × 10^-10 m T = 25\textdegreeC + 273 = 298\textdegreeKR = 0.082 liter-atm/0K-mole n = PN_A / RT = {[(1atm)[6.022 ×10^23(molecules/mole)] [10^3(leter/m^3)]} / {[0.082 (liter-atm/ 0K-mole)] (29 8 \textdegreeK)} n = 2.46 × 10^25 molecules/m^3 Thus,l = \surd2 \pi\sigma^2n = [(1.414) (3.14) (3.61×10^-10 m)^2 (2.46×10^25)]^-1 = 7.02 × 10^-8 m b) P = 10^-3torr= {10^-3torr}/{760 (torr/atm)} = 1.3157 × 10^-6 atm. T = 2980K\sigma = 3.61 × 10^-10 m Here, it is necessary to converttorrunits to units of atmospheres. n = {[(1.3157 × 10^-6atm) [6.022 ×10^23 (molecules/mole)] [10^3(leter/m^3)]} / {[0.082 (liter-atm/ 0K-mole)] (29 8 \textdegreeK)} = 3.24 × 10^19 molecules/m^3 Thus,l = 1/\surd2 \pi\sigma^2n = [(1.414) (3.14) (3.61 × 10^-10 m)^2 (3.24 × 1019/ m^3)]^-1 = 0.053 m = 5.3 cm.

Question:

For the indicator phenolphthalein (In^-),HInis colorless and In^- is red;K_dissis 10^-9. If one has a colorless solution of 50 ml phenolphthalein, how much 0.10 MNaOHsolution would one have to add to make itred: K_W = 10^-14 .

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/Users/wenhuchen/Documents/Crawler/Chemistry/E10-0378.htm

Solution:

To answer this question, write out the dissocia-tion reaction of phenolphthalein in solution. After this, write out the equilibrium constant expression. This allows one to calculate the [H^+]. This informs one of the con-centration of OH^- required and allows the calculation of the amount of NaOHto be added. With this in mind, proceed as follows: An indicator, which changes color, undergoes a dissociation reaction like an acid. For phenolphthalein the reaction is HIn\rightarrow H^+ + In^-. It is given that In^- makes the solution red andHInmakes it colorless. The equilibrium constant expression, which measures the ratio of concentrations of products to re-actants, each raised to the power of its coefficient in the equation, is K_diss= 10^-9 = {[H^+][In^-]}/[HIn] The object is to make a red solution, from an originally colorless one. In general, a specific color of an indicator will show itself when the concentration of the corresponding species is 10 times as great as the concentration of the other species. Since one wants the solution to be red, and In^- = red color, [ln^-]/[HIn] = 10/1. Substituting this into the equilibrium constant expression, 10^-9 = [H^+] (10/1) . Solving for [H^+], [H^+] = 10^-10. From the K_W constant of water, it is known that [H^+][OH^-] = 10^-14 , where K_W is theautodissociationconstant of H_2O. If [H^+] = 10^-10, one can calculate [OH^-] and obtain [OH^-] = 10^-4 M. Thus, to have a red solution, one wants [OH^-] to be 10^-4 M. TheNaOHsolution being added is 0.1 M. (M =molarity= moles/liter.) The volume of the solution is 50 ml. or 0.05 l. (1000 ml = 1 l.) Thus, one needs 10^-4 M or 0.050 l ×10^-4 M = 5.0 ×10^-6 moles. Since volume in liters = number of moles/molarity, andmolarity= 0.1, liters ofNaOHto be added = = [(5.0 × 10^-6)/0.1] = 5.0 × 10^-5 liters.

Question:

When the wavelength of the incident light exceeds 6500 \AA, the emission of photoelectrons from a surface ceases. The surface is irradiated with light of wave-length 3900 \AA. What will be the maximum energy, in electron volts, of the electrons emitted from the surface?

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Solution:

Einstein's photoelectric equation relates the energy of the incident quanta to the maximum energy of the emitted electrons (= W) by the relation hf= h(c/\lambda) = W + W_0 = W + hf_0 =W + h (c/\lambda_0) , where W_0 is the work function of the surface and \lambda_0 the cut-off wavelength. Hence the maximum energy of the emitted photoelectrons in the problem is W =hc[(1/\lambda) - (1/\lambda_0)] = 6.6 × 10^-34 J \textbullet s × 3.0 × 10^8 m \textbullet s^-1 [{1/(3.9 × 10^-7 m)} - {1/(6.5 × 10^-7 m)}] = 19.8 × 10^-19 [(1/3.9) - (1/6.5)] J = [(19.8 × 10^-19 J) / (1.6 × 10^-19 J/eV)] × [(6.3 - 3.9) / (3.9 × 6.5)] = [(19.8 × 10^-19) / (1.6 × 10^-19)] × [(2.6) / (3.9 × 6.5)]eV= 1.27eV.

Question:

A bank of cells having a total emf of 12 V and negligible internal resistance is connected in series with two resistors. A voltmeter of resistance 5000 \Omega is connected across the resistors in turn, and measures 4 V and 6 V, respectively. What are the resistances of the two resistors?

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Solution:

The voltmeter is connected across R_1 as in diagram (a) and since (a) and since the resistance of the voltmeter is in parallel with R_1 the circuit is equivalent to that shown in diagram (b), where that shown in diagram (b), where (1/R) = (1/R_1 ) + (1/5000 \Omega) = [(5000 \Omega + R_1 )/(5000 \Omega R_1 )] R = [(5000R_1 \Omega)/(5000 \Omega + R_1 )](1) Now, the voltmeter measures a 4 V drop across R_1 . By Kirchoff's Voltage Law, the net voltage around the circuit shown in figure (b) is zero. Hence, traversing the circuit as shown \epsilon - IR - IR_2 = 0(2) But we know that the voltage drop across R_1 equals the voltage drop across R, since R_1 and the voltmeter are in parallel. Then IR = 4 V and, from (2) IR_2 = \epsilon - IR = 12 V - 4 V = 8 V Therefore,[(IR_2 )/(IR)] = (8/4) = 2 orR = (1/2) R_2 Using (1)R = [(5000 R_1 \Omega)/(5000 \Omega + R_1 )] = (R_2 /2) Similarly, from diagrams (c) and (d) , showing the second connection of the voltmeter and the equivalent circuit, we have R' = [(5000 R_2 \Omega)/(5000 \Omega + R_2 )]and6 V = I'R_1 = I'R' . R' = [(5000 R_2 \Omega)/(5000 \Omega + R_2 )] = R_1 . Hence, from the two equations obtained, we have (by cross multiplication) 10,000 R_1 \Omega = 5000 R_2 \Omega + R_1 R_2(3) and5000 R_2 \Omega = 5000 R_1 \Omega + R_1 R_2 .(4) Subtracting these equations, 10,000 R_1 \Omega - 5000 R_2 \Omega = 5000 R_2 \Omega - 5000 R_1 \Omega or 15,000 R_1 = 10,000 R_2orR_1 = (2/3) R_2 . Substituting back into equation (3), we get 10,000 [(2/3) R_2 ] \Omega = 5000 R_2 \Omega + [(2/3) R_2 ] (R_2 ) 10,000 \Omega = 7500 \Omega + R_2 R_2 = 2500 \Omega andR_1 = (2/3) R_2 = (2/3) (2500 \Omega) R_1 = 1667 \Omega

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Question:

Two point charges + 2q and \rule{1em}{1pt}q are separated by a distance L. At what point on the line between their centers is the L. At what point on the line between their centers is the electric field zero? electric field zero?

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Solution:

Suppose that this point is a distance x from the + 2q charge, as shown in the figure. By definition, the electric field at a point P is E = K_Eq / r^2, where q is the charge producing the field, r is the distance from q to P, and K_E is a constant (K= 9 × 10^9N \bullet m^2/c^2). The field at x due to 2q is E_+ = K_E [(2q) / (x)^2] The electric field E_\rule{1em}{1pt} due to the negative charge is E_- = \rule{1em}{1pt}K_E [(q) / (x \rule{1em}{1pt} L)^2] The resultant electric field at the point P is then, by the principle of superposition, equal to the sum of the fields due to the individual charges 2q and \rule{1em}{1pt}q. Because we want E_net at P to be zero, we have E = 0 = E_+ + E_\rule{1em}{1pt} = K_E [(2q) / (x)^2]\rule{1em}{1pt}[(K_E q) / (x \rule{1em}{1pt} L)^2] orx^2 = 2(x \rule{1em}{1pt} L)^2 x2= 2x^2 \rule{1em}{1pt} 4xL + 2L^2 x^2 \rule{1em}{1pt} 4xL + 2xL2= 0 x =[{4L \pm \surd(16L^2 \rule{1em}{1pt} 8L^2 )} / 2] = [{4L \pm \surd(2.4)L^2 / 2] = 2 L \pm \surd2 L = 3.41 L or 0.588 L All we have done so far is to find two points along the axis where the fields due to +2q and \rule{1em}{1pt}q are equal in magnitude. We have yet to take account of the vector nature of the fields. At the interior point x = 0.588 L the vectors are in the same direction, so that the fields are added. However at x = 3.41 L the field vectors are in opposite di-rections, so that the resultant field is zero.

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Question:

What are the consequences of intenseinterspecific competition?

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Solution:

The more similar two niches are, the more likely it is that both specieswill utilize at least one common limited resource (food, shelter, nestingsites, etc) . They will therefore be competing for that limited resource. The competitive superiority of one of the rival species may be such thatthe other is driven to extinction. Two species may fluctuate in competitivesuperiority according to the habitats they share, with the result thatone is eliminated in some places and the other is eliminated in other places. One or both of the competing species may evolve in divergent directionsunder the strong selection pressure resulting from their intense competition. Natural selection would favor individuals with characteristics differingfrom those of the other species, because such characteristics wouldtend to minimize competition. In other words, the two species would evolvegreater differences in their niches. Whether any one or a combination of these effects will be the outcomein any given case of intense compe-tition is determined by many complexfactors. If a pair of species occupies identical ecological niches, oneof them will indeed become extinct. This generalization is usually referredto asGause'sprinciple, or the principle of competitive exclusion, whichstates that two different species cannot simultaneously occupy the sameniche in the same place. It is implicit in the definition of niche that no twospecies could ever occupy the same niche. To do so, they would have tobe identical in every respect and hence they would be one species, not two. Thus, the competitive exclusion principle helps us see why the differentspecies living together in a stable community occupy quite distinctniches. Only if organisms living together avoid competition can theysuccessfully coexist. A fit organism is one that avoids competition, andis free from the necessity to struggle with others in its own niche.

Question:

During its lifetime of about 10^10 years, a normal star radiates an energy of about1.0 × 10^52 ergs. What is the energy equivalent inkilowatthours(kwh) ?

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Solution:

This problem illustrates the conversion between watts (w) and ergs/sec (1 w = 10^7 ergs/sec). One kilowatt (kw) is equal to 10^3 watts. Convertingkwto erg/sec, we have 1kw= 10^3 w × 10^7 ergs/sec - w = 10^10 ergs/sec.Also , converting seconds to hours (1 hour = 3600 sec), we have 1kw= 10^10 ergs/sec = 10^10 ergs/sec × 3600 sec/hour = 3.6 × 10^13 ergs/hour. Therefore, 1kw= 3.6 × 10^13 ergs/hour or 1kwh= 3.6 × 10^13 ergs. The total energy radiated by the star is 1.0 × 10^52 ergs. Converting tokwhwe get (1.0 × 10^52 ergs) / (3.6 × 10^13 ergs/kwh) = 7.8 × 10^38kwh. Therefore the total energy radiated by the star is 7.8 × 10^38kwh.

Question:

In contrast to density dependent effects, there are also density independent effects which operate without reference to population size. Elaborate on the major density independent effects and their relationship to population size.

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Solution:

There is no scientific principle which states that populations must be controlled as a function of density. Density effects will always be working in crowded populations, but it is also possible for actual control of population size to be exercised before crowding occurs. Control agents that are not dependent on density can take the form of sudden events that are catastrophic to animal populations. Catastrophic events can solve the problem of overcrowding just as efficiently as a density dependent device. Hurricanes and volcanic explosions can destroy entire populations but these events are too scattered and local to be used as examples for general population controls. More applicable are the common catastrophes which we know as changes in the weather. Nearly all places on the earth suffer seasonal changes from summer to winter, from warm to cold, and from wet to dry. Each of these cyclic changes represents hazards to the animals of each area. Thus, growing populations are frequently cut back, making their normal lives a race to reproduce so quickly that there shall be at least some survivors following the next catastrophe. Unless something happens to prevent the catastrophe from occurring such a popu-lation may never grow large enough to suffer the effects of crowding. Density independent factors may play an important role in limiting some organisms, particularly those with very short life cycles characterized by a growth curve in which the environmental limiting factors do not become effective until after many generations (see fig. 1). A sudden strong density independent limitation then brings growth to an end before density dependent factors can become operative. The density independent effects are definite factors in moving population growth upward or downward, but they cannot hold the population size at a constant level. As a result, populations effected by density independent factors are under control but their numbers fluctuate within wide limits (see fig. 2) and can hardly be described as "in balance."

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Question:

The ease of formation of carbonium ions is 3\textdegree>2\textdegree>1\textdegree>Ch_3+. Explain why.

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Solution:

A carbonium ion is a group of atoms containing a carbon atom with only six electrons around it. Thus the carbonium ion is positively charged. The carbonium ions are classified as primary (1\textdegree), secondary (2\textdegree), and tertiary (3\textdegree) after the carbon bearing the positive charge. For example, where R can be any alkyl group. To understand why the ease of formation is 3\textdegree>2\textdegree>1\textdegree>CH_3+, you need to understand some of the proper-ties of a charged system. According to the laws of electrostatics, the stability of a charged system is in-creased by the distribution of the charge. Alkyl groups, for example, tend to donate electrons, so that any positive charge is dispersed. Thus, the more alkyl groups donating electrons, the greater the distribution of positive charges in a carbonium ion, and, as such, the more stable it is. The greater the stability of a substance, the easier it is to form. In our example, the 3\textdegree carbonium ion has 3 alkyl groups, the 2\textdegree ion has 2 alkyl groups, and the 1\textdegree ion has one alkyl group. Note CH_3+ can be written which shows that no alkyl groups are present. Therefore, 3\textdegree>2\textdegree>1\textdegree>CH_3+ in ease of formation, since this sequence is also the sequence for the number of alkyls that distribute the positive charge, which results in in-creased stability.

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Question:

A student hears the ringing of a bell at the end of the day and begins to pack his books to go home. Trace the path of the stimulus from the source (bell) to its appro-priate center in the brain.

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Solution:

When the bell is struck, it sets off vibrations in the air in the form of sound waves. These sound waves travel through the air and some of them are collected by the pinna. They are then channeled along the auditory canal of the outer ear towards the middle ear. At the inner end of the auditory canal, the sound waves impinge on the tympanic membrane and set the membrane vibrating. These vibrations are first transmitted to the malleus which is in direct contact with the tympanic membrane. Then the vibrations are relayed across the middle ear by the incus and stapes, which are so arranged that they decrease the amplitude but increase the force of vibrations. The stapes transmits the vibrations, via the oval window, to the perilymph in the vestibular canal. This occurs as the vibrating stapes strikes the membrane of the oval window, causing it to oscillate. Since fluids are incompressible, movement of the membrane causes displacement of the perilymph from the vestibular canal to the tympanic canal. (Recall that these two canals are connected at the apex of the cochlea.) The perilymph in the tympanic canal hits the membrane of the round window, resulting in movement of the membrane. Thus vibrations from the bones of the middle ear are transformed into oscillations of the perilymph in the cochlea. The pressure waves of the perilypmh set the thin membranes separating the three canals into vibration, most important of which is the basilar membrane in the cochlea canal, since on it lie the receptor cells for hearing. Vibrations of the basilar membrane cause the sensory hairs of the receptor cells to move up and down against the less movable tectorial membrane. This movement results in the deformation of the hairs, which in some unknown fashion initiates and sends nerve impulses along the sensory neurons, lying at the base of the receptor cells. Axons of the sensory neurons come together to form the auditory nerve which leads the impulses from the organ of Corti to the auditory center in the temporal lobe of the brain. There the impulses are interpreted as the ringing of the bell.

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Question:

List the three primary nitrogenous excretory wastes; tell which animals excrete each, and why. What is uremia?

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Solution:

The three primary nitrogenous waste products elimi-nated from animals are ammonia, urea, and uric acid. Nitrogenous wastes of aquatic animals, both marine and freshwater, are usually excreted as ammonia. Ammonia, a toxic substance, is very soluble in water. A considerable amount of water is needed, but since there is no danger of water loss in aquatic animals, the excretion of ammonia presents no difficulty. Urea is a nitrogenous waste produced in the liver and excreted by the kidneys in mammals and amphibians. It is formed by combining ammonia with carbon dioxide, thus converting ammonia to a relatively non-toxic form. Urea is highly soluble, and requires water for its ex-cretion. However, because it is non-toxic, urea, unlike ammonia, does not require extremely large amounts of water for excretion. Urea may be retained in the body for some time before being excreted. Uric acid is also excreted by terrestrial animals,most notably,the egg-laying or oviparous animals such as birds,snakes,and lizards.Uric acid is a semi-solid,highly insoluble waste product.It is formed in the tissues and in insects is selectively absorbed by the malpighian tubule cells. In birds,it is absorbed by the kidney tubules. The rectum of uric acid-producing animals has powerful water-reabsorptive capacities; the uric acid thus leaves the rectum as a nearly dry powder or hard mass. In its excretion, nearly no water need be lost, and water is highly conserved. Those animals excreting uric acid lay eggs enclosed within a relatively impermeable shell.Excretion of toxic ammonia is out of the question,and urea,if excreted,would build up to harmful concentrations over the period of embryonic development in the egg. Uric acid,on the other hand,being almost solid,can be safely stored in the egg until development is complete. Uremia is a condition where there is a gradual accu-mulation of urea and other waste products in the blood. This condition can result from many different forms of kidney disease.The severity of uremia depends upon how well the impaired kidneys are able to preserve the con-stancy of the internal environment.

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Question:

How can you prove that oxygen is given off by green plants during photosynthesis?

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Solution:

We can show that oxygen is one of the products of photosynthesis by performing a simple experiment. Elodea or some other photosynthetic aquatic plant is placed in a beaker of water and covered with an inverted glass funnel. A rubber tube is attached to the stem of the funnel. This tube is closed off with a pinch cock. The funnel is subsequently raised, creating a column of vacuum in its stem. (See accompanying diagram.) The plant is then exposed to light. The gas given off during the photosynthetic process bubbles upward through the water and fills the column of the stem. To identify the gas, a glowing splinter of wood is applied over the stem of the funnel. Upon removal of the pinch - cock, the glowing splinter bursts into flame. This means that the gas is oxygen, since it supports combustion.

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Question:

Consider again the RCL circuit of the previous problem, or, analogously, an automobile suspension system discussed before. Suppose a voltage source is connected with the circuit [In the case of the automobile sus-pension system, suppose an external force is applied to the system. The road would be such a force.] At time t = 0, S is closed. If E(t) is the voltage at time t, it can be shown that the behavior of the circuit is governed by the following differential equation: Lq̇ ̇+ Rq̇ + (1/C)q(t) = E(t) . where q(t) is the charge on the capacitor, q̇ is the current at time t and q̇ ̇is the rate of change of current. [If E(t) is the external force on the suspension system at time t, then: Mẋ ̇ + Dẋ + Kx(t) = E(t)] If q_0(0) = i_0(0) = 0 [if the suspension system is at equilibrium, so that x_0(0) = i_0(0) = 0], write a digital computer program that models and simulates the behavior of the system from time t = 0 to t = t_f , for the following test inputs: (i) E_1(t) = h where h is constant (ii)E_2(t) =0if t < t1 Hif t1\leq t < t_2 0if t \geq t_2 E_2(t) is a pulse of height h and width (t_2 - t_1). [E_2(t) can be used to simulate a bump in the road.] (See Fig. 1) (iii) E_3(t) = A_1 sin wt where w and A_1 are constants. [E_3(t) could represent a hilly road.] (iv)E_4(t) =0if t < t1 Hif t1\leq t < t_1 + w 0if t1+ w \leq t < t1+ 2w and E_4(t + 2w) = E_4(t) E_4(t) is a pulse train.(See Fig. 2)

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Solution:

Using the variables associated with the automobile suspension system, we call upon the variable v(t) = ẋ again to transform the equation to a simultaneous system of first-order equations to be used in the Euler predictor-corrector program: x = v(t)(1) v̇ = [E(t) - Dẋ ̇- Kx(t)]/M(2) (See Fig. 3) Since E(t) varies with time, it is given a different symbol (trapezoid). Writing equation (2) as a statement function definition, using it at the correct time in the program, substituting the new initial values (T = XCOR = VCOR = 0) and changing I/O statements, complete the solution: (i)REAL K,M,N, T/0.0/, XCOR/0.0/,VCOR/0,0/ F(Y,Z) = (H - D{_\ast}Y - K{_\ast}Z)/M READ K, M, N, TFIN, ACCUR, D,H DT = (TFIN - T)/N Missed segment of the problem is similar to the ones in the previous problems. (ii)REAL K,M,N,T/0.0/, XCOR/0.0/, VCOR/0.0/ F(Y,Z) = (H - D\textasteriskcenteredY - K\textasteriskcenteredZ) /M G(R,S) = - (D\textasteriskcenteredR + K\textasteriskcenteredS)/M READ K,M,N,TF IN, CCUR, D,H,T1,T2 DT = (TF IN - T)/N 50X = XCOR V = VCOR A = G(V, X) IF (T.GE.T1.AND.T.LT.T2)A = F(V, X) 75PRINT T, X T = T + DT IF (T.GT.TFIN) STOP IF (T.LT.T1) GO TO 75 XPRED = X + V{_\ast}DT VPRED = V + A{_\ast}DT 100XDOT = VPRED VDOT = F(V,XPRED) IF (T,GE.T2)VD0T = G(V,XPRED) XCOR = X + 0.5\textasteriskcentered(V + XD0T)\textasteriskcenteredDT : : : Missed segment of the problem is similar to the ones in the previous problems. (iii)REAL K,M,N, T/0.0/, XC0R/0.0/,VC0R/0,0/ F(TIME,Y,Z) = (A1{_\ast}SIN(OMEGA{_\ast}TIME) - D{_\ast}X - K{_\ast}Z)/M READ K,M,N,TFIN,ACCUR,D,A1,OMEGA. DT = (TFIN - T)/N 50X = XCOR V = VCOR A = F(T,V,X) PRINT T,X T = T + DT IF (T.GT.TFIN) STOP XPRED = X + V{_\ast}DT VPRED = V + A{_\ast}DT 100XDOT = VPRED VDOT = F(T,V,XPRED) XCOR = X + 0.5\textasteriskcentered(V + XD0T)\textasteriskcenteredDT : : : Missed segment of the problem is similar to the ones in the previous problems. (iv)REAL K,M,N,T/0.0/,XCOR/0.0/,VCOR/0.0/ F(Y,Z) = (H - D{_\ast}Y - K{_\ast}Z)/M G(R,S) = -(D{_\ast}R + K{_\ast}S)/M READ K,M,N,TFIN,ACCUR,D,H,T1,W, DT = (TFIN - T)/N PER = T1 + 2W CPER = TIME OF COMPLETION OF FIRST PERIOD HPER = T1 + W CHPER = TIME OF COMPLETION OF FIRST HALF-PERIOD TIME = T 50X = XCOR V = VCOR A = G(V,X) IF (TIME.GE.T1.AND.TIME.LT.HPER)A = F(V,X) 75PRINT T,X T = T + DT IF (T.GT.TFIN) STOP TIME = T IF (T.GE.PER) TIME = T - 2\textasteriskcenteredW IF (TIME.LT.T1) GO TO 75 XPRED = X + V{_\ast}DT VRED = V + A{_\ast}DT 100XDOT = VPRED VDOT = F(V,XPRED) IF (TIME. GT. HPER)VDOT = G(V,XPRED) XCOR = X + 0.5{_\ast}(V + XD0T){_\ast}DT : : : Missed segment of the problem is similar to the ones in the previous problems.

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Question:

It has become apparent that a large number of hormones actually have identical initial biochemical actions. This involves the activation of a system containing the enzyme adenyl cyclase, which is found in cell membranes throughout the body. This system has become known as the second-messenger or cyclic AMP system. Using this system; outline the sequence of events involved in producing glucose from glycogen via the action of epinephrine.

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Solution:

Adenyl cyclase is an enzyme found in membranes which, when activated, catalyzes the modification of ATP into cyclic AMP. It is the sole function of the hormone to interact with receptors on the cell membranes and activate the adenyl cyclase. The cyclic AMP is called the second messenger, and it serves to initiate a complex sequence of reactions, leading to the ultimate generation of the final active enzyme. For the case of epinephrine, the following sequence of events occurs: epinephrine, released from the adrenal medulla by stimulation of sympathetic nerves, activates adenyl cyclase on the surface of the liver cell (note that the hormone itself does not gain entry into the cell); adenyl cyclase in turn catalyzes the conversion of ATP into cyclic AMP; the cyclic AMP then stimulates conversion of the inactive enzyme in the cell into an active form; this active enzyme, in turn, catalyzes the critical reaction leading to the breakdown of glycogen. There may appear to be a small inconsistency in this adenylcyclase system. If the generation of cyclic AMP is the common biochemical action of many hormones, why don't all these hormones produce identical effects on different cell types of the body? The reason for this is that the adenyl cyclase systems in different cell types differ in their abilities to be activated by different hormones. There are qualitative differences in membrane receptor sites in different tissues. A second inconsistency which arises is how several hormones, all of which influence cyclic AMP, have different effects on the same cell. One possible explanation is that there are quali-tativelydifferent receptor sites in the cell that respond to different hormones. Within the cell there are separate "compartments" of cyclic AMP, each of which when stimulated gains access to a different intracellular site and catalyzes a reaction there. At least twelve hormones act through intracellular synthesis of cyclic AMP, the most notable exception being the steroid hormones. It is interesting to note that cyclic AMP also plays important roles in non-endocrine mechanisms, such as the control of antibody production and the regulation of vision.

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Question:

How can one demonstrate that amino acids are used tobuild proteinsin cells, while the proteins already in the cell are usedin catabolism?

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Solution:

One way of following the path of the amino acids is by a process calledautoradiography. We can make amino acids radioactive by the incorporationof radioactive isotopes such as tritium (^3H) or carbon-14. We canthen inject theselabelledamino acids into cells in tissue culture, and notechanges in the radioactive level. This is done by exposing the cells to anauto-radiographicemulsion, which upon exposure to any radioactivity in thecells, produces small grains visible microscopically. If we givelabelled aminoacids to the cells in culture and we look for radioactivity at successiveintervals, we should initially see an increase in the number of grainsas thelabelledamino acids are used to build proteins. As time progresses, the amount of radioactive protein should remain dynamically stable: as protein is broken down, new protein is synthesized. If we cease providingthe cells withlabelledamino acids and give them unlabelled aminoacids instead, the amount of radioactive protein in the cells will decreaseand eventually disappear as radioactive protein is degraded and excreted. The disappearance of radioactive protein indicates that protein isbeingcatabolized. (Note: the new protein synthesized is not radioactive sinceunlabelled amino acids are involved.)

Question:

When a swimmer leaves cold water on a warm, breezy day, he experiences a cooling effect. Why?

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Solution:

The equilibrium vapor pressure, is defined as the pressure exerted by the gas of that substance in equilibrium with the liquid state of the same substance. When the swimmer leaves the cold water, he is, coated with water. As the water evaporates a layer of gaseous water molecules form around the swimmer's body. This vapor attains an equilibrium with the atmosphere which hinders further vaporization of the water. However, when the breeze comes, it blows away the vapor above the water, and thereby keeps the partial pressure of the vapor low above the skin. This, then, causes increased evapora-tion so that the partial pressure can be reestablished. This is true because the concentration and vapor pressure of the gaseous water, molecules must be below a certain point for evaporation to occur. Once thisponthas been reached no more liquid molecules of water can go into the gaseous state. For the water to evaporate, it must go from a liquid to a gaseous phase, a process which requires heat. The water removes heat from the swimmer's body. Thus, the swimmer feels a cooling effect.

Question:

Four thieves have just robbed a bank of all its $100 bills. They decide to hide out for the night in an old man's cabin. While the other three are asleep, one of them wakes up and decides to be greedy by dividing the money into four equal piles, hiding one pile for himself and leaving one $100 bill for the old man. Then he recombines the 3 remaining piles and goes back to sleep. Well, it seems that each of the other three thieves has a similar idea. In separate turns, each awakens, divides the remaining money into 4 equal piles, takes a pile for himself, and slips a left over bill to the old man. When they awaken in the morning, they again divide the considerably smaller pile into four equal piles, with a single bill left over for the old man. Write a FORTRAN program to determine the minimum number of bills in the original pile.

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Solution:

First, let us look at an algebraic interpretation of the night's festivities: T = 4N + 1first portion 3N = 4P + 1second portion 3P = 4Q + 1third portion 3Q = 4R + 1fourth portion 3R = 4S + 1final portion To begin, let S = 1 and evaluate the final expression (4S + 1) to see if it is a multiple of 3. If so, the value of R can be substituted in the next from the last equation. This upward climb is continued as long as integers that are multiples of 3 are being obtained. If this condition fails somewhere, S is increased to the next trial value, and the process is repeated until the top equation of the system is reached. To program this brain teaser, two arrays, K and L, are needed. The equations become L (1) = 4K (1) + 1 = 3K (2) L (2) = 4K (2) + 1 = 3K (3) L (3) = 4K (3) + 1 = 3K (4) L (4) = 4K (4) + 1 = 3K (5) T = 4K (5) + 1. The following flowchart will illustrate the program logic. The answer for T is 1021 $100 bills, or $102, 100 (quite a haul!) The components of the arrays are K:80107143191255 L:321429573765 INTERGER T, K, L DIMENSIONK (5), L (4) 5K (1) = 0 10I = 1 K (I) = K (I) + 1 40L (I) = 4\textasteriskcenteredK (I) + 1 K (I+1) = L (I)/3 IF (3\textasteriskcenteredK (I+1) - L (I)) 10, 50, 60 60PRINT "ERROR - TRY AGAIN" GO TO 5 50I = I + 1 IF (I - 5) 40, 90, 60 90J = 4\textasteriskcenteredK (5) + 1 PRINT 100, T, K, L 100FORMAT (I8 // 5I8 // 5I8) STOP END

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Question:

Consider the manometer, illustrated below, first con-structed by Robert Boyle. When h = 40 mm, what is the pressure of the gas trapped in the volume, V_gas . The temperature is constant, and atmospheric pressure is P_atm = 1 atm.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E02-0023.htm

Solution:

We do not need to know any gas law to solve this problem. All we must realize is thatthe pressure exerted on the gas, P_total is equal to the sum of the pressure exerted by the mercury, P_Hg , and the pressure exerted by the air, P_atm . Since 1 mm Hg = 1 torr and 1 atm =760 torr, P_Hg = 40 mm Hg = 40 torr and P_atm = 1 atm =760 torr. Then P_total = P_Hg + P_atm = 40 torr + 760 torr = 800 torr,

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Question:

Using the previous problem write the complete ENVIRON-MENT DIVISION. The SOURCE-COMPUTER is a Honeywell 200, and the system on which the program will be executed is an IBM System/360 Model 40 (IBM-360-40).

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Solution:

Use the system flowchart given in fig. 1 to describe the equipment for a program to randomly update an indexed sequential file. The program uses card input and print output in addition to the random access file. The indexed file requires 2 extra clauses, RECORD KEY and SYMBOLIC KEY if the file is used for random access. The RECORD KEY, C-NR in this program, is used by the computer to search the file for the specific record requested in a program. The SYMBOLIC KEY, SYM-KEY in this program, must be specified if the file is used for output. The variable associated with SYMBOLIC key must be defined in the WORKING STORAGE SECTION. The clause, ACCESS IS RANDOM, sets up the file so that records may be located randomlyby their key, C-NR in this program.

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Question:

(a) 0.20 MHCl, (b) 0.10 MNaOH.

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Solution:

A pH scale has been devised to express the H_3O^+ concentration in solution. By definition, pH = - log [H_3O^+] or [H_3O^+] = 10^-pH It has been shown that water dissociates to H_3O^+ and OH^- ions to a small degree. H_2O + H_2O \rightarrow H_3O^+ + OH^-_ . Theequlibriumconstant is defined as K_W for this reaction and is expressed as [H_3O^+][OH^-]. The H_2O does not appear, since it is presumed to be a constant. From the dis-sociation equation, it can be seen that the concentration of H_3O^+ equals OH^-. By experimentation, K_W has been shown to equal 1.0 × 10^-14. This means that in water, therefore, H_3O^+ and OH^- each have a concentration of 1.0 × 10^-7 M. With this information in mind, one can now solve the problem. (a) The concentration ofHClis 0.20 M. SinceHClis a strong electrolyte, dissociation is complete. Therefore, the concentration of H_3O^+ is also 0.20 M = 2.0 × 10^-1 M. By definition, then pH = - log (2.0 × 10^-1) = 1 - 0.3 = 0.7. (b) The [OH^-] equals the concentration ofNaOH, since it is also a strong electrolyte. One wants the pH, there-fore, employ the expression for K_W. [H_3O^+] = (K_W) / [OH^-] = [(1 × 10^-14)/(0.10)] = 1.0 × 10^-13 M. Therefore, pH = - log (1.0 × 10^-13) = 13.

Question:

Given the consecutive reaction Ak_1\rightarrowBk_2\rightarrowC with k_1 = k_2. Draw a graph for the time variation of the con-centrations of A, B, and C.

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Solution:

To draw this graph, you need to determine the concentration with respect to time of each species in this reaction. This will necessitate writing rate law expressions for each of the species. The rate law states that rate is equal to a proportionality constant k times the concentration of one species or more than one species, each raised to a power. The power is equal to the number of moles reacting. You are told that Ak_1\rightarrowBk_2\rightarrowCwith k_1 = k_2. The rate at which A disappears can be written as {-d[A]} / dt= k_1[A]. A will disappear at a certain rate as given by the rate constant. How long it takes to dis-appear will reflect how much was initially present: the concentration. The rate at which C appears may be written as {-d[C]} / dt=k_2[B]. C forms from B at a certain rate as indicated by k_2. Thus, the amount of B initially present determines the time it take for C to appear. What about the concentration of B? It increases with time as A dis-appears, yet decreases as C appears. The net increase in B will be the differences in these processes. Thus, {-d[B]} / dt = k_1[A] - k_2[B]. If you start with pure A at t = 0, {d [B]} / dt is positive, since k_1[A] >> k_2[B]. After a time, how-ever, A is used up, so that {d [B]} / dt becomes negative. This means the concentration of B will peak sometime along in the reaction. At that instant {d[B]} / dt= 0, so that k_1[A] - k_2[B] = 0, which means [A]/[B] = [k_2]/[k_1] Thus, the behavior of the system depends on k_2/k_1. As k_1 grows small, B becomes less notable. But, you are told that k_2 = k_1, which indicates that k_2/k_1 = 1. [C] will rise as [B] falls. Thus, you have the follow-ing graph. If all reactions proceed to completion C will be the only species present at the conclusion of the process.

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Question:

The outside curve on a highway forms an arc whose radius is 150 ft. If the roadbed is 30 ft. wide and its outer edge is 4 ft. higher than the inner edge, for what speed is it ideally banked?

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Solution:

We wish to relate the velocity of the car to \varphi, the banking angle. Note that the car is undergoing circular motion, hence its acceleration in the x-direction is a = v^2/R , where R is its distance from the center of the circle (see figure). Applying Newton's Second Law, F = ma, to the x circle (see figure). Applying Newton's Second Law, F = ma, to the x component of motion, component of motion, ma = N sin \varphi Buta = v^2/R and mv^2/R = N sin \varphi(1) The acceleration of the car in the y-direction is zero, since it remains on the road. Applying the Second Law to this component of motion, N cos \varphi = mg(2) Dividing (1) by (2), (mv^2/R)/mg = (N sin \varphi)/(N cos \varphi) = tan \varphi tan \varphi = v^2/Rg Hencev = \surd(Rg tan \varphi) Now, note that the width of the road bed is much smaller than the inner radius of the road. Hence, we may approxi-mate R as the inner radius. R \approx 150 ft v = \surd[(150 ft) (32 ft/s^2) tan \varphi] From the figure, sin \varphi = 4/30 cos^2 \varphi = 1 - sin^2 \varphi = (900/900) - (16/900) = 884/900 Hencecos \varphi= \surd884/30 andtan \varphi= (4/30)/(\surd884/30) = 4/\surd884 = .1345 Thereforev = \surd[(150 ft)(32 ft/s^2)(.1345)] v = \surd(645.6 ft^2/s^2) v = 25.41 ft/s

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Question:

Clay contains 30 % AI_2 O_3, 55 % SiO_2, and 15 % H_2 O. What weight of limestone is required per ton of clay to carry out the following chemical change? 6CaCO_3 + AI_2O_3 + SiO_2\rightarrow3CaO\bulletAl_2 O_3 + 3CaO\bulletSiO_2 + 6CO_2. (limestone)

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Solution:

From the coefficients in the reaction equation, one sees that for every mole of AI_2 O_3 and SiO_2 present, moles of CaCO_3 (limestone) are required for the reaction to occur. As such, to determine how much limestone is necessary to react with 1 ton of clay, compute the number of moles of AI_2 O_3 and SiO_2 present. From this figure, the number of moles of CaCO_3 can be found. Because a mole is defined as grams divided by molecular weight (MW), the weight (grams) can be found once the molecular weight is calculated. It is given that 1 ton or 2000 1b of clay will be used. Because clay contains 30 % AI_2 O_3 and 55 % SiO_2 .30 × 2000 = 600 lbs of A1_2 O_3 and .55 × 2000 = 1100 lbs of SiO_2 are present in one ton of clay. To calculate the mole amounts, one must convert lbs to grams. This can be done by using the conversion factor 454 g/lb. Thus, 600 lbs × 454 g/lb = 2724 x 10^2 grams of Al_2 O_3 and 1100 × 454 g/lb = 4994 × 10^2 grams of SiO_2. The MW of Al_2 O_3 = 102 grams/mole. The MW of SiO_2 = 60 grams/mole.Therefore, moles of Al_2 O_3 = [(2724 × 10^2 g ) / (102 g/mole)] = 2.67 ×10^3 moles and moles of SiO_2 = [(4994 × 10^2 g) / (60 g/mole)] = 8.32 × 10^3 moles Notice, in one ton of clay, the mole amounts of SiO_2 and Al_2 O_3 are not equal. The mole amount of limestone re-quired will be six times the mole amount of Al_2 O_3, 2.67 × 10^3 , and not SiO_2. The reason stems from the fact that AI_2 O_3 is the limiting reagent. The amount of any reagent required (or product produced) depends only on the limiting reagent. SiO_2, with a mole of 8.3 × 10^3, is in excess; the amount of limestone required will not depend on it. Consequently, moles of limestone = 6(moles of AI_2 O_3) = 6 (2.67 × 10^3) = 1.60 × 10^4 moles. The molecular weight of CaCO_3 (limestone) = 100 g/mole. Grams of CaCO_3 required per ton of clay = (moles of CaCO_3) (MW of CaCO_3) = (1.60 × 10^4) (100) = 1.60 × 10^6.

Question:

Discuss the modes of asexual reproduction exhibited by Discuss the modes of asexual reproduction exhibited by higher plants.

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Solution:

A number of commercial plants-bananas, seedless grapes, navel oranges, and some varieties of potato, to name just a few-have lost the ability to produce functional seeds and must be propagated entirely by asexual means. Many cultivated trees and shrubs are asexually reproduced from the cuttings of stems, which will develop roots at their tips when placed in moist ground or in water containing a small amount ofindole-acetic acid (auxin). Many plants such as the strawberry develop long, horizontal stems calledstolonsor runners. These grow some distances along and above the ground in a single sea-son and may develop new erect plants at every other mode. Other plants spread by comparable, but underground, stems called rhizomes. Such weeds as witch grass and crab grass spread very fast by means of either runners or rhizomes. Swollen underground stems, or tubers, also serve as an asexual means of reproduction in plants such as the potato. New plants grow out at the buds of the tuber. The stems of raspberries, currants, and wild roses, and the branches of several kinds of trees maydroopto the ground. Adventitious roots and a new erect stem may develop at one of the nodes touching the ground. The stem or branch may lose its connection with the parent plant and grow into a new, independent plant.

Question:

What volume of a block of wood (density = 0.80 g / cm^3), which weighs 1.0 kg, will be above the water surface when . the block is afloat? (Note: A floating object displaces its own weight of water.)

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Solution:

Since a floating object displaces its own weight in water, 1 kg of water is displaced by this block of wood. One can find the volume of the block of wood above the water by solving for the volume of the block and subtracting the volume of 1 kg of water from it. One uses the density to solve for the volume. density = (Weight) / (Volume) = (g / cm^3) Therefore: volume = (Weight / density) Solving for the volume of the woods: 1 kg = 1000 g, Volume = 1000 g /(.80 g/cm^3) = 1.25 × 10^3 Solving for the volume of the water displaced: By definition the density of water is (1.0 g /cm^3). volume = (1000g) / [1.0(g / cm^3)] = 1.00 × 10^3cm^3 volume of wood above water = (volume of wood)- (volume of water) volume of wood above water = (1250 cm^3) - (1000 cm^3) = 250 cm^3

Question:

Determine ∆H\textdegree for the following reaction of burning ethyl alcohol in oxygen: C_2H_5OH(l) + 3O_2(g) \rightarrow 2CO_2(g) + 3H_2O(l) ∆H\textdegree_f of C_2H_2OH(l) = \rule{1em}{1pt} 65.9 Kcal/mole ∆H\textdegree_f ofCO_2(g) = \rule{1em}{1pt} 94.1 Kcal/mole ∆H\textdegree_fofH_2O(l) = \rule{1em}{1pt} 68.3 Kcal/mole

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Solution:

The heat reaction (∆H\textdegree) may be found from the heats of formation (∆H\textdegree_f) by subtracting the sum of the heats of formation of all reactants from the sum of the heats of forma-tion of all products. The heat of formation of all pure elements is zero. When more than one mole of a compound is either reacted or formed, the heat of formation is multiplied by the stoichiometric coefficient for the specific compound in the equation. heat of reaction = [2 × ∆H\textdegree_f of CO_2 + 3 × ∆H\textdegree_f of H_2O] \rule{1em}{1pt} [∆H\textdegree_f of C_2H_2OH + 3 moles × ∆H\textdegree_f of O_2] heat of reaction = [2 moles × (\rule{1em}{1pt} 94.1 Kcal/mole) + 3 moles × (\rule{1em}{1pt} 68.3 Kcal/mole) - 1 mole × (\rule{1em}{1pt} 65.9 Kcal/mole + 3 moles × 0] = \rule{1em}{1pt} 327.2 Kcal.

Question:

Calculate the volume of oxygen necessary to burn completely 1000 cubic feet of butane gas according to the equation 2C_4 H1 0+ 13O_2 \rightarrow 8CO_2 + 10H_2 O

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Solution:

Since volumes are concerned, the procedure for solving this problem is to set up a ratio between the volumes present and the mole requirements. For this re-action, 2molesof butane react with 13 moles of O_2. When two gases are under the same temperature and pressure conditions, a mole of either gas will occupy the same volume. Therefore, since 13/2 times as many moles of O_2 are required, the volume of O_2 must be 13/2 times that of methane. Given the volume of methane is 1000 cu. ft., the volume of O_2 = (13/2) (1000) = 6500 cubic feet.

Question:

A uniform wooden beam, of length 20 ft and weight 200 lb, is lying on a horizontal floor. A carpenter raises one end of it until the beam is inclined at 30\textdegree to the horiz-ontal. He maintains it in this position by exerting a force at right angles to the beam while he waits for his mate to arrive to lift the other end. What is the magni-tude of the force he exerts?

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Solution:

Consult the diagram: AB is the beam and C its midpoint. The weight W acts through C, since the beam is uniform and the two other forces acting on the beam are the force F\ding{217} exerted by the carpenter at B at right angles to AB and a total force P\ding{217} exerted by the floor at A in an unspecified direction. Since the beam is acted on by three forces which maintain it in equilibrium, the lines of action of the three forces must be concurrent. Thus, the direction of force P\ding{217} is from A to the point D at which F\ding{217} and W\ding{217} meet. Since the board is not in motion, the second condition of equilibrium can be applied. Taking moments about point A, we have ^T\sum_A = P × 0 + F × AB - W × AE = 0 F = W(AE/AB) = W[(AC cos 30)/(AB)] = (1/2)W cos 30 = (\surd3/4) W. \therefore F = 86.6 lb.

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Question:

In water vapor at 25\textdegreeC, what is the average speed of a water molecule in meters/sec? Recall that 1 Joule = 1Kg-m^2/sec^2. The Boltzmann constant, k = 1.3806 × 10^-23 J/deg.

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Solution:

Recall, that for all bodies in motion, kinetic energy (K.E.) = (1/2) MV^2, where M = mass of body andV= average velocity of the molecules. Thus, if K.E. and the mass are known, thenVcan be solved. To find K.E., note, that for gases average K.E. is proportional to temperature, This can be written as K.E. = 3/2kT, where k is theBoltzman constant 1.3806 × 10^-23 J/deg and T = temperature in Kelvin (Celsius plus 273\textdegree) . You are given the tem-perature of the system (25\textdegreeC or 298\textdegreeK). Thus, K.E. = 3/2(1.38 × 10^-23)(298) = 6.17 × 10^-21 J/molecule. Since K.E. = (1/2)MV^2, 6.17 × 10^-21 = (1/2)MV^2. If you knew the mass of one water molecule, you could substitute it into this expression and find V. The mass of the H_2O molecule can be found by using the molecular weight (M.W.) and Avogadro's number (6.02 × 10^23/mole). H_2O has an M.W. of 18 grams/mole. Thus, the mass of one molecule = (18 g/mole) / {(6.02 × 10^23 (molecule/mole)} = 2.99 × 10^-23 grams/molecule. = 2.99 × 10^-23 grams/molecule. Substituting and solving for V, V = \surd[2(K.E.)/ M] = \surd[2{6.17 × 10^-21( kg - m^2 / sec^2 - molecule)} / {(2.99 ×10^-23 g/molecule) (.001 kg/g)}] V = 642 m/sec. (Note: .001 kg/g is a conversion factor.)

Question:

Calculate the quantity of heat required to (a) convert a liter of H_2O at 30\textdegreeC to a liter of water at 60\textdegreeC, and (b) heat a 1 kg block of aluminum from 30\textdegree to 60\textdegreeC. Assume the specific heat of water and aluminum is, re-spectively, 1 cal/g\textdegreeC and .215 cal/g\textdegreeC.

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Solution:

When heat is added to a mass of unspecified substance, the temperature will rise. The quantitative relationship between the quantity of heat, expressed in kilocalories, and the rise in temperature - provided there is no phase change - is: quantity of heat = mass × specific heat × ∆T, where ∆T is the change in temperature and the specific heat equals the amount of heat required to raise the temperature of 1 g of any substance by 1\textdegreeC. (a) To find the mass of H_2O, remember that density = mass/volume , and the density of H_2O is one. Thus, mass = (1 liter H_2O) (1000 ml/liter) (1.00 g H_2O)/ml. H_2O). Therefore, quantity of heat= (mass) (specific heat) (∆T) mass= (1 l H_2O) (1000 ml/l) (1 g/ml) = 1000 g specific heat= 1 cal/g\textdegreeC ∆T= (60 - 30)\textdegreeC = 30\textdegreeC quantity of heat= (1000 g)(1 cal/g\textdegreeC)(30\textdegreeC) = 30,000 cal = 30 Kcal (b) A similar calculation can be used to determine the amount necessary to raise the temperature of the aluminum from 30 to 60\textdegreeC. mass= (1 kg Al)(1000 g/kg) = 1000 g specific heat= .215 cal/g\textdegreeC ∆T= (60 - 30)\textdegreeC = 30\textdegreeC quantity of heat= (1000 g)(.215 cal/g\textdegreeC)(30\textdegreeC) = 6450 cal = 6.45 Kcal.

Question:

Find the vapor pressure of CCl_4 (carbon tetrachloride) at 38\textdegreeC, if H^1 = 32,000 (J / mole) and at 23\textdegreeC, the vapor pressure is .132 atm.

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Solution:

To solve this problem, employ the Clausius- Clapeyron equation, which permits evaluation of vapor press-ure in terms of ∆H^1 (.the heat required to transform one mole of liquid to the ideal-gas state) , the original vapor press-ure and the two temperatures. The equation states: log (p_1 / p_2) = (∆H^1 / 19.15)[(1 / T_2) - (1 / T_1)], where p_1 and p_2 are vapor pressures and T_1 and T_2 are the temperatures in Kelvin. Let p_2 be the vapor pressure of CCl_4 at 38\textdegreeC. Substitute the known values for solution, log(.132 / p_2) = [(32,000) / (19.15)] [(1 / 311) - (1 / 296)]. solve for p_2, to obtain p_2 = .250 atm.

Question:

In pea plants, tall plants (T) are dominant to dwarf (t)t yellow color (Y) is dominant to green (y), and smooth seeds (s) are dominant to wrinkled seeds (s). What would be the phenotypes of the following matings? a) Tt Yy Ss×ttyyss b) Tt yy Ss×ttYySs

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Solution:

This is an example of a trihybrid cross, which means a cross involving three different traits. The method of solving follows the same principles as that for any other cross. a) Before we can do the cross, we must determine the gametes involved. We can obtain the total number of different possible gametes from each parent by using a simple mathematical rule. The total number of possible combinations is 2^n, where n is the number of heterozygous traits involved in the cross. For the first parent (Tt Yy Ss), n equals 3 and the total combinations is 2^3 or 8. The genotypes of these eight gametes can be de-termined by using a dichotimous branching system (or forkedline method). By following any set of lines from T (or t) to S (or s) one can determine all possible combinations of factors in the following fashion: Gametes Gametes Y S TYS Y S Tys T s TYs t s tYs y S TyS y S tyS S TYs s tys since the second parent is homozygous (ttyyss) there is only 1 possible gamete type, tys, because n = 0 and 2\textdegree = 1. in the cross: F_1 tys TYS TtYySs TYs TtYyss TyS TtyySs Tys Ttyyss tYS ttYySs tYs ttYyss tyS ttyySs tys ttyyss Phenotypically,TtYySsistall,yellow,smooth; TtYyssistall,yellow,wrinkled; TtyySsistall,green,smooth; Ttyyssistall,green,wrinkled; ttYySsisshort,yellow,smooth; ttYyssisshort,yellow,wrinkled; ttyySsisshort,green,smooth;and ttyyssisshort,green,wrinkled. b)There are two heterozygous traits in the parent with genotype TtyySs, so n equals 2^2. The number of different gametes is or 4. Determining their genotypes: Gametes T-y S TyS s Tys t-y S tyS s tys There are four different gametes from the second parent with genotype ttYySs (2^2 = 4). Gametes Y S tYS t s tYs y S tyS s tys In the cross: F^1 TyS tys Tys tys tYS TtYySS ttYySS TtYySs ttYySs tyS TtyySS ttyySS TtyySs ttyySs tYs TtYySs ttYySs TtYyss ttYyss tys TtyySs ttyySs Ttyyss ttyyss This cross could also be done using the forked line method. View the trihybrid cross as three separate monohybrid crosses. (Since genes segregate independently, we can legitimately do this). Then the genotypic pairs resulting from each cross can occur in any possible combinations with the gene pairs produced in any other cross. The ratios from each cross remain the same as they would if done separately. (Note that this method assumes that you can do the fairly simple monohybrid crosses by inspection, having had sufficient exposure and experience with them.) Summarizing our results: Phenotypes Genotypes Genotypic Frequency Phenotypic Frequency Tall, yellow, smoot TtYySS 1 3 TtYySs 2 1 Tall, yellow, wrinkled TtYyss 1 1 Tall, green, smooth TtyySS 1 3 TtyySs 2 Tall, green, wrinkled Ttyyss 1 1 Short, yellow, smooth ttYySS 1 ttYySs 2 3 Short, yellow, wrinkled ttYyss 1 1 Short, green, smooth ttyySS 1 3 ttyySs 2 Short, green, wrinkled Ttyyss 1 1

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Question:

If a particle of mass 100 g initially had speed 1 × 10^2 cm/s, what would be its kinetic energy and velocity at the end of its 10-cm fall?

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Solution:

The initial kinetic energy is E_i = (1/2) mv^2_i = (1/2) (100 g) (10^4 cm^2/s^2) = 5 × 10^5 ergs During the fall, the particle moves as a result of the gravitational force F_G^\ding{217} acting on it. The work done by F_G^\ding{217} as the particle falls through the height h is, since F_G^\ding{217} is constant and in the same direction as the displacement, W = F_G × h = (100 g) (980 cm/s^2) (10 cm) = 9.8 × 10^5 ergs W is the change in kinetic energy as the particle moves in the direction of the gravitational force. Therefore, the final kinetic energy at the end of the fall is E_f = E_i + W \approx 15 × 10^5 ergs or(1/2) mv^2_f \approx 15 × 10^5 ergs v^2_f \approx (30 × 10^5 ergs)/(100 g) v^2_f \approx 3 × 10^4 cm^2/s^2 v_f \approx 1.73 × 10^2 cm/s This result agrees with what we would calculate from F^\ding{217} = Ma^\ding{217}, but note that we have not specified above the di-rection of the initial speed 1 x 102 cm/s. If it were in horizontal x direction (see the Fig.) it would remain constant and from E_f - E_i = W,we would have (1/2)m(v^2_xf + v^2_yf) - (1/2) m(v^2_xi + v^2_yi) = W But v_xf = v_xi and v_yi = 0, hence (1/2) m(v^2_xf + v^2_yf)^2 - (1/2) mv^2_xi = Wor(1/2)mv^2_vf = W and v_yf^ = \surd(2w/m) = \surd[(2)(9.8 × 10^5 erg)/100 g] v_yf= 139 cm/s. or if v_y^\ding{217} were initially downward in the negative y^\ding{217} direction, we could call upon the familiar relationships for falling bodies: y = y_i + v_yit + (1/2)gt^2 v_yf = v_yi + gt where x_i, v_yi are the initial y position and y velocity of the particle. Hence h = y - y_i = v_yit + (1/2)gt^2(1) v_yf = v_yi + gt(2) where y - y_i is the distance the particle falls through (h). To eliminate t in equation (1), solve (2) for t and substitute in (1). (v_yf - v_yi)/g = t h = v_yi [(v_yf - v_yi)/g] + g/2 [(v_yf - v_yi)/g]^2 orh = 2v_yi [(v_yf - v_yi)/2g] + [(v_yf - v_yi)/2g]^2

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Question:

Two men lift the ends of a 20 ft beam weighing 200 lb onto their shoulders. Both men are of the same height so that the beam is carried horizontally, but one is much the stronger of the two and wishes to bear 50% more of the weight than his mate. How far from the end of the beam should he put his shoulder ?

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Solution:

The beam exerts downward forces on the shoulders of the 2 men. This total downward force is just the weight of the beam and, since the beam is uniform, the weight acts through it's center (see figure). The 2 men exert upward forces R_c\ding{217} and R_m\ding{217}, the former at distance x from one end and the latter at the other end. The beam is in equilibrium vertically since it experiences no motion in that direction. Therefore we can apply the first condition of equilibrium in the vertical direction. \sum F_y = R_m + R_c - 200 lb = 0 R_m + R_c = 200 lb.(1) But we were given in the problem that one man, (the one on the right end) carries 50% more of the weight than his mate. Therefore R_c = (150/100) R_m orR_m = (2/3) R_c(2) Substituting equation (2) in (1), we find (2/3) R_c + R_c = 200 lb (5/3) R_c = 200 lb R_c = 120 lb Note that we solved for R_c\ding{217} since it is this force's distance from end D that we wish to know. Since the beam is in equilibrium rotationally, the second condition of equilibrium can be applied. Taking moments about end A, the magnitude of the force R_m\ding{217} is not needed. We have T \sum_A = (R_m) (0) + Rc(20 ft - x) - (200 lb) (10 ft) = 0 20 ft - x - 2 = (2000 ft-lb)/R_c = (2000 ft-lb)/(120 lb) = (16)(2/3) ft. x = 20 ft - 16 (2/3) ft = 3 (1/3) ft

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Question:

A chemist mixes together 100 ml of 0.5 M sodium acetate, 100 ml of 0.25 M hydrochloric acid (HCl) , and 100 ml of a 1.0 M salt solution. She dilutes this to 1000 ml. Deter-mine the concentrations of all the ions present,un-dissociated acetic acid, and the final pH of the solution, using K_a of acetic acid = 1.7 × 10^-5 andK_w= 1.0 × 10^-14.

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Solution:

To start, determine if any reaction took place in this mixture. If no reaction took place, then the molar concentrations will be numerically the same in 1 liter as the computed moles from the initial concentrations. How-ever, a reaction does take place. Namely, HOAc(acetic acid) + H_2O \rightleftarrows H_3O^+ +OAc^-. The H_3O^+ (fromHCl) reacts with the acetate ion (OAc^-) from the NaOAc(sodium acetate) to produce acetic acid and water. Start the analysis of the reaction by computing the number of moles involved. M =molarity= moles/liter. Thus, in 100 ml of 0.5 MNaOAc, there are 0.5 (100/1000) = 0.05 moles ofNaOAc. Similar calculations show that 0.025 moles ofHCland 0.1 moles ofNaCl(the salt solution) are present. The problem asks one to find the pH, the concentration of the ions present, and the concentration ofundissociatedacetic acid. This necessitates writing the equilibrium constant expression for this reaction, from which these concentrations can be calculated. Thus, write that K_a, the equilibrium constant, equals {[H_3O^+] [OAc^-]} / [HOAc]. Suppose the reaction went to completion. If it did, 0.25moleseach ofHOAcand excessOAc^- would have been produced. One starts with 0.5 moles and only 1 mole ofOAc^- can react per mole of H_3O^+. If there exist 0.25 moles H_3O^+,then, only 0.25molesof theOAc^- can react, leaving 0.25 moles in excess. If the reaction, however, stops x moles short of completion, x moles of H_3O^+ would remain and the excess ofOAc^- would increase from 0.25 to 0.25 + x. TheHOAcwould be reduced to (0.25 - x) moles. Since they are in 1000 ml or 1 liter, these become concentrations that can be substituted into K_a = {[H_3O^+] [OAc^-]}/[HOAc] to give K_a = {x (0.025 + x)}/(0.025 - x). It is given that K_a = 1.75 × 10^-5. Equating, {x (0.025 + x)}/(0.025 - x) = 1.75 × 10^-5. When the K_a for an acid is small, one knows that dissociation is minimal and therefore the concentration of H_3O^+ is very small compared to the concentration ofOAc^- andHOAc. This means that one can assume that their concentrations will not be appreciably changed by the dissociation. Thus, 0.025 - x \allequal 0.025 and 0.025 + x \allequal 0.025. The equation for the K_a can be rewritten as K_a = {x (0.025)} / (0.025) = 1.75 × 10^-5 {x (0.025)} / (0.025) = 1.75 × 10^-5 x = 1.75 × 10^-5. Therefore,[H_3O^+] = 1.75 × 10^-5 M. Since,pH = - log [H_3O^+], pH = - log [1.75 × 10^-5] = 4.76. [OAc^-] = 0.025 + x = 0.025 + 1.75 × 10^-5 \allequal 0.025. [HOAc] = 0.025 - x = 0.025 - 1.75 × 10^-5 \allequal 0.025. The other ions: [Na^+] = 0.150 M fromNaOAcandNaCl. [Cl^-] = 0.125 M fromHClandNaCl. Since [H_3O+] [OH^-] =K_w= 1.0 × 10^-14 and[H_3O+] = 1.7 × 10^-5, [OH^-] = (1.0 × 10^-14) / (1.7 × 10^-5) = 5.9 × 10^-10 M.

Question:

A 1-gram block rests on a frictionless surface and we measure the position of the block to a precision of 0.1 mm. What velocity have we imparted to the block by the act of measuring its position?

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Solution:

If we try to determine the position of a particle at a point x_0 along the x axis as shown in the figure we find that our measurement of the position x_0 cannot be made as precise as required (assuming that it is at rest). As we try to confine the particle to a more exact position, we become less certain about whether it really is at rest at that position. This is a consequence of quantum mechanics and is known as the Heisenberg uncertainty principle. If the uncer-tainty involved in the measurement of position is \Deltax, then the antic-ipated uncertainty in the momentum of the particle will be such that \Deltax \Deltap_x \approx h where h is Planck's constant. Since p_x = mv_x , we have \Deltax m\Deltav_x \approx h or \Deltav_x \approx (h/m \Deltax) = [(6.6 × 10^-27 erg-sec) / {(1 g) × (10^-2 cm)}] = 6.6 × 10^-25 cm/sec This velocity is so small that we must consider the block to be still "at rest". The implications of the uncertainty principle for macro-scopic objects are unimportant, but for microscopic objects they are crucial. That there is any uncertainty at all in the measurement arises from the fact that we can see the block only by virtue of the scatter-ing of photons from the block and this process imparts momentum to the block.

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Question:

A copper strip 2.0cm wide and 1.0mm thick is placed in a magnetic field with B = 1.5T, as in Figure 1. If a current of 200 A is set up in the strip, what Hall potential differ-ence appears across the strip?

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Solution:

The Hall electric field is given by E_H = (jB) / (ne); But E_H = [(V_(x)y) / d], j = (i / A) = (i / dh), and n = (dN_0 / M), where h is the thickness of the strip. Combining these equations gives V_xy = [(iB) / neh] = [{(200A)(1.5T)} / {(8.4 × 10^28/m^3)(1.6 × 10^-19c)(1.0 × 10^-3m)}] = 2.2 × 10^-5v = 22\muV. These potential differences, though quite measurable, are not large.

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Question:

(a)Use a truth table to show that the sum (disregarding the carry) of two binary digits x and y is given by the Boolean expression: (x\bullet \simy) + (\simx \bullet y)(Statement 1) (b)Write a Boolean expression for the carry. (c)Set up a circuit to find the sum and carry of two binary digits, using AND, OR and NOT gates.

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Solution:

Recall that the rule of addition of two binary digits is given in the truth table of Figure 1. First addend Second addend Sum digit Carry digit 0 0 0 0 0 1 1 0 1 0 1 0 1 1 0 1 Since x and y each could be either 0 or 1, there are 2^2 = 4 possible combinations of values of x and y. Evaluating the terms in parentheses of Statement 1, and then Statement 1 itself, the truth table of Figure 2 is created. x y \simx \simy x\bullet \simy \simx \bullet y (x\bullet \simy) + (\simx \bullet y) 0 0 1 1 0 0 0 0 1 1 0 0 1 1 1 0 0 1 1 0 1 1 1 0 0 0 0 0 We see that the entries in the last column of the truth table match the corresponding entries (that is, given the same values for x and y) in the sum-digit column. Therefore, the Boolean sum (that is, disregarding the carry) of x and y can be represented by the logical ex-pression given in Statement (1). Statement (1) is also called the nonequivalence function of x, y, written x (NOT =) y = (x\bullet \simy) + (\simx \bullet y). Note that x (NOT =) y = 1 if and only if x \equiv y = 0, that is, when x \not = y, which is why it is called the nonequivalence function. It is also known as the EXCLUSIVE - OR function. (b)Note that the carry digit is one if and only if both addends, x and y, are 1. But this is the definition of the AND function. Therefore, the carry digit can be represented by x \bullet y. (c)The circuit for the carry is simply the gate repre-sentation for x \bullet y (see Figure 3). Using the rule that oper-ands of the + operation are inputs to an OR gate, operands of the \bullet operation are inputs to an AND gate and the operand of the \sim operation is the input to a NOT gate, we can repre-sent the terms in parentheses as shown in Figure 4. Since each term in parentheses is an operand of the + operation, each of the above gate circuits is an input to an OR gate. Running the carry circuit in parallel to the sum circuit would complete the network. The network in Figure 5 is known as a half-adder. The half-adder has two inputs, corresponding to the two addends, and it will have two outputs, one giving the sum digit and the other giving the carry digit. It performs only half the task for full addition, since it provides neither for a previous carry nor for placing the carry bit to the next column. For complete addition two half-adders are required. Methods for addition may be classified as either parallel or serial. In parallel addition each column is added simultaneously by a separate full-adder. In serial addition each column to be added is transmitted one digit at a time, beginning with the digit in lowest position. The carry bit is retained to be combined with the addition results of the next column. The sum is given as a sequential output.

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Question:

A block hangs at rest from the ceiling by a vertical cord. Find the forces acting on the block and on the cord.

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Solution:

Part (b) of the figure is the free-body diagram for the body. The forces on it are its weight w_1\ding{217} and the upward force T_1\ding{217} exerted on it by the cord. If we take the x-axis horizontal and the y-axis vertical, there are no x-components of force, and the y-components are the forces w_1\ding{217} and T_1\ding{217}. Then, from the condition that \sumF_y = 0, we have \sumF_y = T_1 - W_1 = 0,T_1 = \sumF_y = T_1 - W_1 = 0,T_1 = W_1 from Newton's first law. In order that both forces have the same line of action, the center of gravity of the body must lie vertically below the point of attachment of the cord. Let us emphasize again that the forces W_1\ding{217} and T_1\ding{217} are not an action-reaction pair, although they are equal in magnitude, opposite in direction, and have the same line of action. The weight w_1\ding{217} is a force of attraction exerted on the body by the earth. Its reaction is an equal and opposite force of attraction exerted on the earth by the body. The reaction is one of the set of forces acting on the earth, and therefore it does not appear in the free-body diagram of the suspended block. The reaction to the force is T'_1\ding{217} an equal down-ward force, T'_1, exerted on the cord by the suspended body. T_1 = T'_1 (from Newton's third law). The force T'_1 is shown in part (c), which is the free-body diagram of the cord. The other forces on the cord are its own weight w_2\ding{217} and the upward force T_2\ding{217} exerted on its upper end by the ceiling. Since the cord is also in equilibrium, \sumF_y = T_2 - w_2 - T'_1 = 0 T_2 = w_2 + T'_1 .(1st law) The reaction to T_2\ding{217} is the downward force T'2\ding{217} in part (d), exerted on the ceiling by the cord. T_2 = T'_2(3rd law) As a numerical example, let the body weight 20 lb and the cord weigh 1 lb. Then T1= w_1 = 20 lb, T'_1 = T_1 = 20 lb, T_2= w2+ T'_2 = 1 lb + 20 lb = 21 lb, T'_2 = T_2 = 21 lb.

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Question:

What will be the final concentration of I^- in a solution made by mixing 35.0 ml of 0.100 M NaHSO_3 with 10.0 ml of 0.100 M KIO_3? Assume volumes are additive.

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Solution:

The equation for this reaction is From equation 2 one notices that one mole of each H_2O, HSO^-_3, and I_2 are needed to form 2 moles of I^-. There are 5.0 ×10^-4 moles of H_2O, 1.0 ×10^-4 moles of HSO^-_3, and 5.0 ×10^-4 moles of I_2 present. Thus, the I_2 is the limiting reagent because there is the least of it present to react. Therefore, 2 × 5.0 ×10^-4 moles or 1.0 ×10^-3 moles of I^- formed. If the volumes are additive, the final volume present is 45 ml. Thus, there are 1.0 ×10^-3 moles of I^- present in the final volume of 45 ml of solution.

Question:

Draw a circuit realization of a D (delay) flip-flop, explain the operation of the device.

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Solution:

A circuit realization of a D flip-flop is shown in fig. 1. Note that the circuit is implemented by modi-fying the cross coupled NAND gates. It is important to remember that the clock input can be normally 1 or normally 0; however, in either case data is always clocked to the output Q on a 0 to 1 transition of the clock input C. Inputs Present State Next State D(t) C(t) Q(t) Q(t + E) 0 0\rightarrow0, 1\rightarrow1, 1\rightarrow0 0 0NO 0 0\rightarrow0, 1\rightarrow1, 1\rightarrow0 1 1change 0 0\rightarrow1 0 0Data is 0 0\rightarrow1 1 0clocked in 1 0\rightarrow0, 1\rightarrow1, 1\rightarrow0 0 0No 1 0\rightarrow0, 1\rightarrow1, 1\rightarrow0 1 1change 1 0\rightarrow1 0 1Data is 1 0\rightarrow1 1 1clocked in Fig.2 The characteristics of the device are summarized by the truth table, shown in figure 2, and the state table shown in figure 3.

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Question:

A bag of candies is emptied onto the pan of a spring scale which was originally reading zero. Each piece of candy weighs 1 oz. and drops from a height of 4 ft. at the rate of 6 per second. What is the scale reading at the end of 10 sec. if all collisions are perfectly inelastic?

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Solution:

Each piece of mass m_0 loses\DeltaE_p= m_0gh amount of potential energy while falling through a height h, g being the gravitational acceleration. This loss of poten-tial energy is compensated for by an increase\DeltaE_kin the kinetic energy since the total energy of each piece remains constant during the fall. \DeltaE_k=\DeltaE_p (1/2) m_0v^2 = m_0gh where v is the speed with which each piece strikes the pan. We have v^2 = 2gh = 2 × 32 ft/s^2 × 4 ft = 256 ft^2/s^2 v = 16 ft /s. Since the collision is perfectly inelastic, each time a piece hits the pan, it loses all its momentum and stays on the pan. The mass m striking the pan will exert a force F^\ding{217} on it as a result of the change in its momentum such that the impulse created by the impact equals the momentum change, Ft^\ding{217} =mv^\ding{217}, F^\ding{217} = v^\ding{217} (m/t). Here F is the force acting on the candies due to the pan, and t is the duration of the impact. But, the candy pieces are constantly striking the pan and there is a continuous flux of candies incident on the pan. Hence, if m/t = w/gtis the mass of candies striking the pan per second, F = (16 ft/s) × (6 pieces/s) × (1 oz/piece × {1/16 oz/1b} × {1/32 ft/s^2}) = 6/32 1b = 3 oz. The candies must provide an equal and opposite force F on the pan. After 10 seconds, the force acting on the scale is the continuous force F due to the constant impact plus the weight of the candies already in the pan. The scale reading = 3 oz. + (6 pieces/s) × (1 oz/piece) × 10s = 3 oz. + 60 oz. = 63 oz. = 3 1b. 15 oz.

Question:

A common rocket propellant is a mixture of one part liquid hydrazine (N_2H_4) as fuel and two parts liquid hydrogen peroxide (H_2O_2) as oxidizer. These reactants are hypergolic, that is, they ignite on contact. The reaction proceeds according to the equation N_2H_4(l) + 2H_2O_2 (l) \rightarrow N_2(g) + 4H_2O(g). If reactants are mixed at 25\textdegreeC, what is the heat of reaction?

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Solution:

We will make use of the following expression for the heat of reaction at 25\textdegreeC (298\textdegreeK), ∆H\textdegree_298: ∆H\textdegree_298 = sum of heats of formation of products - sum of heats of formation of reactants. Denoting the heat of formation of species s by ∆H\textdegree_f,s, this becomes ∆H\textdegree_298 = ∆H\textdegreef, N2 (g)+ 4∆H\textdegreef, H2O (g)- [∆H\textdegreef, N2H4 (l)+ 2∆H\textdegreef, H2O2 (l)] From a table of the standard heats of formation at 298\textdegreeK, we find ∆H\textdegreef, N2 (g)= 0 Kcal/mole, ∆H\textdegreef, H2O (g)= -57.8 Kcal/mole, ∆H\textdegreef, N2H4 (l)= 12 Kcal/mole, and ∆H\textdegreef, H2O2 (l)= - 46 Kcal/mole. Then ∆H\textdegree_298= ∆H\textdegreef, N2 (g)+ 4∆H\textdegree_f,H2O (g) - [∆H\textdegreef, N2H4 (l)+ 2∆H\textdegreef, H2O2 (l)] = 0 + 4(- 57.8) - (.12 + 2 (- 46)) = - 151 Kcal/mole.

Question:

K_dissof water for 2H_2O \rightleftarrows H_3O^+ + OH^- changes from 1.0 × 10^-14 at 25\textdegreeC to 9.62 × 10^-14 at 60\textdegreeC. Does the pH of water or its neutrality change when the temperature is increased from 25\textdegreeC to 60\textdegreeC?

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Solution:

To answer this question, it is necessary to find [H^+], since pH = - log [H^+]. To do this, write out the equilibrium dissociation expression for water, which equates the dissociation constant,K_diss, with the ratio of concentrations of products to reactants, each raised to the power of its coefficient in the chemical reaction. ThusK_diss= [H_3O^+][OH^-]. The concentration of water is omitted; it is assumed to be constant. For the chemical reaction one can see that OH^- and H_3O^+ are formed inequimolar amounts, which means [OH^-] = [H_3O^+]. Thus, whenK_diss= 1.0 × 10^-14, one has 1.0 × 10^-14= [H_3O^+][OH^-] = [H_3O^+]^2 = [OH^-]^2. Solving for either, [H_3O^+] = [OH^-] = 1.0 × 10^-7. Thus, pH = - log [H_3O^+] = 7. WhenK_diss = 9.62 × 10^-14 at 60\textdegreeC, 9.62 × 10^-14 = [H_3O^+][OH^-] = [H_3O^+]^2 = [OH^-]^2. Solving for either, [H_3O^+] = [OH^-] = 3.1 × 10^-7. Thus, pH = - log [H_3O^+] = 6.51. Thus, the pH does change. What about neutrality? Neutrality occurs when [H^+] = [OH^-]. In both cases, such a condition exists, so that the neutrality was not altered when the temperature was changed.

Question:

Consider a market model showing a positive relationship between number of new houses sold and washing machines sold. Assuming that the number of households increases exponentially and that washing machines break down, draw a system dynamics model for the above system.

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Solution:

Let H be the number of households, y - the num-ber of houses sold and x - the number of washing machines installed. Since the number of houses sold increases exponentially, the differential equation that gives the rate of change of houses sold over time is y ̇ = k (H - y) .(1) If the washing machines, once installed, never broke down, the rate of growth of washing machines would be ẋ = k_2 (y - x). But washing machines do break down. This means that x, the number of washing machines in use should fall, for every level of x. A new equation needs to be given for x. Assuming that the number of breakdowns is pro-portional to the level of x, we have ẋ = k_2 (y - x) - k_3x.(2) Next, the number of households increases exponentially. This is a more realistic assumption regarding the number of households than assuming that H is constant. The new equation for H is Ḣ = k_4H.(3) A systems diagram for the decay model would trace the interdependencies in the equation system (1) - (3).

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Question:

Describe, using equations, what takes place when the following solutions are prepared: A.A solution containing equal amounts of 0.10M lead nitrate and 0.10M potassium chromate; B.A solution containing 1 mole of sodium chloride and 1 mole of potassium bromide.

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Solution:

In such problems, two processes can occur: mixing of ions in solution and precipitation. A.Both lead nitrate (Pb(NO_3 )_2 ) and potassium chromate (K_2 CrO_4 ) are highlysolube, so that four ions are present in solution: Pb^2+ , K+ , NO-_3 , CrO^2-_4. There are four possible combinations of these ions:Pb(NO_3 )_2 , PbCrO_4 , KNO_3 , and K_2Cr_2 O_4 . Of these, only lead chromate, PbCrO_4 , is insoluble, so that the reaction Pb^2++ CrO_4^2-\rightarrow PbCrO_4 is driven to the right, and yields a precipitate, leaving K+ and NO-_3 ions in solution (along with trace amounts of Pb^2+ and CrO_4^2-, which exist in equilibrium with solid lead chromate). B.Both sodium chloride (NaCI) and potassium bromide (KBr) are highly soluble, giving rise to Na+ , K+ ,Cl-, and Br- ions in solu-tion. The four possible combinations of these ions areNaCI,NaBr,KCl, andKBr, all of which are highly soluble. Therefore no preci-pitate is formed, but only mixing of the ions according to the equation (Na+ ,Cl-) + (K+ , Br-) \rightleftarrows (Na+ , Br- ) + (K+ ,Cl-) .

Question:

Calculate the quantity of heat required to raise the temperature of one gallon of water (3.78 liters) from 10\textdegree to 80\textdegreeC.

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Solution:

The unit of heat is called the calorie. It is defined as the quantity of heat necessary to raise the temperature of one gram of water one degree centigrade. In this problem, one is told that 3.78 liters of water are heated from 10\textdegreeC to 80\textdegreeC. If one subtracts the final tem-perature from the original temperature, one can find the number of degrees the temperature of the water is raised. number of \textdegreeC the temperature is raised = final temperature - original temperature no. of \textdegreeC the temperature is raised = 80\textdegreeC - 10\textdegreeC = 70\textdegreeC. One must now determine the number of grams in 3.78 liters of water. Because 1 milliliter of water weighs one gram, one must determine the number of milliliters in 3.78 liters of water. There are 1000 milliliters in 1 liter; therefore, liters can be converted to milliliters by multiplying the number of liters present by the factor 1000 ml/l liter. 3.78 liters × 1000 ml/l liter = 3780 ml Because number of ml = number of grams for water, there are 3780 g of water present in 3.78 liters. One now knows that 3780 g of water have been raised 70\textdegreeC in tem-perature. Remembering the definition for calorie, one can find the number of calories absorbed in this process by multiplying the number of grams of water by the number of degrees the temperature was raised by the factor 1 calorie/1g \rule{1em}{1pt} 1\textdegreeC, the specific heat of water. number of calories absorbed = 3780 g × 70\textdegreeC × 1 calorie/1g \rule{1em}{1pt} 1\textdegreeC = 265,000 calories There are 1000 calories in 1 kilocalorie, so that calories can be converted to kilocalories by multiplying the number of calories by the factor 1 Kcal/1000 cal. 265,000 cal × 1 Kcal/1000 cal = 265 Kcal. 265 Kcal are absorbed in this process.

Question:

Describe in words the function and the output of the following subroutine. SUBROUTINE GCD (I,J,K) IF (I - J) 1,4,2 1 L = J J = I I = L 2L = I-(I/J){_\ast}J IF(L) 3,4,3 3 I = J J = L GO TO 2 4 K = J RETURN END

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G12-0281.htm

Solution:

The program is written to find the greatest common divisor K of two positive integers I and J. Following the first IF statement, the program pointer will go directly to 1 if the value I - J is negative, to 4 if the value is zero, and 2 if the value is positive. Statement number 1 and the following two statements interchange the values of I and J; therefore prior to statement 2, I is greater than or equal to J. Statement 2 seems useless at first glance. But do not forget that I,J,K and L represent integer variables. Therefore (I/J){_\ast}J may not be equal to I. For example: Let 1 = 4 and J = 3. I/J then will be equal to 1.25, but will be truncated by the computer into integer 1. Then, after multiplying it by J = 3 the computer will get 3 as the result, not 4 as it had originally. In the second IF statement using the same argument as above, the program pointer will go to statements 3 or 4. Statement 3 and the fol-lowing statement replace J with I and L with J. Then the computer goes back to the statement 2 and repeats the calculations. This goes on until I will be divided by J with no remainder left to truncate. In that case value of L will become zero, and the computer will go to statement 4, equate K to J and store this value under K as the desired divisor.

Question:

What is the heat capacity of uranium? Its atomic heat is 26 J/mole and its heat capacity is 238 g/mole.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E14-0483.htm

Solution:

The atomic heat of a solid element at room tem-perature is defined as the amount of heat required to raise the temperature of one mole of an element by one degree Celsius. The heat capacity, or specific heat, is the amount of heat required to raise the temperature of one gram of a substance by one degree Celsius. These two quantities can be related by the equation, atomic heat = atomic weight × heat capacity. Thus, to find the heat capacity of uranium, sub-stitute the given values of atomic heat and weight. heat capacity of uranium= 26 J/moleC\textdegree/238 g/mole =0.11 J/g\textdegreeC.

Question:

A sample of polymer contains 0.50 mole fraction with molecular weight 100,000 and 0.50 mole fraction with molecular weight 200,000. Calculate (a) the number average molecular weight, M_n and (b) the weight average molecular weight, M_w.

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Solution:

Not all particles (or molecules) in polymers have the same weight. Therefore, one seeks some average value and information regarding the distribution of the individual values. (a) The ordinary (un weighted) arithmetic mean is the number average; each item is counted once. The number average molecular weight is equal to the weight of the whole sample divided by the number of molecules in it. M_n= [\sum_i n_iM_i] / [\sum_i n_i] =\sum_i X_iM_i Here, n_i is the number of molecules of molecular weight M_i per gram of dry polymer and X_i is the mole fraction of each component. Solving for M_n: M_n = 0.5(100,000) + 0.5(200,000) = (50,000) + (100,000) = 150,000 (g / mole). (b) In the weight-average M_w, each item counts not as a single unit but in proportion to its weight. The molecular weight is multiplied by the weight (n_i M_i) of material of that molecular weight rather than by the number of molecules. The weight average molecular weight M_w is defined M_w = (\sumin_iM_i^2) / (\sumn_iM_i) Because(\sum_i n_iM_i) / (\sum_i n_i)= \sumiXiM_i; M_w = (\sum_iX_iM_i^2) / (\sum_i X_iM_i) Solving for M_w: M_w = [0.5 (100,000)^2 + 0.5(200,000)^2] / [0.5 (100,000) + 0.5 (200,000)] = [(1.0 × 10^5)^2 + (2.0 × 10^5)^2] / [(1.0 × 10^5) + (2.0 × 10^5)] = [( 5.0 × 10^10) / (3.0 × 10^5) = 1.67 × 10^5 g/mole.

Question:

Explain the use of the terms ortho, meta, and para in systematic nomenclature of benzene hydrocarbons.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E20-0743.htm

Solution:

In establishing a consistent systematic scheme for the naming of benzene hydrocarbons several problems arise. The prefixes ortho, meta, and para (abbreviated o, m, and p) are commonly used to designate the 1,2-, 1,3-, and 1,4-, relationship of substituents on the benzene ring. The problem arises when, for example, the compound named systematically as 1,2-dimethylbenzene is also known as ortho-xylene or ortho- methyltoluene or ortho-dimethylbenzene. Four isomeric compounds are possible with benzene hydrocarbons having the molecular formula C_8H_10. One is obtained by substituting an ethyl group for a hydrogen of benzene, whereas the other three are obtained by sub-stituting methyl groups for two hydrogens.

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Question:

What pigments may be present in plant cells?What are the functions of these pigments?

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Solution:

Chlorophyll a occurs in all photosyntheticeucaryoticcells and is considered to be essential for photosynthesis of the type carried out by plants. It functions in the capture of light energy by either directly absorbing it or receiving it in the form of high energy electrons from the accessory pigments. These accessory pigments, such as chlorophyll b, are found in vascular plants, bryophytes, green algae andeuglenoid algae. Chlorophyll b differs from chlorophyll a structurally and in its absorption spectra. Chlorophyll b shares with chlorophyll a the ability to absorb light energy and produce in the molecule some sort of excited state. The excited chlorophyll b molecule transfers light energy via high energy electrons to a chlorophyll a molecule, which then proceeds to transform it into chemical energy. Since chlorophyll b absorbs light of wavelengths that are different from chlorophyll a, it extends the range of light that the plant can use for photosynthesis.Chorophyllc or d takes the place of chlorophyll b in some algae and plant - likeprotists. Thecarotenoidsare another class of accessory pig-ments. Carotenoids are red, orange, or yellow fat - soluble pigments found in almost all chloroplasts.Carotenoidsthat do not contain oxygen are called carotenes, and are deep orange in color; those that contain oxygen are called xanthophylls, and are yellowish. Like the chlorophylls, the carotenoids are bound to proteins within the lamellae of the chloroplast. In the green leaf, the color of thecarotenoidsis masked by the much more abundant chlorophylls. In some tissues, such as those of a ripe red tomato or the petals of a yellow flower, thecarotenoidspredominate. As accessory pigments, thecarotenoidsfunction in absorbing light not usable by the chlorophylls and in transferring the absorbed energy to chlorophyll a. Another pigment that may be found in plants is the light - sensitive, blue phytochrome. Phytochromes play a fundamental role in the circadian rhythms of plants; they allow the plant to detect whether it is in a light or dark environment.

Question:

Describe the hormonal interactions that control the development and functioning of the breasts.

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/Users/wenhuchen/Documents/Crawler/Biology/F21-0555.htm

Solution:

Perhaps no other process so clearly demonstrates the intricate interplay of various hormonal control mechanisms as milk production. The breasts are composed of numerous ducts which branch through the breast tissue and converge at the nipples. These ducts terminate at their other end within the breast, in saclike structures (alveoli) typical of exocrine glands. It should be noted that an exocrine gland, by definition, secretes material into ducts leading to a specific compartment or surface. The alveoli are the actual milk secreting structures. Both the alveoli and the ducts are lined by contractile cells whose role will be eludicated later. Before puberty, the breasts are small with little internal structure. With the onset of puberty, estrogen and progesterone act upon the ductiletissue and alveoli to produce the basic architecture of the adult breast. In addition to these two hormones, normal breast development at puberty requires prolactin and growth hormone, both secreted by the anterior pituitary. During each menstrual cycle, the morphology of the breast changes in response to fluctuating blood concentra-tions of estrogen and progesterone. The greatest change in breast morphology occurs, however, during pregnancy. The most important hormone which promotes milk production is prolactin. But the onset of lactation requires more than just high prolactin levels. This is seen in that the level of prolactin is elevated and the breasts are enlarged as pregnancy progresses, and yet there is no secretion of milk. The question arises as to what occurs during delivery to permit the onset of lactation. There have been numerous experiments performed with different mammals, which has shown that if the fetus is removed during pregnancy without interfering with the placenta, lactation will not be induced; when the placenta is removed at any stage of the pregnancy, without removal of the fetus, lactation is induced. Present evidence indicates that this inhibitory effect of the placenta is due to its secretion of estrogen and progesterone, which in large concentration appears to inhibit miik release by some direct action on the breasts. Therefore, delivery removes the source of the two hormones, and thereby removes the inhibition to allow lactation to occur. The hypothalamus also influences lactation. The hypo-thalamus releases a Prolactin Inhibitory Factor. This chemi-cal is decreased during suckling; thus increased prolactin allows secretion of milk to occur. While it is still unclear as to what stimulates increased prolactin secretion during pregnancy, it is known what the major factor is in maintaining thesecretion during lactation. It is believed that receptors in the nipples, which are stimulated by suckling, have nervous inputs to the hypothalamus. These inputs ultimately cause prolactin to be released from the anterior pituitary. One final reflex is essential for nursing. The infant cannot suckle milk out of the alveoli where it is produced, but can only remove milk that is present in the ducts. Milk must first move out of the alveoli into the ducts, a process known as milk let-down. This is accomplished by contraction of the contractile cells surrounding the alveoli. The contraction of these cells is under the direct control of the posterior pituitary hormone oxytocin. This hormone is reflexively released by suckling just like prolactin.Oxytocin is also responsible for the contraction of the uterine muscles during delivery, and it is for this reason that many women experience uterine contractions during nursing. Another important neuroendocrine reflex triggered by suckling is the inhibition of follicle stimulating hormone and luteinizing hormone released by the pituitary. Low levels of FSH and LH subsequently prevent ovulation. This inhibition is relatively short-lived in many women, and approximately 50 per cent begin to ovulate despite continued nursing. The end product of all these processes is the milk which contains four major constituents: water, protein, fat, and the carbohydrate lactose. All of these components are very important for the proper development of the infant.

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Question:

How does asexual reproduction take place in the bryophytes ?In the lower vascular plants?

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Solution:

The mosses and liverworts carry out asexual as well as sexual reproduction . The young gametophyte of the moss, theprotonema, is derived from a single spore but may give rise to many moss shoots simply by budding, a process of asexual reproduction. Some liver-worts, such as Marchantia , formgemmaecups on the upper surface of thethallus. Small disks of green tissue, calledgemmae, are produced within these cups. The maturegemmaeare broken off and splashed out by the rain and scattered in the vicinity of the parentthalluswhere they grow into new plants . Like the bryophytes, some lower vascular plants also propagate by vegetative reproduction. In some species of club mosses, small masses of tissue , called bulbils, are formed. These drop from the parent plant and grow directly into new youngsporophyteplants. The ferns reproduce asexually either by death and decay of the older portions of the rhizome and the subsequent separation of the younger growing ends, or by the for-mation of deciduous leaf-borne buds, which detach and grow into new plants .

Question:

You are given a box in which PCl_5(g), PCl_3(g), and Cl_2(g) are in equilibrium with each other at 546\textdegreeK . Assuming that the decomposi-tion of PCl_5 to PCl_3 and Cl_2 is endothermic, what effect would there be on the concentration of PCl_5 in the box if each of the follow-ing changes were made? (a) Add Cl_2 to the box, (b) Reduce the volume of the box, and (c) Raise the temperature of the system.

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Solution:

You are told that the following equilibrium exists in the box PCl_5\rightleftarrows PCl_3 + Cl_2 (all gases) and asked to see what happens to [PCI_5] when certain changes are made. This necessitates the use of Le Chatelier'sprinciple, which states that if a stress is applied to a system at equi-librium, then the system readjusts to reduce the stress. With this in mind, proceed as follows: (a) Here, you are adding Cl_2 to the box. This results in a stress, since one of the components in the equilibrium has its concentration increased. According to LeChatelier'sprinciple, the system will act to relieve this increased concentration of Cl_2 - the stress. It can do so, if the Cl_2 combines with PCl_3 to produce more PCl_5 . In this fashion, the stress is reduced, but the concentration of PCl_5 is increased. (b) When the volume of the box is reduced, the concentration of the species is increased, i.e., the molecules are crowded closer together. Thus, a stress is applied. The stress can only be relieved (LeChatelier's prin-ciple) if the molecules could be reduced in number. Notice, in our equi-librium expression you have 2 molecules, 1 each of PCl_3 and Cl_3 , pro-ducing 1 molecule of PCI_5 . In other words, the number of molecules is reduced if the equilibrium shifts to the left, so that more PCI_5 is pro-duced. This is exactly what happens. As such, the [PCl_5 ] increases . (c) You are told that the decomposition of PCl_5 is endothermic (absorb-ing heat). In other words, if must absorb heat from the surroundings to proceed. If you increase the temperature, more heat is available, and the decomposition proceeds more readily, which means [PCI_5 ] decreases. This fact can also be seen from the equilibrium constant of the reaction, K. This constant measures the ratio of products to reactants, each raised to the power of its coefficients in the chemical reaction. Now, when a re-action is endothermic, K is increased. For K to increase, the re-actant's concentration must decrease. Again, therefore, you see that [PCI_5 ] decreases .

Question:

What does the "pH" of a solution mean? Why is a liquid with a pH of 5 ten times as acidic as a liquid with a pH of 6?

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/Users/wenhuchen/Documents/Crawler/Biology/F01-0020.htm

Solution:

The pH (an abbreviation for "potential of hydrogen") of a solution is a measure of the hydrogen ion (H^+) concentration. Specifically, pH is defined as the negative log of the hydrogen ion concentration. A pH scale is used to quantify the relative acid or base strength. It is based upon the dissociation reaction of water: H_2 O \rightarrow H^+ + OH^-. The dissociation constant (K) of this reaction is 1.0 × 10^-14 and is defined as: K = {[H^+] [OH^-]} / {H_2O}where [H^+] and [OH^-] are the concentrations of hydrogen and hydroxide ions, respectively and [H_2O] is the concentration of water (which is equal to one). The pH of water can be cal-culated from its dissociation constant k: K = 1.0 × 10^-14 = {[H^+] [^-OH]} / {H_2O} = [H^+] [^-OH] Since one H^+ and one ^-OH are formed for every dissociated H_2 O molecule, [H^+] = [^-OH]. 1.0 × 10^-14 = [H^+]^2 ; [H^+] = 1.0 × 10^-7 pH = - log [H^+] = - log(1.0 × 10^-7) = 7 A pH of 7 is considered to be neutral since there are equal concentrations of hydrogen and hydroxide ions. The pH scale ranges from 0 to 14. Acidic compounds have a pH range of 0 to 7 and basic compounds have a range of 7 to 14.

Question:

In what ways do the Upper Paleolithic (Cro-Magnon) and the Neolithic cultures differ?

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/Users/wenhuchen/Documents/Crawler/Biology/F29-0756.htm

Solution:

The Neolithic (or New Stone Age) culture originated somewhere between Egypt and India about 10,000 years ago. It is characterized by implements bearing the marks of careful grinding and polishing and by the beginnings of agriculture and animal husbandry. The earliest animals to be domesticated after the dog (first by Cro-Magnon man) were the pig, sheep, goat, and cow. The horse was not do-mesticated until much later. Our human ancestors gradually changed from wandering hunters and food gatherers of the Cro-Magnon stage to more settled and efficient food producers, raising grain, making pottery and cloth and living in villages. At this time, more diversified tools were made, the hunting men stayed closer to the home and migrated less, people became more self-sufficient and made or obtained what they needed from nearby sources. They no longer lived in small breeding groups but expanded their communities. The increase in food supply enabled an increase in the size of the popu-lation. Neighboring populations interbred and the tendency toward genetic drift was thus greatly reduced. Neolithic people also invented the dugout canoe and the wheel. With the Neolithic age we enter the historical times, for the oldest cultures of Egypt and Mesopotamia were Neolithic. The use of metals began with copper and was followed with bronze. They were used for making tools, vessels and weapons beginning about 4000 B.C. The culture of Cro-Magnon man in comparison to the Neolithic culture may at first seem primitive. Actually, both peoples had about the same intelligence. But culture is an additive phenomenon i.e., one generation will usually build on what is inherited from the previous generations. Conse-quently, culture gets more and more complex and sophisticated. As we can see in the transition from Upper Paleolithic to Neolithic ages, people are living in larger social groups, language is progressing and is an expanding communication device for expressing ideas and experiences, both past and future. The interaction of language and a large social order speed up cultural advancements and the attainment of new knowledge.

Question:

At 273\textdegreeK and 1atm, 1 liter of N_2O_4 decomposes to NO_2 according to the equation N_2O_4\rightleftarrows 2NO_2. To what extent has decomposition proceeded when the original volume is 25 percent less than the existing volume?

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Solution:

Volume is proportional to the number of moles present when Pressure (P) and Temperature (T) are held constant. This can be seen from the equation of state, PV =nRTor V = (nRT/ P ) , where R = universal gas constant, n = moles, and V = volume. When P and T are constant, RT/P is a constant, and V varies directly with n. Thus, in this problem you can discuss volumes in terms of fractions decomposed. Let a be the volume fraction of the original N_2O_4 , that decomposes. Since moles are proportional to volumes and 2 moles of NO_2are produced for every mole of N_2O_4, this fraction, a, results in the production of 2a of N0_2 and 1\Elzbar a of N_2O_4. The total final volume fraction is 1\Elzbar a + 2a = (1 + a); that is, the final volume = (1 + a) times the original volume. The original volume is 1 liter of N_2O_4. The existing volume is (1 + a)(1 liter) = (1 + a) liter. But 1 = .75 (1 + a). Solving, a = .333 .Thus, you have 33% decomposition.

Question:

The pressure in a static water pipe in the basement of an apartment house is 42 lb \textbullet in^-2, but four floors up it is only 20 lb \textbullet in^-2. What is the height between the basement and the fourth floor?

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Solution:

If A is the cross sectional area of the pipe, then the pressure (p_1 = F1/A where is the force exerted) in the pipe at the basement must balance both the pressure P_2 [= F_2/A] in the pipe at the fourth floor and the weight of the water in the column of height h (see the figure). If \rho is the density of water, then by definition \rho = m/v where m is the mass of the water in the column of volume v = Ah. Then m = \rhoAh and the weight of the water is W = mg = \rhoghA. Thus the pressure due to the weight of the water is W/A = \rhogh. Hence P_1 = P_2 + \rhogh (P_1 - P_2)= (42 - 20) lb \bullet in^-2 = 22 lb/in^2 × 144 in^2/1 ft^2 = \rhogh = 1.94 slug \bullet ftr^-3 × 32 ft \bullet s^-2 × h \thereforeh = [(22 × 144)lb \bullet ft^-2]/[1.94 slug \bullet ft^-3 × 32 lb × ft^-3] = 99/1.94 ft = 51.03 ft.

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Question:

A 100-kg man jumps into a swimming pool from a height of 5 m. It takes 0.4 sec for the water to reduce his veloci-ty to zero. What average force did the water exert on the man?

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/Users/wenhuchen/Documents/Crawler/Physics/D06-0318.htm

Solution:

The man's initial velocity (before jumping) is zero. Therefore, as he strikes the water, his velocity v is v^2 = v^2 _0 +2gh, which reduces to v^2 = 2gh v = \surd(2gh) = \surd[2 × (9.8 m/sec^2) x (5 m)] = 10 m/sec Therefore, the man's momentum on striking the water was p_1 =mv = (100 kg) × (10 m/sc) = 1000 kg-m/sec The final momentum was p_2 = 0, so that the average force was F =\Deltap/\Deltat= (p^2 - p^1)/\Deltat= (0 - 1000 kg-m/sec^2)/(0.4 sec) = 2500 N but this is the total force : friction and gravity: F=F_water+ mg = - 2500 - 980 = - 3480 N The negative sign means that the retarding force was directed opposite to the downward velocity of the man.

Question:

A proton (mass 1.67 x 10^-27kg) collides with a neutron (mass almost identical to the proton) to form a deuteron. What will be the velocity of the deuteron if it is formed from a proton moving with velocity 7.0 x 10^6m/sec to the left and a neutron moving with velocity 4.0 x 10^6m/sec to the right? (Figure A)

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Solution:

According to the conservation of momentum we write: m_dv_d = m_pu_p + m_nvn where m_d, m_p, and m_n are the masses of the deuteron, proton, and neutron respectively . Since the masses of the proton and neutron are almost identical: m_p = m_n = 1/2 md Inserting this into the momentum conservation equation (we adopt the convention that velocities to the right are positive): 2m_pv_d = m_pv_p + m_pv_n v_d = (v_p + v_n)/2 = (-7.0 × 10^6m/sec + 4.0 × 10^6m/sec)/2 = -1.5 ×10^6 m/sec(figure B). Thus the neutron moves to the left with speed 1.5 × 10^6 m/sec. In the actual collision a photon is produced and carries off some of the momentum, therefore the velocity calculated above is somewhat too large.

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Question:

What are the main methods used for debugging a program and what are their advantages?

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Solution:

DEBUGGING is the procedure of eliminating bugs (errors) that almost every program has the first time it is run through the computer. These are some of the indications that the program does indeed have mistakes : answers . closer inspection reveals the answers to be wrong. The following is a list of some of the most commonly used debugging methods : assumes the program is fit enough to generate out-put. 4)Planning ahead is based upon occasionally including output statements into the program. Those output statements usually print intermediate results obtained up to that point. This allows the programmer to follow the execution of his program and check the inter-mediate results, accuracy of which is essential. 5) Running the program step by step: When the computer is in its MANUAL condition, pushing the SINGLE STEP key makes the computer perform one instruction. Each time this key is pushed, the computer will perform the instruction and change the lights on the control panel to indicate the address of the next Instruction. In this way, the entire program is performed step by step. This method can only be used when working directly with the central processing unit of the computer.

Question:

The composition of dry air by volume is 78.1% N_2, 20.9% O_2 and 1% other gases. Calculate the partial pressures, in atmospheres, in a tank of dry air compressed to 10.0 atmospheres.

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Solution:

A partial pressure is the individual pressure caused by one gas in a mixture of several gases. The total pressure,P_total, according to Dalton's Laws, is the sum of these individual partial pressure, p_1, p_2 and p_3. i.e. P_total= p_1 + p_2 + p_3. The term dry air indicates that no water vapor is present. The partial pressure is found by multiplying the percent of each gas in the volume by the total pressure. partial pressure = proportion by volume × total pressure P_(N)2 = .781 × 10atm= 7.81atm P_(O)2 = .209 × 10atm= 2.09atm P_other gasses = .010 × 10atm= .1 atm. The total pressure, as required, is 10 atm.

Question:

At what point between moon and earth do the gravitational fields of these two bodies cancel? The earth's mass is 5.98 × 10^24 kg, and the moon's is 7.35 × 10^22 kg. The distance between the centers of the earth and the moon is 3.85 × 10^7 m.

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Solution:

Let the point where the gravitational fields cancel be at a distance R_1 from the earth's center, R_2 from the moon's center. The attraction of the earth at this point will equal that of the moon's: GM\mu/R_1^2 = Gm\mu/R_2^2 where G is the gravitational constant, M the mass of the earth, and m that of the moon, or M/R_1^2 = m/R_2^2. The term on the left side of the equation is called the earth's gravitational field, the term on the right is the moon's. The gravitational field at a point in space is the gravitational force experienced by a unit mass at that point. (It is similar to the electric field, which is the electrostatic force per unit charge experienced due to a particular charged body). Gravitational fields are vector fields and the resultant gravitational field due to two or more masses is calculated by adding the field vectors from all sources at every point in space. The only point in space where the gravitational fields of the earth and the moon cancel, must be collinear to the centers of both bodies since no two vectors cancel if they are not oppositely directed. From the equation above, we get [R_1/R_2]^2 = M/m R_1/R_2 = [M/m]^1/2 = [5.98 × 10^24 kg/7.35 × 10^22]^1/2 = 9. Since the point in question is collinear to both the earth and moon's centers, the distance between the centers must equal R_1 + R_2 (see diagram), thus: R_1/R_2 = 3.65 × 10^8 m R_1 = 3.85 × 10^8 m - R_2 R_1/R_2 = (3.85 × 10^8 m - R_2)/R_2 = {(3.85 × 10^8 m)/R_2} - 1 = 9 R-_2 = (3.85 × 10^7 m) R_1 = 9R_2 = 9(3.85 × 10^7 m) = 34.7 × 10^7 m.

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Question:

If the vapor pressure of CCl_4 (carbon tetrachloride) is .132 atm at 23\textdegreeC and .526 atm at 58\textdegreeC, what is the ∆H' in this temperature range?

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Solution:

Molecules in a vapor exert a pressure that is characteristic of its liquid state. The pressure exerted by this vapor, when in equilibrium with the liquid, is called the equilibrium vapor pressure. The equilibrium vapor pressure can be expressed quantitatively as log P = [(-∆H') / (19.15) T] + C where p = vapor pressure, T = temperature in degrees Kelvin, ∆H' = heat required to transform one mole of liquid to the ideal-gas state, and C = a constant that is characteristic of the liquid. C can be eliminated from the expression by denoting that at two temperature ranges, T_1 and T_2, you have; (log p_1) - (logp_2) = [{(-∆H') / (19.15) T_1} + {(∆H') / (19.15) T_2}] or log (p_1 / p_2) = [(∆H') / (19.15)] [(1 / T_2) - (1 / T_1)]. In this problem, you are asked to find ∆H'. Let p_1 = (.132 atm) at T_1 = 296\textdegreeK and P_2 = (.526 atm) at T_2 = 331\textdegreeK, then substitute these values in the above equation. Therefore, log (.132)/(.526) = (∆H' / 19.15) [ (1 /331) - (1 / 296)] Solving for ∆H', you obtain ∆H' = 32,000 (J / mole).

Question:

An electric field is set up by two point charges q_1 and q_2, of the same magnitude (12 x 10^\rule{1em}{1pt}9 coul; see diagram) but opposite sign, as shown in the figure. What is the electric intensity at points a, b, and c?

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/Users/wenhuchen/Documents/Crawler/Physics/D18-0590.htm

Solution:

By definition, the magnitude of the electric field intensity, E^\ding{217}, is E = kq / r^2 where q is the charge causing the field, and r is the distance from q to the point at which we wish to cal-culate E^\ding{217}. (k = 9 × 10^9 N \textbullet m^2 / c^2). If the field is due to more than one charge, then the total field at a point is the vector sum of the fields due to each charge at that point. At every point, the intensity due to the positive charge is directed radially away from that charge, and the intensity due to the negative charge is radially inward toward the charge. At point a, the intensity due to each charge is directed toward the right. The resultant intensity E^\ding{217}_a is also toward the right and its magnitude is the arith-metic sum of the individual intensities. Hence, E_a = (kq_1) / (r_1^2) \rule{1em}{1pt}(kq_2) / (r_2 ^2) E_a = [9 × 10^9 (N \textbullet m^2 / c^2)][{(12 × 10^\rule{1em}{1pt}9c ) / (36 × 10^\rule{1em}{1pt}4 m^2)} \rule{1em}{1pt} {(\rule{1em}{1pt}12 ×10^\rule{1em}{1pt}9 c) / (16 ×10\rule{1em}{1pt}4m^2)}] E_a= 9.75 × 10^4 N / Ctoward the right. At point b, the intensity set up by q_1 is directed toward the left and that set up by q_2 is toward the right. The magnitude of the first is greater than that of the second because q_1 is closer to b than q_2. The resultant intensityE^\ding{217}_b is toward the left and its magnitude is the difference between the individual intensities. Therefore, E_b = \rule{1em}{1pt}[(kq_1) / (4cm)^2 ] \rule{1em}{1pt} [(kq_2) / (14cm)^2 ] Eb=[{\rule{1em}{1pt}9 × 10^\rule{1em}{1pt}9(N \bullet m^2 / C)^2}] [{(12 × 10^\rule{1em}{1pt}9c)/( 16 × 10^\rule{1em}{1pt}4m^2)} \rule{1em}{1pt} {(12 × 10^\rule{1em}{1pt}9c) / (196 × 10^\rule{1em}{1pt}4 m^2)}] E_b = \rule{1em}{1pt}6.2 × 10^4 N/C, which is toward the left. At point c, the individual intensities have the directions shown and the resultant intensity E^\ding{217}_c is their vector sum. Note, from the figure, that the vertical components of E^\ding{217}_1 and E^\ding{217}_2 will cancel, and their horizontal components will reinforce. Hence E_c = │E^\ding{217}_1│ cos 60\textdegree + │E^\ding{217}_2│cos 60\textdegree = [│E^\ding{217}_1│ / 2 ]+ [│E^\ding{217}_2│/ 2] = (1 / 2) [│{kq_1 / (10cm)^2}│] + (1 /2) [│{kq_2 / (10cm)^2}│] = (1 / 2)[│({9 × 10^\rule{1em}{1pt}9(N \bullet m^2 / C^2)} {(12 × 10^\rule{1em}{1pt}9c)}) / (100 × 10^\rule{1em}{1pt}4 m^2)│] +1 / 2[│({9 × 10^\rule{1em}{1pt}9(N \bullet m^2 / C^2)}{(\rule{1em}{1pt}12 × 10^\rule{1em}{1pt}9c)}) / (100 × 10^\rule{1em}{1pt}4 m^2)│] = 1.08 × 10^4 N / C towards the right.

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Question:

Explain what is meant by Initialization or Initial Attributes. Explain how a variable is initialized. Also, explain how initialization is done, and what is stored in the computer memory by the following statement: DCL A FIXED(3,2) INITIAL(1\textbullet5), B FIXED(2,1) INIT (2\textbullet25) , C FLOAT(4) INIT(170), J INIT(10010B), P INIT(-11E2).

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G13-0320.htm

Solution:

It is often necessary that a variable appearing in a program have a value before it is used in expressions in the program. This value is given to the variable by explicit-ly assigning an initial value or initial attribute to the variable. The assigning of an initial attribute is called Initialization. Initialization can be done by means of a statement as follows: A = 3 \textbullet 2; B= - 4E - 1; C = 2 In the above, A is given an initial value of 3\textbullet2. B is ini-tialized to a floating point value of -0\textbullet4, and the initial value of C is 2. Initialization can be done anywhere within the body of the program, but it must be done before the variable appears in an expression in the program. A good place to do the initialization of the variables in a program is immediately following the DECLARE statements. A better and shorter way is to initialize the variables simultaneously as they are declared, that is, do the initial-ization in the DECLARE statement itself. This is done as il-lustrated below: DCL A FIXED(3,2) INITIAL(1\textbullet5), B FIXED(2,1) INIT (2\textbullet25), C FLOAT(4) INIT(170), J INIT(10010B), P INIT( - 11E2) ; In the above, INIT is a short form for INITIAL. A is de-clared as FIXED(3,2) with an initial value of 1\textbullet5. Hence, the initial value stored in memory for A is 1\textbullet50. B is de-clared as FIXED(2,1) with an initial value of 2\textbullet25. Therefore, the value stored for B is as follows: 2\textbullet2. Notice the 5 is truncated because the declaration FIXED(2,1) specifies only one position to the right of the decimal. C is declared as FLOAT (4) with an initial value of 170. Therefore, the value stored for C is as follows: 0\textbullet1700E + 03. If a PUT DATA(C) statement were used in the program before the value of C was changed, the print out for C would be as follows: C = 1\textbullet700E + 02. For J, except for the initial value, no other attributes are declared. Hence, by Default Rules, J receives the following attributes: BIN REAL FIXED 15,0). The initial value of J is 10010B, which is stored in memory. A PUT DATA(J) statement would print out as below: J = 000000000010010B For P, by Default Rules, the attributes are as follows: DEC REAL FLOAT(6). Hence, the value of P stored in memory is as follows: -0\textbullet110000E + 04. A PUT DATA(P) statement will cause a print out as follows: P = - 1\textbullet10000E + 03

Question:

The air in a tank has a pressure of 640 mm of Hg at 23\textdegreeC. When placed in sunlight the temperature rose to 48\textdegreeC. What was the pressure in the tank?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E02-0036.htm

Solution:

The law of Gay-Lussac deals with the relation-ship existing between pressure and the absolute temperature (\textdegreeC + 273\textdegree) , for a given mass of gas at constant volume. The relationship is expressed in the law of Gay-Lussac: volume constant, the pressure exerted by a given mass of gas varies directly with the absolute temperature. That is: P \alpha T(volume and mass of gas constant). The variation that exists between pressure and tem-perature at different states can be expressed as (P_1 / T_1 ) T_1 ) = (P_2 /T_2 ) = (P_2 /T_2 ) where P_1 = pressure of original state, T_1 = absolute tem-perature of original state , P_2 = pressure of final state, and T_2 = absolute temperature of final state . Thus this problem is solved by substituting the given values into Gay-Lussac's Law. P_1 = 640 mm HgT_1 = 23\textdegreeC + 273\textdegree = 296\textdegreeK P_2= ?T_2 = 48\textdegreeC + 273\textdegree = 321\textdegreeK Substituting and solving, [(640 mm)/(296\textdegreeK)] = [(P_2 )/(321\textdegreeK)] P_2 = 640 mm × [(321\textdegreeK)/296\textdegreeK)] = 694 mm of Hg.

Question:

A very large conducting plate has a surface charge density of 10^-2 coul/m^2. What is the energy density of the field produced by this charge distribution?

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/Users/wenhuchen/Documents/Crawler/Physics/D17-0564.htm

Solution:

The electric field due to the evenly charged sheet can be obtained by applying Gauss' Law at the sur-face of the sheet (see figure). By the definition of a conductor, the electric field at the surface of the con-ductor must be at right angles to it. (The y components of E^\ding{217} due to 2 symmetrically placed elements of the con-ductor's surface cancel, whereas the x components re-inforce. ) If we allow the Gaussian surface S to be very close to the surface of the conductor and neglect the fringing at the edges, we may use Gauss' Law, \epsilon_(0)s \intE^\ding{217} \bullet dS^\ding{217} = Q,or\epsilon_0AE = \deltaA, where A is the total surface area of the conductor. Therefore, E = \delta/\epsilon_0 where \delta is the surface charge density. The energy density of the field E is, by definition, u = \epsilon_0E^2/2 oru = (\epsilon_0/2) (\delta/\epsilon_0- )^2 = \delta^2/2\epsilon_0 But\epsilon_0 = 1/4\piK_E, where K_E = 9 × 10^9 [(N\bulletm^2)/c^2]. Henceu = (\delta^2/2) 4\piK_E = 2\piK_E \delta^2 In this problem u = 2\pi × [9 × 10^9 {(N\bulletm^2)/c2}] × 10^-4m (c^2/m^4) u = 56.52 × 10^5 (N/m^2) Because 1 Joule = 1N \bullet m u = 56.52 × 10^5 Joule/m^3.

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Question:

Write a computer program in FORTRAN to find the correlation coefficient r of the two variables. What is r for the data below? X Y 30 .9 20 .8 10 .5 30 1.0 10 .8

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G19-0490.htm

Solution:

Answering the second question first, the correlation coefficient r of the given data is r = ([n \sumxy - (\sumx) (\sumy)] / [\surd{n\sumx^2 - (\sumx)^2}\surd{n\sumy^2 - (\sumy)^2}])(1) = [5(86) - (100) (4.0)] / [\surd{5(2400) - (100)^2}\surd{5(3.34) - 4^2}] = .80 To understand what r is and what it measures, suppose that a regression line is drawn through the given data. Also, letybe drawn through the points. The regression line and the sample average line are predictors of how y varies when x varies. Note that (y -y) = (y - y˄) + (y˄ -y˄) (see Fig. 1) The explained variation of y is defined as \sum(y˄ -y)^2 . The unexplained variation is then \sum(y - y˄)^2 . The coefficient of variation, r^2 is defined as r^2 = (Explained Variation of y) / (Total Variation of y) or, r^2 = {\sum(y˄ -y)^2} / {\sum(y -y)^2}(2) The correlation coefficient r is the square root of equation (2). The computational formula for r, (1), follows from (2). Thus, r measures how much of the variation in y is explained by the regression line, in this case - 80%. This means that 20% of the variation is due to chance or other factors The program looks as follows: CPROGRAM TO FIND CORRELATION COEFFICIENT DIMENSION KF(2\O), A(1\O\O), S(1\O\O), R(1\O\O,1\O\O) 1READ (5,3) NV, NS, KF 3FORMAT (2I5)/2\OA4) DO 5 I = 1,NV A (I) = 0 DO 5 J = 1,NV 5R(I,J) = 0 DO 15 J = 1,NS READ (5,KF) (S(J), J= 1,NV) DO 15 J = 1,NV A(J) = A(J) + S(J) DO 15 K = J,NV 15R(J,K) = R(J,K) + S(J) \textasteriskcentered S(K) T = NS DO 2\O I = 1,NV A (I) = A(I)/T 20S(I) = SQRT(R(I,I)/T - A(I) \textasteriskcentered\textasteriskcentered 2) DO 25 I = 1,NV DO 23 J = I,NV IF (S(I) \textasteriskcentered S(J) .EQ. 0.0) GO TO 23 R(J, I) = (R(I,J)/T - A(I)\textasteriskcentered A(J))/S(I)\textasteriskcenteredS(J)) 23R(I,J) = R(J,I) 25R(I,I) = 1.0 WRITE (6,3\O) (A(I), I = 1,NV) 3\OFORMAT (8H1 × MEANS., (/1 OF 10.4)) WRITE (6,31)(S(I), I = 1,NV) 31FORMAT (8H SIGMAS., (1 1 OF 10.4)) DO 4\O K = 1,NV, 1\O L = K + 9 IF (NV - L) 32, 33, 32 32L = NV 33WRITE (6,35) K,L 35FORMAT (14H1R MATRIX. COL, I3, 3H TO I3) DO 4\O I = 1,NV 4\OWRITE (6,45) I, (R(I, J) , J = K,L) 45FORMAT (I5, 10F 10.4) STOP END

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Question:

A chemist dissolvesPbSin water . TheK_spis found to be 1 × 10-^28 . What is themolarityof each ion in solution ?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E11-0395.htm

Solution:

The key to answering this question is the determination of the equilibrium equation. SincePbSis a salt, it dissociates into ions in solution. Upon inspection of the oxidation states of the periodic table, the equilibrium will be found to be PbS(s) \rightleftarrows Pb^2+ (Aq) + S^2- (Aq) . The presence of the equilibrium derives from the fact that a saturated salt solution's ions will exist in equilibrium with any additional salt. This additional salt is called the solid phase.K_spmeasures this equilibrium and can be expressed as K_sp= [Pb^2+] [S^2-] . This, equals 1 × 10-^28 . Therefore, you have 1 × 10-^28 = [Pb^2+] [S^2-]. The question asks for the concentrations represented on the right side. From the equilibrium equation, you see that whatever the concentration of S^2-, it will be equal to Pb^2+ . Therefore, you can represent both by x . As such, you now have 1 × 10-^28 = x \textbullet x 1 × 10-^28 = x^2 . Solving for x you obtain 1 × 10-^14 M. This means, therefore, that the concentration of each ion in solution is 1 × 10-^14 M .

Question:

A block of steel 2 ft. square and 1/4in. thick is to be compressed .01 in. in length by application of a force F to faces A_1 and A_2. If Young's modulus for steel is 29 × 10^6 lb. in.^2, find F.

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/Users/wenhuchen/Documents/Crawler/Physics/D08-0351.htm

Solution:

Young's modulus is defined as Y = (STRESS)/(STRAIN)(a) where STRESS = (NORMAL FORCE)/(AREA) = F/(24 × 1/4) = F/6 and STRAIN = (CHANGE IN LENGTH)/(ORIGINAL LENGTH) = .01/24 = 1/2400 Substituting our values in (a), 29 × 10^6 = (f/6)/(1/2400) Thus F = 72,500 lb.

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Question:

What is the output of the following program? main ( ) { intx = 5; inty = 5; printf("%d%d\textbackslashn", x++, x); printf("%d%d\textbackslashn", ++y, y); }

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G17-0424.htm

Solution:

Here it is important to distinguish between the two styles of increment operations. 1) x++: Use variable x first and then increment afterwards,eg. Let z = x++; means z = x; x = x + 1; 2) ++y: Increment the variable y first and then use it afterwards.eg. Let z = ++y; means y = y + 1; z = y; Therefore the output is as follows: 56/\textasteriskcentered 1stprintf\textasteriskcentered/ 65/\textasteriskcentered 2ndprintf\textasteriskcentered/

Question:

(a) Compute the maximum height attained by a projectile launched with velocity of magnitude v_o directed at an angle \texttheta_0 to the horizontal, using the principle of con-servation of energy. (b) What is the magnitude of the projectile's velocity u^\ding{217}, when it has reached half its greatest height?

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/Users/wenhuchen/Documents/Crawler/Physics/D05-0238.htm

Solution:

(a) At the start of the motion, the projectile possesses kinetic energy of amount (1/2) mv^2_0. At its greatest height, h, it possesses kinetic energy due to the com-ponent of its velocity in the x-direction (v_O cos \texttheta) only, (1/2) m(v_0 cos \texttheta_0)^2 and also potential energy due to its increased height, mgh. By the principle of conserva-tion of energy, (1/2)mv^2_0 = (1/2)mv^2_0 cos^2 \texttheta_0 + mgh \therefore h = [v^2_0(1 - cos^2 \texttheta_0)] / (2g) = (v^2_0 sin^2 \texttheta_0) / (2g) (b) By the principle of conservation of energy, the sum of the projectile's potential and kinetic energies at half its greatest height must equal its initial kinetic energy at the time of launching. (The potential energy at the point of launching is also zero.) At half the greatest height, the potential energy possessed by the projectile is mg(h/2) = mg [(v^2_0 sin^2 \texttheta_0) / (4g)] = (1/4) mv^2_0 sin^2 \texttheta_0. and its kinetic energy is (1/2)mu^2. Thus: (1/2) mv^2_0 = (1/2) mu^2 + (1/4) mv^2_0 sin^2 \texttheta_0 u = \surd[v^2_0 - (1/2) v^2_0 sin^2 \texttheta_0]

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Question:

Explain how peristalsis moves food through the digestive tract .

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/Users/wenhuchen/Documents/Crawler/Biology/F17-0425.htm

Solution:

In each region of the digestive tract, rhythmic waves of constriction move food down the tract. This form of contractile activity is called peristalsis, and involves involuntary smooth muscles. There are two layers of smooth muscle throughout most of the digestive tract. Circular muscles run around the circumference of the tract while longitudinal muscles traverse its length. Once a food bolus is moved into the lower esophagus, circular muscles in the esophageal wall just behind the bolus contract, squeezing and pushing the food downward. At the same time, longitudinal muscles in the esophageal wall in front of the bolus relax to facilitate movement of the food. As the bolus moves, the muscles it passes also contract, so that a wave of contraction follows the bolus and constantly pushes it forward. This wave of constriction al-ternates with a wave of relaxation. Swallowing initiates peristalsis and once started, the waves of contraction cannot be stopped voluntarily. Like other involuntary responses, peristaltic waves are controlled by the autonomic nervous system. When a peristaltic wave reaches a sphincter, the sphincter opens slightly and a small amount of food is forced through. Immediately afterwards, the sphincter closes to prevent the food from moving back. In the stomach, the waves of peristalsis increase in speed and intensity as they approach the pyloric end. As this happens, the pyloric sphincter of the stomach opens slightly. Some chyme escapes into the duodenum but most of it is forced back into the stomach (see figure). This allows the food to be more efficiently digested. There is little peristalsis in the intestine, and more of a slower oscillating con-traction. This is why most of the 12-24 hours that food requires for complete digestion is spent in the intestine.

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Question:

Although different simulation studies require variations in procedure, certain basic steps are usually followed. State the stepsof the general procedure.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G24-0569.htm

Solution:

1)Definition of the problem. 2)Planning the study (considering the exogenous factors that willaffect the system; deciding if such factors will be included in the study; etc.) 3)Formulation of a mathematical model. Simplifying assumptions aboutthe system components and their behavior are usually made at this stage . 4)Gathering of relevant data. 5)Construction of a computer program for the model. 6)Validation of the model. 7)Design of experiments. 8)Execution of simulation run and analysis of results. In common use, there are two different types of simulation, continuous anddiscrete. Continuous simulation deals with variables (such as velocity, current, work, etc.)thattake on real values and change continuously with time. These changes are defined as being smooth andnonabrupt. In discretesimulation, the model deals with random events that occur at isolatedtimes. For example, the arrivals of motorists at a gas station couldbe considered random, and the sim-ulation model might have to determinethe average waiting time per customer under various organizationsof attendants and gas pumps. This process of using random variablesis called a stochastic process. There exists a problem-oriented programming language called GPSS (General Purpose System Simulation) which can be used for solving the simulationproblems.

Question:

Differentiate between the monocots and dicots.

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/Users/wenhuchen/Documents/Crawler/Biology/F08-0204.htm

Solution:

The two classes of angiosperms, the monocotyledons and dicotyledons, differ in eight respects. First, the monocot embryo has one cotyledon (seed leaf) while the dicot has two. The cotyledon of the monocots functions generally in food absorption while that of the dicots can function both in absorption and storage. Second, the endosperm typically persists in the mature seed of the monocots, while it is usually absent from the mature seed of the dicots. Third, the leaves of the monocots have parallel veins and smooth edges; dicot leaves have net-like veins and lobed or indented edges. Fourth, cambium or meristematic cork and vascular tissue is usually present in the stems of dicots but absent in the monocots. Fifth, the vascular bundles of xylem and phloem are scattered throughout the stem of monocots. In the dicots they occur either as a single solid mass in the center of the stem or as a ring between the cortex and pith. Sixth, the flowering parts of monocots - petals, sepals, stamens, and pistils - exist in threes or multi-ples of three , whereas dicot floral parts usually occur in fours, fives or multiples thereof. Seventh, the monocots are mostly herbaceous plants while many dicots are woody plants. Finally, the roots of monocots are typically fibrous and adventitious (outgrowths of the stem) whereas the root system of dicots usually consist of one or more primary tap roots and numerous secondary roots.

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Question:

A certain boat can move at a speed of 10 mi/hr in still water. The helmsman steers straight across a river in which the current is 4 mi/hr. What is the velocity of the boat?

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/Users/wenhuchen/Documents/Crawler/Physics/D01-0017.htm

Solution:

The boat has a speed of v_b = 10 mi/hr per-pendicular to the river due to the power of the boat. The current gives it a speed of v_c = 4 mi/hr in the direction of flow of the river. The boat's resultant velocity (having both magnitude and direction) can be found through vector addition. v\ding{217} = v\ding{217}_b + v\ding{217}_c The magnitude of the velocity which is the speed of the boat is found using the Pythagorean theorem (see figure). v = \surd(v^2_b + v^2_c ) = \surd[(10)^2 + (4)^2] = \surd116 = 10.8 mi/hr. The angle \texttheta, which determines the direction of the velocity is, \texttheta = tan^-1 (v_c/v_b) = tan^-1 (4/10) = 22\textdegree

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Question:

In a cloud chamber, a proton crosses at right angles a uniform magnetic field of 1.0 weber/m^2 . From the photograph of the proton's path the radius of curvature is measured as 0.22 m. Calculate the speed of the proton.

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/Users/wenhuchen/Documents/Crawler/Physics/D21-0720.htm

Solution:

The centripetal force acting on the proton causes it to curve. This force is due to the magnetic deflecting force Bqv where q, the charge on the proton, is 1.6 × 10^-19 coulombs. The force can also be represented in terms of the radius of curvature and the velocity of the proton as (mv^2 /r), with the mass of the proton, m, equaling 1.67 × 10^-27 kg. Setting the two expressions equal and solving for the velocity, Bqv = (mv^2 /r) v = (Bqr/m) = [{(1.0 weber/m^2 ) (1.60 × 10^-19 coul) (0.22 m)} / (1.67 × 10^-27 kg)] = 2.1 × 10^7 (m/sec). (Since this speed is less than a tenth the speed of light, we are justified in neglecting any relativistic change in the mass of the proton.)

Question:

Two substances A and B have the same 0.132-atm vapor press-ure at 15\textdegreeC, but A has a heat of vaporization of 25,000 (J / mole), whereas B, 35,000 (J / mole). What will be the respective normal boiling points?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E07-0264.htm

Solution:

The vapor pressure, heat of vaporization and absolute temperature are related in the Clausius-Clapeyron equation. This equation is written: ln (p_1 / p_2) = (∆H / R) [(1 / T_2) / (1 / T_2)] where p_1 is the initial pressure, p_2 is the final pressure, ∆H is the heat of vaporization when T_2 is the boiling point, R is the gas constant, (8.314 J / mole \textdegreeK), T_1 is the initial temperature and T_2 is the final temperature. To solve for the boiling point of A, note that at the boiling point of a liquid the vapor pressure is equal to the atmospheric pressure. Atmospheric pressure is 1 atm. P_1 = 0.132 atmT_1 = 15\textdegreeC + 273 = 288\textdegreeK p_2 = 1 atmT_2 = ? ∆H = 25,000 (J / mole)R = 8.314 (J /mole \textdegreeK) In (0.132 / 1) = [(25,000 J/mole) / (18.314 J/mole \textdegreeK)] {(1 / T_2)-(1/288\textdegreeK)}] -2.025 = (3.00 × 10^3\textdegreeK) (1 / T_2) - (3.47 × 10^-3 /\textdegreeK) -6.75 × 10^-4 / \textdegreeK = (1 / T) - (3.47 × 10^-3 /\textdegreeK) 2.795 × 10^-3 / \textdegreeK = 1/T 357.78 = T The boiling point of A in \textdegreeC is (357.78 - 273) = 84.78\textdegreeC Solving for the boiling point of B: P_1 = 0.132 atmT_1 = 15\textdegreeC + 273 = 288\textdegreeK p_2 = 1.0 atmT_2 = ? ∆H = 35,000 (J / mole) R = 8.314 (J / mole \textdegreeK) In (0.132 / 1.0) = [(35,000 J/mole)] / (8.314 J/mole \textdegreeK)] [(1/T_2) - (1/288\textdegreeK)] -2.025 = 4.21 × 103 \textdegreeK (1 / T_2) - 3.47 × 10^-3 / \textdegreeK) -4.81 × 10^-4 / \textdegreeK = (1 / T_2) - 3.47 × 10-3/ \textdegreeK 2.989 × 10^-3 / \textdegreeK = (1 / T_2) 334.56\textdegreeK = T_2 The boiling point of B is equal to 334.56\textdegreeK or 334.56 -273 = 61.56\textdegreeC.

Question:

Distinguish between the two types of proteins produced by ribosomes.

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/Users/wenhuchen/Documents/Crawler/Biology/F24-0630.htm

Solution:

Ribosomes are the cellular organelles that provide the sites for protein synthesis. The two types of proteins that are synthesized on the ribosomesare structural proteins and functional proteins. Structural proteins are actualstructualcomponents of the cell. They constitute part of the cell protoplasm, and as such, are generally insoluble. They are an important component of cell membranes. Some structural proteins, such as collagen, or keratin are specialized for strength and support. Structural proteins help to form the cell, and the organism of which the cell is a part, therefore, struc-tural proteins contribute to both cell and body growth. Functional proteins are responsible for the control of cell activity, such as hormone production or nutrientmetablism. Some of these proteins are known as enzymes their function is to catalyze chemical reactions in the cell. Others function as carrier molecules. Still others function as regulatory molecules which control the synthesis of other proteins. Most functional proteins are soluble in the cell, since their action depends on their interaction with various molecules throughout the cell. Some are soluble in cell membranes. The distinction between structural and functional proteins, then, is based on what they do in the cell.The former are involved in the structure of the cell or organism, whereas the latter deal with its functions.

Question:

Explain the difference between truncation and rounding. How can one operation be made to perform the function of the other?

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G01-0015.htm

Solution:

Consider the real number 5.6158 whose integer part is 5 and fractional part is .6158. This number can be converted into a whole number (integer) in one of two ways. First, the fractional part can be chopped off to give 5, and this is called truncation. Second, the integer whose value is closest to 5.6158 is 6, and this process is known as rounding. Suppose the computer is capable of truncation only. How can we accomplish rounding? Just add .5 to every number to be rounded before the truncation. The truncated increased value will be equal to rounded initial value. For example: 5.6158 ROUNDED= 6 5.6158 + .5000 = 6.1158 TRUNCATED = 6 Suppose the computer is capable of rounding only, and we need to truncate. In that case, simply subtract .5 from every number to be truncated before the rounding. The rounded de-creased value will be equal to truncated initial value. For example: 5.6158 TRUNCATED = 5 5.6158 - .5000 = 5.1158 ROUNDED = 5

Question:

Standing waves are produced by the superposition of two waves of the form y_1= 15 sin (3\pit \rule{1em}{1pt} 5x)(1) y_2 = 15 sin (3\pit + 5x)(2) Find the amplitude of motion at x = 21.

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Solution:

We note that the two waves are of the form y = A sin(wt \pmkx) where y is the displacement of the wave at position x and time t. A is the amplitude of the wave (i.e., the maximum dis-placement) , w is the angular frequency (= 2\pif) and k is the angular wavelength (= 2\pi/\lambda) of the wave. Both waves have the same characteristics (i.e., the same amplitude, frequency and wavelength). They differ only in that they travel in opposite directions. The negative sign in wave (1) indicates that it is travelling to the right. Wave (2) is travelling to the left. The resultant wave produced by the superposition of these two waves can be found as follows We use the relationships sin(\alpha + \beta) = sin \alphacos\beta +cos\alpha sin \beta sin(\alpha \rule{1em}{1pt} \beta) = sin \alphacos\beta \rule{1em}{1pt}cos\alpha sin \beta Therefore sin (\alpha + \beta) + sin (\alpha \rule{1em}{1pt} \beta) = 2 sin \alphacos\beta(3) Thus, comparing (3) with (2) and (1), y = y_1 + y_2 = 2 (15) sin (3\pi)tcos(5) x = 30 sin 3\pitcos5x . This wave pattern is called a standing wave (as opposed to a travelling wave). The wave remains in one location; or alternatively, the energy associated with the wave is not transferred from one location to another. With x = 21, 5x = 105 radians = 33.4\pi radians. Nowcos33.4\pi =cos(.4\pi + 33\pi) =cos.4\pi =cos72\textdegree = 0.309. Thus, x = 21, y = 30 sin 3\pitcos5x y = (30) (.309) sin 3\pit y = 9.27 sin 3\pit. The amplitude of this wave is the maximum value of y. This maximum value is 9.27.

Question:

From what you have learned of the processes of heredity andevolution, how do you think new species arise with respectto mutation? Which is more important, the accu-mulationof small mutations or a few mutations with largephenotypic effects? Give the reasons for your answer.

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Solution:

Most adaptive mutations produce barely distinguishable changes in thestructure and function of the organism in which they occur. On the otherhand, mutations with the potential to produce large phenotypic effects, necessary to elicit major changes in one jump, are practically alwayslethal or at least detrimental to survival. Major mutations are deleteriousbecause they constitute a great disturbance to the delicately balancedgenetic systems in which they arise. Therefore, major genotypic changesthrough mutation are not likely to be important in evolution. In contrast, the accumulation of small mutations over the course of manygenerations has significant evolutionary value. This has been supportedby fossil evidence, laboratory experiments and field studies. Evolutionary changes appear to arise as a result of many small changes thathave accumulated over time, ultimately producing a change in the compositionof a given population's gene pool. An example of this can be seenin the evolution of the orchid. The modern orchid flower, in both shapeand color, resembles the female wasp. Male wasps, thinking it is a female, are attracted to the flower and stimulated to attempt copulation. In doingso, they become coated with the sticky pollen grains of the flower andcarry them along to the next orchid. Any mutation that gave the orchid anysimilarity, however slight, to the female wasp would increase its chancesof being pollinated. Thus, such mutations would give those orchidsbearing them a selective advantage over orchids without them. The increased rate of pollination of the mutated forms, would promote the propagationof the mutant gene. It is easy to see how, over time, such mutationswould tend to be propagated and accumulate to the point where theorchid flower is today.

Question:

A scientifically minded Romeo has found a method of sending secret messages to a beautiful Juliet who is immured in the top floor of a castle 50 ft from the ground. Romeo places two light metal rods (too light to use for climbing up)against her windowsill, and between the rods he mounts a wire 10 cm long, to which is attached the message and a magnet so placed that the wire is permanently in a magnetic field of strength 0.049 Wb \textbullet m^-2, at right angles to the plane of the rods. When he passes a current of 10 A up one rod, through the connecting wire and back down the other rod, the message, wire, and magnet travel at uniform speed up the rods. The moving assembly weighs 0.25 kg. Neglecting friction, calculate what the length of the rods must be.

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Solution:

From figure A, we see that the magnetic field must be at right angles to the plane of the rods and acting downward. The magnetic field vector is per-pendicular to both the direction of current and to the force. The right hand rule determines that the direc-tion is downward. The magnitude of the force experienced by the wire and attachments is \vertF^\ding{217}\vert = lIB sin \varphi where \varphi is the angle between the direction of I and B \textbullet \varphi = 90\textdegree, and F = lIB = 0.1 m × 10 A × 0.049 Wb \bullet m^-2 = 0.49 N . Considering fig. B, we see that the forces acting on the wire and attachments are three in number: the weight acting vertically downward, the force P acting up the plane of the rods, and N, the normal reaction of the rods on the wire acting at right angles to the plane of the rods. Since the assembly moves up the rods at uniform speed, F = mg sin \texttheta . \thereforesin \texttheta = [(0.049 N) / (0.25 kg × 9.8 m \textbullet s^-2 )] = 0.02 . From the diagram, h/L = sin \texttheta = 0.02 . \thereforeL = (h/0.02) = (50 ft/0.02) = 2500 ft.

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Question:

The "law of cooling" states that the temperature of a body, warmer than the surrounding substance, decreases at a rate proportional to the difference in temperature between the body and the surrounding substance. If the difference in temperature at time t is D(T), the constant of proportionality is r (the so-called "rate of cooling") and D(0) = D_0, use the modified Euler's method to simulate this system in FORTRAN, from time t = 0 to t = t_f . Compare the approxi-mations with the exact values.

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Solution:

If Ḋ is the rate of change of temperature difference, a process diagram looks as shown in Fig. 1. The corresponding equation is: Ḋ = -r \textasteriskcentered D(T)(1) The minus sign Is included because the temperature difference decreases at the rate r. The "goal" of this system is to reach a state of zero temperature difference (i.e., equilibrium). The exact solution of equation (1) is: D(T) = D_0e^-rt Note that t \ding{217} \infty \Rightarrow D(T) \ding{217} 0 as expected. The program looks as follows: DATA. T/0.0/ COMMON R READ, N, TFIN, ACC UR, R, DO PRINT,T,DO REALN = N DT = TFIN/REALN D = DO DO 10 I = 1,N T = T + DT CALL MEULER(T,D,ACCUR,DT) EXACT = DO \textasteriskcentered EXP(-R\textasteriskcenteredT) ERROR = ABS((EXACT-D)/EXACT) \textasteriskcentered100. PRINT, T, D, EXACT, ERROR 10CONTINUE STOP END FUNCTION G(W) COMMON R G = R\textasteriskcenteredW RETURN END

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Question:

What is the adaptive importance of display in aggressive behavior?

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Solution:

It is necessary to distinguish aggressive behavior between members of the same species from violent predatory behavioral patterns, which are usually directed at members of different species. Aggressive behavior is usually exhibited between same-species males in defending their territory or establishing their status in a social order. The critical factor in aggressive behavior is that most of the fighting consists of display. Aggressivedisplay between fishes. Male pike blennies (Chaenopsis ocellata ) defend territories on the sea floor with highly ritualized exchanges that include lowering of the gill case, raising the fins, gaping of the jaws, and changes of color. Winner is determined by size and aggressiveness. Display consists of ritualized, highly-exaggerated mo-vements or sounds which convey the attack motivation of the contestants. In attempting to appear as formi-dable as possible, the animal often changes the shape of certain body parts in an attempt to make itself appear as large as possible. The adaptive significance of display is that the two contestants are rarely seriously hurt. Aggression is a type of bluff. The contestant's goal is to repel his opponent to a point outside of his territorial boundary or to convince his rival that a certain female is "taken." Aggressive behavior is therefore highly ritualized, with each movement conveying a certain message. It seldom results in bodily harm, even if physical contact is made. The animals can tell without serious fighting which one would be the probable winner if they were to actually fight. It would be biologically deleterious if during every aggressive contest between rivals, the loser were to be killed. Display behavior allows the "would-be loser" to escape and establish a territory or acquire a mate elsewhere. Survival and reproduction are therefore preserved.

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Question:

Prove that the virtual image observed in a plane mirror is the same distance behind the mirror as the object is in front of the mirror.

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Solution:

As shown in the figure, let OA be the ray of light that strikes normal to the reflecting surface, while OB represents the ray that strikes the mirror at point B. The law of reflection states that the angle of incidence i equals the angle of reflection r, i = r. Rays DB and OA are extended back through the mirror to form triangle AIB, where point I is the apparent position of the image. Angle r must equal a; therefore, we obtain a = r = i. Angles b and c are equal since they are both right angles. Therefore, we have shown that the two triangles OAB and IAB are congruent, that is, coincide perfectly when super-imposed because they share a common side, AB. We conclude that OA = IA and that the virtual image appears as far behind the mirror as the object is in front of the mirror.

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Question:

Discuss the role of pheromones in chemical communication.

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Solution:

In general, communication between organisms can be defined as actionon the part of one individual that alters the behavior of another individual. Communication need not necessarily be visual or auditory, nor mustit always be confined to the present. If it is chemical, it obviously has theability to span time. Specific chemicals involved in social communicationare called pheromones, by analogy with hormones. Pheromones are small quantities of specific chemical substances released byan organism which act upon another organism at a distance from their pointof release and in very specific ways. Pheromones can be classified in two groups: those that possess releasereffects, which elicit immediate behavioral responses, and those thathave primer effects, which work by altering the physiology and subsequentbehavior of the recipient. Releaser pheromones are used by a variety of animals for different purposes, such as the attraction of mates, individual recognition, or trail or territorialmarking. The females of many moth species secrete pheromonesto attract males over large areas. The area within which the substanceis dense enough to be detected by a male moth is called the activespace. When a sexually active male flies into the active space, it fliesupwind until it reaches the pheromone-emitting female. This particular pheromoneis very potent since only one molecule of it is needed to stimulatea receptor on the male's antennae. The sugarglider ,asquirrel- likemarsupial, uses its own glandular secretions to mark its territorial boundaries. Male sugar gliders also mark their mates with pheromone, for recognitionand in order to warn other males that their mate is claimed. Ants use pheromones to lay down a trail from a recent food source to their nest. The pheromone, which is laid from the tip of their stinger, evaporates toform an active space. This enables other ants to rapidly find the food source. Primer pheromones change the physiology of the recipient by activatingcertain endocrine glands to secrete hormones. These hormones subsequentlycause changes in the behavioral patterns of the recipient. For example, social insects such as ants, bees and termites use primer pheromonesin caste determination .In termites the king and queen secreteprimer pheromones which are ingested by workers, preventing themfrom developing reproductive capabilities. In mammals, the estrous cycleof female mice can be initiated by the odor of a male mouse. In addition, the odor of an unfamiliar male mouse can block the pregnancy of afemale that has been newly impregnated by another male. This enables thestrange mouse to mate with the female. However, if the olfactory bulbs ofthe female are removed, the pregnancy block will not occur. This signifiesthat pheromones, unlike hormones, have an external rather than internalsource.

Question:

Steroid hormones do not act through the adenyl cyclase system. What is the mechanism of steroid hormone action?

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Solution:

Steroid hormones serve to increase the synthesis of proteins in target cells. The mechanism by which steroid hormones do this has been called the mobile-receptor model. The first step in this mechanism is the entry of the hormone into the cytoplasm. Steroid hormones are lipid soluble and can readily cross cell membranes. Once within the cell, the hormone binds to a soluble hormone-specific cytoplasmicprotein, known as the receptor. The hormone-receptor complex then moves into the nucleus and combines with specific proteins associated with DNA in the chromosome. The molecular interaction triggers off the specific RNA synthesis which results in increased protein synthesis by the cell. The thyroid hormones (amino acid derivatives) also act via direct diffusion into target cells. The distinction to be made is that thyroid hormones enter the nucleus and bind to a nuclear receptor. Diagram showing mobile-receptor model of asteriod hormone.

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Question:

Give a list and briefly define the primitive (built in) functions of SNOBOL IV language.

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Solution:

ANY -matches any character appearing in its argument. APPLY -creates and executes a function call. ARBNO -takes a value of pattern structure that matches zero or more consecutive occurrences of strings matched by its argument. ARG -takes a value of the indicated argument from programmer- defined function. ARRAY -creates an array of variables. BACKSPACE - backspaces the indicated record. BREAK -matches a string up to indicated character. CLEAR -sets the values of all natural variables to the null string. CODE -converts a string of characters into object code. COLLECT -returns as value the amount of storage available after regeneration. CONVERT - converts a given object into given data type. COPY -produces a copy of an array, DATA -defines a new data type. DATATYPE - identifies the type of given data. DATE -gives the date of program execution. DEFINE -defines a programmer defined function. DETACH -removes any input and output association from the argument. DIFFER -succeeds if two given arguments are not identical. DUMP -lists all natural variables and their values in the program if argument \not = 0. DUPL -repeats the first argument the number of times specified by second argument. ENDFILE -closes the data set, indicated by argument. EQ-predicate "EQUAL". EVAL -evaluates an unevaluated argument. FIELD -gives function a value equal to the name of a given field of the programmer defined data type, specified in the argument. GE-predicate "GREATER THAN, or EQUAL TO" GT-predicate "GREATER THAN" IDENT -succeeds if two given arguments are identical. INPUT -specifies a programmer - defined input. INTEGER -succeeds if the value of argument is an integer. ITEM -assigns values to the specified members of the specified arrays. LE -predicate "LESS THAN, or EQUAL TO". LEN -returns a pattern structure that matches any string of the specified length. LGT -succeeds if the first argument is lexically greater than the second. LOCAL -gives function a value equal to the name of a given local variable from a specified defined function. LT -predicate "LESS THAN". NE -predicate "NOT EQUAL". NOTANY -matches any character not appearing in the argument. OPSYN -provides synonyms for existing functions, or operators. OUTPUT -specifies a programmer formatted output of a pro-grammer defined data. POS -succeeds if the cursor's position is just to the right of the one, specified by the integer argument. PROTOTYPE - returns a value of the dimension or range of an array. REMDR -returns an integer that is the remainder of dividing the first argument by the second. REPLACE -replaces character, given in the argument, by another in the specified string. REWIND -repositions the data set associated with the given number to the first file. RPOS -succeeds if the cursor's position is just to the left of the one, specified by the integer argument. RTAB -matches all the characters of a string up to, but not including the one specified in the argument, from the right end of a subject string. SIZE -returns an integer equal to the length of a given string. SPAN -matches a run of characters given in the argument. STOPTR -cancels a single trace association. TAB -matches all the characters of a string up through the one, specified in the argument. TABLE -creates a table of variables. TIME -gives the elapsed time of program execution (in MSEC.) TRACE -traces a specified data segment. TRIM -removes trailing blanks from a string in the argument. VALUE -assigns a given value to a specified variable.

Question:

(a) What is the acceleration of gravity on the moon's surface, g_m (as compared to the acceleration of gravity on the earth surface,g_e) if the diameter of the moon is one-quarter that of the earth's and its density is two-thirds that of the earth? (b) An astronaut, in his space suit and fully equipped, can jump 2 ft vertically on earth using maximum effort. How high can he jump on the moon?

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Solution:

(a) On the earth's surfaces: G is the Universal Gravitational Constant, M_e is the earth's mass and R_e is the radius of the earth. The volume of the earth is v_e= (4/3)\piR3_e If\rho_eis the earth's density, then, by definition of density M_e =\rho_ev_e= (4/3)\piR^3_e\rho_e Hence, by Newton's Law of Universal Gravitation g_e=GM_e/R2_e = (4\pi/3)[(GR^3_e\rhoe)/R2_e] = (4/3)\pi\rho_eGR_e(1) On the surface of the moon, we have a similar relation for the acceleration of gravity g_m =GM_m/R2_m = (4/3)\pi\rho_mGR_m(2) Therefore, upon division of equation (1) by equation (2), g_e/g_m =\rho_eR_e/\rho_mR_m= 3/2×4/1 = 6; that is, g_m = (1/6)g_e. Here, use was made of the given ratios\rho_e/\rho_mand D_e/D_m (D_e and D_m are the diameters of the earth and the moon respectively). (b) When he jumps at maximum effort, the astronaut can launch himself upward with an initial velocity v_0. On earth, whenv_final= 0, then, using the kinematic relation for constant acceleration, v2_final = v2_0 + 2as 0 = v2_0 - 2g_e×2 ft where the negative sign forg_eindicates that it is in the downward direction. \therefore v2_0 = 2g_e×2 ft = 12 g_m×2 ft. Similarly, on the moon, when he starts with the same initial velocity, he jumps the distance y vertically, where 0 = v^2_0 - 2g_m × y. \therefore y = v^2_0/2g_m = (24g_m/2g_m)ft = 12 ft.

Question:

Suppose a fisherman was interested in trying to extract the maximum number of fish from a pond or lake. Why would it not be to his economic advantage to reduce the fish populations by excessive harvesting?

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Solution:

The ideal economic strategy for the fisher- man to follow would be to catch only enough fish so as to keep the population at a level of optimal yield. If the population is allowed to reach the maximum that the environment can support, or if it is exploited to the point where the organisms become scarce, the yield will decline. This phenomenon exists because the greatest amount of increase in population size does not occur either when the population is very low or when the population has reached the carrying capacity of the environment. The point of greatest growth and hence the maximum replacement capacity of a population occurs at a point midway on the exponential part of an S-shaped growth curve, that is, half way between the baseline and the maximum carrying capacity (see figure). This point corresponds to the point of steepest rise on the S- shaped curve. At this point, the greatest number of new individuals are added to the population in a given amount of time. This maximum rate of population growth is referred to as the optimal yield. Fish populations, as well as other animal populations that man harvests, can be obtained with much less effort at a point of optimal yield than populations reduced by overfishing. Moreover, the greater the exploitation of populations, the more the population will come to consist of younger and smaller individuals, which are usually the least valuable commercially. The proportion of large fish caught is higher when the population is at a point of optimal yield. Consequently, as the fish populations are reduced, more expensive and sophisticated ships and trawling techniques are required to produce the same amount of yield.

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Question:

Explain the various features of DO statements used for iter-ation .

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Solution:

By means of condition testing, one can execute only a single statement. There is a DO statement which al-lows a group of statements to be executed on condition test-ing. This is called a THEN DO statement. It is illustrated by the following example. IF A>0THEN DO ; SUM = SUM + X ; A = A + 1 ; B = SUM/Y ; PUT LIST(SUM,B) ; END ; Observe that four statements are executed when the condition tested is true as compared to only one statement that would be executed without the use of the DO statement. This DO statement allows a grouping of statements. The DO statement must have as its last statement an END statement. The DO and END statement pair do not influence the execution of the PL/I machine but has an editing function in the program. The pair identifies the loop more explicitly. Besides the simple DO group, there is also an iterative DO group. The format of the iterative DO statement is as follows: DO CONTROL VARIABLE=INITIAL VALUE TO LIMIT VALUE BY INCRE- MENT VALUE. Control variable is a variable name whose value will change according to control specifications. The variable is fol-lowed by an assignment symbol (=), which in turn is followed by a control specification. The control specification may consist of up to three elements as follows: a) Initial value, which is the value assigned to the control variable initially. The initial quantity must always be specified in the DO statement. Also, this initial quan-tity must be the first item to be specified on the right hand side of the (=) sign in the DO statement. b) The TO clause describes the limiting value to be used by the loop control variable. Once the value of the control variable exceeds the limiting value, no statement within the loop is executed and the next statement immedi-ately outside the loop is executed. c) The BY clause describes the increment value to be used in the loop control mechanism. Every time the loop is executed the value of the control variable is incremented by the increment value. ' The new value of the control variable is tested against the limiting value and the loop is exe-cuted if the control variable's value is less than or equal to the limiting value. For example, consider the following: DO I = -5 TO 5 BY 2; STATEMENT 1; STATEMENT 2; \textbullet \textbullet \textbullet STATEMENT N; END; STATEMENT X; is the control variable, its initial value is -5. As -5 is less than the limiting value of 5, statements 1 to n are executed, then the value of I is incremented by the incre-ment value 2. Hence, the value of the control variable is -3. Still it is less than 5, and again the loop is executed. Once again I is incremented by 2, tested against the limiting value of 5, and so on. When 1 = 5, still I is not. greater than the limit value of 5 and therefore the loop is executed once again. Now 1 = 7 which is greater than 5 and exit from the loop to statement X is performed. With regard to DO loops, the points to remember are as follows: a) If the increment value is positive, the limiting value must be greater than the initial value. b) If the increment value is negative, the limiting value must be less than the initial value. c) The BY and TO clauses are optional and may be omitted. i) When the BY clause is omitted: If no BY clause is found, but the TO clause is present then, by default the PL/I machine assumes BY 1 (increment by one). That is, PL/I assumes that the follow-ing statements DO I = 1 TO 70 BY 1,and, DO I = 1 TO 70; are completely equivalent. ii) When the TO clause is omitted: If the DO statement has the form given below: DO X = 2 BY 3, it can be seen that the TO clause is omitted. But there are no default rules to supply a TO clause in PL/I. Instead, the PL/I machine proceeds without testing the limiting value when the TO clause is omitted. Hence, a mechanism must be provided within the body of the loop to exit out of the loop, otherwise infinite looping will result. e.g.LOOP:DO I = 1 BY 1; SUM = SUM + I; IF SUM>10 THEN GOTO LABEL; END LOOP; LABEL:PUT LIST(I,SUM); END; In the above, exit from the loop is achieved when the value of SUM exceeds 10. If the TO clause and the BY clause are both omitted, for instance, DO J = 3. The DO statement will be executed only once as a result of the above. The While Clause In the Specification of a DO Group: The flexibility and power of the DO statement is fur-ther enhanced by using a WHILE clause in the control speci-fication of the DO statement. The format is as follows: DO WHILE (CONDITION) STATEMENT 1 STATEMENT 2 \textbullet \textbullet \textbullet STATEMENT N END In the above, processing of the DO group is terminated as soon as the CONDITION specified in the WHILE clause is not satisfied anymore. There are various formats and features of DO statements used for iteration. The DO statements are extensively used for iteration purposes as compared to GOTO statements. For a DO statement, the following segmented flowchart symbol is used:

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Question:

The kangaroo rat never drinks water, yet it survives in some of the hottest and driest regions of the world. Explain how it obtains and conserves water.

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Solution:

Animals which live in the desert have limited means of obtaining water. For these animals, water retention and conservation are very crucial. The kangaroo rat can survive without ever drinking water, and in this respect is fairly unique. This animal is found in theSoutwestern deserts of the United States. It lives on dry seeds and obtains all of its water from carbohydrate metabolism. The excretory organs of the kangaroo rat are modified to conserve as much water as possible. Its kidneys are remarkably efficient in concentrating the urine, allowing very little water loss. As was noted earlier, a concentration gradient is achieved in the interstitial medium surrounding the loop ofHenleand the collecting duct in the mammalian kidney. The longer the loop, the greater the gradient can become, and therefore, the greater is the amount of water that can be reabsorbed. In the kangaroo rat, the loops are extremely long and thus a large amount of water is retained by the body and very little excreted.

Question:

A man holds a ball of weight w = 1/4 lb at rest in his hand. He then throws the ball vertically upward. In this process, his hand moves up 2 ft and the ball leaves his hand with an upward velocity of 48 ft/sec. Find the force P\ding{217}with which the \ding{217} man pushes on the ball.

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Solution:

During the action of throwing the ball, it experiences a net force (see figure) equal to the force of the hand, P, less the gravitational force of the earth on the ball, or mg. Hence, F_net = P - mg By Newton's Second Law, F = ma, the ball's acceleration, a, is given by a = (F_net/m) = (P - mg)/m This acceleration is delivered to the ball over a distance s = 2 ft, and accelerates the ball from rest (v_0 = 0) to v_f = 48 ft/sec. Hence, by our kinematics equations (a = constant if P is constant) v2_f - v2_0 = 2as v2_f = 2[(P - mg)/m]s = (2ps/m) - (2mgs/m) v2_f + 2gs = (2Ps/m) P = m/2s(v2_f + 2gs) P = (mv2_f/2s) + mg To calculate the mass of the ball from its weight, w, note that w = mg andm = w/g ThereforeP = (wv2_f/2gs) + w = w [(v2_f/2gs) + 1] P = (1/4 lb)[{(48 f/s^2)} / {(2)(32 f/s^2)(2 f)} + 1] P = (1/4 lb)(18 + 1) = 4.75 lb.

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Question:

What is the energy of a photon of blue light whose frequency is7 × 101 4Hz and the energy of a photon for FM electromagneticradiation if the frequency is 1 × 10^8 Hz?

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Solution:

We shall use an approximate value for h of 6.6 × 10^-34 J s. The energyof a light quantum(photon) is given by E =hf wheref is the frequency of the light. Therefore, we have bluelight E =hf= (6.6 × 10^-34 Js) (7 × 10^14 Hz) = 4.6 × 10^-19 J FM waves E =hf= (6.6 × 10^-34 Js) (1 × 10^8 Hz) = 6.6 × 10^-26 J Notice that the higher the frequency of the electro-magnetic radiation, the greater the energy of the photon.

Question:

A wire of length 1 m is moving at a speed of 2 m \bullet s^-1 perpendicular to a magnetic field of induction of 0.5 Wb\bulletm^-2. What is the potential' difference induced between the ends of the wire? The ends are joined by a circuit of total resistance R = 6 \Omega. At what rate is work being done to keep the wire moving at constant velocity?

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/Users/wenhuchen/Documents/Crawler/Physics/D22-0752.htm

Solution:

Since the direction of motion and the magnetic induction are at right angles, the induced emf will have magnitude given by Faraday's law \epsilon = [(d\varphi)/(dt )](1) where the flux, \varphi = \int B^\ding{217} \bullet dA^\ding{217}, is evaluated over the area enclosed by the circuit, B^\ding{217} is the constant magnetic induction passing through the loop of area A. The angle between the direction of the vectors B^\ding{217} and dA^\ding{217} is 0\textdegree. Thus \varphi = BA (for A = \int dA). However, A = lx, where l is the length of the wire, and x is the position of the wire within the magnetic field (see the figure). Hence, (1) becomes \epsilon = Bl (dx/dt) = Bl v where v is the velocity of the wire.Hence, \epsilon = vBl, = 2 m \bullet s^-1 × 0.5 Wb \bullet m^-2 × 1 m = 1 V. The power dissipated in the external circuit is P = (\epsilon^2 /R) = (1^2 V^2)/(6\Omega) =(1/6) W, and this must equal the rate at which work is being done to keep the wire moving at constant velocity. If the closed system is the loop, magnetic field, and the agent moving the loop, then the energy given up by the agent is used to heat the resistance. Only an energy exchange occurred, and not a change in energy, as required by the principle of conservation of energy. This can be checked by considering that a current I = (\epsilon/R) = (1/6)A will flow in the wire in the direction shown in the diagram for only in this direction will the current create a flux which will oppose the flux of the magnet-ic field in accordance with Faraday's Law. Using the right hand rule, if the fingers of the right hand curl in the direction of the current about the loop, then the thumb will point in the direction of the magnetic field created by this current. In this case, the created B points out of the page, opposite to the direction of the given field. When a current flows in a wire in a magnetic field, the wire experiences a force of magnitude F = IlB, which will be in this case in the direction shown. The work done per second to keep the wire traveling at uniform speed v is F (x/t) = Fv = IlvB = (1/6)A × 1 m × 2 m \bullet s^-1 × 0.5 Wb \bullet m^-2 = (1/6) W, in agreement with the result obtained above.

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Question:

A piece of ice floats in a vessel filled with water. Will the water level change when the ice melts, if the final temperature of the water remains 0\textdegree C?

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Solution:

The temperature of the system remains constant, therefore we do not expect any thermal expansion of the water. Now, suppose that somehow we were able to take the ice out of water while preserving the geometry of the water as shown in figures A and B. The volume V_c of the cavity that is going to result is the volume of the ice which was submerged in water. In order to keep the level of the water the same, the pressure we must exert on the walls of this cavity must be equal to that exerted by the weight of the ice. Therefore, we can fill the cavity with water whose weight is equal to that of ice in order to ac-complish this. But, we already stated that this amount of water will have exactly the same volume as the submerged part of the ice, i.e., the volume of the cavity (Fig. C). As a result, we see that the level remains the same if the cavity is filled up by water. When ice melts, the situation is equivalent to what has been described so far since ice becomes water upon melting and effectively fills up the volume vacated by the ice in water.

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Question:

Explain why tetraethyl lead is added to gasoline.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E31-0928.htm

Solution:

To explain the function of this additive in gasoline, consider how an automobile uses gasoline. The composition of gasoline varies according to the time of year, the refinery at which it was processed and the climate. Basically, gasoline is composed of hydrocarbon isomers containing between seven and ten carbon atoms. In cold climates, hydrocarbons such as butanes are used, which have a lower boiling point, so that the volatility of the gasoline is increased. The increased volatility allows the engine to start more easily. In warmer climates, higher-boiling components are used so that the gasoline's volatility is decreased; this minimizes such problems as vapor lock. The octane rating is a measure of the manner in which gasoline burns in the engine. The fuel should be consumed at the end of the piston cycle and the gasoline should burn such that there is steady pressure on the piston. If these conditions are not met as closely as possible, the automobile engine will knock or ping. This can eventually damage some engine parts. The quantitative measure of a fuel's ability to knock resulted in the formation of octane rating. By definition, an octane rating of 100 is the highest premium gasoline, while an octane of zero is the lowest. The octane rating of a part-icular gasoline can be raised by blending in compounds of relatively high octane rating. The only problem is that such compounds are extremely expensive. The discovery of additives solved the problem of improving octane rating. Tetraethyl lead (as pictured above) is one of these additives that can improve the octane rating without using the expensive high-octane hydrocarbons.

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Question:

What are the evolutionary advances shown in the anatomy of the proboscis worms? How do these animals compare with the round worms?

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Solution:

The proboscis worms (or nemerteans) are a relatively small group of animals. Almost all are marine forms, excepting a few which inhabit fresh water or damp soil. They have long, narrow bodies, either cylin-drical or flattened, and range in length from 5 cm. to 20 meters. Their most remarkable organ is the proboscis, a long, hollow, muscular tube which they evert from the anterior end of the body to capture food. A mucus secreted by the proboscis helps the worm in catching and retaining the prey. There are three important evolutionary advances achieved by the proboscis worms. First, these animals have a complete digestive tract, with a mouth at one end for ingesting food, an anus at the other for egesting feces, and an esophagus and intestine in between. This is in contrast to the coelenterates and planarians, in which food enters and wastes leave by the same opening. As in the flatworms, water and metabolic wastes are eli-minated from the proboscis' body by flame cells. The second advance exhibited by the proboscis worms is the separation of digestive and circulatory functions. These organisms have a primitive circulatory system consisting of only three muscular blood vessels which extend the length of the body and are connected by transverse vessels. Hemoglobin, the oxygen-carrying protein present in higher animals, is found in the red blood cells of this rudimen-tary circulatory system. In the absence of a heart, the blood is circulated through the vessels by the movements of the body and the contractions of the muscular blood vessels. Capillaries have not developed in these animals. The third advance in the anatomy of the proboscis worms is seen in the structure of the nervous system. There is a primitive "brain" at the anterior end of the body, consisting of two groups of nerve cells, called ganglia, connected by a ring of nerves. Two nerve cords run poste-riorly from the brain. Compared to the proboscis worms, the round worms (or nematodes) are a far more numerous and ecologically diver-sified group than the proboscis worms. There are about 8000 species of round worms, all with a similar basic body plan. They have elongate, cylindrical, threadlike bodies which are pointed at both ends. Their many habitats include the sea, fresh water, the soil, and other animals or plants which they parasitize. This last characteristic is in sharp contrast to the proboscis worms, none of which is parasitic and none is of economic importance. Common parasitic nematodes are the hookworm, trichina worm, ascaris worm, filaria worm and guinea worm, which all utilize man as the host. Because of their parasitic exis-tence, these worms are covered with a protective cuticle and have only longitudinal muscles for simple bending movements. Contrary to the proboscis worms, which have cilia all over the epithelium and the lining of the diges-tive tract, none of the nematodes has any cilia at all. However, like the proboscis worms, the round worms have evolved a complete digestive system, a separate circulatory system, and a nervous system composed of a "brain" and nerve cords.

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Question:

Given the ground-state oxygen atom, tabulate each of the electrons by its quantum number.

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Solution:

Begin by writing the electronic configuration of the ground-state oxygen atom. An electron is in its ground state when it is in its lowest energy level, this, is the normal state of the electrons. The atomic number of oxygen is 8. Thus, the oxygen atom has 8 electrons. The electronic configuration of the oxygen atom is 1s^2 2s^2 2p^4. The quantum numbers are N, l, M, and s. N is the principal quantum number which, in the electronic con-figuration, is the integer in front of thesubshells(s and p) and corresponds to the energy level occupied by the electron. The other quantum numbers can be found once N is determined, l is equal to 0, 1, 2, . . . N - 1, depending upon N. M is equal to + l to - l and S is equal to +1/2 or - 1/2. l is the orbital quantum number which denotes thesubshelland angular shape of electron dis-tribution. M is the magnetic quantum number and s denotes the spin of electron. All four are needed to describe each electron. To tabulate the electrons according to their quantum numbers proceed as follows: the electron conf-iguration is 1s^2 s^2 2p^4. Take the first shell, or N = 1, 1s^2. The superscript indicates that 2 electrons are present in this shell. Thus, 2 sets of quantum numbers are needed. For both electrons, N = 1. Because l = 0 . . . n - 1, l = 0 for both electrons in the s subshell . M + l to - l, thus, M = 0 for both also, The four quantum numbers for each electron must be different, Therefore, one electron has a spin of + 1/2 and one has a spin of - 1/2. The sets of quantum numbers for these two electrons are 1, 0, 0, + 1/2 and 1, 0, 0, - 1/2. For the second shell, N = 2 There are twosubshells, s and p. Consider the ssub-shellfirst. In the 2s^2 orbital, N = 2 and there are two electrons to be described. For both electrons l = 0, because both are in the ssubshell, M = 0, because M = + l to - l. For one electron s = +1/2, for the other s = -1/2. The sets of quantum numbers for these two electrons are 2, 0, 0, + 1/2 and 2,0, 0, -1/2, in the 2p^4 orbital there are 4 electrons to be described. For an, N = 2,and l = 1 (l is always equal to 1 for a psubshell), M = + l to - l. Thus, one electron has M = + 1, the second electron has M = 0, and the third electron has M = - 1. The fourth electron will have an M = - 1, 0, or + 1. Whatever it is, it will differ in spin from the electron with the same M. The spins of the others can be + 1/2 or - 1/2. The sets of the quantum numbers for the first three electrons in a 2p orbital can be written (2, 1, -1, + 1/2) , (2, 1, 0. + 1/2) , and (2, 1, 1, + 1/2) , The fourth might be written (2, 1, - 1, - 1/2), (2, 1, 0 - 1/2), or(2, 1, 1, - 1/2).

Question:

Complete the following nuclear equations. (a) _7N^14 + _2He^4\rightarrow_8O^17 + ..... (b) _4Be^9 + _2He^4\rightarrow_6C^12 + ..... (c) _15P^90\rightarrow_14Si^30 + ..... (d) _1H^3\rightarrow_2He^3+ .....

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Solution:

The rules for balancing nuclear equations are: (1) the superscript assigned to each particle is equal to its mass number and the subscript is equal to its atomic number or nuclear charge; (2) a free proton is the nucleus of a hydrogen atom, and is therefore written as _1H^1; (3) a free neutron has no charge and is therefore assigned zero atomic number. Its mass number is one and its notation is _0n^1; (4) an electron, \beta^-, has zero mass and its atomic number is - 1, hence the notation _-1e^0; (5) a positron has zero mass and its atomic number is + 1, hence the notation _+1e^0; (6) an alpha particle (\alpha-particle) is a helium nucleus, and is represented by _2He4or \alpha; (7) Gamma radiation (\gamma) is a form of light, and has no mass and no charge; (8) in a balanced equation, the sum of the sub-scripts must be the same on both sides of the equation; the sum of the superscripts must also be the same on both sides of the equation. In equation (a), _7N^11 + _2He^4\rightarrow_8O^17 + ..., the sum of the subscripts on the left is (7 + 2) = 9. The subscript of one of the products is 8, thus, the other product must have a subscript or net charge of 1. The sum of the super-scripts on the left is (14 + 4)= 18. The superscript of one of the products is 17, thus the other product on the right must have a superscript or mass number of 1. The particle with a + 1 nuclear charge and a mass number of 1 is the proton, _1H^1. In equation (b) , _4Be^9 + _2He^4\rightarrow_6C^12 + ..., the nuclear charge of the second product particle (that is, its subscript) is (4+2) -6=0. The mass number of the particle (its superscript) is (9 + 4) - 12 = 1. Thus, the particle must be the neutron, _0n^1. In equation (c),is _15P^30 _14Si^30 + ..., the nuclear charge of the second particle is 15 - 14 = + 1. Its mass number is 30 - 30 = 0. Thus, the particle must be the positron, _+1e^0. In equation (d), _1H^3\rightarrow_2He^3 + ..., the nuclear charge of the second product is 1 - 2 = - 1. Its mass number is 3 - 3 = 0. Thus, the particle must be a \beta^-1 or an electron, _-1e^0.

Question:

Examine the following COBOL program and describe in words the function of each section.

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Solution:

We have the first three divisions (IDENTIFICATION DIVISION, ENVIRONMENT DIVISION, and DATA DIVISION) of the program LIST- REPORT given. The IDENTIFICATION DIVISION gives the program name, who wrote it and when it was written, and may include remarks about the program. The ENVIRONMENT DIVISION consists of two sections. In the CONFIGURATION SECTION the compiling (source) and executing (object) computers are specified. In this program both compilation and execution of the program is done on an IBM-370. The INPUT-OUTPUT section specifies the interface between the program files and the actual external devices. In this program the actual file asso-ciated with the identifier TAPE-FILE is the UT-S-TAP01 tape reader. Similarly, CARD- FILE is associated with the UR-S-CAR01 card reader and PRINT-FILE is associated with the UR-S-PRINT01 printer. The DATA DIVISION is also composed of two sections. The File Section describes the structure of the files used in the program. Every file selected in the INPUT- OUTPUT SECTION of the Environment Division must have an FD (File description) entry in this section. Every FD contains: 1) a file-name 2) a LABEL RECORDS clause. The label record can be either STANDARD or OMITTED. OMITTED is used when there is no label record, STANDARD is used when the label is a standard system label. 3) A description of the record associated with the file-name. As an example of an FD, the file TAPE-FILE will be used from our program. In this file the label records are standard. The statement BLOCK CONTAINS 10 RECORDS. means that the tapereader UT-S-TAP01 sendsdata to the computer in the form of Data Blocks, each block containing 10 records of the type described in TAPE-RECORD. The WORKING STORAGE SECTION describes the independent or group variables which will be used for data storage and manipulation in the PROCEDURE SECTION of a program.

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Question:

Consider the simple regression model. When there is a large number of observations, storage in single subscripted arrays can create problems. Devise a flowchart and write the program for performing the linear regression without using arrays.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G19-0485.htm

Solution:

Let (x_i,y_i) be data points. The simple regression problem is to find numbers\alpha˄, \beta˄ such that \alpha˄ ^n\sum_i=1 [y_i - (\alpha˄+ \beta˄x_i)]^2 \alpha˄ is minimized. The solution is given by the formulae \beta˄^ = {[\sum(x -x) (y -y)] / [\sum(x -x)^2]}\alpha˄=y- \beta˄x \alpha˄ \alpha˄ where x= (1/n) ^n\sum_1 x_i,y= (1/n) ^n\sum_1 y_i. One method of performing linear regression on a computer is to read and store the data (x_i,y_i) in arrays. The sums of squares and means are then found by operating with these arrays. However, the regression formulae can be rewritten so that \sum(x-x)^2, \sum(y-y)^2 , \sum(x-x) (y-y) can be expressed in terms of ^n\sum_1 x_i , ^n\sum_1 y_i , ^n\sum_1 y^2_iand^n\sum_1 x^2_i . These latter sums can be computed directly from the observations without storing them in arrays. To clarify, consider \sum(x-x)^2 : ^n\sum_i=1 (x_i -x)^2 = ^n\sum_1 x^2_i - 2x^n\sum_1 x_i + ^n\sum_1x^2(1) But 2x^n\sum_1 x_i = 2xnx(^n\sum_1 x_i = nx= 2nx^2 . Also, ^n\sum_1x^2 =x^2 + x^2 +...+ x^2= nx^2 \downarrow n times Thus, (1) is equal to ^n\sum_1 (x_i -x)^2 = ^n\sum_1 x^2_i - nx^2.(2) To compute (2), all we need is ^n\sum_1 x^2_i andx. No array is necessary for either of these since^n\sum_1 x^2_i can be computed directly by squaring each observation, as it is read in, and summing. Similarly,x= (1/n) \sumx_i can be computed directly. From (1) and (2) above, \sum (y_i -y)^2 = ^n\sum_1 y^2_i - ny^2.(3) To simplify \sum(x-x) (y-y) consider the following: ^n\sum_1 (x_i -x) (y_i -y) = x_1 y_1 - x_1y-xy_1 +xy + x_2 y_2 - x_2y_2 -xy_2 +xy+...+ x_n y_n - x_ny-xy_n +xy = ^n\sum_1 x_i y_i -y^n\sum_1 x_i -x^n\sum_1 y_i + nxy(4) But -y^n\sum_1 x_i = -ynx= -x^n\sum_1 y_i . Thus (4) equals ^n\sum_1 (x_i -x) (y_i -y) = ^n\sum_1 x_1 y_1 - nxy(5) The program looks as follows: READ, N SX = 0 SY = 0 SXX = 0 SYY = 0 SXY = 0 J = 1 1IF (J.GT.N) GO TO 2 READ, X,Y SX = SX + X SY = SY + Y SXX = SXX + X\textasteriskcenteredX SYY = SYY + Y\textasteriskcenteredY SXY = SXY + X\textasteriskcenteredY J = J + 1 GO TO 1 2XM = SX/N YM = SY/N SXX = SXX - N\textasteriskcenteredXM\textasteriskcentered\textasteriskcentered2 SXY = SXY - N\textasteriskcenteredXM\textasteriskcenteredYM SYY = SYY - N\textasteriskcenteredYM\textasteriskcentered\textasteriskcentered2 B = SXY/SXX A = YM - B\textasteriskcentered XM SD = SYY - B\textasteriskcenteredSXY PRINT, A,B,SD STOP END

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Question:

Calculate the voltage (E) of a cell with E\textdegree = 1.1 volts, If the copper half-cell is at standard conditions but the zinc ion concentra-tion is only .001 molar. Temperature is 25\textdegreec . The overall reaction is Zn + Cu^+2 \rightarrow Cu + Zn^+2.

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Solution:

To solve this problem requires a relation between the cell potential (voltage), E, and the standard cell potential, E\textdegree . The equation used is the Nernst equation, which states E = E\textdegree -(RT / nF) In(Q), where n is the number of mole of electrons transferred, R is the constant 8.314 joules per degree, F is 96,500 coulombs, and Q is the equilibrium constant expression. In other words, Q = [product] / [reactant] You are told the concentration of Zn^+2 is .001. = [Zn^+2] / [Cu^+2 ]= .001 / 1 Standard conditions indicate that the concentration of the substance is 1M and at a temperature of 25\textdegree c . Thus [Cu^+2] = 1 and there-fore, from the reaction equation, it can be seen that 2 electrons are transferred. Thus, n = 2. Substituting these values into the Nernst equation, one obtains E = 1.1 - {[8.314(298)] / [2(96,500)]} log (.001 / 1) = 1.1 - [.0295(-3)] = 1.19 volts.

Question:

Why must mosses and liverworts (phylumBryophyta) always live in close association with the water?

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Solution:

The bryophytes are generally considered to be primitive land plants because they are relatively ill-adapted to the terrestrial environment compared to the higher land plants. Mosses, for instance, are found frequently on stream banks or moist roadsides. Liverworts, lacking a cuticle , are not as well protected againstdessicationas are the mosses, and are even more restricted in their distribution - the majority of them grow in moist, shady localities, and some are even true aquatic plants. The fact that mosses and liverworts always live in close association with water can be explained by their anatomical structures and reproductive mechanism. They have rhizoids which are simple filaments of cells or cellular projections performing the function of water absorption. The rhizoids are, however, not efficient absorbers and in relatively dry areas cannot withdraw adequate materials from the ground. In addition, mosses and liverworts lack vascular tissues to conduct water, minerals, and organic substances to different parts of the plant. Therefore transport of materials in the mosses and liverworts depends to a great extent on simple diffusion and active transport through the leaves and plant axis. Furthermore, mosses and liverworts are small plants with a high surface to volume ratio. This means that evaporation can cause a rapid dehydration of inner as well as outer tissues. These disadvantages can be avoided in a moist habitat where evaporation by the atmosphere is slower and sufficient water is available through diffusion to compensate for water lost through the surface. A moist habitat also favors the reproductive process of the bryophytes . The gametophyte plant produces flagellated sperm which can swim to the egg only when water is present. In order to reproduce successfully , mosses and liverworts must grow in close proximity to water.

Question:

(1) Cadmium and Hydrogen (2) Silverand Hydrogen (3) Cadmium and silver, using the following data: Reaction E\textdegree volts Cd \rightarrow Cd+^2 + 2e^- +.403 H_2 \rightarrow 2H^+ + 2e^- 0.00 Ag \rightarrow Ag^+ +e^- -.799

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Solution:

You are asked to calculate the standard cell potential (∆E\textdegree) for each of the given pairs. To do this, you must realize that in such cells, you have 2 half-reactions. Namely, an oxidation re-action (loss of electrons) and a reduction reaction ( gain of elec-trons). The sum of these half- reactions yields the overall reaction and the ∆E\textdegree of the whole cell. Thus, to find the ∆E\textdegree for each of these pairs, you need to know the E\textdegree of the half- reactions. There is one other important fact to be kept in mind. In cells, if the reaction is to proceed spontaneously, ∆E\textdegree must have a positive value. This means, therefore, that you must choose the half- reactions such that their sum always gives a positive ∆E\textdegree You are told the E\textdegree of the oxidation half-reaction for each element in each pair. The reduction half-reaction is the reverse of the oxidation reaction for each element, with a change in sign of E\textdegree . For example, if you have A \rightarrow A^+1 + e^- with an E\textdegree = B for oxidation, the reduction is A^+1 + e^-\rightarrow A with a E\textdegree = -B. With this in mind, the procedure is as follows: (1) Cadmium (Cd) and hydrogen (H_2). The reaction for this cell must be the sum of the oxidation and reduction such that the ∆E\textdegree is positive. This can only occur if the anode (oxidation) has the higher oxidation potential. Thus, you calculate ∆E\textdegree as + (.403) - (0.000) = + .4303v . Similarly, for (2)and (3), a positive ∆E\textdegree can only be obtained with the anode having the higher oxidation potential. Thus, for (2), E\textdegree =(0.000) - (-.799) = +( .799v.) For (3), E\textdegree = (.403) - (-.799) = +1.202v.

Question:

What is the period of a small oscillation of an ideal pendulum of length l, if it oscillates in a truck moving in a horizontal direction with acceleration a? (see figure a).

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Solution:

Let the equilibrium position be given by the angle \alpha. In this position, the net force on m is along the horizontal axis and equals ma. The angle \alpha is determined by the equations T sin \alpha = ma(1) T cos \alpha = mg(2) where g is the gravitational acceleration. When the pendu-lum is displaced by a small amount \texttheta, it will perform a simple harmonic motion around the equilibrium position. The force on m along the horizontal x-axis is T sin(\texttheta + \alpha), as shown in Fig. b. Newton's second law for the motion along this axis is m[d^2x/dt^2] = -T sin (\texttheta + \alpha)(3) where x is the distance from the vertical. For small \texttheta, we can expand sin(\texttheta + \alpha) as sin(\texttheta + \alpha) \approx sin \alpha + \texttheta cos \alpha. Substituting in (3) m[d^2x/dt^2] \approx -T sin \alpha\Elzbar \textthetaT cos \alpha.(4) Combining (1), (2) and (4), we get m[d^2x/dt^2] \approx -ma - \textthetamg d^2x/dt^2 \approx -a - \textthetag.(5) \texttheta and l are related geometrically as x = l sin (\texttheta + \alpha) = l sin \alpha + \textthetal cos \alpha therefore d^2x/dt^2 \approx l cos \alpha (d^2\texttheta/dt^2) Substituting in (5), l cos \alpha (d^2\texttheta/dt^2) \approx -\alpha - \textthetag d^2\texttheta/dt^2 \approx - [g/(l cos \alpha)][\texttheta + (a/g)](6) If we make the following substitution in (6) \Phi = \texttheta + (a/g) we get d^2\Phi/dt^2 = - [g/(l cos \alpha)]\Phi. This is the differential equation for a simple harmonic motion. Therefore its solution is \Phi = A sin(\omegat + B) where A and B are the constants of integration and \omega = \surd[g/(l cos \alpha)]. The expression for \texttheta is \texttheta = A sin(\omegat + B) - (a/g). The boundary conditions are \texttheta = 0 at t = 0 and \texttheta = \texttheta_max at t = \tau/2 (where \tau is the period). Determine the constants A, B: \cyrchar\cyrii) t = 0: A sin B = a/g \cyrchar\cyrii\cyrchar\cyrii) t = \tau/2: A sin[(\omega\tau)/2 + B] - a/g = \texttheta_max . The period of motion is \tau = 2\pi/\omega = 2\pi\surd[(l cos \alpha)/g].

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Question:

What is eutrophication? What problems are associated with it? Why would a reduction of phosphate ions from waterways alleviate eutrophication in lakes and streams?

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Solution:

The enrichment of water with nutrients is a naturally occurring biological process called eutrophication. It is caused by excess nutrients in water that enable algae to grow to great abundance. The problem of massive algae growth occurs when it dies. As it dies, it sinks to the bottom of lakes and rivers and begins to decay. This decay-ing process consumes oxygen thereby deleting the oxygen content of the water. Fish and other aquatic life dependent on oxygen suffocate. Therefore, the decaying of excessive amounts of algae can lead to the death of marine life. This is one major problem associated with eutrophication. A second problem also results from the depletion of oxygen. Remember that anaerobic bacteria, i.e. bacteria that can live without oxygen, still function effectively under the conditions described above. These bacteria feed on decaying algae. As a result of their metabolic process, they produce compounds such as hydrogen sulfide, which cause the putrid odor associated with decaying organic matter. The water becomes foul smelling and foul tasting. A reduction of phosphate would alleviate eutrophica-tion because it is one of the major nutrients of algae. A smaller amount of phosphate in the water would limit the growth of algae.

Question:

Minimize the given minterm function f(A,B,C,D) via the Karnaugh map. f (A., B, C, D) = \summ(0,1,3,8,9,11,13,14)

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G03-0056.htm

Solution:

We are given the function f(A,B,C,D) = \summ(0,1,3,8,9,11,13,14) which consists of minterms to be minimized. The first step in obtaining the solution is to plot the function on the K-map. In figure # 1 a K-map for four variables is given. These maps are standard for each problem, however they change size according to the number of variables in the function. To plot the given function into the K-map, 1's are put into the boxes which number matches with the given minterm variable. The final K-map with the variables plotted looks as shown in figure 2. Secondly, the number of adjacencies for each minterm is counted. The number of adjacencies for each minterm is shown in the lower right-hand corner of the minterm block of figure 3. Now the map is ready for the simplification process. This is done by choosing first the minterm m_14 which has no adjacencies, as in figure 4. This minterm must be taken as a group itself; hence the first term in the minimized function is ABCD since minterm box #14 is not covered by variable D. Next, minterms with one adjacency are examined; m_13 the only one. Consequently, m_13 has one way of being grouped and that is with m_9 .This is illustrated in figure 5. Note that minterms m_13 and m_9 have the following variables: m_13 = ABCD , m_9 = ABCD , therefore, B (in m_13), andB(in m_9) may be cancelled according to Boolean algebra theorems. Continuing the solution; four minterms with two adjacencies are taken into account. These are m_0,m_3, m_8 and m_11 . One of these is picked to group at random. If m_3 is picked (it is adjacent to m_1, and m_11), since m_9) is also available (any minterm may be used as many times as it is needed), a large group of four minterms may be formed. Note that m_1 =A B CD m_3 =A BC D m_11= ABC D m_9 = AB CD in these four minterms A, and -Aterms also C, andCterms cancel each other, using Boolean theorems a\bulleta= 0 , therefore the minimized form of minterm m_1,m_3,m_11, and m_9 isBD. The function now becomes f(A,B,C,D) = ABCD+ ACD +BD. Finally there are still two minterms which have not been accounted for. These two, m_0 and m_8, can be grouped with m_1 and m_9 respectively, to form a last group of four minterms as shown in figure 6. This last group adds the final product termBCto_the function f(A,B,C,D), and hence: f(A,B,C,D) = ACBD+ ACD +BD +BC.

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Question:

Find Antilog2.3625.

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Solution:

By definition, b = Antilog a, Is equivalent to log b = a. Let N = Antilog 2.3625. Therefore, log N = 2.3625. The characteristic is 2. Hence, the number that corresponds to the mantissa 0.3625 will be multiplied by 10^2 or 100. In a table of four-place common logarithms, the mantissas 0.3617 and 0.3636 are those given that are closest to 0.3625. Therefore, since the mantissa 0.3625 does not appear in the table, the number that corresponds to this mantissa will be found through inter-polation. The following proportion is now established: d / 0.1 = .0008 / .0019 Cross multiplying, .0019d =(.01) (.0008) d = .01 [(.0008) / (.0019)] = 1 × 10^\rule{1em}{1pt}2[(8 × 10^\rule{1em}{1pt}4) / (1.9 × 10^\rule{1em}{1pt}3)] = (8 × 10^\rule{1em}{1pt}2+(\rule{1em}{1pt}4)) / (1.9 × 10^\rule{1em}{1pt}3) = (8 × 10^\rule{1em}{1pt}6) / (1.9 × 10^\rule{1em}{1pt}3) = (8/1.9) × (10^\rule{1em}{1pt}6 /10^\rule{1em}{1pt}3) = 4.2 × 10^\rule{1em}{1pt}6\rule{1em}{1pt}(\rule{1em}{1pt}3) = 4.2 × 10^\rule{1em}{1pt}3 d = .0042 Hence, x = 2.30 + 0.0042 = 2.3043 \approx 2.304. Therefore, Antilog 2.3625 = N= 2.304 ×10^2 = 2.304 ×100 = 230.4

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Question:

An observer moves past a meter stick at a velocity that is one-half the velocity of light. What length does he measure for the meter stick?

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/Users/wenhuchen/Documents/Crawler/Physics/D32-0931.htm

Solution:

Einstein stated two postulates which form the basis for the theory of special relativity. They are From these postulates, it is found that the length of an objectviewed by an observer travelling with velocity v with respect to the object is less in the direction of motion than the length seen by a person stationary with respect to the object. The apparent shortened length l ' of the meter stick as it moves with respect to the observer (Lorentz contraction) is given by l ' = [ l \surd{1 -(v/c )^2} =l \surd( 1 - \beta^2) where l is the length seen by a person stationary with respect to the object and \beta is v/c. For this problem, \beta = v/c = 0.5c/c = 0.5 and the length l ' measured by the observer is l ' = { l ' \surd(1 - \beta^2)} = (10cm) × \surd{1 - (0.5)2} = (100cm) × {\surd0.75} = 86.6 cm

Question:

A converging lens with a focal length of 3 m forms an image of an object placed 9 m from it. Find the position of the image and the magnification.

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/Users/wenhuchen/Documents/Crawler/Physics/D28-0875.htm

Solution:

The simple lens equation for the converging lens of this problem is 1/f= 1/p + 1/q where f, p and q are respectively the focal length of the lens and the distances of the object and the image from the lens. The image is real and inverted (see the figure). Substituting the given values in the above equation, we get 1/q = 1/f - 1/p = 1/3m - 1/9m = (2/9)m-1and q = (9/2)m = 4.5m Since the value of q is positive, the image occurs on the right side of the right side of lens. The magnification M is M = q/p = 4.5m/9m = 0.5 so the image is one-half as high as the object.

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Question:

One problem concerning space flight is the removal of carbon dioxide (CO_2, molecular weight = 44.0 g / mole) emitted by the human body (about 924 g of CO_2 per day). One solution is to react CO_2 with sodium hydroxide (NaOH, molecular weight = 40.0 g / mole) to form water and sodium carbonate (Na_2CO_3, molecular weight = 94 g / mole) according to the followingreation: 2NaOH + CO_2 \ding{217} NaCO_3- + H_2O. How muchNaOHmust be carried on board a space capsule to remove the CO_2 produced by an astronaut on a 10 day flight?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E27-0902.htm

Solution:

Two moles ofNaOHare required to remove one mole of CO_2, as indicated by the coefficients in the reaction. The number of moles of CO2 produced is calculated from the mass emitted by the astronaut during the flight. From this, the number of moles ofNaOHrequired for the removal of he CO_2 is calculated, and this number of moles is then con-verted to grams. At the rate of 924 g per day, in 10 days the astronaut will produce 924 g / day × 10 days = 9240 g of CO_2. Dividing this mass by the molecular weight of CO_2, it is calculated that the astronaut will produce a total of [9240 g / (44.0 g / mole)] = 210 moles of CO_2. The number of moles ofNaOHrequired to remove 210 moles of CO_2 is 2 × 210 moles = 420 moles. Multiplying this by the molecular weight, the minimum amount ofNaOHthat must be carried on board is 420 moles × 40 g / mole = 16,800 g = 16.8 kg.

Question:

Illustrate transfer of control in a PL/I program, usingGOTO statements.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G13-0340.htm

Solution:

The PL/I machine has an implicitly defined sequential execution. The execution of statements is normally a sequential process, however, a specialstatement, the GOTO statement can alter the processing sequenceof the PL/I ma-chine. This GOTO statement discontinues the normalsequen-tial execution and processing resumes at a point that is in-dicatedby a label. Examples of well formed GOTO statements are given below: 1) SUM:A= B + C; \textbullet \textbullet \textbullet \textbullet GOTO SUM; 2)GOTO NEXT; \textbullet \textbullet \textbullet \textbullet NEXT:X=Y\textasteriskcenteredZ; 3) JUMP:GOTO TERMINATE; \textbullet \textbullet \textbullet \textbullet TERMINATE:END; The following are a few examples of not well-formed GOTO statements. a)JOE:PROCEDURE OPTIONS (MAIN); \textbullet \textbullet \textbullet \textbullet GOTO JOE; The above is wrong due to the fact that the control can only be transferred withinthe body of the main program. In other words, GOTO cannot transfercontrol to the starting point of a procedure as above. b) GOTO END_1 \textbullet \textbullet \textbullet \textbullet END END_1 The above is wrong because there is no label END_1 anywhere in the bodyof the program. END_1 is used as a part of an-other statement, but END_1 is not a label of any statement in the program body.

Question:

What would be the simplest and safest reliable method of distinguishing between the following pairs of materials? (a) Water and rubbing alcohol. (b) Gasoline and kerosene, (c) Baking soda and baking powder.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E31-0926.htm

Solution:

(a) Pour to a depth of about one-half inch of each compound into two separate test tubes. Cover each of the test tubes with your thumb and shake them vigorously. There will be an outward pressure on your thumb from the contents of one of the test tubes. This tube contains the rubbing alcohol. This pressure is called vapor pressure. It is proportional to the ability of a liquid to vaporize or evaporate. Vapor pressure is a measure of the physical property called volatility. Rubbing alcohol is more volatile than water and will have a greater vapor pressure and lower boiling point. (b) Gasoline and kerosene are both products of the distillation of crude oil. Distillation is a process used to separate mixtures of liquids. When a liquid is distilled it is boiled in a round bottomed container at the end of a glass enclosed tube. The various compounds in the mixture burn at their own distinctive boiling points. Thus, first the lowest boiling substance will boil and evaporate, then as the mixture becomes hotter the next lowest and so on until no compounds are left in the container. As each com-pound boils its vapors enter the tube or column where they are condensed and eventually collected. A different vessel is used to collect each fraction with a distinct boiling point and hence the mixture becomes separated into its various components. The more times a fraction is distilled the more pure a sample made. This is the best method for separating gasoline and kerosene. Gasoline burns over a range from 40\textdegree to 200\textdegreeC and kerosene from 175\textdegree to 275\textdegreeC. They can be easily separated by distillation. The accompany-ing figure shows a fractional distillation of crude oil. (c) The difference between baking soda and baking powder is that baking soda is a compound and baking powder is a mixture. A compound is a homogeneous substance composed of two or more elements, the proportions of which are fixed and invariable by weight. A mixture is a substance made of two or more elements or compounds in a nonhomogeneous combination. Under a magnifying glass or microscope one will not be able to distinguish the components of a compound whereas one can readily identify the components of a mix-ture. Baking powder is a mixture of baking soda, an acid substance and starch, while baking soda is made up of a single compound, sodium hydrogen carbonate (NaHCO_3).

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Question:

Five resistors, each of resistance 10 \Omega, are connected to form a letter H, a 2-V cell of internal resistance 1.86 \Omega being connected across the upper ends and an ammeter of resistance 5 \Omega across the lower ends. What current passes through the ammeter?

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/Users/wenhuchen/Documents/Crawler/Physics/D20-0681.htm

Solution:

Because the 5 \Omega resistance of the ammeter and the 10 \Omega resistance of the lower branch of the circuit in figure (A) are in series, figure (A) is equivalent to figure (B). This follows because the equivalent resistance of n resistors in series is equal to the sum of their individual resistances. The resistances of 10 \Omega and 25 \Omega are in parallel. Hence the equivalent resistance is R, where (1/R) = (1/10 \Omega) + (1/25 \Omega) = [(35 \Omega)/(250 \Omega^2)] \thereforeR = (50/7) \Omega = 7(1/7) \Omega = 7.14 \Omega. The circuit is therefore equivalent to the one shown in diagram (C). It is now possible to find the current I_0 in the battery circuit, for, by Ohm's Law I_0 = (\epsilon/R) = [(2 V) / {(10 + 10 + 7.14 + 1.86) \Omega}] = (2/29) A.(1) This current splits up into currents I_1 and I_2 through the lower parts of the circuit, as shown in diagrams (A) and (B). Using figure (B), note that branches CD and EF are in parallel. Therefore, the voltage drops across CD and EF are equal. Using Ohm's Law V_CD = V_EF or(10 \Omega)I_2 = (25 \Omega)I_1 (I_1 /I_2 ) = (R_2 /R_1 ) = [(10 \Omega)/(25 \Omega)] = (2/5) Since no charge can accumulate in the circuit, then I_1 + I_2 = I_0. Therefore, I_1 + (25/10) I_1 = I_0 I_1 = (10/35) I_0 = (10/35) × (2/29) A = 0.0197 A, where we have used (1). This is the current flowing through the ammeter.

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Question:

Two small loudspeakers A and B vibrate in phase at 800 cps, and are set apart by 10 ft. The speakers emit spherical waves which obey the inverse square law with respect to intensity. The intensity varia-tion (for distances away from the speakers greater than a few inches) is I_A = 60/R^2_A I_B = 30/R^2_BThe units are arbitrary; R_A is the distance from speaker A, R_B is the distance from speaker B. Find the point of minimum intensity along the line joining the speakers. Find the first three points of zero intensity that are nearest to the speakers.

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Solution:

Assume that points of minimum and zero intensity occur only at points of destructive inter-ference. Destructive interference of two waves occurs when the waves are 180\textdegree out of phase. The resultant amplitude at the point of interference will be the difference of the amplitude of the two waves at the point. The speed of sound is 1200 ft/sec. Consider point C (in the figure) such that R_A =CAR_B =CB Wave cancellation occurs when the two waves are 180\textdegree out of phase. Since the two waves originate with a 0\textdegree phase difference, one wave must undergo a 180\textdegree phase shift (relative to the second wave) for destruct-ive interference to occur. In our case, the phase change is effected by letting the two waves move through different path lengths before they meet and interfere. Destructive interference will then occur when this path difference is an odd number of half wavelengths. This will insure a phase shift of 180\textdegree. Therefore, │R_A - R_B│ = (2n - 1) \lambda/2n = 1,2,3,... Now, if \lambda and f are the wavelength and frequency of the sound wave, and c is its velocity, \lambdaf = c \lambda = c/f = (1200 ft/see)/(800 sec^-1) = 1.5 ft We are also given thatAB= 10 ft = R_A + R_B, when C is on the line joining AB ; thus │R_A - R_B│ = │R_A -(10 - R_A)│ = │2R_A - 10│ and cancellation will occur for 2R_A - 10 = (2n - 1) \lambda/2n = 1, 2, 3, 4, ... = (1) 1.5/2 = 0.75if n = 1 = (3) 1.5/2 = 2.25if n = 2 = (5) 1.5/2 = 3.75if n = 3 = (7) 1.5/2 = 5.25if n = 4 and so forth. The resultant intensity at the point of destructive interference is The resultant intensity at the point of destructive interference is I_C = │I_A - I_B│ = │[60/R^2_A] - [30/R^2_B]│ This follows from the very nature of destructive inter-ference; the intensities of the interfering waves tend to try to cancel one another.

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Question:

A recipe for artificial sea water is NaCl, 26.518 g/kg; MgCl_2, 2.447 g/kg; MgSO_4, 3.305 g/kg; CaCl_2, 1.141 g/kg; KCl, .725 g/kg; NaHCO_3, .202 g/kg; NaBr, .083 g/kg. All of these salts are dissolved in water to make a total weight of 1000 g. What are the chlorinity and salinity of the solution?

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Solution:

The chlorinity of a solution is defined as the percent of Cl^- ions in the solution by weight. The salinity is a measure of the salt content of seawater. To solve this problem, the number of moles of Cl^- must be determined. Once this value is known the weight of Cl^- can be determined and thus the percentage of Cl^- can be found. The molecular weights of the salts containing Cl^- are 58.5 g/mole for NaCl, 95.3 g/mole for MgCl_2, 111 g/mole for CaCl_2, and 74.5 g/mole for KCl. Since there are 1000 g in 1 kg of sea water, the weights of the salts are 265.18 g for NaCl, 24.47 g for MgCl_2, 11.41 g for CaCl_2, and 7.25 g for KCl. Thus, moles of Cl^- from NaCl= 26.518 g / (58.5 g/mole)= .453 moles moles of Cl^- from MgCl_2 = [2(2.447 g)] / [95.3 g/mole = .0514 moles moles of Cl^- from CaCl_2 = [2(1.141 g)] / [111 g/mole]= .0206 moles moles of Cl^- fromKCl= (.725 g) / (74.5 g/mole)= .00973 moles Total number of moles of Cl^- = .535 moles The total weight of Cl^- in the sea water is (.535 moles) (35.5 g/mole) = 19.0 g. Thus the chlorinity of the sea water is [(19.0 g Cl^-) / (1000 g sea water)] × 100 = 1.9% chlorinity. [(19.0 g Cl^-) / (1000 g sea water)] × 100 = 1.9% chlorinity. This value of chlorinity is average for sea water. To find the salinity of the sea water use the formula: % salinity = 1.805 (% Cl) + 0.03 % = (1.805) (1.9%) + 0.03 % = 3.433 %. = (1.805) (1.9%) + 0.03 % = 3.433 %.

Question:

Plant tissues that are neither considered surface tissues nor vascular tissues, are referred to as fundamental tissue. Describe the various types of fundamental tissues.

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Solution:

Fundamental tissues make up most of the plant body. Examples include the soft parts of the leaf, the pith (central core of stems and roots), the cortex (outer area of stems and roots), and the soft parts of flowers and fruits. The chief functions of the fundamental tissues are the production and storage of food. Some fundamental tissues may also function in physical support of the plant body. There are four types of fundamental tissues, each composed of a single type of cell. One type is parenchyma tissue, which occurs in roots, stems and leaves. Parenchyma consists of small cells with a thin cell wall and a thin layer of cytoplasm surrounding a large vacuole. The cells are loosely packed, resulting in abundant spaces in the tissue for gas and nutrient exchange. Most of the chloroplasts of leaves are found in these cells. A second type is collenchyma tissue. It contains cells that are generally more elongate and their cell walls are irregularly thickened compared to the cell walls of parenchyma. The corners of their cell walls are thickened to provide the plant with support. These tissues occur just beneath the epidermis of stems and leaf stalks. Sclerenchyma tissue, a third type, also functions in support. Most mature cells of this tissue are dead. They have very thick cells walls to provide support and mechanical strength. The cell wall is usually impregnated with an additional tough substance called lignin. The cell wall may be so thick that the internal space of the cell is nearly obliterated. Sclerenchyma is found in many stems and roots. Sclerenchyma cells are divided into two cate-gories, fibers and sclereids. Fibers are elongate cells with tapered ends. Fiber cells are tough and strong, but flexible. Sclereids are more irregularly shaped cells. Stone cells are rounded sclereids. They- are found in the hard shells of nuts and in seeds. Endodermis, the fourth type of fundamental tissue, occurs in a layer surrounding the vascular core of roots. Endodermal cells have cell walls thickened with lignin and suberin which are chemical substances that make the cells waterproof. The endodermal layer is one cell thick, and the cells are compactly arranged. This cell layer is called the Casparian strip. The Casparian strip regulates the entry of water into the vascular tissues of roots.

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Question:

Using methyl orange as an indicator, a solution of unknown pH was matched to a Bjerrum wedge at a point where the acid wedge was 40 percent of the total thickness of the combined wedges. This point is indicated by the arrow in the diagram. Determine the pH of the solution.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E10-0377.htm

Solution:

From the Bjerrum wedge, the ratio of the concentration of methyl orange indicator in basic solution ([Ind^-]) to the concentration in acid solution ([HInd]) will be determined. This ratio will then be used to find the pH of the unknown solution. A Bjerrum wedge is a glass box divided into two wedge-shaped compartments by a glass plate placed diagonal-ly in the box (see diagram, above). In one compartment is placed an acidic solution of the chosen indicator (HInd) and into the other a basic solution of the indicator (Ind^-). The concentration of indicator is the same in both wedges. A view from the top will produce a continuum of color varying from that of pure HInd on one side to that of pure Ind^- on the other. A solution of indicator of the same concentration as the Bjerrum wedge is prepared using a solution of unknown pH. The resulting solution is then placed in a glass box having a thickness equal to that of the Bjerrum wedge. The boxes are placed side- by-side and the color of the unknown solution is matched to a color on the Bjerrum wedge. At the point where the two colors match, let the acidic wedge be x percent of the total thickness of the box. Then the basic wedge is 100 - x percent of the total thickness. The ratio [lnd^-]/[HInd] is then [Ind^-]/[HInd] = [(100 - x)/x] . In this problem, the acidic wedge is 40 percent the total thickness, hence [Ind^-]/[Hind] = [(100 - 40)/40] = (60/40) . To make use of this ratio, we need an expression relating the pH to [lnd^-]/[HInd]. Consider the dissociation of acidic indicator: HInd + H_2O \rightleftarrows H_3O^+ + Ind^-. The equilibrium constant, K_a , for this reaction is K_a = [H_3O^+] × ([Ind^-] / [HInd]) . Taking the logarithm of both sides and multiplying by negative one gives - log K_a = - log [H_3O^+] - log ([Ind^-]/[Hind]) . - log K_a = pK_aand- log [H_3O^+] = pH. Hence, pK_a = pH - log ([Ind^-]/[Hind]) , or,pH = pK_a + log ([Ind^-]/[Hind]) . For methyl orange, pK_a = 3.7. Using the value of [lnd^-]/[HInd] obtained above, we get pH = 3.7 + log [60/40] = 3.9 as the pH of our unknown solution.

Question:

What advantages and disadvantages do the fungi have in comparison withchlorphyll-bearing plants in terms of survival?

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Solution:

The chlorophyll-bearing plants require no organic food source. Their sole needs are light, carbon dioxide, water, and inorganic minerals. Plants are able to grow in any environment where these needs are met. The ocean is a habitat very favorable to green plants (e.g. the green algae), but not to fungi. Since fungi do not photosynthesize, they are able to grow in the dark. Green plants cannot do this. Fungi do not require light penetration and they can survive with very thick, tough cell walls. The cell walls of fungi sometimes contain cellulose, but chitin is usually their most im-portant constituent. Chitin, a polysaccharide' of acetyl glucosamine, is also the principal constituent of the exoskeleton of arthropods-insects, lobsters, shrimp, and crabs. The strong cell wall of fungi permits them to grow in environments where no plants or other organisms can grow. Certain fungi are extremely resistant toplasmolysis(cell shrinkage in hypertonic medium), and can grow in concentrated salt solutions and sugar solutions. Certain fungi are able to withstand high con-centrations of toxic substances. An example is the ethanol-producing yeasts, which grow at extremely high concentrations of ethanol. Other organisms are killed at these ethanol concentrations. Fungi are less adept at surviving than green plants in that fungi require an organic food source, and are only able to grow where large quantities of foodstuffs are present. Fungi cannot, capture their food, they must obtain nutrition by growing directly on or within their food source. For these reasons, they are not commonly found in ocean habitats.

Question:

A luminous object and a screen are placed at a fixed distance D apart. Show that if a converging lens of focal length f, where f < D/4, is inserted between them it will produce a real image of the object on the screen for two positions separated by a distanced ={ \surd(D - 4f)}, and that the ratio of the two image sizes for these two positions of the lens is(D - d)^2 / (D + d)^2.

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Solution:

In both cases, 1/s + 1/s' = 1/f where s is the object distance, and s' is the image distance. ss' / (s + s') = for ss'/D = f. Sinces + s' = D (see figure) ss' = fD(1) ands + s' = D(2) letd = s - s' . d^2 = (s - s')2= (s -s')^2 - 4ss' = D2- 4fD. Therefore d = \surdD(D - 4f). Also, from Eqs. (1) and (2), s + (fD/s) = Dor s^2 - Ds + fD = 0. s= [D\pm \surd(D2- 4fD)] / 2 = (D \pm d) / 2 This is a general formula for s. It has 2 roots, one of which must be s_1, and the other of which must be s2.(see figure ). s_1 =(1/2)(D-d)and S_2 = (1/2)(D + d) From (2) , s'_1= D - s'1= (1/2)(D-d)ands'2= D - s_2 = (1/2)(D - d). Therefore S_2 - S_1 = d, and the two positions of the lens are separated by the distance d = \surd[D(D - 4f)]. Also, the magnifications for the 2 positions are m_1 = -(s'1/ s_1) = y-1/ yand m2= -(s'_2 / s_2) = y2/ y. \thereforey_2 / y = m_2 / m1= (s_1 s'_2)/(s'_1 s_2) ={1/4(D - d)^2} / {1/4D + d)^2} =(D - d)^2 / (D + d)^2 . Note that this method only works iff < D/4; for otherwise d^2 is negative and d is thus imaginary.

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Question:

A 40-lb stone is carried up a ramp, along a path making a 30\textdegree angle to the horizontal, to the top of a building 100 ft high. How much work is done? (Neglect friction.)

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Solution:

Work is defined as the component of the force in the direction of the displacement multiplied by the displace-ment, for constant forces. In mathematical terms, W = F^\ding{217} \textbullet S^\ding{217} = FS cos \texttheta where \texttheta is the angle between the force and the displacement. We may compute the length of the ramp because from the figure (part a), sin 30\textdegree = (1/2) = 100/(length ot ramp) Therefore, length of ramp = 200 ft. Since the stone is being carried up the ramp the force is upwards and we see that the angle \texttheta is 60\textdegree (see figure (b)). Hence the work is W = (40 lb)(200 ft)cos 60\textdegree = (40)(200)(1/2) ft-lbs = 4000 ft-lbs. As a check we can use the fact that this work done must equal the change in the potential energy of the stone. ∆PE = (weight) (∆ height) = (40 lb) (100 ft) = 4000 ft-lbs.

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Question:

Besides temperature, what other physical conditions must be taken into account for the growth of bacteria?

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Solution:

Although all organisms require small amounts of carbon dioxide, most require different levels of oxygen. Bacteria are divided into four groups according to their need for gaseous oxygen. Aerobic bacteria can only grow in the presence of atmospheric oxygen .Shigelladysenteriaeare pathogenic bacteria (causingdysentery) which require the presence of oxygen. Anaerobic bacteria grow in the absence of oxygen. Obligate anaerobes grow only in environments lacking O_2.Clostridiumtetaniare able to grow in a deep puncture wound since air does not reach them. These bacteria produce a toxin which causes the painful symptoms of tetanus (a neuromuscular disease). Facultative anaerobic bacteria can grow in either the presence or absence of oxygen.Staphylococcus,a genera commonly causing food poisoning , is a facultative anaerobe. Microaerophilic bacteria grow only in the presence of minute quantities of oxygen.Propionibacterium, a genus of bacteria used in the production of Swiss cheese, is a microaerophile. The growth of bacteria is also dependent on the acidity or alkalinity of the medium. For most bacteria, the optimum pH for growth lies between 6.5and7.5, although the pH range for growth extends from pH 4 to pH 9. Some exceptions do exist, such as thesulfer-oxidizing bacteria:Thioba- cillus thiooxidansgrow well at pH 1. Often the pH of the medium will change as a result of the accumulation of metabolic products. The resulting acidity or alkalinity may inhibit further growth of the organism or may actually kill the organism. This phenomenon can be prevented by addition of a buffer to the original medium. Buffers are compounds which act to resist changes inpH.During the industrial production of lactic acid from whey by Lactobacillusbulgaricus, lime, Ca(OH)_2, is periodically added to neutralize the acid. Otherwise, the accumulation of acid would retard fermentation.

Question:

Give a brief history of the six generations of computers. What do you think the next generation has in store for us?

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G01-0007.htm

Solution:

The first generation, starting with the ENIAC machine that filled an entire room, is based on "valve technology," where vacuum tubes were used as the basic electronic units to regulate the flow of electricity in much the same way as the flow of water in pipes. The memory was based on a primitive mercury delay line. ENIAC and UNIVAC I (Universal Automatic Computer) were the forerunners of this generation. The second generation was based on "transistor" technology. With transistors replacing the vacuum tubes, these computers were less expensive , more reliable, more compact, and more pow-erful. With this generation , we also begin to see the "symbolic" processing of letters, that is to say text, as well as numbers. The third generation was based on "integrated circuits," also loosely referred to as "chips" that were theby productof space technology . This technology was introduced by the Depart-ment of Defense and the fledgling aerospace program. The fore-runners of this generation were IBM 360 and 370 mainframe com-puters, Honeywell 200 series mainframes, and Digital Equipment Corporation (DEC) PDP-8 and PDP-11 series. This generation of computers normally used ferrous magnetic core memories. The fourth generation was based on LSI (Large Scale Inte-gration) and VLSI (Very Large Scale Integration) technology. This technology used very small area to pack hundreds of tran-sistors and circuitry. This technology for processor (e.g. CPU: Central Processing Unit) design was coupled with solid-state semiconductor memories (generally using CMOS manufacturing tech-nology) to yield low cost, compact systems with impressive ex-ecution speeds and primary memory capacities. The fifth generation, a term coined by the Japanese, is not as well defined , except that it heavily depends upon "symbolic computing" where a computer is a symbol-processing engine rather than just a number- cruncher . It is characterized by heavy use of parallelism and knowledge- based system software technologies. The sixth generation, with the advent of "neural networks" and neurocomputing , will be computers that are capable of func-tioning very much like biological brains: Very good in practical tasks that require approximations such as scene analysis and cognition, speech synthesis, etc., but not toogoodat number crunching operations such as multiplying two 18-digit numbers in nanoseconds. We can venture to state the seventh generation as "molecu-lar computing " where the architecture of the computer will uti-lize a large number of molecules that are capable of self-organization and intelligent behavior . In other words, these comput-ers will be "chemical" in nature rather than "electronic." It is too early to give any details regarding this generation ! So, a plausible answer to the second part of the question is that the sixth generation will be neural computing machines and the seventh molecular computing systems.

Question:

One method for determining the molecular weight of large, biologically important molecules, is by measuring the density by standard procedures and determining the average volume occupied by a single molecule by X-ray crystallographic analysis. If a biochemist measures the density of a sample of deoxyribonucleic acid (DNA) as 1.1 g/cm^3 and X-ray analysis of the same sample estimates the volume of a single DNA molecule as 0.91 × 10^-15 cm^3, what is the molecular weight of this type of DNA?

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Solution:

The number density (molecules/cm^3) is the reciprocal of the volume of a single molecule and is re-lated to the density (g/cm^3) by the following formula: number density = 1 / (volume per molecule) = (density / molecular weight) × Avogadro's number The validity of this formula is readily seen by considering the dimensions of the quantities involved. Dividing the density (g/cm^3) by the molecular weight (g/mole) gives a quantity with units of (g/cm^3) / (g/mole) = mole/cm^3 (molar density) . Multiplying this number by Avogadro's number (molecules/mole) gives a quantity with dimensions of (mole/cm^3 ) × (molecules/mole) = molecules/cm^3 (number density). Solving for the molecular weight in the above ex-pression we obtain molecular weight = density × Avogadro's number × volume per molecule = 1.1 g/cm^3 × 6 × 10^23 molecules/mole ×0.91 × 10^-15 cm^3/molecule = 6 × 10^8 g/mole.

Question:

Write a PL / I program to initialize, search, insert and delete items in a one-way list.

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Solution:

To simulate a one-way list in PL / I we shall use two dimensional array. The first column stores the Data and second column the pointers to next OWL element. ONE-WAY LIST / \textasteriskcentered OWL: TWO DIMENSIONAL ARRAY \textasteriskcentered / / \textasteriskcentered 1st ENTRY OP OWL IS DATA, 2nd ENTRY IS THE POINTER \textasteriskcentered / / \textasteriskcentered TO THE NEXT OWL ELEMENT, HEAD IS PNTR \textasteriskcentered / / \textasteriskcentered TO THE HEAD OP LIST \textasteriskcentered / /\textasteriskcentered INITIALIZE \textasteriskcentered/ DCL OWL (10, 2) FIXED (5); DCL HEAD FIXED (5) INIT (\cyrchar\CYRF); / \textasteriskcentered INITIAL CHECK AND FIRST INSERT \textasteriskcentered / IF HEAD = \cyrchar\CYRF THEN DO; / \textasteriskcentered OWL IS EMPTY \textasteriskcentered / HEAD = J; / \textasteriskcentered J CAN HAVE A VALUE FROM 1 TO 10 \textasteriskcentered / OWL (J, 1) = X; / \textasteriskcentered INSERT X IN OWL (J, 1) \textasteriskcentered / OWL (3, 2) = \cyrchar\CYRF; / \textasteriskcentered REARRANGE THE PNTR \textasteriskcentered / / \textasteriskcentered SEARCH AND INSERT \textasteriskcentered / / \textasteriskcentered SEARCH OWL ELEMENT KEY AGAINST THE INSETING ELEMENT, INSERT \textasteriskcentered / / \textasteriskcentered AFTER KEY REARRANGE THE POINTERS \textasteriskcentered / NEXT = HEAD; DO WHILE OWL (NEXT, 1)\rceil = KEY; / \textasteriskcentered SEARCH FOR KEY \textasteriskcentered / NEXT = OWL (NEXT, 2); / \textasteriskcentered GOES ON MATCHING DOWN \textasteriskcentered / IF NEXT = \cyrchar\CYRFTHEN [go to error routine]; / \textasteriskcenteredTHE LIST, \cyrchar\CYRF PTR INDICATES LAST ELEMENT \textasteriskcentered / END; OWL [Insert, 2] = OWL [NEXT, 2]; / \textasteriskcentered PTR OF KEY ELEMENT IS TO BE ASSIGNED TO INSERT \textasteriskcentered / OWL (NEXT, 2) = INSERT / \textasteriskcentered NOW OWL (NEXT, 2) POINTS TO INSERT \textasteriskcentered / / \textasteriskcentered SEARCH AND DELETE \textasteriskcentered / Initially NEXT points to the top element and there is no previous pointer. But as the search proceeds NEXT points out the next element and PREV points to the element having pointer NEXT. We want to delete B. The goal is to make PREV point to C and if the match is successful for the 1^st element A we want to change the value of the head. NEXT = HEAD; PREV = 0 DO WHILE (OWL (NEXT, 1)\rceil = KEY) ;/ \textasteriskcentered SEARCH \textasteriskcentered / PREV = NEXT NEXT = OWL (N, 2)/ \textasteriskcentered LOOPING TO SEARCH NEXT ELEMENT \textasteriskcentered/ IF NEXT = \cyrchar\CYRF THEN [go to error routine];/ \textasteriskcentered NO MATCH \textasteriskcentered / END; IF PREV=0 THEN HEAD = OWL (NEXT,2) / \textasteriskcentered MATCH SUCCEEDS AT FIRST ELEMENT \textasteriskcentered / ELSE OWL (PREV, 2) = OWL (NEXT, 2); / \textasteriskcentered MATCH SUCCEEDS IN THE LIST END;AND REARRANGE POINTERS \textasteriskcentered /

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Question:

A cylinder rests on a horizontal rotating disc, as shown in the figures. Find at what angular velocity, \omega, the cylinder falls off the disc, if the distance between the axes of the disc and cylinder is R, and the coefficient of friction \mu = D/h , where D is the diameter of the cylinder and h is its height.

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Solution:

The centripetal force that keeps the cylinder at rest on the disc is the frictional force f. According to Newton's third law of motion, the cylinder reacts with an equal and opposite force, F, which sometimes is referred to as the centrifugal force, F = M \omega^2R where M is the mass of the cylinder. The cylinder can fall off either by slipping away (Fig. A) or by tilt-ing about point P (Fig. B), depending on whichever takes place first. The critical angular speed, \omega_1, for slipping occurs when F equals f; F = f M \omega^2_1 R = \mugM where g is the gravitational acceleration. Hence \omega_1 = \surd(\mug/R). F tries to rotate the cylinder about P, but the weight W opposes it. The rotation becomes possible, when the torque created by F is large enough to take over the opposing torque caused by W; F (h/2) = W(D/2) F = W(D/h) M \omega^2_2 R = Mg (D/h) giving\omega_2 = \surd(D/hR) Since we are given that \mu > D/h , we see that \omega_1 > \omega_2 and the cylinder falls off by rolling over at \omega = \omega_2 .

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Question:

The following reaction is performed at 298\textdegreeK. 2NO (g) + O_2 (g) \rightleftarrows 2NO_2 (g). The standard free energy of formation of NO (g) is 86.6 kJ/mole at 29 8\textdegreeK; find the standard free energy of formation of NO_2 (g) at 298\textdegreeK.k_p= 1.6 × 101 2.

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Solution:

Free energy, ∆G\textdegree, is a measure of the useful work that a system can perform. There is a quantitative relation between standard free energy and the equilibrium constant at a constant temperatures: ∆G\textdegree = - RT Ink_p, where R = the universal gas constant, which equals 8.31 J/mole deg, T = temperature in Kelvin (Celsius plus 273\textdegree) andk_p= the equilibrium constant for the reaction. You have all the information to find the ∆G\textdegree for the en-tire reaction. Therefore, to find ∆G\textdegree_NO2, you (1) find ∆G\textdegree for the reaction. (2) Use the fact that ∆G\textdegree _react = 2∆G\textdegree_NO2 - 2∆G\textdegree_NO - ∆G\textdegree_O2 to find ∆G\textdegree NO2 . ∆G\textdegree = - RT Ink_p = - (8.31 J mol^-1 deg^-1) (298) In 1.6 × 101 2 = - (8.31 J mol^-1 deg^-1) (298) (2.303) log (1.6 × 101 2) = - 69.6 kJ/mole. ∆G\textdegree _react = 2∆G\textdegree_NO2 - 2∆G\textdegree_NO - ∆G\textdegree_O2 The ∆G\textdegree for the entire reaction must just be the ∆G\textdegree of the product minus the ∆G\textdegree's of the reactants. ∆G\textdegree _react = 2∆G\textdegree_NO2 - 2∆G\textdegree_NO - ∆G\textdegree_O2 You have calculated ∆G\textdegree overall and are given ∆G\textdegree_NO. ∆G\textdegree_O2 is zero, since the ∆G\textdegree of any element in its standard state is taken as zero. To find ∆G\textdegree_NO2, therefore, substitute in these values and solve for it. Thus, ∆G\textdegree_NO2 = (1/2) [∆G\textdegree + 2∆G\textdegree_NO] = (1/2) (- 69.6 + 2 × 86.6) = 51.8 kJ/mole.

Question:

Assuming complete ionization, calculate (a) the pH of 0.0001 NHCl, .(b) thepOHof 0.0001 N KOH.

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Solution:

(a) pH is defined as the negative log of the hydrogen ion concentration. pH = - log [H^+] The normality of an acid is defined as the number of equivalents of H^+ per liter of solution. The ionization ofHClcan be written HCl\leftrightarrows H^+ +Cl^- This means that there is one H^+ for everyHCl, and that the concentration of H^+ equals the concentration of completely ionizedHCl. [H^+] = [HCl] We are told that [HCl] = 0.0001 N = 1 × 10^-4 N. Therefore, [H^+] = 1 × 10^-4 N. We can now solve forpH.Note: In this problem, normality =molarity (concentration), since equi-valent weight = M.W. pH = - log [H^+] pH = - log (1 × 10^-4) = 4. The pH of this solution is 4. (b) ThepOHis defined as the negative log of the OH^- ion concentration. pOH= - log [OH^-]. The ionization of KOH can be stated KOH \leftrightarrows K^+ + OH^- Therefore, one OH^- is formed for every KOH, and when KOH is completely ionized, their concentrations are equal. [KOH] = [OH^-] We are told that [KOH] = 0.0001 N, thus [OH^-] = 0.0001 N (again, normality =molarity.) Solving forpOH: pOH= - log [OH^-] = - log (0.0001) = - log (1 × 10^-4) = 4. ThepOHof this solution is 4.

Question:

What must be the length of a simple pendulum that will have a period of 1 sec at the surface of the Earth?

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Solution:

In the simple pendulum figure, the torque about point 0 is \cyrchar\cyrt = - mglsin \texttheta, wherel is the length of the string, and the negative sign is introduced to show that the, torque acts to decrease \texttheta. For \texttheta less than 30\textdegree, \texttheta \cong sin \texttheta is a good approxima-tion, thus: \cyrchar\cyrt = -mg l\texttheta We note that this equation resembles the standard form for simple harmonic oscillators: F = -kx where mglis analogous to k, and \texttheta to x. We find the expression for the period T of a simple pendulum by analogy with the equation for that of the harmonic oscillator: T = 2\pi\surd(m/k) thus: T = 2\pi\surd(I/mgl) = 2\pi\surd(ml^2/mgl) = 2\pi\surd(l/g) where I = ml^2 is the moment of inertia of the mass m about point 0. From the above equation: l = T^2g/4\pi^2 l = [(1 sec)^2 (9.8 m/sec^2)] / [4(3.14)^2] = 0.248 m

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Question:

Electrons with energies of 1 MeV are injected into the vacuum tube of a betatron of radius 1 m, through the center of which is a magnetic flux which changes at the rate of 100 Wb \textbullet s^-1 . The electrons make 250,000 re-volutions before being ejected from the betatron. What is their final energy?

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Solution:

A betatron (see Fig. A) basically consists of an evacuated tube in the shape of a circular ring with a magnetic field passing through the center of the ring. The magnetic field is slowly increased. This induces an increasing tangential electric field around the circumference of the ring by Faraday's Law. An electron placed in the ring will be accelerated around the ring by this electric field. The tube is evacuated to pre-vent collisions between the accelerating electrons and air particles. The force due to the changing magnetic flux acting on the electron is found using Faraday's law and the definition of emf \epsilon. \epsilon \equiv^-\int_C E^\ding{217} \textbullet dl^\ding{217} = - (d\varphi/dt) \equiv - \varphi(1) The line integral is taken about the contour C defined by the circumference of the circular ring. E^\ding{217} is the electric field vector induced by the changing magnetic field and dl^\ding{217} is an element of arc length (see figure B). \varphi is the rate of change of magnetic flux through the contour. The magnitude of E^\ding{217} is constant at all points on the contour. E^\ding{217} is also parallel to dl^\ding{217} at any point on the contour. Thus (1) becomes E \int_C dl = E(2\pir) = \varphi(2) where r is the radius of the circular contour within the evacuated tube. But by definition of electric field, E is equal to the force per unit charge, or E = (F/q) The work, then, done on the electron (of charge e) by the induced electric field in moving the electron once around the ring is the product of the force act-ing on it times the distance it moves. Or W = Fs = eE(2\pir) = e\varphi Here, we used equation (2). This is equal to the energy gained by the electron in one revolution. Making use of the electronvolt-joule conversion factor, W = e\varphi Joules × [(1 eV)/(1.6 × 10-1 9J)] = (1.6 × 10-1 9\varphi)J × [(1 eV)/(1.6 × 10^-19 J)] = \varphi eV The total energy the electrons acquire while in the betatron is thus the energy gained per revolution times the number of revolutions performed ( = 250, 000). \thereforeW = 100 × 250, 000 eV = 25 MeV. The final energy is thus E = 1 MeV + W = 26 MeV.

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Question:

Briefly explain thesignificance of the ASSEMBLER language.

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Solution:

Any high level-language, such as BASIC, FORTRAN, COBOL, and others, is a very convenient and relatively simple set of instructions that a programmeruses to record a program needed to be executed. However, theselanguages are not readable by a computer, and have to be translatedinto an intermediate language, which, in turn, is changed into binarycode before the actual execution pro-cess starts. This intermediate language is called the ASSEMBLER language. It is not a mandatory rule to use a high level-language for writing a program. A program can be written in assemb-ler language. This saves a computersome execution time by eliminating the process of translation of thehigh-level in-structions into assembly instructions. It adds, however, the procedureof checking the input assembly program for syntax errors. But theloss of time from this check is minimal compared to the gain of time fromeliminating the translation step. Being the next step from high-level language coding to the actual executionprocess, the assembler language is called a low-level programminglanguage.

Question:

Calculate the value of the universal gravitational constant, G, assuming that the earth's radius is 6.4 × 10^6 m the earth has an average density of 5.5 × 10^3 kg/m^3. The volume V of a sphere is V = (4/3)\pir^3 , where r is the radius of the sphere.

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Solution:

Begin by computing the volume of the earth: V = (4/3)\pir^3 = (4/3\pi)(6.4 × 10^6m)^3 = 1.1 × 10^21m^3. Since Density = Mass/Volume, we have Mass = Density × volume or, the total mass of the earth, m_2 = (1.1 × 10^21m^3)(5.5 × 10^3 kg/m^3) = 6.0 × 10^24 kg. If an object of mass m_1 is placed on the Earth's sur-face, the gravitational force of attraction (W) between it, and the Earth (i.e., its weight) is found by use of Newton's second law. W = m_1g where g is the acceleration of the object due to the gravi-tational force (i.e., acceleration of gravity). According to the Universal Law of Gravitation, however, the gravitational force of attraction between m_1 and m_2 is W = [Gm_1m_2/d_2] where d is the radius of the Earth. Then [Gm_1m_2/d^2] =m_1g After rearranging, G = (gd^2/m_2) = [{(9.8 m/s^2)(6.4 × 10^6 m)^2}/(6.0 × 10^24 kg)] = 6.7 × 10^-11 Nm^2/kg^2 Since the value of the earth's radius is more accurately known today, Newton's estimate of G differed from this calculated value.

Question:

A 5 -kg block slides down a frictionless plane inclined at an angle of 30\textdegree with the horizontal as shown in figure (a). Calculate the amount of work W done by the force of gravity for a displacement of 8 m along the plane.

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Solution:

We will solve this problem first by the dynamics method and then by the energy method. The formula for calculating work is: W = F^\ding{217} \textbullet s^\ding{217} = F s cos \texttheta where \texttheta is the angle between the force F^\ding{217} and the dis-placement s^\ding{217} of the mass in question. We see from figure (a) that the angle between F^\ding{217} and s^\ding{217} is 60\textdegree: W = Fs cos 60\textdegree = (1/2) mgs = (1/2)(5 kg) (9.8 m/sec^2) (8 m) = 196 kg-m^2/sec^2 = 196 Joules Another way to solve this problem is to calculate the difference in gravitational potential energy that the block goes through as it slides 8 m down the incline. We know that this equals the amount of work that gravity does on the block. As the block slides 8 m down the incline it falls through a vertical height ∆h (see figure (b)) : ∆h = 8(sin 30\textdegree)m = 4m The gravitational potential energy difference that the block experiences is: W = ∆E_p = mgh_2 - mgh1= mg(h_2 - h_1) = mg∆h = (5 kg) (9.8 m/sec^2) (4 m) = 196 Joules

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Question:

Describe the phenomena of the critical angle in optics. What is the critical angle for a glass-air interface, if the index of refrac-tion of glass with respect to air is 1.33?

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Solution:

Consider two media,1 and 2,such that the index of refrac-tion of 1 with respect to 2 is less than unity, as shown in Fig. A. That is, medium 1 is "denser", and the angle of refraction will be greater than the angle incidence. In general part of the incident ray is reflected and part re-fracted. As the angle of incidence is increased, the angle of refraction will increase until r = 90\textdegree. At the critical angle ,i_c (see Fig. B), we have (sin i_c/sin 90\textdegree) = sin i_c = n_21 At the critical angle, and for values of i greater than i_c, refraction cannot occur and all the energy of the incident beam appears in the reflected beam. This phenomenon is called total internal reflec-tion. The index of refraction of air with respect to glass is (1/1.33) = 0.75. The critical angle for a glass-air interference is therefore sin i_c = 0.75 or i_c = 48.6\textdegree Thus, for angles of incidence \geq 48,6\textdegree, total internal reflection will occur for a glass-air combination similar to the one shown in the figure.

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Question:

At 395 K, chlorine and carbon monoxide react to form phosgene, COCI_2 , according to the equation CO (g) + Cl_2 (g) \rightleftarrows COCI_2 (g). The equilibrium partial pressures are PCI_2 = 0.128atm, P_CO = 0.116atm, and P(COCl)2= 0.334atm. Determine the equilibrium constantK_pfor the dissociation of phosgene and the degree of dissociation at 395 K under a pressure of 1 atm.

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Solution:

After obtaining an expression forK_pfor the dissociation of phosgene, the degree of dissociation under 1atmof total pressure will be obtained by combiningK_pwith Dalton's law of partial pressures. The dissociation of phosgene may be written as COCI_2 (g) \rightleftarrows CO (g) + Cl_2 (g) . By definition,K_pis the product of the partial pressures of the products divided by the product of the partial pressure of the reactants. Hence, K = { P_CO P(Cl)2} / {P(COCl)2} = {(0.116atm) (0.128atm)} / {.334atm} = 0.0444 atm. K_pfor the dissociation of phosgene is thus 0.0444 atm. Let \alpha denote the fraction of the original number of moles of phosgene that decomposed. Then, 1 - \alpha is the fraction of the original number of moles of phosgene re-maining. From thestoichiometryof the dissociation re-action, one mole of CO and one mole of Cl_2 are formed for every mole of phosgene that decomposes. Thus, \alpha moles of phosgene decomposes to \alpha moles of CO and \alpha moles of CI_2. From Dalton's law of partial pressures, P_(C0Cl)2 = X_(C0Cl)2 P_T P_CO = X_CO P_TandP(Cl)2= X(Cl)2P_T where X(COCl)2is the mole fraction of phosgene, X_CO is the mole fraction of CO, X(Cl)2is the mole fraction of CI_2 , and P_T is the total pressure. Mole fraction may be defined as the number of moles of that particular substance divided by the total number of moles present. Since the total number of moles present after \alpha moles of phosgene decomposes is (1 - \alpha)(from remaining phosgene) + \alpha (from CO) + \alpha (from Cl_2 ) = 1 + \alpha we have P(COCl)2= X(COCl)2P_T= [(1 -\alpha) / (1 + \alpha)] P_T P_CO = X_CO P_T= [\alpha / (1 + \alpha)] P_T and P(Cl)2= X(Cl)2P_T = [\alpha / (1 + \alpha)] P_T Substituting these into the expression forK_p, we obtain K_p= {P_CO P(Cl)2} / {P(COCl)2} = {([\alpha / (1 + \alpha)] P_T ) ( [\alpha / (1 + \alpha)] P_T )} / ( [ (1 -\alpha) / (1 + \alpha) ] P_T ) = {\alpha^2 / [(1 + \alpha) / (1 - \alpha) ]} P_T = { \alpha^2 / (1 -\alpha^2 ) } P_T , Now, the total pressure is P_T = 1atmandK_phas been determined as K_p= 0.0444 atm. Hence , K_p= {\alpha^2 / (1 -\alpha^2)} P_T , 0.0444atm= {\alpha^2 / (1 -\alpha^2)} × 1atm, 0.0444-0.0444 \alpha^2 = \alpha^2 , or\alpha = (0.0444) / (1.0444)^1/2 = 0.206. The degree of dissociation of phosgene is equal to the fraction, \alpha, of original moles that have dissociated. Hence, the degree of dissociation of phosgene at 395 K under a pressure of 1atmis 0.206.

Question:

Examine the following program and explain the usage of Subprograms (i.e., Subroutines) COMPLEX X1, X2 COMM0N/QUADR/X1, X2, A,B,C DATA INPUT /60/ 7STOP END SUBROUTINE ROOTS COMPLEX X1, X2, W COMMON/QUADR/X1, X2, C1, C2, C3 W = C2 {_\ast}{_\ast} 2 - 4.0 {_\ast} C1 {_\ast} C3 W = W {_\ast}{_\ast} 0.5 X1 = (-C2 + W) / (2.0 {_\ast} C1) X2 = (-C2 - W) / (2.0 {_\ast} C1) RETURN END

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Solution:

This is a complete program which consists of a MAIN PART and a SUBPROGRAM. The MAIN part reads in data for the coefficients of a quadratic equation and then calls upon a subprogram to find its roots. The following points are significant in examining the program. common block named QUADR. QUADR - X1, X2, C1, C2, and C3. Getting into a more detailed explana-tion of Subprograms: Every subroutine program begins with a statement of the form SUBROUTINE NAME. There should be at least one blank space between the word SUBROUTINE and its name. The name of the subroutine must not appear anywhere else within the subprogram. Every subprogram has END as its last statement, and must contain at least one RETURN statement in order to continue the execution of the MAIN program from the next statement after the one that called this subroutine. A subroutine is entered from another program unit (main program or a different program) by means of a CALL statement. The general form of the CALL statement is: CALL name of the subroutine A subroutine may assign values to one or more of its arguments in order to "return" results to the main part of the program. Furthermore, there is the possibility of storing results in COMMON AREA so that values obtained in a subprogram could be available to other program units.

Question:

Write a routine that implements shell sorting. Assume there is an array of integers and size of array is known in the main program. The input parameters to the routine are a) array. b) size of array.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G17-0439.htm

Solution:

The shell sorting algorithm 1) Compares elements that are far apart and then compares adjacent elements. 2) It uses another: routine, values _ swaped ( ) where the values sent are interchanged. 3) The logic used is that number of elements in an array is divided by 2, then element {0] and element [no. of ele-ments/2] are compared. The difference between [no. of elements/2 - 0] is a constant for the first pass and is referred in the program as difference. shell _ sort (array, number _ of _ elements) int array [ ]; int number _ of _ elements; { int i; int difference = number _ of _ elements/2; int swap _ occurred; /\textasteriskcenteredTRUE if a swap occurred \textasteriskcentered/ do do { swap _ occurred =0; for (i = 0; i < number _ of _ elements - difference; i++) if (array [i] > array [i + difference]). { values _ swaped (&(array[i]), &(array [i + difference])); swap _ occurred = 1; } } while (swap _ occurred); while ((difference/=2) > 0); /\textasteriskcenteredmeaning (difference = difference/2) >0 \textasteriskcentered/ } values _ swaped (value1, value2) int \textasteriskcenteredvalue1 = \textasteriskcenteredvalue2; { int temp; /\textasteriskcenteredstore temporary values \textasteriskcentered/ temp = \textasteriskcenteredvalue1; \textasteriskcenteredvalue1 = \textasteriskcenteredvalue2; \textasteriskcenteredvalue2 = temp; } } C REFERENCE GUIDE C REFERENCE GUIDE Arithmetic Operators + addition. - subtraction. \textasteriskcentered multiplication. / division. % modulus. Operators unique to C ++ increment. -- decrement. Assignment operators operator = eg. a\textasteriskcentered = 5; a = a \textasteriskcentered 5; a% = 5; a = a % 5; Bitwise Operators &bitwise AND. !bitwise OR. ^bitwise exclusive - or. <>shift right. \simone's complement. Break Statement: The break statement is used within loops and the switch statement. Within A loop the break statement causes the execu-tion of the loop to cease and execution of the program to con-tinue at the statement that follows the loop. Format is break ; Global Variables: In C the use of global variables is done usingexternstatement. It is done as follows type variable _ name; /\textasteriskcenteredoutside of main \textasteriskcentered/ main ( ) { extern type variable _ name; /\textasteriskcenteredextern tells the compiler that the \textasteriskcentered/ /\textasteriskcenteredvariable is a global one\textasteriskcentered/ Relational Operators: greater than <=less than or equal to >=greater than or equal to Relational operators unique to C ==equal to !=not equal to Ternary Operator : ?:conditional operator. It is a conditional operator eg. conditional - expression: expression 1? expression 2: expression 3 Conditional expressions group right to left. They operate as follows a) If expression 1 (i.e. expression to be evaluated) is TRUE, expression 2 will be performed. If it is FALSE expression 3 will be performed. Logical AND Operator: logical - AND - expression expression 1 && expression 2 Logical OR Operator: logical-OR-expression expression 1 \vert\vert expression 2. Type Specifiers: type - Specifiers are char short int long unsigned float double stifuct -or-union specifier typedef name.

Question:

The plates of a parallel plate capacitor are 5 mm apart and 2m^2 in area. The plates are in vacuum. A potential dif-ference of 10,000 volts is applied across the capacitor. Compute (a) the capacitance, (b) the charge on each plate, and (c) the electric intensity.

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/Users/wenhuchen/Documents/Crawler/Physics/D18-0606.htm

Solution:

(a)C_0 = \epsilon_0(A / d) = 8.85 × 10^\rule{1em}{1pt}12 × [2 / (5 × 10^\rule{1em}{1pt}3)] = 3.54 × 10^\rule{1em}{1pt}9 farad = 3.54 × 10^\rule{1em}{1pt}3 \muf = 3540 \mu\muf. (b)The charge on the capacitor is q = CV_ab = 3.54 × 10^\rule{1em}{1pt}9 × 10^4 = 3.54 × 10^\rule{1em}{1pt}5 coulomb. (c)The electric intensity is E = (1 / \epsilon_0) \sigma = (1 / \epsilon_0) (q / A) = (36\pi × 10^9) × (1.77 × 10^\rule{1em}{1pt}5) = 20 ×10^5 volts/meter. The electric intensity may also be computed from the poten-tial gradient. E = V_ab / d = 10^4 / (5 × 10^\rule{1em}{1pt}3) = 20 × 10^5 volts / meter.

Question:

A laboratory technician has a solution of 0.1 M NaC_2H_3O_2. Find its pH and the percent hydrolysis. AssumeK_w= 1.0 × 10^-4 andK_diss= 1.8 × 10^-5.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E12-0414.htm

Solution:

NaC_2H_3O_2 is a salt that exists as ions Na+ and C_2H_3O_2^- in water. Since water is a weak acid and weak base, which dissociates to H^+ and OH^- ions, it can react with ions from weak acids and bases. Such a process, termed hydrolysis, occurs in this problem. Because the problem asks for pH, one must determine the concentrations of the H_3O^+ or OH^- in this problem. This means that the hydrolysis reaction must be written and an equilibrium constant set up. The net hydrolysis reaction may be written as . C_2H_3O_2^- + H_2O \rightleftarrows HC_2H_3O_2 + OH^-. TheK_hydfor this reaction is {[OH^-] [HC_2H_3O_2]} / [C_2H_3O_2^-] =K_w/K_diss. However,K_hydis defined asK_w/K_diss_' whereK_wis the dissociation constant for water andK_dissis the dissociation constant for the acid formed. One can equate the two to obtain {[OH^-] [HC_2H_3O_2]} / [C_2H_3O_2^-] =K_w/K_diss. But the value ofK_wandK_dissis known. Therefore, {[OH^-] [HC_2H_3O_2]} / [C_2H_3O_2^-] = (1.0 × 10^-14)/(1.8 × 10^-5) = 5.6 × 10^-10. If one can find [OH^-], one can determine [H_3O^+] from [H_3O^+] [OH^-] =K_w= 1.0 × 10^-14. Therefore, let x = moles/liter of OH^-. From the hydrolysis reaction, one can see that for each mole of OH^- generated, a mole of HC_2H_3O_2 is also generated. Therefore, let x = [HC_2H_3O_2] also. If one starts with 0.1 M of C_2H_3O_2^-, and x moles/liter reacts to form OH^- (and C_2H_3O_2^-), then, at equilibrium, one has 0.1 - x moles/liter of C_2H_3O_2^- left. Substituting these values into the previous expression: (x \bullet x) / (0.1 - x) = 5.6 × 10^-6. Solving for x, x = 7.5 × 10^-4 M. (assuming in this calculation that x is small, therefore approximating 0.1- x as 0.1.) To find [H^+], solve [H_3O^+] = K_W / [OH^-] = (1.0 × 10^-14)/(7.5 × 10^-4) = 1.33 × 10^-11. Since pH = - log [H_3O^+] pH = - log [1.33 x 10^-11] = 10.88. The percent hydrolysis of this reaction is the number of moles of hydrolyzed acetate ion, divided by the number of moles available for hydrolysis, times 100. % Hydrolysis = {(Moles C_2H_3O_2^- hydrolyzed) / (Moles C_2H_3O_2^- available) × 100 = [(7.5 × 10^-4) / 0.1] × 100 = 0.75%.

Question:

Distinguish between crystalline and amorphous solid substances, using some specific examples. To what extent is the distinction useful?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E06-0214.htm

Solution:

Crystalline substances can be generally thought of as species composed of structural units with specific geometric patterns. The accompanying drawing of sodium chloride would be an example of such a pattern. The important point is that there exists a regularity in the arrangement of structural units. Structures with regularity generally show a sharp and characteristic melting point, which is the case with crystalline substances. Amorphous substances, however, tend to be shapeless and without definite order. That is, you have a randomness. For example, glassy or glasslike materials such as Plexi-glas and silicate glasses. In substances with a general lack of order, the melting points vary over a range or temperature interval. For amorphous substances, this is exactly what you find. It would, however, beincorrectto state categorically that amorphous substances are without ANY order. For they do tend to have short range order even though they do contain long-range randomness.

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Question:

Under standard temperature and pressure conditions, compare the relative rates at which inert gases,Ar, He, and Kr diffuse through a common orifice.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E03-0094.htm

Solution:

This problem involves the application of Graham's law of diffusion. It states that the relative rates at which gases will diffuse will be inversely proportional to the square roots of their respective densities or molecular weights. That is, rate \alpha 1/ \surdmass. Thus, to compare the rates of diffusion ofAr, He, and Kr, look up their weights in the Periodic Table of Elements and substitute this value into 1/ \surdM \alpha\mu_m, where M = mass of that element and\mu_m= rate. Therefore,\mu_Ar: \mu_He :\mu_Kr = 1/ \surdM_Ar: 1/ \surdM_He: 1/ \surdM_Kr = 1/ \surd39.95 : 1/ \surd4.003 : 1/ \surd83.80 = .1582 : .4998 : .1092.

Question:

Two different molecules, A and B, were reacted to give products according to the equation A + B \rightarrow products. The rate of reaction was measured for fixed concentrations of A and B and the following data was obtained: Experiment [A] A (moles/liter) [B] B (moles/liter) Rate (moles^2/liter^2 - sec) 1 1.0 1.0 0.05 2 1.0 2.0 0.10 3 3.0 1.0 0.45 Write an expression for the rate of reaction.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E13-0443.htm

Solution:

The rate of reaction is equal to some rate constant, k, multiplied by the concentrations of A and B raised to appropriate powers. That is, rate = k [A]^m [B]^n where the exponents m and n are to be determined. Comparing experiment 2 with experiment 1, we see that when the concentration of B is doubled from 1.0 M to 2.0 M, the concentration of A being held constant at 1.0 M, the rate of reaction doubles from 0.05 to 0.10 moles^2/liter^2 - sec, Hence, the rate of reaction is directly proportional to the concentration of B, and the exponent of [B] in the rate expression is n = 1. If we were to triple [B], holding [A] constant, the rate would triple, and so on. Comparing experiment 3 with experiment 1, we see that when the concentration of A is tripled from 1.0 M to 3.0 M, the concentration of B being held constant at 1.0 M, the rate of reaction is multiplied by 9 (9 × 0.05 = 0.45 moles^2/liter^2 - sec). Since the factor by which the rate is multiplied (9) is the square of the factor by which the concentration of A is multiplied (3), the exponent of [A] in the rate expression must be m = z. Thus if [A] is doubled, [B] being held constant, the rate is multiplied by 4. Substituting the exponents of [A] and [B] determined above into the rate expression gives rate = k [A]^2 [B].

Question:

What is meant by translocation? What theories have been advanced to explain translocation in plants? Discuss the value and weakness of each.

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Solution:

Translocation is the movement of nutrients from the leaves where theyare synthesized to other parts of the plant body where they are neededfor a wide variety of metabolic activities. Translocation takes place throughthe sieve tubes of the phloem. Experimental data indicates that the high rate at which translocation occurs cannot be attributed to simple diffusion. Three theories have been offered to explain the mechanism of translocation. Each of them has its own value and weakness, but the first one presented (pressure-flow theory ) holds the most support. The pressure-flow theory, which had been widely accepted in the past , proposes that the nutrient sap moves as a result of differences in turgor pressure. Sieve tubes of the phloem in the leaf contain a high concentration of sugars, which results in high osmotic pressure into the cells. This osmotic pressure causes an influx of water, and build up ofturgor pressure against the walls of the sieve tube cells. In the roots however, sugars are constantly being removed. Here, a lower concentration of solute is present resulting in a lowering of the osmotic pressure. There is consequently a lowerturgorpressure exerted against the sieve tube walls of the root as compared to the leaf. This difference inturgorpressure along the different regions at the phloem is believed to bring about the mass flow of nutrients from a region of high turgor pressure - such as the leaves (where photosynthe-sis produces osmotically -active substances like glucose) -toregions of lowerturgor pressure - such as the stem and roots. The fluid containing the nutrients is pushed by adjacent cells, along the gradient of decreasingturgor, from the leaves to the roots. This theory predicts that the sap should be under pressure as it moves down the phloem, and this is experimentally verified. But there are problems with this hypothesis. Under some conditions, sugar is clearly transported from cells of lesserturgorto cells of greaterturgor. In addition , this theory fails to explain how two substances can flow along the phloem in different directions at the same time, a situation observed to occur by some investigators. Another theory proposes thatcyclosis, the streaming movement evident in many plant cells, is responsible for translocation. According to this theory, materials pass into one end of a sieve tube through the sieve plate and are picked up by the cytoplasm which streams up one side of the cell and down the other. At the other end of the sieve tube, the material passes across the sieve plate to the next adjacent sieve tube by diffusion or active transport. This theory is able to account for the simultaneous flowing of nutrient saps in different direc-tions. However, it is attacked by some investigators on the basis thatcyclosishas not been observed in mature sieve cells. A third theory proposes that adjacent sieve-tube cells are connected bycytoplasmictubules, in which sugars and other substances pass from cell to cell. The movement of these substances is powered by the ATP from the mitochondria-like particles that are believed to lie within these connecting tubules. This theory is, however, weak since it is supported by few experimental findings and much of its content is based on speculation.

Question:

Find the tension in the cable shown in Figure A. Neglect the weight of the wooden boom.

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/Users/wenhuchen/Documents/Crawler/Physics/D02-0025.htm

Solution:

Take the directions of the tensions in the cable and the boom to be as shown in the force diagram(fig. B). We assume at this point, that the given directions are correct. However, the forces may turn out to point in the opposite direction. If this is the case, our solutions for the tensions will be negative. We can thus correct ourselves at the end of the problem. The first condition of equilibrium yields \sumF_x = T_2 cos 60\textdegree - T_1 cos 30\textdegree = 0(1) \sumF_y = T_2 sin 60\textdegree - T_1 sin 30\textdegree - 2000 = 0(2) We wish to find T_1, the tension in the cable. Solving for T_2 in terms of T_1 in equation (1) gives T_2 = (T_1 cos 30\textdegree)/(cos 60\textdegree) Substituting this in equation (2), [(T_1 cos 30\textdegree)/(cos 60\textdegree)] sin 60\textdegree - T_1 sin 30\textdegree = 2000 Solving for T_1: T_1(cos 30\textdegree tan 60\textdegree - sin 30\textdegree) = 2000 T_1 = 2000/(cos 30\textdegree tan 60\textdegree - sin 30\textdegree) = 2000/ [(0.8660)(1.7321) - (0.5000)] = 2000/(1.5 - 0.5) = 2000 lb Since our answer is positive, the force acts in the direction assumed in the beginning.

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Question:

Of the following two pairs, which member will more likely deviate from ideal gas behavior? (1) N_2 versus CO, (2) CH_4 = versus C_2 H_6 .

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/Users/wenhuchen/Documents/Crawler/Chemistry/E02-0074.htm

Solution:

Characteristics of ideal gases include: (1)Gases are composed of molecules such that the actual volume of the molecules is negligible compared with the empty space between them. (2) There are no attractive forces between molecules. (3) Molecules do not lose net kinetic energy in collisions. (4) The average kinetic energy is directly proportional to the absolute tem-perature . From these assumptions came the ideal equation of state: PV = nRT, where P = pressure, V = volume, n = moles, R = universal gas constant, and T = temperature. No gas is ideal; they don't obey these assumptions absolutely. To reflect these limitations the ideal gas law can be written as: [P + (n^2 a/V^2 )] (V - nb) = nRT. The term V = nb can be thought of as representing the free volume minus the volume occupied by the gas molecules themselves. The magnitude of (b) is pro-portional to a gas molecule's size. Thus, the greater the size of a gas molecule, the greater the nb is, and, the greater the deviation, since the volume is assumed to equal exactly V in the ideal Gas Law. V - nb is smaller than V. The greater nb, the smaller it becomes. The term n/V, a concentration term, when squared, gives probability of collisions, (a) gives a measure of the cohesive force between molecules. Real molecules do have an attraction for each other. The greater the value of (a), the larger n^2 a/V^2 becomes, and the larger (P + n^2 a/V^2) gets. The ideal gas law assumes a value of ex-actly p. Thus, as (a) increases, the more deviation. To determine which gas deviates the most, look at the size of the gas molecules and their dipole moment. The dipole moment is an indication of unlike charges separated by a given amount of distance. Unlike charges on separate molecules will be attracted to each other. Thus, a higher dipole moment indicates greater cohesive attraction among molecules and greater deviation. Proceed as follows: (1)N_2 versus CO. The CO gas has a net dipole moment, while N_2 does not. This is because O is more electro-magnetic than C. In N_2 both of the atoms are the same and thus their electronegativities are equal and no net dipole moment exists. Thus, CO deviates to a greater extent. (2)CH_4 versus C_2 H_6. C_2 H_6 is a much larger molecule and occupies more volume. Therefore, it is more likely to deviate from ideal gas behavior.

Question:

A student sets up a Young's experiment using a sodium vapor lamp as a light source and placing his slits 1 m from a screen. He is unsure of the slit separation and, standing beside the slits, he varies the separation and finds the interference pattern vanishes if the slits are too far apart. If the angular resolution of his eye is 1 minute of arc, how far apart are the slits when he just cannot see the interference fringes ?

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Solution:

The position of the mth bright fringe of the fringe system from the center O is given by y_m = mR\lambda/d , where \lambda is the wavelength used (see figure (A)). Similarly, for the ( m + 1) th fringe, y_m+1 = [{(m + 1)R\lambda} / d] Thus, the fringe separation is ym+1- ym= R\lambda/d , a quantity independent of mand thus constant throughout the fringe system. The angle subtended by this fringe separation at the student's eye is (see figure (B)) \texttheta = [(R\lambda/d)/R] = \lambda/d rad. Here, we have approximated the arc shown in figure (B) (dotted line) by the straight line distanceAB. But the student cannot see the fringes as distinct unless the angle neighboring fringes subtend at his eye \geq1 minute of arc. But 1 minute of arc is 1/60\textdegree = \pi/60 × 180 rad. Hence, \texttheta_min = \lambda/d_max = [\pi/(180 × 60)] rad \therefored_max = [(180 × 60 × 5.89 × 10^-5) / (\pi rad)] cm = 2.025 mm.

Question:

A worker hangs a uniform bar of mass 12 kg horizontally from the roof of a laboratory by means of three steel wires each 1 mm in diameter. Two of the wires are 200 cm long and one, by an oversight, 200.05 cm long. The long wire is fastened to the middle of the bar, the others to the two ends. By how much is each wire stretched, and how much of the weight does each wire carry? Young's modulus for steel = 2.0 × 10^12 dynes \textbullet cm^-2.

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Solution:

If the bar is hanging horizontally (see the figure), two of the wires will be extended ∆1 and one ∆1 - 0.05 cm. Now the formula for Young's modulus is Y = stress/strain = (F_n/A)/(∆1/1_0), where A is the cross sectional area of the steel wire, 1_0 is the length of the wire with no stress acting on it, and F_n is the (stretching) force the bar exerts on the wire (equal in magnitude to the. force the wire exerts on the bar). Thus, F_n = (AY∆1)/(1_0) . Hence two wires exert upward forces on the bar of magni-tude F_n, and one wire exerts a force of magnitude F'_n = [AY(∆1 - 0.05 cm)]/[1_0 + 0.05 cm] Since the ratio 0.05/1_0 = .00025 is so small, we may ignore 0.05 in comparison with 1_0 in the denominator of the expression for F'_n. Then because thebar is in equilibrium, we obtain bar is in equilibrium, we obtain W = 2F_n + F'_n = (AY/1_0)(2∆1 + ∆1 - 0.05 cm). Therefore ∆1 = (1/3)[(1_0w)/AY) + 0.05 cm] However A = \piR^2 = \piD^2/4, where R and D are the radius and diameter, respectively, of the wire. We are given D = 1 mm = 1 mm × (1cm/10 mm) = 0.1 cm. Hence ∆1 = (1/3)[{(200 cm × 12 kg × 981 × 10^3 dynes)/(\pi/4 × 10^-2 cm^2 × 2 × 10^12 dynes \textbullet cm^-2)} + 0.05 cm] = (1/3)(0.15 + 0.05)cm = 0.0667 cm = 0.667 mm. Thus two of the wires are stretched by 0.667 mm and the other by (0.667 - 0.05)mm = 0.167mm. Also F_n/F'_n = [∆1/(∆1 - 0.05 cm)] = (0.667 mm)/(0.167 mm) = 4; 12 kg = W = 2F_n + F'_n = 9F'_n. \thereforeF'_n = 1(1/3) kgandF_n = 5(1/3) kg. \thereforeF'_n = 1(1/3) kgandF_n = 5(1/3) kg.

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Question:

The energy of the doubly charged a-particles of mass 6.64 × 10^-27 kg emitted from The is 6.048MeV. What is their velocity and what magnetic field applied per-pendicular to a collimated beam of such particles would bend the beam into a circle of radius 40 cm?

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Solution:

The energy of the \alpha-particles is 6.048MeV = 6.048MeV× 1.602 × 10^-13 J/ 1MeV= 9.689 × 10^-13 J. The energy of the particle refers to its kinetic energy or KE = 1/2 mv^2 = 9. 689 × 10^-13 J, orv = [\surd{(19.378 × 10^-13 J)/(6.64 × 10^-27 kg)}] = 1.709 × 10^7 m \bullet s^-1 . In the magnetic field the magnetic force supplies the centripetal force necessary to keep the particles moving in a circle. This centripetal force, F, due to the magnetic field has magnitude F =qvBsin \texttheta where is the angle between v^\ding{217} and B^\ding{217}. Since we are given that this angle is 90\textdegree, then sin \texttheta = 1. The centripetal acceleration of a particle with velocity v about a circle of radius R is a = (v^2 /R). By Newton's second law, F = (mv^2 /R) =qvBor v = (qBR/m). Therefore B = (mv/qR). The \alpha-particles carry twice the electronic charge (for they contain two positively charged protons and two neutral neutrons). \therefore b = [(6.64 × 10^-27 kg × 1.709 × 10^7 m \bullet s^-1 )/(2 × 1.602 × 10^-19 C × 0.40 m)] = 0.885Wb\textbullet m^-2 .

Question:

A capacitor with capacitance C and initial charge q_0 and an inductor with inductance L are connected in series. At time t = t_0 , the switch Sis closed. If q(t) is the charge on the capacitor at time t it can be shown that the behavior of the circuit is governed by the following differential equation: Lq̇ ̇+ (1/C)q(t) = 0 where q̇ ̇= d^3q/dt^3 . If i_0 is the initial q̇ ̇ current, construct a model of this circuit and devise a digital program which simulates the behavior of the circuit from time t = t0to t = t_f .

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Solution:

Following the steps of the general procedure outlined in the two previous problems: 1.If the given differential equation can be solved, and samples of the values q assumes from t = t0to t = t_f, recorded, we will have a model which effectively simulates the given system, i.e., the circuit. 2.The assumption that L and C are linear and time-invariant has already been included in the model, through the derivation of the given equation. Similarly, the assumption that all the capacitance of the circuit is considered "lumped" at C and all inductance "lumped" at L has already been made. 3.Circuit resistance is neglected for simplicity. 4.The given differential equation is a dynamic mathematical model of the system. 5.The input data is t_0, t_f, q(t_0) = q_0, i(t_0) = i_0, L and C. 6.The equation given here and the differential equation of the previous two problems appear to be similar. In fact, using the equivalences: q \rightarrow x, i = q̇ \rightarrow v = ẋ,q̇ ̇\rightarrow a = v̇ = ẋ ̇, L \rightarrow M, (1/C) \rightarrow K q̇ ̇ the differential equation of harmonic motion can be constructed from the equation of this problem. Hence, the programmer needs only to substitute his input data as follows: USE THE VALUE(1/C) AS K USE THE VALUEL AS M USE THE VALUEq_0 AS XCOR USE THE VALUEi_0 AS VCOR

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Question:

Explain what is meant by reduction of vectors. Also explain the rule for APL arithmetic. Evaluate the expression: 6 + ((5 + 2 × 11 - 2) - 3 + 2) \div 2 × 3.

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Solution:

Reduction of a vector, in APL, is equivalent to placing a dyadic operator between each element of the vec-tor and evaluating the resulting expression. Examples of dyadic operators are plus +, minus -, multiply ×, divide \div, exponentiation \textasteriskcentered, etc. The symbol for reduction is a right-tilting slash, e.g.,/. The use of the symbol is illustrated below for a vector V. +/V, -/V, ×/V, \div/V, \textasteriskcentered/V, etc. If V is defined by V \leftarrow 1 2 3 then each of the above re-ductions is equivalent to the following, respectively. +/V is equivalent to 1 + 2 + 3 -/V is equivalent to 1 - 2 - 3 ×/V is equivalent to 1 × 2 × 3 \div/V is equivalent to 1 \div 2 \div 3 \textasteriskcentered/V is equivalent to 1 \textasteriskcentered 2 \textasteriskcentered 3 The rule for APL arithmetic is a right to left rule. That is, the arithmetic operations are performed on a string of operators from right to left. Items enclosed within parentheses are evaluated first, and the whole parenthesized content acts as a single opera-tor. The given expression evaluates as follows: Expression = 6 + ((5 + 2 × 11 - 2) - 3+ 2) \div 2 × 3 = 6 + ((5 + 2 × 9) - 3+ 2) \div 2 × 3 = 6 + ((5 + 18) - 3+ 2) \div 2 × 3 = 6 + ((23) - 3+ 2) \div 2 × 3 = 6 + (23 - 3+ 2) \div 2 × 3 = 6 + (23 - 5) \div 2 × 3 = 6 + (18) \div 2 × 3 = 6 + 18 \div 2 × 3 = 6 + 18 \div 6 = 6 + 3 = 9Answer Note that the innermost pair of brackets is evaluated out first using the right to left rule inside the brackets. Then the outer pair of brackets is evaluated using the right to left rule inside the brackets. Then, after all the brackets have been eliminated, the whole expression is evaluated using the right to left rule. Also note that the effect of a reduction1; operation performed on the vector V is to reduce it to a scalar quantity, which is the evaluated value of the resulting expression. Now, a vector has a dimension of 1. And, a scalar has a dimension of 0. Thus, the reduction operation has reduced the dimension of the variable V from 1 to 0. In general, it is found that for an n-dimensional ar-ray, a reduction operation reduces the dimension of the ar-ray by 1, from n to n-1. Finally, note that the arithmetical operations are not ordered in a hierarchy unlike FORTRAN, where multiplications and divisions are performed before additions and subtrac-tions .

Question:

A rope C helps to support a uniform 200-lb beam, 20 ft long, one end of which is hinged at the wall and the other end of which supports a 1.0-ton load. The rope makes an angle of 30\textdegree with the beam, which is horizont-al. (a) Determine the tension in the rope, (b) Find the force F\ding{217} at the hinge .

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Solution:

Since all the known forces act on the 20-ft beam, let us consider it as the object in equilibrium. In addition to the 200- and 2000-lb forces straight down, there are the pull of the rope on the beam and the force F which the hinge exerts on the beam at the wall. Let us not make the mistake of assuming that the force at the hinge is straight up or straight along the beam. A little thought will convince us that the hinge must be pushing both up and out on the beam. The exact direction of this force, as well as its magnitude, is un-known. The second condition for equilibrium is an ex-cellent tool to employ in such a situation for if we use an axis through the point 0 as the axis about which to take torques, the unknown force at the hinge has zero moment arm and therefore, causes zero torque. The remark-able result is that we can determine the tension T in the rope without knowing either the magnitude or the direction of the force at 0. (a) The torques about an axis through O are (200 lb) (10 ft)= 2000 lb-ft counterclockwise (2000 lb)(20 ft)= 40,000 lb-ft counterclockwise The moment arm of T is OP = (20 ft)(sin 30\textdegree) = 10 ft. Thus we have T (20 ft) sin 30\textdegree= T(10 ft)clockwise Since the beam is in equilibrium, the torque about any point of the beam is zero. We can then say -T (10 ft) + 2000 lb-ft + 40,000 lb-ft + F(0 ft) = 0 orT = 4200 lb = 2.1 tons The trick just used in removing the unknown force from the problem by taking torques about the hinge as an axis is a standard device in statics. The student should always be on the lookout for the opportunity to sidestep (temporarily) a troublesome unknown force by selecting an axis of torques that lies on the line of action of the unknown force he wishes to avoid. (b) Using the first condition for equilibrium, \sumF_x = 0 = T cos 30\textdegree - F cos \texttheta \sumF_y = 0 = T sin 30\textdegree + F sin \texttheta + W - 20001b The above two equations can be solved simultaneously, since there are two unknowns. Substituting numerical values, F cos \texttheta= T cos 30\textdegree = (4200 lb)(0.866) = 3640 lb F sin \texttheta=-W + 2000 lb - T sin 30\textdegree = 200 lb + 2000 lb - (4200 lb)(0.500) = 100 lb Dividing the first equation by the second, (F cos \texttheta)/(F sin \texttheta)= cot \texttheta = 3640/100 = 36.4 \texttheta\approx 1.5\textdegree Since \texttheta is almost zero, we have from the first equation F cos 1.5\textdegree \approx F cos 0\textdegree = F = 3640 lb.

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Question:

During run of the Millikan oil - drop experiment, the following charges were observed on five different oil droplets: 1.44 × 10^-1 8, 2.56 × 10-1 8, 4.80 × 10-1 9, and 9.60 × 10-1 9 coulomb. Does this support the charge of an electron as being 1.60 × 10-1 9coulomb?

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Solution:

This experiment, illustrated schematically in the above diagram, was first performed in 1909 by Robert A. Millikan. Oil droplets are sprayed between a top plate having a positive charge and a bottom plate having a negative charge. A beam of X-rays is simultaneously di-rected at this spray. The X-rays will knock electrons out of molecules in the air and these electrons will be picked up by the oil droplets, which will acquire a charge that is an integral multiple of the electronic charge. This occurs because each oil droplet will pick up an integral number of electrons. By adjusting the charge on the plates, the upward pull on the electrons due to attraction by the positive plate on the top and repulsion by the negative plate on the bottom can be balanced against the downward pull of gravity. The forces are balanced when the oil drop is stationary, as viewed through a telescope. By equating the gravitational and the electrostatic forces one can determine the charge on an oil droplet. It is given that the following charges were observed on oil droplets: 1.44 × 10^-18, 2.56 × 10^-18, 4.80 × 10^-19, and 9.60 × 10^-19 coulomb. If the elementary electronic charge is 1.60 × 10^-19 coulomb, each of these observed charges must be an integral multiple of the elementary charge. Hence, each of these charges must be exactly divisible by 1.60 × 10^-19 coulombs. [(1.44 × 10^-18 coulomb)/(1.60 × 10^-19 coulomb)] = 9.00, [(2.56 × 10^-18 coulomb)/(1.60 × 10^-19 coulomb)] = 16.00, [(4.80 × 10^-19 coulomb)/(1.60 × 10^-19 coulomb)] = 3.00and [(9.60 × 10^-19 coulomb)/(1.60 × 10^-19 coulomb)] = 6.00. Each of the observed charges is an integral multiple of 1.60 × 10^-19 coulomb, supporting the contention that 1.60 × 10^-19 coulomb is the charge of an electron.

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Question:

The atomic radius of hydrogen is 0.037 nm, Compare this figure with the length of the first Bohr radius. Explain any differences.

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Solution:

The first Bohr radius, r, is equal to (N^2/Z) a_0 where a_0 is .05292 nm, N is an integer, and Z the atomic number. For hydrogen, Z = 1. For the first Bohr radius N = 1. Thus r = [(1)^2/(1)] a_0 = .05292 nm. Thus, the atomic radius of hydrogen (.037 nm) is consider-ably smaller than the first Bohr radius. This can be ex-plained by the fact that the atomic radius is calculated from the bond length of a H_2 molecule. There is consider-able orbital overlap in bonded atoms, which means the bond length is shorter than the sum of the radii of lone atoms.

Question:

Viruses do not have any ribosomes which are the structures needed for protein synthesis. How are they able to syn-thesize their protein coats during their replication? In your explanation, describe the life cycle of a typical virus.

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Solution:

Viruses consist of a nucleic acid core (either DNA or RNA single or double stranded) surrounded by a protective protein coat, which may be further enveloped. There are no mitochondria, ribosomes, or any of the other organelles normally found in eucaryotic cells. How does the virus acquire the energy and the structural machinery (ribosomes) for making proteins? Since viruses have no independent metabolic activity, they are incapable of reproduction by fission, budding, or other simple means of reproduction. Instead, they replicate by inserting their nucleic acid into a functional host cell. The host either cell provides both the energy and structural machinery necessary for the production of new viral components. After these components are assembled, they are released from the host cell and a new cycle of replication begins. The T_4, bacteriophage (a virus which attacks bacteria) may be employed to demonstrate viral replication. T_4 is one of the bacterial viruses which multiplies inE.colicells. The general structure of a bacteriophage, or simply phage, is shown in Figure 1. The replication process begins when the phage par-ticle collides with a sensitive bacterium and the phage tail fibers specifically attach to the bacterial cell wall. Host cells have specific receptor sites to which the phage attaches. An enzyme in the phage tail then breaks down a small portion of the cell wall, creating a small hole in the bacterium. The spikes are believed to help anchor the phage to the bacterium. The tail sheath then contracts, driving the core of the tail into the cell (see Figure IB). The viral DNA is then injected into the bacterium through the enzymatically weakened cell wall. The protein coat is left outside the bacterium. (Most animal viruses are completely engulfed by the cell through a process known as pinocytosis, so that no viral substance remains outside the cell.) The free viral chromosome (DNA molecule) serves as a template to direct the synthesis of new viral components. The DNA serves as a template for its own replication and for the viral-specific RNA necessary for the synthesis of its specific proteins. The viral-specific mRNA molecules attach to the host ribosomes, where viral protein synthesis takes place. Sometimes viral infection does not affect normal cell metabolism, but more often, most of the host cell's metabolism is directed toward the synthesis of new molecules needed for new viral particles. In some cases, all DNA and RNA synthesis on the host chromosome is inhibited and pre- existing bacterial RNA templates (mRNA) are broken down, so that all protein synthesis occurs only on new viral RNA templates.E.coli ribosomes may become modified, so that they work better with T_4, mRNA than with their own mRNA. This causes more viral molecular synthesis and less bacteri-al molecular synthesis. During infection, therefore, both host nucleic acid synthesis and protein synthesis are usual-ly suppressed. Once all the viral products are synthesized (there are over 40 of them), they interact in a definite sequence or morphogenetic pathway to produce the mature viruses. The complete infective phages that are produced must now be released from the bacterium, which has a rigid cell wall. To ensure release, many phages have a gene which codes for lysozyme, an enzyme which degrades bacterial cell walls. Lysozyme begins to be synthesized when the coat proteins appear, so that its accumulation causes rupture of the bacterial cell wall when the phages are mature and ready for release. Usually several hundred viral particles are released from each bacterial cell. These new viral particles then infect otherE.coli, replicating themselves and lysing the bacteria. The life cycle of a T_2 phage is summarized in Figure 3.

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Question:

A particle which is performing simple harmonic motion passes through two points 20 cm apart with the same velocity, taking 1 s to get from one point to the other. It takes a further 2 s to pass through the second point in the opposite direction. What are the period and amplitude of the motion?

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Solution:

From the equations of simple harmonic motion, the velocity at any displacementx has a value v = \omega\surd(A^2 - x^2), where \omega is the angular frequency of the motion and A its amplitude. Thus v can be the same at two points only if the displacements of these points are \pm x. If the origin of the time scale is taken at the mean position of the motion, as shown in the rotor diagram, and if the second point is reached t_0 thereafter, the displacement at the first point is - x = A sin \omega(t_0 - 1 s)(1) at the second point + x = A sin \omegat_0,(2) and at the second point on the return journey, + x = A sin \omega(t_0 + 2 s)(3) Hence, using equation (1) and (2) sin \omegat_0 = x/A = - sin \omega(t_0 - 1 s) Since - sin \omega(t_0 - 1 s) = sin [-\omega (t_0 - 1 s)] = sin \omegat_0 t_0 = -(t_0 - 1 s) or t_0 = 1/2 s. Also, using equations (2) and (3), sin \omega(t_0 + 2 s) = x/A = sin \omegat_0. Since, in general, sin \texttheta_1 = sin \texttheta_2 implies \texttheta_2 = \pi - \texttheta_1 then, in this particular case \omega(t_0 + 2 s) = \pi - \omegat_0 or \pi/\omega = 2t_0 + 2 s = 3 s \therefore \omega = (\pi/3) rad\bullets^-1. Also sin \omega(t_0 - 1 s) = sin[\pi/3 rad\bullets^-1×-1/2 s] = - sin \pi/6 = -1/2. But sin \omega(t_0 - 1 s) = - x/A = -(0.1/A)m. \therefore 1/2 = (0.1/A)morA = 0.2 m.

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Question:

A chemist reacts metal "B" with sulfur and obtains a compound of metal and sulfur. Assuming metal "B" weighed 2.435 g (MW = 121.75 amu), and the compound weighs 3.397g, what is the simplest or empirical formula of the compound? The atomic weight of sulfur is 32.06 g/mole.

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Solution:

To calculate the simplest formula, one must know the mole ratio of the elements that reacted. A mole is defined as the weight in grams of a substance divided by its atomic weight. The number of moles of "B" = (2.435g) / (121.75 g/mole) = .0200 moles of "B" . The number of grams of sulfur that reacted must be 3.397 - 2.435 = .962 g. The increase in weight can only be derived from the addition of sulfur. The number of moles of sulfur is (0.962 g) / (32.06g/mole) = .0300 moles of sulfur. The ratio of "B" to sulfur moles is .0200 B /.0300 S = 2:3. Therefore, the simplest formula must be B_2S_3 .

Question:

How does nutrient procurement by green plants differfrom thatof animals, fungi, andprotista?

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Solution:

All green plants areautotrophs, subsisting exclusively on inorganic substancestaken from their en-vironment. The necessary molecules, being smalland soluble, are able to pass through cell membranes. Autotrophic organismstherefore do not need to digest their nutrients before taking theminto their cells. Green plants are photosynthetic, taking in carbon dioxideand water from their environment, and using the energy from sunlightto synthesize carbohydrates from them. 93% of the dry weight of plant was originally carbon dioxide in the air. Plants also need to synthesize proteins. Protein synthesis requires otherelements, such as nitrogen and sulfur. When rain water dissolves thesenutrients in the soil, the solution is taken up by the root system and transportedto various parts of the plant via the vascular system. Minerals requiredby plants are also transported up the plant body via the root and vascularsystems. Some plants supplement their protein diet by trapping insects(the Venus flytrap is an example). Many fungi such as mushroom, cannot produce their own carbohydratesfrom the air and water. They areheterotrophs: they must obtainthe carbohydrates they require from the surroundings. This is usuallysupplied by dead plant matter, but plant matter is mainly cellulose andother polysaccharides which must be first digested. Specialized secretorycells in these organisms release the enzymes for digestion into thesurrounding organic matter. The products of this extracellular digestion arethen directly absorbed through the cells surfaces or some-times by specializedabsorbing structures. Almost all animals are dependent on plant matter, or other animals thatfeed on plant matter, for food. They too areheterotrophs. Without plants, there would be virtually noheterotrophson earth. Like the mushroom, most animals digest the majority of their foodextracellularly. The simple products are then taken into the cells where they are further degradedby intracellular enzymes. Digestion is unimportant in green plantsbut fungi and animals have di-gestive process of the extracellular type. Protistaareautotrophsorheterotrophs. Protozoa are heterotrophic, nonphotosyntheticprotista. Protozoa includes theparemeciumand amoeba, in which digestion is intracellu-lar.

Question:

Write a program that performs matrix multiplication on a) array matrix [2] [4] and array vector [4], b) Result of multiplication is to be stored in array result [ ]. c) Print the result of multiplication i.e. print each row of result [ ]. d) Assume column vector contains {12 3 4} and matrix contains values [0 1 2 3] and [1 2 3 4].

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Solution:

# include < stdio. h > # define NUM _ C0L 4 /\textasteriskcenteredno. of columns\textasteriskcentered/ # define NUM _ R0WS 2 /\textasteriskcenteredno. of rows\textasteriskcentered/ main ( ) { int matrix [NUM _ R0WS][NUM _ COL]; /\textasteriskcenteredcreates a 2 × 4.array of type integer\textasteriskcentered/ int vector [NUM _ COL]; int result [NUM _ ROWS]; /\textasteriskcenteredTo store the result of multiplication \textasteriskcentered/ int i, j; /\textasteriskcenteredsubscripts or counters \textasteriskcentered/ /\textasteriskcentered1. Initialize the matrix, vector and result\textasteriskcentered/ for (i = 0; i < NUM _ ROWS; i++) /\textasteriskcenteredSet result to zero\textasteriskcentered/ result [i] = 0; for (i = 0; i < NUM _ C0L; i++) /\textasteriskcenteredset vector to 1 2 3 and 4 \textasteriskcentered/ vector [i] = i + 1; for (i = 0; i < NUM _ R0WS; i++) for ( j = 0; j < NUM _ COL; j++) /\textasteriskcenteredset matrix to 0 1 2 3\textasteriskcentered/ /\textasteriskcentered1 2 3 4\textasteriskcentered/ matrix [i] [ j] = i + j; /\textasteriskcentered2. Perform matrix multiplication \textasteriskcentered/ for (i = 0; i < NUM_R0WS; i++) for ( j = 0; j < NUM_COL; j++) result [i] = result [i] + matrix [i] [ j] \textasteriskcentered vector [ j]; printf ("The result of matrix multiplication = "); /\textasteriskcenteredPrint each row of result [ ]\textasteriskcentered/ for ( j = 0; j < NUM _ ROWS; j++) printf ("%d", result [ j]); /\textasteriskcenteredput newline character\textasteriskcentered/ put char ('\textbackslashn'); }

Question:

You want to plate out 50g of copper from an aqueous solution of CuSO_4. Assuming 100% efficiency, how many coulombs are required?

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Solution:

In this question, you are dealing with the phenomenon of electrolysis. When an electric current is applied to a solution containing ions, the ions will either be reduced or oxidized to their electronically neutral state. To answer this question, you must realize that Cu^2+ ions exist in solution. To plate out copper, 2 electrons must be added to obtain the copper atom, Cu^0. Since the Cu^2+ must gain electrons, it must be reduced. The amount of electricity that produces a specific amount of reduction (or oxidation) is related by q = nF (Faraday's Law). Where q = the quantity of electricity in coulombs, n = number of equivalents oxidized or reduced and F = Faradays. The number of equivalents equals the weight of material oxidized or reduced (m) divided by the gram- equivalent weight of the material (M_eq) i.e., N = (M) / (M_eq). A faraday = 96,490 coulombs or one mole of electrons. Since copper ion requires two electrons for reduction, the gram- equivalent weight is one half of the atomic weight or 31.77g-equiv. You have, therefore, q = [(50) / (31.77)] (96,490) = 1.52 × 10^5 coul. required.

Question:

Find an expression for the hydrolysis constant of the bicarbonate ion HCO_3^-,K_h(HCO_3^-), using only the first dissociation constant for H_2CO_3, K_(a)1 (H_2CO_3), and the water constant, K_W .

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Solution:

The solution to this problem is a direct application of the definitions ofK_h, K_(a)1 , and K_W . The removal of one proton may be represented by K_a1, H_2CO_3 \rightleftarrows H^+ + HCO_3^- Hence, the first dissociation constant is K_(a)1 (H_2CO_3) = {[H^+][HCO_3^-]}/[H_2CO_3] The water constant is given by K_W = [H^+][OH^-]. Hydrolysis of HCO_3^- proceeds according to the reaction HCO_3^- + H_2O \rightleftarrows H_2CO_3 + OH^-. We are trying to determine an expression for the hydrolysis constantK_h(HCO_3^-) in terms of K_(a)1 (H_2CO_3) and K_W . By definition, K_h(HCO_3^-) = {[H_2CO_3][OH^-]} / [HCO_3^-] . Hence,K_h(HCO_3^-) = {[H_2CO_3][OH^-]} / [HCO_3^-] = ({[H_2CO_3][OH^-]} / [HCO_3^-]) × [H^+]/H^+] = ([H_2CO_3] / {[H^+] [HCO_3^-]}) × [H^+] [OH^-] = [(1)/{K_(a)1 (H_2CO_3)}] × K_W, or,K_h(HCO_3^-) = [(K_W)/{K_(a)1 (H_2CO_3)}] .

Question:

In man, two abnormal conditions, cataracts in the eye and excessive fragility of the bones, seem to depend on separate dominant genes located on different chromosomes. A man with cataracts and normal bones, whose father had normal eyes, married a woman free from cataracts but with fragile bones. Her father had normal bones. Their daughter marries a man with normal eyes and bones. If these two people have a child, what is the probability that it will have both cataracts and fragile bones?

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Solution:

First, let us determine the possible genotypes of the individuals involved. Let C represent the dominant allele for cataracts, c the recessive allele for normal eyes, B the dominant allele for fragile bones, and b the recessive allele for normal bones. The man has cataracts, but he must be heterozygous for that trait (Cc) since his father had normal eyes (cc). Since the man has normal bones, his genotype for that trait is bb. The woman whom he married has normal eyes (cc). Since she has fragile bones, and her father had normal bones (bb), she must be heterozygous for that trait (Bb). The daughter has one of four possible genotypes: CcBb, Ccbb, ccBb, and ccbb. Her husband, who is normal for both traits involved, is ccbb. In order for their child to express both abnormalities, its genotype must be CcBb (since the daughter's husband can transmit only recessive alleles). This means that the daughter must donate both dominant alleles. Looking at the cross outlined above, we see that there is only one genotype possible in which the daughter carries both dominant alleles, namely CcBb. The probability of this occurring is 1/4 (one out of four possible genotypes). Given this, there is a 1/2 probability that the daughter will donate the B, rather than the b allele, to her child, and a 1/2 probability that she will donate the C, rather than the c allele. The probability that she will both possess and subsequently donate both dominant alleles is equal to the product of the separate probabilities of each of the three events, namely 1/4 × 1/2 ×1/2 or 1/16 probability that a child having both cataracts and fragile bones will be produced.

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Question:

The gypsy moth produces a natural attractant, C_18H_34O_3. If a female moth is trapped behind a cellophane screen containing a pinhole and the carbon dioxide she produces diffuses through the pinhole at the rate of 1millimicromole per 90 seconds, what quantity of attractant will diffuse through the orifice in the same amount of time (90 seconds)?

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Solution:

The rate of escape of gases through an orifice is inversely proportional to the square root of the density of the gas. This principle is known as Graham's Law. If two gases are compared at the same temperature and pressure (as in this problem), then u_1 / u_2 = \surd(d_2 / d_1 ) where u is the velocity (or the rate at which the gas diffuses out of the orifice) and d is the density of the particular gas. To solve this problem, this equation is modified somewhat to avoid using densities in the expression (since the densities are unknown). However, the density is defined as mass per unit volume, and since the volume per mole of both gases, C_18H_34O_3 and CO_2 , are the same, Graham's Law becomes u_1 / u_2 = \surd(d_2 / d_1 ) = \surd[(m_2 / V) / (m_1 / V)] = \surd(m_2 / m_1 ), where V is volume and m_1 and m_2 are the masses of the gases. Another modification is necessary before solving this problem: the mass of any substance is its molecular weight times the number of moles of that substance. However, assume that 1 mole is involved of both substances and thus molecular weights of the gases (which can be determined) is substituted. Graham's Law then becomes u_1 / u_2 = \surd(M_2 / M_1 ) u_1 for CO_2 = 10-^9 mole/90 sec = 1.1 × 10-^11 moles/sec. Note: 1milli micromole = 10-^9 mole. M_1 for CO_2 = 44 g/mole M_2 for C_18H_34O_3 = 298 g/mole. Thus, solve for u_2. (1.1 × 10-^11moles/sec) / u_2 = \surd[(298 g/mole) / (44 g/mole)] = 2.6 u_2 = 4.2 × 10-^12 moles/sec. 1picomole= 10-^12 moles. u_2 = 4.2 × 10-^12 moles/sec = 4.2picomolesper sec u_1 = 1.1 × 10-^11moles/sec = 11.0picomolesper sec. The CO_2 diffuses 1millimicromole within 90 sec ([ 0.011millimicromole/sec] [90 sec] = 1millimicromole) and C_18H_34O_3 diffuses (0.0042millimicromoles/sec) (90 sec) = 0.38millimicromoles within the same time as the CO_2.

Question:

The deviation of the second-order diffracted image formed by an optical grating having 5000 lines/cm is 32\textdegree. Calculate the wavelength of the light used.

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Solution:

An optical grating is a transparent piece of glass or plastic on which there are closely spaced parallel scratches. These scratches behave like opaque barriers and the spaces between slits like narrow slits. Each adjacent pair of slits produces a double-slit diffraction pattern so that this problem can be analyzed using double-slit interference. Point P_n in the diagram receives light from both slits. The path lengths from slits S_1 and S_2 to P_n differby an amount very close to ∆L. If the two waves are in phase as they leave slits Si and S2, then they will also be in phase at P_n if ∆L is an integral multiple of the wavelength of the light. That is, if ∆L = n\lambda then rein-forcement occurs at P_n and a bright band occurs there. If b is small compared to L, then the angle \texttheta in the small right triangle is nearly equal to \texttheta in the larger right triangle. Therefore, we can say sin \texttheta_N = N\lambda/b orb sin \texttheta_N = N\lambda where \texttheta_N is the deviation of the Nth order diffracted image and b is the distance between slits. b = 1/5000 cm = 0.00020 cm \lambda = [(b sin \texttheta_N)] / [N] = [(0.00020 cm × 0.53) / (2)] = 0.000053 cm = 5300 A

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Question:

Write a program in FORTRAN that prints a calendar for any month of the year from March 1600 to year N.

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Solution:

To grasp a thorough theoretical knowledge of the procedure behind this problem, one needs to know something about congruence and modulo arithmetic. Here, we will try to minimize the theoretical discussion and simply state some pertinent theorems. We will look at the Gregorian calendar, first implemented on October 15, 1582. Let N be any year later than 1600. Then, between the year 1600 exclusive, and the year N inclusive, there are [N/4] - 400 years divisible by 4 To prove this let y be any year such that 1600 < y \leq N, Year y is divisible by 4 only if y = 4x, where x is an integer. Thus, the number of years divisible by 4 is the same as the number of integers x satisfying 1600 < 4x \leq N, or 400 < x \leq N/4. There are clearly [N/4] - 400 of such integers. [N/400] - 4 years divisible by 400 [N/100] - 16 century-beginning years [N/100] - [N/400] - 12 century-beginning years that are not leap years [N/4] - [N/100] + [N/400] - 388 leap years [A/4] + [B/4] + 24A - 388 leap years, where A and B are integers governed uniquely by the restrictions N = 100A + B and 0 \leq B < 100 (Brackets are used to express the greater integer part of the division.) These relations tell us that leap years, containing 366 days, are those divisible by 4, and that century-beginning years are leap years only if divisible by 400. Thus, 1600, 2000, 2400,... are century-beginning leap years. Since February is a troublesome month, especially during leap years, we rearrange our calendar so that March is the first month of the year N. January and February of year N become the eleventh and twelfth months of year N - 1. Then, April through the following February become the second through twelfth months of year N. This places February at the end of the year, a manipulation which will make the program simpler to follow. We must also consider the days of the week. Starting with Sunday, we number the days 0,1,2,3,4,5,6, ending with Saturday. The theorem which gives the weekday number of theKthday of month M of year N (known as Zeller's congruence) is K + [2.6M - 0.2] + B + [B/4] + [A/4] + 2A(mod 7). To understand the derivation of this formula, consider the following: Since March 1, 1600 fell on a Wednesday, its weekday number is W = 3. Since 365 \equiv 1 (mod 7) (365 - 1 = 364 is divisible by 7), March 1, 1601 has weekday number 4. March 1, 1602 has number 5 and March 1, 1603 has number 6. But March 1, 1604 has number 1, not 0 because it is a leap year. In general, the weekday number W_N of March 1 of year N is: W_N = W + t + e (mod 7), where W = 3(1) where t is the total number of years and e is the number of leap years between 1600 exclusive and N inclusive. Now, t = N - 1600, N = 100A + B, and, e = [A/4] + [B/4] + 24A - 388. Thus, equation (1) is: W_N = 3 + 100A + B - 1600 + [B/4] + [A/4] + 24A - 388 (mod 7) or, after simplification, W_N = 3 + B + [B/4] + [A/4] - 2A (mod 7).(2) Equation (2) gives the weekday number of March 1 for any N. Since April 1 occurs 31 days after March 1, and 31 \equiv 3 (mod 7), its week-day number is 3 more than the weekday number of March 1, i.e., 6. Continuing for the other months we obtain the sequence of constants W: 3,6, 8, 11, 13, 16, 19, 21, 24, 26, 29, 32 \uparrow\uparrow MarApr .... Reverend Zeller noticed that the arithmetic function 1 + [2.6m - 0.2] assumes exactly this sequence of integers for m = 1,2,...,12. Hence, the weekday number of the first day of any month, m, of the Nth year is given by W_N \equiv 1 + [2.6m - 0.2] + B + [B/4] + [A/4] - 2A (mod 7)(3) Finally, if, instead of the first day, we consider thekthday, Zeller's congruence is obtained; W_N \equiv k + [2.6m - 0.2] + B + [B/4] + [A/4] - 2A (mod 7). Take for example November 1, 2060. K is 1, M is 9 (according to our renumbering scheme), B is 60, and A is 20 (since N = 2060, and N = 100A + B). Plugging in the values, we get 1 + 23 + 60 + 15 + 5 - 40 = 64. To complete the problem, we find that 64 \equiv 1 (mod 7) since 64 - 1 is divisible by 7. Hence, 1 corresponds to Monday, and the answer is obtained. We present the program forthwith, high-lighting control sections with comments. CCALENDAR PROGRAM INTEGER BLANK, DATE, YEAR, RMTHS CTHE ELEMENTS OF DATE, ALONG WITH THE CCHARACTER BLANK, ARE USED TO PRINT CALENDAR CIF FIRST DAY OF MONTH IS NOT SUNDAY. DIMENSION RMTHS (12), DATE (6) DATA RMTHS/'JAN ', 'FEB ', 'MAR ', 'APR, 'MAY', 'JUN', 1'JUL ', 'AUG ', 'SEP ', 'OCT', 'NOV ', 'DEC' / DATA BLANK, DATE/' ', '1', '2', '3', '4', '5', '6'/ CINPUT MONTH (JAN = 1, FEB = 2,..., DEC = 12) AND YEAR 10READ (5,100) MONTH, YEAR 100FORMAT (I2, I4) CDO WHILE YEAR GREATER THAN ZERO IF (YEAR.LE.O) GO TO 99 CNOW, MONTHS ARE RENUMBERED SUCH THAT CM = 1 FOR MAR, M = 2 FOR APR, AMD SO ON UNTIL CM = 10 FOR DEC. FOR JAN AND FEB, M = 11 AND M = 12 COF PRECEDING YEAR. 20NEWYR = YEAR IF (MONTH - 2) 25, 30, 35 CJANUARY SECTION 25M = 11 NEWYR = NEWYR - 1 CNDIM EQUALS NUMBER OF DAYS IN MONTH NDIM = 31 GO TO 40 CFEBRUARY SECTION 30M = 12 NEWYR = NEWYR - 1 CDETERMINE WHETHER NYEAR IS A LEAP YEAR IF (YEAR.NE. (YEAR/4) \textasteriskcentered4) GO TO 33 IF (YEAR.NE. (YEAR/100)\textasteriskcentered100) GO TO 34 IF (YEAR.EQ. (YEAR/400) \textasteriskcentered400) GO TO 34 CIF YEAR IS NOT A LEAP YEAR 33NDIM = 28 GO TO 40 CIF YEAR IS A LEAP YEAR 34NDIM = 29 GO TO 40 CNOW COMPUTE MARCH THROUGH DECEMBER M = MONTH - 2 CCASE STRUCTURE TO DETERMINE DAYS IN MONTH GO TO (36, 37, 36, 37, 36, 36, 37, 36, 37, 36), M 36NDIM = 31 GO TO 40 37NDIM = 30 CNOW CALCULATE WEEKDAY NUMBER FOR THE CFIRST DAY OF THE MONTH 40NC = NEWYR/100 ND = NEWYR - 100\textasteriskcenteredNC NW = 1 + (26\textasteriskcenteredM-2)/10 + ND + ND/4 + NC/4 - 2\textasteriskcenteredNC CDO LOOP REDUCES WEEKDAY NUMBER MODULO 7 CTO ONE OF THE VALUES 0, 1, 2,...,6. DO 41 I = 1,7 NWKDAY =I - 1 NDIFF = NW - NWKDAY IF (NDIFF.EQ.(NDIFF/7)\textasteriskcentered7) GO TO 45 41CONTINUE CWRITE HEADINGS 45WRITE (6, 101) RMTHS (MONTH), YEAR 101FORMAT (1HO, 10X, A4, I4) WRITE (6, 102) 102FORMAT (1HO, 27HSUN MON TUE WED THU FRI SAT/) IF (NWKDAY. EQ.O) GO TO 52 CTHIS SECTION PRINTS THE FIRST WEEK OF THE CMONTH IF SUNDAY IS NOT THE FIRST DAY LIM = 7 - NWKDAY WRITE (6, 103) (BLANK, I = 1, NWKDAY), (DATE(I), I = 1, LIM) 103FORMAT (7(3X,A1)) LIM = LIM + 1 GO TO 53 52LIM = 1 CFINISH PRINTING CALENDAR 53WRITE (6, 104) (I, I = LIM,NDIM) 104FORMAT (7I4) GO TO 10 CEND DO-WHILE 99STOP END

Question:

A chemist has an 18% solution and a 45% solution of a disinfectant. How many ounces of each should be used to make 12 ounces of a 36% solution?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E33-0946.htm

Solution:

Let x = Number of ounces from the 18% solution And y = Number of ounces from the 45% solution (1) x + y = 12 (2) .18x + .45y = .36(12) = 4.32 Note that .18 of the first solution is pure disinfectant and that .45 of the second solution is pure disinfectant. When the proper quantities are drawn from each mixture the result is 12 gallons of mixture which is .36 pure disinfectant, i.e., the resulting mixture contains 4.32 ounces of pure disinfectant. When the equations are solved, it is found that x = 4 and y = 8.

Question:

The faint light sometimes seen over Marshland at night, the "will-o'-the-wisp", is believed to come about as a result of the burning of a compound of phosphorus (P) and hydrogen (H) . What is the formula of this compound?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E17-0613.htm

Solution:

To find the formula of this compound, it is necessary to determine the valence of the elements from which it is composed. The valence of an element is the number of electrons that are involved in chemical bonding. To find the valence of phosphorus and hydrogen, consider their atomic number and electronic configuration. Hydrogen: Atomic number= 1. Electronic configuration= 1s1 Phosphorus: Atomic number= 15. Electronic configuration= 1s^22s^22p^63s^23p^3. The outer electrons are 1s^1 for hydrogen, and 3p^3 for phosphorus. It takes one additional electron to fill hydrogen's s orbital; its valence is one. It takes 3 more electrons to fill phosphorus' p orbital; its valence is three. Elements react with the purpose of filling all their orbitals with the maximum number of electrons by either a transfer of electrons or by sharing electrons. It would take three hydrogen atoms to complete phosphorus' outer orbital. In turn, each electron of phosphorus would serve to complete the outer orbital of each hydrogen atom. This can be pictured in an electron-dot formula as shown above. In this figure, the X's represent the outer electrons of hydrogen and dots represent the electrons in the outer shell of phosphorus. The formula of this compound is, thus, PH_3.

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Question:

A radio telescope sited on the edge of a cliff over-looking the sea operates on a wavelength of 100 m. A radio star rises above the horizon and is tracked by the telescope. The first minimum of the received signal occurs when the star is 30\textdegree above the horizontal. Explain why the minimum occurs and determine the height of the cliff, assuming that radio wave suffer a phase change of \pi on reflection at a water surface. (See figure.)

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/Users/wenhuchen/Documents/Crawler/Physics/D29-0900.htm

Solution:

The radio telescope is receiving signals direct from the star (ray XBC) and also by reflection from the water surface, (YAC) which is acting like a plane mirror. In this case the source of radiation is far enough away to be considered at infinity and the rays are descending on the earth in a parallel beam. When the path difference between rays YAC and XBC (see figure) is such that they reach the telescope exactly out of phase, destructive interference occurs and the signal at C will be a minimum. Consider the diagram. If AB is drawn perpendicular to XC, it will also be perpendicular to YA, and AB is a plane wave front of the incoming beam. Thus the disturbances at A and B have the same phase at all times. A phase change of it occurs at A and this is equivalent to an increase in path of \lambda/2. Thus the path difference between the two rays is \delta = AC + (\lambda/2) - BC. But by the laws of reflection \angleYAZ = \angleCAO = 30\textdegree, and therefore \angleBAC = 180\textdegree - 90\textdegree - 30\textdegree - 30\textdegree = 30\textdegree. \delta = AC + \lambda/2 - AC sin 30\textdegree = \lambda/2+ AC(1 - sin 30\textdegree) Since sin 30\textdegree = Y/AC, AC = Y/sin 30\textdegree and \delta = \lambda/2 + [y/(sin 30\textdegree)] (1- sin 30\textdegree) = \lambda/2 + y. In order for minima to occur, the path difference \delta must equal an odd number of half wavelengths, or\delta = (2n + 1)\lambda/2n = 0, 1, 2, ... Then(2n + 1) \lambda/2 = \lambda/2 + y. If n = 0, \delta = \lambda /2, y = 0, and this corresponds to the center of the interference pattern being a minimum. Since this is the first minimum the telescope receives, C must be at the first minimum above 0 for the situation shown in the diagram. (Where it a minimum other than the first, a previous minimum must have occurred for an angle lower than 30\textdegree. But we are told that this doesn't happen.) Therefore, n = 1 and \delta must be 3\lambda/2. Hence,3\lambda/2 = \lambda/2 + y and y = \lambda = 100 m.

Question:

What electric field intensity is just sufficient to bal-ance the earth's gravitational attraction on an electron? If this electric field were to be produced by a second electron how far away must it be put?

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/Users/wenhuchen/Documents/Crawler/Physics/D18-0599.htm

Solution:

The gravitational force on the electron is mg and the force due to the field E is eE. Thus if they are to balance (see the figure) eE = mg E = mg / e = {(9.1 × 10^\rule{1em}{1pt}31 kg)(9.8 m/s^2)} / (\rule{1em}{1pt}1.6 × 10^\rule{1em}{1pt}19 c) = \rule{1em}{1pt}5.57 × 10^\rule{1em}{1pt}11 N/coul If this field is supplied by another electron a distance R away (see figure) then by definition, the electric field due to this electron at R is E = ke / R^2 where k = 9 × 10^9 (N \bullet m2/ c^2), and e is the signed charge of an electron, Hence, solving for R, R^2 = ke / E. Using the previous calculation of E, R^2 = [{9 × 10^9 (N \bullet m^2 / c^2)} (\rule{1em}{1pt}1.6 × 10^\rule{1em}{1pt}19 c) ] / (\rule{1em}{1pt}5.57 × 10^\rule{1em}{1pt}11 N / c) R^2 = 25.8 m^2 R = 5.1m

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Question:

A grain of sand has a mass of 10^-5 gram. Through what heightmust it fall from rest in order to have a kinetic energy equalto the energy of a quantum of visible light with a frequencyof 5 × 10^14 sec^-1 ?

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/Users/wenhuchen/Documents/Crawler/Physics/D33-0972.htm

Solution:

The energy of the quantum is given by the product of its frequency, v, and Planck's constant h. \epsilon =hv = 6.625 × 10^-27 × 5 × 10^14 = 3.31 × 10^-12erg When the grain of sand falls through a height H from rest, its final kinetic energy, (1/2)mv^2 , is equal to its decrease in gravitational potential energy, mgH. (1/2)mv^2= mg H Therefore mg H =hv H = (hv/mg) = [(3.31 × 10^-12)/(10^-5 × 980)] = 3.38 × 10^-10 cm The grain of sand would have to fall through a height of only 3.38 × 10^-10 cm.The diameter of an atom is about 10^-8 cm.

Question:

Show that the angular momentum of a particle about a point 0 will be equal to zero if any one of the following conditions applies? (a) P^\ding{217} = 0 (b) the particle is at point 0 (c) P^\ding{217} is parallel (or anti-parallel) to r^\ding{217}

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/Users/wenhuchen/Documents/Crawler/Physics/D37-1068.htm

Solution:

In general, for a particle of mass m and velocity v^\ding{217} = (v1i˄ + v_2 \^{\j} + v_3 k ̂) located a distance r from point 0, r^\ding{217} = (r_1 i˄ + r_2 \^{\j} + r_3 k ̂); angular momentum L^\ding{217} = r^\ding{217} × mv^\ding{217}. In scalar form, this equation becomes L = r m v sin \texttheta where \texttheta is the angle between r^\ding{217} and v^\ding{217}. a) P\ding{217}= m v^\ding{217}. If P^\ding{217} = 0 then m v must be 0. L = r (0) sin \texttheta L = 0 b) If the particle is located at point 0 then r^\ding{217} = 0 and r = 0. (0) (m v) sin \texttheta = 0 L = 0 c) If P^\ding{217} is parallel or anti-parallel to r^\ding{217} then v^\ding{217} is parallel or anti- parallel to r^\ding{217} since P^\ding{217} has the same direction as v^\ding{217} because m is a scalar. If v^\ding{217} is parallel or anti-parallel to r^\ding{217} then \texttheta = 0\textdegree or 180\textdegree. In either case, sin \texttheta = 0. r m v (0) = 0 L = 0

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Question:

The diffusion coefficient for serum globulin at 20\textdegree in a dilute aqueous salt solution is 4.0 × 10^-11 m^2 s^-1. If the molecules are assumed to be spherical, calculate their molecular weight. Given \etaH_2O = 0.001005 Pascal sec at 20\textdegree andѵ= 0.75 (cm^3/g) for the protein.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E22-0820.htm

Solution:

For spherical molecules, the molecular weight M is related to the diffusion coefficient D by the follow-ing equality D = [(RT) / (N_A6\pi\eta)][(4 \pi N_A) / (3Mѵ)]^1/3 where R is the gas constant in joules (8.314 J/mole \textdegreeK), N_A is Avogradro's Number (6.02 × 10^23), \eta is the viscosity of the solvent, andѵthe partial specific volume of the protein. Solving for M; D = 4.0 × 10^-11 (m^2/s) = [(8.314 J/mole \textdegreeK) (293\textdegreeK)] / [(6.02 × 10^23/mole) 6\pi (.001005 Js/m^2)] × [4\pi(6.02 × 10^23 mole)] / [3 × m × 0.75 × 10^-6 m^3/s]^1/3 4.0 × 10^-11 m^2/s = (2.14 × 10^-19 m^2/s) × (3.36 × 10^30/M)^1/3 1.87 × 10^8 = (3.36 × 10^30/M)^1/3 6.53 × 10^24 = 3.36 × 10^30/M M = 5.15 × 10^5 g/mole.

Question:

Spring-Heel Jack was a legendary English criminal who was never captured because of his ability to jump over high walls and other obstacles which his pursuers were unable to scale. It is believed that he had a powerful spring attached to each shoe for this purpose. Assuming that he weighed 150 lb and that his springs were com-pressed by 1 in. when he stood on them, by how much did he need to keep his springs compressed on one of his operations in order to be ready to clear a 10-ft wall in the event of an emergency?

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/Users/wenhuchen/Documents/Crawler/Physics/D05-0220.htm

Solution:

The figure shows an idealized drawing of Spring-Heel Jack. When Jack stands up, the springs are compressed a distance x. When in equilibrium, the net force on Jack is zero. Hence, 2kx= mg ork= mg/2x where m is his mass and k is the spring constant. Thus k= [150 lb]/[(2)(1 in)] = (75 lb)/(1/2 ft) = 900 lb/ft. If Jack wishes to clear a height h while remaining erect, the potential energy stored in the springs must have been sufficient to raise his 150-lb weight through a vertical distance h. But if x was the compression of each spring, then by conservation of energy 1/2 kx^2 + 1/2 kx^2= mgh x^2= mgh/k \thereforex^2= (150 × 10 ft \bullet lb)/(900 lb \bullet ft^-1) = 20/12 ft^2. x= \surd(1.67) ft = 1.29 ft.

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Question:

Two electrochemical cells were fitted with inert platinum electrodes and connected in series. The first cell contained a solution of a metal nitrate; the second cell, metal sulfate. After operation of the cell for a prescribed period of time, it was observed that 675 mg of the metal had been deposited in the first cell and 76.3 ml of hydrogen gas (25\textdegree and 1 atm) evolved in the second. Determine the equivalent weight of the metal.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E16-0564.htm

Solution:

Equivalent weight may be defined as that amount of substance which one mole of electrons will reduce or oxidize. Since the electrodes are connected in series, the same current passes through both cells. Thus, equal numbers of equivalents of the metal (m) and hydrogen must be liberated. The cathode reactions for the two cells are: first cell: M^n+ + ne^- \rightarrow M(s) ; second cell: 2H^+ + 2e^- \rightarrow H_2(g), where n, is the number of electrons needed to form the M(s) from the metal ion. One can solve for the number of moles of H_2 formed by use of the Ideal Gas Law. n = (PV/ RT) where n is the number of moles, R is the gas constant, 0.082 . liter-atm / mole\textdegreeK, T is the absolute temperature, V is the volume and P is the pressure. One is given the temperature in \textdegreeC, to convert to \textdegreeK add 273 to the temperature in \textdegreeC . T = 25\textdegree + 273 = 298\textdegree K . Now, solving for n: n = [(1 atm)(.0763 liters)] / [(0.082 liter-atm / mole\textdegreeK) (298\textdegreeK)] = 3.12 × 10^-3 moles. From the half equations, one can see that there are two equivalents for each mole of H_2 formed. Thus, there must be 2 × (3.12 × 10^-3) moles of equivalents of the metal present. Hence, equivalents of M = 2 × 3.12 × 10-^3 = 6.24 × 10^-3 equiv. One is given that 675mg or .675 g of metal are formed, thus, in .675 g of the metal there are 6.24 × 10^-3 equiv . The weight of one equivalent can be found by dividing .675 g by 6.24 × 10^-3 equiv . equiv. wt. =(.675g) / (6.24 × 10^-3 equiv.) = 108.2 g / equiv.

Question:

Explain the importance and structure of the endoplasmic reticulumin the cell.

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/Users/wenhuchen/Documents/Crawler/Biology/F02-0037.htm

Solution:

The endoplasmic reticulum is responsible for transporting certain moleculesto specific areas within the cytoplasm. Lipids and proteins are mainlytransported and distributed by this system. The endoplasmic reticulumis more than a passive channel for intracellular transport. It containsa variety of enzymes playing important roles in metabolic processes. The structure of the endoplasmic reticulum is a complex of membranesthat traverses the cytoplasm. The membranes form interconnectingchannels that take the form of flattened sacs and tubes. When the endoplasmic reticulum hasribosomesattached to its surface, werefer to it as rough endoplasmic reticulum and when there are no ribosomesattached, it is called smooth endoplasmic reticulum. The rough endoplasmicreticulum functions in transport of cellular products; the role ofthe smooth endoplasmic reticulum is less well known, but is believed to beinvolved in lipid synthesis (thus the predominance of smooth end reticuluminhepatocytesof the liver). In most cells, the endoplasmic reticulum is continuous and interconnectedat some points with the nuclear membrane and sometimes, with the plasma membrane. This may indicate a pathway by whichmaterials synthesized in the nucleus are transported to the cytoplasm. In cells actively engaged in protein synthesis and secretion (suchasacinarcells of the pancreas), rough endoplasmic reticulum is abundant. By a well regulated and organized process, protein or polypep-tidechains are synthesized on theribsosomes. These products arethen transported by the endoplasmic reticulum to other sites of the cell wherethey are needed. If they aresecre-toryproducts, they have to be packagedfor release. They are carried by the endoplasmic reticulum to theGolgi appar-atus, another organelle system. Some terminal portions of theendoplasmic reticulum containing protein molecules bud off from the membranesof the reticulum complex, and move to the Golgi apparatus in theform of membrane-bounded vesicles. In the Golgi apparatus, the proteinmolecules are concentrated, chemically modified, and packaged sothat they can be released to the outside byexocytosis. This process is necessarybecause some proteins may be digestive enzymes which may degradethe cytoplasmanslysethe cell if direct contact is made.

Question:

How would one expect the bond strength of NO to compare with that of O_2?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E18-0660.htm

Solution:

The molecular orbital is the sum of the atomic orbitals. For example, when two atomic orbitals are combined, two molecular orbitals are formed. One orbital is called a bonding orbital and the other is an antibonding orbital. The bonding orbital is at a lower energy level than the antibonding orbital. If possible, electrons seek out the bonding orbital rather than the antibonding orbital. The orbitals in the production of H_2 from H atoms can be visualized as shown in the accompanying figure. The greater the number of antibonding orbitals in a molecule, the weaker the bond. Bond order is a quantity that indicates the strength of bonding orbitals. It is defined as half the number of bonding electrons minus half the number of antibonding electrons. Thus, the higher the bond order, the stronger the chemical bond. To compare the bond strength of NO with O_2, compare their bond orders. To do this, consider the total number of valence electrons in each element. Valence electrons are the outer electrons, which participate in bonding. For NO, the total number of valence electrons is 3 + 4 = 7. There exist 3 bonding p orbitals, which accommodate 6 electrons. The 1 unpaired electron must be in an anti-bonding orbital. This means that the bond order of NO is (1/2) (6) - (1/2) (1) = 3 - 0.5 = 2.5. In O_2 there is a total of 8 valence electrons. This means 2 electrons must be in antibonding orbitals, since the 3 bonding orbitals can accommodate only 6 electrons. The bond order of O_2 = (1/2) (6) - (1/2) (2) = 2. The bond order of NO is higher, which means, its chemical bond is stronger than that of O_2.

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Question:

An experimenter crossed a stock of corn homozygous for the linked recessive genes colorless (c), shrunken (s) and waxy (w) to a stock homozygous for the dominant alleles of these genes (C, S, W) and then backcrossed the F_1 plants to the homozygous recessive stock. The progeny were as follows: Phenotype (all are -c-s-w) Number 1 CSW 17,959 2 csw 17,699 3 Csw 2,009 4 cSW 2,024 5 CSw 4,455 6 csW 4,654 7 CsW 288 8 cSw 312 49,400 Draw a linkage map for the three genes.

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/Users/wenhuchen/Documents/Crawler/Biology/F25-0679.htm

Solution:

In order to draw a linkage map for the three genes, we must determine the distances between the genes in map units and the order of the genes on the chromo-some. Let us first determine the parental type. We know that since the F_1 plants resulted from a cross between a homozygous dominant (CC, SS, WW) and a homozygous recessive (cc, ss, ww) parent, they are heterozygous (CcSsWw). They were crossed with homozygous recessive plants (ccssww) Therefore the parental types are CcSsWs ( 1 ) and ccssww ( 2 ). The other six phenotypes are recombinants. To determine the distances between the genes, let us determine the RF for two genes at a time. Consider a cross-over anywhere between c and S. The number or recombinants (from F_2) = 2009 3 + 2024 4 + 288 7 + 312 8 = 4633. RF = (4633) / (49400) = 0.094 An RF of 0.094 gives a separation of 9.4 units between C and S. Consider a cross-over between S and W. The number of recombinants = 4455 5 + 4654 6 + 288 7 + 312 8 = 9709. RF = (9709) / (49400) = 0.196 That is, S and W are 19.6 units apart. Consider a cross-over between C and Wx. The number of recombinants = 2009 3 + 2024 4 + 4455 5 + 4654 6 = 13142. RF = (13142) / (49400) = 0.266 That is, C and W are 26.6 map units apart. To construct a linkage map, we choose randomly C and Wx as our reference points. S is 19.6 units away from w. The only way s can be 9.4 units away from C is for s to lie between C and W. On a genetic map, we have: However, the sum of the map distances between C' and S, and between S and W = 9.4 + 19.6 or 29.0 unit. This sum is greater than the distance between C and W (26.6 units). How do we explain this discrepancy? The explanation lies in the fact that more than one cross-over can occur between two genes, expecially if they are far apart. In this case, there is a cross-over between C and S and between S and W making two cross-overs between C and Wx. Diagrammatically, this can be rep- resented as follows: The result of two cross-overs between two genes is called a double recombinant. Since single cross-overs are quite infrequent, the frequency of double recombinants is even lower. The double recombinants are usually identified by their lowest frequency in the F_2. From F_2, the number of double recombinants = 288 7 312 8 or 600. Percentage of = [(number of double cross-over)/(total number of progeny) × 100 = [(600)/(49400)] × 100 = 1.2% The cause of the discrepancy between the map distances can be traced to the fact that when we calculate the RF between C and S and between S and W, we include the number of double recombinants into the number of recombinants each time. (Refer to previous calculations). However, when we determine the RF between C and W, we neglect the double cross-overs because they do not affect the parental types CW and cw. To compensate for the double cross-overs, we have to add to the map distance between C and W twice the perc.entage of double cross- overs. This is because each double cross-over is the result of two separate events, a cross between c and S and a cross between S and W. In essence, four more crosses have occured between c and w that had not been considered before, having a total frequency of: (288)/(49400) (between c and s)(288)/(49400)(between s and w) (312)/(+49400) (between c and s)(312)/(+49400) (between c and w) or 2 × 1. 2. Thus the distance between c and w is 26.6 + (2 × 1.2) or 29.0 units. After the correction is made, everything falls in place. Note that the recombination frequency between each pair of genes is increased by the frequency of the double cross-overs.

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Question:

Convert the following sequential machine which uses T type flip-flops into one using S-C type flip-flops only.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G04-0088.htm

Solution:

Step I. First, write out all the outputs and inputs of all the gates and flip flops in terms of x and the flip-flop outputs y_1, y'_1, y_2, y'_2 as follows: Hence, we find that T_1 = [(xy_1)' \textbullet y'_2]' (from the figure). \therefore T_1 = (xy_1) + y_2 (using De Morgan's laws). Also, T_2 = (xy_2)' = x' + y'_2 (by De Morgan's laws) and,z = (y'_1 y_2)' = y_1 + y'_2 (by DeMorgan's laws). Step II, Now, we plot each equation T_1, and T_2 into a Karnaugh Map for y_1y_2 and x, as follows: Now, in order to convert a T flip-flop into an S-C flip-flop, we make the following observations: i) The excitation equations of the T and S-C type flip-flops, for y_1\rightarroware: \rightarrow y_1\rightarrow= T_1 \textbullet y'_1 + T'_1 \bullet y_1 y_1\rightarrow= T_1 \textbullet y'_1 + T'_1 \bullet y_1 \rightarrow and, and, y_1\rightarrow= S_1 \textbullet y'_1 + C' \textbullet y y_1\rightarrow= S_1 \textbullet y'_1 + C' \textbullet y \rightarrow 1 ii) On putting y_1 = 0 (and hence y_1' = 1), in both the above equations, we get: y_1\rightarrow= T_1 \bullet 1 + T'_1 \bullet 0 = T_1 \rightarrow and, y_1\rightarrow= S_1 \bullet 1 + C'_1 \bullet 0 = S_1. \rightarrow Hence we see that in our Karnaugh Map for T_1, the first two rows (viz., for y_1y_2 = 00 and 01, i.e., for which y_1 = 0), correspond to T_1 or S_1. iii) Similarly, on putting y_1 = 1 (and hence y'_1 = 0) in the above equations, we get: y_1\rightarrow= T_1 \bullet 0 + T'_1 \bullet 1 = T'_1, \rightarrow and, y_1\rightarrow= S_1 \bullet 0 + C'_1 \bullet 1 = C'_1. \rightarrow \therefore (y_1\rightarrow)' = T_1 , \rightarrow and, (y_1\rightarrow)' = C_1 (complementing both sides in the above equation). \rightarrow Hence, we see that the last two rows of our Karnaugh Map for T_1, viz., y_1y_2 = 11 and 10 are complements of y_1\rightarrow, and correspond to the \rightarrow remainder of T_1 in a T flip-flop, or to C_1 in an S-C flip-flop. Step III. In view of the observations i), ii), and iii), we can now split up T_1 into its components for S_1 and C_1. The fig. (3a) for T_1 leads to fig. (3b) for y_1\rightarrowby complementing the entries \rightarrow For y_1y_2 = 11 and 10 from fig. (3a), as explained in observation iii) before. Fig. (3b) is now split up into two parts corresponding to y_1 = 0 and y_1 = 1 in fig. (3c) and fig. (3d), respectively. Note that fig. (3d) uses the complements of the entries in fig. (3b). We could have split up fig. (3a) directly into the figs. (3c) and (3d) without worrying about the fig. (3b) for y_1\rightarrow, and the complements. But this \rightarrow intervening step was used in the example only to illustrate the reasoning in-volved. Step IV. Now complete the partial Karnaugh Maps of figs. (3c) and (3d) by making use of any special properties of the type of flip-flop used. For example, for changing over to J-K flip-flops, all the remaining entries would be don't care conditions, "d". But, an SC type flip-flop, must satisfy the condition S\bulletC = 0. Hence, fill up the empty blanks such that the product of corresponding blanks in the Maps for S_1 and C_1 = 0. This is done as follows, in fig. (4): Where a zero already exists in one of the partial Karnaugh Maps, a 'd' can be entered in the corresponding square in the other Karnaugh Map. This is because the product of S_1 = 0 and C_1 = 0 or 1 will always be 0. For example, note that a 0 already exists in the Karnaugh Map for S_1 in the square for y_1y_2 x equal to 00 0. Hence, a 'd' can be entered in the corresponding square for y_1y_2 x equal to 00 0 in the Map for C_1, because S_1 \textbullet C_1 will equal 0. Similarly, a 'd' is entered in the square for y_1y_2 x equal to 00 1 in the Map for C_1 because there is a 0 in the square for y_1y_2 x equal to 00 1 in the map for S_1. But, the squares for y_1y_2 x equal to 01 0 and 01 1 are entered with zeroes in the Map for because there are 1's in the corresponding squares in the Map for S_1. Similarly, fill up the lower half of the Map for S_1 with d's or 0's according to whether there are 0's or 1's in the corresponding squares in the Map for C_1. Step IV. We now write out equations for S_1 and C_1 S_1 = y'_1y_2(from fig. (4a).) C_1 = y'_2 + y_1 x . (from fig. (4b)) We repeat the Steps III to V for T_2 to get S_2 and C_2. \therefore S_2 = y'_1x' + y'_1y'_2 = y'_1 \bullet (x' + y'_2)(from fig. (5d).) (from fig. (5d).) and C2= y'_1x' + y'_1y'_2 = y'_1 \bullet (x' + y'_2)(from fig. (5e).) C2= y'_1x' + y'_1y'_2 = y'_1 \bullet (x' + y'_2)(from fig. (5e).) (from fig. (5e).) Step VI. Step VI. We now draw a circuit diagram with two SC type flip-flops, and draw necessary input gates to satisfy the inputs S_1, C_1, S_2, C_2 of the flip flops obtained above.

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Question:

A piece of iron of mass, M = 20g is placed in liquid air until thermal equilibrium is achieved. When it is quickly taken out and placed in water at 0\textdegreeC, a coating of ice of mass, M = 5.22g forms on it. The mean specific heat,c_iof iron over the range of temperature of the experiment is 0.095 cal/g -\textdegreeC and the heat of fusion L, of water is 80 cal/g. What is the temperature of the liquid air?

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Solution:

The iron is initially at the same temperature T as the liquid air. When placed in water, it takes heat from the water until its temperature reaches 0\textdegreeC. The amount of heat Q lost by the water can be found from the amount of ice formed as a result of this transfer of heat. Q = ML = (5.22gr)(80cal/gr) = 417.6 cal. The heat acquired by the iron as its temperature is raised by 0\textdegreeC - T\textdegreeC = - T\textdegreeC, must be equal to the heat Q lost by the water in turning to ice. Then, if M is the mass of the iron, Q = -C_iMT 417.6 cal = -(0.095 cal/gr-0C) (20gr) T giving T = -[417.6 / (20 × 0.095)] 0C = -220\textdegreeC

Question:

What is meant by the term tonus, or tone?

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Solution:

The term tonus refers to the state of sustained partial contraction present in skeletal muscles as long as the nerves to the muscle are intact. Unlike skeletal muscle, cardiac and smooth muscle exhibit tonus even after their nerves are cut. Tonus is a mild state of tetanus. It is present at all times and involves only a small fraction of the fibers of a muscle at any one time. It is believed that the individual fibers contract in turn, working in relays, so that each fiber has a chance to recover completely while other fibers are contracting before it is called upon to contract again. A muscle under slight tension can react more rapidly and contract more strongly than one that is completely relaxed, because of changes in the elastic component in the latter.

Question:

Compare the types of reactions in which the energy of ATP is made available for biosynthetic processes.

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Solution:

ATP has two high energy phosphate bonds and can transfer energy to another molecule in a variety of ways. The most common way is the transfer of the terminal phosphate group to the energy-requiring molecule while liberating ADP (adenosine diphosphate). This is known as a phosphorylation reaction, and is usually catalyzed by enzymes called kinases. The hydrolysis of ATP, in which the terminal phosphate group is transferred to water, releases energy (∆G0^' = -7 kilocalories/mole). By coupling the energy-yielding hydrolysis of ATP to the reaction of a phosphate group with a given compound, one can "capture" part of the energy of hydrolysis in the phosphorylated compound, as seen below: The phosphorylated compound is now "activated," that is, it is of a higher energy content than before, and as such, is able to undergo reactions that were previously unfeasible. A second type of reaction involves the transfer of AMP with the liberation of the last two phosphate groups as pyrophosphate (PPi). The energy released when ATP is split in this manner (-10 kcal/mole) is substantially greater than when a single phosphate group is cleaved off. A compound activated by AMP transfer can thus be of a significantly higher energy than its unactivated form. An example of this type of activation is the activation of an amino acid. Here, AMP is transferred to the COO- group of the amino acid. The activated amino acid is then transferred to the terminus of a transfer RNA molecule for use in protein synthesis, forming a new high energy bond to the tRNA and releasing AMP. Usually, the phyrophosphate group initially generated is quickly cleaved to form two orthophosphate groups: PPi + H2O \textemdash\textemdash> Pi + Pi∆G0^' = - 4.6 kcal/mole Note that the reaction releases a good deal of energy. Although this may Note that the reaction releases a good deal of energy. Although this may appear wasteful, the reaction is highly useful in that it provides an exergonic "push" to the initial reaction, assuring that it will indeed occur to completeness. A third type of reaction utilizing the energy of ATP involves the transfer of the last two pyrophosphate groups and the liberation of AMP. Again, this results in a highly activated compound (R\simP\simP) . The \sim \sim intermediate compounds in the synthesis of cholesterol and its derivatives are activated in this manner. The final major type of reaction involves the transfer of the adenosine group with the liberation of pyrophosphate and inorganic phosphate. This reaction is important in the activation of the amino acid methionine so that its methyl (CH3) group can be transferred to other compounds.

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Question:

Write a program to read input variables length, width, and height and compute the area and the volume using those variables. At the end of the program print values found.

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Solution:

To write this program we need four independent data items: INPUT-VARIABLE (which is nine numeric charac-ters long), VOLUME (which is nine numeric characters long), AREA (which is six numeric characters long), and PERIMETER (which is four numeric characters long). We also need one group item subdivided into three elementary data items: DIMENSIONS subdivided into LENGTH, WIDTH, and HEIGHT. Each of these elementary data items is three numeric characters long, with the decimal point at the rightmost end. Note the character types used in COBOL language are shown in figure 1. Data Type Picture Char. Allowable Characters Alphanumeric Alphabetic Numeric Numeric X A 9 V Letters, digits, spaces, special characters Letters and spaces only Digits 0-9 only Implied decimal point Arithmetic Operations: The formats for arithmetic operations are very straight-forward, and the four standard ADD, SUBTRACT, MULTIPLY, and DIVIDE statements are very similar. Either numeric data-names or numeric literals can be specified in these four statements. ADD statement format: SUBTRACT statement format: MULTIPLY statement format: DIVIDE statement format: COMPUTESTATEMENT: This statement has several uses. It can be used in much the same way as .the other arithmetic statements we have seen so far. Compute statement format Relational Operator Symbol GREATER THAN NOT GREATER THAN LESS THAN NOT LESS THAN EQUAL TO NOT EQUAL TO > NOT> < NOT< = NOT= Fig. 2 format for GO TO statement: GO TO statement specifies a paragraph name to which control is transferred, format for IF statement: the IF statement allows execution of statement - 1 if the condition is true. Relation Condition Operators are shown in fig. 2 Data values can be moved around in the Procedure section. One way to do this is to use the MOVE statement. This statement causes the data to be copied from one data location to another. format for MOVE statement: Using the data, information, and the statements given we write the program as shown in fig. 3.

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Question:

A 5000-cm^3 container holds 4.90 gm of a gas when the pressure is 75.0 cm Hg and the temperature is 50\textdegree C. What will be the pressure if 6.00 gm of this gas is confined in a 2000-cm^3 container at 0\textdegree C?

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Solution:

From the ideal gas law, (P_1V_1)/(m_1T_1) = (P_2V_2)/(m_2T_2) Note we can use masses instead of number of moles since they are Note we can use masses instead of number of moles since they are proportional. (P_1 × 2000 cm^3)/(6.00 gm × 273\textdegree K) = (75.0 cm Hg × 5000 cm^3)/(4.90 gm × 323\textdegree K) P_1 = 194 cm Hg

Question:

Solve 6x^2 - 7x - 20 = 0.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E33-0973.htm

Solution:

6x^2 - 7x - 20 = 0 is not factorable. Therefore, find the roots of the quadratic equation ax^2 + bx + c using: x = [{\rule{1em}{1pt}b \pm \surd(b^2 - 4ac)} / {2a}], where a = 6, b = \rule{1em}{1pt}7, c = \rule{1em}{1pt}20. x = [{7 \pm \surd(49 - 4(6) (\rule{1em}{1pt}20)} / {12}], x = [{7 \pm \surd(49 - 4(6) (\rule{1em}{1pt}20)} / {12}], x = [{7 \pm \surd529} / {12}] = [{7 \pm 23} / {12}]. Therefore,x_1 = (7 + 23) / 12 = 30/12 = 5/2 x_2 = (7 - 23) / 12 = \rule{1em}{1pt} (16/12) = \rule{1em}{1pt}(4/3).

Question:

Calculate the potential energy of the charge distribution shown in the diagram.

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Solution:

The potential energy of any two charges q_i and qjis U_ij = K_E [(q_iq_j) / R_ij ] where Rijis the distancebetween the two charges, ad K_E = 9 × 10^9 (N \bullet m^2 / c^2). The potential energy of the charge distribution is the sum of the poten-tial energies of every possible pair of charges within the distribution. Hence, U = U_12 + U_13 + U_23 = K_E[{(q)(2q) / (L)}+ {(\rule{1em}{1pt}q)(\rule{1em}{1pt}q) / (d)}+ {(+2q)(\rule{1em}{1pt}q) / (L)}] Since d^2 =L^2 + L^2 we have d = \surd2L and U = K_E[{(\rule{1em}{1pt}2q^2)/(L)} + {(q^2)/(\surd2L)} \rule{1em}{1pt} {(2q^2)/(L)}] = \rule{1em}{1pt}3.29 K_E(q^2 / L) joules If q = e and L = 1A\textdegree = 10^\rule{1em}{1pt}8 m, then U = \rule{1em}{1pt}3.29 × 9 × 109(nt \rule{1em}{1pt} m^2 / coul^2) × (1.6 × 10^\rule{1em}{1pt}19 coul)^2 / 10\rule{1em}{1pt}8 m = -7.58 × 10^-20 J The negative sign Indicates that work would be required to disassemble the charge distribution (i.e., work oust be done against attractive electrical forces).

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Question:

For the reaction 2NO(g) + H_2 (g) \rightarrow N_2O(g) + H_2O(g) at 1100\textdegreeK, data, as shown in the following table, were obtained. Find the rate law and the numerical value of the specific rate constant. Initial Pressure of NO,atm Initial Pressure of H_2,atm Initial rate of pressure decrease,atm/min 0.150 0.400 0.020 0.075 0.400 0.005 0.150 0.200 0.010

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/Users/wenhuchen/Documents/Crawler/Chemistry/E13-0444.htm

Solution:

The rate of a homogeneous reaction, i.e. one which occurs in only one phase, depends on the concentration of reactants in that phase. This reaction is a homogeneous reaction in the gaseous phase. It can be investigated kinetically by following the change in pressure of the gaseous mixture as the reaction proceeds. The pressure drops because 3 moles of gaseous reactants are converted to two moles of gaseous products. Since the reactants are being used up during the course of the reaction, their concentrations and their rate if reaction are constantly changing. The concentrations and rates listed are those at the very beginning of the reaction, when little change has occurred. From the data given, one can see that when the initial pressure of NO is halved with the pressure of H2 remaining constant, the initial rate is quartered . When the pressure of H_2 is halved and the NO remains constant, the initial rate is halved . When the pressure of H_2 is doubled and NO remains constant, the rate doubles. We conclude that the rate is proportional to [H_2]. When the H_2 pressure is kept constant and the NO concentration is doubled, the rate is quadrupled, this means that the rate is proportional to [NO]^2. The equation for the rate can thus be written Rate = k [H_2][NO]2 where k is the rate constant. Using the values in the first trial of the table one can solve for k. 0.020atm/min = k (0.400atm) (0.150atm) ^2 (0.020atm/min) / {(0.400atm)(0.150atm)^2} = k k = 2.22 atm^-2 min^-1.

Question:

Suppose you learned that "shmoos" may have long, oval or round bodies and that matings of shmoos resulted in the following: a) long × oval gave 52 long :48 oval; b) long × round gave 99 oval;and c) oval × oval gave 24 long :53 oval : 27 round. What hypothesis about the inheritance of shmoo shape would be consistent with these results? Assume that shmoos are diploid. (Note that the shmoo is a hypotheti-cal genetic organism.)

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Solution:

In this problem we have to decide what type of inheritance is used to determine a shmoo's body shape. From the data given in (b), we observe that the off-spring of that cross possess an intermediate characteristic rather than a parental one; that is, oval instead of long or round. (Note that an oval, geometrically, is somewhere in between round and long.) This would immediately suggest the concept of codominance. Codominance is defined as the expression in the heterozygote of neither the dominant nor recessive phenotype, but something intermediate between the two. Let us test our hypothesis. Let A be the gene for long body in shmoos, and B be the gene for round body in shmoos. According to our hypothesis of codominance, AB would then be the genotype of an oval- bodied shmoo. Outlining the given crosses in the usual way: The ratio of long to oval offspring is 1:1, which agrees with the data in (a). All the offspring in F_1 are oval-shaped; this is in agreement with the data in (b). the ratio predicted in F_1 tallies with the data in (c). Thus, body shape in shmoos is determined by means of the genes for long body and short body, and these two genes are codfaninant alleles.

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Question:

Trace the path of an anesthetic, such as ether, from the ether cone over the nose to the cells in the brain.

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Solution:

The ether enters the body through the external nares, or nostrils, and would then enter the nasal cavities. The sense organs for olfaction are located here, and the individual would at this point smell the ether. The air containing the ether would be warmed, moistened and fil-tered here before passing via the internal nares into the pharynx. The pharynx is where the paths of the digestive and respiratory systems cross. From there the ether would pass through the glottis into the larynx, the trachea, the bronchi, the bronchioles and ultimately would reach the alveoli. At the alveoli, the ether will diffuse into the blood capillaries surrounding the alveoli, due to the con-centration gradient of ether that exists across the alveolar membrane. Diffusion is possible because of the extreme thinness and moistness of the membranes separating the pulmonary air from the blood in the capillaries. Electron micrographs have shown that these membranes, the alveolar epithelium and the endothelium of the capillary, are each one-cell thick. Once within the capillary, the ether will follow the circulatory routes to the heart. It is here that the blood containing the ether will be pumped to the brain, where the action of the gas will take effect.

Question:

The muscle fibers innervated by a motor unit will contract if a stimulus of sufficient magnitude is propagated or it will not contract at all. This is known as the all-or- none law. If this is true, how can muscle force be graded? For example, we use the same muscles to lift a one ounce object that we use to lift a 20 lb. object.

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Solution:

The total tension that a muscle can develop depends upon two factors: 1) the number of muscle fibers in the muscle bundle that are contracting at any given time and 2) the amount of tension developed by each contracting fiber. The number of fibers in a muscle that are contractiong at any given time depends upon the number of motor neurons to the muscle that are being stimulated. Recall that each motor neuron innervates several muscle fibers, forming a motor unit. The number of motor units that are activated is determined by the activity of the synaptic inputs to the motor neurons in the brain and spinal cord. With proper stimulation by the brain and spinal cord, more motor units may be activated at any one time, thus increasing the number of contracting fibers, and therefore, the strength of the muscle bundle. The process of increasing the number of active motor neurons and thus the number of active motor units is known as recruitment. The number of muscle fibers associated with a single motor neuron varies considerably in different types of muscle. In muscles of the hand, for example, which are able to produce very delicate movements, the size of the individual motor units is small. In the muscles of the back and legs, each motor unit contains hundreds of muscle fibers. The smaller the size of the motor units, the more precisely the tension of the muscle can be controlled by the recruitment of additional motor units. In addition to the variability of the number of active motor units, the tension produced by individual fibers can be varied. This can be accomplished by increasing the frequency of action potentials, resulting in stronger contractions, such as in summation or tetanus.

Question:

A chemist possessesKCl, PH_3, GeCl_4, H_2S, andCsF. Which of these compounds do not contain a covalent bond?

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Solution:

A covalent bond is defined as one in which electrons are shared. The stability of covalent bonds in molecules depends on the difference in electronegativity values of the two atoms which make up the molecule. Electronegativity refers to the tendency of an atom to attract shared electrons in a chemical bond. If theelectronegativitydifference of two elements is greater than 1.7, an ionic bond is formed; if it is less than 1.7, a covalent bond is formed. To solve this problem, consult a table ofelectro-negativityvalues, and compute theelectronegativitydifference of the atoms in each of the given compounds. Proceed as follows: Electronegativity value Difference KCl K = 0.8Cl= 3.0 2.2 PH_3 P = 2.1P = 2.1 0 GeCl_4 Ge = 1.8Cl=3.0 1.2 H_2S H = 2.1S = 2.5 0.4 CsF Cs = 0.7F = 4.0 3.3 Thus, onlyKClandCsFexceed 1.7. They possess ionic bonds. The remainder of the molecules possess covalent bonds.

Question:

A horizontal capillary tube closed at one end contains a column of air imprisoned by means of a small volume of water. At 7\textdegreeC and a barometric pressure of 76.0 cm of mercury, the length of the air column is 15.0 cm. What is the length at 17\textdegreeC if the saturation pressures of water vapor at 7\textdegreeC and 17\textdegreeC are 0.75 cm and 1.42 cm of mercury, respectively?

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Solution:

Since the tube is horizontal and the pressure at the open end of the water column is always atmospheric, the pressure at the closed end of the water column is also always atmospheric. The pressure in the moist air is made up of the partial pressures of air and of water vapor. When the equilibrium between the liquid and gas phases of a liquid is reached in a closed volume, the pressure of the vapor acting on the liquid equals the saturated vapor pressure. The eva-poration process effectively stops once this pressure is attained by the vapor. The air and water vapor act on the liquid surface independently, therefore the pressure on the inner surface of the water is the sum of the pressures p and p_v, due to the air and vapor in the tube, respectively (as shown in the figure). p_a = P + p_v. Hence, for the two cases, the air pressure inside is p = (76.00 - 0.75) cm Hg at 7\textdegreeC, andp = (76.00 - 1.42)cm Hg at 17\textdegreeC. Applying the gas law to the air alone, since the air and water vapor exert effects independent of one another, pV/T = p'V pV/T = p'V '/T where p, V, T are the pressure, volume and temperature (in \textdegreeK) of the air. or(75.25 cm × 15 cm × A)/(280 \textdegreeK) = (74.58 cm × yA)/(290\textdegreeK), where A is the cross-sectional area of the tube and y is the length of the column at the temperature of 17 \textdegreeC. Hence y = [(290 × 75.25 × 15)/(280 × 74.58)] cm = 15.68 cm.

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Question:

What is implied by the theory ofuniformitarianism?

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Solution:

The concept ofuniformitariansism, first developed by James Hutton in1785, states that the geologic forces at work in the past were the same asthose of the present. After carefully studying the erosion of valleys by riversand the formation of sedimentary deposits at the river mouths, Hutton concluded that the processes of erosion, sedimentation, disruption andup-lift, over long periods of time, could account for the formation of the fossil-bearingrock strata which had interested many scientists. Since the geologic factors operating on earth remain uniform throughoutthe ages, the rock strata bearing fossils of varying life forms reflectthe gradual changes in the structures and living habits of organisms throughtime. The theory ofuniformitarianismtherefore implies that animalsand plants continually undergo a process of organic evolution; theirremains being fossilized in the earth's crust in layers by gradual depositionof sediment.Uniformitarianismalso implies that the earthcan notbe only a few thousand years in age, as was widely accepted to be truetwo centuries ago. It must be much older, at least old enough for the processof organic evolution and the geological evolution of the earth's crust, both very slow, and gradual processes, to have occurred.

Question:

What is the structure of a sponge? How do sponges obtain food and water?

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Solution:

The most primitive metazoans belong to the phylum Porifera and are commonly called the sponges. Sponges may be radially symmetrical, but more commonly they are asymmetrical animals, consisting of loose aggregations of cells which are poorly arranged into tissues. The surface of a sponge is perforated by small openings, or in current pores, which open into the interior cavity, the atrium or spongocoel. The atrium opens to the outside by a large opening at the top of the tube--shaped sponge, called the osculum. There is a constant stream of water passing through the incurrent pores into the atrium, and out through the osculum. The body wall is relatively simple. The outer surface is covered by flattened polygonal cells called pinacocytes. The pores are guarded by cells called porocytes, which are modified pinacocytes, and are shaped like tubes extending from the outside of the sponge to the spongocoel. The pore, or ostium, can be regulated by contractions in the outer end of the porocyte. The mesohyl, a layer consisting of a gelatinous protein matrix, contains skeletal material and amoeboid cells, and lies directly beneath the pinacoderm. The skeletal material may be calcareous or siliceous spicules, or protein spongin fibers, or a combination of the last two. The composition, size and shape of these spicules form the basis for the classification of species of sponges. Though the spicules are located in the mesohyl, they frequently project through the pinacoderm. The skeleton, whether composed of spicules or spongin fibers, is secreted by amoebocytes called sclerocytes. Amoedboid cells in the mesohyl include the archaeocytes. These calls are capable of forming other types of cells that are needed by the sponge. The inner side of the mesohyl, lining the atrium, is a layer of cells called choanocytes, or collar cells. The characteristic shape of a choanocyte is ovoid, with a flagellum surrounded by a collar. The choanocytes are responsible for the movement of water through the sppnge and for obtaining food. The current is produced by the beating of the flagella of the choanocytes. The flagella are oriented toward the spongocoel, and beat in a spiral manner from their base to their tip. As a result of this, water is sucked into the spongocoel through the incurrent pores. Sponges feed on very fine particulate matter. Only particles smaller than a certain size can pass through the pores. Food particles are brought inside the spongocoel by water currents. Larger food particles are phagocytized by amoebocytes lining the inhalant chamber. Particles of bacterial size or below are probably removed and engulfed by the choanocytes. It is thought that the amoebocytes act as storage centers for food reserves. Egested wastes leave the body in water currents. There are three basic structural types of sponges. The simplest, called the asconoid type, has been described earlier when basic sponge structure was discussed. The syconoid and leuconoid types are more complex, having thicker mesohyls and a greater number of water channels, but their structures can be seen to be just complexes of asconoid sponges attached to one another.

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Question:

Suppose we have a list consisting of people's names, sexes, eye colors, heights, weights, and telephone numbers. Sup pose in addition that we wish to look up the telephone num-bers of three particular people, namely Robert J. Coyle, Mary Holmes, and Victor E. Lee. Write a computer program in PL / I to accomplish this.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G07-0155.htm

Solution:

We use mnemonic notation for the alphanumeric data in the problem by using identifiers like NAME, SEX, and COLOR. We abbreviate phone numbers as NUMBER, and we desig-nate the height of the person by H and the weight by W. The program proceeds as follows. We start by declaring our four character strings and then transmitting the data to the program. Notice that in the GET LIST statement, we transmit to the program all 6 variables, even though we want only the name and phone number of that person. In the next three statements in the program, we make tests to determine if the name is Coyle, Holmes, or Lee. Note that the names must be put in quotation marks. If any of the tests yield TRUE, we go to PRINT where the desired message is printed. If any of the tests fail, we drop through to the next step. / \textasteriskcentered PHONE NUMBERS \textasteriskcentered / DECLARE NAME CHARACTER (15), SEX CHARACTER (7), COLOR CHARACTER (6), NUMBER CHARACTER (9); START: GET LIST (NAME, SEX, COLOR, H, W, NUMBER); IF NAME = 'ROBT J. COYLE' THEN GO TO PRINT; IF NAME = 'MARY HOLMES' THEN GO TO PRINT; IF NAME = 'VICTOR E. LEE' THEN GO TO PRINT; GO TO START; PRINT: PUT EDIT ('PHONE NO. OF', NAME, 'IS', NUMBER) (A (20), A (15), A (3), A (9)); PUT SKIP (2); GO TO START;

Question:

How does a worker bee communicate to other bees the distance and direction of food from the hive?

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/Users/wenhuchen/Documents/Crawler/Biology/F31-0798.htm

Solution:

After one bee has discovered a food source, many bees soon appear. The first bee must somehow communicate to the others the location of the new food source. It has been determined that when a worker bee discovers food, it returns to the hive, feeds several other bees, and performs a dance on the surface of the honeycomb. If the food source is near the hive, the bee performs a "round" dance. This consists of a re-peating pattern of circling, first to the right and then to the left: The round dance excites the hive mates in the dancer's vicinity and they soon leave the hive in search of the food. This dance does not indicate direction or exact distance. It simply indicates that the food source is a short distance away. With increasing distance of the source from the hive, the round dance is gradually transformed into a "waggle" dance, which gives information about distance and direction. In the waggle dance, the bee repeatedly runs in a figure- eight pattern, wagging its abdomen from side to side as it traverses the "waist\textquotedblright of the figure eight: The number of turns per unit time is inversely proportional to the distance of the food source. The faster the bee runs (or turns), the closer the food source. For honey-bees, an estimate of distance is based on the amount of energy expended traveling the round trip. When conditions exist such as headwinds, uphill flights, clipped wings, or added weight to the bees, excessive distances will be reported. The waggle dance also indicates direction. The location of the food relative to the sun is indicated by the direction of the straight portion of the waggle dance. For instance, if the bee runs straight up the vertical comb, the food lies in the direction of the sun; if it runs down, the food is found in the opposite direction from the sun. A straight run at an angle indicates that the food source is to be found at that angle to the sun. Therefore a run 30\textdegree to the left of vertical indicates a food source 30\textdegree to the left of the sun. (See figure 3) If the bee should perform the waggle dance on a horizontal surface outside the hive, it simply dances in the direction of the food source relative to the sun. Waggle dance of bees: B and C show the same dance, performed inside and out-side the hive respectively, for locating the same food source in A.

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Question:

In the design of a modern steel bridge, provisions must obviously be made for expansion. How much does this amount to in the case of a bridge two miles long which is subjected to temperatures ranging from - 40\textdegreeF to + 110\textdegreeF, assuming an average expansion coefficient of .000012/\textdegreeC?

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/Users/wenhuchen/Documents/Crawler/Physics/D12-0450.htm

Solution:

By definition of the coefficient of linear expansion \alpha =\Deltal/ (l_0 \DeltaT) where\Deltal/l_0 is the fractional change in length of an object due to a temperature change \DeltaT. In our case \Deltal = \alpha l_0 \DeltaT \Deltal = (.000012 \textdegreeC^-1)(2 miles)[110\textdegreeF - (- 40\textdegreeF)] \Deltal = (.000012 \textdegreeC^-1)(2 miles) (150\textdegreeF) Since 150\textdegreeF = (5/9) \textbullet 150\textdegreeC = (750\textdegreeC/9) \Deltal = (1.2 × 10 ^-5) (2 miles) (750/9) \Deltal = .002 miles

Question:

Array A contains all odd integers from 1 to 100 (e.g. 1,3,5,7,.......99) in random order, and is declared as follows : TYPEARRAYTYPE = ARRAY [1..50] of integer; VARA :ARRAYTYPE; Write a short procedure that will accept A and return an array with the same elements in sorted, increasing order.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G16-0403.htm

Solution:

The key here is to recognize the relationship be-tween the indices and the elements of the array. Thus, given an element e, we can compute its correct place in the array - let's call it X, from the following formula: X: = ediv 2 + 1; This is, then, the procedure: PROCEDURE SORT (A: ARRAYTYPE; VAR B:ARRAYTYPE ); VARi, X:integer ; BEGIN FORi: =1 to 50 DO BEGIN X: = A [i] div 2 + 1; B[x]: = A[i] END END; Let's simulate this procedure for an Array of 5 elements con-taining odd integers from 1 to 10 in random order: A 7 9 3 5 1 Wheni= 1,x= A[1] div 2 + 1 = 7 div 2 + 1 = 4. B[4]: = A [1] = 7 The array B now looks: 7 at step 2 ,i= 2, x = A[2] div 2 + 1 = 5 B[ 5]: = 9.Array B now looks: 7 9 At step 3,i= 3, x = 3 div 2 + 1 = 2 B[ 2] : = 3 3 7 9 Fori=4, x=5 div 2+1 = 3 B[ 3] : 5 3 5 7 9 Finally , fori= 5, x = 1 div 2 + 1 = 1 B[ 1] : = 1.Now array B is in sorted increasing order: 1 3 5 7 9 The procedure works exactly like this for larger numbers.

Question:

The pilot of an airplane flying on a straight course knows from his instruments that his airspeed is 300 mph. He also knows that a 60-mph gale is blowing at an angle of 60\textdegree to his course. How can he calculate his velocity relative to the ground?

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/Users/wenhuchen/Documents/Crawler/Physics/D03-0084.htm

Solution:

Relative to an observer on the ground, the airplane has two velocities, one of 300 mph relative to the air and the other of 60 mph at an angle of 60\textdegree to the course, due to the fact that it is carried along by the moving air mass. To obtain the resultant velocity, it is therefore necessary to add the two components by vector addition. In the diagram, A\ding{217} represents the, velocity of the air-craft relative to the air, and B\ding{217} the velocity of the air relative to the ground. When they are added in the normal manner of vector addition, C\ding{217} is their resultant. The magnitude of C\ding{217} is given by the trigonometric formula known as the law of cosines (see figure). C^2 = A^2 + B^2 - 2AB cos \texttheta. But A = 300 mph, B = 60 mph, and \texttheta = (180\textdegree - 60\textdegree) = 120\textdegree. Therefore C^2 = (300 mph)^2 + (60 mph)^2 - 2 × 300 mph × 60 mph (- 1/2) = 111,600 (mph)^2; \thereforeC= 334 mph. Also, from the addition formula for vectors, we have sin \alpha = (B/C) sin \texttheta = (60 mph)/(334 mph) × (\surd3/2) = 0.156. \therefore\alpha = 9\textdegree.

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Question:

(a) ^3\surd\rule{1em}{1pt}511(b) ^4\surd(81/16)(c) ^3\surd\rule{1em}{1pt}16 \div ^3\surd\rule{1em}{1pt}2.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E33-0955.htm

Solution:

(a) By the law of radicals which states that ^n\surdab = ^n\surda ^n\surdb where a and b are any two numbers, ^3\surd\rule{1em}{1pt}512 = ^3\surd8(\rule{1em}{1pt}64) = ^3\surd8 ^3\surd\rule{1em}{1pt}64. Therefore, ^3\surd\rule{1em}{1pt}512 = ^3\surd8 ^3\surd\rule{1em}{1pt}64 = (2) (\rule{1em}{1pt}4) = \rule{1em}{1pt}8. The last result is true because (2)^3 = 8 and (\rule{1em}{1pt}4) = \rule{1em}{1pt}64. (b) By another law of radicals which states that ^n\surd(a/b) = (^n\surda) / (^n\surdb) where a and b are any two numbers, ^4\surd(81/16) = (^4\surd81) / (^4\surd16). Therefore, ^4\surd(81/16) = (^4\surd81) / (^4\surd16) = (3/2). The last result is true because (3)^4 = 81 and (2)^4 = 16. (c) By the law of radicals stated in example (b) ^3\surd\rule{1em}{1pt}16 \div ^3\surd\rule{1em}{1pt}2 = (^3\surd\rule{1em}{1pt}16) / (^3\surd\rule{1em}{1pt}2) = ^3[\surd(\rule{1em}{1pt}16) / (\rule{1em}{1pt}2)] = ^3\surd8 = 2. The last result is true because (2)^3 = 8.

Question:

Silver, atomic weight 107.9 and density 10.5 gm/cm^3, has one free electron per atom. Calculate the Fermi energy of the electrons.

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/Users/wenhuchen/Documents/Crawler/Physics/D33-0975.htm

Solution:

The Fermi energy is given by \epsilon_F = (h^2/8m) [3N/\piV]^2/3(1) where V is the molar volume of silver, and N is the number of free electrons in 1 mole of silver. Sincedensity = (mass/volume) density= [{molar mass (M)}/{molar volume (V)}] Hence, V = (M/density) But M is the mass of an element in atomic mass units. Therefore V = [(107.9 amu)/(10.5 g/cm^3) = [(107.9 g/mole)/(10.5 g/cm^3)] = 10.26 cm^3/mole = 1.026 × 10^-5 m^3/mole. In order to find N, note that each silver atom con-tributes one electron, (that is, its free or valence electron). Since there are Avogadro's number, (6.02 × 10^23) of particles in a mole of a substance, the mole of silver will contain 6.02 × 10^23 free electrons. These electrons are distributed into energy states accord-ing to Fermi-Diracstatistics (in other words, each energy state can contain only one particle). The energy levels are filled by placing the electrons first into the lowest energy levels, and then progressively filling the higher levels. The highest energy level into which the last electron is placed is called the Fermi energy. Substitut-ing the values of N and V into (1) and using the given data, \epsilonF= [{(6.63 × 10^-34 J\textbullets)^2} / {(8)(9.11 × 10^-31kg)] [(3 × 6.03 × 10^23elec/mole)/ (\pi×1.026×10^-5 m^3/mole)]^2/3 = 8.85 × 10^-19 Joules Since1 eV = 1.6 × 10^-19 J \epsilon_F = [(8.85 × 10^-19 J)/(1.6 × 10^-19 J/eV)] = 5.54 eV We could interpret this as a maximum kinetic energy of the electrons. Then the maximum velocity would be (1/2) mv^2 = 8.85 × 10^-19 J v^2 = [(2 × 8.85 × 10^-19 J)/(9.11 × 10^-31 kg)] = 1.94 × 10^12 m^2/sec^2 v = 1.4 × 10^6 m/sec = 0.005 c where c = speed of light.

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Question:

(a) 2-hexanone; (b) 2-methylbutanal ; (c) O-methylbenzaldehyde; (d) methyl phenyl ketone.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E20-0744.htm

Solution:

The aldehydes are derivatives of hydrocarbons whose molecules have a double bond to oxygen in place of two hydrogens at the end of a chain. Those derived from alkane hydrocarbons have the general formula RCHO, where R is the hydrogen for the first member of the series and an alkyl group for higher homologs. The longest chain containing the -CHO group is considered the parent compound and is named by changing the -e ending of the corresponding alkane to -al. The ketones are derivatives of hydrocarbons whose molecules have a double bond to oxygen in place of two hydrogens at a position other than the end of the carbon chain. Those derived from alkane hydrocarbons have the general formula RR'C = 0, where R and R' are alkyl groups. The longest chain containing the carbonyl group is considered the parent compound and is named by changing the -e ending of the corresponding alkane to -one. Where it is necessary, the locations of the carbonyl carbon and attached groups are indicated by numbers. Thus, compound (a) is a ketone whose longest chain has 6 carbons to which oxygen is attached on carbon number 2. The formula is Compound (b) is an aldehyde whose longest chain has 4 carbons to which a methyl group is attached to carbon 2 and whose end carbon atom is -CHO. The formula is Compound (c) is an aldehyde derived from benzene by replacing a hydrogen with thegroup. This compound also has a methyl group attached to the ortho position on the benzaldehyde (C_6H_5CHO) ring. The structure can be written as shown in Figure A. Compound (d) is an aromatic ketone (any compound that contains benzene derivatives is called aromatic), derived from phenols. Phenols are hydroxy derivatives of aromatic compounds whose general formula is ArOH (Ar is an aromatic group. Phenyl is C_6H_5- and is attached to In this case. This formula can be written asor as shown in the figure B.

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Question:

For a T type (toggle) clocked, master-slave flip-flop, give a) the block diagram,b) the timing diagram, c) the Kar-naugh Map,d) the excitation equation, e) the state table,andf) the state diagram.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G04-0080.htm

Solution:

a) The block diagram is shown in figure 1. The slave gets a clock pulse which is a complement of the master's clock pulse. The input to the slave is taken from the x-output of the master. (It could be taken from the x'-output, too.) b) The timing diagram is shown in Figure 2. Note that the transitions take place in x at the leading edge of the clock pulse, while transitions in Y take place at the trailing edge of the clock pulse. This is becauseof the Inverter which is incorporated inside the master of the Inverter which is incorporated inside the master slave flip-flop. c)The Karnaugh Map of the T flip-flop is shown in figure3. The Karnaugh Map of the T flip-flop is shown in figure d) The excitation equation of the T-flip-flop is obtained from the Karnaugh Map as follows: y\rightarrow= Ty' + T'y. \rightarrow e) The State Table, shown in figure 4, is obtained from the excitation equation: Ty y\rightarrow \rightarrow REMARKS 00 01 0 1 No Change 10 11 1 0 FLIPS f) The state diagram, drawn from the Karnaugh Map is shown in fig. 5.

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Question:

Howarethe type of muscle tissues differentiated?

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/Users/wenhuchen/Documents/Crawler/Biology/F02-0052.htm

Solution:

The cells of muscle tissue have great capacity for contraction. Muscles are able to perform work by the summed contractions of their individualcells. The individual muscle cells are usually elongate, cylindricalor spindle-shaped cells that are bound together into sheets or bundlesby connective tissue. Three principal types of muscle tissue are found in vertebrates. Skeletal or striated muscle is responsible for most voluntary movements. Smooth muscle is involved in most involuntary movements of internal organs, such as the stomach. Cardiac muscle is the tissue of which much ofthe heart wall is composed; it is also involuntary. Most skeletal muscle, as the name implies, is attached to the bones ofthe body, and its contraction is responsible for the movements of parts ofthe skeleton. Skeletal muscle contraction is also involved in other activitiesof the body, such as the voluntary release of urine and feces. Thus the movements produced by skeletal muscle are primarily involved withinteractions between the body and the external environment. Skeletal muscle reveals a striated appearance, and is therefore alsoreferred to as striated muscle. These striations are actually due to the regulararrangement of thick and thinmyofilamentsin individual muscle fibercells. Skeletal muscle is an exception to the common observation thateach cell contains only one nucleus: each skeletal fiber cell is multinucleated: it has many nuclei. Skeletal muscle can contractcery rapidlybut cannot remain contracted; the fibers must relax before the next contrac-tioncan occur. Smooth muscle can be classified as visceral smooth muscle. It is foundin the walls of hollow visceral or-gans, such as the uterus, urinary bladder, bronchioles and much of the gastrointestinal tract. Vascular smoothmuscle refers to that smooth muscle in the walls of blood vessels. Unlike skeletal muscle cells that are cylindrical, multinucleate, and striated, smooth muscle cells are spin-dle-shaped,uninucleate, and lack striations. This lack of striations accounts for its smooth appearance. They havea slower speed of contraction, but can remain contracted for a longer periodof time than the striated muscle cells. Cardiac muscle has properties similar to those of both skeletal and smoothmuscles. Like skeletal muscle cells, cardiac muscle cells are striated. Like smooth muscle fibers, cardiac muscle fibers are uninucleatedand are designed for endured contraction rather than speedy orstrong contraction. Cardiac and smooth muscles are not voluntarily controlledbut have spontaneous activities, and are regulated by the autonomicnervous system. The skeletal muscle, being voluntary, is controlledby the somatic nervous system. Comparision of the types of muscle tissue Skeletal Smooth Cardiac Location Attached to skeleton Walls of visceral organs. Walls of blood vessels Wails of heart Shape of fiber Elongate, cylindrical, blunt ends. elongate , spindle shaped, pointed ends. Elongate, cylindrical, fibers branch and fuse. Number of nuclei per fiber Many One Many Position of nuclei Peripheral Central Central Cross striations Present Absent Present Speed of contractions Most rapid Slowest Intermediate Ability to remain contracted Least Greatest Intermediate Type of control Voluntary Involuntary Involuntary

Question:

A hoist raises a load of 330 pounds a distance of 15 feet in 5 seconds. At what rate in horsepower is the hoist working?

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/Users/wenhuchen/Documents/Crawler/Physics/D05-0293.htm

Solution:

Power is equivalent to work per unit of time. In 5 seconds, the hoist does W = Fs = 330 lbs × 15 ft = 4950 ft-lbs of work. The rate at which work is done (the power) is: (4950 ft-lb)/(5 sec) = 990 ft-lbs/sec Since there are 550 ft-lb/sec per horsepower, [(990 ft-lbs/sec)/(550 ft-lbs/sec-hp)] = 1.8 hp

Question:

Determine the value of x such that 10^x = 3.142.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E33-0969.htm

Solution:

The statement 10^x = 3.142 is equivalent by de-finition to log_10 3.142 = x. Thus we must find log 3.142, using the following interpolation: We set up the proportion, (.002 / .01) = x / .0014 Cross multiply to obtain, .01x = .0000028 x = .00028 x \approx .0003 Thus log 3.142 = .4969 + .0003 = 0.4972 Therefore x = log_103.142 = 0.4972.

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Question:

It is known that the sun radiates approximately 3.8 × 10^26 J of energy into space each second. Determine how much mass must be converted into energy each second, and how many years these thermonuclear reactions may continue at this rate if the sun's total mass is 2 × 10^30 kg .

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/Users/wenhuchen/Documents/Crawler/Physics/D32-0936.htm

Solution:

The mass, m, converted to energy, E, in each second may be obtained from the relationship, E = mc^2 m = E / c2= (3.8 × 10^26 J) / {(3 × 108m/s)^2} = 4.2 × 109kg Since 1 year equals 3.1 × 107s, the mass lost in 1 year is, m = (4.2 × 10^9 kg/s) (3.1 x 10^7 s/year) . =1.3 × 10 kg/year The time t required to consume the sun's mass is, t = sun's mass / mass loss per year = (2 × 10^30 kg) / (1.3 × 10^17 kg/year) = 1.5 10^13 years

Question:

What volume (in cc) is occupied by a block of woodof dimensions25.0 m × 10.0 cm × 300 mm. All edges are 90\textdegree toone another.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E01-0008.htm

Solution:

Since all of the edges are 90\textdegree to one another, one knows that the blockis a rectangular solid. The volume of a rectangle is equal to the lengthtimes the width times the height. If one wishes to find the volume in cubiccentimeters, the lengths of all of the sides must be first expressed in centimeters. There are 100 cm in 1 m; thus, to convert meters to centimeters, thenumber of meters must be multiplied by 100 cm/1 m. 25.0m × 100 cm/1 m = 2500 cm. There are 10 mm in 1 cm; thus, to convert milli-meters to centimeters, multiply the number of milli-meters by 1 cm/10 mm. 300 mm × 1 cm/10 mm = 30 cm Solving for the volume: volume= 2500 cm × 10.0 cm × 30 cm = 7.50 × 10^5 cc.

Question:

The flywheel in a delivery truck is mounted with its axis vertical, and thus acts as a stabilizing gyroscope for the truck. Calculate the torque that would have to be applied to it when it is rotating at full speed to make it precess in a vertical plane.

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/Users/wenhuchen/Documents/Crawler/Physics/D07-0346.htm

Solution:

Figure (a) shows the situation. The flywheel is to precess about axis AA' as shown in figure (b). Originally, the angular momentum vector L^\ding{217} is as shown in figure (b), position 1. After a time ∆t, L^\ding{217} has the new value L(t + ∆t), and the angular momentum vector has gone through an angle ∆\varphi, as shown in the vector diagram. Note that we have neglected the fact that there is a component of L^\ding{217} along AA' due to the precession of the flywheel. This approximation is valid if the rate of precession, w_p , small. By the relation (dL^\ding{217}/dt) = \cyrchar\cyrt^\ding{217}(1) where \cyrchar\cyrt^\ding{217} is the net torque on the flywheel, we see that if we are to change L^\ding{217}, and thereby cause precession to occur, we must exert a torque \cyrchar\cyrt^\ding{217}. Now, the most efficient way for the torque to cause precession is if it acts in a direction perpendicular to L^\ding{217}, as shown in figure (b). As a result of (1), dL^\ding{217} is also perpendicular to L^\ding{217}, and the length of L^\ding{217} doesn't change. Hence ∆L/∆t \approx ɩ If ∆L^\ding{217} is small, and \cyrchar\cyrt^\ding{217} is always perpendicular to L^\ding{217}, \vert∆L^\ding{217}/L^\ding{217}\vert = ∆L/L = ∆\varphi or\tau \approx L(∆\varphi/∆t) = L w_p \tau \approx L w_p(2) butL = I w_s(3) where I is the moment of inertia of the flywheel about its symmetry axis and w_s is the spin angular velocity of the disc. Using (3) in (2) and w_s is the spin angular velocity of the disc. Using (3) in (2) \tau\approx I w_s w_p AlsoI = (1/2) Mr^2 where M and r are the mass and radius of the disc. Final-ly \tau \approx (1/2) Mr^2 w_s w_p

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Question:

Calculate the percentage composition of aluminum sulfate, Al_2 (SO_4)_3.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E04-0145.htm

Solution:

Because the formula of a compound is constant, the percentage composition of each element present can be calculated by using the parts by weight of each element in one molecular weight of the compound. The molecular weight of the compound can be calculated by adding to-gether the weights of the various elements contained in the compound. There are 2 atoms of aluminum present, and the molecular weight of aluminum is 27, so the contribution of the aluminum to the compound's total molecular weight is found by multiplying27 by 2. atoms of aluminum weigh 2 × 27 = 54 The weight contributions of the other elements present can be calculated in the same way. The molecular weight of sulfur is 32, and there are 3 sulfur atoms present in the compound. 3 atoms of sulfur weigh 3 × 32 = 96 The molecular weight of oxygen is 16, and there are 12 oxygen atoms present in the compound. 12 atoms of oxygen weigh 12 × 16 = 192 The weights of the three elements contained in the compound are added together to find the total molecular weight of the compound. 2 atoms of aluminum=54 3 atoms of sulfur=96 12 atoms of oxygen=192 molecular weight of A1_2(SO_4)_3 342 The percentage composition of each element present can now be calculated byconsidering what fraction by weight each element is of the total compound. The weight of two atoms of aluminum is 54 so the fraction by weight of aluminum in the compound can be found by dividing 54 by 342, the molecular weight of the compound. The percentage is then found by multiplying this fraction by 100. percentage composition of aluminum = [(weight of aluminum in compound) / (molecular weight compound)] × 100 percentage composition of aluminum = (54/342) × 100 = 15.8 %. The same method can be applied to the sulfur and oxygen. percentage composition of sulfur = (96/342) × 100= 28. 1% percentage composition of oxygen =(192/342) × 100 = 56.1% If the percent compositions of all the elements in a compound are addedtogether they will equal 100.0. percentage composition of aluminum= 15.8 percentage composition of sulfur= 28.1 percentage composition of oxygen=56.1 100.0

Question:

The period of a compound pendulum is 2 s on the earth's surface. What is its period if it is aboard a rocket accelerating upward with an acceleration of 4.3 m\bullets^-2?

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/Users/wenhuchen/Documents/Crawler/Physics/D09-0385.htm

Solution:

When on the surface the compound pendulum has a period T = 2\pi\surd(I_0/Mgh), where M is the mass of the pendulum, I is its moment of inertia about the axis of rotation, h is the distance between the center of mass and the axis of rotation and g is the accelera-tion of gravity, as shown in the figure. Under acceleration a^\ding{217} upward, the forces acting on the body when it is displaced through an angle \texttheta are those shown in the figure, the weight Mg^\ding{217} downward and the vertical force F^\ding{217} and horizontal force R^\ding{217} exerted by the pivot on the pendulum. Consider the linear accelerations, horizontal and vertical, acting on the body at its center of mass and the rotational acceleration about the center of mass. Applying Newton's Law to the horizontal and vertical motions, we get R = Ma_H(1) F - Mg = Ma_v(2) The equation of motion for the rotation is Net torque = Rh cos \texttheta - F h sin \texttheta = I_G \alpha(3) where I_G is the moment of inertia with respect to the center of mass, and \alpha is the angular acceleration. The point O thus has an acceleration, a_H horizontally, an acceleration a_v vertically, and a further linear acceleration a_v = h\alpha at right angles to OG, due to the rotation about G. But the point O does not move sideways, only upward with an accelera-tion a. Thus, there is no net horizontal acceleration a_h + h \alpha cos \texttheta = 0(4) Substituting (4) in (1), R = - Mh \alpha cos \texttheta The upward acceleration is a = a_v - h \alpha sin \texttheta ora_v = a + h \alpha sin \texttheta(5) Substituting (5) in (2) F = Mg + M (a + h \alpha sin \texttheta) = M (g + a + h \alpha sin \texttheta) and finally equation (3) becomes I_G \alpha= Rh cos \texttheta - Fh sin \texttheta = - Mh^2 \alpha cos^2 \texttheta - Mh^2 \alpha sin^2 \texttheta - M(g +a)h sin \texttheta = - Mh^2 \alpha (cos^2 \texttheta + sin^2 \texttheta)- M(g +a)h sin \texttheta = - Mh^2 \alpha - M(g +a)h sin \texttheta or(I_G + Mh^2) \alpha = I_0\alpha = - M(g +a)h sin \texttheta where we used the equation I_0 = I_G + Mh^2. But the angle \texttheta is small, and sin \texttheta can be replaced by \texttheta, \alpha = [(d^2\texttheta)/(dt^2)] = - [{M(g +a)h}/I_0-] sin \texttheta \approx - [{M(g +a)h}/I_0] \texttheta. This is the differential equation of a simple harmonic motion, therefore it follows from the theory of simple harmonic motion that the period of oscilla-tion is T' = 2\pi\surd[I_0/{- M(g +a)h}] If T is the period of the motion when the pendulum is at rest on the earth T'/T = [2\pi\surd{I_0/(M(g +a)h)}] /[2\pi\surd{I_0/(Mgh)}] = \surd[g/(g + a)] = \surd[(9.8 m\bullets^-2)/(9.8 m\bullets^-2 + 4.3 m\bullets^-2)] = 0.83 HenceT' = 0.83 T = 0.83 ×2 s = 1.67 s. Note that the result can be obtained more quickly if the idea of an accelerated frame of reference is applied. An observer in the rocket considers the point of support of the pendulum at rest relative to himself. To explain the observed equilibrium he finds it necess-ary to postulate a force Ma acting downward on the body in addition to the weight Mg^\ding{217}. Hence the pendulum actsas if the weight were M(g^\ding{217} + a^\ding{217}) instead of Mg^\ding{217}, from which the formula follows immediately. Mg^\ding{217}, from which the formula follows immediately.

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Question:

20REM INTREST PROBLEM 30INPUT P, I, N 40LET A = P\textasteriskcentered (1 + I) \textasteriskcenteredN 50PRINT "P = "; P, "I = ";I, "N = "; N, "A = "; A 60END

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G25-0595.htm

Solution:

When a computer program is written, it usually contains two different types of errors. The first kind are grammatical (or syntactical) errors, the second kind are logical errors. Syntactical errors are caused by incorrectly written program statements. These errors are easily detected since a program containing such errors usually cannot be executed. In addition, the computer prints out a diagnostic message that identifies the location and nature of each error. When the program contains a logical error the computer correctly carries out the given instructions, but the wrong instructions produce incorrect results. Turning to the given problem, the FORTRAN program has the following errors: 1) FORMAT has been misspelled as FORNAT in statement 200. Since FORNAT is not part of the compiler's vocabulary, an error message is printed. The program cannot be executed because the nature of the input has not been specified (this is, of course, done by the FORMAT statement). 2) In FORTRAN, instructions must correspond exactly to a given prototype. Any deviation results in an error message. Here, the READ statement has a comma after it. In FORTRAN, READ is followed by a comma only in the unformatted input statement. Since the READ statement in the given program is formatted, there should be no comma after READ. 3) The statement A = (P\textasteriskcentered(l + I) \textasteriskcentered\textasteriskcenteredN is missing a right parenthesis. Some compilers (PL / I for example) may make the insertion, but FORTRAN is strict. An error message is sent out, and the program 'bombs'. 4) The WRITE (6,20) P,I,N,A statement contains a reference to a statement labelled 20 (the alleged FORMAT statement). But since the corresponding FORMAT statement is labelled 200, the compiler is stymied. It prints an error message. Note that all of the above errors were syntactical errors. Consider the BASIC program for solving the compound interest problem. Statement 40 is 40 LET A = P\textasteriskcentered(1 + I)\textasteriskcenteredN(1) Algebraically, the formula for the amount A from a given principal P held for N years at interest I is A = P(1 + I)^N. But statement 40 says A = P(1 + I)N Hence it should be changed to A = P\textasteriskcentered(1 + I) \uparrow N.(2) Comparing (1) and (2) one is tempted to think that the \uparrow was mistyped as \textasteriskcentered and that this is a syntax error. But the compiler will accept (1) as a valid BASIC statement. The CPU will execute it and print the results which, unfortunately, are useless to the programmer. Hence this is a logical, not a syntactical error.

Question:

The actual physical distances between linked genes bear no direct relationship to the map distances calculated on the basis of crossover percentages. Explain.

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/Users/wenhuchen/Documents/Crawler/Biology/F25-0681.htm

Solution:

In certain organisms, such asDrosophilia, the actual physical locations of genes can be observed. The chromosomes of the salivary gland cells in these insects have been found to duplicate themselves repeatedly without separating, giving rise to giant bundled chromo-somes, called polytenechromosomes. Such chromosomes show extreme magnification of any differences in density along their length, producing light and dark regions known as banding patterns. Each band on the chromosome has been shown by experiment to correspond to a single gene on the same chromosome. The physical location of genes determined by banding patterns gives rise to a physical map, giving absolute distances between genes on a chromosome. Since crossover percentage is theoretically directly proportional to the physical distance separating linked genes, we would expect a direct correspondence between physical distance and map distance. This, however , is not necessarily so. An important reason for this is the fact that the frequency of crossing over is not the same for all regions of the chromosome . Chromosome sections near thecentromereregions and elsewhere have been found to cross over with less frequency than other parts near the free end of the chromosome. In addition, mapping units determined from crossover percentages can be deceiving. Due to double crossing over (which results in a parental type ), the actual amount of crossover may be greater than that indicated by recombinant type percentages. However, crossover percentages are nevertheless invaluable because the linear order of the genes obtained is identical to that determined by physical mapping.

Question:

Describe the structures found in a cross section of the mammalian spinal cord.

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/Users/wenhuchen/Documents/Crawler/Biology/F20-0507.htm

Solution:

The mammalian spinal cord extends from the base of the brain and is tubular in shape. Along with the brain, the spinal cord makes up the central nervous system of all vertebrates. It has two very important functions: to transmit impulses to and from the brain, and to act as a reflex center. The spinal cord is protected by the vertebral column. The vertebral column is composed of segments of bone, called the vertebrae, connected to each other by cartilage. The cartilage gives it flexibility, while the bone gives it strength. It is possible to anatomically divide the spinal cord into five different regions using the vertebrae as a guide. These are (1) cervical region, (2) thoracic region, (3) lumbar region, (4) sacral region, and (5) coccygeal region. The regions of the spinal cord send out nerves which innervate different parts of the body (see Figure 1). The spinal cord is surrounded by three membranes: the dura mater, arachnoid, and pia mater. It is a hollow, tubular structure. The hollow portion, called the spinal lumen, runs through the entire length of the spinal cord and is filled with spinal fluid. A cross-section shows 2 regions, an inner mass of gray matter composed of nerve cell bodies, and an outer mass of white matter made up of bundles of axons and dendrites (see Figure 2). The coloration of the white matter region is due to the presence of white myelinated fibers. The nerve cells of the gray matter lack myelin, which is white, and hence the "natural" gray color of nerve cells is seen. Four protuberances of the gray matter are noted. The two anterior processes are called the ventral horns, and the two posterior ones are called the dorsal horns. The axons and dendrites of the white matter carry impulses between lower levels and higher levels of the spinal cord, or between various levels of the spinal cord and the brain. The spinal cord receives sensory fibers from peripheral receptors at the dorsal root. The cell bodies from which the sensory fibers arise are located in a cluster, called the dorsal root ganglion, in the dorsal root. These sensory fibers pass into the dorsal horns of the gray matter, where they synapse with interneurons in the dorsal horns and/or motor neurons in the ventral horns. Axons from the motor neurons leave the spinal cord at the ven-tral root and soon join the sensory fibers - together they constitute the spinal nerve. Spinal nerves arising from the spinal cord branch to supply various parts of the body except the head, part of the neck, the thorax, abdomen, and the upper and lower extremities.

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Question:

What is the physiological basis of allergies? What measures can be taken to decrease the effect of allergies?

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/Users/wenhuchen/Documents/Crawler/Biology/F14-0367.htm

Solution:

Although immune reactions generally have a protective role in the body, they can sometimes cause damage to the body itself. Such an adverse immunological response, called an allergy or hypersensitivity reaction, results from an acquired reactivity to an antigen. The reactivity can result in bodily damage upon subsequent re-exposure to that specific antigen. The allergic response can be elicited by a variety of antigens, including pollen, dust, certain foods and bee stings. Initial exposure to the antigen leads to some anti-body synthesis (the primary response), but also to the formation of more lymphocytes which act as memory cells. Subsequent exposure to the same antigen elicits a more vigorous response (secondary response); many memory cells are stimulated to differentiate into plasma cells which produce the corresponding antibody. These particular antigens stimulate production of IgE immunoglobinswhich circulate in the blood and bind to mast cells (found in connective tissue) andbasophils(a type of white blood cell), When the antigen forms a complex with theIgEattached to the mast cell, release of histamine and other chemicals from the mast cell results. The symptoms of the allergy result from the effects of these chemicals. For example, when a previously sensitized person inhales pollen, the antigenic part of the pollen combines with theIgE-mast cell complex in the respiratory passages. The histamine released causes increased mucous secretion, in-creased blood flow and constriction of smooth muscle lining the passages. These physiological changes cause the familiar symptoms of running nose, congestion and difficulty of breathing. In some people, a very severe allergic reaction called anaphylaxis can occur which may have systemic (non-localized) symptoms. People have died from anaphylaxis resulting from a bee sting. To decrease the effect of allergies, one can ad-minister antihistamines. These only provide incomplete relief, however, since other chemicals are also released. A therapy called desensitization can help allergic people. In this process, the antigen is injected repeatedly into the person in small but increasing dosages. The theory behind desensitization is that this procedure inducesIgGsynthesis against the antigen. When the antigen is sub-sequently presented, it is more likely to becomplexedby IgG antibodies than byIgEantibodies bound to mast cells.

Question:

How does the term "growth" as used in bacteriologydiffer from the same term as applied to higher plants and animals?

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/Users/wenhuchen/Documents/Crawler/Biology/F05-0132.htm

Solution:

When a small number of bacteria are transferred into the proper medium and incubated under the appropriate physical conditions, a tremendous increase in the number of bacteria results in a short time. As applied to bacteria and microorganisms, the term "growth" refers to an increase in the entire population of cells. When we speak of the growth of plants and animals, we usually refer to the increase in size of the individual organism. The growth of bacteria involves the increase in numbers of cells over the initial quantity used to start the culture (called the inoculum) . Some species of bacteria require only a day to reach their maximum population size, while others require a longer period of incubation . Growth can usually be determined by measuring cell number, cell mass, or cell activity.

Question:

A book publisher is in the business of producing volumes in mathematics, science and the humanities. He basically uses two kinds of labor a) skilled for writing the books b) unskilled to correct the completed volumes. His average costs per book for each type of labor and each type of book is given below: Expenses Mathematics Science Humanities Skilled Unskilled 17.00 6.00 16,00 5.00 12.00 4.00 Write a program in Basic to print the costs of skilled and unskilled labor needed to produce M books in math, S books in science, and H books in humanities, if M = 1, 2, 3, 4, 5; S = 2M + 1; H = 2M - 2.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G18-0452.htm

Solution:

We can use matrices to facilitate the solution of this problem │17.00 (M) + 16.00(2M + 1) + 12.00 (2M - 2)│(1) │6.00 (M) + 5.00(2M + 1) + 4.00(2M - 2)│ If we multiply (1) by the vector │M│ │2M + 1│we obtain: │2m - 2│ │175001620012400 ││M│ │620055004900││2M + 1 │ │2M - 2 │ =│17500M + 16200(2M+1) + 12400(2M-2)│(2) │6200M +5500(2M+1) + 4900(2M-2)│ Adding the two rows of (2) together gives the total cost of skilled and unskilled labor. Note that costs of skilled labor comprise the first row of the result in (2) , while costs of unskilled labor comprise the second row. 10DIM A (1, 2), L(1): REM L(0), L(1) DENOTE SKILLED, UN-SKILLED LABOR 15FOR I = 0 TO 1 20FOR J = 0 TO 2 25READ X 30LET A (I, J) = X 35NEXT J 40NEXT I 45PRINT "MATH", "SCIENCES", "HUMANITIES", "SKILLED LA-BOR", "UNSKILLED TOTAL COSTS" 50PRINT 55FOR M = 1 TO 5 60LET S = 2\textasteriskcenteredM-1 65LET H = 2\textasteriskcenteredM-2 70FOR I = 0 TO 1 75LET L(I) = A(I,0)\textasteriskcenteredM + A(I,1)\textasteriskcenteredS + A(I,2)\textasteriskcenteredL 80NEXT I 85PRINT M, S, H, L(0), L(l); ""; L(0)+L(1) 90NEXT M 95PRINT 100STOP 105DATA 17, 16, 12, 6, 5, 4 110END

Question:

(a) Establish the numerical relationship between the frequencyѵ_min sec^-1 of an emitted x-ray and the minimum voltage V applied to an x-ray tube. (b) What is the minimum voltage which must be applied to the x-ray tube to produce x-rays of wavelength 10^-8 cm?

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/Users/wenhuchen/Documents/Crawler/Physics/D35-1038.htm

Solution:

(a) The minimum potential required to boost the speed of the electron between the plates of an x-ray tube, to a value sufficient to excite the target atoms on the screen, is determined by the wavelength of the emitted photons (x-rays). The kinetic energy of the electron should therefore be at least as large as the energy of the outgoing photon. The electron, when moving between the plates, acquires the kinetic energy KE =eV where V is the voltage across thetuve. KE = V(volts) × 1.60 × 10^-19 joule = 1.60 × 10^-12 V erg. (This could also have been obtained directly from the knowledge that 1eV is 1.60 × 10^-12 erg.) Therefore, the minimum energy of an electron being equivalent to its minimum frequency multiplied by Planck's constant h, we find E =hѵ_m= 1.60 × 10^-12V ѵ_m= [(1.60 × 10^-12 V) / (6.625 × 10^-27)] s^-1 ѵ_m= (2.41 × 10^14 V)s^-1 . This is the required relationship betweenѵ_mand V. (b) A wavelength, \lambda, of 10^-8 cm corresponds to a frequency ѵ = c/\lambda = [(3 × 10^10) / 10^-8] see^-1. = (3 × 10^18) see^-1. In order to produce this frequency, the voltage across the tube must be at least V_min= ѵ / (2.41 × 10^14 V \bullet s) = [(3 × 10^18) / (2.41 × 10^14)]V = 12,500 volts.

Question:

Two long-winged flies were mated. The offspring consisted of 77 with long wings and 24 with short wings. Is the short- winged condition dominant or recessive?What are the genotypes of the parents?

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/Users/wenhuchen/Documents/Crawler/Biology/F25-0651.htm

Solution:

When we are not told which of the charac-teristics is dominant and which is recessive, we can deduce it from the ratio of phenotypes in the progeny. We know that 77 flies have long wings and 24 have short wings. This gives us an approximate ratio of 3 long-winged flies to every 1 short- winged fly (77)/(24) \sim (3/1) As previously noted, the three-to-one ratio signifies that dominant and recessive characteristics are most likely involved. Moreover, because there are three long-winged flies to every short-winged one, it suggests that short-winged is the recessive characteristic, and long-winged is dominant. We cannot immediately conclude that both the long--winged parents are homozygous. In fact they are not, because if they were, no short-winged offspring could have resulted in the cross. So the presence of short-winged flies (homozygous recessive) in the progeny suggests that both parents carry the recessive gene and are thus heterozygotes. Let L be the gene for long wings in flies and l be the gene for short wings in flies. In the cross between two long-winged heterozygous parents: F_11 LL:2 Ll:1 ll long wingshort wing The phenotypes of the F_1 show the three-to-one ratio of long-winged flies to short-winged flies, which concurs with the data given. Therefore the genotypes of the parents are the same, both being heterozygous (Ll).

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Question:

Using the S-R (set-reset) flip-flop, design a memory cell with the following inputs: Data-in, Read/Write, Select. Output: Data-out.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G05-0094.htm

Solution:

The memory cell, which is the basic building block of the memory unit in a computer, is capable of storing one bit of information. The memory cell is shown in Figure 1. The memory cell is used to store information (Write) or recall information (Read). To store information, connect the Select line to logic level 1, the Data-in line to the logic level to be stored, and the Read/Write line to logic level 0. Under these conditions gates G_1 and G_2 are enabled and gate G_3 is disabled. The values of the S and R inputs of the flip-flop are the Data-in and Data-in, respectively. Since the S and R inputs are complements of each other, the output of the flip-flop, Q, has the same logic level as the S input. Hence data has been written into the memory cell. After writing in the data, the Select line is connected to logic level 0. This disables all the gates so that changes in the Data-in line no longer affect the inputs to the flip-flop. Also, when the Select line is 0, both inputs to the flip-flop are 0 and the flip-flop is in its "memory" state, that is Q retains the previous value of S. To read data from the memory cell, the Select line and the Read/Write line are connected to logic level 1. Under these conditions gates G_1 and G_2 are disabled and G_3 is enabled. Hence, the value of Q appears at the Data-out line. After reading the data the Select line is connected to logic level 0.

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Question:

A 60.0-lb block rests on a smooth plane inclined at an angle of 20\textdegree with the horizontal. The block is pulled up the plane with a force of 30.0 lb parallel to the plane. What is its acceleration?

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/Users/wenhuchen/Documents/Crawler/Physics/D04-0117.htm

Solution:

Here three forces are acting on the block. Its weight W is 60 lb downward. The force of the plane on the block is a thrust N normal to the plane. There is a pull P parallel to the plane. The force acting on the block can be resolved into forces acting normal and parallel to the plane. The weight of the block may be resolved into components of 60.0 lb × cos 20\textdegree normal to the plane and 60.0 lb × sin 20\textdegree parallel to the plane. Since there is no motion in the direction perpendicular to the plane, forces in that direction cancel each other. Therefore, the normal component of the weight is balanced by the force N. Parallel to the plane, taking the direction of P as positive, the sum of the forces is F= 30.0 lb - 60.0 lb × sin 20o = 30.0 lb - (60.0 × 0.342)lb = 9.5 lb m= W/g = 60.0 lb/(32 ft/sec^2) = 1.87 slugs From F = ma, a= F/m = (9.5 lb) / (1.87 slugs) = 5.1 ft/sec^2. Note that if the angle were 30\textdegree, the component of the weight down the plane would be equal to the force up the plane and there would be no unbalanced force acting on the block. Hence it would not be accelerated. If the angle were greater than 30\textdegree, the block would be accelerated down the plane.

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Question:

Write a function to compute the Legendre polynomials.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G16-0413.htm

Solution:

For positive values of n, the recursive definition to compute the Legendre polynomials is given by 1ifn = 0 P_n(X) =xifn = 1 ((2n - 1)xP_n-1 (x) - (n - 1)P_n-2(x)) /_nif n>1 we obtain: FunctionP(n:integer; x:real) :real; BEGIN IF n = 0 THEN p: = 1 ELSE IF n = 1 THEN p:=x ELSEp:=((2\textasteriskcenteredn-1)\textasteriskcenteredx\textasteriskcenteredp(n-1,x) - (n-1)\textasteriskcenteredp(n-2,x)) / _n END; In most of the cases, however, the recursive solution is not the best because a simple solution may be obtained by itera-tion. The Legendre polynomials are more efficiently computed in this way: FUNCTIONp(n:integer; x:real) :real; VAR prev, this, next :real; count: integer; BEGIN IF n = 0 THEN p: =1 ELSE IF n=1 THEN p: =x ELSE BEGIN prev: =1; this:=x; FOR COUNT: = 2 TO n DO BEGIN next:= ((2\textasteriskcenteredcount-1)\textasteriskcenteredx\textasteriskcentered this -( count - 1)\textasteriskcenteredprev) / count; prev:= this; this :=next END; {for} p: =next END END; {p}

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Question:

A flat bottom swimming pool is 8 ft. deep. How deep does it ap-pear to be when filled with water whose refractive index is 4/3?

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/Users/wenhuchen/Documents/Crawler/Physics/D27-0857.htm

Solution:

In order to see why we would expect to observe a different depth for the pool when it is filled with water, examine the figure. If no water is in the pool, light coming from a point S on the bottom of the pool will travel directly to the observer's eye. If the pool is filled with water, light emanating from point S will be refracted at P, as shown. Upon reaching the observer's eye, the light appears to be coming from Q and he perceives the depth of the pool to be the dis-tance OQ, rather than the actual depth OS. Our problem is to find the distance d. Note that, from the figure, tan \varphi_1 = (OP/d) tan \varphi_2 = (OP/8 ft). Hence (tan \varphi_1/tan \varphi_2) = (OP/d) \textbullet (8 ft./OP) = (8 ft./d) and d = [{(8 ft)tan \varphi_2}/tan \varphi_1](1) From Snell's Law, n_1 sin \varphi_1 = n_2 sin \varphi_2 where n_1 and n_2 are the indices of refraction of air and water, re-spectively. Therefore (n_1/n_2) sin \varphi_1 = sin \varphi_2(2) To calculate the tangents in (1), we must also know cos \varphi_1 and cos \varphi_2. These we may find by observing that cos \varphi = \surd(1 - sin^2 \varphi)(3) Using (2) in (3) cos \varphi_2 = \surd(1 - sin^2 \varphi_2) cos \varphi_2 = \surd[1 - (n_1/n_2)^2 sin^2 \varphi_1(4) cos \varphi_1 = \surd(1 - sin^2 \varphi_1) Hence tan \varphi_1 = (sin \varphi_1/cos \varphi_1) = {sin \varphi_1/\surd(1 - sin^2 \varphi_1)}(5) and using (2) with (4) tan \varphi_2 = (sin \varphi_2/cos \varphi_2) = [{(n_1/n_2) sin \varphi_1} / \surd{1 - (n_1/n_2)^2 sin^2 \varphi_1}](6) Substituting (5) and (6) in (1) d = (8 ft)[{(n_1/n_2) sin \varphi_1} / \surd{1 - (n_1/n_2)^2 sin^2 \varphi_1}] \textbullet [ \surd{1 - sin^2\varphi_1} / (sin \varphi_1)] d = (8 ft)(n_1/n_2) \surd[(1 - sin^2 \varphi_1) / {1 - (n_1/n_2)^2 sin^2\varphi_1}](7) Now, since we don't know the angle \varphi_1., we make an approximation. Sup-pose \varphi_1 is very small. (This means that the observer is looking almost directly down into the pool.) Then sin \varphi_1 \approx 0 and the square root in (7) becomes 1. Therefore d = (8 ft)[n_1/n_2] = (8 ft)[1/(4/3)] = 6 ft. The pool appears to be 6 ft. deep.

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Question:

The density of a 25.0 % sugar solution is 1.208 g/ml. What weight of sugar would be contained in 1.00 liter of this solution ?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E04-0156.htm

Solution:

When a solution is said to be 25.0% sugar , it means that 25.0%of the weight of the solution is made up by the sugar . Thus one can determine the weight of the sugar in this solution by multiplying the total weight of solution by .25 .Here one is not given the weight of the solution but the densityand the volume , The density is the weight 1 ml of the solution , thus the weight of 1.0 littler or 1000ml is equal to the volume 100ml times the density (1.208g/ml). weight of solution = 1000ml × 1.208 g/ml = 1208g. The weight of the sugar in the solution can now be found . weight of sugar = .25 × weight of solution = .25 × 1208 g = 302 g.

Question:

A solid cylinder of radius R rolls on a flat surface. Find the moment of inertia I_s of the cylinder about its line of contact with the surface.

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/Users/wenhuchen/Documents/Crawler/Physics/D04-0179.htm

Solution:

The definition of moment of inertia for a continuous mass distribution is dependent upon whichaxis we wish to calculate the moment axis we wish to calculate the moment of inertia about. In this case, we want to calculate I_s , the moment of inertia about an axis parallel to the symmetry axis of the cylinder, but tangent to the surface of the cylinder. To simplify the required integrations, we may equivalently calculate the moment of inertia, I_O, of the cylinder about its symmetry axis, and then employ the parallel axis theorem to find I_s. The moment of inertia I_o is (with reference to the figure) defined as: I_o = \intr^2 dm(1) where r is the perpendicular distance between the symmetry axis of the cylinder and the mass element dm. Also note that dm, the mass contained in a differential volume dv, is the mass per volume contained in the cylinder times the volume dv, or dm = [M/(\piR^2H)] r dr d\texttheta dz(2) where M is the total mass of the cylinder, and \piR^2H is the volume of the cylinder. Combining (2) and (1), we obtain I_o = \int[M/(\piR^2H)] r^3 dr d\texttheta dz(3) where the integral is over the volume of the cylinder. Performing the integral: I_o = [M/(\piR^2H)] ^H/2\int_-H/2^2\pi\int_0^R\int_0 r^3 dr d\texttheta dz I_o = [M/(\piR^2H)] [(R^4)/4] ^H/2\int-H/2^2\pi\int_0 d\texttheta dz I_o = [M/(\piR^2H)] [(R^4)/4] ^(2\pi)^H/2\int_-H/2 dz I_o = [M/(\piR^2H)] [(R^4)/4] (2\pi) (H) I_o = (1/2) M R^2(4) This is the moment of inertia about the symmetry axis. To find the moment of inertia about the axis tangent to the surface of the cylinder, use the parallel axis theorem. Mathematically, (see the diagram) I_s =I_o + M\eta^2(5) where \eta is the separation of the 2 axes. Rewriting this using (4); and noting that \eta = R, we have I_s = (1/2) MR^2 + MR^2 orI_s = (3//2) MR^2 .

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Question:

The Pauli exclusion principle is the main reason why atoms do not collapse to a point.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E18-0647.htm

Solution:

The Pauli exclusion principle states that no two electrons in the same atom can be completely identical; that is, have the same values for all four quantum numbers. The quantum numbers measure the energy level of the electron, the order of increasing distance of the average electron distribution from the nucleus, the angular shape of electron distribution, (the electronic magnetism) , and the two possible orientations of electron spin. An electron in an atom is completely described by its four quantum numbers. Inherent in this description is the point of loca-tion of an electron at a specific time. If two electrons have the same four quantum numbers, they would have, the same point of location at a specific time. This means, therefore, that all the electrons would collapse to a point at a specific time. The Pauli exclusion principle states that no two electrons can have the same four quantum numbers and therefore the atom cannot collapse to a point.

Question:

How far does a wooden (spherical) ball of specific gravity 0.4 and radius 2 feet sink in water?

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/Users/wenhuchen/Documents/Crawler/Physics/D10-0399.htm

Solution:

By "Archimedes' principle" we know that the ball will sink until it displaces a weight of water equal to the entire weight of the ball: (weight of ball) = (weight of displaced water) We know that: (weight of ball) = (volume) (specific gravity) (density of water) where the density of water w is measured in pounds per cubic foot. We must first calculate the volume of the part of the sphere immersed in water when the sphere sinks to a depth of h, and then solve for h from the given information (figure A). Finding this volume is equivalent to finding the volume generated by rotating the area between the curves x^2 + (y - R)^2 and y = h, about the y-axis, ( see figure (B)) We note that the general formula for a circle is (x - h)^2 + (y - k)^2 = r^2, where (h, k) are the coordinates of its center and r is the length of its radius. In this case the coordinates of the center are (0, R), and r = R. Recall that (see figure (C)) : dV = \pi x^2 dy = \pi[R2- (y - R)^2]^2 dy = \pi(2Ry - y^2)dy where dV is the differential cylindrical volume element, x being its radius and dy its height. Finally, we integrate dV between y = 0 and y = h. V = \pi\int^h_0 (2Ry - y^2)dy = \pi[2R (y^2/2) - (y^3/3)]^h_0 = \pi[2R (y^2/2) - (y^3/3)]^h_0 = \pi [Rh^2 - (h^3 / 3) ] = \pi [Rh^2 - (h^3 / 3) ] = \pi h^2(R - h/3) = \pi h^2(R - h/3) Hence, from the equation for the weight of the ball: (weight of ball) = (volume of submerged portion)(density of water) [(4/3) \piR^3](0.4)w = \pi h^2(R - h/2)w (4/3)(2)^3(0.4)= h^2(2 - h/3) h^3 - 6h^2 + 12.8 = 0 We find by synthetic division that h \approx 1.75 feet.

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Question:

Write a complete IDENTIFICATION DIVISION for a COBOL program which calculates the weekly paychecks of the employees of a company.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G11-0251.htm

Solution:

IDENTIFICATION DIVISION is the first part of any COBOL program. Its main purpose is to identify the program to the computer, and then it can also be used to explain to the reader what the program does. The first statement in this part of the program is the IDENTIFICATION DIVISION. In format form the statement starts from the eighth column and ends with a full stop mark (.). PROGRAM-ID. This is the second mandatory statement of IDENTIFICATIONDIVISION; it also starts from the eighth column and ends with a full stop mark. Leaving one empty space and continuing on the same line program-name follows the PROGRAM-ID statement. Program name could be any name except the reserved words such as IF, GO TO, DATA, and a hyphen must be used to separate two or more words. So, call the program PAYROLL-LIST. The third, fourth and fifth statements are not manda-tory. It is up to the programmer to decide whether the name of the author, date-written, and any extra remarks would be included or not. Following those steps, the IDENTIFICATION DIVISION, takes the complete form of fig. 1. All of the actual remarks in the REMARKS, section must come after margin B.

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Question:

A constant horizontal force of 10 N is required to drag an object across a rough surface at a constant speed of 5 m/sec. What power is being expended? How much work would be done in 30 min?

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/Users/wenhuchen/Documents/Crawler/Physics/D05-0280.htm

Solution:

Power is the rate of doing work, P = (∆W)/(∆t) = (F∆s)/(∆t) P = (∆W)/(∆t) = (F∆s)/(∆t) . (Note that in this problem the work reduces to the force multiplied by the distance the object is moved.) But ∆s/∆t is just the velocity. Therefore, P= Fv = (10 N) × (5 m/sec) = 50 J/sec = 50 W W= Pt = (50 W) × (1/2hr) = 25 W-hr. The work, of course, is done against the force of sliding friction.

Question:

The spherical atomic nucleus of potassium has a radius of about 4 × 10^-13 cm.; 6.02 × 10^23 of these nuclei weigh about 39 g. Calculate the approximate density of the nucleus of potassium in tons per cubic centimeter. Volume of a sphere = 4/3 \pir^3, where r is the radius.

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Solution:

The density is defined as the mass of a substance divided by its volume. To calculate the density one must know the weight and volume of the substance. Here one is given the weight in grams but the problem asks for the density in tons, thus one must convert grams to tons. The volume of the nuclei is not given, but can be found by using the radius of a nucleus, the formula for the volume of a sphere and the number of nuclei present. 1) Converting grams to tons There are 454 g in one pound and 2000 lb in one ton, therefore, grams can be converted to tons by multiplying the number of grams by the following conversion factor 1 lb/454 g × 1 ton/2000 lb number of tons = 39 g × 1 lb/454 g × 1 ton/2000 lb = 4.3 × 10^-5 tons. 2) Calculating volume of nuclei The volume of a sphere is equal to 4/3 \pir^3 where r is the radius of the sphere. One is given that the radius of a potassium nuclei is 4 × 10^-13 cm. volume of 1 nuclei = 4/3 \pi (4 × 10^-13 cm)^3 = 4/3 (3.14)(64 × 10^-39 cm^3) = 2.6 × 10^-37 cm^3 The total volume of the nuclei is equal to the number of nuclei times the volume of one nucleus total volume = no. of nuclei × volume of one nucleus total volume of potassium nuclei = 6.02 × 10^23 × 2.6 × 10^-37cm^3 = 1.6 × 10^-13cm^3 3) The weight in tons and the volume in cubic centi-meters is now known. The density in tons per cm^3 can be calculated density = [no. of tons (= mass)] / [no. of cm^3 (=voIume)] density = (4.3 × 10^-5 tons) / (1.6 × 10^-13 cm^3) = 2.7 × 10^8 tons/cm^3.

Question:

A sign of a good computer programmer is the ability to write a program which minimizes storage usage. The IBM 360/370 computers provide us with the DSECT and CSECT statements . How do they enable us to becomebetter programmers ?

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Solution:

The DSECT (Dummy Section) is actually an indica-tor that tells the assembler that the following code defines a dummy section. This facility enables us to define and use many variables in our program without actually allocating storage space for them all at one time. This may sound somewhat strange but the procedure is rather simple. The DSECT is an assembler instruction, and it generates no ma-chine code. It tells the assembler that the storage defini-tions which follow forms a dummy section . The CSECT (Control Section) instruction indicates the end of the dummy section. For the DSECT a substitute base register is used. The register to be used is generally specified in a 'USING' statement at the beginning of the program. When the DSECT is encountered, the displacements of the labels within it are assigned, not from the beginning of the program, but from the DSECT statement. The CSECT instruction indicates the end of the dummy section. With this, the substitute base register is restored (freed), the location counter pro-ceeds as if the DSECT instruction was not there. It has the value it had previous to encountering the DSECT instruction . Instead of having storage allocation for the 'running- life' of the program , the symbols of a DSECT are allocated space as they are encountered in the body of the program. This space is then freed when usage is completed. By placing a label in the label field of the DSECT instruction, the dummy section can be named. The label field of the SCECT statement must contain the name of the program, this name is taken from the label field of the start instruc-tion. Coding of the section is generally done as follows USINGCRDMAP, 5 APTOTSUM, SLSREG CRDMAPDESECT SLSCARDDSOCL80 SLSNBRDSCL4 SLSNAMDSCL20 SLSREGDSCL2 SLSQTYDSCL6 DSCL48 PROGNAMECSECT The 'USING' statement is generally coded after the pro-gram's base register 'USING' statement. The AP statement adds a number previously placed in SLSREG to TOTSUM. The DSECT is usually coded for repetitious procedures. In many instances it is used to replace explicit coding. This is done because DSECT is considered easier to code and there are less changes of addressing errors. It is important to note that since DSECT is a dummy section only storage space can be defined in it and not con-stants. Thus, only the 'DS' statement is used for the labels.

Question:

The cross-over percentage between linked genes A and B is 40%, between B and C, 20%, between C and D, 10%; between A and C, 20%; between B and D, 10%. What is the sequence of genes on the chromosome?

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Solution:

This question allows us further practice in ordering genes. Again we rely on the visual method because it is the most convenient. From the data, we know that A and B are 40 map units apart (recall that map units are directly proportional to cross-over per-centage) . B and C are 20 units apart. In order to determine whether C is to the right or left of B, we look at the map distance between A and C. This distance is 20 units. With this information, we now know that C is to the left of B because this is the only way that A and C can be 20 units apart. If C were to the right of B, then the distance between C and A would be 60 units, which is not the case. C and D are 10 units apart. We must determine whether D is to the right or left of C. We know that B and D are also separated by 10 units. Since B and C are 20 units apart, D must be between B and C. Note: when ordering genes, we have first to establish two points (genes) on the chromosome. Using these as reference points, we make comparisons with other genes. In the case above, our reference points were A and B and we then proceeded to compare these points with C and D. Although it is helpful to pick the two most widely separated points as reference points, the choice is actually completely arbitrary. We could just as well have started with B and D as our reference points. The answer obtained would still be the same.

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Question:

What function does oxygen serve in the body?

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Solution:

Oxygen is necessary for the process of cellular respiration. This process involves the catabolism of glucose to result in the production of energy, in the form of ATP, for utilization by the cell. Cellular respiration can be thought to have two parts. The first series of reactions can take place without the presence of oxygen, while the second series of re-actions is dependent on oxygen. In the second reaction sequence, oxygen acts as the final hydrogen acceptor. This occurs in the electron transport system within the mitochondria. The oxygen combines with hydrogen, and is converted to a molecule of water. If anaerobic con-ditions existed (absence of oxygen), no ATP formation will occur in the mitochondria. It should be noted however that two ATP per glucose oxidized are still produced by the first series of reactions, collectively called glycolysis. Glycolysis will occur regardless of the availability of oxygen, and is restricted to the cytoplasm; it does not occur in the mitochondria. The amount of ATP produced via glycolysis is very small com-pared to that produced by the second reaction sequence. The second reaction sequence, which occurs only in the mitochondria, is collectively referred to as oxidative phosphorylation. Since the mitochondria provide about 95% of all the ATP formed in the cell, the absence of oxygen results in less energy production. This energy is required to maintain cell structure and function, and the cell will ultimately die without it. Roughly 99% of all the molec-ular oxygen consumed by a cell is used for cellular respiration.

Question:

A 1-\muF capacitor charged to 200 V and a 2-\muF capacitor charged to 400 V are connected; the positive plate of each is connected to the negative plate of the other. Find the difference of potential and charge on each capacitor and the loss of energy that has taken place.

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Solution:

The charge Q on a charged capacitor of capacitance C is Q = CV where V is the potential difference between the plates of the capacitor. Thus the charge on the 1-\muF capacitor before connection is 1 \muf × 200 V = 200 \muC and on the 2-\muF capacitor is 2\muf × 400 V = 800 \muC. Fig. A depicts the situation before connection of \muf × 400 V = 800 \muC. Fig. A depicts the situation before connection of the two capacitors. If the two capacitors are now connected positive to negative, a charge of + 800 \muC and one of - 200 \muC are joined, as are charges of - 800 \muC and + 200 \muC. The situation just after the connection is depicted in Figure B. The composite capacitor thus has charges of \pm 600 \muC on its plates, since charge cannot be created or destroyed, but only neutralized. These charges will be shared between the individual capa-citors, with \pm Q_1 on the first capacitor, and \pm Q_2 on the second, since the charge flows from one to on the second, since the charge flows from one to the other until a common potential V_0 is achieved. Fig. C depicts this situation. Thus Q_1 = C_1 V_0 and Q_2 = C_2 V_0, where Q_1 + Q_2 - 600 \muC. Thus combining the above three equations (Q_1 + Q 2 ) = (C_1 + C_2 )V_0or ) = (C_1 + C_2 )V_0or 600 × 10^-6 C = (1 × 10^-6 F + 2 × 10^-6 F) V_0or V_0 = [(600 × 10^-6 C)/(3 × 10^-6 F)] = 200 V. \thereforeQ_1 = 200 \muC,Q_2 = 400 \muC. The initial energy of the charged capacitors minus the final energy is W = (1/2) C_1 V^2 _1 + (1/2) C_2 V^2 _2 - (1/2) C_1 V^2 _0 - (1/2) C_2 V^2 _0 = (1/2) × 10^-6 F × (200)^2 V^2 + (1/2) × 2 × 10^-6 F × (400)^2 V^2 - (1/2) × 10^-6 F × (200)^2 V^2 = 12 × 10^4 × 10^-6 J = 0.12 J. This energy is lost as heat due to the transient current in the connecting wires when the two capacitors are joined.

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Question:

Input records for a sales agency include, along with identifying information for each salesman, commission sales for each of the last six months. Write a procedure division to read these cards, select the largest of six commission payments, and print out the salesman's name and his highest commission payment over the months. The commission is described as COM OCCURS 6 TIMES PIC 9(4)V99. Start your program with a flowchart.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G11-0270.htm

Solution:

This problem requires the computer to read a set of cards, find the largest commission paid to each salesman, and print the particular salesman's name and commission. The flowchart can be started by the read statement, that is a command should be put into the computer to cause it to read the set of variables. To prevent an error a conditional statement should also be used immediately so that when there are no more variables to read the process will come to halt. A complete flowchart of the problem is given in fig. 1. It should be noted that each programmer has his or her own methods in programming. Therefore flowcharts also vary from person to person, depending on the method of approach to the problem. In the flowchart given, the programmer uses a set of conditional statements to find the largest commission and once it is found, the card or the group of variables is taken out of the deck and certain information contained in it is printed. Once the flowchart is done, the rest of the work is purely mechanical, and depends on the ability to use the technical knowledge related to the particular programming language, COBOL in this case. In Fig. 2 is the procedure division of the program. Note that adding the remaining three divisions IDENTIFICATION, ENVIRONMENT, and DATA would make this program complete and ready to run. The use of the OCCURS clause sets up a table of 6 com elements. To index a certain element a statement of the form: is used. As an example, the statement COM (subs) where subs = 4 gives com the value of the fourth com element in the table.

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Question:

When the higher vertebrates left the water to take on a terrestrial existence, they had to adopt mechanisms to conserve body water. One of these was the evolution of a concentrated urine. How is this achieved physiologically?

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/Users/wenhuchen/Documents/Crawler/Biology/F18-0442.htm

Solution:

The higher vertebrates have evolved a final urine having a greater osmolarity than blood plasma. This is advantageous in that water can be retained within the body making the organism less likely to suffer dehydratipn. In the absence of a special renal mechanism, the urine would be highly dilute and severe water loss from the organism could become fatal. The loop of Henle is a hairpin sequent of the renal tubule having an ascending limb and a descending limb. The connecting tubule lies in close proximity and parallel to the loop of Henle so that the three parts of the renal tubule can interact with one another (see figure). The walls of these three portions differ in their permeabilities to substances such as sodium and water. Both the anatomical and physiological aspects of the renal tubule determine the concentration of the urine excreted. The ascending limb of the loop of Henle actively pumps sodium ions out of its tubular space into the interstitial fluid, with chloride following passively due to electrostatic attraction. The ascending limb is impermeable to water and thus the osmolarity of fluid inside the ascending limb decreases as the loop ascends and a concentration gradient is established between the loop and the interstitialfluid. The descending limb is permeable both to water and sodium, so that some of the sodium and chloride ions diffuse passively into its tubular space. The cycling of sodium from ascending limb to interstitial fluid to descending limb results in the establishment of a concentration gradient of sodium and chloride in the tissue fluid surrounding the loop, with the lowest concentration near the cortex and the highest con-centration deep in the medulla. (See accompanying figure) . This mechanism of flow in the mammalian kidney, called the countercurrent flow, permits the concent-ration gradient in the interstitial fluid to be main-tained; this gradient is essential in establishing the final concentration of the urine. With thisin mind, we can understand how the final urine becomes highly concentrated relative to the blood. The urine that leaves the ascending limb is not more concen-trated than the glomerularfiltrate since it has been di-luted in the ascending loop. But as the urine flows through the collecting tubule, it is essentially flowing through a concentration gradient in the interstitial fluid from the cortex to the medulla. Moreover, the collecting tubule is permeable only to water. Since the osmolarity of the inter-stitial fluid is lowest near the cortex and increases toward the medulla, the urine flowing down the collecting tubule will lose water during it scourse to the medulla. The final urine that emerges from the collecting tubule is substan-tially hypertonic to blood and remains in this same concen-tration until it is excreted to the outside.

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Question:

What is the apparent loss of weight of a cube of steel 2 in. on a side submerged in water if the weight density of H_2O is 62.4 lb/ft3?

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/Users/wenhuchen/Documents/Crawler/Physics/D10-0398.htm

Solution:

The situation is as depicted in the diagram. Assume that a diver's hand exerts an upward force H on the steel, which is just large enough to keep the block in equilibrium. In this case, from Newton's Second Law, the net force on the block is zero and B + H - mg = 0(1) where B is the buoyant force of the water on the steel. By Archimedes' Principle, B is equal to the weight of water displaced by the block. Hence, B = \rho_w gV(2) B = \rho_w gV(2) where V is the block's volume, and \rho_w is the density of water. Therefore, solving (1) for H, and using (2), H = mg - B = mg - \rho_wgV But m may be written as m = \rho_sV(3) where \rho_s is steel's density. Finally, then H = \rho_sgV - \rho_wgV But H is the apparent weight of the steel, since this is the force we exert on the block to keep it in equi-librium. The weight of the block outside the water is given by (3) as mg = \rho_sgV The difference between "apparent" and "actual" weights is then ∆ = H - mg = (\rho_sgV - \rho_wgV) - \rho_sgV \vert∆\vert = \rho_wgV which is B. Hence, \vert∆\vert = (62.4 lb/ft^3) (2 in.)^3 = (62.4 lb/ft^3) (1/6 ft)^3 \vert∆\vert = (62.4/216) lb = .29 lb.

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Question:

Discuss feedback control involved inglycolysisand the TCA cycle.

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/Users/wenhuchen/Documents/Crawler/Biology/F03-0083.htm

Solution:

Glycolysisand the TCA cycle are both regulated by means of allostericcontrol of the enzymes involved.Aliosteryinvolves regulation of theactivity of an enzyme by the binding of a molecule at a site other than theactive site. The molecule that is bound changes the conformation of theenzyme, thereby changing its activity. If the molecule regulates an enzymaticreaction involved in its own synthesis, this regulation is called feedbackcontrol. Inglycolysis, one control point occurs at the conversion of fructose- 6-phosphate to fructose 1,6-diphosphate, catalyzed by the enzyme phosphofructokinase. This enzyme can beallostericallyinhibited by either ATP or citrate.This explains why the rate ofglycolysisdecreases as soon asaerobic respiration begins in yeast; once the electron transport chain beginsto operate, much more ATP is formed per glucose molecule. This causesan accumulation of ATP in more than sufficient amounts for cellularmetabolism. This large amount of ATP inhibits phosphofructokinaseand therefore inhibitsglycolysis. Sinceglycolysisis inhibited, so is ATP production, for the end products ofglycolysiswould normallyenter the Krebs cycle in order to produce more ATP. This makes sense, since the cell no longer needs to utilize glucose at such a high rate inorder to generate sufficient energy. The cell needs only to allow glycolysisto proceed at a rate such that the amount ofglycolyticproducts enteringthe Krebs cycle will yield an appropriate supply of ATP. Feedback inhibitionprovides the mechanism for this control. The inhibition by citrate, aKrebs cycle intermediate, works much the same way. In the Krebs cycle, a major control point occurs at the conversion of isocitrateto \alpha-ketoglutarate, catalyzed by the enzymeisocitrate dehydrogenase. This enzyme isallostericallyinhibited by either ATP or NADH.If too much ATP is being synthesized, or too much NADH producedto be oxidized by the electron transport chain, the reaction is blockedand the whole cycle ceases. Some citrate is accumulated which actstoallostericallyinhibitphospho-fructokinase, thus shutting down glycolysisas well. Withglycolysisshut down,pyruvatewill not build up or beconverted to lactate or ethanol (which would normally occur if only the TCAcycle werestopped). Allostericcontrol allows for a high degree of efficiency and cooperationinglycolysisand the TCA cycle. In addition, thereareother controlpoints in both these pathways and in other pathways as well, all servingto keep the cell's metabolic processes in balance.

Question:

How do the fork ferns differ from the true ferns? What unusual feature is present inPsilotum?

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Solution:

The fork ferns (Psilophyta) and true ferns (Pterophyta) are classified separately primarily because of the difference in their leaf structures . The leaves of the fork ferns are small, simple, scale--like, and are considered mere epidermal outgrowths ratherthentrue leaves. The leaves of the true ferns are relatively large, elaborate, vein-containing, and usually compound, being finely divided intopinnae. The fork and true ferns also differ in that the well- developed roots present in the true ferns are lacking in the fork ferns. The latter have instead, numerous unicellular rhizoids which grow off the rhizome. The rhizome of the fork ferns is usually found in association with a fungus. Sporangia of the fork ferns are borne in axils of some of the scale-like leaves; sporangia of most true ferns are, however, carried on the undersurfaces of the leaves. Psilotum, one of the two existing genera of fork ferns, is of particular interest to botanists because both its gametophyte and sporophyte have vascular tissues. Whereas the gametophyte of the bryophytes , seed plants, and almost all lower vascular plants lack a vascular system, the gametophyte ofPsilotumhas a stele complete with xylem and phloem surrounded by an endodermis.

Question:

How are oxidation and reduction reactions related? How does an oxidase enzyme differ from a dehydrogenase enzyme?

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Solution:

An oxidation reaction involves the removal of electrons from an atom or molecule. The reverse process, called a reduction reaction, involves the addition of electrons to an atom or molecule. Every oxidation reaction must be accompanied by a corresponding reduction reaction. The electrons given off by the substance oxidized (also called the reductant or reducing agent), must be accepted by the substance reduced (also called the oxidant or oxidizing agent). A simple example of an oxidation reaction is the oxidation of iron: Fe\textdegree \textemdash\textemdash\textemdash> Fe^2+ + 2e- This involves the actual removal of electrons from the iron atom. The reverse process is a reduction reaction. Fe^2+ + 2e- \textemdash\textemdash> Fe\textdegree A not-so-obvious oxidation reaction is the oxidation of methane (CH_4): CH_4 + 2O_2 \textemdash\textemdash> CO_2 + 2H_2O This involves the formation of bonds in which electrons are drawn toward a more electronegative (electron attracting) atom or atoms which is essentially the loss of electrons required in oxidation. Both carbon and hydrogen are considered to be oxidized, since in the new the electrons are removed by oxygen farther from the C and H atoms then they were in the C-H bonds of methane. Oxygen is strongly electronegative and is a common agent in oxidation reactions. Oxidation and reduction processes are important as a means of liberating and storing energy. This can be seen in the oxidation of succinate to fumarate by succinic dehydrogenase. The removal of two hydrogen atoms (with their corresponding electrons) from succinate forms fumarate. Succinate is therefore oxidized. The two hydrogen atoms are transferred to a molecule of flavin adenine dinucleotide (FAD) to form FADH_2. FAD is therefore reduced. This coupling of an oxidation to a reduction reaction traps the energy given off during the oxidation of succinate and uses it to synthesize FADH_2, an energy- requiring reduction process. FADH_2 can subsequently under-go an oxidation reaction and release this temporarily stored energy to another molecule. Oxidation processes thus liberate energy, while reduction processes act to store chemical energy. Most organisms obtain energy by enzymatic reactions which involve the flow of electrons from NADH (reduced nicotinamide adenine dinucleotide) to oxygen, the final electron acceptor. The electrons are transferred by a system of enzymes located in the mitochondria called the electron transport system. An oxidase is an enzyme which removed electrons, while a dehydrogenase is an enzyme which removes hydrogen ions (H+) and their associated electrons. Cytochrome oxidase, the final enzyme in the electron trans-port chain, transfers electrons directly to O_2 to form H_2O. Succinate dehydrogenase, in the citric acid cycle, transfers electrons and hydrogen ions from succinate to FAD, yeilding fumarate and FADH_2 as a result.

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Question:

With a certain current in circuit 1 of the figure, a flux of 5 × 10^-4 weber links with circuit 2. When circuit 1 is opened the flux falls to zero in 0.001 sec. What average e.m.f. is induced in circuit 2 ?

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Solution:

The induced e.m.f. \epsilon in circuit 2 opposes the reduction of the magnetic flux \varphi linking it by setting up its own flux in the direction of \varphi. (see figure) . The average rate of decrease of flux in circuit 2 is [(∆\varphi)/(∆t)] = [(5 × 10^-4 w)/(.001 sec)] = 0.5 weber/sec . The average induced e.m.f. is therefore, by faraday's Law, \epsilon̅ = - [(∆\varphi)/(∆t) = - 0.5 volt since 1 volt = 1 weber/sec .

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Question:

An oil drop carries a net charge of three times the electronic charge and has a radius of 10^\rule{1em}{1pt}4 cm. What is its terminal velocity when it falls between two horizontal plates kept at a potential difference of 1000 V and 2 cm apart, the positive plate being upper-most? The densities of the oil and of air are 800 kg\bulletm^\rule{1em}{1pt}3 and 1.29 kg\bulletm^\rule{1em}{1pt}3 and the visocity of air is 1.80 x 10^\rule{1em}{1pt}s N\bullets\bulletm^\rule{1em}{1pt}2.

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Solution:

Between two parallel plates, separated by a distance d and maintained at a difference of potential V, the electric intensity E is E = V/d. The electro-static force acting on the drop is upward, and of magnitude F_E = qE = qV/d. This follows from the defintion of E (q = 3e^\rule{1em}{1pt}).Three other forces are acting on the drop; its weight downward and two upward forces, the viscous retarding force f, due to the surrounding air, and the buoyant upthrust B (see figure). When the terminal velocity is achieved, the forces balance, by definition of terminal velocity. The magnitude of the viscous retarding force is given by Stoke's law: f = 6\pi\etarv\bullet r is the radius of the drop, \rho its density, \eta the viscosity of air and v its terminal velocity. The buoyant force is equal to the weight of air dis-placed by the drop (Archimede' s principle). The volume of the (spherical) drop is 4/3 \pir^3 and the density of air is \sigma. The weight of the dis-placed air (equal to B) is then (4/3\pir^3)\sigmag = B. The weight of the drop is W = mg = (4/3\pir^3)\rhog, where \rho is oil's density. Therefore W = B = f + FE (4 / 3)\pir^3 \rhog = (4 / 3)\pi r^3 \sigmag + 6\pi\etarv + qV / d,(1) Hence, v = [(4 / 3)\pir^3g(\rho \rule{1em}{1pt} \sigma) \rule{1em}{1pt} (qV / d)] /6\pi\etar = [(4 / 3)\pi× 10^\rule{1em}{1pt}18m^3 ×9.8m \bullet s^\rule{1em}{1pt}2 (800 \rule{1em}{1pt} 1\bullet 29)kg \bullet m^\rule{1em}{1pt}3 {-(3 ×1 \bullet 6× 10^\rule{1em}{1pt}19 C×103v) / 0.02 m)}] / [6\pi×1.80×10\rule{1em}{1pt}5N\bullets\bullet m^\rule{1em}{1pt}2×10^\rule{1em}{1pt}6 m] = [{(3.27 \rule{1em}{1pt} 2.40)×10^\rule{1em}{1pt}14 } / (3.40 × 10^\rule{1em}{1pt}10 )] m \bullet s^\rule{1em}{1pt}1 = 2.56 × 10\rule{1em}{1pt}5m \bullet s^\rule{1em}{1pt}1. In the classical Millikan oil drop experiment, the same set up is used to determine the electronic charge. A microscope is used to find the terminal velocity v of the drop. Equation (1) is then used to solve for q.

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Question:

The accompanying figure shows the wavelength ranges for the different colors in the visible spectrum. If potassium- containing materials emit light of frequency 7.41 × 10^14 sec-^1, what color flame would you expect to see when you heat a potassium compound such as potassium chloride in a Bunsen burner flame?

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Solution:

The color that you can expect will depend on the wavelength of the light emitted, as indicated by the diagram. Given the frequency of the light emitted, the wavelength can be determined. The frequency (ѵ) , wavelength (\lambda), and speed of light (c) are related by the equation ѵ\lambda = c. c = 3.00 × 10^10 cm/sec. Solving for \lambda: (7.41 ×10^14) \lambda = 3.00 × 10^10. \lambda = (3.00 × 10^10) / (7.41 × 10^14) = 4.05 × 10^-5 cm. To convert this into \AA; remember that there are 10^8 \AA per centimeter. Consequently, 4.05 × 10-^5 cm = (4.05 × 10-^5) (10^8) = 4050 \AA. According to the diagram, this \lambda corresponds to the violet region.

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Question:

Give a pseudo code to do a tree search, using an ALGOL- like language.

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Solution:

We make the following assumptions: Each node n of the tree has 3 fields 1st field is: Data [n], i.e., the number or data in the sort list. 2nd field is: Left [n] , i.e., pointer to the root of the left sub-tree. (the pointer = 0 if there is no left sub-tree) 3rd field is: Right [n], i.e., pointer to the root of the right sub-tree (the pointer = 0 if there is no right sub-tree) The placement of the different data items in the tree is as follows: If the first data item is x, then, if the second data item in the list is greater than x it is placed in the left subtree, and if the second item is less than x it is placed in the right sub-tree. i.e., Data [n] < Data [x] for all nodes x which are elements of the left sub- tree (n), and, Data [n] > Data [x] for all nodes x which are elements of the right sub-tree (n) We are now ready to form a strategy to search a tree, as follows: 1) Begin the search of the sort tree at the root for a node whose data field contains the argument, i.e., the value or number which we want to search 2) If there is such a node, set m to point to it and set a variable 'treesearch' to the value 'True' 3) Otherwise, if there is no such node then set the variable 'treesearch' to 'false'. The ALGOL-like pseudo-code is given below:-- procedure Treesearch (root, arg, m) / \textasteriskcentered arg is the data we are searching for, and m is a pointer to the node we are on \textasteriskcentered / if root = 0 then Treesearch : = false / \textasteriskcentered we are checking if there is still any root at all \textasteriskcentered / else begin d: = Data [root] if arg = d then begin m: = root Treesearch: = True end; else if arg > d then Treesearch: = Treesearch (left (root), arg, m); else Treesearch: = (right (root), arg, m) end;

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Question:

It was once thought that a neutron is made up of an electron and proton held together by Coulomb attraction. Assume the neutron radius is 10^-15 m. (a) According to the uncertain-ty principle, find\Deltapfor such an electron, (b) The lowest average momentum such an electron could have would be (1/2)\Deltap. What would be the corresponding energy? (c) What is the electrostatic potential energy of an electron 10^-13 cm from a proton? (d) From the calculations in (b) and (c) does it appear likely that a neutron could be made up of an electron and a proton?

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Solution:

(a) The uncertainty principle states that\Deltap\Deltax\cong (h/4\pi) , where h is Planck's constant,\Deltapis the un-certainty of the momentum of a particle, and\Deltaxis the un-certainty in the position of the particle. If we assume that the electron can be found anywhere within the neutron, then\Deltaxwill be of the order of the neutron radius 10^-15 m. p can then be found using the uncertainty prin-ciple . \Deltap= (h/4\pi\Deltax) = [(6.63 × 10^-34 Joule-sec)/(4\pi 10^-15 m)] = 5.28 × 10^-20 kg-m/sec (b) We must find a relation between energy E and momentum p. Since E = mc^2 = [m_0/\surd{1 - (v^2/c^2)}]^c(2) where m_0 is the rest mass of the electron, and v its speed, we have, upon squaring both sides of the equation, E^2[1 - (v^2/c^2)] = (m_0 c^2)^2 E^2 = [(E^2 v^2)/c^2] + (m_0 c^2)^2 = (E^2/c^4)c^2v^2 + (m_0 c^2)^2 = [(m^2 c^2)/c^4]c^2 v^2 + (m_0 c^2)^2 E^2 = p^2 c^2 + (m_0 c^2)^2.(1) Here, we used the definition of momentum as p =mv. We have m_0 = 9.11 × 10^-31 kg, c = 3 × 10^8 m/sec and p = (\Deltap/2) = [(5.28 × 10^-20)/2] (km/sec) = 2.64 × 10^-20 (km/sec). Upon substitution into (1), the result becomes E = \surd[(2.64 × 10^-20 (k-m/sec)^2 {3 × 10^8 (m/sec)}^2 + (9.11 × 10^-31 kg)^2 {3 × 10^8 (m/sec)}^4] = 7.46 × 10^-12 Since1eV= 1,6 × 10^-19 J = 7.46 × 10^-12 J × [(1eV)/(1.60 × 10^-19 J)] = 4.67 × 10^7eV. (c) By the definition of V V = [1/(4\pi\epsilon_0)] [{(e)(e)}/r] = [[(9.0 × 10^9 {(N - m)/coul^2} {(1.60 × 10^-19 coul)^2}] / {10^-13 cm (1m/10^2 cm)}] = 2.30 × 10^-13 J = 2.30 × 10^-13 J × [1eV/(1.60 × 10^-19 J)] = 1.44 × 10^6eV. (d) No. The energy of the electron is much greater than the energy available to bind the electron to the proton to form the neutron.

Question:

Associate a medicinal property for each of the following families of compounds: (a)Phenothiazines, (b) Barbiturates, and (c) Amphetamines.

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Solution:

(a)Phenothiazinesare one major type of tranquilizers used to treat psychoses. In general, tran-quilizers are compounds which act as depressants of the central nervous system with highly selective action on brain function. (b) There exist drugs that induce relaxation and sleep, which are sometimes used to treat patients with organic and emotional disorders. The barbiturate family is the most common and widely used for this sedative and hypnotic purpose. (c) Amphetamines belong to the class of anti-depressant drugs. These drugs, as their name implies, are used to treat depression disorders, one of the conditions associated with mental illness. They stimulate the central nervous system, and are used to bring about an increase in alertness, elevation of blood pressure and increase in heart action.

Question:

An LED used to display a decimal number has the following configuration. Design a decoder which will accept a four bit binary number, whose value will range from 0_10 to 9_10, and decode it to a seven bit number to be accepted by the LED. the LED will then display the decimal form of the four bit number.

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Solution:

Begin designing the decoder by listing a truth table. The inputs to the decoder will be A, B, C and D with A the most significant bit. The outputs are a, b, c, d, e, f and g. decimal number Input ABCD Output abcdefg 0 0000 1111110 1 0001 0110000 2 0010 1101101 3 0011 1111001 4 0100 0110011 5 1010 1011011 6 0110 1011111 7 0111 1110000 8 1000 1111111 9 1001 1111011 10 1010 XXXXXXX 11 1011 XXXXXXX 12 1100 XXXXXXX 13 1101 XXXXXXX 14 1110 XXXXXXX 15 1111 XXXXXXX Notice that all of the inputs greater than 9_10 have don't-care outputs. This is possible because the inputs will not be greater than 9_10. The don't care states will help in minimizing the expressions. Evaluate the sum of products form for each output, a(A, B, C, D) = \sum (0, 2, 3, 5, 6, 7, 8, 9) + d b(A, B, C, D) = \sum (0, 1, 2, 3, 4, 7, 8, 9) + d c(A, B, C, D) = \sum (0, 1, 3, 4, 5, 6, 7, 8, 9) + d d(A, B, C, D) = \sum (0,2, 3, 5, 6, 8, 9) + d e(A, B, C, D) = \sum (0, 2, 6, 8) + d f(A, B, C, D) = \sum (0, 4, 5, 6, 8, 9) + d g(A, B, C, D) = \sum (2, 3, 4, 5, 6, 8, 9) + d where d represents the don't care conditions d(A, B, C, D) = \sum10, 11, 12, 13, 14, 15 Construct a 4 variable K-map to minimize each Boolean expression. Note: X corresponds to don't care conditions The logic diagram of figure 1 is constructed from the minimized expressions a through g.

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Question:

As you know 9 bits can address 512 words. Show how on a PDP-8 machine with 4096_10 words in memory, any word can be addressed using 9 bits (extend the memory addressing capa-bilities) .

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Solution:

When there were 512_10 words in memory, 9 bits were required to address any word location in the memory. Hence the Instruction Counter register (IC) and Memory Address Register (MAR) were to store 9 bits. The word length we want is 12 bits long therefore, the Memory Buffer Register (MBR) is 12 bits long. The scheme was as follows: When a word had an instruction in it, the first three bits specified operation code and the remaining 9 bits were sufficient. Now as there are 4096 words IC and MAR have to be of 12 bits. The MBR is of 12 bits as word length is 12 bits. Direct addressing of any word in the 4096 words will use a total 12 bits of word and no room (bits) is left for the OP-CODE. The scheme to extend the memory addressing capabilitiesis that is that the 4096_10 words memory is divided into pages. Each page has 128_10 (200_8) words. This is a logical division and not a physical division of memory as shown in Fig. 1. As each page has 128_10 words, 7 bits can address any word on the same page, but as there are 32 pages we need 5 bits to address a page. 5 bits for page + 7 bits for words on page = 12 bits and we end up still with the same problem of having only a total of 12 bits where the first 3 bits are for the OP-CODE and only 9 bits for address. To solve this, two bits, bit 3 and bit 4 are used as I bit and P bit respectively - as illustrated in Fig. 2 When the P bit is 1, the word is on the current page, where the instruction was, and the next address is found by concatenating the 7 address bits and the first five bits of the instruction counter. For example, if we are on page 2 with the IC pointing to 0410_8, the instruction at this location is 1352: 1352 = 001011101010. The first three bits 001 specify the ADD instruction to the accumulator ; the address of the operand is found by observing bits. Bit 4 = 1 i.e. P bit is 1, hence the operand is on the current page and its address is: first 5 bits IC concatenate 7 bits of address 0410_8 = 000100001000 Address = 000101101010 = 0552_8 and 552_8 is the word which is on page 2. If P bit = 0, the page referred is always Page 0 and the address is formed by 5 zeros followed by a 7 bit address in MBR. As an example, assume we are in page 2 with IC point-ing to 0410_8. Now suppose the instruction at this location was 1152_8 i.e. 1152_8 = 001001101010. Since P bit = 0, it references page 0, the OP-CODE 001 part specifies the ADD instruction and the address of the operand is 000001101010 = 0152_8 where this a word on page 0. By doing the above we have extended our addressing capabilities to address 256_10 words, 128 word on page 0 + 128 words on any current page. Yet the memory has 4096 words. A mechanism to address from one page to another is established using bit 3 as an indirect bit. When I = 0 it is Direct address and the address is ob-tained, depending on the P bit and the 7-address bits. When I = 1 it is Indirect address, i.e. the address specified by the address bit is not the address of the op-erand but the contents of the given address (location) are the address of the operand. For example, suppose data 25_10 is stored in location 100_10 and 100_10 is stored at 15_10. Then indirect addressing will be ADD I15 I being present, data at location 15 i.e. 100 is not to be added to accumulator; rather, the data of 100 has to be used as an address. At this address data = 25 is stored; this is to be added to accumulator. Now in the PDP-8 when I bit = 1 there are two condi-tions of the P bit. If I = 1, P = 0 then the page is zero page. On page zero, go to the location specified by 7 - address bits. The contents of this location, a 12 bit word, is the address of the operand. If I = 1, P = 1 then the page is the current page as given by instruction counter IC. Suppose we are on page 2 where the IC is pointing to 0410_8 brought in the MBR instruction1752_8, 1752_8 = 001111101010 OP-CODE = 001 is the ADD instruction. I bit = 1, P bit = 1 As in indirect addressing, the location specified by address bits has the address of the operand we need to find the lo-cation specified. As P = 1 the address of the location = first 5 bits of IC concatenated with 7 address bits = 000101101010 = 0552_8. Now location 0552_8 has the full 12 bit address of operand. These 12 bits can be the location of any of the 4096 words in memory. Thus by using the concept of I bit and P bit the addressing capability of memory is increased.

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Question:

Modify the program of the previous problem to handle overflowif the capacity of the tank is finite.

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Solution:

Each time VOL is computed, check to see if it exceeds the given capacityCAP. If it does, indicate this and stop the simulation: DATA T/0.0/ READ N, TFIN, Q, VOL, CAP PRINT T, VOL REALN = N DT + (TFIN - T)/REALN DO 50 I = 1,N T = T + DT VOL = VOL + Q {_\ast} DT IF(VOL.GT. CAP) GO TO 60 PRINT T, VOL 50CONTINUE STOP 60PRINT 'OVERFLOW' STOP END

Question:

Two gases,HBrand CH_4, have molecular weights 81 and 16, respectively. TheHBreffuses through a certain small opening at the rate of 4 ml/sec. At what rate will the CH_4 effuse through the same opening?

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Solution:

The comparative rates or speeds of effusion of gases are inversely proportional to the square roots of their molecular weights. This is written rate_1 / rate_2 =\surdMW_2 /\surdMW_1 For this case =rate_HBr/ rate_(CH)4 =\surdMW_(CH)4 / \surdMW_HBr One is given therate_HBr, MW_(CH)4 andMW_HBrand asked to find rate _(CH)4 Solving for rate_(CH)4 : rate_HBr/ rate(CH)4=-\surdMW_(CH)4/ -\surdMW_HBrrate_HBr= 4 ml/sec. rate_(CH)4 = ? (4 ml/sec) / (rate_(CH)4) = \surd16 / \surd81MW_(CH)4 = 16 MW_HBr= 81 rate_(CH)4 = [(4 ml/sec) × \surd81] / \surd16 = [(4 ml/sec) × 9] / 4 = 9 ml/sec.

Question:

What velocity must a rocket be given so that it escapes the gravitational field of the earth?

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/Users/wenhuchen/Documents/Crawler/Physics/D05-0251.htm

Solution:

In order to escape the earth's gravitational field, the rocket must travel an infinite distance from the earth. The change in potential energy, ∆V, of the rocket as it goes from the earth's surface to infinity is ∆V =GM_Em_R[1/R_E - 1/r] =GM_Em_R[1/R_E - 1/\infty] = (GM_Em_R)/(R_E). The gravitational acceleration at the surface of the earth is written as g = (GM_E)/(R^2_E) = 9.8 m/sec^2. Rewriting ∆V so that g appears in the expression, we have ∆V = (GM_E)/(R^2_E) ×R_Em_R=gR_Em_R. This is the potential energy that the rocket must acquire if it is to escape the earth's pull. This energy comes from the conversion of kinetic energy to potential energy. There-fore the initial kinetic energy given to the rocket must at least equal, this change in potential energy. Equating the two energy expressions, with R_E = 6.36 × 10^6 meters, ∆V = gR_Em_R= (1/2) m_Rv^2 v^2 = 2g R_E = 2 × 9.8 × 6.36 × 10^6 v= 1.12 × 10^4 m/sec. This minimum velocity needed to escape the earth's gravita-tional field is known as the escape velocity.

Question:

Explain how the following procedures are useful in debugging a program: a) tracing a function b) Stop control.

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Solution:

When a function returns improper or unexpected results, the programmer can investigate what is happening in each instruction of his program by tracing its execution, using the trace function. Consider the following program: \nablaDOOMED [1]S \leftarrow X1, X2, ..... , XN [2]\DeltaS \Delta \Delta [3]S[\Delta] \Delta [4]INVS \leftarrow \divS [5]PS1S2 \leftarrow ((S(I) - S(I - 1)) \div (INVS(I) - INVS(I - 1)) [6]\rightarrow [8] × ʅ(PS1S2 < 0) [7]'THE NEGATIVE PRODUCTS ARE:'; PS1S2 [8]\rightarrow [10] × ʅ(PS1S2 < 100) [9]'THE POSITIVE PRODUCTS LESS THAN 100:'; PS1S2 [10]'THE POSITIVE PRODUCTS GREATER THAN 100:'; PS1S2 [11]\nabla If we run this program through, only negative products are printed. To see what is wrong we can put the function in trace mode, i.e. T\DeltaDOOMED \leftarrow 1 2 3 4 5 6 7 8 9 10 Where 1,2...,10 are the statement numbers. But the program is quite long. The mistake probably lies in statement [5]. Hence, instead of trying to trace the action of the system through the whole function we can concentrate on [5]. Key in TADOOMED \leftarrow [5]. The output is DOOMED DOOMED[5]-XIX2 DOOMED[5]-X3X2 DOOMED[5]-X4X3 \bullet\bullet \bullet\bullet \bullet\bullet DOOMED[5]-XNXN-1 Since all the products are negative, there is something wrong with instruction [5]. Converting into algebraic form: PS1S2 = [X_1 -X_2] / [(1/X_1) - (1/X_2)] = [X_1 - X_2] / [(X2- X_1) / (X_1X_2)] = [(X1- X_2) / (X_2 - X_1)] (X_1X_2) = [(X1- X_2) / -(X_1 - X_2)] (X_1X_2) = -(X1X_2). Statement [5] always results in a negative product. To en-sure correct calculations we can change [5] to [5]PS1S2 \leftarrow ((S(I) - S(I - 1)) \div (INVS(I - 1) - INVS(I)). b) The stop control mode halts program execution at a specific instruction so the programmer can examine it. For example, suppose we wish to inspect [6] of DOOMED. To activate the stop control mode, key in S\DeltaDOOMED \leftarrow 6 DOOMED The system responds with [6] \rightarrow [8] × ʅ(PS1S2 < 0). To end the stop control mode, key in S\DeltaDOOMED \leftarrow 10.

Question:

In the reaction, CaCO_3(s) \rightarrow CaCO(s) + CO_2 (g) at 950\textdegreeC and CO_2 pressure of 1 atm, the ∆H is found to be 176 kJ/mole. Assuming that the volume of the solid phase changes little by comparison with the volume of gas generated, calculate the ∆E for this reaction.

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Solution:

To solve this problem, you must know that ∆E, the change in energy, can be related quantitatively to the change in enthalpy, ∆H, the pressure (P) and the change in volume (∆V), by the formula ∆E = ∆H - P∆V. This equation means that if heat is added to a system at constant pressure, part of it goes into increasing the internal energy of the system (∆E) and the rest is used to do work on its surroundings, P∆V. Therefore, the total of both ∆E and P∆V is the heat content of the system, ∆H. To solve the problem, therefore, you need only substitute in the values for P, ∆V and ∆H. ∆V = V_products - V_reactants There is little change in the volume of the solid by comparison with the gas. This means, therefore, that ∆V is approximately equal to V_gas \bullet V_gas can be found from the equation of state; PV = nRt or V = nRt/P, where n = number of moles, P = pressure, R = universal gas constant and t = temperature in Kelvin (Celsius temp + 273 ). From the reaction, 1 mole of CO_2 gas is produced, thus n = 1. R is defined as being equal to 8.31 J mol^-1 deg^-1. You are given P and t. Therefore V = [1(8.13) (1223)] / [1] = 10.2 kJ Recalling, ∆E = ∆H - P∆V, you have ∆E = 176 - 1(10.2) = 166 kJ.

Question:

In the circuit shown in Fig. 1, find V_1 and the voltage v across both sources if i_s is given as 12A.

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Solution:

In Fig. 2 observe that i = i_s + 0.6v_1 = i_1 + i_2. Since i_1 and i_2 both flow through a branch with the same equivalent resistances, they must be equal. Thus we have i = i_s + 0.6v_1 = 2i_1 = 2i_2. We can write v_1 = i_1R where R = 2-\Omega. Since i = 2i_1 we have i1= (1/ 2)i = (1 / 2) (i_s + 0.6 v_1) v_1 = i_1R = (1 / 2) (i_s + 0.6 v_1)R Substituting 2-\Omega of R and 12A for i_s, solve for v_1 v_1 = (1 / 2)(12 + 0.6v_1) 2 v_1 = (12 + 0.6v_1) 0.4v_1 = 12 givingv_1 =12 / 0.4 = 30V. v is found by calculating i and the equivalent resistance R_eq across the terminals of the two sources From Fig. 3 we find v = R_eq , i = 7.5(30) = 225V.

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Question:

Calculate ∆H\textdegree_r for the combustion of methane, CH_4. The balanced reaction is CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(l) ∆H\textdegree_f in Kcal/mole = -17.89 , 0 , - 94.05 , 2(-68.32)

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Solution:

∆H\textdegree_r is the Standard Enthalpy Change of Reaction. Standard conditions are defined as 25\textdegreeC and 1 atm. Enthalpy is the heat content of a system. If the overall change in enthalpy is negative, then heat is given off to the surroundings and the reaction is called exothermic. When the change is positive, heat is absorbed and the reaction is endothermic. Endothermic compounds are often unstable and can sometimes explode. An endo-thermic compound, however, is a more efficient fuel because, upon combustion, it yields more heat energy. ∆H\textdegree_r is calculated using the enthalpies of formation, ∆H\textdegree_f. The sum of enthalpies of formation of products minus the sum of the enthalpies of formation of reactants, where each product and reactant is multiplied by its molar amount in the reaction as indicated by the coefficients, gives the value of ∆H\textdegree_r. In other words, ∆H\textdegree_r = \sum∆H\textdegree_f _products - \sum∆H\textdegree_f _reactants ∆H\textdegree_f for elements is always zero. In this reaction, therefore, ∆H\textdegree_f for O_2 is zero. \sum∆H\textdegree_f _products = - 94.05 - 136,64 \sum∆H\textdegree_f _reactants = - 17.89 Kcal/mole ∆H\textdegree_r = - 94.05 - 136.64 - (- 17.89) = - 212.8 Kcal/mole CH_4, burned This reaction is exothermic, because the ∆H\textdegree_r = a negative value.

Question:

A missile is fired radically from the surface of the earth (of radius 3.4 × 10^6 m) at a satellite orbiting the earth. The satellite appears stationary at the point where the missile is launched. Its distance from the center of the earth is 25.4 × 10^6 m. Will the missile actually hit the satellite?

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Solution:

To an observer outside the planet (see fig. A), the earth and the missile (of mass m) on the sur-face are rotating about the axis of the planet with angular velocity \omega_0. When the missile is fired radical-ly from the surface, its distance from the center of the earth increases and thus its moment of inertia (I = mr^2) about the rotation axis increases also. There are no forces with a moment about the rotation axis acting on the missile (the gravitational force of attraction acting on it is exerted along the rotational axis and has no moment). The net torque \cyrchar\CYRG acting on the missile is then zero. According to the rigid body analogue of Newton's second law, if L is the magnitude of the angular momentum of the missile, then \cyrchar\CYRG = (dL)/(dt) But\cyrchar\CYRG = 0 and0 = (dL)/(dt) or L = constant at all times. But L = mvr where v is the tangential velocity of the missile. Since v = \omegar where \omega is the angular velocity of the missile, then L = m\omegar^2 = I\omega. Thus since L and m are constant at all times, as the missile moves farther away from the earth and closer to the satellite (i.e., r increases) then \omega must decrease. We are given that the satellite appears stationary at the point where the missile is fired. Thus the radius vector passing through the launching pad and the satellite continues to rotate with angular velocity \omega_0. The missile has an angular velocity \omega which drops more and more from the value \omega_0 as the missile rises. At the height of the satellite, the moment of inertia of the rocket about the axis of rotation is I_1 = mr^2_s = m × (25.4 × 10^6 m)^2 ,whereas, at the launching pad, its moment of inertia is only I_2 = mr^2_e = m × (3.4× 10^6 m)^2 .Thus, finally, if \omega is the satellite's angular velocity at height r_s, and \omega_0 its angular velocity at launching (equal to that of the earth) then I_1\omega= L = I_2\omega_0 \omega/\omega_0= (I_2)/( I_1) = (mr^2_e)/(mr^2_s) = [(3.4× 10^6 m)^2]/[(25.4 × 10^6 m)^2 ] = 0.018. The missile thus moves further and further from the vertical as it rises and will miss the satellite (unless the missile is fitted with a homing device). To an observer on the planet, the departure of the missile from the vertical is, of course, also observed and is explained in terms of the Carioles force associated with a rotating frame of reference.

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Question:

Outline a breeding procedure whereby a true breeding strain of red cattle could be established from a roan bull and a white cow.

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Solution:

This problem deals with establishing a pure line having a trait which is codominant. A pure line is a strain of organisms that when inbred, always produce offspring having the same phenotype. They are homozygous and can never be heteroyzgous because when heterozygotes are inbred, they produce offspring of more than one phenotype, (which is not true breeding). Color in cattle is determined by two alleles which are codominant. Neither allele is dominant or recessive to the other. In the heterozygous state, each allele acts to produce its own gene product independent of the other, and the total product is a sum of the contributions from each allele. Phenotypically, this usually results in a trait which is intermediate between the two dominant alleles. In this case, red and white color are codominant characteristics, and roan color is the intermediate phenotype of a heterozygote. Let R respresent the gene for red color in cattle and W the gene for white color. By definition then, red cattle have the genotype RR, and white cattle have the genotype WW, and roan cattle are RW. (Note that we no longer use capital and lower case forms of the same letter to represent the alleles, since neither is dominant or recessive). Now since red cattle by definition must be homozygous, any red cattle produced will breed true when inbred. Therefore, in order to establish a pure line of red cattle from a roan bull and a white cow, all we need do is obtain male and female red cattle among the off-spring. Then these red cattle should only be bred among themselves. In the cross between the roan bull and the white cow: No red cattle were obtained in the F_1. The only way to obtain red cattle would be to cross two of the roan cattle from the F_1. In the mating then: In the F_2, 1/4 of the offspring will be red, and thus homozygous for red color. They will always breed true.

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Question:

A plane parallel plate capacitor consisting of two metal circular plates 5 cm in radius separated 1 mm in air, is charged to 300 stat-volts, whereupon it is connected in parallel to another similarly charged capacitor (positive terminals connected together and negative terminals connected) (see Figure A). How much energy would be re-- leased if the combinations were discharged by a short circuit?

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Solution:

For a plane parallel plate capacitor: C = KA / 4\pid This result can be obtained as follows. According to the definition of capacitance C, C = Q / V where Q is the total charge on one plate, and V is the potential difference between the plates. According to Gauss's law, if the Gaussian Surface is constructed as shown in Figure B, then (K / 4\pi) \varphiE= (K / 4\pi)EA = Q(1) in the CGS system. This relation holds because the electric field E is a constant in the parallel plate capacitor. Q is the charge enclosed by the Gaussian Surface. It is also the total charge on either plate, due to the construc-tion of the surface. Also, for a parallel plate capacitor V = Ed(2) Therefore combining (1) and (2), we get the result for the capacitance of the parallel plate capacitor. \therefore C_1 =\pir^2 / 4\pid = 5^2 / (4 × 1 mm × 1cm / 10 mm) = 25 / .4 = 62.5 stat\rule{1em}{1pt}farads. Recalling that the energy stored in a capacitor is W = (1 / 2)QV = (1 / 2) CV^2 = (1/2)(Q^2 / C) and choosing the second form because C and V are known W_1 = (1 / 2)C_1V_1^2 = (1 / 2)(62.5)(300)^2 = 31.25(90,000) =2,820,000 ergs But the total energy stored in the two capacitors is W_1 + W_2 = 2W_1 \thereforeW = 2(2,820,000) = 5,640,000 ergs =.564 joulesAns. This is the energy available and able to be released if the combination were discharged by a short circuit. Solution in Mks Units: Data: plate radius = 5 × 10^\rule{1em}{1pt}2 m d =10^\rule{1em}{1pt}3 m V = 9 × 10^4 volts C = (1 / 4\pik)(A / d) = (\pi25 × 10^\rule{1em}{1pt}4) / (4\pi9 × 10^9 × 10^\rule{1em}{1pt}3) = 6.94 x 10^11 farads But 1 farad = 9 × 10^11 stat\rule{1em}{1pt}farads \thereforeC = 6.94 × 10^\rule{1em}{1pt}11 × 9 × 10^11 = 62.5 stat-faradsCheck. W = (1 / 2) C_1V_1^2 = (1 / 2)6.94 x 10^\rule{1em}{1pt}11 × 81 × 108 = 28.1 × 10^\rule{1em}{1pt}2 joules But 1 joule = 10^7 ergs. \thereforeW = 28.1 × 10^\rule{1em}{1pt}2 × 10^7 = 2.81 x 10^6 ergs Total W = 2 × 28.1 × 10^\rule{1em}{1pt}2 = .562 joules = 5.62 × 10^6 ergsCheck.

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Question:

(a) Find the current in a circuit consisting of a coil and capacitor in series, if the applied voltage is v = 110 volts, frequency f = 60 cycles/sec; the inductance of the coil is L = 0.80 henry; the resistance of the coil is R = 50.0 ohms; and the capacitance of the capacitor is C = 8.0 \muf. (b) Find the power used in the circuit.

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Solution:

(a) The coil can be represented by an inductance and resistance in series (see figure A). The effective impedance of the inductor and capacitor depend on frequency. Since the inductor and capacitor cause the current to either lead or lag the voltage by a phase shift of 90\textdegree, the impedance Z is the vector addition of their combined reactance and the pure resistance (see figure B). X_L = 2\pifL = 2\pi(60) (0.80) ohms = 300 ohms X_C = (\rule{1em}{1pt}1/2\pifC) = {\rule{1em}{1pt}1/(2\pi × 60 × 8.0 × 10^\rule{1em}{1pt}6)} ohms = \rule{1em}{1pt} 330 ohms X_L + X_C = 300 - 330 = \rule{1em}{1pt} 30 ohms Z = \surd{R^2 + (X_L + X_C)^2} = \surd{(50)^2 + (300 - 330)^2} ohms = \surd{50^2 + (\rule{1em}{1pt} 30)^2} ohms = 58 ohms I= (V/Z) = (110 volts/58 ohms) = 1.9 amp (b) The power is equal to P = I^2R or P = VI cos \textphi where is the angle between the impedance and the pure resistance in the circuit, (see figure b) cos \textphi = (R/Z) = (50 ohms)/(58 ohms) = 0.86 P= VI cos \textphi = 110 volts × 1.9 amp × 0.86 = 180 watts. Using P = I^2R as a check, P = I^2R = (1.9 amp)^2 × 50 ohms = 180 watts.

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Question:

Determine the power required to maintain a temperature difference of 20\textdegree C between the faces of a glass window area 2 m^2 and thickness 3 mm. Why does a much lower power suffice to keep a room with such a window at a temperature 20\textdegreeC above the outside? The thermal conductivity of glass is 25 × 10^-4 cal\bullets^-1\bulletcm^-1\bulletC deg^-1.

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Solution:

The equation appropriate to thermal conductivity is dQ/d\cyrchar\cyrt = - k A dt/dx(1) where dQ/d\cyrchar\cyrtis the rate of heat transfer across a cross section of area A of a slab of material of thermal conductivity k. dt/dx is the temperature gradient in the material. Now, if we assume that the heat transfer is occurring as a steady state process, the temperature of each point of the slab will be time independent. If this is the case, dQ/d\cyrchar\cyrtis the same at all cross- sections of the slab. However, by (1), this means that dt/dx is constant at all cross-sections. Hence dt/dx decreases linearly along the slab. Using the figure, dt/dx = (t_2 - t_1)/(x_2 - x_1) But x_2 - x_1 = L, the slab thickness, whence dt/dx = (t_2 - t_1)/L anddQ/d\cyrchar\cyrt= -k A [(t_2 - t_1)/L] Since t_2 < t_1 , we may rewrite this as dQ/d\cyrchar\cyrt= k A [(t_1 - t_2)/L] In this equation, t_1 and t_2 are the temperatures of the slab at positions x_1 and x_2. (See figure.)In this particular case, t_1 = t_2 + 20\textdegreeC and dQ/d\cyrchar\cyrt= 25 × 10^-4 cal\bullets^-1\bulletcm^-1\bulletC deg^-1 × 2 × 10^4 cm^2 × 20 C deg/0.3 cm = 3.33 × 10^3 cal \bullet s^-1. Here we have used the fact that 2 m^2 = 2 × 10^4 cm^2 and3 mm = .3 cm These conversions follow from the definitions of a cm. and mm. Since 1 cal = 4.19 Joules dQ/d\cyrchar\cyrt= 3.3 × 10^3 cal\bullets^-1 = (3.3 × 10^3 cal\bullets^-1)(4.19 J\bulletcal^-1) dQ/d\cyrchar\cyrt= 13.95 × 10^3 J\bullets^-1. Thus 3.33 × 10^3 cal = 13.95 × 10^3 J are required per second to replace the lost heat. The power required is thus 13.95 kilowatts (kW). A much smaller power than this is required in practice because the inner surface of the window and the air in contact with it drops in temperature because of the heat loss through the glass. Heat is conducted from the rest of the room through air, the thermal conductivity of which is very low. The inner surface of the window is thus not maintained at a temperature 20\textdegreeC above the outside. A similar effect will occur on the outside of the window. The temperature difference across the window may well drop to only a few degrees, in which case only a fraction of the above power needs to be supplied, giving a much more reasonable figure for the heat that needs to be supplied per second.

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Question:

Aspaceship moving away from the Earth at a velocity v_1 = 0.75 c with respect to the Earth, launches a rocket (in the direction away from the Earth) that attains a velocity V_2 = 0.75 c with respect to the spaceship. What is the velocity of the rocket with respect to the Earth?

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Solution:

Newtonian relativity states that the laws of mechanics are the same in all inertial reference frames but that the laws of electrodynamics are not. Since such a theory was not acceptable, the theory of relativity was developed and resolved this problem. One of the statements of this new theory was that the velocity of light is independent of any relative motion between the light source and the observer. Experiments have shown this to be true. In support of this statementEinstein showed that ordinary mechanical velocities do not add algebraically. Instead, for the addition of two velocities v_1 and v_2 , the sum is V = [ (v_1 + v_2) / {1 + (v_1v_2 / c^2)}] For small velocities (v < < c), this reduces to V = v_1 + v2, as in Newtonian mechanics, since the term v_1v_2 / c^2 is very small compared to one and can be ignored. Since the velocities in this problem are large, the velocity of the rocket with respect to the Earth is V = [ (v_1 + v_2) / {1 + (v_1v_2 / c^2)}] = (0.75c + 0.75c) / [1 + {(0.75c)(0.75c) / c^2 }] = 1.5c / (1 + 0.5625) = 0.96c Therefore, in spite of the fact that the simple sum of the two velocities exceeds c, the actual velocity relative to the Earth is slightly less than c .

Question:

Why would one call a flock of birds a \textquotedblleftsociety\textquotedblright?

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Solution:

A society is a group of animals that belongs to the same species and is organized in a cooperative manner. The animals are usually bound together by re-ciprocal communication, which leads to the cooperative behavior. A flock of birds can be considered an elementary form of society called a motion group. Motion groups in-clude schools of fish and herds of mammals. Birds within a flock communicate with one another in order to stay together as they move from place to place. The coopera-tive behavior displayed by a flock of birds enables them to more efficiently detect and avoid predators. It is more likely that a flock of birds will detect a predator than a solitary bird. Many birds have special signals, such as alarm cries and wing movements, that seem to alert the entire group to the presence of a predator. Membership in a group also affords protection once the predator attacks the group. For example, starlings usually fly in a wide, scattered pattern, but assemble quickly in tight formation when attacked by a falcon, their natural predator. The falcon usually attacks by swooping at high speed into the group. By assembling into a tight group, the starlings make it difficult for the falcon to hit one bird without collid-ing into others and possibly injuring itself. Other cooperative behavior patterns which may be performed by a flock of birds are the discovery of food and the coordination of reproductive activities.

Question:

Discuss the citric acid cycle as the common pathway of oxidation of carbohydrates, fatty acids and amino acids.

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/Users/wenhuchen/Documents/Crawler/Biology/F03-0075.htm

Solution:

The oxidation of the above substances all yield products which either enter or are intermediates of the citric acid (also known as Krebs and TCA, tricarboxylic acid) cycle. Glucose, a carbohydrate, undergoes a series of reactions in glycolysis to be oxidized to pyruvic acid. Pyruvic acid reacts with coenzyme A to yield acetyl coenzyme A. It is acetyl coenzyme A which enters the citric acid cycle (see Figure). Acetyl coenzyme A is also formed from the oxidation of fatty acids. A repeating series of reactions cleaves the long carbon chain of a fatty acid into mole-cules of acetyl coenzyme A. Finally, amino acids are metabolized by one of several reactions to products capable of participating in the reactions of the TCA cycle. The first step in the oxidation of all amino acids is a deamination reaction by which an amino group (\ElzbarNH_2) is removed from the amino acid. For three amino acids, this is the only reaction required for conversion into a compound entering the TCA cycle directly. Alanine undergoes deamination to form pyruvate which is converted into acetyl coenzyme A. The deamination of glutamic acid yields \alpha-ketoglutaric acid,while that of aspartic acid yields oxaloacetic acid. Both these compounds are intermediates of the TCA cycle. The TCA cycle is thus the final common pathway by which carbohydrates, fatty acids and amino acids are metabolized.

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Question:

A small metallic ball is charged positively and negatively in a sinusoidal manner at a frequency of 60 cps. The maximum charge on the ball is 10^-6coul. What is the displacement current due to this alter-nating charge? What would the displacement current be if the charging frequency were 10^6 cps?

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Solution:

The charge on the ball can be written as q = q_0 sin\omegat where q_0 is the maximum charge on the ball (10^-6coul), \omega = 2\pif (f = 60 cps) and the sine function expresses the sinusoidal variation of the charge. According to Gauss's law, the total electric flux through a Gaussian surface about the sphere is given by \varphi_E= \int E^\ding{217} \bulletds^\ding{217} = q/\epsilon_0 = q_0 sin\omegat/\epsilon_0 The displacement current i_0 is defined as i_0 \equiv \epsilon_0 [(d\varphiE)/dt] = \omegaq_0 sin\omegat=dq/dt(1) It is called a displacement current because even though there is no actual movement of charge from one position to another, the mathematical form of i_0 has the form of an actual charge flow. (See equation (1)). The maximum value of the displacement current occurs when \vertcos\omegat\vert = 1. Then i_(D)max = \omegaq_0 = 2\pifq_0 At 60 cpsi_(D)max = 2\pi × 60 sec^-1 × 10^-6coul = 3.77 × 10^-4 amp If the frequency were 10^6 cps, then i_(D)max = 2\pi × 10^6 sec^-1 × 10^-6coul = 6.28 amp

Question:

Explain how the planarian changes its body shape.

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/Users/wenhuchen/Documents/Crawler/Biology/F11-0284.htm

Solution:

In the planarian, three kinds of muscle fibers can be differentiated: an outer, circular layer of muscle just beneath the epidermis; an inner, longi-tudinal layer; anddorsoventralmuscles that occur in strands. Contraction of the longitudinal muscles causes constriction of the body, whereas contraction of the circular muscles causes an elongation of the body. Other alterations of body shape are produced by contraction of the dorsoventral muscle strands. Thus, by coordinated contraction and relaxation of these muscles, the body shape of a planarian can be varied.

Question:

Calculate the normality of a solution containing 2.45 g of sulfuric acid in 2.00 liters of solution. (MW of H_2SO_4 = 98.1.)

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Solution:

Normality is defined by the following equation. Normality = [(grams of solute)/(equivalent weight × liters of solution)] In this problem, one is given the grams of solute (H_2SO_4,) present and the number of liters of solution it is dissolved in. One equivalent of a substance is the weight in which the acid contains one gram atom of re-placeable hydrogen. This means that when the acid is dissolved in a solution, and it ionizes, that in one equivalent weight, one hydrogen atom is released. The equivalent weight for acids is defined as: equivalent weight = [(MW)/(no. of replaceable H)] When H_2SO_4 is dissolved in a solution, there are two replaceable H as shown in the following equation: H_2SO_4 \leftrightarrows 2H^+ + SO-_4 This means that in calculating the equivalent weight for H_2SO_4 the molecular weight is divided by 2. equivalent weight of H_2SO_4 = [(98.1 g)/(2 equivalents)] = 49 g/equiv The normality can now be calculated. Normality = [(grams of solute)/(equiv. wt × liters ofsoln)] Normality = [(2.45 g)/(49 g/equiv × 2.0 liters)] = 0.025 equiv/liter.

Question:

What is the illumination 5 feet from a 75 candlepower lamp?

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Solution:

Illumination is defined as flux divided by area [E = (F/A)] . Making the simplifying assumption that our lamp is a point source, the total flux that it 4\piI where I is intensity of the source .If we construct a sphere enveloping the point source with the point source at its center , and having a radius of 5 ft., we canfind the illumination at this distance. Hence , E = F/A =4\piI /4\pir2= I / r2 And Illumination= 75 candlepower / (5 × 5) ft^2 = 3 foot-candles .

Question:

A solution of 10.0 g of HF in 500 g H_2O freezes at - 1.98\textdegreeC. Calculate the degree of ionization of HF. (M.W. HF = 20.0 The Freezing point depression of H_2O is 1.86\textdegree.)

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Solution:

When an acid is dissolved in solution, it dissociates into ions. For example, when one mole of HF is dissolved in H_2O, one mole of H^+ and one mole of F^- are present after complete dissociation. The freezing point of a solution is dependent on the number of particles in the solution, which means that the freezing point of a solution will be lowered more by a compound which ionizes than by the same amount of a compound which does not ionize. The degree of ionization of a compound is a measure of what percent of the compound is ionized when it is placed in a particular solution. In this problem, one is told that when 10 g of HF is added to 500 g of H_2O the original freezing point of the water (0\textdegree) is lowered to - 1.98\textdegreeC. The freezing point depression is related to the concentration of particles in the solution by the statement freezing point depression =molality× freezing point constant The freezing point constant of water is 1.86\textdegree. This means that one mole of a substance (except those substances which ionize) dissolved in 1000 g of water lowers the freezing point 1.86\textdegree. Since HF ionizes, we cannot use itsmolalityin this equation. You can use the effectivemolality, which is the sum of themolalitiesof H^+, F^- and HF. Themolalityof H^+ and F^- will be equal to the degree of ionization of the HF. The concentrations of H^+ and F^- will be equal because, when HF ionizes, one H^+ and one F^- will be formed. This reaction is given by the equation HF \leftrightarrows H^+ + F^- To solve for the degree of ionization of HF, one must: (a) find the molalityof HF as if it were not ionized, (b) define a variable for the molalitiesof H^+ and F^- ; here, x will be used, (c) find themolalityof HF after ionization is taken into account, (d) find the effectivemolalityof the species, (e) use the effectivemolalityin the freezing point depression equation to solve for x, themolalityof H^+ and F^- . (a) Themolalityof HF before ionization. Themolalityis defined as the number of moles of solute in 1000 g of solvent. The number of moles of HF can be found by dividing the number of grams present by the molecular weight. number of moles = [(number of grams)/(molecular weight)] moles of HF = [(10.0 g)/(20.0 g/mole)] = 0.5 moles Themolalitycan now be found by dividing the number of moles of HF by the number of kg of water present. molality= [(no. of moles of HF)/(no. of kg of H_2O)] molality= [(0.5moles)/(0.500 kg)] = 1.0 moles/kg (b) x =molalityof H^+ =molalityof F^- (c) After ionization of xmolalof H^+ and F^-, themolalityof HF will be 1.0 - x. (e) The effectivemolalityof the species will be equal to the sum of themolalitiesof all of the species present. The effectivemolality=molalityof H^+ +molalityof F^- +molalityof HF effectivemolality= x + x + (1.0 - x) = 1.0 + x. To solve for x, the concentration of H^+ and F^-, the effectivemolality of the species will be used in the freezing point depression equation. freezing pointdepreesion= eff.molality× freezing pt constant The freezing point depression in this case is 1.98\textdegree and the freezing point constant of water is 1.86\textdegree. Solving the equation 1.98\textdegree = (1.0 + x) × 1.86\textdegree 1.0 + x = (1.98\textdegree/1.86\textdegree) x = .06 The percent of ionization is themolalityof H^+ and F^- divided by the molalityof the unionized HF multiplied by 100. degree of ionization = [(molalityof ion)/(molalityof unionized species)]× 100 degree of ionization =(0.06/1.00) ×100 = 6% Hence, the HF in this solution is 6 % ionized.

Question:

For the general reaction A + B \rightarrow C + D, what is the effect on the number of collisions between A and B of (a) tripling the concentration of each, (b) quadrupling the concentration of each?

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Solution:

For any given chemical reaction, the number of collisions is proportional to the rate. The Law of Mass Action states that the speed of a chemical reaction is proportional to the product of the concentrations of the reacting molecules. The number of collisions in the above reaction depends upon the concentration of A and B. If the concentration of A is tripled the number of collisions is tripled, then if the concentration of B is tripled the number of collisions will triple again. Thus the number of collisions is increased 3 × 3 times or 9 times. When the concentrations of A and B are quadrupled a similar method is used. The number of collisions is increased 4 × 4 or 16 times.

Question:

How many Btu are required to raise the temperature of 2 pounds of water from 32\textdegreeF to 212\textdegreeF?

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Solution:

The temperature rise is 180 degrees Fahrenheit. The number of Btu necessary is 2 × 180\textdegree = 360 Btu, since one Btu is required to raise the temperature of one pound of water one degree Fahrenheit.

Question:

A group of students were invited to taste phenylthiocarbamide (PTC). The ability to taste PTC is inherited by a single pair of genes and tasting (T) is dominant to non-tasting (t). Among 798 students, 60.4 percent were tasters, a) Calculate the allelic frequency of T and t. b) How many of the students were TT? Tt? tt?

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Solution:

Putting this problem in terms of the Hardy- Weinberg principle, let p represent the frequency of the T allele and q the frequency of the t allele. We know that p^2 + 2 pq + q^2 = 1 describes the distribution of phenotypes in the population, where p^2 equals the frequency of TT, 2 pq equals the frequency of Tt, and q^2 equals the frequency of tt. Note that the sum of ail proportions, or frequencies, equals 1. a) We are told that 60.4% of our given population are tasters. Since tasting is dominant to nontasting, tasters can have one of two genotypes, TT and Tt. This means that the sum of the frequencies of the TT and Tt genotypes is .604. Since all frequencies total 1, we can infer that the frequency of nontasters (tt) in the sample of students is 1- .604 or .396. Thusq^2 = .396, and q = \surd.396 or .629. Since p+q = 1, p = 1-.629 or .371. The frequency of T is therefore .371, and that of t, .629. b) frequency TT = p^2 = (.371)^2= .137 frequency Tt = 2p = 2(.371)(.629)= .467 frequency tt = q^2 = (.629)^2= .396 1.0 1.0 To convert the actual number of students carrying each genotype, we multiply the corresponding genotypic frequencies (or proportions) by the total sample size (i.e., 798): TT = (.137) (798) = 109 Tt = (.467) (798) = 373 tt = (.396) (798)=316 798

Question:

The average concentration of particulate lead in a city with highly polluted air is 5\mug/m^3. If an adult respires 8500 liters of air daily and if 50 per cent of the particles below 1\mumin size are retained in the lungs, calculate how much lead is absorbed in the lungs each day. Of the particu-late lead in urban air, 75 per cent is less than 1\mumin size.

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Solution:

To solve this problem (1) determine the total percentage of particulate lead remaining in the lungs, (2) determine the amount of lead absorbed into the lungs. The problem states 75 % of particulate lead is less than 1\mumin size, therefore, 25 % of the lead is greater than 1\mumin size and does not remain inside the lungs. Of the 75 % particulate lead less than 1\mumin size, only 50 % of this is retained in the lungs. The total percentage of lead retained is 37.5 %. The total volume of air-intake is 8500 liters, so that total volume of air intake = (8500) (10^-3 m^3/liter) = 8.5 m^3 The total amount of lead-intake is amount of lead = (8.5 m^3)(5 \mug/m3) = 42.5\muglead. The total amount of lead absorbed is amount of lead absorbed = (42.5\mug)(0.375) = 15.9\mug.

Question:

How was it proven that DNA replication is semi-conservative?

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Solution:

Three possible models v/ere considered to account for DNA replication, the process by which the two complementary strands of the double helical DNA replicate to form new complementary strands. In the first model, called conservative replication, the two strands of parental DNA replicate to yield the unchanged parental DNA and a newly synthesized DNA. The second model, semi-conservative replication, claims that the replication of one DNA molecule yields two hybrids, each composed of one parental strand and one newly synthesized strand. Finally, the third model, dispersive replication, has the parental strand breaking at intervals. These parental segments then combine with new segments to form the daughter strands(see Figure 1). Meselson and Stahl's classic experiments proved that DNA replicates by a semi-conservative mechanism. They grew E. coli cells for several generations on a medium which contained "heavy" nitrogen, ^15N. Thus, all the nitrogen bases of the E. coli DNA were labeled with ^15N instead of the normal or "light" nitrogen, ^15N. The ^14N - DNA of these cells could be separated from the ^14N - DNA by equilibrium centrifugation in a cesium-chloride gradient . because the ^15N - DNA has a significantly greater density than the ^14N - DNA. Meselson and Stahl took the "heavy" (^15N containing) cells and placed them in a 11 light" (lf\textasteriskcenteredN containing) medium. The cells were allowed to grow for several generations and were analyzed at various generation times. From the samples collected, the DNA was examined by determining its buoyant density with cesium-chloride density gradients. It was found that after one generation, the isolated DNA showed only a single band falling midway between the places where the heavy and light bands lie in a cesium-chloride density gradient, This is to be expected if the first generation DNA is a hybrid of one parental strand and one newly synthesized strand. After two generations, the isolated DNA exhibited two bands, one equal to the hybrid density of the first generation and one having a density equal to that of normal "light" DNA. After many generations passed, a \textquotedblleftlight" density band of DNA predominated. These observations are consistent with those expected from the hypothesis of semi-conservative replication but not with the other two models (see Figure 2) .

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Question:

What is meant by a gene pool, balanced polymorphism, and genetic load?

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Solution:

Population geneticists visualize a population of organisms as a collection of all the genes possessed by the individuals of the population. Such a sum total of genes is termed a gene pool. Each allele in the gene pool has a certain probability of being expressed in the population. The frequency of occurrence of a given genotype in the population is given by the product of the frequencies of its alleles in the gene pool. A gene pool is said to be in equilibrium if the frequency of each individual allele remains constant from generation to generation; that is to say, when there is no change in the genetic composition of the population over a period of time. The term polymorphism refers to the existence in the same interbreeding population of two or more distinct phenotypic forms of a genetically determined trait. The human blood groups 0, A, B, and AB are a classic example of polymorphism. Balanced polymorphism is a state in which the different forms of the polymorphic genotype are maintained together in equilibrium in the population over a period of time, Such a balance can be achieved and maintained through a variety of means, involving, what are often complex genetic-environmental interactions. One condition which produces balanced polymorphism is heterozygote superiority or heterosis. Here the heterozy-gote (Aa) has a survival advantage over both the dominant (AA) and recessive (aa) homozygotes. Thus both alleles are maintained in the population, and neither can eliminate the other. A classic example of this is sickle-cell anemia in a malarial environment. Contrary to what one might expect, the sickle-cell trait is maintained at a relatively high frequency in the population despite its obvious harmful effects. Individuals carrying the recessive sickle-cell gene in the heterozygous state have resistance to malarial infection. When both malaria and sickle-cell anemia are strong selection factors, the heterozygote has superiority, since normal homozygotes are subject to malaria and those homozygous for the sickle-cell gene have sickle-cell anemia. Thus the recessive gene is maintained along with the normal gene, and their traits are kept in balance. Often, a given species maintains different forms in the population because one or the other is favored at varying times in the year. The polymorphism gives the species an overall survival ability throughout the year. In geographic polymorphism, the species population may maintain a variety of forms, each of which is advantageous under specific environment conditions; one form may do better in one area, another in a different area. On the whole, the species is thus able to maintain itself in a variety of areas and environmental conditions, allowing it to become more widespread. Balanced polymorphism is important in evolution because it maintains a certain amount of variability in the population to permit further adaptive changes. However, although a certain amount of variability is advantageous, too great a genetic variety in the population can be harmful. Geneticists reason that different forms have differing effects upon the overall fitness of a population. Given that there is one optimal phenotype, the greater the number of other suboptimal types present, the less the average fitness of the population will be. The presence of these suboptimal forms imposes a condition of genetic load in terms of a reduction in average population fitness. It is possible that the population could be driven to extinction as a result of too heavy a genetic load.

Question:

Find the logarithm of 30,700.

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Solution:

First express 30,700 in scientific notation. 30,700 = 3.07 × 10^4. 4 is the characteristic. To find the mantissa, see a table of common logarithms of numbers The mantissa is 4871. Thus log 30,700 = 4 + .4871 = 4.4871.

Question:

Plot the following function on a K-map and simplify in SOP and POS forms. F(A,B,C,D) = \sum(0,2,5,8,10,13,14,15) + X(1,11,12). By using the simplified expression, determine the output when a redundant input occurs.

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Solution:

The K-map is shown in figure 1. The "X" mark at locations one, eleven, and twelve denotes the redundancies. First simplify in the SOP form. Minterms at 13,14, and 15 and the redundancy at 12 combine to form AB. Minterms at 0,2,8, and 10 combine to formBD. Minterms at 5 and 13 combine to form BCD. The SOP form is: F = AB +BD+ BCD To simplify F into the POS form, simplifyFin SOP form and then use DeMorgan's Law. From the K-map it is seen that the 0's at 6 and 7 combine to formABC. The 0 at 4 and the redundancy at 12 combine to form BCD. The 0's at 3 and 9 and the redundancies at 1 and 11 combine to formBD. The SOP ofFis F=ABC + BCD+BD Apply DeMorgan's Law toFgives the POS form of F. F = (A+B+C) (B+C+D) (B+D) The table of fig. 2 shows the output when redundancies occur.

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Question:

Calculate the electrostatic force on q1due to q_2(F_12) and the force on q_2 due to q_1( F_21 ) for the case illustrated in the figure.

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Solution:

Using Coulomb's law and doing the calculation in the CGS system, of units we have F_12 ={(+3 atat C)(+5 stat C)} / (5cm)^2 = + 0.6 dyne (to the left) F_21 = {(+5 stat C) (+3 stat C)} / (5 cm)^2 = + 0.6 dyne (to the right) Both forces have the same magnitude (Newton's third law for electrostatis forces) and are repulsive.

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Question:

Produce a circuit for coding a binary coded decimal digit input into the Hamming code.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G06-0132.htm

Solution:

A binary coded decimal digit has four input lines D_0,D_1,D_2,D_3. The number K of parity bits required with 4 information input lines = 3 (see the previous problem). Let these 3 parity check bits be C_1, C_2, C_3 Hence, the coded information will be of the type:C_1 C_2 D_3 C3D_2 D_1 D_0 C_1 C_2 D_3 C3D_2 D_1 D_0 (see problem 3 of this chapter). And, let us name these 7 lines as: d_1 d_2 d_3 d_4 d_5 d_6 d_7 Now, by the theory of Hamming code, one must have an even parity between C_1, D_0, D_2, D_3; C_2, D_0, D_1, D_3; andC_3, D_0, D_1, D_2. In other words, one must ensure that the exclusive-or sum of each of the four variables in the above 3 sets equals to 0. i.e.,C_1 \oplus D_0 \oplus D_2 \oplus D_3 = 0, C_2 \oplus D_0 \oplus D_1 \oplus D_3 = 0, C_3 \oplus D_0 \oplus D_1 \oplus D_2 = 0, The circuit for generating C_1, C_2, and C_3 can nowbe conceived by noting that the parity check bit should be a 1 if the exclusive-or of the other noting that the parity check bit should be a 1 if the exclusive-or of the other three variables = 1 (i.e., there is an odd number of variables which are equal to 1), and the parity check bit should be = 0 if the exclusive-or of the other three variables = 0 . Hence, the circuit can be drawn as shown in Fig. 1. Note in the above figure, that all gates A to F are 2-input exclusive-or gates. Gate A gives the exclusive-or of D_2 and D_3.Output of gate A is 1 if only of D_2 or D_3 is a 1 . This output is again exclusive-ored with D_0 .If the output of A was a 1, then, if D_0 is 1, the output of B will be 0, i.e., C_1 = 0, as required. In each case it can be verified that the check bits C_1, C_2 ,C_3 take on values 0 or 1 so as to maintain an even-parity in their respective groups.

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Question:

Explain the operation of NOR gates, and generate OR, NOT, and AND operations using NOR gates. Draw symbol diagrams, and explain the functions of each system with the help of truth tables.

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Solution:

The NOR gate is derived by combining the functions of an OR operator followed by a NOT operator. The key to remembering the function of a NOR gate is the first row of the truth table of fig. 1; ab a + b f(a, b) 00 01 10 11 0 1 1 1 1 0 0 0 Fig. 1f(a, b) =a+b The output of a NOR gate is 1. if and only if both inputs are simultaneously 0. The block diagram of a NOR gate is shown in fig. 2. NOR gates may also be used to generate AND, OR, and NOT operations. As seen in fig. 3, an OR gate can be constructed using two NOR gates, When the inputs a and b into gate 1 are both 0, the output of gate 1 (which is the input of gate 2) is 1. Therefore, the output of gate 2 is 0, following the truth table given for NOR gates. When one or both inputs are 1 the output of the OR gate is 0. The truth table of fig. 3(a) illustrates this principle throughly; ab F(a, b) 00 01 10 11 0 1 1 1 Fig, 3(a)f(a, b) = a + b Construction of a NOT gate using NOR gates: Operation of the circuit in fig. 4 is illustrated by its truth table given in fig. 5; a f(a) 0 1 1 0 Fig. 5 Fig. 6 shows the construction of an AND gate using NOR gates. 1st and 2nd NOR gates act as NOT gates and supplya, andbinto the third NOR gate. The output of this gate is then;by the use of DeMorgan's Law.

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Question:

What is the rotational inertia about an axis through the What is the rotational inertia about an axis through the center of a 25-kg solid sphere whose diameter is 0.30 m?

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Solution:

For a rigid body rotating with angular speed \omega about a fixed axis, the kinetic energy is K = (1/2) mv^2 = (1/2) m(\omegar)^2. Each particle of this body can be considered as contributing to the total kinetic energy. The angular velocity, \omega, of all the particles is the same, but their distance r from the axis of rotation varies. Therefore, the total kinetic energy can be written as K = (1/2) (m_1r^2_1 + m_2r^2_2 + ... ) \omega^2 = (1/2)\sum(m_ir^2_i)\omega^2 where the summation is taken over all the particles in the rigid body. The rotational inertia, I, is defined as I = \sum m_ir^2_i. As can be seen from the above equations, the ro-tational energy of a body, for a given angular speed \omega, depends on the mass of the body and the way that mass is distributed around the axis of rotation. Since most rigid bodies are not composed of discrete point masses but are continuous distributions of matter, the summation for I in the above equation becomes an integration. Let the body be divided into infinitesimal elements of mass dm at a distance r from the axis of rotation. Then the rotational inertia is I = \int r^2 dm where the integral is taken over the whole body. For a solid sphere of radius R, dm = \rho dV where \rho is the density of the sphere and dV is an infinitesimal volume. For dV, take a circle of radius l and of thickness r d\texttheta, where l is the distance from the axis of rotation. We have dV= (2\pil) (dr) (rd\texttheta) = 2\pi (r sin \texttheta) r drd\texttheta = 2\pi r^2 sin \texttheta dr d\texttheta \rho= m/v = m/[(4/3)\piR^3] Then I= \int l^2 dm = \int (r sin \texttheta)^2 \rhodv = _\texttheta=0\int^\pi _r=0\int^R (r^2 sin^2 \texttheta) [m / (4/3)\piR^3](2\pi r^2 sin \texttheta dr d\texttheta) = _\texttheta=0\int^\pi _r=0\int^R {(3m)/(2 R^3)} r^4 sin^3 \texttheta dr d\texttheta = _\texttheta=0\int^\pi ({(3m)/(2 R^3)} (r^5/5) sin^3 \texttheta]^R_r=0 ) d\texttheta = 3/10 m R^2^\pi\int_\texttheta=0 sin^3 \texttheta d\texttheta = 3/10 m R^2^\pi\int_\texttheta=0 sin \texttheta (1 - cos^2 \texttheta) d\texttheta Let x = - cos \texttheta and dx = sin \texttheta d\texttheta. Then I= 3/10 m R^2^\pi=-cos\pi\int_x=-cos0\textdegree- (1 - x^2) dx = 3/10 m R^2 {x - (x^3/3)}]^1_-1 = 3/10 m R^2 [1 - (1/3) + 1 - (1/3)] = [(3/10) mR^2][4/3] = 2/5 mR^2 For the given sphere, the mass is 25 kg and the radius is 0.30 m/2 = 0.15 m. Its rotational inertia is then I= (2/5)mR^2 = (2/5) (25 kg) (0.15 m)^2 = 0.22 kg-m^2.

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Question:

The Berkeley synchrocyclotron was designed to accelerate protons to a kinetic energy of 5.4 x 10^-11 J, which corresponds to a particle speed of about 2 × 10^8 m/s. At this speed the relativistic mass increase is important and must be taken into account in the design of the accelerator. Calculate the percentage increase in the proton mass en- countered in this instrument.

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/Users/wenhuchen/Documents/Crawler/Physics/D32-0943.htm

Solution:

The relation between relative mass, m, and rest mass, m_0, is given by m = m_0 / \surd(1 - v2/ c^2)' = m_0 / \surd[1 - {(2 × 10^8) / (3 × 10^8)}^2] = m0/ \surd{1- (4/9)} = m_0 / {\surd(5/9)} = 3m0/ \surd5 = 1.34 m_0 . The percentage increase in mass is the fractional increase in mass times 100, or [(m - m_0) / m_0] × 100 = [(1.34m_0 - m_0 ) / m_0] × 100 = 34 per cent. The proton mass at a speed of 2 × 10^8 m/s is 34 per cent greater than its rest mass.

Question:

An enquiring physics student connects a cell to a circuit and measures the current drawn from the cell to be I_1. When he joins a second, identical cell in series with the first, the current becomes I_2 . When he connects the cells in parallel, the current through the circuit is I_3. Show that the relation he finds between the currents is 3I_2 I_3 = 2I_1 (I_2 + I_3).(See figures (a), (b) and (c)).

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Solution:

Let the emf of any of the cells be \epsilon and its internal resistance be r. Let the external circuit have a resistance R. When a single cell is used (see figure (a)) R and r will be in series. Hence, the equivalent circuit resistance R_eq is R_eq = R + r Ohm's Law yields I_1 = [(\epsilon_net )/(R_net )] = [(\epsilon)/(R_eq )] = [(\epsilon)/(R + r)] If two identical cells are connected in series, their emf's act in the same sense. Hence, again using Ohm's Law (see figure (b)) I_2 = [(\epsilon_net )/(R_net )] = [(2\epsilon)/(R + r)] When the cells are connected in parallel, since they are identical, by the symmetry of the arrangement identical currents I_0 must flow through each cell. Further, since no charge accumulates at point A in this circuit, (by conservation of charge) I_3 = I_0 + I_0 = 2I_0 Considering the passage of current through either cell, we have V_AB = \epsilon - I_0 r = \epsilon - (I_3 /2)r. This follows because, each charge passing through \epsilon is raised in potential an amount \epsilon. However, by Ohm's Law, each charge also loses potential l_0 r by crossing the battery's internal resistance. When we consider the passage of current through the external circuit, then V_AB = I_3 R, by Ohm's Law. Hence I_3 R = \epsilon - (I_3 r/2) Rewriting the three equations obtained, we find that R + r = (\epsilon/I_1 ), R + 2r = (2\epsilon/I_2 ), and R + (r/2) = (\epsilon/I_3 ) We must eliminate all resistance variables to obtain the given formula. Solving the first equation for r, and substituting this value in the second equation, we obtain r = (\epsilon/I_1 ) - R R + 2 [(\epsilon/I_1 ) - R] = (2\epsilon/I_2 ) or- R = (2\epsilon/I_2 ) - (2\epsilon/I_1 ) R = 2\epsilon [(1/I_1 ) - (1/I_2 )](1) Inserting the calculated value of r in the third equation R + (1/2) [(\epsilon/I_1 ) - R] = (\epsilon/I_3 )or(1/2) R = (\epsilon/I_3 ) - (1/2) (\epsilon/I_1 ) R = \epsilon [(2/I_3 ) - (1/I_1 )](2) Dividing equations (1) and (2) gives 1 = [{2[(1/I_1 ) - (1/I_2)]} / {(2/I_3) - (1/I_1)}] (2/I_3 ) - (1/I_1 ) = (2/I_1 ) - (2 /I_2 ) or(2/I_3) = (3/I_1) - (2/I_2) Multiplying both sides by I_1 I_2 I_3 2I_1 I_2 = 3I_2 I_3 - 2I_1 I_3 or3I_2 I_3 = 2I_1 (I_3 + I_2 ) \therefore2I_1 I_2 - I_2 I_3 = 2I_2 I_3 - 2I_1 I_3 .\therefore3I_2 I_3 = 2I_1 (I_2 + I_3 )

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Question:

Theketoneacid (CH_2CO_2H)_2CO undergoes a first-order decomposition in aqueous solution to yield acetone and carbon dioxide: (CH_2CO_2H)_2CO \rightarrow (CH_3)_2CO + 2CO_2 (a) Write the expression for the reaction rate, (b) The rate constant k has been determined experimentally as 5.48 × 10^-2/sec at 60\textdegreeC.Calculate t_1/2 at 60\textdegreeC. (c) The rate constant at 0\textdegreeC has been determined as 2.46 × 10^-5/sec. Calculate t_1/2 at 0\textdegreeC. (d) Are the calculated half-lives in accord with the stated influence of temperature on reaction rate?

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Solution:

(a) For a chemical decomposition, the rate of the reaction is equal to the product of the rate constant (k) and the concentration of the compound decomposing. Thus, Rate = k [(CH_2CO_2H)_2CO] (b)Because the rate is only proportional to [(CH_2CO_2H)_2CO], the reaction is first-order. For a first- order reaction,. the half-life (t_1/2) is related to k by the following equation: t_1/2 = 0.693 / k Solving for t_1/2 : t_1/2 = (0.693) / (5.48 × 10^-2 /sec) = 12.65 sec. (c) One can solve for t_1/2 at 0\textdegreeC using the same equation. t_1/2 =(0.693) / (2.46 × 10^-5/sec)= 2.82 × 10^4 sec. (d) In general, the speed of a chemical change is approximately doubled for each ten degrees rise in tem-perature. The temperature rises 60\textdegree, from 0\textdegreeC to 60\textdegreeC, therefore, the rate should double six times or the ratio of the t_1/2, at 0\textdegreeC to the t_1/2 at 60\textdegreeC is 2^6. 2^6 = 64. (t_1/2 0\textdegree) /(t_1/2 60\textdegree) = (2.82 × 10^4 sec) / (12.65 sec) = 2.23 × 10^3 This is much greater than the expected ratio of 64.

Question:

Explain why the hydrolysis of an ester to yield alcohol and acid is best carried out in basic rather than acidic solution. Use equations to illustrate your answer.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E21-0774.htm

Solution:

The best way to solve this problem is to consider what products will be formed with acid and base. The general formula of an ester is . The carboxylic acid and alcohol are generated. But esters are generated from carboyxlic acid and alcohols by the use of H+. In other words, there exists nothing to prevent the reaction from going back in the other direction. That is, you have the following equilibrium: Let us consider the reaction with base. With the basic reaction, you generate an alcohol and a carboxylate anion, not an acid. This is the key. If the base is KOH, is produced. If NaOH you have Due to the fact that the carboxylate anion, which forms a salt, is generated, the reaction is prevented from going in the opposite direction. As such, your yields upon hydrolysis are greater with the base than with acid.

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Question:

A neutron traveling at a speed of 2 × 10^3 m/s collides with a nucleus and rebounds with a speed of 3 × 10^2 m/s. (a) Determine the work done by the force of interaction between the nucleus and neutron and (b) estimate the strength of the force if the distance over which the collision occurred is 2 × 10^5 m. The mass of a neutron is 1.67 × 10^-27 kg.

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/Users/wenhuchen/Documents/Crawler/Physics/D05-0275.htm

Solution:

The mass of the neutron, m = 1.67 × 10^-27 kg, its initial speed v_0 = 2 × 10^3 m/s, its final speed, v = 3 × 10^2 m/s, and the distance over which the particles interacted, d = 2 × 10^-15 m, are the known observables. The work done by the force of the interaction of the particle, W, and the force F are the unknown observables. (a) Assume that the nucleus experiences no change in velocity due to the collision with the neutron. Then the total change in energy of the system is due only to the change in the kinetic energy of the neutron. This change in energy is equal to the work W. W = ∆KS = (1/2) m(v^2 - v^2_0) W = (1/2) (1.67 × 10^-27 kg) (3 × 10^2 m/s)^2 - (1/2) (1.67 × 10^-27 kg)(2 × 10^3 m/s)^2 = 7.5 × 10-2 3J - 3.34 × 10^-2 1 J = - 3.27 × 10-2 1J The minus sign indicates that the work done by the force decreased the kinetic energy of the neutron. (b) Although it is not stated that the force is constant during the collision, assume that it is and also assume that it is parallel to the displacement vector. Using these assumptions, the force can be determined from F = W/d = (-3.75 × 10-2 1Nm)/(2 × 10^-15m) = - 1.64 × 10^-6 N

Question:

A cart of mass 5 kg moves horizontally across a frictionless table with a speed of 1 m/sec. When a brick of mass 5 kg is dropped on the cart, what is the change in velocity of the cart?

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/Users/wenhuchen/Documents/Crawler/Physics/D06-0300.htm

Solution:

Assume that the brick has no horizontal velocity when it is dropped on the cart. Its initial horizontal momentum is therefore zero. Since no external horizontal forces act on the system of cart and brick, horizontal momentum must be con-served. We can say, for the horizontal direction, m_cv_ci+m_bv_bi=m_cv_cf+m_bv_bf Since the final velocities of the brick and cart are the same, m_cv_ci= (m_c +m_b)v_f Substituting values, v_f= (m_cv_ci)/(mc+m_b) = [(5 kg) (1 m/sec)]/[(5 kg + 5 kg)] = .5 m/sec The change in velocity of the cart is v_f-v_ci_ = (0.5 - 1.0)m/sec = - 0.5 m/sec.

Question:

At t_1 = 0 an automobile is moving eastward with a velocity of 30 mi/hr. At t_2 = 1 min the automobile is moving north-ward at the same velocity. What average acceleration has the automobile experienced?

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Solution:

Since velocity is a vector quantity, vector addition must be used to solve this problem. Geometrical-ly, when two vectors are added, the tail of the second vector is placed at the head of the first and the re-sultant vector is drawn from the tail of the first to the head of the second. To find the difference between the two velocities, we write v_2\ding{217} - v_1\ding{217} = \Deltav\ding{217} Changing the expression above into one including only addition: v_2\ding{217} = \Deltav\ding{217} + v_1\ding{217} This is shown in the accompanying vector diagram. The magnitude of \Deltav is (refer to the figure and use the Pythagorean theorem) \Deltav = \surd[(30 mi/hr)^2 + (30 mi/hr)^2] = \surd[1800 (mi/hr)^2] = 42.4 mi/hr The magnitude of the average acceleration is a= (\Deltav/\Deltat) = [(42.4 mi/hr)/60 sec] = 0.71 (mi/hr)/sec The direction of \Deltav, and hence the direction of a is, from the figure, in the direction northwest.

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Question:

Explain and exemplify internal and external parasitism.

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Solution:

Parasites can be classified as external or internal parasites. The formerlive on the outer surface of their host, usually feeding on the hair, feathers, scales or skin of the host or sucking its blood. Internal parasites maylive in the various tubes and ducts of the host's body, such as the digestivetract, respiratory passages or urinary ducts. They may also bore intoand live embedded in tissues such as muscle or liver. However, pathogenslike viruses and some bacteria are believed to actually live insidethe individual cells of their host. External parasitism may have evolved from what was once a commensalrelationship. In some cases, internal parasitism may have evolvedfrom external parasitism. External parasites might have wandered intoone of the body openings of their host, such as the mouth nasal openings, or anus. Most of them cannot survive in the different environmentthey encounter there. But it is possible that in the course of thousandsor millions of years some wander-ers might have had the geneticconstitutions enabling them to survive in the new habitat they foundinside the host's body. From such a beginning, the specializations thatordinarily characterize the more advanced internal parasites could haveevolved. Other internal parasites may have evolved from free living formsthat were swallowed or inhaled. Internal parasitism is usually marked by much more extreme specializationsthan external parasitism because the habitats available insidethe body of a living organism are completely unlike those outside it. The unusual internal conditions have resulted in adaptations different from thoseseen in free-living forms. For example, internal parasites have frequentlylost individual organs or whole organ systems that would be essentialto a free-living species. Tapeworms, for instance, have no digestivesystem. They live in their host's intestine, where they are bathed byproducts of the host's digestion, which they can absorb directly across theirbody wall without having to digest anything themselves. Loss of certain structures is not the only sort of special adaptation commonlyseen in internal parasites. They often have body walls highly resistantto the de-structive enzymes and antibodies of the host. Moreover, theyfrequently possess specialized structures such as the specialized headsof tapeworms with hooks and suckers that enable them to anchor themselvesto the intestinal wall. One of the most striking of all adaptations of internal parasites concernstheir life cycles and repro-ductive capability. Individual hosts do notlive forever. If the parasitic species is to be perpetuated, a mechanism isneeded for leaving one host and penetrating another. Thus, at some pointin their life cycle, all internal parasites must try to move from one hostindi-vidual to another. But this is seldom simple. Rarely can a parasite movedirectly from one host individual to another. It is not unusual for a life cycleto include two or three intermediate hosts and a free living larval stage. But such a complex development makes the chances that any one larvalparasite will encounter the right hosts in the right sequence exceedinglypoor. The vast majority die without completing their life cycle. As a result, internal parasites characteristically produce huge numbers of eggs. Although they may be deficient in some organs through specializationthey usually have extremely well-developed reproductive structures. In fact, some internal parasites seem to be little more than a sac of highly efficient reproductive organs.

Question:

Ribonuclease is 1.65% leucine and 2.48% isoleucine by weight. Calculate its minimum molecular weight.

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Solution:

If Ribonuclease is 1.65% by weight leucine, .0165 times the minimum molecular weight of ribonuclease is equal to the molecular weight of leucine. (MW of leucine = 130) MW of Leu = MW of Ribonuclease × .0165) MW of Ribonuclease = (130) / (.0165) = 7.89 × 10^3 If Ribonuclease is 2.48% by weight isoleucine (MW = 131), one can solve for the molecular weight of ribonuclease using a similar equation as the one used for leucine. MW of Ribonuclease= (MW of Ile) / (.0248) = (131/.0248) = 5.29 × 10^3 Therefore, to cover both conditions, the minimum molecular weight must be 7.89 × 10^3.

Question:

Discuss the possible sequence of events in the evolution of photosynthesis.

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Solution:

There are two types ofphotophosphorylationcyclic andnoncyclic. It has been proposed that cyclicphotophosphorylationis the more primitive of the two processes and perhaps was the earliest form of energy production. It might have developed at a time when the earth's atmosphere contained little or no oxygen gas. Cyclicphotophosphorylation probably served as a means of synthesizing ATP for primitive organisms living in an anaerobic environment. The next evolutionary step was per-haps the development ofnoncyclicphotophosphorylation, such as that found in certain bacteria today. In these bacteria, NADP^+ molecules are the electron acceptors but the electron donors that restore the excited chlorophyll molecules to ground state are molecules such asthiosulfateor succinate , not water. The third step, achieved by green plants, was probably the ability to use water molecules innoncyclic photophosphorylation . By evolving such an ability, green plants were able to live nearly everywhere and were not restricted to places where thiosulfate orsuccinateare found. Then, as plants spread and proliferated, they released into the atmosphere oxygen gas, a by - product of their noncyclic photophosphorylation. The accumulation of oxygen made possible the further biochemical evolution of animals and other organisms that use aerobic respir-ation (oxygen) for survival and active growth.

Question:

What potential difference V in an electron gun is required to accelerate an electron that was initially at rest to a final speed of 1 × 10^7 m/s?

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Solution:

The electron starts from rest at the negative plate and ac-celerates straight across to the positive plate and then escapes through a small hole to form the electron beam. (See figure). By the work- energy theorem the change in kinetic energy of the electron is equal to the work done on the electron by the net force acting on it. Looking at the diagram, the net force on the electron is that due to the elec-tric field E^\ding{217}. By definition, E^\ding{217} =F^\ding{217}/q where F^\ding{217} is the force on charge q. Since q = \rule{1em}{1pt}e, the electronic charge, F^\ding{217} = qE^\ding{217} = \rule{1em}{1pt}eE^\ding{217} Hence, the work done on the electron is W= ^b\int_a F^\ding{217}\bullet dr^\ding{217} = ^b\int_a \rule{1em}{1pt}eE^\ding{217} \bullet dr^\ding{217} where the integral is evaluated over a path between points a and b. Using the work-energy theorem, (1/2) mv^2 _b \rule{1em}{1pt} (1/2)mv2a= \rule{1em}{1pt}e^b\int_a E^\ding{217}\bullet dr^\ding{217}. Here, v_b and vaare the electron speeds at points a and b, and m is its mass. Since the electron starts from rest, va= 0 and 1/2 mv^2b= \rule{1em}{1pt}e^b\int_aE^\ding{217}\bullet dr ^\ding{217}(1) By definition, \rule{1em}{1pt} ^b\int_a E^\ding{217}\bullet dr ^\ding{217} = v_b\rule{1em}{1pt}va(2) where v_b ,v_aare the potentials at points b and a. Then, using (2) in (1) 1/2 mv2_b = e(v_b\rule{1em}{1pt}_ v_a) Hence v_b\rule{1em}{1pt}_ va= mv2b/ 2e Using the given speed and the known values of electron mass and charge, v_b\rule{1em}{1pt}_ va= [(9.11 x 10^\rule{1em}{1pt}31kg) (10^14 m^2/ s^2)] / [2(1.6× 10\rule{1em}{1pt}19coul)] v_b\rule{1em}{1pt}_ va= (9.11 x 10^\rule{1em}{1pt}17 joules) / (3.2 × 10^\rule{1em}{1pt}19 Coul) But1 (Joule / coul) = 1 volt v_b\rule{1em}{1pt}_ va= 2.8 × 102volts

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Question:

Liquid naphthalene normally freezes at 80.2\textdegreeC. When 1 mole of solute is dissolved in 1000 g of naphthalene, the freezing point of the solution is 73.2\textdegreeC. When 6.0 g of sulfur is dissolved in 250 g of naphthalene, the freezing point is 79.5\textdegreeC. What is the molecular weight of sulfur?

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Solution:

In order to determine the molecular weight of sulfur, we must determine how many moles of sulfur corre-sponds to 6.0 g. When 1 mole of solute is added to 1000 g of naphthalene, the normal freezing point is lowered by 7\textdegreeC, from 80.2\textdegreeC to 73.2\textdegreeC. This corresponds to a concentration of 1molal(1 mole solute / 1000 g solvent) = (1molal). The ratio of sulfur to naphthalene in the sulfur-naphthalene solution is (6.0 g sulfur) / (250 g) naphthalene = 0.024. Thus, adding 6.0 g of sulfur to 250 g of naphthalene is equivalent to adding 24 g of sulfur to 1000 g of naphthalene (because 24 g sulfur / 1000 g naphthalene is also 0.024). The observed freezing point depression in the sulfur- naphthalene solution is 0.7\textdegreeC (from 80.2\textdegreeC to 79.5\textdegreeC). This is the freezing point depression one would obtain by adding 24 g of sulfur to 1000 g of naphthalene. But this is one- tenth the lowering one would obtain by adding 1 mole of sulfur (the lowering would then be 7.0\textdegreeC), hence 24 g of sulfur must correspond to one-tenth of a mole. Thus, the apparent molecular weight of sulfur is (24 g / 0.1 mole) = (240 g / mole).

Question:

Iron (III) oxide is reacted with carbon monoxide in a blast furnace to produce iron. The balanced reaction is: Fe2O_3 + 3CO + 2Fe + 3CO_2 What volume of CO at STP is required to completely use up 31.94 kg of iron oxide? (MW of Fe_2O_3 = 159.7, MW of CO = 28.)

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Solution:

From thestoichiometryof the above reaction, one sees that 1 mole of Fe_2 O_3 requires 3 moles of CO to react. Thus, three times as many moles of CO will react with a given number of moles of Fe_2 O_3 . Once the number of moles of CO is known, one can use the fact that 1 mole of any gas at STP has a volume of 22.4 l to determine the volume of CO required. The number of moles of Fe_2 O_3 available for the re-action is the weight of Fe_2 O_3 in grams divided by its molecular weights : moles Fe_2 O_3 = [(weight Fe_2 O_3) / (MW Fe_2 O_3)] = [(31940 g) / (159.7 g/mole)] = 200 moles. The number of moles of CO needed to completely react is 3(200 moles) = 600 moles. The volume at STP of 600 moles of CO is the number of moles of CO times the volume of 1 mole: volume CO at STP = (moles CO) (22.4 l/mole) = (600moles) (22.4 l/mole) = 13.440 l.

Question:

Two vessels, filled with liquids at temperatures T_1i and T_2i are joined by a metal rod of length L, cross-section A and thermal conductivity k. The masses and specific heats of the liquids are m_1, m_2 and c_1, c_2 respectively. The vessels and rod are thermally insulated from the surrounding medium. What is the time t required for the temperature difference to be halved?

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Solution:

Let the vessels have instantaneous temperatures T_1 and T_2 (as shown in the figure) with T_1 > T_2. After they are joined by the metal rod, heat flows from vessel 1 to vessel 2 at a rate heat/sec = dQ/dt = kA[(T_1 - T_2)/L].(1) As heat is transferred, the temperatures of the vessels tend to equalize. Therefore the rate of heat transfer de-creases in time. The temperature variations dT_1 and dT2caused by the transfer of heat can be obtained by noting that the same heat dQ which leaves vessel 1 must enter vessel2. Therefore dQ = c_1m_1(-dT_1) = c_2m_2dT_2 where (-dT_1) indicates a decrease in T_1. Substituting for dQ in (1), we get [-c_1m_1dT_1]/dt = [{kA(T_1 - T_2)}/L]and(2) [c_2m_2dT_2]/dt= [{kA(T_1 - T_2)}/L](3) Let us rearrange equations (2) and (3) as follows dT_1/dt = - [kA(T_1 - T_2)] / [c_1m_1L_1],(4) dT_2/dt =[kA(T_1 - T_2)] /[c_2m_2L_2].(5) Then, if we subtract (5) from (4), we obtain the rate of change of the temperature difference T_1 - T_2 with time: (dT_1/dt) - (dT_2/dt)= [d(T_1 - T_2)]/dt = -(kA/L)[(1/c_1m_1) + (1/c_2m_2)](T_1 - T_2). The solution of the above differential equation gives the time dependence of (T_1 - T_2): (T_1 - T_2) = (T_1 - T_2)_initial e-(kA/L)[(1/c1m1) + (1/c2m2)]t where (T_1 - T_2)_initial is the temperature difference at t = 0. The time t required for (T_1 - T_2) to equal one half of its initial value (T_1 - T_2)_initial is e-(kA/L)[(1/c1m1) + (1/c2m2)]t= [(T_1 - T_2)(t)]/[(T_1 - T_2)_initial] = (1/2). Taking the natural logarithm of both sides of the above equation, we get -(kA/L)[(1/c_1m_1) + (1/c_2m_2)]t = 1n (1/2) = - 1n 2. ort = L/[kA(1/c_1m_1) + (1/c_2m_2)] ln 2 = (Lc_1c_2m_1m_2)/[kA(m_1c_1 + m_2c_2)] 1n 2.

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Question:

A specific enzyme is found in the endoplasmic reticulum of a cell. The code for this enzyme is contained in a molecule of DNA in the nucleus. Start with this DNA and trace all steps involved in the production of the enzyme and its movement to the endoplasmic reticulum.

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Solution:

This question deals with a series of processes beginning with transcription (production of mRNA), followed by translation (production of a polypeptide chain) and ending with the passage of the polypeptide into the endoplasmic reticulum. Inside the nucleus, transcription of the DNA is mediated by RNA polymerase, an enzyme composed of five polypeptide subunits comprising the core, plus a loosly bound protein known as the sigma factor. The DNA undergoes a localized unfolding in the vicinity of the gene to be transcribed. The weak hydrogen bonds are broken and the two strands separate. Only one of the strands within a particular genome serves as a template for transcription by RNA polymerase. The sigma factor is responsible for re-cognizing the promoter region and attaches the enzyme core, to the DNA where transcription is initiated. Without the sigma factor RNA polymerase would bind at any place along the template and transcription would proceed on both strands. Once the core has been properly attached, the sigma factor can be released to become associated with another core polymerase molecule. Several core molecules may be trans-cribing a certain genome simultaneously. RNA polymerase selects precursor ribonucleotides complementary to the DNA template. That is, adenine on the template is paired with uracil, cytosine is paired with guanine, and thymine with adenine. The precurser ribonucleotides consist of a base attached to a ribose tri-phosphate molecule. Other enzymes catalyze phosphate bond formation between adjacent ribonucleotides, and the new strand of mRNA is held together by these bonds in the back-bone. The polymerization of the mRNA is driven by the energy of hydrolysis of the triphosphate of the precursor ribonuc-leotide. The direction of transcription is specific. RNA polymerase reads from the 5' end of the DNA, with subsequent synthesis of mRNA in the 3' to 5' direction. The newly synthesized single-stranded mRNA peels away from the template allowing a new RNA polymerase molecule to attach or the DNA strand to reunite. Now the synthesized mRNA has to move into the cytoplasm where protein synthesis occurs. The mRNA is able to diffuse into the cytoplasm because the nuclear membrane surrounding the nucleus has numerous rather large openings (nuclear pores) which allow the mRNA to pass through. Once the mRNA gets into the cytoplasm, a ribosome attaches to the 5' end of the mRNA and translation begins. Several ribosomes may bind sequentially one mRNA strand, forming a polyribosome or polysome. The sequence of genetic information present in the mRNA directs the order of in-corporation of amino acids into the protein chain. But before the amino acids can line up in the correct sequence, several things must happen. Amino acids have to be oriented in the correct sequence on the ribosome. This is achieved by transfer RNA, or tRNA, whose function is to transport amino acids to the ribosome and translate the genetic code of the mRNA. To be transported, an amino acid must first be activated. This occurs when the amino acid, which is bound to a loading enzyme, joins with a molecule of ATP, forming the activated amino acid.The activated amino acid is then transferred to the correct tRNA, which carries it to the ribosome, (see fig. 1) The high energy bond originally derived from the activation by ATP is retained in the amino acid \sim tRNA complex. The activating enzyme is now free to bind another amino acid. The amino acid must be transferred to a tRNA molecule which is specific for the amino acid carried by the activating enzyme. Therefore, an activating enzyme must be able to re-cognize both a specific amino acid and a tRNA specific for that amino acid. There are at least twenty different ac-tivating enzymes and each one is specific for one amino acid and tRNA. There are also at least 20 different tRNA's, each one is specific for one amino acid or the DNA strands to reunite. The correct sequencing of the amino acids is specified by the codon, a sequence of three bases on the mRNA. Each group of 3 bases codes for an amino acid. The tRNA has an anticodon, also a nitrogenous base triplet, which is comple-mentary code to the codon specific for the amino acid carried by that tRNA. It is through the complementation of the codon and anticodon that the right order of amino acids is obtained. As each amino acid gets into the right position in the sequence, an enzyme, peptidyl transferase, forms the peptide bond between two adjacent amino acid residues, (see fig.2). The reaction is driven by the splitting of the high energy bond between the amino acid and the tRNA. As more and more amino acids are linked together, a polypeptide chain is formed. When translation is complete, this chain forms the enzyme that the DNA coded for. Ribosomes are composed of two subunits, the 30S and 50S subunits. These two subunits will dissociate upon reaching the stop codon at the 3' end of the mRNA and the newly synthesized polypeptide chain is released. The sub-units can reassociate at the 5' end of the mRNA, forming a functional 70S complex, and translation of the mRNA may proceed again. The formation of the polypeptide begins with the amino terminal end. In bactelrial polypeptides, the amino terminal residue is always N-formy1-methionine, in which a hydrogen atom on the amino group of methionine has been replaced by a formyl group. The function of this modified residue is not entirely clear, but it may be involved in the protection of the growing polypeptide from digestion by various peptidases. To get the enzyme into the endoplasmic reticulum, transport across the membrane of the endoplasmic reticulum has to occur. The actual mechanism of this transport is not known but there is a fairly sound hypothesis to account for it. This hypothesis states that when the enzyme is first formed on the ribosome, it has not assumed its three- dimensional shape, but remains an unfolded chain. In this conformation, the unfolded chain is transported across the membrane into the interior of the endoplasmic reticulum. We have to recall the fact that polyribosomes are associated with the endoplasmic reticulum. So the polypeptide chain made on theribosome is next to the membrane of the endo-plasmic reticulum. Once it gets across the membrane and reaches the interior, the polypeptide chain becomes folded into its normal three-dimensional configuration. Because it isnow folded, it can no longer squeeze out through the membrane the way it came in. A summary of the transfer of genetic information ispresented below:

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Question:

Given that ∆H\textdegree_CO2(g) = \rule{1em}{1pt} 94.0, ∆H\textdegree_CO(g) = \rule{1em}{1pt} 26,4, ∆H\textdegree_H2O(l) = 68.4 and ∆H\textdegree_H2O(g) = -57.8 Kcal/mole, determine the heats of reaction of (1) CO(g) + (1/2)O_2 (g) \rightarrow CO_2 (g), (2) H_2(g) + (1/2) O_2 (g) \rightarrow H_2O(l) and (3) H_2O(l) \rightarrow H_2O(g).

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Solution:

To solve this problem, you need to know what heat of reaction means and how to determine it quantitati-vely. The heat of reaction may be defined as the heat released or absorbed as a reaction proceeds to completion. It is measured quantitatively from the heats of formation of products minus heats of formation of reactants. Namely, heat of reaction, ∆H\textdegree = ∆H\textdegree _products - ∆H\textdegreereactants. (Note: By convention, the ∆H\textdegree of any element is defined as being equal to zero.) You proceed as follows: (1) ∆H\textdegree = ∆H\textdegree_CO2(g) -∆H\textdegree_CO(g) - (1/2) ∆H\textdegree_O2(g) = - 94 - (- 26.4) - (1/2) (0) = - 67.6 Kcal/mole. (2) ∆H\textdegree = ∆H\textdegree_H2O(l) - ∆H\textdegree_H2(g) - (1/2) ∆H\textdegree_O2(g) (2) ∆H\textdegree = ∆H\textdegree_H2O(l) - ∆H\textdegree_H2(g) - (1/2) ∆H\textdegree_O2(g) = - 68.4 - 0 - (1/2) (0) = - 68.4 Kcal/mole. (3) ∆H\textdegree = ∆H\textdegree_H2O(g) -∆H\textdegree_H2O(l) = - 57.8 - (- 68.4) = + 10.6 Kcal/mole.

Question:

The density of a liquid can be measured by means of a "pyknometer" (see the figure). The pyknometer is a glass vessel with a ground glass stopper having a narrow hole along its axis. If you are given the mass, m_p, of the empty pyknometer, the mass m_p+l, of the pyknometer when filled with the liquid and the mass m_p+w of the pykno-meter filled with distilled water (all masses being de-termined at the same temperature), calculate the density of the liquid.

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Solution:

Let V be the volume of liquid contained by the pyknometer. Then V = m_\omega/\rho_\omega = m_l/\rho_l(1) where \rho_\omega and \rho_l are the water and the liquid densities respectively, and m_\omega and m_l are the masses of the water and the liquid in the pyknometer. Since m_\omega = m_p+\omega - m_p m_\omega = m_p+\omega - m_p m_l = m_p+l - m_p equation (1) becomes V = (m_p+\omega - m_p)/(\rho_\omega) = (m_p+l - m_p)/(\rho_l), giving \rho_l in terms of \rho_\omega: \rho_l = [(m_p+l - m_p)/(m_p+\omega - m_p)] \rho_\omega.

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Question:

Develop a flow-chart for evaluating the Fourier coefficients of the periodic waveform f(t) in the given figure. The values measured for f(t) are f(0) = 0, f(0.05) = 0.25, f(0.1) = 0.3, f(0.15) = 0.5, f(0.2) = f(0.25) = f(0.3) = 1.0, f(0.35) = 0.55, f(0.4) = 0.3, f(0.45) = 0, f(0.5) = -0.25, f(0.55) = f(0.6) = -0.3, f(0.65) = -0.2, f(0.7) = -0.1, f(0.75) = f(0.8) = f(0.85) = f(0.9) = f(0.95) = 0 .

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Solution:

The waveform in the figure can be expressed as a Fourier series: f(t) = a0 + ^\infty\sum_n=1 (a_n cos n\omegat + b_n sin n\omegat)(1) where, a0 = (1/T) ^T\int0 f(t) dt(2) a_n = (2/T) ^T\int0 f(t) cos n\omegat dt(3) b_n = (2/T) ^T\int0 f(t) sin n\omegat dt.(4) Equations (2)-(4) are the Fourier coefficients. We approximate the integrals by dividing f(t) into small pulses of width ∆T and summing: a0 = (1/T) ^K\sum_m=1 f(m∆T) ∆T(5) a_n = (2/T) ^K\sum_m=1 f(m∆T) cos (n(2/T)m∆T) ∆T(6) b_n = (2/T) ^n\sum_m=1 f(m∆T) sin (n(2/T)m∆T) ∆T(7) Equations (5) - (7) must be evaluated for each (harmonic) value of n. The period T of the waveform is broken into (K-1) parts such that (K-1)∆T = T. Then using the values found for f(m∆T), m = 1 to K and setting a limit for n, say 20, a computer program can be developed for evaluating the Fourier coefficients from the flow-chart. Note, finally, that a subroutine can process the coefficients found by the program into magnitude and phase form for the plotting of spectral lines.

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Question:

At the melting point of a solid (or the freezing point of a liquid), the free energies of the solid state and the liquid state are equal, ∆G = 0. Likewise, at the boiling point of a liquid, where there is an equilibrium between the liquid and vapor phases, the free energy is equal in the two states. Calculate the change in entropy for the following process at 0\textdegreeC if the heat of fusion of H_2 O = 80 cal/g. H_2 O (s) \rightarrow H_2 O (l).

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Solution:

The free energy, ∆G, can be expressed as ∆G = ∆H - T∆S, where ∆H, in this case, is the heat of fusion, T is the absolute temperature and ∆S is the change in the entropy. One is given that ∆G = 0 for this process and thus, ∆H - T∆S = 0. One is told that the process takes place at 0\textdegreeC. To use the free energy equation, the tem-perature must be in \textdegreeK. To convert from \textdegreeC to \textdegreeK, add 273 to the temperature in \textdegreeC, T = 0\textdegree + 273 = 273\textdegreeK ∆H is given as 80 cal/g. When solving for the entropy, convert ∆H to cal/mole. There are 18 g of H_2 O in one mole, thus the ∆H can be converted by multiplying it by 18 g/mole ∆H = 80 cal/g × 18 g/mole = 1440 cal/mole It is now possible to solve for ∆S. 0 = ∆H - T∆S∆H = 1440 cal/mole 0 = 1440 cal/mole - 273\textdegreeK × ∆S - 1440 cal/mole = - 273\textdegreeK × ∆S [(- 1440 cal/mole)/(- 273\textdegreeK)] = ∆S = 5.27 cal/mole - \textdegreeK

Question:

Manufacturers use soda-lime glass for containers. This type of glass can be prepared by melting sodium carbonate (Na_2CO_3), limestone (CaCO_3), and sand (SiO_2) together. The equation for this reaction is Na_2CO_3 + CaCO_3 + 6SiO_2 \ding{217} Na_2O\bulletCaO\bullet6SiO_2 + 2CO2 soda-lime glass If a manufacturer wanted to make 5000 wine bottles, each weighing 400 g, how many grams of sand would be required?

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Solution:

From the reaction equation, it can be seen that 6 moles of SiO_2 are consumed for every mole of soda-lime glass formed. Thus, one can determine the amount of sand needed to form the 5000 bottles once the number of moles of glass that makes up these bottles is found. To solve: (1) determine the weight of the bottles and the number of moles of soda-lime glass that this weight is equivalent to. (2) Calculate the number of moles of sand needed and from this its weight. Solving: (1) weight of bottles = (400 g / bottle) (5000 bottles) = 2.0 × 106 g. The molecular weight of soda-lime glass is 478 g / mole. number of moles = [(2.0 × 10^6 g) / (478 g / mole)] = 4.18 × 10^3 moles of glass. (2) For each mole of glass formed, 6molesof sand must react. Thus, no. of moles of sand = 6 × 4.18 × 10^3 = 2.51 × 10^4 moles. The molecular weight of sand is 60.1 g / mole. Hence, the amount of sand needed = (2.51 × 10^4 moles) (60.1 g / mole) = 1.51 × 10^6 g.

Question:

"Since plants cannot move about as freely as animals, they have no need for a skeletal system." Do you agree or disagree with this statement? Justify your answer.

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Solution:

A skeletal system not only provides for locomotion but also for support. While it is true that most algae are small, aquatic, and have little need for a skeletal system, the land plants do need some skeletal structures to hold their leaves in position to receive sunlight and to help the stems stand erect on land. The land plants meet these requirements in one of two ways: either the cell wall can be very thick, as in the woody stems of trees and shrubs, and serve directly for the sup-port of the plant body, or the cell wall can be rather thin yet provide support indirectly by way of turgor pressure. Trees and shrubs have in their stems tough, woody cells called tracheids and vessels, whose structures are adapted for support in addition to conduction of water and minerals. These cells secrete a very thick wall made up largely of cellulose. Cellulose is a polysaccharide of repeating molecules of glucose attached end to end. The cellulose molecules are bound together in a complex matrix to form long, thin microfibrils. The microfibrils intertwine to form fine threads or fibrils, which may in turn coil around one another like strands in a cable. Cellulose molecules arranged in such an organized pattern form fibers as strong as steel cables of equivalent thickness. The cell walls of tracheids and vessels are also impregnated with lignin which adds to the strength of cellulose. Lignin is a highly branched, rigid organic molecule of repeating subunits. It is often deposited between the cellulose microfibrils in the walls of tracheids and vessels when the protoplast has ceased to function, and serves as a strengthen-ing element because of its regular, rigid strucure and its resistance to decomposition. In addition, land plants have phloem cells which contain thick fibers in their walls to help support the trunk. Non - woody plant cells are hot equipped with thick walls but manage to support their shape and form with the aid of turgor pressure. Turgor pressure is the pressure exerted by the contents of the cell (such as organic mate-rials and inorganic salts dissolved in water) against a cell wall which is slightly elastic and stretchable. Turgor can be maintained by a plant cell because it is naturally in a hypotonic medium. In such a medium, water tends to enter the cell, passing osmotically from a region of higher water concentration to a region of lower water concentration. The pressure resulting from a difference in the concentration of water molecules between two re-gions is known as osmotic pressure. The influx of water causes the cell to swell and the cell contents to exert a hydrostatic pressure, or turgor pressure, against the cell wall. Equilibrium is reached when the two opposing forces of hydrostatic pressure (built up in response to cell wall resistance) and osmotic pressure are equalized. Under such opposing pressures, the plant cell remains stiff or turgid The importance of turgor pressure in supporting the form of the plant body can be demonstrated by placing a crisp lettuce leaf in salt water. The lettuce leaf, after being in the salt water for some time, becomes limp and flaccid. In this hypertonic solution, water follows the osmotic gradient from the leaf into the salt solution. Without the accumulation of water inside the cell, no turgor pressure can be exerted against the walls of the cell. As a result, the cell loses its stiffness and be-comes flaccid. For although the cells of the leaf still have their cellulose walls, these walls are alone unable to provide sufficient support. (See Figure.)

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Question:

Pilocarpineis a drug that stimulates the nerve endings of parasympathetic nerves. What effects would you expect this drug to have on (a) the digestive tract, (b) the iris of the eye, and (c) the heart rate?

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Solution:

The parasympathetic system consists exclusively of motor fibers originating from the brain and emerging via the third, seventh, ninth and tenth cranial nerves, and of fibers originating from the sacral region of the spinal cord and emerging by way of the spinal nerves in that region. The parasympa-thetic system therefore has nerves from two separate regions - the brain and the sacral region of the spinal cord. (a) The ninth cranial nerve, theglossopharyngeal, innervates the salivary gland. It stimulates the salivary gland to release digestive enzymes . A more important cranial nerve, thevagus, sends branches to the stomach, duodenum and the pancreas. This nerveenchancesmotility and peristalsis of the digestive tract. (Motility and peristalsis of the digestive tract should not be confused with each other. Motility of the tract causes food to mix and churn but no movement of it. Peristalsis moves the food down the digestive tract). It also stimulates the secretion of digestive enzymes by digestive glands. For example, stimulation of the pancreas by the vagusnerve causes secretion of the pancreatic juice. Thus the glossopharyngeal andvagustogether act to promote digestion. When pilocarpine is present, the parasympathetic fibers are stimulated. The result is that digestion is facilitated or enhanced. (b) The third cranial nerve supplies the muscles of the iris which controls the size of the pupil of the eye. Excitation of this nerve causes constriction of the pupil, by contraction of the circular smooth muscle. Sincepilocarpineexcites the parasympathetic system, its presence would cause the pupil to reduce in size. (c) The parasympathetic branch supplying the heart is part of the vagus nerve. Its action is to weaken and slow down the heart beat. Pilocarpine intro-duced into the blood stream would activate the parasympa-thetic nerve endings, causing the heart beat to slow and weaken .

Question:

Explain and discuss why the branch and link register (BALR) instruction is usually one of the most important instruc-tions of the IBM/360-370 computers?

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Solution:

The branch-and-link register instruction has the register-to-register (RR) format, and as such, it has a single register for each of its operands. When this instruc-tion is executed, it places the address of the next instruc-tion in storage in the register specified as the first op-erand, and then branches of the address are specified in the second register. To illustrate the function of this instruc-tion, let us assume that the instruction is stored in bytes 5000 and 5001 (the instruction is 2 bytes long). As the first operand we have Register 3 and as the second - Regis-ter 8, which contains the binary equivalent of 2048. When this instruction is executed, the address 5002 is placed in Register 3 and the computer then goes to location 2048, the address specified by Register 8. Execution now proceeds from this location. However, if a zero is specified as the second operand of the BALR instruction, no branching takes place. The address of the next instruction in memory is placed in op-erand 1 as usual, but since the operand 2 is zero, no branch-ing takes place. Instead, processing of the next sequential instruction occurs. For example, let's use the addresses of Fig. 1 for Fig. 2. On execution, address 5002 is placed in Register 3, and pro-cessing of the next sequential instruction begins. As we can see this instruction also begins at location 5002. It is this feature of the BALR instruction that makes it one of the IBM/360-370 computers' most important instructions. By this procedure a base address is loaded into the base regis-ter of the program. In the example (2), Register 3 is the base register. If a program is loaded into storage and a proper address is not loaded into the base register, the program will not be executed as intended. As a result, a BALR instruction with zero as its second operand is normally the first executable instruction of a program.

Question:

A myelinated nerve can conduct impulses of velocities of up to 100 meters per second, whereas unmyelinated nerve conducts at velocities of 20 to 50 meters per second. Explain this difference.

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Solution:

The big discrepancy between the rates of conduction of a myelinated and an unmyelinated nerve lies in the morphology of the nerves. Myelinated nerves are bundles of axons, each enveloped in a sheath of fatty substance, known as myelin. Axons of unmyelinated nerves have no myelin sheath, and are for this reason sometimes referred to as bare axons. A myelin sheath around a neuron is a highly insulating covering which prevents the flow of ions between the cell interior and the exterior of the membrane. Hence no action potential will be generated at the myelinated regions of an axon. More importantly, the insulating sheath prevents the loss of charges when Na^+ ions flow from one region to another. At successive nodes of Ranvier there are gaps in the sheath where there is no insulation. At these points, free ionic communication between the inside and outside of the membrane is possible. Thus, when Na^+ ions arrive at the gaps, the membrane is depolarized and an action poten-tial is set off. Therefore, impulses are generated only at the nodes, and nerve impulses leap from one node to the next. This kind of conduction, by "jumping" from node to node, is called saltatory transmission. The "jumping" is responsible for the more rapid impulse conduction by myelinated nerves, as compared with unmyelinated nerves. The latter lack the extensive insulation that makes saltatory conduction possible. In myelinated fibers the key permeability changes take place at the nodes where the insulationg myelin sheath is interrupted. Between the nodes, the flow of Na^+ ions carries the impulse which "jumps" from node to node along the axon.

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Question:

Consider the experimental apparatus, illustrated in figure A which is used to measure the deflection of an electron. The source of electrons is a heated filament and the detector consists of a screen painted with a material which phosphoresces when struck by energetic particles. The magnetic field is confined to the spherical region. During a particular run of this experiment, an electron enters the magnetic field at a distance of 30 cm away from the detecting screen and travels 25 cm after leaving the field. If the strength of the magnetic field is 0.18 g/coulomb-sec and the beam deflection on the screen is 4.0 mm, what is the velocity of the electrons in the beam?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E18-0664.htm

Solution:

We are looking for a relationship between the magnetic field strength H, the deflection of the electron and the electron velocity v. Such a relationship is pro-vided by the equation of motion of an electron in a magnetic field, v = Her/m, where e is the electronic charge (e = 1.602 × 10^-19 coulomb), r is the radius of curvature, H is the strength of the magnetic field, and m is the electron mass (m = 9.11 × 10^28 g). In order to apply this equation to solve for v, we must first determine r. In the following discussion to determine r, we will make reference to figure B (which is not drawn to scale for reasons of clarity). In the problem, we are given that the distance from the point at which the electron beam leaves the field to the deflecting screen is 25 cm. The point at which the electron leaves the field is D. H and D lie on the same vertical line, hence the distance from D to the screen is the same as the distance from H to the screen, or HJ = 25 cm. Since the electron de-flection is small, we can assume that the path of the electron is not far from the path the electron would have in the absence of a field. Hence, we can assume that HI is much smaller than CI and HJ, and that FG is much smaller than FB and BG. If we make this assumption that HI is negligible, then IJ = HJ - HI \cong HJ = 25 cm. The distance from the point at which the electron enters the field (point C) to the screen is 30 cm, or CJ = 30 cm. Then Cl = CJ - IJ = 30 cm - 25 cm = 5 cm. Denoting the center of the circle containing the magnetic field by G, we have CG = GI = (1/2)CI = 5 cm/2 = 2.5 cm. Using our assumption on the smallness of HI, we obtain GH = GI - HI \cong GI = 2.5 cm. In summary, so far we have obtained the following distances: CG = GI \cong GH = 2.5 cm,IJ \cong HJ = 25 cm. The strategy from this point on will be as follows. We will use the similarity of triangle GJE and triangle GHD to determine HD. Then, using the similarity of tri-angle CHD and triangle CFB, we will obtain the ratio BF/CF, and from this, the radius of curvature, r. There is a theorem from elementary plane geometry which states that when a triangle is cut by a line parallel to one of its sides, the two resulting tri-angles are similar. Since HD and JE are both drawn vertically, and are therefore parallel, triangle GJE is similar to triangle GHD, Hence [(GH)/(GJ)] = [(HD)/(JE)]or HD = JE × [(GH)/(GJ)] = 4 mm × [(2.5 cm)/(27.5 cm)] = 0.4 cm × [(2.5 cm)/(27.5 cm)] = 0.036 cm. Next we note that, using the theorems of plane geometry, it can be shown that triangle CHD is similar to CGB. Then [(HD)/(CG)] = [(CD)/(BG)] Because r = BG, we solve this ratio for BG. We have al-ready determined HD and CG and can find CD by using the Pyhthagorean Theorem, thus, we can find BG. Solving for CD: CD^2 = CH^2 + HD^2 = (5.0 cm)^2 + (0.036 cm)^2 = 25.001 cm^2; CD = 5.00 cm. Solving for BG; BG = [(CD \textbullet CG)/(HD)] = [(5.0 cm) (2.5 cm)] / (0.036 cm) = 347.2cm. We can now substitute r into the equation of motion of an electron in a magnetic field. We thus obtain the electron velocity, v = (Her/m) = [(0.18 g/coulomb-sec × 1.602 × 10^-19 coulomb × 347.2 cm)] /(9.11 × 10^-28 g)] = [(1.10 × 101 0cm/sec)].

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Question:

What is the approximate boiling point at standard pressure of a solution prepared by dissolving 234 g ofNaClin 500 g of H_2O?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E07-0246.htm

Solution:

The boiling point of a solution is acolligativeproperty. This means that it depends on the ratio of number of solute particles to number of solvent particles ormolality.Molalityis defined as the number of moles of solute divided by the number of kilograms of solvent. When water is the solvent the boiling point constant, K_b,0 is 0.52\textdegree. This means that when one mole of solute is in1 kg of H_2O the boiling point of water is raised 0.52\textdegree. boiling pt elevation =kb×molalityof solute One can find the boiling point elevation once themolalityof the NaCl is found. To determine themolalityone must first determine the number of moles present. This is done by dividing the number of grams present by the molecular weight. (MW ofNaCl= 58.5.) no. of moles = (234g) - (58.5) = 4.0 moles There are 4.0 moles ofNaClin 500 g, or 0.5 kg, of H_2O. Themolalitycan now be found. Molality = (moles of solute) / (kg of solvent) molality = (4.0moles) / ( 0.5 kg) = 8.0m NaCl is an electrolyte which means that when it is placed in H_2O it dissociates into its ions. NaClt Na+ +Cl" This means that for everyNaClpresent, there are two ions formed. There are thus twice as many moles of particles as molecules ofNaCl present. The effectivemolalityis, therefore, twice the truemolality. effectivemolality= 2 × 8.0 m = 16.0 m One can now solve for the boiling point elevation, boiling point elevation = (0.52\textdegree) × (16.0 m) = 8.32\textdegree The boiling point of the water is raised 8.32\textdegree. It was originally 100\textdegreeC, therefore the new boiling point is 100\textdegreec +8.32\textdegree = 108.32\textdegreeC.

Question:

What is a polypeptide chain and how is it related to the proteins in a cell?

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/Users/wenhuchen/Documents/Crawler/Biology/F24-0623.htm

Solution:

A polypeptide chain consists of linked units called amino acids. Each amino acid consists of a carboxyl group (C terminal), an amino group (N-terminal), and a side chain which varies with each amino acid. Structurally an amino acid is represented as follows: Amino acids can be linked together by peptide bonds, which form between the N terminal of an amino acid on a C terminal. An example of a peptide bond is illustrated below: When many amino acids are linked together by peptide bonds, we have a chain of amino acids known as a polypeptide chain. Proteins are actually-polypeptide chains. A protein molecule may be composed of one such chain, usually folded and cross-linked within itself to form a more compact and stable structure than a very long expanded chain. Some protein molecules are made of combinations of two or more polypeptide chains. The human protein hemoglobin, for instance, is composed of four polypeptides, two identical alpha chains and two identical beta chains. Each alpha chain has 141 amino acids, and each beta chain has 146. An important distinction must be made between poly-peptides and proteins at the genetic level. A given gene can code for only one polypeptide. Many proteins, however, are composed of a number of polypeptides. When a protein is composed of more than one polypeptide chain, each chain may be coded for and made separately. These chains aggregate only after their individual syntheses to form the final protein. In addition, some proteins exist in their functional form only after certain changes have taken place in the original polypeptide, such as the cleavage or loss of certain portions of the chain. Thus a given gene does not necessarily code directly for an active, functioning protein.

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Question:

What roles does glucose play in cellmetabolism ?

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Solution:

Glucose, a six carbon monosaccharide, is the primary source of energyfor all cells in the body. The complete oxidation of one molecule of glucoseyields 36 ATP molecules in a cell. The energy stored in the form ofATP can then be used in a variety of ways including the contraction of a muscleor the secretion of an enzyme. The supply of glucose to all cells mustbe maintained at certain minimal levels so that every cell gets an ade-quateamount of glucose. Brain cells, unable to store glucose, are the firstto suffer when the blood glucose concentration falls below a certain criticallevel. On the other hand, muscle cells are less affected by changes inthe glucose level of the blood because of their local storage of glucose asglycogen. In certain cells, glucose can be converted to glycogen and can be storedonly in this form, because glucose will diffuse into and out of a cell readily. Glycogen is a highly branched polysaccharide of high molecular weight, composed of glucose units linked by \alpha-glycosidicbonds. Once the conversionis complete, glycogen remains inside the cell as a storage substance. When glucose is needed, glycogen is reconverted to glucose. In addition to being stored as glycogen or oxidized to generate energy, glucose can be converted into fat for storage. For example, after a heavymeal causes the supply of glucose to exceed the immediate need, glucosemolecules can be transformed into fat which is stored in the liver and/oradipose (fat) tissue. This fat serves as an energy supply for later use.

Question:

An athlete in his run-up for a pole vault can achieve a speed of 30 ft/s. What is the maximum possible record for the pole vault likely to be?

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/Users/wenhuchen/Documents/Crawler/Physics/D05-0236.htm

Solution:

At the end of the run with a velocity of 30 ft/s, the athlete possesses kinetic energy of amount E_k = (1/2) × m × 30^2 ft^2/s^2, where m is his mass. By causing rotation about the end of his pole he transforms this kinetic energy into potential energy. The mass of the pole is negligible in comparison with that of the man and need not be considered. Further, the pole must not be made of a material which can boost the athlete's energy by its elastic springiness. The most favorable case occurs when the athlete plans his jump in such a way that, as he clears the bar, he has negligible kinetic energy left. Thus if h is the height by which the athlete's center of gravity alters in the jump, we have, by conservation of energy, (1/2)m × 30^2 ft^2/s^2 =mgh.\therefore h = (30^2 ft^2/s^2)/(2g) = 14 ft (3/4) in. But if the athlete is very tall, his center of gravity may be as much as 3 ft 9 in. from the ground during his run-up. Hence the final height of his center of gravity above the ground is maximally 17 ft 9 3/4 in. His center of gravity is located inside his body and the bar must be lower than his center of gravity by roughly half the thickness of his body. If we assume a reasonably thin athlete, a minimum of 4(1/2) in. must be subtracted from the height previously mentioned to allow for clearance. The maximum possible record for the pole vault would appear to be 17 ft 5(1/4) in. (as measured by the height of the bar). The present world record is 17 ft 4 in. (It should be noted that fiber-glass poles do not meet the conditions about elasticity stated above.)

Question:

Why do ecologists not attempt to divide the ocean intobiome typeslike those on land? How do they insteaddistinguish thedifferent parts of the ocean?

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Solution:

Because of the vastness and relative uniformity of the ocean, and thecomparatively rapid mixing of the organisms within it, the various regionsof the ocean cannot be divided into distinct biome types as are terrestrialregions. Even though the ocean is charac-terized by many differenttypes of habitats, these habi-tats are all interconnected, and some organismsmay move freely from one to the other. It seems more appropriate, therefore, to describe an ocean byzonationrather than by biometypes. The oceans and seas cover 71 per cent of the earth's surface, but, exceptfor the shallow water of the margins, they are far less complex in structureand productivity than the land. The continental margin gradually slopesseaward to depths of 150 to 200 meters and then slopes steeply to depthsof 3000 meters or more. This slope is known as the continental shelf. The continental shelf then gives way to the abyssal ocean floor. The accompanying figure depicts thezonationin an ocean. Waters overthe continental shelf comprise theneriticzone, and these beyond the edgeof the shelf comprise the main part of the ocean basin, called the oceaniczone. The floor of the ocean basin, called the abyssal plain, rangesfrom 3000 to 5000 meters in depth and may be marked by such featuresas sea mounts, ridges and trenches. The oceanic zone is further distinguishableinto four zones. The upper part of the open ocean into whichenough light can penetrate is called theeuphoticzone. Below this zoneis thebathyalzone, which ranges to a depth of perhaps 2,000 meters. Below thebathyalzone, the waters over the abyssalplain .(ocean floor) form the abyssal zone. Trenches and valleys of the ocean floor form thetidal zone, which is generally below 6,000 meters. Theeuphoticzone, usually about 100 meters in thickness, has enoughpenetrated sunlight effective for photosynthesis. Below this is a transitionzone where some light penetrates. From the transition zone downto the ocean floor, total darkness prevail and this region constitutes theaphoticzone. The exact boundary between the lighted zone and the deeplightless zone at any given locality depends upon the intensity of the sunlight, which in turn depends upon the latitude, and upon the turbidity of thewater. Usually no light penetrates below 600 meters. The physical environment of the ocean is most diverse in the shallowwater along its margin. Here in theneritic(coastal) zone exists the greatestvariation in temperature, salinity, lightintesityand turbulence. Also, here, sunlight can penetrate to the bottom of the water body. For this reason, here occurs the most luxuriant forms of both plant and animal life. The edge of the ocean, which rises and falls with the tide, is the intertidal zoneor littoral zone. It extends out into the water to the depths at which thewater is no longer stirred by tides and waves. The region after the intertidalzone is called thesubtidalzone and farther on is the lowerneritic region. In general, the density of living organisms in the open ocean is muchless than that found in theneriticregion, but because of its large areaand volume, the total biomass in the open ocean is greater.

Question:

The following reaction 2H_2S(g) \rightleftarrows 2H_2(g) + S_2(g) was allowed to proceed to equilibrium. The contents of the two-liter reaction vessel were then subjected to analysis and found to contain 1.0 mole H_2S, 0.20 mole H_2, and 0.80 mole S_2 . What is the equilib-rium constant K for this reaction?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E09-0307.htm

Solution:

This problem involves substitution into the equilibrium constant expression for this reaction, K = {[H_2 ]^2 [S_2]} / [H_2S]^2 . The equilibrium concentration of the reactant and products are [H_2S] = 1.0 mole/2 liters =0.50M, [H_2] = 0.20 mole/2 liters = 0.10M, and [S_2] = 0.80 mole/2 liters = 0.40M , Hence, the value of the equilibrium constant is K = {[H_2 ]^2 [S_2 ]} / [H_2S ]^2 = {(0.10)^2 (0.40)} / (0.50)^2 = 0.016 for this reaction.

Question:

A sample of gas occupies 14.3 l at 19\textdegreeC and 790 mm Hg. Determine the number of moles of gas present, what volume will this same amount of gas occupy at 190\textdegreeC and 79.0 mm Hg?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E02-0055.htm

Solution:

This problem involves an application of the ideal gas equation, PV = nRT, where p = pressure, V = volume, n = number of moles, R = gas constant, T = absolute temperature, we will solve this equation for n and then use this value of the number of moles, and the ideal gas equation, to obtain V at the new temperature and pressure. For our gas, we initially have V = 14.3 l, T = 19\textdegreeC = 292.15\textdegreeK, P = 790 mm Hg × 1 atm/760 mmatm/760 mm Hg = 1.039 atm and R = 0.0821 l-atm/\textdegreeK-mole. Solving the ideal gas equation for n, we obtain n = (PV/RT) = [(1.039 atm × 14.3 l)/(0.0821 l-atm/\textdegreeK-mole × 292.15\textdegreeK)] = 0.6194 mole. Under the new conditions, we have T = 190\textdegreeC = 463.15\textdegreeK and P = 79.0 mm Hg = 79.0 mmHg × 1 atm/760 mm Hg = 0.1039 atm. Solving the Ideal gas equation for V, the new volume is V = (nRT/P) = [(0.6194 mole × 0.0821 l-atm/\textdegreeK-mole × 463.15\textdegreeK) / (0.1039 atm)] = 226.68 l.

Question:

An ionic bond is established between positive ion A and negative ion B. How would one expect the strength of the bond to be affected by each of the following changes: (a) Doubling the charge on A, (b) Simultaneously doubling the charge on A and B, (c) Doubling the radius of B, and (d) Simultaneously doubling the radius of A and B?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E17-0618.htm

Solution:

When two ions come together to form an ionic bond, energy is released. The more energy released, the stronger the ionic bond. The amount of energy released is determined by the equation E = 1.44 (q_1q_2 / r) where E = energy in electron volts, q_1 = charge on the positive ion, q_2 = charge on the negative ion, and r = the distance between the nuclear centers. If one changes one parameter, another must adjust to maintain this e-quality. Solving: (a) If the charge on A, the positive ion, is doubled, the energy released is doubled. That is, 2E = 1.44 [(2q_1q_2)/ r] is equivalent to E = [(1.44 q_1q_2) / r]. If E is doubled, the strength of the chemical bond is doubled. (b) Using a similar line of reasoning, it is found that if one doubles both the positive and negative charges, four times E is released. The strength of the bond is, thus, 4 times as great as the original bond. (c) If one increases the radius of the ions then, the distance between their nuclei must also increase. To maintain the equality, E must decrease because r is inverselypropotionalto E. Therefore, the bond strength decreases. When the ionic radius of one ion is doubled, E is halved. (d) Both radii are doubled. E is diminished by a factor of 4 and the strength of the bond is decreased by a factor of 4.

Question:

An Inclined plane 5 meters long has its upper end 1 meter above the ground. A load of 100 newtons is pushed up the plane against a force of friction of 5 newtons. What is the effort, the work input, the work output, the AMA, the IMA, and the efficiency?

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Solution:

We first construct a diagram (see diagram). A vector is drawn vertically downward representing the force of gravity. Orienting our axes so that the y-axis coincides with the inclined plane, we resolve the gravitational force into its x and y components. This gives us a right triangle in which angle A' is equal to angle A, which the inclined plane forms with the ground, since they both are complements of the same angle. Therefore, since both large and small right triangles have equal angles, they are similar, so that we can write a pro-portion : 1 m/5 m = x/(100 nt) x = 20 nt Therefore, the force that the load exerts parallel to the inclined plane is 20 nt. This, plus the frictional force, is the effort. Hence, Effort = Gravitational force along plane + frictional force E = 20 nt + 5 nt = 25 nt Work is force × distance when the force is in the direction of the distance. Therefore, the work input (W_i) is: W_i = 25 nt × 5 m = 125 joules AMA (actual mechanical advantage) is: AMA = R/E where R is resistance (load), and E is effort. Then, AMA = (100 nt)/(25 nt) = 4 The IMA (imaginary mechanical advantage) is the ratio of the length of the plane to its height: IMA = (5 m)/(1 m) = 5 Efficiency is output work over input work. Efficiency= (W_O)/(W_i) = (100 joules)/(125 joules) = (AMA)/(IMA) = 4/5 = 0.80 = 80%

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Question:

Perform the following conversions from a) 1 0 0 1 1 0 1 0 b) 1 0 1 \textbullet 1 0 1 c) 0 \textbullet 0 0 1 1 base 2 to base 8, 10, and 16 .

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Solution:

a) To convert base 2 to base 8, we use the triad method where three digits are taken together and converted to equivalent decimal value; this results in base 8 conversion Similarly for hexadecimal conversion 4 bits are taken together. Zeros are assumed for most significant grouping. For binary to decimal conversion 10011010 1×2^7 + 0×2^6 + 0×2^5 + 1×2^4 + 1×2^3 + 0×2^2 + 1×2^1 + 0×2^0 128 + 0 + 0 + 16 + 8 + 0 + 2 + 0 = 154_10 b)i) 101 \textbullet 101_2 = 5 \textbullet5_8 ii) (0101/5)\textbullet (1010_2/A) = 5 \textbullet A_16 iii) 101 \textbullet 101_2 (1×2^2 + 0×2^1 + 1×2^0) \textbullet (1×2^-1 + 0×2^-2 + 1×2^-3) = 4 + 0 + 1 + 0.5 + 0 + 0.125 = 5 \textbullet 625_10 = 0 \textbullet 1875_10

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Question:

A boy leaning over a railway bridge 49 ft high sees a train approaching with uniform speed and attempts to drop a stone down the funnel. He releases the stone when the engine is 80 ft away from the bridge and sees the stone hit the ground 3 ft in front of the engine. What is the speed of the train?

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Solution:

Applying the equation applicable to uniform acceleration, x - x_0 = v_0t + (1/2) at^2, to the dropping of the stone 49 ft from rest under the action of gravity, we can find the time t the stone is in motion. The initial velocity of the stone v_0 is zero. The distance the stone travels, x - x_0 = 49 ft. Therefore, 49 ft = 0 + (1/2) (32 ft/sec^2) (t^2) \thereforet = \surd[(2 × 49 ft)/(32 ft/s^2)] =7/4 s. In the time of 7/4 s it takes the stone to drop, the engine has moved with uniform speed u a distance of (80 - 3) ft. \thereforeu = d/t = (77 ft)/(7/4 sec) = 44 ft/sec = 30 mph.

Question:

Why is the drop in blood pressure important in retaining fluid within the capillaries?

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Solution:

As the blood pressure through the capillaries, there is a pressure resulting from the beating of the heart, which tends to force the plasma through the capillary membrane and into the interstitial fluid. This pressure is called the capillary. or hydrostatic pressure. There are also two other factors determining whether fluid will move out of the blood or not: 1) the interstitial fluid pressure and 2) the plasma colloid osmotic pressure. The interstitial fluid pressure is the hydrostatic pressure exerted by the interstitial fluid. It has been measured to be about - 7 mm Hg , and so tends to move fluid outward through the capillary membrane. The plasma colloid osmotic pressure arises from the proteins found in the plasma. The plasma proteinsdo not diffuse readily through the capillary membrane. Therefore, there is a concentration gradient with four times as much protein in the plasma as in the interstitial fluid. An osmotic pressure develops since water tends to pass through the capillary membrane from the interstitial fluid (which has a higher concentration of water and a lower concentration of protein molecules) to the blood plasma. The plasma colloid (\textquotedblleftcolloid\textquotedblright because the protein solution resembles a colloidal solution) osmotic pressure is approximately 23.5 mm Hg. The pressure, unlike the interstitial fluid pressure, opposes the capillary pressure by tending to move fluid inward through the capillary. An analysis of the force at the arteriole end follows: Forces moving plasma outward: (mm Hg/25) Capillary pressure25 Interstitial fluid pressure7 Total outward pressure32 Forces moving plasma inward: Plasma colloid osmotic pressure23.5 Total inward pressure23.5 Net outward force32 - 23.5 = 8.5 At the arteriole end, then, there is a net outward filtration pressure of about 8.5 mm Hg. If this force were constant throughout the length of the capillary, there would be a large net loss of fluid outward. There is, however, a drop in capillary pressure (to 9 mm Hg) at thevenuleend due to the filtration of fluid along the capillary. Therefore netreabsorptionof the lost fluid occurs at thevenuleend as follows: Forces moving plasma outward: mm Hg Capillary pressure9 (Negative) interstitial fluid pressure7 Total outward pressure16 Forces moving plasma inward: Plasma colloid osmotic pressure23.5 Total inward pressure23.5 Net inward force23.5 - 16 = 7.5 At thevenuleend, there is areabsorptionpressure of 7.5 mm Hg This reabsorptioncauses about nine-tenths (7.5/8.5) of the fluid that had filtered out at the arteriole end to be recovered. The slight excess of filtration will be recovered by the lymphatic system (see next question).

Question:

A boy drops a rubber ball from a height of 4 feet and lets it bounce. If the ball loses 10% of its energy, due to friction of com-pression on each bounce, how many bounces will it take before the ball will only rise 2 feet above the ground?

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Solution:

When the boy first lets go of the ball, its energy is purely potential and is given byE_p=mghwhere h is the ball's original height above the ground. (In this problem h = 4 feet). When the ball hits the ground its energy is purely kinetic. This is also the case when the ball just leaves the ground and begins its upward flight. As it rises its energy gradually changes from kinetic to potential. We note from the above equation that the potential energy is directly pro-portional to the height the ball rises. Thus, each time the ball bounces and loses 10% of its energy, it rises to 9/10 of its previous height. After the first bounce the ball rises (9/10) 4 feet, then after 2 bounces it rises (9/10) (9/10) 4 feet = (9/10)^2 4 feet. We can see that after n bounces maximum height of ball = (9/10)^n 4 feet Thus, we set the expression for maximum height equal to 2 feet and solve for n: (9/10)^n 4 feet = 2 feet (9/10)^n = 1/2 n log (9/10) = log (1/2) n[log 9 - log 10] = - log 2 n[0.9542 - 1.0000] = - 0.3010 - 0.0458n = - 0.3010 n = 6.55 We must round this off to 7 since physically we cannot have a fraction of a bounce.

Question:

What is the evidence that the chemical compounds of a cell arein a "dynamic state"?

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Solution:

The term "dynamic state" is meant to indicate that the chemical compoundsof a cell are cons-tantly being broken down and synthesized. At one time, it was thought that since an adult organism fails to change in outwardappearance over a short period of time, its molecular constitution isalso unchanging. That is, the cellular compounds are stable and remain inthe cell for long periods of time without being degraded and replaced, theonly exception being those compounds used as "fuel" for energy, whichby necessity must undergo catabolism. However, experimental evidence has shown that this belief in the staticstate of the cell's compounds is false. The cell's compounds are constantlychanging - being broken down and replaced by newly synthesizedcompounds. Radio- activelylabelledamino acids, fats, and carbohydratescan be injected into laboratory animals in order to demonstratethe dynamic nature of the cell's environment. If the cell were static, thelabelledmolecules would only be broken down for fuel. However, it has been observed that thelabelledamino acids are incorporatedinto proteins and that thelabelledfatty acids into fats(triglycerides). The marked proteins and fats are then degraded by the bodyto be replaced by the incorporation of new amino acids and fatty acids. All these metabolic processes occur without any visible change in bodysize. The only exception to the dynamic state of the cell is its DNA molecules, which contain genes, the body's basic units of genetic infor-mation. DNA is extremely stable in order that the here-ditary characteristics, for which it codes, have little or no chance of changing. Not only is the cell dynamic on the molecular level, but on the cellularlevel as well. Indeed, 2.5 million red blood cells alone are producedand destroyed per second in the human body.

Question:

A copper bar is 8.0 ft long at 68\textdegreeF and has a linear expansivity of 9.4 × 10^\rule{1em}{1pt}6/F\textdegree, What is the increase in length of the bar when it is heated to 110\textdegreeF?

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Solution:

Change in this object's dimensions is pro-portional to the change in temperature and the original length. Therefore the change in length of the bar is \DeltaL = L_0\alpha\Deltat= (8.0 ft)(9.4 × 10^\rule{1em}{1pt}6/F\textdegree)(110\textdegreeF \rule{1em}{1pt} 68\textdegreeF) = 0.0032 ft.

Question:

Given that the masses of a proton, neutron, and electron are 1.00728, 1.00867, and .000549amu, respectively, how much missing mass is there in ^19_9F (atomic weight = 18.9984)?

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Solution:

The total number of particles in ^19_9F and their total weight can be calculated. The amount of missing mass in ^19_9F will be the difference of this calculated weight and the given atomic weight of ^19_9F. The subscript number 9, in ^19_9F, indicates the atomic number of fluorine (F) . Because the atomic number equals the number of protons, there are 9 protons in F. The superscript, 19, indicates the total number of particles in the nucleus. Since the nucleus is composed of protons and neutrons, and there are 9 protons, there are 10 neutrons present. In a neutral atom, the number of electrons equals the number of protons. Thus, there are 9 electrons. The total number of particles is, thus, 28. The mass and quantity of each particle is now known. Calculating the total weight contribution of each type of particle: Protons:9 × 1.00728= 9.06552 Neutrons:10 × 1.00867= 10.0867 Electrons:9 × 0.000549=0.004941 Total mass= 19.1572. It is given that the mass of the fluorine atom is 18.9984. Therefore, the missing mass is 19.1572 - 18.9984 = 0.1588amu.

Question:

In aquatic plants, which haveprotoplasmsthatare hypertonic in comparison to the water in which they live, the amount of water in the plant cells remains at a constant level . Explain how this is regulated.

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Solution:

Plants living in fresh water have a protoplasm more concentrated in solutes than their surroundings i.e. their surroundings are hypotonic. Accordingly by osmosis, water tends to enter the cells of these plants in order to equalize the concentration of solutes inside and outside the cells. Although plant cells do not have contractile vacuoles (theosmoregulatory organelle in amoeba) to pump out excess water, they do have a firm cellulose wall to prevent undue swelling. As water enters a plant cell, an internal pressure known asturgorpressure is built up within the cell which counterbalances the osmotic pressure and prevents the entrance of additional water molecules. Hence theturgorpressure is pressure that is generated against the cell wall to regulate the influx of water. Thus water plant cells are protected against excessive swelling and bursting.

Question:

A chemist wants to produce chlorine from molten NaCl. How many grams could be produced if he uses a current of one amp for 5.00 minutes?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E16-0577.htm

Solution:

When electrical energy is used to produce a chemical change, it is called electrolysis. Here, electricity is being used to produce chlorine from NaCl. Michael Faraday discovered the laws that illustrate this process quantitatively. You must make use of them to answer this question. You know that chlorine exists as Cl_2 gas. In NaCl, however, it is in the form of the Cl^- ion. Therefore, to obtain Cl_2 from NaCl, you must cause the following reaction to occur: 2Cl^- \rightarrowCl_2 + 2e^- A loss of 2 electrons is involved. Faraday's law states that the amount of gas produced is proportional to the amount of electricity. One mole of electrons is defined as the faraday (F) and equals 96,500 coulombs of electricity. From the equa-tion,2Cl^- \rightarrowCl_2 + 2e^-, you see that 1 mole of Cl_2 is produced for every 2 faradays of electricity. We can determine the amount of Cl_2, by relating it to the number of Faradays present. First determine the number of coulombs. You used one ampere, which is just equal to 1 coulomb/sec, for 5 minutes. Therefore, number of coulombs = (1.00 coulomb/sec) (5.00 minutes)(60 sec/min) = 300 coulombs. Since there are 96,500 coulombs per faraday, then the number of faradays = (300 coul) / (96500coul/F) = 0.0311F. Recalling that 1 mole of Cl_2 is obtained for every two faradays, you have (0.00311 / 2) = .00155 moles of chlorine formed, which translates into 0.111g of Cl_2. Atomic weight Cl_2 = 70.9 g/mole. wt. of Cl_2=(mole)(g/mole) weight of Cl_2 = .00155 moles × 70.9g/mole = 0.111g .

Question:

The teeth of different species of vertebrates are special-ized in a variety of ways and may be quite unlike those of man in number , structure, arrangement, and function. Give examples .

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/Users/wenhuchen/Documents/Crawler/Biology/F19-0481.htm

Solution:

The teeth of snakes, unlike those of man, are very thin and sharp, and are usually curved backward. These teeth do not serve the function of mechanical breakdown of food; they serve in capturing prey. Snakes swallow their prey whole. The teeth of carnivorous mammals, such as cats and dogs, are more pointed than those of man; the canines are longer, and the premolars lack flat, grinding surfaces, being more adapted for cutting and shearing. On the other hand, herbivorous animals such as cows and horses may lack canines, but have very large, flat premolars and molars. Notice that sharp pointed teeth, poorly adapted for chewing, seem to characterize meat eaters, whereas broad flat teeth, well-adapted for chewing, seem to characterize vegetarians. In vegetarians, the flat teeth serve to break up the indigestible cell walls of the ingested plant, allowing the cellular contents to be exposed to the action of di-gestive enzymes . Animal cells do not have the indigestib-le armor of a cell wall and can be acted upon directly by digestive enzymes; hence, there is no need for flat, grinding teeth.

Question:

Write a program which reads an input text and prints out the number of occurences of each adjacent letter pair which appears within the text.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G16-0418.htm

Solution:

PROGRAM NUMBEROFOCCURENCESOFADJACENTPAIRS (INPUT, OUTPUT ) ; (\textasteriskcentered This program, reads an input text and \textasteriskcentered) (\textasteriskcentered prints out the number of occurences of \textasteriskcentered) (\textasteriskcentered each adjacent letter pair which appears within \textasteriskcentered) (\textasteriskcentered the text. This program assumes that the \textasteriskcentered) (\textasteriskcentered letters A to Z are 26 consecutive values \textasteriskcentered) (\textasteriskcentered of the type char.\textasteriskcentered) TYPE letter = 'A'. . 'Z'; VAR COUNT: Array [Letter, Letter] of o. . Maxint; FIRST, SECOND ; LETTER ; Thisch, Lastch : char; Last was letter : Boolean; BEGIN \textasciigraveFor First: ='A' To 'Z' DO For second: ='A' To 'Z' DO COUNT [First, second]:= o; While not EOF (input) DO BEGIN Last was letter : = False; (\textasteriskcentered At the start of a line \textasteriskcentered) While not EOLN (Input) DO BEGIN Read (Thisch); write (Thisch) If (Thisch>='A') and (Thisch<='Z') Then if last was letter Then count[Lastch, Thisch]:=count[Lastch, Thisch] + 1 Else lastwasletter: = True Else lastwasletter:=False; lastch:=Thisch End; Read1n; write1n End; write1n; write1n; write1n ('occurences of letter pairs......'); For First: ='A' to 'Z' DO For second : = 'A' to ' Z ' DO IF COUNT [First, second]<>o Then write1n (First, second, , count[First, second]:4) End.

Question:

A car battery supplies a currentI of 50 amp to the starter motor. How much charge passes through the starter in (1/2) min?

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/Users/wenhuchen/Documents/Crawler/Physics/D20-0652.htm

Solution:

Current (I) is defined as the net amount of charge, Q, passing a pointper unit time. Therefore, Q = It = (50 amp) (30 sec) = 1500 coul .

Question:

A chemist mixes nitric oxide and bromine to form nitrosy1 bromide at 298\textdegreeK, according to the equation 2NO_(g)+ Br_2(g) \rightleftarrows 2NOBr_(g) . Assuming K = 100, what is the quantity of nitrosy1 bromide formed, if the reactants are at an initial pressure of 1atm? R = 0.0821 liter-atm./mole\textdegreeK.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E09-0310.htm

Solution:

You are given the equilibrium constant for this reaction and asked to calculate the quantity of nitrosy1 bromide produced. The first step is to write out the equilibrium expression and equate it with the given value. For the general reaction , xA+yB\rightarrowzC, K is defined[C]^2 / { [B]^y [A]^x } , where the brackets represent concentrations. For this reaction , K = [NOBr]^2/ { [NO]^2[Br_2 ] } = 100 . To find out how muchNOBris produced, you would have to know how many moles of NO and Br_2 were reacted. Once this is known, you can find the number of grams produced. You know that the equi-librium expression is based on concentration of reactants and prod-ucts. Concentration is expressed in moles per liter. This means that if the volume of theNOBrand its concentration is known, you can find moles, since concentration X volume (liters) = moles. Let us represent the concentration as (N / V) = (moles) / (Volume) . Thus, the equilibrium expression becomes K = [(N_NOBr/ V)^2] / [(N_NO / V)^2 (N_(Br)2 / V)]. Let x = moles ofNOBrformed. Then, x moles of NO and x/2molesof Br_2 are consumed, since the coefficients of the reaction show a 2:2:1 ratio amongNOBr:NO :Br_2 . The equilibrium expression becomes 100 = (N^2_NOBrV) / (N^2 _NO N_(Br)2 ) = {x^2 V} / { (2\Elzbar x)^2 (1\Elzbar .5x) } . If x moles ofNOBrform, and you started with 2 moles of NO, then, at equilibrium, you have left 2\Elzbarx moles of NO . You started with only 1 mole of Br_2 and (1/2)x moles of itformNOBr; thus you have 1\Elzbar.5x moles left. Therefore, you need to determine only the volume to find the quantity NOBrformed. V can be found from the equation of state, PV = NRT , where P = pressure, V = volume, N = moles, R = universal gas constant, and T = temperature inkelvin(celsiusplus 273\textdegree). You are told that the reactants are under a pressure of 1 atm. at 298\textdegree K . N = 3, since the coefficients inform you that a relative sum of 3 moles of reagents exist. You know R. Thus , V = NRT / P = { (3) (.0821) (298) } / {1} = 73.4 liters . Now that V is known, the equilibrium expression becomes { x^2 (73.4)} / {(2\Elzbar x)^2 (1\Elzbar 0.5x)} = 100 . Solving for x, you obtain x = .923 moles = moles ofNOBrformed. Molecular weight = 110. Grams produced = .923 × 110 = 101.53g.

Question:

If a nuclear reactor is designed to deliver 1 MW. of heat energy continuously, how many fission events must occur each second to sustain this power level? How much uranium-235 would be consumed each year?

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Solution:

It is known that the amount of energy released per fission event is about 200 MeV. In joules, this energy becomes E = (2.0 × 10^8 ev) × (1.6 × 10^-19 J/ev) = 3.2 × 10^-11 J. The total number, n, of nuclei undergoing fission each second to produce 1 MW = 10^6 W of power is n = power / Energy per fission = (1.0 × 106J/sec) / (3.2 × 10^-11 J/nuclei) = 3.1 × 10^16 nuclei/sec. Since 1 year = 365 days/year × 24 hrs/day × 60 min/hr × 60 sec/min 7 \approx 3.1 × 10^7 sec . , the total number N of uranium-235 atoms consumed in one year is N = (3.1 × 10^7 sec/year ) × (3.1 × 10^16 nuclei/sec) = 9.6 × 10^23 nuclei/year. The mass m of one uranium-235 atom is approximately m = (2.35 × 10^2 nucleons) × (average mass of one nucleon) = (2.35 × 10^2 nucleons) × (1.67 × 10^-27 kg/nucleon) = 3.92 × 10^-25 kg. The total mass M of, uranium fuel consumed in one year is M = mn = (3.92 × 10^-25 kg/atom) × (9.6 × 10^23 atoms/year) = 3.8 × 10^-1 kg/year.

Question:

How is the respiratory quotient obtained and how can it be usedto determine when a person is in an advanced state of starvation?

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Solution:

The respiratory quotient (RQ) is a quantitative measure of the type offoods being used by the body. If we measure both the oxygen consumptionand carbon dioxide release by the body, we can determine whethercarbohydrates, fats, or proteins are being used for respiratory needs. This measure, called the respiratory quotient, is the ratio of carbon dioxideemitted to the amount of oxygen taken in: RQ = {CO2[produced]} / {O2[consumed]} During aerobic respiration, one molecule of glucoseutilisessix moleculesof oxygen and releases six molecules of carbon dioxide. This 1:1 ratio gives carbohydrates a respiratory quotient of 1.0. Fats, however, havea lower respiratory quotient. This indicates that more oxygen is necessaryto break down a fat than a carbohydrate molecule. Oxygen is neededto convert the fat (or fatty acid) into intermediate products before it canenter the respiratory process. Free fatty acids are first formed from fatsenzymatically, and are then oxidized to acetyl-CoA, which enters the TCA cycle.The overall equation forpalmiticacid (a 16-carbon fatty acid) oxidationcan be written: palmitate+ 23 O2 + 129 Pi + 129 ADP \textemdash> 16 CO2 + 145 H2O + 129 ATP The respiratory quotient is therefore (16/23) = 0.70 This is typical for most fats. The RQ of proteins lies between those of carbohydrates and fats- usuallyabout 0.80. The oxidation of two molecules ofalaninerequires six moleculesof oxygen and liberates five molecules of carbon dioxide accordingto the following equation: 2 C3H7O2N + 6 O2 \textemdash\textemdash> (NH2)2 CO + 5 CO2 + 5 H2O. This gives a respiratory quotient of 5/6, or 0.80 for proteins. Normally, if an individual is consuming all three types of food, his RQ will be about .85. After a meal heavy in carbohydrates, however, a person'sRQ will approach 1.0. If he stops eating, his RQ will decrease to .70 as his body uses its fatty deposits as an energy source. Further starvation(about 2 weeks) will tend to increase his RQ to 0.80 as the fatty depositsbecome exhausted and the proteins of the body are used. This is dangerousand can result in consumption and atrophy of muscles and organs. Therefore, a person in an advanced state of starvation will have anRQ of about 0.80.

Question:

Write a BASIC program to compute the inverse and the transpose of a 3×3 matrix A. Apply the program to the case where A =\mid 1-12 \mid \mid 201 \mid \mid 031 \mid

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G20-0502.htm

Solution:

The inverse of a matrix A is another matrix B such that, when multiplied by A, gives an identity matrix. The statement in BASIC which results in forming of an inverse matrix B of matrix A is as follows: MATB = INV (A) A transpose C of the matrix A is a matrix with rows identical to the respective columns of A, and columns identical to the respective rows of A =\mid 123 \mid \mid 7110 \mid \mid-645 \mid Then the transpose of A is: \mid 17-6 \mid \mid 2114 \mid \mid 305 \mid The corresponding BASIC statement is: MAT C = TRN(S) After generating the inverse and the transpose matrices of A, the following program also checks if B is really an inverse by multi-plying A and B to form matrix I. If I comes out to be an identity matrix, then B is indeed an inverse of A. The program looks as follows: 100DIM A (3, 3), B (3, 3), C (3, 3), I (3, 3) 110MAT READA 120MAT B = INV (A) 121PRINT 122PRINT 130MAT PRINT B: 140MAT C = TRN (A) 141PRINT 150MAT PRINT C: 160MAT I = A\textasteriskcenteredB 161PRINT 170MAT PRINT I; 800DATA 1, -1, 2, 2, 0, 1, 0, 3, 1 999END

Question:

In BASIC the RND function is used to generate a random number between 0 and 1. What statements would you use to: (a) generate integral random number between 0 and X - 1? (b) generate integral random number between 1 and X? (c) simulate the toss of a cubical die? (d) simulate a binary random choice taking values 1 or 2?

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G10-0242.htm

Solution:

(a) 1\O LET T = INT(X\textasteriskcenteredRND) (b) 2\O LET V = INT(1 + X\textasteriskcenteredRND) (c) 3\O LET W = INT(1 + 6\textasteriskcenteredRND) (d) 4\O LET N = INT(1 + 2\textasteriskcenteredRND)

Question:

The solubility product constant of magnesium hydroxide is 8.9 × 10-^12 , calculate its solubility in (a) water and (b) .05M NaOH.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E11-0389.htm

Solution:

Whenever an ionic solid is placed in water, an equilibrium is established between its ions and the excess solid phase. The solu-bility constant,K_sp, measures this equilibrium. For the general re action AxBy\rightleftarrowsxA+ +yB- , the K_sp is defined as being equal to [A+]^x [B-]^y . The concentration of solid is always constant, no matter how much is in contact with the ions. This means the solid phase will not appear in the equilibrium constant expression. For this problem, part (a), you have Mg(OH)_2(s) \rightleftarrows Mg^2+ + 2OH- . The K_sp= [Mg^2+] [OH- ]^2 = 8.9 × 10-^12 . You are asked to find these concentrations. From the chemical equa-tion, you find 2 moles of OH- will be generated per mole of Mg^2+ . Thus, at equilibrium, if the concentration of Mg^2+ = x (mol/liter), then the concentration of OH- = 2x(mol/litei). (Note: The dissociation of water contributes some OH-, but this amount is very small, and can be ignored.) Thus, you have [Mg^2+] [OH- ]^2 = x(2x)^2 = 8.9 × 10-^12 =K_sp. Solving for x, you obtain x = 1.3 × 10-^4 mol/liter. Thus, 1.3 × 10-^4 mole of Mg(OH)_2 dissolves per liter of water, producing a solution of 1.3 × 10-^4 M Mg^2+ and 2.6 × 10-^4 M OH- . To find the solubility in .05MNaOH, part (b), perform the same process, except you must realize that theNaOHsupplies OH- in addition to the OH- that dissolves from the salt. This means, as such, [Mg^2+ ] [OH- ]^2 =K_sp= (x)(2x + .050)^2 = 8.9 × 10-^12 . If you solve for x, you obtain 3.6 × 10-^9 mol/liter.

Question:

Obelia, a marine hydrozoan, exists in a polypoid and medusoid form. Explain how these stages are formed in the life cycle of the organism.

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/Users/wenhuchen/Documents/Crawler/Biology/F11-0280.htm

Solution:

The coelenterate Obelia is a marine colonial consisting of a group of polypoid individuals living together in close association. Within the colony are two types of polyps serving different functions. There are nutritive (or feeding) polyps and reproductive polyps. All the polyps of the colony are derived from an ancestral hydralike polyp by asexual budding. These buds fail to separate and the new polyps remain attached by hollow stemlike connections to the colony. Each reproductive polyp has a central core, the blastostyle. When mature, the blastostyle buds off an entirely new body form, an umbrella- shaped medusa. In the process of budding, the medusa breaks free of the blastostyle and swims out through the opening of the blastostyle called the gonotheca. The medusa of Obelia has a fringe of tentacles on its edge, and hanging down from the center, similar to the handle of an umbrella, is the manubrium containing the mouth. The gastrovascular cavity extends from the mouth in the manubrium into four radial canals, on which are located the reproductive organs, or gonads. Medusae also reproduce sexually. There are two types of medusae, each bearing either testes, the male gonads which pro-duce sperm, or ovaries, the female gonads which produce eggs. Sperm and eggs are released into the surrounding sea water. Fertilization takes place in the sea water, and a zygote is formed. Following early embryonic cleavage, a free-swimming larva, called a planula, is formed which is ciliated and lacking a gastrovascular cavity or mouth. After a free-swimming existence lasting for several hours to several days, the planula attaches to an object and develops into a polypoid colony, thus completing the life cycle. The alternation of a polypoid and medusoid stage is an example of "alternation of generations", or meta-genesis. Asexually and sexually reproducing stages al-ternate with each other. It must be noted that it is different from the alternation of generations that occurs in plants in that both polyp and medusa are diploid, as in all multicellular animals, the only haploid stage in the life cycle are the gametes.

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Question:

What power is needed to move a 3000-lb car up an 8.0\textdegree incline with a constant speed of 50 mi/hr against a frictional force of 80 lb?

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/Users/wenhuchen/Documents/Crawler/Physics/D05-0288.htm

Solution:

In order to just be able to move the car up the incline at constant velocity, there must be no net force on the car. Looking at the figure, we see that the net force on the car, acting down the plane, is F_net = mg sin 8\textdegree + f. If there is to be no resultant force on the car, we must act on it with a force equal in magnitude to F_net, but acting up the plane. Therefore, F_app = mg sin 8\textdegree + f = (3000 lb) (.139) + 80 lb = 497 lb Now, the power expended in moving an object equals the time rate of change of the work done on the object. Hence, P = dW/dt ButW = \intF_app^\ding{217} \bullet ds^\ding{217} P = d/dt [\intF_app^\ding{217} \bullet ds^\ding{217}] In our case, F_app^\ding{217} is constant, and P = [d/dt] F_app^\ding{217} \bullet\intds^\ding{217} = [d/dt] F_app^\ding{217} \bullet s^\ding{217} Since F_app^\ding{217} is constant in time also, we obtain P = F_app^\ding{217} \bullet ds^\ding{217}/dt = F_app^\ding{217} \bullet v^\ding{217} Since ds^\ding{217}/dt is defined as the velocity of the object we're moving. In this problem, F_app^\ding{217} and v^\ding{217} are parallel and in the same direction and P = F_app^\ding{217} \bullet v^\ding{217} = F_app v = (497 lb) (50 mi/hr) But 1 mi/hr = 1.48 ft/sec andP = (497 lb) (50) (1.48 ft/sec) = 3.7 × 10^4 ft \textbullet lb/sec

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Question:

A 3200-lb car traveling with a speed of 60 mi/hr rounds a curve whose radius is 484 ft. Find the necessary centripetal force.

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Solution:

A Centripetal force is a force which results when a particle executes circular motion with constant speed. It is called centripetal because it points to the center of the circle. Note that although the speed of the particle is constant, its velocity is not, because the latter is continually changing in direction. As a result, the centripetal force is responsible for changing the velocity of the particle. Using Newton's Second Law, we may write F = ma(1) where F is the net force acting on the mass, m. Because this is uniform circular motion, a = v^2/R , where v is the speed of the particle in a circular orbit of radius R. Inserting this result in (1), F = mv^2/R(2) Equation (2) gives the centripetal force needed to accelerate m. In order to use this formula, we must trans-form the weight, mg, given in the question as 3200 lb., into a mass by dividing by 32 ft/sec.^2 . Then, using (2) F = [(3200/32)sl] [(88 f/s)^2/(484 ft)] = 1600 lb.

Question:

Write a BASIC program to search a data statement for a given name which is typed in when the program is executed.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G10-0235.htm

Solution:

This problem introduces the notion of string variables and of comparisons between strings. String variables are denoted by any valid variable name with $ at the end, for example: A$, B$, C1$,..., etc. One can test strings for equality by means of the ' = ' and '< >' operators. When the search is finished, if the user wishes to search for another name, the RESTORE statement is used. 1\O REM PROGRAM SEARCHES THE DATA STATEMENT 2\O REM FOR A CERTAIN NAME 3\O PRINT "TYPE NAME, THEN HIT RETURN" 4\O INPUT A$ 5\O FOR J = 1 TO 5 6\O READ B$ 7\O IF B$ = A$ THEN 11\O 8\O NEXT J 9\O PRINT "CAN'T FIND NAME" 1\O\O GO TO 12\O 11\O PRINT B$; 'WAS FOUND" 12\O PRINT "TO CONTINUE TYPE YES; ELSE - NO." 13\O INPUT C 14\O IF C$ < > "YES" THEN 18\O 15\O DATA "JOHN", "BOB", "JILL", "MARY", "FRED" 16\O RESTORE 17\O GO TO 3\O 18\O END

Question:

Suppose the coefficient of friction between a horizontal surface and a moving body is \mu. With what speed must a body of mass m be projected parallel to the surface to travel a distance D before stopping?

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/Users/wenhuchen/Documents/Crawler/Physics/D02-0053.htm

Solution:

In this problem, we actually want to describe the motion of m, for we want it to travel a distance D before stopping. Our task is to determine the initial velocity which the mass must have in order for this to be possible. We are seeking to describe the properties of the motion of m, such as its acceleration. Therefore, we apply Newton's Second Law, F = ma, to both the vertical and horizontal directions for the mass m (see the free body diagram). In the horizontal direction, Newton's Law becomes F_x = ma_x(1) where F_x represents the sum of all forces acting horizontally on m, and a_x represents the resulting acceleration in the x direction due to these forces. From the figure, we see that f, the force of friction, is the only horizontal force acting on m. Therefore, substitut-ing into equation (1) we find - f = ma_x(2) where the minus sign appears because f acts in the negative x direction. (We are taking a_x to be positive in the positive x direction). Writing Newton's Second Law for the vertical direction, we find F_y = ma_y(3) where F_y is the sum of all forces acting on m in the y direction, and a_y is the resulting acceleration of m in the y direction. The only forces which act on m in the vertical direction are N, the normal force of the surface which pushes on m, and mg, the weight of the mass, which points downward. Substituting into equation (3), we obtain N - mg = ma_y(4) where the minus sign indicates that the two forces point in opposite directions. Furthermore, note that a_y must be zero since the mass never rises off the surface on which it slides. Substituting this into equation (4), we have N = mg(5) Now, the frictional force law is given by f = \muN(6) where \mu is the coefficient of sliding friction between m and the surface, and N is the magnitude of the normal force. Substituting equation (5) into equation (6), we obtain f = \mumg(7) Inserting this into equation (2), -\mumg = ma_x(8) Solving for a_x a_x = -\mug(9) Note that because \mu and g are constants, a_x is also constant. Hence we may use the kinematical equations for constant acceleration to describe the position of the mass. The equation needed is V^2_f - V^2_o = 2a_x(x_f - x_o)(10) where x_o and v_o are the initial position and velocity of m, respectively, and x_f, v_f are the final position and velocity of m, respectively. For this problem: x_o = 0x_f = D v_o = ?v_f = 0(11) v_f is 0 because the mass is at rest after travelling to its final position, which is D. Substituting these values into equation (10), 0 - v^2_o = 2a_x(D - 0)(12) or-v^2_o = 2a_xD(13) Then, substituting equation (9) into equation (13), we obtain v^2_o = 2\mugD orv_o = \surd(2\mugD)(14) for the initial velocity of m needed so that it may travel a distance D before stopping.

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Question:

In the Citric Acid Cycle (also known as the Krebs Cycle) , citrate is converted to isocitrate. From the algebraic sign of ∆G\textdegree for the isomerization, what is the favored or spontaneous direction of the reaction? Calculate ∆G\textdegree for the following reaction and explain what bearing this reaction will have on the isomerization of citrate and the operation of the Krebs Cycle: Isocitrate^3- + (1/2) O_2 (g) + H^f \rightleftarrows \alpha-ketoglutarate^2- + H_2O (l) + CO_2 (g) ∆G_f in (Kcal / mole) are for citrate^3- = - 279.24; isocitrate^3- = -277.65; H^f = 0, O_2 = 0; \alpha-ketoglutarate^2- = -190.62; H_2O(l) = - 56.69; CO_2(g) = - 94.26.

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Solution:

The Krebs Cycle is a series of oxidations aided by enzymes that leads to citric acid and eventually to CO_2, H_2O and ATP. To find ∆G of the isomerization, set up the equation citrate^3- \rightleftarrowsisocitrate^3- and use the Gibbs free energy of formation values given. ∆G\textdegree_reaction= \sumG\textdegree_f,_products - \sum∆G\textdegree_f,_reactants = (-277.65 Kcal/mole) - (- 279.24 Kcal/mole) = + 1.59 Kcal/mole. Thus, isomerization of citrate to isocitrate is not a spontaneous process, since ∆G\textdegree for the reaction is positive. To find ∆G of the following reaction: Isocitrate^3- + (1/2)O_2 (g) + H^f \rightleftarrows \alpha-ketoglutarate^2- + H_2O (l) + CO_2 (g) Use ∆G\textdegree_reaction = \sum∆G\textdegree_f,_products - \sum∆G\textdegree_f,_reactants = [∆G\textdegree_f,CO_2 + ∆G\textdegree_f,H_2O + ∆G\textdegree_f,_ketoglutarate] - [∆G\textdegree_f, H^f + ∆G\textdegree_f, (1/2O_2) + ∆G\textdegree_f,_Isocitrate] = (- 94.26 - 56.69 - 190.62) - (0 + 0 - 277.65) = -63.92 (Kcal /mole), ∆G\textdegree for the oxidation and decarboxylation of isocitrate is -63.92 (Kcal / mole); this large free energy decrease will favor the constant removal of isocitrate and permit the cycle to move forward despite the non-spontaneity of the isomerization.

Question:

State what is wrong with the following flowchart entries, and suggest better ways of writing them:

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Solution:

a) When writing flowcharts, it is bad practice to use the equal sign to mean "store the integer 7 in location B." You should use equal signs only in decision boxes when comparing two quantities. Instead, a better way of writing this box requires an arrow; thus we have b) As written, this decision box is meaningless. When the question "Is Q odd?" is asked, an appropriate answer would be "yes" or "no", or perhaps "true" or false". Equal signs should be used when two things are compared, not when you question the status of one item. c) This box has a trivial error, but one that should be clarified. The author of this statement probably meant "store the integer 3 in location N." To correct this, write the following: d) The colon is used in flowcharts to compare two items. The pos-sibilities are M > N, M \geq N, M = N, M < N, or M \leq N, and these cases should be taken into consideration. One possible interpretation is the following:

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Question:

Three researchers studied 1 mole of an ideal gas at 273\textdegreeK in order to determine the value of the gas constant, R. The first researcher found that at a pressure of 1 atm the gas occupies 22.4 l. The second researcher found that the gas occupies 22.4 l at a pressure of 760 torr. Finally, the third researcher reported the product of pressure and volume as 542 cal. What value for R did each researcher determine?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E02-0050.htm

Solution:

This problem is an application of the ideal gas equation, PV = n RT, where P = pressure, V = volume, n = number of moles, R = gas constant, and T = absolute tem-perature. Specifically, we are trying to determine R from the relation R = PV/nT. All three researchers worked with one mole of gas (n = 1) at T = 273\textdegreeK. Their results are as follows: First researcher: P = 1 atm, V = 22.4 l. R = (PV/nT) = [(1 atm × 22.4 l)/(1 mole × 273\textdegreeK)] = 0.0821 [(l-atm)/(\textdegreeK-mole)] Second researcher: P = 760 torr = 760 torr × [(1 atm)/(760 torr)] = 1 atm, V = 22.4 l. R = (PV/nT) = [(1 atmt × 22.4 l)/(1 molex 273\textdegreeK)] = 0.0821 [(l-atm)/(\textdegreeK-mole)] Third researcher: PV = 542 cal = nRT = 1 mole (R)(273\textdegreeK) R = (PV/nT) = [(542 cal)/{(1 mole) (273\textdegreeK)}] = 1.99 cal/mole \textdegreeK.

Question:

How does a facultative anaerobe differ from anobligatory anaerobe?

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Solution:

Organisms that can liveanaerobicallyare divided into two groups. The obligatory, or strict, anaerobe cannot use oxygen and dies in the presenceof oxygen. These include denitrifying bacteria of the soil, which areresponsible for reducing nitrate to nitrogen, and methane-forming bacteria, which produce marsh gas. Some obligatory anaerobes are pathogenicto man; these includeClostridiumbotulinum, responsible for botulism, a fatal form of food poisoning;Clostridiumperfringens, which causesgas gangrene in wound infections; and Clostridiumtetani, which causesthe disease tetanus. The facultative anaerobes can live either in the presence or absenceof oxygen. Under anaerobic conditions, they obtain energy from afermentation process; under aerobic conditions, they continue to degradetheir energy sourceanaerobically(viaglycolysis) and then oxidizethe products of fermentation using oxygen as the final electron acceptor. Yeast will grow rapidly under aerobic conditions but will still continueto live when oxygen is removed. It reproduces more slowly but maintainsit-self by fermentation. Winemakers take advantage of this behaviorby first aerating crushed grapes to allow the yeasts to grow rapidly. They then let the mixture stand in closed vats while the yeasts convertthe grape sugaranaerobicallyto ethanol.

Question:

Describe the various types and functions pertaining to the epithelial tissues of animals.

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Solution:

Epithelial tissues form the covering or lining of the internal and external body surfaces. Epithelial tissues include, for example, the outer part of the skin, the linings of the digestive tract, the lungs, the urogenital tract, the covering of the body cavity and so forth. The cells that make up the epithelial tissues are packed closely together, thus providing a continuous protective barrier between the underlying cells and the outside world. This protection keeps the body safe from mechanical injury, from harmful chemicals and bacteria, and from drying. Epithelial cells may have other functions, such as absorption, secretion, and sen-sation. Types of epithelial tissue A. squamous epitheliumB. cuboidal epithelium C. columnar epitheliumD. ciliated columnar epithelium E. sensory epitheliumF. glandular epithelium The surface of the epithelial cell that is exposed to air or fluid often becomes specialized. This surface commonly bears cilia, hairs or fingerlike processes. It may also be covered with waxy or mucous secretions. The opposite surface rests upon other cell layers. Epithelial cells are generally grouped into four categories according to their shape and function: squamous, cuboidal, columnar, and glandular epithelial cells. Squamous cells are much broader than they are thick and have the appearance of thin, flat plates. They are found on the surface of the skin and the lining of the mouth, esophagus and vagina. Cuboidal epithelium (cube-shaped cells) is found in kidney tubules. The cells of columnar epithelium are long and thin, resembling pillars or columns. The stomach and intestines are lined with columnar eoithelium. Some columnar cells have cilia on their surfaces. The function of the cilia is to move sub-stances past the cell. The respiratory tact is character-ized by ciliated columnar epithelium. Finally, a gland may consist of one or many epithelical cells. Exocrine glands have ducts, while endocrine glands do not. Examples of secretions from glandular epithelium include wax, sweat and milk. Epithelial tissue may further be classified as simple or stratified. Simple epithelium is one cell layer thick. Stratified epithelium consists of several layers of cells. Both simple and stratified epirhelium may be squamous, cuboidal, or columnar.

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Question:

The principle of negative feedback, where an increase in the output of a system acts on the system to halt its further production, is manifested in the menstrual cycle. Where?

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/Users/wenhuchen/Documents/Crawler/Biology/F21-0557.htm

Solution:

The first event in a menstrual cycle following the menstrual phase (flow phase) is an increase in the secretion of FSH by the anterior pituitary as a result of stimulation by the hypothalamus. This hormone stimulates the development of the follicles in the ovaries, which contain the eggs. After some time, one of the follicle gains predominance. At this point, only this one follicle continues to grow. Under the combined influence of both FSH and LH, the latter of which is maintained, at a low level in the blood by the follicular cells, the cells surrounding the egg within the follicle begin to secrete estrogen. As more and more estrogen is produced, the increased level of this hormone in the blood exerts an inhibitory effect on the FSH-stimulating center in the hypothalamus. This results in a decrease in the level of FSH produced by the anterior pituitary. This is the negative feedback mechanism between estrogen and FSH. High concentration of a secreted substance (estrogen) turns off the original stimulus for its production (FSH), so that an excessive amount of the secreted substance (estrogen) would not be made. This occurs roughly 8 to 10 days following the menstrual phase, or 13 to 15 days after the start of the cycle. While the high level of estrogen inhibits FSH production, it activates the LH-stimulating center in the hypothalamus, ultimately inducing a surge of LH into the circulatory system, which initiates ovulation. After ovulation the ovarian follicle with the egg discharged develops into the progesterone-secreting corpus luteum. The corpus luteum promotes the plasma progesterone level to such a state that progesterone begins to exert ah inhibitory effect on the LH-stimulating center in the hypothalamus. This is the negative feed-back mechanism between progesterone and LH. The high progesterone concentration suppresses further release of LH so that another ovulation is prevented; hence no other corupus luteum is formed to contribute more progesterone to the circulatory system.

Question:

How many moles are contained in 196 g of H_2SO_4?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E04-0117.htm

Solution:

To calculate the number of moles of a given compound from a certain number of grams of that compound one should use the following relation: number of moles =(numberof grams) / (molecular weight) When determining the molecular weight of a compound one must first know the atomic weights of the elements within the compound. One then determines the molecular weight by adding together the weights of the individual elements making up the compound. The molecular weights of the elements contained in H_2SO_4 are MW of H = 1, MW of S = 32, and MW of O = 16. If there is more than one atom of a particular element, the molecular weight of the element must be multiplied by the number of atoms present in the compound. Calculating the molecular weight of H_2SO_4 H 2 1.0 =2 S 1 32 =32 O 4 16 =64 Total =98 The molecular weight of H_2SO_4 if is thus 98 g. The number of moles of H_2SO_4 in 196 g can now be calculated. number of moles = (196 g) / (98 g/mole) = 2.0 moles.

Question:

Write the reaction equation for the formation of the corresponding ether from methanol. How many ethers are possible when a mixture of CH_3OH and CH_3CH_2OH reacts with the acid H_2SO_4?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E21-0776.htm

Solution:

The formation of ethers from alcohols requires acid. In acid solution, the hydroxyl group of an alcohol is protonated to form the oxonium ion : A free electron pair on the oxygen of another molecule of alcohol reacts with the oxonium ion to form an ether. This is accompanied by water loss : If you had a mixture of CH_3OH and CH_3CH_2OH, three ethers would result. Upon the addition of acid, methanol molecules can react with other methanol molecules to form methyl ether. It can also react with ethanol (CH_3CH_2OH) to form methyl ethyl ether. The ethanol molecules can react with other ethanol molecules to form ethyl ether. (Ethanol can react with meth-anol, but this yields methyl ethyl ether.) Thus, there are 3 possible ethers: methyl ether, ethyl ether and methyl ethyl ether.

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Question:

For the block and tackle shown in the figure (a) find the displacement ratio. (b) What force, F, must be exerted on the free end of the rope to lift a 200 lb load?

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/Users/wenhuchen/Documents/Crawler/Physics/D02-0045.htm

Solution:

(a) When F\ding{217} pulls down the rope by an amount L, pulley 2 moves up by (1/2)L (as shown in the figure) since the shortening of the rope is shared by the two segments of rope that hold the pulley. Therefore, the ratio of the displacement of load to the displacement of the rope is [(1/2)L]/L = (1/2). (b) From the figure, we see that the load is held up by a force 2T\ding{217} where T\ding{217} is the tension in the rope. Hence, in order to lift the load, the minimum tension should satisfy W= 2T orT= (1/2)W where W is the weight of the load. F\ding{217} is equal to T\ding{217} as long as the rope does not break since the stress in the rope is caused by the action of F\ding{217}. We have F= T = W/2 = (200 lb)/2 = 100 lb.

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Question:

Explain why the following statements or expressions are illegal in Pascal.

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Solution:

a) Program name is an identifier. It must start with a letter and continue with letters or digits only. No hyphen is allowed. b) 2.5 is an invalid label. A label must be an integer. c) Constants are defined, not assigned. The correct form is CONST K = 10.5; d) 'New' is a reserved word in Pascal. e) Theindextypeof an array must be a finitesubrange. 'Integer' is not a subrange. f)Anindextypemay not be asubrangeof real numbers. There are infinitelymany real numbers between 3.5 and 8.5. g)Each record definition must end with 'END' statement. h)illegalassignment. Should be y: =7.5; i )Function odd(x) returns abooleanvalue, while p is declared as an integer. Its type must be changed tobooleanfor the statement to be correct.

Question:

Use the quadratic formula to solve for x in the equation x^2 - 5x + 6 = 0.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E33-0972.htm

Solution:

The quadratic formula, x = [{\rule{1em}{1pt}b \pm\surd(b^2 - 4ac)} / {2a}], is used to solve equation in the form ax^2 + bx + c = 0. Here a = 1, b = \rule{1em}{1pt}5, and c = 6. Hence x = {\rule{1em}{1pt} (\rule{1em}{1pt}5) \pm\surd[(\rule{1em}{1pt}5)^2 - 4\bullet1\bullet6)]} / {2 \bullet 1} = [{5 \pm\surd(25 - 24)} / {2}] = [(5 \pm\surd1) / 2] = [(5 \pm1) / 2] = [(5 + 1) / 2]or[(5 \rule{1em}{1pt} 1) / 2] = 6/2or4/2 = 3or2 Thus the roots of the equation x^2 - 5x + 6 = 0 are x = 3 and x = 2.

Question:

H. Hertz produced radio waves whose wavelength was about 3 m. What was the frequency of the oscillating electric charges responsible for this electromagnetic radiation?

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/Users/wenhuchen/Documents/Crawler/Physics/D25-0809.htm

Solution:

Hertz produced radio waves by use of an induction coil. This alternating current generator was connected in series with a metal loop containing a gap in it (see the figure). A second loop (the receiver) with a similar gap was placed a few feet away from the first loop (the transmitter). The induction coil created an arc discharge across the gap of the transmitting circuit. A similar discharge appeared across the gap of the receiving circuit. Apparently, the disturbance at the first circuit was transmitted to the second circuit. This disturbance was shown to propagate with the speed of light c. The effect of the AC generator is to cause the electric charge within the wire in the transmitting circuit to oscillate about an equilibrium position. An oscillating charge undergoes acceleration and emits electromagnetic radiation with a frequency equal to the oscillating frequency of the charge. The electro-magnetic wave disturbance is transmitted through space. When it reaches the. receiving circuit, it causes the charge in the wire to oscillate. An AC current is then created. An arc discharge is then developed across the gap due to this current. If the wavelength of the wave is given, the frequency of the wave can be found by use of the following relation. c = f\lambda Since the value of the wavelength, \lambda = 3m, and the speed of light, c = 3 × 10^8 m/s, are known, the frequency is then f = (c/\lambda) = {(3 × 10^8m/s)/3m} = 1 ×10^8 Hz This frequency falls within the present-day FM radio band. The frequency of the source (i.e. the oscillating charge) is the same as that of the wave, so the current (i.e. the movement of charge) is oscillating at a frequency of 1 × 10^8 Hz also.

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Question:

Change the Subroutine of the previous problem into a FUNCTION SUBPROGRAM and explain the procedure.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G12-0282.htm

Solution:

In FORTRAN, there is another kind of subprogram, called the FUNCTION SUBPROGRAM. The Function Subprogram may replace the subroutine if the result obtained by the SUBROUTINE is a single value. In order to get a function subprogram from the subroutine of the previous problem, 3 changes have to be made. FUNCTIONIGCD(I,J) that is, the word SUBROUTINE is replaced by FUNCTION. Parameter K is not being used anymore because the name of the function will be assigned the value of the greatest common divisor. The divisor value obtained by the subprogram is an integer; therefore the name of the FUNCTION subprogram must begin with the corresponding letter (i.e., I,J,...,N). time it appears in the statement, i.e., statement 4, (K = J), has to be changed into IGCD = J. parameters . It was observed in the Subroutine program GCD that the values of I and J were changed, therefore some modifications must be made before this subroutine can become a function subprogram. The simplest modification would be to assign the values of I and J to two independent variables, say M and N, and then use them in place of I and J. Therefore, the modified FUNCTION SUBPROGRAM will take the following form: FUNCTION IGCD (I,J) M = I N = J IF (M - N) 1,4,2 1 L = N N = M M = L 2 L = M - (M/N){_\ast}N IF( L)3,4,3 3 M = N N = L GO TO 2 4 IGCD = N RETURN END

Question:

Calculate, E\textdegree, ∆G\textdegree, and K for the following reaction at 25\textdegreeC. (1/2) Cu (s) + (1/2) Cl_2 (g) = (1/2) Cu^2+ + Cl^-.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E15-0553.htm

Solution:

The magnitude of the standard electrode potential, (E\textdegree) is a measure of the tendency of the half-reaction to occur in the direction of reduction. The standard EMF (Electromotive Force) of a cell is equal to the standard electrode potential of the right - hand electrode minus the standard electrode potential of the left - hand electrode (by convention) , E\textdegree = E\textdegree _right - E\textdegree _left The reaction taking place at the left electrode is written as an oxidation reaction and the reaction taking place at the right electrode is written as a reduction re-action. The cell reaction is the sum of these two reactions. Thus, Oxidation:(1/2) Cu (s) = (1/2) Cu^2+ + e Reduction:(1/2) CI_2 (g) + e = Cl^- Since the oxidation reaction is to occur at the left electrode and the reduction at the right electrode, the cell is written as Cu \vert Cu^2+ \vert \vert Cl^- \vert Cl_2 (g) \mid Pt E\textdegree = E\textdegreeCl- \mid Cl2 \mid pt- E\textdegree _Cu^2_+\midCu E\textdegree can be found on any Standard Electrode Potential Table. E\textdegreeCl- \mid Cl2 \mid pt= 1.360 E\textdegree _Cu^2+ \midCu= 0.337; Thus, E\textdegree = 1.360 - 0.337 E\textdegree = 1.023 V Now that E\textdegree is known, ∆G\textdegree, standard Gibbs free energy, i.e., the energy available to do useful work, can be calcul-ated from ∆G\textdegree = - n F E\textdegree, where n = number of electrons transferred and F = the value of a Faraday of electricity, (23, 060 cal/V-mole). From the redox half-reactions written, one sees that 1 electron is being transferred. Consequent-ly, n = 1. Substituting these values one obtains ∆G\textdegree = - (1) [23, 060 (cal/V-mole)] (1.023 V) = - 23, 590 cal. The equilibrium constant, K, can be calculated using the Nernst equation for unit activity. In other words, E\textdegree = .0591/N log K. Substituting the previously calculated values for E\textdegree and n, 1.023 V = .0591 log K Solving,log K = 17.31orK = 2 × 101 7.

Question:

At constant volume, the heat capacity of gas differs from the heat capacity at constant pressure. Why?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E14-0482.htm

Solution:

Heat capacity is defined as the quantity of heat that will result in a temperature rise of 1\textdegreeC per mole of material. When heat is added at constant volume, all the heat goes into increasing the kinetic energy of the mo-lecules. However, with constant pressure, the volume can expand. This means the added heat must do two things. It must be able to increase the kinetic energy of the molecules AND do work against the pressure to expand the volume. The constant pressure case would require more heat. In fact, the difference between C_p andC_v, where C_p = heat capacity at constant pressure, andC_v= heat capacity at constant volume, is the universal gas constant R. In summary, then, the difference stems from the fact that C_p must increase kinetic energy and do work to expand the volume, whileC_vonly has to increase the kinetic energy.

Question:

What difference can be observed in carbohydrate metabolism between liver and skeletal muscle cells?

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Solution:

Both liver cells and muscle cells have the ability to store excess glucose as glycogen. Glucose is not transformed directly into glycogen. The first step in this conversion involves the change of glucose into glucose 6-phosphate. This ATP-dependent reaction is catalyzed by hexokinase. After several reversible intermediate steps, glucose 6-phosphate molecules are con-densed to form the polysaccharide glycogen. The major difference in carbohydrate metabolism between liver and muscle cells is the presence of the enzyme, glucose 6-phosphatase in liver cells. This enzyme converts glucose 6-phosphate to free glucose. The unphosphorylated glucose can then enter the circula-tory system. The presence of glucose 6-phosphatase allows the liver cells to break down glycogen into glucose molecules and release the latter into the bloodstream. Thus, one of the functions of the liver is to regulate glucose level in the blood by releasing or storing the monosaccharide as the situation demands. Muscle cells are unable to release free glucose and thus cannot regulate the blood glucose level because it lacks glucose 6-phosphatase. However, muscle cells can still store glucose as glycogen due to the presence of hexokinase.

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Question:

The owner of a car and a helpful passer-by attempt to pull the former's car from the field Into which it has skidded. They attach two ropes to the front of the chassis symmetrically, each rope being 1 ft from the center point, C, and exert pulls of 200 lb and 150 lb in parallel directions, both at an angle of 30\textdegree to the horizontal (see the figure). To what point of the chassis must a tractor be attached and what horizontal force must it exert to produce an equivalent effect?

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Solution:

The resultant force of the two pulls exerted by the men must be R^\ding{217} of magnitude (200 + 150) lb = 350 lb, in the same direction as either of the forces, i.e., at 30\textdegree to the horizontal. Only the horizontal component of this force is doing useful work in pulling the car from the field. This component has magnitude R cos 30 = 350 lb × \surd3/2 = 303.1 lb. This is the force that the tractor must exert. The point of attachment of the tractor must be the point O at which the line of action of the resultant R^\ding{217} cuts the front of the chassis. The forces that the two men exert on the chassis produce a net torque \cyrchar\cyrt about the center point C. The point of action O ofR^\ding{217} must lie at a distance from C, along the front of the chassis, such that R^\ding{217} produces a net torque equal to \cyrchar\cyrt: \cyrchar\cyrt = (150 lb)(1 ft) - (200 lb)(1 ft) = (350 lb)x x = (-50 ft-lb)/(350 lb) = - (1/7)ft where x is the distance of O from C. All counterclock-wise torques are taken as positive. Since x is negative, we see that R produces a clockwise torque about C. This tells us that O must be to the left of C (above C in the figure). Thus the point of attachment of the tractor is 6/7 ft from A, that is, 1/7 ft from the center point of the front of the chassis.

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Question:

Sunlight is the ultimate energy source on earth. Energy from sunlight is not returned to its source but is transformed to other forms of energy which are closely tied together in an energy cycle. Describe the energy cycle.

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Solution:

The energy cycle starts with sunlight being utilized by green plants on earth. The Kinetic energy of sunlight is transformed into potential energy stored in chemical bonds in green plants. The chemical bonds are synthesized by the process of photosynthesis. The potential energy is released in cell respiration and is used in various ways. Thus, fundamental to the energy cycle is the ability of energy to be transformed. Inorganic (nonliving) matter in the ecosystem is closely tied to organic (living) matter in the energy cycle. For example, photosynthesis requires carbon dioxide from the air and water and minerals from the soil to occur besides sunlight. These are the nonliving components of photosynthesis. Chlorophyll in green plants captures the sunlight, and organic substances (i.e., glucose) are generated from inorganic ingredients via a series of enzymatic reactions in the plant cell. Chlorophyll, enzymes, and other cellular components form the living part of photosynthesis. In the energy cycle (Figure), some of the food synthesized by green plants are broken down by the plants for energy, and consequently carbon dioxide and water are released. These again become available for green plants in capturing more energy of the sunlight. Some of the synthesized compounds are used in building the bodies of the plants and are hence stored as potential energy until the plants die. The bacteria and fungi of decay break down the bodies of the dead plants, using the liberated energy for their own metabolism. In these processes, carbon dioxide and water are released, and the minerals go back into the soil. These substances are thus recycled. Animals which feed on plants utilize a part of the energy from the food in cell respiration, with a release of carbon dioxide and water which are again recycled. Some of the minerals in the plant food are excreted by the animals and are thus available to be reused. Animals which feed on other animals utilize some energy from the food in building their own bodies. They break down some of their stored food to yield energy for daily activities such as locomotion. Food degradation is accompanied by release of carbon dioxide and water, which are returned to the ecosystem. When animals die, their bodies decay and all of the materials that were used in the construction are restored to a state which can be reused by the action of the bacteria and fungi of decay. It must be remembered that at no point in the cycle is energy destroyed. The energy from sunlight is not destroyed but is transformed into heat, chemical or mechanical energy.

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Question:

The development of a parasitic mode of existence is a remarkable example of an adaptation that has evolved to permit one organism to exist at the expense of an-other. Among the flatworms the flukes are parasitic. Describe the body structure of a fluke, its modifica-tions for a parasitic existence, and its life cycle.

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Solution:

The adult fluke is roughly one inch long, and its body is quite flat. The worm is covered by a cuticle that is secreted by the underlying cells. The cuticle is one of its adaptation for parasitism, for it protects the worm from the enzymatic action of the host. With the development of the cuticle, the cilia and sense organs so characteristic of other flatworms have disappeared. The mouth is located at the anterior end and is surrounded by an oral sucker. Another sucker, the ventral sucker, is located some distance posterior to the mouth. The ventral sucker is used to attach the fluke to the body of its victim, the host. The anterior sucker, with the aid of the muscular pharynx, is used to withdraw nutrients from the host. The mouth opens into a muscular pharynx, and just back of this, the alimentary tract branches into two long intestines that extend posteriorly almost the entire length of the body. By far the most complicated aspect of these worms is their reproductive cycle. The fluke is hermaphroditic, that is, it contains both male and female sexual organs. There is no copulatory structure in the flukes and re-production occurs through self fertilization. Following fertilization, the eggs are expelled from the fluke, which at this point is within the host's body; they then exit from the host's body in the feces. Hatching of the eggs in the oriental liver fluke occurs only when the eggs are eaten by certain species of freshwater snails, and takes place in the digestive system of the snail. Here, the egg hatches into a ciliated, free-swimming miracidium (see Figure). Within the digestive system of the snail, which is called the first intermediate host, the miracidium develops into a second larval stage, called a sporocyst. Inside the hollow sporocyst, germinal cells give rise to a number of embryonic masses. Each mass develops into another larval stage called a redia, or daughter sporocyst. Germinal cells within the redia again develop into a number of larvae called cercariae. The cercaria possesses a digestive tract, suckers and a tail. The cercaria is free swimming and leaves the snail. If it comes in contact with a second intermediate host, an invertebrate or a vertebrate, it penetrates the host and encysts. The encysted stage is called a metacercaria. If the host of the metacercaria is eaten by the final vertebrate host, the metacercaria is released, migrates, and develops into the adult form within a characteristic location in the host, usually the bile passages of the liver. Man, the final host, usually gets the fluke by eating a fish which contains the en-cysted cercaria.

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Question:

Body temperature on the average is 98.6\textdegreeF. What is thison (a)theCelsius scale and (b) the Kelvin scale?

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Solution:

(a) One converts \textdegreeF to \textdegreeC by using the following equation. \textdegreeC = 5/9 (\textdegreeF\Elzbar32) \textdegreeC = 5/9 (98.6\Elzbar 32) = 5/9 (66.6) = 37.00\textdegreeC. (b) \textdegreeC can be converted to \textdegreeK by adding 273.15 to the Celsius temperature. \textdegreeK = \textdegreeC + 273.15 \textdegreeK = 37.0 + 273.15 = 310.15\textdegreeK.

Question:

In the figure \epsilon_1 = 12 volts, r_1 = 0.2 ohm; \epsilon_2 = 6 volts, r_2 = 0.1 ohm; R_3 = 1.4 ohms; R_4 = 2.3 ohms; compute (a) the current in the circuit, in magnitude and direction, and (b) the potential difference V_ac .

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Solution:

(a) From conservation of energy we know that the sum of the changes in potential (voltage changes) around any closed loop must equal zero. Therefore the current (which is conventionally taken as the flow of positive charge) is in a clockwise direction because \epsilon_1 > \epsilon_2 ; but let us choose it as going counterclockwise to show that it doesn't make any difference. Note that each battery has an internal re-sistance. Starting at point A and traversing counterclock-wise yields the following equation: - \epsilon_1 - ir_1 - iR_4 - iR_3 - ir_2 + \epsilon_2 = 0 i (r_1 + r_2 + R_3 + R_4 ) = \epsilon_2 - \epsilon 1 i = [(\epsilon2- \epsilon_1 )/(r_1 + r_2 + R_3 + R_4 ) = (-6/4) = -1.5 amp. The negative value for the current merely means that we chose the wrong direction (i.e., the current flows clockwise) (b)We may use either a clockwise or counterclockwise path from point A to point C to find V_AC . Since we are only concerned with differences in potential, let us assume a zero potential of A and then traverse the loop clockwise from A to C. Taking into account the clockwise flow of cur-rent this yields, -\epsilon_2 - ir_2 - iR_3 = -6 - (1.5) (.1) - (1.5) (1.4) = -8.25 volts This means that the potential is 8.25 volts lower at point C than at point A. If we go from point A to point C in a counterclockwise direction we must remember to use the negative value for the current. This path yields: -\epsilon_1 - ir_1 - iR_4 = -12 - (-1.5) (.2) - (-1.5) (2.3) = -12 + .3 + 3.45 = -8.25 volts. Again, we see that the potential at point C is 8.25 volts lower than that at point A.

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Question:

What are the common editing procedures in APL? Give examples.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G14-0381.htm

Solution:

Errors have to be corrected before a program be-comes operational. Editing procedures help in making the corrections. These procedures are: i ) Open a function, display all lines of the func-tion, then close the function. ii) Open a function, replace a line that contains an error with a new line, then close the function. iii) Open a function, list all its instructions and remain in definition mode. iv) While in programming mode, direct program control to a specific line number. v) While in programming mode, direct program control to a specific line and then eliminate that line. vi) Insert an instruction between two other instruc-tions. To illustrate these procedures, consider the following program: \nablaTRIANGLE [1]'ENTER HEIGHT' [2]H \leftarrow [] [3]'ENTER BASE' [4]B \leftarrow [] [5]A \leftarrow .5 × H × B [6]'THE AREA IS:';A i) To display the total program (assuming it has been keyed in) key in \nablaTRIANGLE [ [] ]\nabla. This command causes a listing of the program to be made at the users terminal. Note that [6] has an error. ii) 'THEAREA...' should be 'THE AREA...'. To make this correction, type in \nablaTRIANGLE [6] 'THE AREA is:'; A\nabla. iii) To remain in definition mode, type \nablaTRIANGLE [ [] ]. (the \nabla has not yet been added). iv) Suppose we want to add more information so as to calculate the perimeter of the triangle. Then [3] and [4] can be deleted. Key in [3] and depress ATTN and RETURN keys. This action deletes [3]. Similarly, key in [4] and depress ATTN and RETURN keys. The program now looks like this: \nablaTRIANGLE [ [] ] \nablaTRIANGLE [1]'ENTER HEIGHT' [2]H \leftarrow [] [3]A \leftarrow .5 × H × B [4]'THE AREA IS:'; A [5] v,vi) Of course, the program does not make any sense right now. We must key in the three sides, replace B in [3] by the side corresponding to B and add a line for cal-culating the perimeter. To add the three sides: \nablaTRIANGLE[2.1] [2.1]'ENTER LEFT SIDE' [2.2]L \leftarrow [] [2.3]'ENTER BASE' [2.4]B \leftarrow [] [2.5]'ENTER RIGHT SIDE' [2.6]R \leftarrow [] [2.7]P \leftarrow L + B + R [2.8]\nabla Note that there is no need to change B. However, since P should be printed, add another line after [6]: \nablaTRIANGLE[6.1] [6.1]'THE PERIMETER IS:';P [6.2]\nabla The command NUM will renumber the statements. The fin-al result is: \nablaTRIANGLE [1]'ENTER HEIGHT' [2]H \leftarrow [] [3]'ENTER LEFT SIDE' [4]L \leftarrow [] [5]'ENTER BASE' [6]B \leftarrow [] [7]'ENTER RIGHT SIDE' [8]R \leftarrow [] [9]P \leftarrow L + B + R [10]A \leftarrow .5 × H × B [11]'THE AREA IS:'; A [12]'THE PERIMETER IS:'; P \nabla

Question:

Describe the output of the following program. L = 0 DO 1 K = 1, 25

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G12-0277.htm

Solution:

This problem is a good example of the usage of the CONTROL STATEMENT "DO". DO statements are always of the form DO ri= j, k, or DO ri= j, k, l, where r-statement number,i- variable name,j,k, l,-variable names or positive integers. A DO statement is interpreted as follows: do all FORTRAN statements down to and including the one numbered r, first using the value ofi= j, theni= j + l,i= j + 2l, and so on, untilibecomes larger than k. If l is missing, it is assumed that l = 1. Therefore, the program takes L = 0 and K = 1 as initial values and then, according to DO statement, executes the formula L = L + K{_\ast}K 25 times with the value of K being increased by 1 each time. The program then writes the value of L which is finally obtained. Arithmetical execution of the program is as follows: L = 0 + 1 \bullet 1 = 1 L = 1 + 2 \bullet 2 = 5 L = 5 + 3 \bullet 3 = 14 L = 14 + 4 \bullet 4 = 30 L = 30 + 5 \bullet 5 = 55 L = 55 + 6 \bullet 6 = 91 L = 91 + 7 \bullet 7 = 140 L = 140 + 8 \bullet 8 = 204 L = 204 + 9 \bullet 9 = 285 L = 285 + 10 \bullet 10 = 385 L = 385 + 11 \bullet 11 = 506 L = 506 + 12 \bullet 12 = 650 L = 650 + 13 \bullet 13 = 819 L = 819 + 14 \bullet 14 = 1015 L = 1015 + 15 \bullet 15 = 1240 L = 1240 + 16 \bullet 16 = 1496 L = 1496 + 17 \bullet 17 = 1785 L = 1785 + 18 \bullet 18 = 2109 L = 2109 + 19 \bullet 19 = 2470 L = 2470 + 20 \bullet 20 = 2870 L = 2870 + 21 \bullet 21 = 3311 L = 3311 + 22 \bullet 22 = 3795 L = 3795 + 23 \bullet 23 = 4324 L = 4324 + 24 \bullet 24 = 4900 L = 4900 + 25 \bullet 25 = 5525 5525 is the desired result, which will be printed.

Question:

A hydrogen atom consists of a proton and an electron separated by about 5 × 10^-11 m. If the electron moves around the proton in a circular orbit with a frequency of 10^13 sec^-1 , what is the magnetic field at the position of the proton due to the moving electron?

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/Users/wenhuchen/Documents/Crawler/Physics/D21-0712.htm

Solution:

The motion of the electron is equivalent to an electric current. That is, the charge e moves by a point on the orbit in time T, where T is the period of revolution. Thus the equivalent current is i = [(charge pass a point)/(time)] = e/t = ef . The magnetic field at the center of the circular loop of radius R is obtainedfrom the Biot - Savart Law; dB^\ding{217} = (\mu_0i / 4\pi) [(dl^\ding{217} × R^\ding{217})/R^3](1) where the permeability constant\mu_0 = 4\pi × 10[-7\omega/(a-m)] . As shown in the diagram, all the Infinitesimal contributions to B^\ding{217} from the infinitesimal circuit elements are in the direction per-pendicular to the plane of the orbit. In this case we may neglect the vector nature of (1) and obtain dB = (\mu_0i / 4\pi) [(dlR sin 90 o )/(R^3)] = (\mu_01 / 4\pi) (dl/R^2) since R^\ding{217} and dl^\ding{217} are perpendicular to each other. Therefore, the total magnetic field is the sum of the infinitesimal contributions dB or B = \intdB = (\mu_0i / 4\pi) (1/R^2 ) \intdl = (\mu_01 / 4\pi) [(2\piR)/R^2] = (\mu_0 / 2) (i/R) = 2\pi × 10[-7\omega/(a-m)] × [(1.6 × 10^-19 coul × 10^13 sec^-1 ) / (5 × 10^-11 m)] = 2.0 × 10^-2 (\omega/m^2) .

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Question:

What are the five principles of evolution?

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/Users/wenhuchen/Documents/Crawler/Biology/F27-0721.htm

Solution:

There are now five principles of evolution on which all biologists agree. First, evolution occurs more rapidly at some times than at others. At the present geological time it is occurring rapidly, with many new forms appearingand many old ones becoming extinct. Secondly, evolution does notproceed at the same rate in all types of organisms. For example, the lampshellshave remained unchanged for at least five hundred million years, while several species of man have appeared and become extinct in thepast few hundred thousand years. Thirdly, most new species do not evolvefrom the most advanced and specialized forms already living, but fromrelatively simple, unspecialized forms. Thus, mammals did not evolve fromthe large, specialized dinosaurs but from a group of rather small and unspecializedreptiles. Likewise, the vascular plants did not derive from thebryophytes but from lower and more primitive plants, like green algae. Fourthly, evolution is not always from the simple to the complex. There are manyexamples of "regressive" evolution, in which a complex form has givenrise to simpler ones. For instance, many wingless birds have evolvedfrom winged ones, and legless snakes from reptiles with appendages. This is because mutations occur at random, and if there is someadvantage for a species in having a simpler structure, any mutations thathappen to occur for such conditions will tend to accumulate by natural selection. Those mutations that result in a complicated delirious structure willbe selected against. Fifthly, evolution occurs in populations (not in individuals) by the processes of mutation, migration, natural selection, and geneticdrift.

Question:

Find thedeBrogliewavelength corresponding to an electron with energy 1, 10^4 , or 10^5eV. Neglect any corrections connected with the theory of relativity.

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/Users/wenhuchen/Documents/Crawler/Physics/D31-0921.htm

Solution:

DeBroglieknew that light has a dual, wave-particle nature given by the relationship \lambda = h/p which relates the wavelength \lambda of a light wave with the momentum p of its photons through Planck's constant h.DeBrogliereasoned that matter also has a wave-particle nature, and that the wavelength of the matter wave was given by the same equation used for light with p the momentum of the particle of matter. We then have \lambda = h/mv The energy given for the electron is its kinetic energy K. This is equal to K = (1/2) mv^2 orv = \surd(2K/m) We can therefore say \lambda = [h/{m\surd(2K/m)}] = \surd[h^2/(2Km)] where m is the mass of the electron. Substituting values, we find \lambda = \surd[{6.6 × 10^-34 joule/sec^2} / {(2) × (KeV) × (1.6 × 10^-19 joule/eV) × (9.1 × 10^-31 kg)}] = \surd(150/K) 10^-10 m = \surd(150/K) A where K is in election volts and \lambda in Angstroms. ForK1= 1eV \lambda_1 = \surd(150/1)A = 12.2 A ForK_2 = 10^4eV \lambda_2 = \surd(150/10^4)A = 0.122 A ForK_3 = 10^5eV \lambda_3= \surd(150/10^5)A = 0.039 A

Question:

A brass plug has a diameter of 10.000 cm at 150\textdegreeC. At what temperature will the diameter be 9.950 cm?

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Solution:

It is observed experimentally that when a sample is exposed to a temperature change \DeltaT, the sample experiences a change in length \DeltaL, proportional to \DeltaT and L, the original length of the sample. This (approximate) result may be written as: \DeltaL =\alphaL\DeltaT(1) where \alpha is the coefficient of linear thermal expansion. Solving (1) for \DeltaT, we find that \DeltaT = (\DeltaL/ \alphaL \alphaL )(2 )(2 ) is the change in temperature required to implement a change in length \DeltaL. Substituting the given data into (2), we obtain: \DeltaT = [{(9.950 - 10.00) cm.}/{(19 x 10^\rule{1em}{1pt}6 /\textdegreeC)(10.000 cm.)}] T= \rule{1em}{1pt} 260\textdegree C But we want the final value of T, not \DeltaT. Since \DeltaT =T_f\rule{1em}{1pt} T_O T_f = \DeltaT + T_O T_f = 150\textdegree C \rule{1em}{1pt} 260\textdegree C = \rule{1em}{1pt} 110\textdegree C. This is the value to which the temperature must be lowered in order to shrink the diameter of the plug to 9.950 cm.

Question:

Phosphate can be removed from sewage effluent by adding magnesium oxide and ammonium ion according to the re- action 5H_2O + H_2PO^-_4 + MgO + NH^+_4 \rightarrow Mg(NH_4)PO_4 \bullet 6H_2O. The magnesium ammonium phosphate hexahydrate (Mg(NH_4)PO_4 \bullet6H_2O) precipitates out. In theory, how much of this could be produced if a city processes 5 million gal. (41.7 million lb.) daily of wastewater containing 30 ppm H_2PO^-_4? (1 pound = 453.59 g.)

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/Users/wenhuchen/Documents/Crawler/Chemistry/E24-0845.htm

Solution:

The best way to approach this problem is to calculate how much H_2PO^-_4 is present in grams. Once the weight in grams is known one can calculate from the stoichiometry of the equation, the number of moles of Mg(NH_4)PO_4 \bullet 6H_2O that will precipitate. The wastewater has 30 ppm (parts per million) of H_2PO^-_4. In other words, 30 lbs. of H_2PO^-_4 are present in one million lbs. of wastewater. The city produces 41.7 million lbs. of wastewater. Thus, 30(41.7) = 1,251lbs. ofH_2PO^-4should be present. Because 1 pound = 453.59 grams; 1,251 lbs = 567,454 grams. Compute how many moles of H_2PO^-_4 are present. A mole is defined as grams (weight)/molecular weight (M.W.). The M.W. of H_2PO^-_4 = 96.97 g/mole. The number of moles of H_2PO^-4is =(567,454 g)/(96.97 g/mole) = 5852 moles. According to the reaction equation, one mole of H_2PO^-4generates one mole of precipitate. The M.W. of the precipitate = 245 g/mole. Recalling the definition of a mole, the amount of Mg(NH_4)PO4\bulletH_2O = (5852) (245) = 1.433 × 10^6 g = 1.433 × 10^3 kg.

Question:

Calculate the theoretical quantity of chlorine obtainable by the electrolysis of 2.0 kg of a 20% sodium chloride solution. What other products would be obtained and what would be the weight of each?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E16-0560.htm

Solution:

The equation for this reaction is as follows: 2NaCl + 2H_2O (electric / current)\rightarrow 2NaOH + H_2 + Cl_2 . One mole of Cl_2 will be formed from every 2 moles of NaCl reacted. Therefore, to find the quantity of Cl_2 formed one must first know the amount of NaCl reacted. This can be determined by solving for the weight of NaCl in the solution and dividing this number by the molecular weight to find the number of moles present. Since the solution is 20% NaCl, it means that 20% of its weight is made up by NaCl. weight of NaCl = 0.20 × 2.0 kg = 0.4 kg . Therefore, there are 0.4 kg or 400 g of NaCl present. One can now solve for the number of moles present by dividing 400g by the molecular weight (MW of NaCl = 58.5) no. of moles = (400g) / (58.5 g /mole) = 6.84 moles . From the equation one sees that (1 /2) of this amount is equal to the number of moles of Cl_2 formed. no. of moles of Cl_2 = (1 / 2) × 6.84 moles = 3.42 moles. The weight of this number of moles of Cl_2 will be equal to the mole-cular weight of Cl_2 times the number of moles. (MW of Cl_2 = 71.0) weight of Cl_2= 3.42 moles × (71.0g / mole) = 242.8g. From the reaction one can also see that if 2 moles of NaCl react, 2 moles of NaOH are formed. Here, 6.84 moles of NaCl is reacted, therefore, 6.84 moles of NaOH are formed. The weight of this quantity is equal to the number of moles × the molecular weight. (MW of NaOH = 40). weight of NaOH = (40g / mole) × 6.84 moles = 273.6g. It is also seen from the reaction, that if 2 moles of NaCl are reacted, 1 mole of H_2 is formed. Thus, if 6.84 moles of NaCl are reacted, 3.42 moles of H_2 are formed. The weight of H_2 can then be found. (MW of H_2 = 2) weight of H_2 = (2g / mole) × 3.42 moles = 6.84g .

Question:

Two conductors of the same length and material but of different cross-sectional area are connected (a) in series, and (b) in parallel. When a potential differ-ence is applied across the combinations, in which conductor will the heating be greater in each case?

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/Users/wenhuchen/Documents/Crawler/Physics/D20-0651.htm

Solution:

The resistance of each conductor has the form R =\rhol/A. That is, R is directly proportional to the length of the conductor and inversely proportional to the cross-sectional area of the conductor. \rho is the constant of proportionality (the resistivity). Since the resistivity and length are the same in each case then R_1 = \rho l/A_1 and R_2 = \rho l/A_2 or R_1/R_2 = A_2/A_1 . (a)When the conductors are in series, the same current passes through each, by definition of a series connection. Hence, the ratio of the heating produced (or the power developed) in the wires is (H_1 /H_2 ) = [(i^2 R_1 )/(i^2 R_2 )] = (R_1 /R_2 ) = (A_1 /A_2 ) . The heating is thus greater in the conductor with the smaller cross-sectional area. (b)When the conductors are in parallel, different currents pass through them but the potential difference across each is the same, by definition of a parallel connection. Hence, (H_1' /H_2' ) = [(V^2 /R_1 )/(V^2 /R_2 )] = (R_1 /R_2 ) = (A_1 /A_2 ) . We made use of the alternate form of power development, P = IV = I^2 R = V^2 /R . In this case, the heating is greater in the conductor with the larger cross-sectional area.

Question:

A free particle, which has a mass of 20 grams is initially at rest. If a force of 100 dynes is applied for a period of 10 sec, what kinetic energy is acquired by the particle?

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/Users/wenhuchen/Documents/Crawler/Physics/D05-0229.htm

Solution:

In order to calculate the kinetic energy we must compute the final velocity acquired by the particle: v = at +v_Owherev_Ois the initial velocity. Since we are told that initially the particle is at rest, the expression for velocity becomes v = at. Now, from Newton's Laws we know that a = F/m. Substituting this for a yields v = [F/m]t = [(100 dynes) / (20 g)] × (10 sec) = 50 cm/sec Then, KE = (1/2)mv^2 = 1/2 × (20 g) × (50 cm/sec)^2 = 25,000 ergs How much work was done by the applied force? The dis-tance moved is s = (1/2)at^2 = (1/2)[F/m]t^2 = 1/2 × [(100 dynes) / (20 g)] × (10 sec)^2 = 250 cm so that the work done is W = Fs since the force and displace-ment are in the same direction. W = (100 dynes) × (250 cm) = 25,000 ergs Thus, the work done is transformed entirely into the kinetic energy of the particle.

Question:

What structural advantages enable mitochondia to be efficient metabolic organelles?

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Solution:

Mitochondria are called the "powerhouses" of the cell because they are the major sites of ATP production in the cell. They contain the enzymes involved in both the Krebs (citric acid) cycle and the electron transport system, which work together to furnish 34 of the 36 ATPs produced by the complete oxidation of glucose. Another 2 ATPs are produced by glycolysis, which occurs in the cytoplasm, making the total 36 ATPs /oxidized glucose. Pyruvate from glycolysis traverses both the outer and inner mitochondral membranes easily because the membranes are totally permeable to pyruvate. Once inside the mitochondrion's matrix (see figure) pyruvate is converted by the enzyme pyruvate dehydrogenase to acetyl coenzyme A. The coenzyme can then enter the Krebs Cycle which occurs in the matrix where all the enzymes that catalyze the cycle's reactions are present (except succinic dehydrogenase which is located on the inner membrane). Conveniently bound to the surface of the cristae, the greatly folded inner membrane of the mitochondria, are the enzymes of the electron transport system. Since the cristae jut into the matrix, the distance needed to travel by FADH_2 and NADH, generated by the TCA cycle, to the enzymes of the electron transport system is reduced, and, since the surface area of the inner membrane is so greatly enhanced by its cristae, the probabilities of NADH and FADH_2 encount-ering the electron transport system's enzymes which are con-cerned with ATP production become sharply increased. Also cont-ributing to the efficiency of this organelle's ATP producing ability is the nature of the cristae's enzymes: they are actually multi-enzyme complexes which group together a number of sequentially acting enzymes responsible for electron transport and oxidative phosphorylation. These assemblies enhance the efficiency of respiration, since the product of one reaction is located near the enzyme of the subsequent reaction.

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Question:

To what volume must a liter of oxygen be expanded if the mean free path of the molecules is to become 2m? The molecules of oxygen have a diameter of 3 A. Assume that the gas starts at STP.

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Solution:

The mean free path L is the average distance between successive collisions of a gas molecule and is given by L = 1/(\pi\surd2nd^2) = (0.707)/(\pind^2) where n is the number of molecules per unit volume and d is the molecule's diameter. If L is 200 cm, then n = (0.707)/(L\pid^2) = [0.707]/[200 cm × \pi(3 × 10^-8)^2 cm^2]. The gas equation ispV= n_0RT, where n_0 is the number of moles present in the volume V. But n_0/V is the number of moles of gas per unit volume. Multiplying it by Avogadro's number, N_0, (the number of molecules in one mole of a substance) yields the number of molecules of oxygen in a unit volume. Therefore, (n_0/V)N_0 = nandn_0/V = n/N_0. Substitution into the gas equation gives P = (n/N_0)RT = [0.707 × 8.3 × 10^7 dynes \textbullet cm \textbullet mole^-1 \textbullet K deg^-1 × 273 K deg]/[200 cm × (3 × 10^-8)^2 cm^2 × 6.02 × 10^23 mole^-1)] = 0.047 dyne \bullet cm^-2. But since the temperature remains unchanged, the expansion takes place according to Boyle's law. P_1V_1 = P_2V_2 The gas starts at STP (standard temperature and pressure). This corresponds to a temperature of 0\textdegreeC and a pressure of 1 atm. Thus 1 liter changes to a volume V_2 while the pressure changes from 1atmto 0.047 dyne \bullet cm^-2 . To keep the units consistent, use is made of the fact that 1atm= 1.013 × 10^6 dynes \bullet cm^-2 and 1 liter = 10^3 cm^3 . Then 1.013 × 10^6 dynes \textbullet cm^-2 × 10^3 cm^3 = 0.047 dyne \textbullet cm^-2 × V2 \thereforeV_2 = [(1.013 × 10^9)cm^3]/[0.047] = 2.155 × 10^10 cm^3.

Question:

Describe the mode of action of enzymes. What factors affect enzyme activity?

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/Users/wenhuchen/Documents/Crawler/Biology/F03-0067.htm

Solution:

An enzyme (E) combines with its substrate (S) to form an intermediate enzyme-substrate complex (ES), which then decomposes into reaction products (P) and the free enzyme, as seen in the equation below. E + S \textemdash> ES\textemdash> E + P enzyme-substrate complex An enzyme causes the substrate upon which it is acting to be much more reactive than when it is free. One postulate accounting for this is that the enzyme holds the substrate in a position which strains and weakens the substrate's molecular bonds. This weakening of the bonds within the substrate makes them easier to cleave and results in a general lowering of the energy of acti-vation of the reaction. This postulate is extremely simplistic - the actual forces at work are much more numerous and complex. When the substrate binds to the enzyme, it combines with only a relatively small part of the enzyme molecule the active site. Information about the active site, such as its location and the nature and sequence of amino acids in it, provides an indication of the mechanism of binding and catalysis. The binding of the substrate to the enzyme's active, site depends on many forces: hydrogen bonding, the interaction of hydrophobic (water-repelling) groups, and the electrostatic interaction between charged groups on the amino acids. Many active sites also contain metal ions which aid in binding the substrate or expediting the catalytic reaction by withdrawing or stabilizing electrons. For example, the enzyme carboxypeptidase, which hydrolyzes polypeptide bonds of proteins in food, contains a zinc atom in its active site. The electrophilic (electron-attracting) zinc atom coordinates electrons from the carbonyl of the peptide bond, weakening the bond for attack by a specific amino acid of the enzyme at the active site. Such a mechanism, however, is beyond the scope of elementary biology and one would require a good course in biochemistry to understand fully. Some enzymes, the regulatory or allosteric enzymes, have two binding sites: an active site and a regulatory site Regulatory enzymes are a key controlling factor in me-tabolic pathways. If the end product of a pathway is in excess, it inhibits the action of the regulatory enzyme by binding to its regulatory site. The end product shuts off the catalytic activity of the active site by altering the arrangement of the enzyme's polypeptide chains, thus deforming and inactivating the enzyme (see diagram below) . This feedback mechanism is known as end-product inhibition and is important in preventing the accumulation of unwanted substcincss. There are several factors which affect enzyme activity, one of which is temperature. High temperatures of 50\textdegreeC or above can inactivate or denature most enzymes. Upon denaturation, the structure of the enzyme is per-manently altered resulting in an irreversible loss of activity. When most organisms are exposed to high temp-eratures, death occurs due to enzyme inactivation and the resulting loss of metabolic activity. Enzymes are usually not denatured by freezing, but their activity is decreased or disappears. This loss of activity is tempo-rary and the activity reappears upon exposure to normal temperatures. Most enzymes have an optimal temperature range. At temperatures below 50\textdegreeC, enzymatic reactions double for each temperature rise of 10\textdegree. Enzymes are also affected by and can be denatured by changes in pH. Although most enzymes have an optimum pH a- round neutrality, (pH 7) some require an acidic medium and others require an alkaline medium. For example, although both pepsin and trypsin works optimally at pH 8.5. The de-pendence of enzyme activity on pH is explained by the pre-sence of ionizable groups on the protein molecules of the enzyme. pH controls, in part, the number of positive and negative charges on the enzyme molecule, which consequently affect activity. A final factor affecting enzyme activity is the presence of enzyme poisons. Cytochrome oxidase, an enzyme involved in respiration, is inactivated by minute amounts of cyanide. Death from cyanide poisoning thus results from the inhibition of cytochrome enzymes. Other enzyme poisons include iodoacetic acid, flouride and lewisite.

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Question:

How can the stack be used with subroutines?

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G07-0138.htm

Solution:

The stack can be used to store return addresses and any other contents of registers that need to be saved before going to a subroutine. Consider a computer with ac-cumulator ACC and general purpose registers X and Y. The program in fig. 1 shows how the stack is used to call a sub-routine which will store the sum of the contents of the ACC and register X into register Y. The main program includes steps 14 through 18. and subroutine 1 starts at step 57 and stops at step 60. At location 15 the ACC contents is pushed into the stack. Note that the value of the ACC remains unchanged even though it was pushed on the stack. During the Jump to Subroutine 1 the return address is stored in the stack. At location 58 the sum of [ACC] + [X] is stored in the ACC. At location 60 the return address is taken off of the stack and put back in the PC, thus the computer has "returned" to the main pro-gram. After returning, the ACC is popped, hence the ACC con-tains its previous value. As you can see stacks are a valuable tool for storing registers while they are temporarily being used for other purposes.

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Question:

Write a FORTRAN program that generates the prime numbers in the 1-2000 interval, using an ancient method known as the SIEVE of ERATOSTHENES.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G22-0547.htm

Solution:

A prime is a number which has only 2 divisors: 1 and itself. There is an ancient method for finding the primes from 2 (the smallest prime) to some number N, called the SIEVE of ERATOSTHENES. This method works as follows: First, the integers from 2 to N are written down. Then, the multiples of the smal-lest available integer, 2, greater than this integer, are crossed out. This step, obviously, eliminates all the even numbers except 2. The procedure is then repeated for the next smallest available integer-3, and then next - 5, because 4 is not available, since it was eliminated during the first step, and so on... Thus, the initial list is scanned repeatedly, and the appropriate integers are crossed off, until the next least remaining integer considered is greater than \surdN. At this point, all the remaining integers are indeed all the primes between 2 and N. The program is created based upon the above discussion. First, an array of integers from 2 to 2000 is generated. Next, this array is scanned repeatedly, deleting the appropriate integers by replacing them with zeros. At the point when the considered integer is greater than \surd2000 , all the nonzero elements of the array are placed left justified within the array and then printed. The program looks as follows: DIMENSION N(2000) K = 2000 RK = K KROOT = SQRT (RK) KK = K - 1 CGENERATE ARRAY OF INTEGERS FROM 2 TO 2000 DO 10 I = 1,KK 10N(I) = I+1 J = 1 CSECTION OF REPEATED ELIMINATION PROCESS 15INC = N(J) JJ = J + INC DO 20 L = JJ, K, INC 20N(L) = 0 25J = J + 1 IF (N(J).LE.O) GO TO 25 IF (N(J).LE.KROOT) GO TO 15 CPLACE THE PRIMES LEFT JUSTIFIED CIN THE ARRAY N. J = 0 DO 45 I = 1, KK IF (N(I).LE.O) GO TO 45 J = J + 1 N(J) = N(I) 45CONTINUE COUTPUT THE PRIMES WRITE (6,50) (N(I), I = 1,J) 50FORMAT (1H, 10I6) STOP END

Question:

How much heat is required to change 25 kg of ice at -10 \textdegreeC to steam at 100\textdegreeC?

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/Users/wenhuchen/Documents/Crawler/Physics/D13-0471.htm

Solution:

We have 4 separate situations which we must consider in this problem. First, the ice is heated to its melting point, which involves a change in temperature. Next, the ice changes state to form water during which there is no change of temperature. Then, as heat is added the water reaches its boiling point and any further addition of heat serves only to finally change the state of the water to steam and will not raise the temperature of the boiling water. Note that the specific heat of ice is different from that of water. Heat to raise temperature of ice to its melting point =m_ic_i∆t_i= 25 kg (0.51 Cal/kgC\textdegree) [0 - (-10\textdegreeC)] = 128 Cal. Heat to melt ice = m_1L_v = 25 kg (80 Cal/kg) = 2000 Cal. Heat to warm water to its boiling point =m_wc_w∆t_w= 25 kg(1.0 Cal/kg C\textdegree) (100\textdegreeC - 0\textdegreeC) = 2500 Cal Heat to vaporize water =m_wL_v= 25 kg(540 Cal/kg) = 13,500 Total heat required128 Cal 2,000 2,500 14,000 19,000 Cal Note that in this summation 128 is negligible and may be disregarded, since there is a doubtful figure in the thou-sands place in 14,000.

Question:

A charge of 1 C flows from a 100-V outlet through a 100-W light bulb each second. How much work is done each second by this electrical source?

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/Users/wenhuchen/Documents/Crawler/Physics/D17-0561.htm

Solution:

The voltage, V = 100 V, and the charge, q =1 C, are given. By definition of potential V = W/q where W is the work done in moving the charge q through a potential difference V. Then W = qV = (1C)(100 V) = 1 × 10^2 J Alternatively, since the power rating of the light bulb is given as 100 watt, it dissipates 100 joules of energy per second. Therefore, for each second, the source must supply 100 joules of energy (or do 100 joules of work per second) to provide the energy dissipated by the light bulb resistance.

Question:

What is the power factor of a circuit if the following meter readings prevail? Ammeter reads .15 amperes, volt-meter reads 115 volts, and wattmeter reads 15.9 watts.

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/Users/wenhuchen/Documents/Crawler/Physics/D24-0796.htm

Solution:

The ammeter and voltmeter read root mean square (RMS) values of the actual current and voltage. If V = V0cos\omegat andI = I0cos(\omegat+ \rho) are the actual current and voltages through the circuit, then the instantaneous power (i.e. the instantaneous rate at which energy flows in the circuit) is P = VI = V0I0cos\omegatcos(\omegat+ \rho) = V0I0 (cos^2\omegatcos\rho -cos\omegatsin\omegatsin \rho) Nowcos\omegatsin\omegat= (1/2) sin 2\omegat, and the time average of sin 2\omegat is zero (since its graph is symmetric about the horizontal axis). The average of cos^2\omegatis 1/2 , for its graph is symmetric about the value of 1/2. This is to say that its value is greater than 1/2 as often as it is less than 1/2 , and by the same amount. Therefore the average power is: P= V0I0 (cos^2\omegat\leftrightarrowcos\rho\rule{1em}{1pt}cos\omegatsin\omegatsin \rho) = V0I0 [(1/2)cos\rho - 0] cos \rho is called the power factor of the circuit. It is a measure of the phase relation between the current and voltage in the circuit. According to the definition of the RMS value of a function, I_RMS = \surd[(1/2\pi) ^2\pi\int_0 I^2dt] = \surd[(1/2\pi) ^2\pi\int_0 I^20 cos^2 (\omegat+ \rho)dt] = (1/\surd2)I0 Similarly, V_RMS = \surd[(1/2\pi) ^2\pi\int_0 V^2dt] = \surd[(1/2\pi) ^2\pi\int_0 V^20 cos^2\omegatdt] = (1/\surd2) V0 HenceP= 1/2 (\surd2 I_RMS) (\surd2 V_RMS)cos\rho = I_RMS V_RMScos\rho The wattmeter reads the average power of a circuit. Then given I_RMS = .15 amp, V_RMS = 115 vP= 1.59 watt, the power factor is cos\rho = [P/(I_RMS V_RMS] = {15.9w/[(.15 a) (115 v)]} = 0.92

Question:

How many nuclei are there in 1kg of aluminum?

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/Users/wenhuchen/Documents/Crawler/Physics/D31-0911.htm

Solution:

The mass m of a single aluminum atom ^27_31A1, is 26.98153amu. 1a.m.u. corresponds to the mass of a proton or 1.66 × 10^-27 kg, therefore m = [{26.98amu} {1.66 x 10^-27(kg/amu)}] = 4.48 x 10^-26kg This mass value is also approximately equal tothe mass of an aluminum nucleus since the electron mass is so much smaller than the mass of the atom. Therefore, the number of aluminum nuclei is N = [{1.0 kg} / {4.48 x 10^-26 (kg/nuclei)}] = 2.23 x 10^-25 nuclei.

Question:

A uranium nucleus under certain conditions will spontaneously emit an alpha particle, which consists of two pro-tons and two neutrons. If the nucleus was initially at rest and the speed of the emitted alpha particle is 2 × 10^7 m/s, what is the "recoil" speed of the nucleus? The nuclear mass is 3.9 × 10\Elzbar25kg and the alpha-particle mass is 6.7 × 10\Elzbar27 kg.

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/Users/wenhuchen/Documents/Crawler/Physics/D34-1024.htm

Solution:

The "nuclear\textquotedblright force responsible for the particle emission is not explained, but this information is not needed if it is recognized that the total momentum of the system is zero. The mass of the recoiling nucleus m_1 = 3.9 ×10\Elzbar25kg; the mass of the alpha particle m_2 = 6.7 × 10\Elzbar27kg ; and the velocity of the alpha particle v_2 = 2 × 10^7 m/s are the known observables. As shown in the figure, the momentum of the initial state is zero, since the uranium atom is initially at rest. Therefore, the vectorial sum of the momenta of the frag-ments in the final state is also zero. (We are viewing the \alpha-decay of the uranium atom from the center of mass frame of the motion.) The speed of fragment x is expected to be smaller than that of the \alpha-particle since x is more massive than \alpha. Both speeds are therefore non-relativistic and we are allowed to use the non-relativistic formulae. From the conservation of momentum, we have P^\ding{217}_f = P^\ding{217}_i = 0 P^\ding{217}_1 + P^\ding{217}_2 = 0 orm_2v^\ding{217}_2 + m_1 v^\ding{217}_1 = 0 Thereforev^\ding{217}_1 =\Elzbar (m_2 / m_1 ) v^\ding{217}_2 . We see that v^\ding{217}_1 and v^\ding{217}2have opposite directions. The magni-tude of v^\ding{217}_1 is V_1 = (m_2 v_2 / m_1 ) = [{(6.7 × 10\Elzbar27kg)(2 × 10^7 m/s.)} / (3.9 ×10\Elzbar25kg )] = 3.4 × 10^5 m/s After emission, the alpha particle and nucleus move apart as free particles according to Newton's first law because the force causing the separation quickly becomes negligible.

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Question:

Compute the electric field and the electric potential at point P midway betweentwo charge, Q_1 = Q_2 = +5 statC, separated by 1 m.

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/Users/wenhuchen/Documents/Crawler/Physics/D18-0586.htm

Solution:

The magnitude of a test charge is +1 unit of charge. The forces on a test charge placed midway between two identical charges Q_1 and Q_2 are F_1^\ding{217} = (Q_1 / r^2_1 ) ȓ_12 F_2^\ding{217} = (Q_2 / r^22)ȓ21= (\rule{1em}{1pt}Q_2 / r^22)ȓ_12 where the unit vector ȓ_12points from to Q_1 to Q_2, r_1, r_2 are the distances between the test charge and and Q_1 and Q_2, F^\ding{217}_1 and F^\ding{217}_2 are equal in magnitude but opposite in direction. The net force is therefore zero at P, F = F_1 + F_2 = 0. The force on a unit charge gives the electric field strength at that point, then the electric field E^\ding{217} at P is also zero. Although the electric field at P is zero, this does not imply that the electric potential is also zero. The total potential \varphi_E _total is the sum (the algebraic sum since potential is a scalar) of the potentials due to Q_1 and Q_2: \varphi_E,1 = (Q_1 / r_1) = 5 / 50 = 0.1 statV \varphi_E,2 = Q_2 / r_2 = 5 / 50 = 0.1 statV Therefore, \varphiE, total= \varphi_E,1 + \varphiE,2= 0.2 statV Notice that if either Q_1 or Q_2 is changed from +5 stateC to \rule{1em}{1pt}5 stateC, the electric potential will vanish but the electric field will not. Therefore, the fact that either the field or the potential is zero in any particular case does not necessarily mean that the other quantity will also be zero; each quantity must be calculated separately.

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Question:

What is the structure of a Pascal program?

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G16-0395.htm

Solution:

A Pascal program consist of the following sections outputfiles. Programexample (input, output): (All underlined words are reserved words and are required in a given situation .) 2.Label Declarations (optional) Specifies labels used in the program. Label 5 , 20; 3.Constant definitions (optional) Defines constants used in the program. CONSTmax = 80, tax = 0.0825; 4.TYPE Definitions (optional) Defines data structures used in the program TYPE fixed= integer ; string= PACKED ARRAY [1 . .MAX ]of CHAR 5.Variable Declarations Declares variables used in the program and their type. VAR X :integer ; TEST :BOOLEAN ; NAME :STRING ; 6.Subroutine Declarations (optional) Declares function and procedures internal to the mainprogram . Each subroutine may contain including othersubroutines . PROCEDURESexample 1 (varparam1: real); FUNCTIONEXAMPLES 2 (param1: integer): integer; 7.Main body of the program always has the following form: BEGIN \bullet \bullet \bullet \bullet \bullet Some Other Statements \bullet \bullet \bullet END .

Question:

Calculate ∆H for the reaction at 25\textdegreeC. CO(g) + (1/2)O_2 (g) \rightarrow CO_2 (g)∆E = - 67.4Kcal.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E14-0508.htm

Solution:

The change in enthalpy (∆H) is defined by the first law of thermodynamics as the sum of the change of the internal energy (∆E) and the work done by a particular system. Work is defined as the product of the pressure (P) and the change in volume (∆V) of a system, where only pressure-volume work is done. Much of the experimentation done by chemists is pressure-volume work because the work is usually done in an open container in the laboratory. The pressure, in this case, is constant and the volume of the materials is allowed to change. ∆H = ∆E + P∆V In this problem, one is given ∆E but not P or ∆V, so that one must calculate P∆V in some other way. The Ideal gas Law is defined as: P∆V = ∆nRT, where P is the pressure, ∆V is the change in volume, ∆n is the change in the number of moles reacting, R is the gas constant (1.99 cal/mole-\textdegreeK), and T is the absolute tem-perature. Here, one knows R and can calculate T and ∆n. Therefore, ∆nRTcan be substituted for P∆V in the equation to find ∆H. ∆n is found by subtracting the number of moles reacting from the number of moles formed as products. Here, 1 mole of CO and 0.5 mole of O_2 form 1 mole of CO_2. ∆n = 1 mole CO_2 - (1 mole CO + 0.5 mole O_2) = - 0.5 mole. The absolute temperature is found by adding 273 to the temperature in \textdegreeC. T = 25 + 273 = 298\textdegreeK. Therefore, ∆nRT= (- 0.5 mole) × 1.99 cal/mole-\textdegreeK × 298\textdegreek = - 0.3 Kcal. One is now ready to find ∆H, via substitution. ∆H = ∆E + ∆nRT ∆E = - 67.4 Kcal ∆nRT= - 0.3 Kcal ∆H = - 67.4 Kcal + (- 0.3 Kcal) = - 67.7 Kcal.

Question:

A steel bar, 20 ft long and of rectangular cross-section 2.0 by 1.0 in., supports a load of 2.0 tons. How much is the bar stretched?

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/Users/wenhuchen/Documents/Crawler/Physics/D08-0347.htm

Solution:

The Young's modulus of the metal bar is the ratio of longitudinal stress , F/A, to tensile strain ∆L/L (see the figure) Y = (F/A)/(∆L/L) = (FL)/(A∆L) Here, A is the bar's cross-sectional area. Therefore, the elongation ∆L of the bar is ∆L = (∆FL)/YA ∆F = 2.0 tons = 2.0 ton × (2000 lb)/(1 ton) = 4000 lbs. A= (2.0 in. × 1.0 in.) = 2.0 in.^2. Young's modulus for steel is 29 × 10^6 lb/in.^2. ∆L = [(4.0 × 10^3 lb)(20 ft)]/[(29 × 10^6 lb/in.^2)(2.0 in.^2)] = 0.0014 ft = 0.017 in.

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Question:

Balance the following by filling in missing species and proper coefficient: (a) NaOH + _____\rightarrow NaHSO_4 + HOH; (b)PCI_3 +__ HOH \rightarrow _____ + 3HCl, (c) CH_4 + ____ \rightarrowCC1_4 + 4HCl.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E05-0167.htm

Solution:

To balance chemical equations you must remember thatALLatoms (and charges) must be accounted for. The use of coefficients in front of compounds is a means to this end. Thus, (a)NaOH + ____ \rightarrow NaHSO_4 + HOH On the right side of the equation, you have 1 Na, 3 H' s, 5 O' s, and 1 S. This same number of elements must appear on the left side. However, on the left side, there exists only 1 Na, 1 0, and 1 H. You are missing 2 H' s, 1 S, and 4 O' s. The missing species is H_2 SO_4, sulfuric acid. You could have anticipated this since a strong base (NaOH) reacting with a strong acid yields a salt (NaHSO_4) and water. The point is, however, that H_2 SO_4 balances the equation by supplying all the missing atoms. (b)PC1_3 + 3HOH \rightarrow ____ + 3HCl. Here, the left side has 1 P, 3Cl' s, 6 H' s, and 3 O' s. The right has 3 H' s and 3 C1' s. You are missing 1 P, 3 O' s and 3hydrogens. Therefore, P (OH)_3 is formed. (c)CH_4 + ____ \rightarrow CCl_4 + 4HCl Here, there are 1 C, 8Cl's, and 4 H's on the right and 1 C and 4 H' s on the left. The missing compound, therefore, contains 8 C1' s and thus it is 4 CI2. One knows that it is 4 CI_2 rather than CI_8 or 8CI because elemental chlorine gas is a diatomic or 2 atom molecule.

Question:

In over 90% of cases, the concentration of ethanol (C_2H_5OH, density = 0.80 g/ml) in blood necessary to produce intoxication is 0.0030 g/ml. A concentration of 0.0070 g/ml is fatal. What volume of 80 proof (40% ethanol by volume) Scotch whiskey must an intoxicated person consume before the concentration of ethanol in his blood reaches a fatal level? Assume that all the alcohol goes directly to the blood and that the blood volume of a person is 7.0 liters.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E08-0273.htm

Solution:

The difference between a fatal concentration of ethanol and an intoxicating concentration of ethanol is 0.0070 g/ml - 0.0030 g/ml = 0.0040 g/ml. We must determine the total volume of ethanol in the blood which corresponds to a concentration of 0.0040 g/ml and then calculate the volume of Scotch whiskey which will provide this concentration. The total mass of ethanol in blood needed to raise the concentration from an intoxicating to a fatal level is equal to concentration of ethanol × volume of blood = 0.0040 g/ml × 7 liters = 0.0040 g/ml × 7000 ml = 28 g of ethanol. Dividing this mass by the density, we obtain the corresponding volume of ethanol, or 28 g/0.80 g/ml = 35 ml of ethanol. The amount of Scotch whiskey that must be consumed must provide 35 ml of ethanol. But the Scotch whiskey is only 40% ethanol, or 0.40 ml ethanol/ml Scotch. Let v denote the volume of Scotch in ml. Then [ratio of ethanol to Scotch ] × volume of Scotch = volume of ethanol 0.40 ml ethanol/ml Scotch × v ml Scotch = 35 ml ethanol or, v = [(35 ml ethanol)/(0.40 ml ethanol/ml Scotch)] 88 ml Scotch. Thus, under our assumptions, an intoxicated person must drink 88 ml of 80 proof Scotch whiskey (about 3 ounces) before ethanol reaches a fatal level in his blood.

Question:

A force F = 10 newtons in the +y-direction is applied to a wrench which extends in the +x-direction and grasps a bolt. What is the resulting torque about the bolt if the point of application of the force is 30 cm = 0.3 m away from the bolt?

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Solution:

Torque is calculated from the relation: \cyrchar\cyrt^\ding{217} = r^\ding{217} × F^\ding{217} where \cyrchar\cyrt stands for torque, F stands for the force, and r denotes the distance from the origin, about which the torque is calculated, of the point of application of the force. In this problem we use the bolt as our origin about which we calculate the torque (see the Figure above). Then, \cyrchar\cyrt^\ding{217} = 0.3 m \^{\i} x 10N\^{\j} = 3 N\bulletm ( \^{\i} × \^{\j}) = 3 N \bullet m k˄ where \^{\i}, \^{\j}, and k˄ are the unit vectors in the +x, +y, and +z directions respectively.

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Question:

What arrangement of the integers 1, 2, 3, 4, and 5 in a five- element array will yield the longest sort time (also known as the worst-case sort time) for the insertion sort? Gen-eralize the answer you obtain for N integers.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G07-0147.htm

Solution:

As developed earlier, in the insertion sort method, the list is scanned until an element is found to be out of order. The unordered element is stored temporarily, and the scan reverses direction until it reaches the place where the unordered element should be. The rest of the list's elements are moved up one place, and the sort continues. If the list of integers is presented initially in re-verse order, the insertion sort will have to make the great-est number of comparisons. Let us walk through the steps that must be taken: pass after 1st pass4, 5, 3, 2, 1 after 2nd pass4, 3, 5, 2, 1 after 3rd pass3, 4, 5, 2, 1 after 4th pass3, 4, 2, 5, 1 after 5th pass3, 2, 4, 5, 1 after 6th pass2, 3, 4, 5, 1 after 7th pass2, 3, 4, 1, 5 after 8th pass2, 3, 1, 4, 5 after 9th pass2, 1, 3, 4, 5 after 10th pass1, 2, 3, 4, 5 Ten comparisons must be made to get the list in order. We can generalize from this result to obtain a worst-case equation for the insertion sort. The maximum number of comparisons for the insertion sort is N (N - 1) / 2 This can be seen by writing the series in the order given and in the reverse order: (N - 1) + (N - 2) + (N - 3) + . . . + 3 + 2 + 1 1 + 2 + 3 + . . . + (N - 3) + (N - 2) + (N - 1) Adding term by term, we find that twice the sum of the series is N + N + N + ... + N + N + N Since N-l terms are in the series above, we can say that twice the sum of the series is N (N - 1), so, dividing by 2, the sum of the series is N (N -1) / 2.

Question:

What fraction of the total space in a body-centered cubic unit cell is unoccupied? Assume that the central atom touches each of the eight corner atoms of the cube.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E06-0220.htm

Solution:

To determine the percent of the unit cube that is unoccupied by the atoms making up the lattice, one must subtract the volume taken up by the atoms from the volume of the cube. This volume is then divided by the volume of the cube and multiplied by 100 to find the percent of unoccupied space. 1) Determining the volume of the atoms: One must first determine the number of atoms contributing to the unit cube. In this lattice there are 8 corner atoms and 1 atom in the center. Corner atoms contribute 1/8 of their volume to the cube. The atom in the center contributes its entire volume. no. of atoms in unit cube = (1/8 atom / corner × 8 corners) + 1 atom (in center) = 2 atoms. Thus, the volume in the unit cube taken up by the atoms is equal to the volume of two atoms. The radius of these atoms is taken to be 1. The volume of a sphere is 4/3 \pir^3. volume of 2 atoms = 2 × 4/3 \pi(1)^3 = 8.38 2) Volume of the cube: The corner atoms are assumed to be touching the central atom. The diagonal of the cube can be visualized as shown in Figure 2. BecauseABis shown to be 4r or 4 the side of the cubeACcan be found using the Pythagorean Theorem. OnceACis known the volume of the cube can be found. From geometry, it is known that CB=AC× \surd2 AB^2 =AC^2 +CB^2AB= 4 4^2 =AC^2 + 2(AC)^2CB=AC× \surd2 4^2 = 3AC^2 4/\surd3 =AC The volume of the cube is equal to the length of the side cubed. volume of cube = [4/(\surd3)]^3 3) The space in the cube is equal to the volume of the spheres subtracted from the volume of the cube. vol. of space = vol. of cube - vol. of spheres vol. of space = 12.32 - 8.38 = 3.94 percentage of cube taken up by space = [3.94/12.32] × 100 = 32%.

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Question:

Using the tables of standard electrode potentials, arrange the fol-lowing substances in decreasing order of ability as reducing agents: Al, Co, Ni, Ag, H_2, Na .

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Solution:

The tables of standard electrode potentials list substances according to their ability as oxidizing agents. The greater the stan-dard electrode potential, E\textdegree, of a substance, the more effective it is as an oxidizing agent and the less effective it is as a reducing agent. From the table of standard electrode potentials, Al^3+ + 3e^-\leftrightarrowsAl(s) E\textdegree = -1.66v Co^2+ + 2e^-\leftrightarrowsCo(s) E\textdegree = -0.28v Ni^2+ + 2e^-\leftrightarrowsNi(s) E\textdegree = -0.25v Ag^+ + e^-\leftrightarrowsAg(s) E\textdegree = +0.80v 2H^+ + 2e^-\leftrightarrowsH_2(g) E\textdegree = 0v Na^+ + e^-\leftrightarrowsNa(s) E\textdegree = -2.71v Thus, in increasing ability as oxidizing agents, Na^+ < Al^3+ < Co^2+ < Ni^2+ < H^+ < Ag^+ But if Na^+ has a greater tendency to oxidize (gain electrons) than Al^3+ then, from looking at the reverse reactions, the "conjugate oxidant" Na must have a greater tendency to reduce (lose electrons) than the "conjugate oxidant" AI . Thus, Na, is a better reducing agent than Al, and so on. The substances, in order of decreasing ability as reducing agents, are therefore Na > Al > Co > Ni > H_2 > Ag .

Question:

What are the functions of the stem? How are stems and roots differentiated?

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/Users/wenhuchen/Documents/Crawler/Biology/F10-0246.htm

Solution:

The stem is the connecting link between the roots, where water and minerals enter the plant, and the leaves, where organic foodstuffs are synthesized. The vascular tissues of the stem are continuous with those of the root and the leaves and provide a pathway for the transport of materials between these parts. The stem and its branches support the leaves so that each leaf is exposed to as much sunlight as is possible. The stem of a flowering plant also supports flowers in the proper orientation to enhance reproduction and later, seed dispersal. Along the stem are growing points where theprimordiaof leaves and flowers originate. Some stems have cells which carry out photosynthesis, others have cells specialized for the storage of starch and other nutrients. Roots and stems are structurally quite different. Stems have an epidermis covered with a protective layer ofcutinwhile the epidermis of roots has nocutinbut gives rise tohairlikeprojections called the root hairs. Stems, but not roots, have nodes, which are junctions where leaves arise. Stems may have lenticels for "breathing" while roots do not. The tip of a root is always covered by a root cap whereas the tip of a stem is naked unless it terminates in a bud. Thedicotstem typically contains separate rings of xylem and phloem, with the xylem central to the phloem, whereas in the roots, phloem tubes lie between the arms of the star-shaped xylem tissues.

Question:

It has been determined experimentally that two elements, A and B react chemically to produce a compound or compounds. Experimental data obtained on combining proportions of the elements are: Grams of A Grams of B Grams of compound Experiment 1 06.08 04.00 10.08 Experiment 2 18.24 12.00 30.24 Experiment 3 3.04 2.00 5.04 (a)Which two laws of chemical change are illustrated by the above data? (b) If 80 g of element: B combines with 355 g of a third element C, what weight of A will combine with 71 g of element C? (c) If element B is oxygen, what is the equivalent weight of element C?

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Solution:

(a) If one adds the weight of A to the weight of B and obtains the weight of the compound formed, the Law of Conservation of Mass is illustrated. This law states that there is no detectable gain or loss of mass in a chemical change. Using the data from the experiments described, you find the following: For experiment 16.08 g A + 4.00 g B = 10.08 g of compound 6.08 + 4.00 = 10.08 For experiment 218.24 g A + 12.00 g B =30.24 g of compound 18.24 + 12.00 = 30.24 For experiment 303.04 g A +2.00 g B =5.04 g of compound 3.24 +2.00 = 5.04 From these calculations one can see that the Law of Conservation of Mass is shown. Another important law of chemistry is the Law of Definite Proportions. This law is stated: When elements combine to form a given compound, they do so in a fixed and invariable ratio by weight. One can check to see if this law is adhered to by calculating the ratio of the weight of A to the weight of B in the three experiments. If all of these ratios are equal, the Law of Definite Proportions is shown. For experiment 16.08 g A /4.00 g B = 1.5 For experiment 218.24 g A / 12.00 g B =1.5 For experiment 303.04 g A /2.00 g B =1.5 The Law of Definite Proportions is illustrated here. (1) From the Law of Definite Proportions, one can find the number of grams of B that will combine with 71 g of C. After this weight is found, one can find the number of grams of A that will react with 71 g of C by finding the amount of A that reacts with that amount of B. It is assumed that the amount of A that reacts with 71 g of C is equal to the amount of A that will react with the amount of B that reacts with 71 g of C. The amount of A that will react with this amount of B can be found by remembering from the previous section of this problem, that A reacts with B in a ratio of 1.5. (1) Finding the amount B that would react with 71 g of C. One is told that 80 g of B reacts with 355 g of C. By the Law of Definite Proportions, a ratio can be set up to calculate the number of grams of B that will react with 71 g of C. Let x = the number of grams of B that will react with 71 g of C. (80g B) / (355 g C) = (x g B) / (71 g C) (80g B) / (355 g C) = (x g B) / (71 g C) x= (71 × 80) / (355) = 16. x= (71 × 80) / (355) = 16. 16 grams of B will react with 71 g of C. (2) It is assumed that the same amount of A that will react with 16 g of B will react with 71 g of C. Therefore, using the fact that the ratio of the amount of A that reacts to the amount of B is equal to 1.5 (this fact was obtained in part (1)), one can calculate the amount of A that will react with 71 g of C. Let x = the number of grams of A that will react with 16 g of B . (xg A) / (16g B) = 1.5 x = 16 × 1.5 = 24 g. 24 g of A will react with 16 g of B or 71 g of C. (c) In finding the equivalent weight of C when B is taken to be oxygen, the Law of Definite Proportions is used again. The equivalent weight of oxygen is 8. Knowing that 16 g of B react with 71 g of C, one can set up the following ratio x = weight of C if the weight of B is taken to be 8. (71 g C) / (16 g B) = (xg C) / (8g B) x = (8 × 71) / (16) = 35.5 g. The equivalent weight of C when B is taken to be oxygen is 35.5 g.

Question:

If an electron is projected into an upward electric field with a horizontal velocity v_0, find the equation of its trajectory.

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/Users/wenhuchen/Documents/Crawler/Physics/D19-0627.htm

Solution:

The direction of the field is upward (see the figure) and the force on e is eE^\ding{217}, where e is the electronic charge, and E^\ding{217} is the electric field intensity. Since e < 0 the force on the electron points downward. The initial velocity is along the positive x-axis. The only force is in the - y- direction, therefore the acceleration along the x-axis is zero. The y acceleration is a^\ding{217}_y = F^\ding{217}y/m = eE^\ding{217}/m,(1) wheree < 0. After a time t, the position of the electron will be x = x_0 + v(0)xt + 1/2 a_xt2 y = y_0 + v(0)yt + 1/2 a_yt2 (x_0, y_0) and (v_(0)x, v_(0)y) are the components of the initial position and velocity of the particle. If we take x_0 = 0, y_0 = 0 as the starting point of the particle, and note that a_x = 0, v_(0)y = 0, we may write x = v_(0)xt y = 1/2 a_yt^2(2) substituting (1) in (2) x = v(0)xt y = [eE^\ding{217} / 2m] t^2(3) Writing E^\ding{217} in terms of a unit vector (\^{\j}) in the direction of the y - axis E^\ding{217} = E \^{\j} Substituting this in the equation for y appearing in (3) for, we obtain y = (eEt^2 / 2m)\^{\j}(4) Solving the x equation of (3) for t, and sub-stituting this in (4), t = x / v(0)x y = (eEx2/ 2mv_(0)x^2)\^{\j}(5) Note that e < 0, and since all the other quantities in (5) are positive, y < 0 which is the equation of a parabola. The motion is the same as that of a body projected horizontally in the earth's gravitational field. The deflection of electrons by an electric field is used to control the direction of an electron stream in many electronic devices, such as the cathode-ray oscilloscope.

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Question:

In the casualty department of a hospital, it is necessary to raise or lower the examination table without disturbing the patient. This is accomplished by mounting the table on a screw jack which has a pitch of (1/2) in. The raising of the table is accomplished by applying a force of 12.5 lb tangentially at the end of a lever 12 in. long and rotating the lever in a circle. Find the efficiency of this machine if patient and table together have a weight of 480 lb.

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Solution:

When the lever is rotated through one complete circle, the table is raised by one pitch of the screw. The work done on the machine by the operator is the force applied times the distance traveled in the direction of the force. Thus W_1= F × 2\piR, where R is the radius of the circle swept out by the lever. W1= 12(1/2) lb × 2\pi × 1 ft = 25\pi ft \textbullet lb. The table and patient acquire additional potential energy, since their height above the ground is increased. The additional energy is their combined weight times the extra height. Thus W_2 = 480 lb × (1/24) ft = 20 ft \textbullet lb. The efficiency of the machine is the energy gained by the table divided by the energy supplied. Thus the efficiency is E = W_2/W_1 = (20 ft \bullet lb)/(25\pi ft \bullet lb) = 0.255or E = 25.5%.

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Question:

Outline a method, giving the precautions to be aware of, by which an enzyme could be used to measure the amount of a given substrate.

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/Users/wenhuchen/Documents/Crawler/Biology/F03-0069.htm

Solution:

An enzyme catalysed reaction can be represented by the equation: E + S \rightleftharpoons ES \rightleftharpoons E + P By increasing the concentration of substrate and keeping the concentration of enzyme constant, we can increase the rate of catalysis: the above reaction will shift toward the right. We can react a fixed amount of enzyme with different substrate concentrations to determine the corresponding reaction velocity. The reaction velocity must be an initial velocity, that is, the rate must be measured very soon after the start of the reaction. If the rate is measured over a long period, the measured rate may not be the true value since the reaction would tend to slow down because the amount of substrate becomes used up as the reaction proceeds. The plot of reaction velocity V, as a function of the substrate concentration [S], for a constant amount of enzyme is shown below: We now know that a specific substrate concentration corresponds to a specific rate. We can then react an unknown amount of substrate with the same amount of enzyme used to determine our original plot. By measuring the initial rate of the reaction, we can then read the corres-ponding concentration of substrate from our plot. For example, if we measured a rate V_1 we know that the 'unknown' concentration of substrate is S_1 . An important precaution must be noted: at very large substrate con-centrations, the plot is no longer linear. The rate increases very slowly at large [S] until it reaches a maximum velocity. The slow-down of rate increase occurs when the amount of enzyme is limiting. At this point, (point E_2 on the plot) most of the catalytic sites on the enzyme molecules are filled, so increasing the amount of substrate present has little effect on increasing the rate. Therefore, at very high [S], the rate becomes independent of the amount of substrate present. It is therefore critical that the unknown concentration of substrate be low enough to validly correspond to a specific rate. One can measure the initial rate of reaction bystopping the reaction after a short period of time, usually by adding certain inhibitors or poisons. To the reaction mixture may then be added a substance which forms a colored complex with one of the reaction products. The denser the color, the more product formed within the measured time interval, and therefore, the higher the reaction rate. The density is related to how much light the colored sample absorbs when a beam of light is shone on the sample. (An instrument called a spectrophotometer is used to measure absorbance of light by a liquid sample.) The measured absorbance is compared to the absorbance of known amounts of product. Hence, when the absorbance measured is known, we will know how much product was formed by the reaction and since we know the time the reaction was stopped, we know how much product is produced per unit time, that is, the rate. The substrate concentration can then be read off the graph from its corresponding velocity.

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Question:

In what respects is Euglena like a plant and in what respects is it like an animal?

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Solution:

Euglena is a single-celled organism. Like plants, Euglena contains chlorophyll in chloroplasts and is capable of photosynthesis. Like animals, Euglena lacks a cell wall, is capable of active motion and ingests organic compounds as food. Botanists classified Euglena and its relatives as algae (phylum: Euglenophyta). Zoologists classified the euglenoids in the phylum Protozoa (class Phytomastigophora). Euglenoids are also often classified as a separate phylum in the kingdom Protista. Most euglenoids are capable of photosynthesis, but require certain organic nutrients as well as in-organic salts. Some are completely autotrophic, and some do not possess chlorophyll and are completely hetero- trophic. Of the heterotrophs, some are saprophytic, feeding on dead organic materials, and others are holozoic, capturing and ingesting other organisms. The euglenoids all have two flagella, by means of which they are capable of active motion. The euglenoid cell is delimited by a cell membrane within which there is a series of flexible, interlocking proteinaceous strips termed the pellicle. The flexible pellicle permits Euglena to change its shape, providing an alternate means of locomotion for mud-dwelling forms. As is true for protozoans, but not for plants, an area of the plasma membrane is specialized for the ingestion of food. This area is near the base of the flagella and is called the gullet. Unlike higher plants, but similar to many algae, the euglenoids have a stigma or an eyespot that functions as a light detector. The non--photosynthetic euglenoid species lack the stigma. Unique to the euglenoids is the storage product paramylum. This compound is a carbohydrate polymer unlike plant starch or animal glycogen. The paramylum is produced in an organelle called a pyrenoid. The cell division cycle of Euglena is also different from that of plants or animals. During division the nucleolus does not dis-appear, and the nuclear membrane remains intact. The euglenoids have a mixture of plant and animal characteristics, in addition to having their own unique features. They have no close relatives among the algae and few among the protozoans.

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Question:

Write a pseudocoded program to find an approximation of pi using the Monte Carlo method. This method is an example of a system which uses random digits to solve a problem that does not, by itself, have ran-domness.

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Solution:

First, we must assume that there is a subprogram called RANDOM available in the user's compiler. Different compilers have different means of obtaining random digits, so we omit a discussion of the actual process of randomization. Monte Carlo methods typically use random digits to solve problems which have no inherently random properties. For the approximation of pi, we need the dart board analogy. We have a square board with an inscribed circle, as in Fig. 1. Suppose we have monkeys that have been trained to toss darts at the board. Now, assuming all the darts land somewhere on the board, and assuming that a dart can land anywhere with equal likelihood, we ask: what is the probability that a dart will land inside the circle? If you remember the equations for the areas of a circle and a square (A = \pi(L/2)^2 and A = L^2, respectively), you can discover that the probability equals the ratio of the circle's area to the square's area, or \pi/4, which is about 0.7854. For our example, we do not know \pi. However, if we could get the monkeys to throw a large number of darts at the board, we could compute the proportion of hits falling inside the circle. From this proportion, we can find an approximation for \pi. The following pseudocode generates randomly the X- and Y- coordinates of each "dart" thrown at the board. Since these coordinates lie between 0 and 1, darts appear only in the positive quadrant of the square (the top right corner). However, the full figure can be built from four identical pieces, so the result remains the same. We just multiply \pi/4 by 4 to get the desired result. Also note that if we increase the number of darts thrown, the degree of deviation from the expected result decreases. This axiom is called in statistics the "law of large numbers". We are testing for 4-digit accuracy in this problem. Our output consists of the number of darts, the approximate value of \pi, and the deviation from the expected result. INTEGER REAL C END PROGRAM I, M, N, KOUNT APPROX, X, Y N \leftarrow 0 KOUNT \leftarrow 0 OUTPUT MESSAGE 'INPUT M: NUMBER OF DARTS TO BE TOSSED' INPUTM DO WHILE M > 0 DO FOR I \leftarrow 1 TO M X \leftarrow RANDOM Y \leftarrow RANDOM CHECK TO SEE IF THE DART HAS LANDED INSIDE CIRCLE IF(X\textasteriskcenteredX) + (Y\textasteriskcenteredY) < 0 THEN KOUNT \leftarrow KOUNT + 1 END IF - THEN END DO FOR I N \leftarrow N + M APPROX \leftarrow FLOAT (KOUNT) / FLOAT (N) \textasteriskcentered4.0 OUTPUT N,APPROX, APPROX - 3.1416 OUTPUT MESSAGE 'IF YOU WISH TO TOSS MORE DARTS, INPUT M' INPUT M END DO WHILE

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Question:

How are fungi important to man?

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Solution:

Fungi are divided into four groups:Oomycetes(Oomycota)- the egg fungi,Zygomycetes(Zygomycota)\rule{1em}{1pt}Zygo-spore-forming fungi, Ascomycetes (Ascomycota)-the sac fungi, andBasidiomycetes (Basidiomycota)-the club fungi. Fungi are both beneficial and detrimental to man. Bene-ficial fungi are great importance commercially.Ascomycetes, or sac fungi, are used routinely in food production. Yeasts, members of this group, are utilized in liquor and bread man-ufacture. All alcohol production relies on the ability of yeasts to degrade glucose to ethanol and carbon dioxide, when they are grown in the absence of oxygen. Yeasts used in alco-hol production continue to grow until the ethanol concentration reaches about 13 percent. Wine, champagne, and beer are not concentrated any further. However, liquors such as whiskey or vodka are then distilled, so that the ethanol concentration reaches 40 to 50 percent. The different types of yeasts used in wine production are in part responsible for the distinctive flavors of different wines. Bread baking relies on CO_2 pro-duced by the yeasts which causes the dough to rise. Yeasts used in baking and in the brewing of beer are cultivated yeasts, and are carefully kept as pure strains to prevent contamination. Sac fungi of genusPenicilliumare used in cheese production. They are responsible for the unique flavor of cheeses such as Roquefort and Camembert. The medically important antibiotic penicillin is also produced by members of this genus. CertainAscomycetesare edible. These include the delicious morels and truffles. The club fungi, orBasidiomycetes, are of agricultural importance. Mushrooms are members of this group. About 200 species of mushrooms are edible while a small number are poi-sonous. The cultivated mushroom, Agaricus campestris, differs from its wild relatives, and is grown commercially. Fungi are often of agricultural significance in that they can seriously damage crops. Members of theOomycetes, also known as water molds, cause plant seedling diseases, downy mildew of grapes, and potato blight (this was the cause of the Irish potato famine). Mildew is a water mold that grows parasitically on damp, shaded areas. Rhizopusstolonifer, a member of theZygomycetes, is known as black bread mold. Once very common, it is now controlled by refrigeration and by additives that inhibit mold growth. TheAascomyceteClaviceps purpurea causes the disease ergot, which occurs in rye and other cereal plants and results in ergot poi-soning of humans and livestock. This type of poisoning may be fatal. The disease caused in humans is called St.Vitus's dance Visual hallucinations are a common symptom of this disease. Lysergic acid is a constituent of ergot and is an intermediate in the synthesis of LSD. The "dance macabre" of the Middle Ages is now believed to have been caused by ergot poisoning. Basidiomycetes are also responsible for agricultural dam-age. Certain club fungi are known as smuts and rusts. Smuts damage crops such as corn, and rusts damage cereal crops such as wheat. Bracket fungi, another type of club fungi, cause enormous economic losses by damaging wood of both living trees and stored lumber. Fungi are also important to man because of the diseases they cause in man and livestock.Candidaalbicanscauses a throat and mouth disease, "thrush", and also infects the mucous membranes of the lungs and genital organs. Many skin diseases are caused by fungi, including ringworm and athlete's foot.

Question:

What is the rotational inertia of a 50-lb cylindrical flywheel whose diameter is 16 in.?

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Solution:

To find the rotational inertia of the flywheel, consider a mass element consisting of a thin cylindrical shell of radius r, thickness dr, and length L, as shown in the figure. Then dV= (2\pir) (dr) (L) density,\rho= m/V = m/(\piR^2L) anddm= \rhodV = m/(\piR^2L)(2\pir dr L) = (2m/R^2) r dr The moment of inertia is given by I = \intr^2 dm = \int(r^2)[(2m/R^2) r dr] = 2m/R^2 ^R\int_r=0 r^3 dr = (2m/R^2)(r^4/4)]^R_r=0 = (1/2)mR^2 For the given cylindrical flywheel, m= w/g = (50 lb)/(32 ft/sec^2) = 1.6 slugs R= 8.0 in = 2/3 ft I = (1/2)mR^2 = (1/2)(1.6 slugs) (2/3 ft)^2 = 0.35 slug-ft^2

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Question:

Obtain the 15's and 16's complement of the following hexadecimalnumbers: a) FFOb) 1234c) ABCDd) 0000

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G02-0032.htm

Solution:

Hexadecimal numbers arise frequently in computer applications because16 is the fourth power of 2. This means that each hexadecimal digitcan be represented uniquely by four bits. a) The 15's complement of FFO is found by subtracting it from FFF. FFF - FFO 00F Since radix complement = radix complement minus one + 1, the 16's complementhere is: 00F + 1 =010 . b) The 15's complement of 1234 is: FFFF - 1234 EDCB The16 'scomplement is then EDCB + 1, or EDCC. To make this more comprehensible, convert into the decimal system: 10000_16, = 16^4 + 0 × 16^3 + 0 × 16^2 + 0 × 16^1 + 0 × 16^0 = 65,536_10 1234_16 = 1 × 16^3 + 2 × 16^2 + 3 × 16^1 + 4 × 16^0 = 4660_10 Then the 16's complement (decimal equivalent) is: 65,536 - 4,660 60,876 Now, EDCC_16 = 14 × 16^3 + 13 × 16^2 + 12 × 16^1 + 12 × 16^0 = 60,876_10as above. Thus EDCC_16 is the 16's complement of 1234_16. c) The radix minus one complement of ABCD is: FFFF - ABCD 5432 The radix complement of ABCD is 5433, d) The radix complement of 0000 is 10000 while the radix complement minus one is FFFF.

Question:

One pair of genes for coat color in cats is sex-linked. The gene B produces yellow coat, b produces black coat, and the heterozygous Bb produces tortoise-shell coat. What kind of offspring will result from the mating of a black male and a tortoise-shell female?

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Solution:

Sex determination and sex linkage in cats is similar to that found in man, and indeed, in most animals and plants that have been investigated. So for cats, we can assume that if a gene is sex-linked, it is carried on the X chromosome. We can also assume that the Y chromosome carries few genes, and none that will mask the expression of a sex-linked gene or an X chromosome. Female cats, like female humans, are XX, and male cats are XY. Let X^B represent the chromosome carrying the gene for yellow coat, and X^b represent the chromosome carrying the gene for black coat. The male parent in this problem is black, so his genotype must be X^bY. The female is tortoise-shell, which means that she is carrying both the gene for yellow color and the gene for black color. Her genotype is X^BX^b. In the cross: Phenotypically, the offspring consist of: 1/4 tortoise-shell females (X^BX^b), 1/4 black females (X^bX^b), 1/4 yellow males (X^BY),and 1/4 black males (X^bY). Note that there can never be a tortoise-shell male, because a male can carry only one of the two possible alleles at a time.

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Question:

Black-headed gulls remove broken eggshells from their nests immediately after the young have hatched. Explain the biological significance of this behavior.

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Solution:

Behavior that contributes to the survival and reproduction of the animal is adaptive. Since those animals which demonstrate adaptive behavior survive longer and successfully raise more offspring, it is more likely that the behavior patterns of these animals will be continued throughout further generations. The eggshell - removing habit of Black-headed gulls is such adaptive behavior. The gulls do not remove only broken shells, but also any conspicuous object placed in the nest during the breeding season. The more conspicuous the object, the more likely it is to be removed. The gulls seem to be discarding conspicuous objects as a defense mechanism against visual predators. When inves-tigators placed conspicuous objects along with eggs in nests, these nests were robbed of eggs (by other gulls) more often than nests having only eggs. Thus eggshell- removing behavior is significant in that it reduces the chances of a nest being robbed, thus enhancing the survi-val of offspring. The gulls that evoke this behavioral pattern will successfully raise more offspring than those who do not, and thus this behavioral pattern will tend to be passed on.

Question:

A refrigerator which has a coefficient of performance A refrigerator which has a coefficient of performance one-third that of a Carnot refrigerator is operated between two reservoirs at temperatures of 200\textdegreeK and 350\textdegreeK. It absorbs 500 J from the low-temperature reservoir. How much heat is rejected at the high- temperature reservoir?

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Solution:

The coefficient of performance of a Carnot refrigerator is defined as the ratio of the heat extracted from the cold source and the work needed to run the cycle. Hence, E_c= Q_1/(Q_2 - Q_1)orequivalentlyT_1/(T_2 - T_1) where Q_1 is the heat absorbed at temperature T_1 and Q_2 is the heat rejected at the higher temperature T_2. Here all temperatures T are to be expressed in Kelvin degrees. The actual refrigerator has thus a coefficient of performance Q_1/(Q_2 - Q_1) = (1/3) [T_1/(T_2 - T_1)] or(Q_2 - 500 J)/(500 J) = [{3(350 - 25 0)\textdegreeK}/{250\textdegreeK}] \thereforeQ_2 = 1100 J.

Question:

A long solenoid of length լ and cross-sectional area A is closely wound with N_1 turns of wire. A small coil of N_2 turns surrounds it at its center, as in the figure. Find the mutual inductance of the coils.

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/Users/wenhuchen/Documents/Crawler/Physics/D22-0749.htm

Solution:

If a current I in a coil is changing at a rate [(dI)/(dt)], then a voltage across the coil will be induced in the direction to oppose the change, V_coil = - L [(dI)/(dt)] where L is the self inductance of the coil. Similarly, if the time varying magnetic flux of one coil links another coil, then it will induce a voltage in the second coil equal to V'_coil= -M [(dI)/(dt)](1 ) where I is the current in the first coil. The constant M is the mutual inductance of two coils. Its value is determined only by the geometry and positioning of the coils. In our problem, the magnetic field in the solenoid (away from the edges) is given by B = [(\mu_0 N_1 I_1 )/(լ)] where I_1 is the current in the solenoid coil. The flux \varphi through the N_2 coil is BA, where A is the area of the second coil. Then the voltage induced in the second coil by I 1 is, by Faraday's Law, V1 2= - N_2 [(d\varphi)/(dt)] = - [(\mu_0 N_1 N_2 )/(լ)] [dI/dt] Therefore we see by comparison with (1) that the mutual inductance of this system is M = [(\mu_0 N_1 N_2 )/(լ)] . If լ = 0.50 m, A = 10 cm^2 = 10^-3 m^2 , N_1 = 1000 turns, N_2 = 10 turns, M = [{(4\pi × 10^-7 henry/m) (10^-3 m^2 ) (1000 turns) (10 turns)} / (.5 m)] \approx 25 -10^-6 henry \approx 25 \muh .

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Question:

50 ml of rhubarb juice is titrated against 0.25 NNaOH. 20 ml ofNaOHsolution is required for neutralization. Assuming the acidity of the juice is due to oxalic acid (H_2C_2O_4) determine (a) the weight of oxalic acid per liter of juice, (b) normality of the juice.

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Solution:

(a) For the neutralization to occur, there must be the same number of OH^- ions as there are H^+ ions present. H^+ + OH^- \rightleftarrows H_2O The normality of a base is defined as the number of equivalents of base in one liter of solution. An equi-valent is the weight of an acid or base that produces one mole of H^+ or OH^- ions, respectively. WhenNaOH ionizes, there is 1 OH^- ion formed by eachNaOHthat ionizes. NaOH\leftrightarrows OH^- + Na^+ Therefore, in a 0.25 NNaOHsolution there are 0.25 equivalents of OH^-, which, in this case is also the number of moles. 20 ml is 0.02 liters. The number of moles ofNaOHpresent will equal the number of liters of solution times themolarityof the solution. As pre-viously indicated, normality equalsmolarity. Thus, no. of moles = no. of liters × normality = 0.02 liters × (.25 moles/liter) = 0.005 moles. Thus, there must be 0.005 moles of H^+ ions from the oxalic acid to neutralize theNaOH. When oxalic acid ionizes, there are 2 H^+ ions formed for each molecule ionized. H_2C_2O_4 = 2H^+ + C_2O_4 Therefore, each mole of oxalic acid ionizes 2 moles ofNaOH. Thus, for neutralization to occur, only one half as much oxalic acid is needed as NaOH. There are 0.005 moles ofNaOH, present, therefore, 0.0025 moles of oxalic acid is needed to neutralize it. One now knows that there are 0.0025 moles of oxalic acid in the 50 ml of rhubarb juice. One can find the number of moles in 1 liter by multiplying 0.0025 moles/50 ml by the conversion factor 1000 ml/1 liter. no. of moles/liter of oxalic acid = [(0.0025moles)/(50 ml)] × (1000 ml/1 liter) = 0.05 moles/liter. Since there are 0.05 moles of H_2C_2O_4, in 1 liter of juice, one can find the weight of H_2C_2O_4 in this quantity of juice by multiplying 0.05 moles by the MW of H_2C_2O_4. (MW = 90). weight of H_2C_2O_4, in 1 liter = 0.05 moles × 90 g/moles = 4.5 g (b) For titrations, the following relation is found: N_acidV_acid=N_baseV_base whereN_acidis the normality of the acid,V_acidis the volume of the acid, N_baseis the normality of the base, andV_baseis the volume of the base. Here, one is givenN_base,V_base, andV_acid. One is asked to findN_acid. N_b= .25 NN_aV_a=N_bV_b V_b= 20 mlN_a × 50 ml = .25 N × 20 ml N_a = ? V_a= 50 mlN_a = [(0.25 N × 20 ml)/(50 ml)] = 0.10 N.

Question:

Discuss the various ways in which energy is expended within the human body.

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Solution:

Energy is primarily expended during the anabolic reactions of the cell; that is, energy is required for the synthesis of macromolecules. Some organisms ingest simple chemical substances and combine them to form the complex molecules needed for the building of new protoplasm. Others, the higher organisms, ingest complex organic materials, degrade them, and use the simple products to synthesize the complex organic compounds that they need. In both cases, growth results. Energy is also expended during motion. Motion includes the contraction of muscle tissue, which allows an animal to either move itself (locomotion) or part of its body (the bending of a limb). Muscle contraction also occurs in the diaphragm during breathing, in the heart and in the walls of the digestive organs. Energy is also required during movement of materials across the cell membrane. During active transport, molecules move from a region of low concentration to one of high concentration,a process which requires energy. Active transport is thought to involve a carrier molecule driven by the hydrolysis of ATP, the energy-releasing source. Another energy-requiring process is heat production. During metabolic activity, the heat given off is important to "warm-blooded" animals (homeotherms) in keeping body temperature constant despite changes in evironmental temperatures. Even "cold-blooded" animals(poikilotherms) whose body temperature is determined by the environmental temperature, must release heat in order to survive extremely cold conditions. Homeotherms, which include birds and mammals, generally eat more food per body weight than do poikilotherms, which include fish, amphibians, reptiles, and invertebrates. This fact reflects the need to furnish more energy to maintain body temperature. Both anabolism and muscle contraction require the energy obtained from the splitting of the energy-rich terminal phosphate group of ATP, which has the structure: ATP is produced during oxidative phosphorylation - the process by which the flow of electrons in the electron transport system of respiration transfers the released energy to be stored in ATP. However, certain compounds, such as the hormone thyroxine, can uncouple phosphory-lation from the electron flow so that the energy is not trapped as ATP, but is released as heat. This mechanism is important when more heat is needed by an organism than is provided by its normal metabolic activity.

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Question:

Explain what occurs during the human "birth process" and when these phenomena take place.

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Solution:

The actual process of birth is preceded by the gestation period. The gestation period or duration of pregnancy varies in different species. It normally lasts about 280 days in humans (the time of the last menstrual period to the birth of the baby). Little is known about the factors that initiate the birth process, or parturition, after gestation is complete, but it is believed that hormon-al factors play an important role. It has been found ex-perimentally in mice that the level of progesterone in the blood declines sharply, while that of estrogen rises short-ly before parturition. The start of parturition is marked by a long series of involuntary uterine contractions called "labor pains" Labor is divided into three periods. The first lasts about twelve hours. The fetus is moved down toward the cervix which becomes dilated to enable the fetus to pass through. The amnion ruptures, releasing the amniotic fluid through the vagina. The second period is the birth, where the fetus passes through the cervix and vagina and is "delivered." This stage takes about twenty minutes to an hour, and consists of combined involuntary uterine contractions and voluntary contractions of the abdomen by the mother. The last stage of labor begins after birth and lasts for ten to fifteen minutes. The placenta and the fetal membranes are loosened from the uterine lining by another series of contractions and expelled as the after--birth. Sometimes, a woman will require more help than usual during childbirth. Drugs such as oxytocin and prosta-glandins will help increase the uterine contractions. Special forceps can be used by an obstetrician to pull the infant through the birth canal. When the canal through which the baby passes is too small, a caesarean delivery is performed. This is an operation in which the abdominal wall and uterus are cut open from the front and the baby is removed.

Question:

Calculate the volume of O_2 necessary to burn 50 liters of CO completely. The balanced reaction is: 2CO + O_2 \rightarrow 2CO_2 Also, calculate the volume of CO_2 formed.

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Solution:

This is a volume-volume problem and the technique shown applies only to gases. It is assumed that all the gases must be at the same temperature and pressure. No molecular weights or molecular volumes are needed. One solves this problem by setting up a ratio between the actual volumes present and the mole volumes of the re-acting compounds as shown in the equation below. [(volume CO present) / ( mole volume CO)] = [(volume O_2 needed)/(mole volume O_2 )] The mole ratio for this reaction is 2 : 1, therefore, the mole volume ratio is 2 : 1. Substitute, to obtain [(50 liters) / (2 liters)] = [(volume O_2 ) / (1 liter)] volume O_2 = 25 liters. To find the volume of CO_2 produced, set up a similar proportion between CO and CO_2 or O_2 and CO_2. The mole ratio for CO and CO_2 is 1 : 1 and for O_2 and CO_2 is 1 : 2. There-fore use the equation [(volume reactant) / (mole volume reactant)] = [(volume CO_2 ) / (mole volume CO_2 )] The values for CO as the reactant are: (50/2) = [(volume CO_2 ) / (2)] volume CO_2 = 50 liters. The values of O_2 as the reactant are: (25/1) = [(volume CO_2 ) / (2)] volume CO_2 = 50 liters.

Question:

When mercury is vaporized at its boiling point at standard pressure, the entropy change is 20,7 cal/mole - \textdegreeK, Determine the boiling point of Hg if the heat of vaporization is 65 cal/g.

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Solution:

When a process occurs at constant temperature, the change in entropy, ∆S, is equal to the heat absorbed divided by the absolute temperature at which the change occurs, i.e. , ∆S = (∆H/T), where ∆H is the heat of vaporization (in this, case) and T is the absolute temperature. T will be equal to the boiling point of mercury in this problem. Using this equation, one can solve for T after either converting ∆H to cal/moles or converting ∆S to cal/g. Here, one will convert ∆H from cal/g to cal/mole by multiplying ∆H by the molecular weight of Hg. (MW of Hg = 200.6.) ∆H in cal/moles = 65 cal/g × 200.6 g/mole = 13039 cal/mole One can now solve for T. T = [(∆H)/(∆S)]∆H = 13039 cal/mole ∆S = 20.7 cal/mole - \textdegreeK T = [(13039 cal/mole)/(20.7 cal/mole - \textdegreeK)] = 630 \textdegreeK T in \textdegreeC = 630 - 273 = 357\textdegreeC.

Question:

Calculate the weight of iron in 350 pounds of Fe_2O_3. First, calculate the percent of iron in the compound.

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Solution:

The percent composition of iron in the compound Fe_2O_3 can be calculated by first finding the molecular weight of the compound. After the molecular weight of the compound is found, the percent of this weight that is made up by the iron is found by dividing the weight of the iron by the molecular weight of the compound and then multiplying this fraction by 100. The molecular weight of iron is 55.8, so that two atoms of iron weigh 2 × 55.8 = 111.6. The molecular weight of oxygen is 16.0, so that 3 atoms of oxygen weigh 3 × 16.0 = 48.0 The molecular weight of Fe_2O_3 is found by adding the weights of the iron and oxygen together. weight of iron=111.6 weight of oxygen=48.0 molecular weight of Fe_2O3=159.6 The percent composition of iron = {(weight of iron in compound ) / (molecular weight of compound)} × 100 percent composition of iron = (111.6 / 159.6 ) × 100 = 69.9 Since the percent composition by weight of iron in the compound is known, the weight of iron in 350 pounds of this compound can now be found. Because the percent compo-sition of iron in this compound is 69.9, the percent of iron in 350 pounds of Fe_2O_3is also 69.9 . Weight of iron in 350 lb of Fe_2O3= 350 × 0.699 = 245 lb.

Question:

Many organs develop in the embryo without having to function at the same time, but the heart must function while undergoing development. Describe the steps in the formation of the heart.

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Solution:

In its early stage of development, the heart consists only of a simple tube, formed from the fusion of two thin-walled blood vessels located below the head. In this early condition, the human heart resembles a frog heart. A frog heart consists of four "chambers" arranged in series: the sinus venosus, which receives blood from the veins, an atrium, a ventricle, and the arterial cone, which leads to the aortic arches. Since the tube grows faster than the points to which its front and rear ends are attached, it bulges to one side. The ventricle then twists in an S-shaped curve down and in front of the atrium, coming to lie ventral to it as it exists in the adult. Gradually, the sinus venosus becomes incorporated into the atrium as the atrium grows around it, and most of the arterial cone is merged with the wall of the ventricle. The heart becomes partitioned into a right side and a left side in all air-breathing vertebrates. This takes place rather early in development, long before the lungs have begun to function at the time of birth. No such division occurs in the hearts of vertebrates which breathe in water. Their hearts pump just one kind of blood, namely, "venous blood" (blood low in oxygen). Aeration takes place in the gills, and only "arterial blood" (blood rich in oxygen) is distributed to the body. The advent of lungs necessitated a radical change in the arrangement of the circulation. The heart now receives two sorts of blood: venous blood from the body, and arterial blood from the lungs. It is impor-tant that the two streams be kept separate so that the venous blood is sent to the lungs for aeration and only the arterial blood is distributed to the body. The heart begins separating into four chambers at an early stage. By the end of the second month of de-velopment, the two ventricles are completely separated. The septum separating the atria, the foramen ovale, however, is not completely fused until the fetus is born. If this septum remains open at birth, the baby is termed a "blue baby\textquotedblright due to the flow of deoxygenated blood through the body. Surgical correction is performed for this condition.

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Question:

Determine the de Broglie wavelength of an electron (mass = 9.11 × 10^-28 g) having a kinetic energy of 100eV.

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Solution:

Using the value for the kinetic energy of the electron, we will calculate its momentum and substitute this into the de Broglie equation to get the wavelength of the corresponding matter wave. The kinetic energy, k, of a particle of mass, m, moving with velocity v is k = (1/2) mv^2 and its momentum, p, is p =mv. Kinetic energy and momentum are related as follows: k = (1/2) mv^2 = (1/2m) m^2 v^2 = (1/2m) (mv)^2 = (p^2/2m) = k or p = (2mk)^1/2. The kinetic energy of our electron is k = 100eV = 100eV× 1.602 × 10^-12 erg/eV= 1.602 × 10^-10 erg. Hence, its momentum is: p = (2mk)^1/2 = (2 × 9.11 × 10^-28 g × 1.602 × 10^-10 erg)^1/2 = (.2919 × 10^-36)^1/2 erg^1/2 g^1/2 = (.2919 × 10^-36)^1/2 (g - cm^2/sec^2)^1/2 g^1/2 = 5.403 × 10^-19 g-cm/sec. The de Broglie equation is: \lambda = h/mv= h/p, where h is Planck's constant (h = 6.626 × 10^-27 erg/sec) = 6.626 × 10^-27 g-cm^2/sec. Hence, the de Broglie wave-length of a 100eVelectron is: \lambda = (h/p) = [(6.626 × 10^-27 g-cm^2/sec)/(5.403 × 10^-19 g-cm/sec)] = 1.226 × 10^-8 cm = 1.226 × 10^-8 cm × 10^8 \AA/cm = 1.226 \AA.

Question:

Suppose a 15-g bullet is fired into a 10-kg wooden block that is mounted on wheels and the time required for the block to travel a distance of 45cm is measured. This can easily be accomplished with a pair of photocells and an electric clock. If the measured time is 1 sec, what is the muzzle velocity of the bullet?

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Solution:

In this example, the bullet comes to rest in the block and imparts its momentum to the block. Since the block travels 45 cm in one second, the recoil velocity of the block is 45 cm/sec, and from momentum conservation we have m_1v_1 = m_2v2 (Here, we do not have a negative sign because both velo-cities are in the same direction. Also, we take m_2 to be 10 kg, that is, we neglect the mass of the bullet embedded in the block.) Then, v_1 = (m_2v_2)/(m_1) = [(10^4g ) × (45 cm/sec )]/15g = 3 × 10^4 cm/sec = 300 m/sec \cong 985 ft/sec

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Question:

The hydrogen-filled dirigible "Hindenberg" had a gas volume of 1.4 × 10^8 liters. Assuming that it was filled to a pressure of 1 atm at 0\textdegreeC, how much energy was released when it burned to gaseous products at a reaction temperature of 25\textdegreeC? The standard heat of formation of water vapor at 25\textdegreeC is - 57.8 Kcal/mole.

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Solution:

The solution to this problem involves calcu-lation of the heat of reaction for the combustion of hydrogen and the number of moles of hydrogen in the dirigible. The combustion reaction is written H_2 (g) + (1/2) O_2 (g) \rightarrowH_2O (g) . The heat of reaction at 25\textdegreeC (298\textdegreeK) is written ∆H\textdegree_298 = sum of heats of formation of products - sum of heats of formation of reactants = ∆H\textdegreef, H2O- [∆H\textdegreef, H2+ ∆H\textdegreef, O2] Since the heats of formation of O_2 and H_2 are zero by definition, ∆H\textdegree_f, O2 = ∆H\textdegreef, H2= 0 Kcal/mole, the heat of reaction is equal to the heat of formation of water vapor: ∆H\textdegree_298 = ∆H\textdegreef, H2O=- 57.8 Kcal/mole. Since ∆H\textdegree_298is negative, the reaction is exothermic and - 57.8 Kcal are released per mole of water produced. But since one mole of water is formed per mole of hydrogen burned, - 57.8 Kcal are released per mole of hydrogen burned. To calculate the number of moles of hydrogen present in the dirigible, we make use of the fact that at 0\textdegreeC and 1 atm pressure, one mole of gas occupies 22.4 l (molar volume of gas). Then number of mole of H_2 = (volume of dirigible) / (number of moles of H_2) = (1.4 × 10^8l) / (22.4 l/mole) = [(1.4 × 10^8) / (22.4)] moles The heat released is then the absolute value of the heat of reaction (per mole) multiplied by the number of moles of gas. heat released = \vert∆H\textdegree_298\vert × number of moles of H_2 = 57.8 Kcal/mole × [(1.4 × 10^8_ ) / (22.4)] moles = 360 × 10^6 Kcal. (We use the absolute value of ∆H\textdegree298because the word "released\textquotedblright already takes into account the fact that the re-action is exothermic and therefore the sign of ∆H\textdegree_298 is negative.) Note that we have assumed that all of the hydrogen is burned.

Question:

Why is it that animals and plants of England and Japan are very similar despite the fact that they lie on nearly the opposite sides of the world?

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Solution:

The observation that animals and plants of England and Japan closely resemble each other is explained by the fact that England and Japan belong to the same biogeographic realm. A realm is a geographic division separated from other realms by major physical barriers, and characterized by the existence of certain unique organisms. Inside a realm there may be variations of climate and topography, but organisms are nevertheless able to pass more or less freely from one region of the realm to another. It must be understood that some species have more ready physical access to a region suitable for habitation than others. What may be an effective barrier for one species may be a possible but difficult route for another or an easily crossed part for a third. Thus, keeping in mind the fact that different species have varying access to widely separated habitats, like England and Japan, and realizing that historically, geogrophic and environmental conditions were different than they are today, we can begin to understand the underlying concept to this question. England and Japan both belong to the Palearctic realm (see Figure). By definition then, the animals of these two countries are free to migrate from one country to the other, either by way of the sea or the air. Though plants themselves cannot migrate, their seeds can be carried by wind, water or migratory animals to distant locations. In other words, there are no absolutely impassable physical barriers between England and Japan, and genes can flow from one country to the other. Therefore England and Japan share many common groups of animals and plants.

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Question:

Memory Reference Instructions: 013415 I OP CODE ADDRESS Register-Reference Instruction (Indirect): Input Output Instruction: The computer operates in four different cycles: Fetch Cycle (read instruction) Indirect cycle (read address of operand) Execute cycle (read operand) Interrupt where each cycle will execute no more than four steps. Design a block diagram of the control unit for this com-puter.

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Solution:

The control unit is shown in Figure 1. The F and R registers are used to denote which of the four cycles the computer is operating in. The 2 × 4 decoder associated with these registers generates four control signals C_0 - C_3. Each cycle has its own control signal specified by the table of Figure 2. Registers FR Decoder Output Computer Cycle 00 C_0 Fetch Cycle 01 C_1 Indirect Cycle 10 C_2 Execute Cycle 11 C_3 Interrupt Cycle Each cycle is divided into four steps. Each step is designated by the output of the two-bit binary step counter, SC. The counter starts at 00 when the computer enters a cycle and the computer stays in this cycle until the counter reaches 11. At this point, the computer enters another cycle and the counter returns to 00. Each step is decoded into one of four timing signals, t_0 - t_3. The S bit is used to start and stop the computer. The user starts the computer by putting a logic level 1 in the S register. This enables the SC decoder and the timing signals. The user stops the computer by putting the S register to 0. This disables the decoder and all the timing signals t_0 - t_3-. The OPR register contains the three bit op-code specified in the instruction format. Each of the eight possible op-codes is decoded into eight separate control signals q_0 - q_7. Each signal specifies one of eight instruc-tions from the instruction set of the computer. The I bit is bit 1 of the instruction format. It is used to designate whether the instruction is: (1) In-direct addressing (I = 1), when q_7 = 0 (op-code \not = 111); (2) Register reference (I = 0), or I/O (I = I), when q_7 = 1 (op-code = 111). All of these signals (C_0 - C_3; t_0 - t_3; q_0 - q_7; I) go to the control logic section which then carries out the actual Instruction through the control functions.

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Question:

The arithmetic logic unit (ALU) in Figure 1 can add the contents of A and B, complement the contents of A or B, and logically AND the contents of A and B. Show how the ALU can: a)Increment b)Decrement c)Subtract d)Logically OR e)Shift left f)Clear

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Solution:

(a) The contents of register A can be incre-mented by clearing register B and letting C_i = 1. When the addition operation is performed the value of sum be-comes A + B + C_i. = A + C_i. = A + 1. Thus, A is incremented. (b)The contents of A can be decremented by adding nega-tive ones. Negative one in two's complement is all 1's; thus, to decrement A: let the contents of B be all 1's, C_i = 0, and add. (c)Subtraction is simply adding a negative number. Thus, to form A - B, get the two's complement of B and add to A. The two's complement of a number is found by complementing the number and adding one. Thus, to subtract B from A: complementBthrough the ALU using theBfunction input. The complement function will place the result (B) in storage area SUM. Put the contents of SUM back into B. Then, to complete the complementing process, add 1 to the new B by zeroing A and making C_i = 1. Again, this result will be placed in SUM. This new number in SUM now repre-sents the fully complemented B (or -B). With this, we can store the complement in Register B. Let us now place the number which B (or more precisely, the contents of B) is to be subtracted from in Register A. Hence, by adding the contents of the two registers, we obtain the difference of the two numbers, represented in two's comple-ment form. (d)Logically ORing A and B can be carried out using DeMorgan's Law. A + B =A ̅ \textbullet B ̅ Thus, to OR A and B: complement A and put the result back into A, complement B and put the result back into B, AND A and B and put back in A, then complement A. (e)Shifting-left is the same as multiplying by two in base two, just like shifting left is the same as multi-plying by ten in base ten. Instead of multiplying by two, however, just add A to A. Thus, to shift the con-tents of register A left: put the contents of A into B (registers A and B now hold the same numbers) , let C_i = 0 and add. (f)Clearing a register (making it all zero's) is done by the following formula: 0 = AA ̅ Thus, to clear register A:complement A through the ALU and put back into B, AND registers A and B, put sum (which is zero) into A.

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Question:

Discuss the role of hemoglobin in the transport of oxygen and of carbon dioxide.

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Solution:

Hemoglobin is the pigment in red blood cells that is responsible for transporting nearly all the oxygen and some of the carbon dioxide in the body. In arterial blood, oxygen is present in two forms. It is either physically dissolved in the blood plasma or chemically bound to hemoglobin. Because oxygen is relatively in-soluble in water, and blood plasma is comprised primarily of water, only 3 ml. of oxygen can be dissolved in 1 liter of plasma at normal oxygen pressure (100 mm Hg on average). Hence, approximately 2% of the oxygen in the blood is dissolved in the plasma; the rest is transported via hemo-globin. In the lungs, oxygen enters the capillaries, and diffuses into the red blood cells where it binds to hemoglobin. Four oxygen molecules are attached to four iron atoms in a single hemoglobin molecule (see Figure). The chemical reaction between oxygen and hemoglobin is usually written as O_2 + Hb \rightleftarrows HbO_2 (oxyhemoglobin) The reaction goes to the right in the lungs and to the left in the body tissues. Hemoglobin combined with oxygen (HbO_2) is called oxyhemoglobin; when not combined (Hb), it is called reduced hemoglobin or deoxyhemoglobin. Because there is a finite number (four) of binding sites for oxygen on the hemoglobin molecule, there is a maximum amount of oxygen which can combine with hemoglobin. When hemoglobin exists as both Hb and HbO_2, it is said to be partially saturated. When it exists as only HbO_2, it is said to be fully saturated. The combination of oxygen with hemoglobin and its release from oxyhemoglobin are controlled by the concen-tration of oxygen present and, to a lesser extent, by the amount of carbon dioxide present. For example, the percentage of saturated hemoglobin undergoes little change between the lungs and the arteries, because the oxygen concentration in the lungs is very similar to that of the arteries. However, in the tissues, the concentration of oxygen is low, so by the law of mass action, the reaction will go to the left. As a result, oxygen will break off from the oxyhemoglobin to diffuse into the tissues. Carbon dioxide reacts with water to form carbonic acid, H_2CO_3, which then dissociates: CO_2 + H_2O \rightleftarrows H_2CO_3 \rightleftarrows H^+ + HCO_3^- (bicarbonate) An increase in the concentration of CO2drives the reaction to the right, increasing the H^+ concentration and hence the acidity of the blood. The oxygen carrying capacity of hemoglobin decreases as the blood becomes more acidic. Thus, more O_2 is released to the tissues under conditions of increasing acidity. This shows that the combination of hemoglobin and oxygen is controlled, in part, by the amount of CO_2 present. This results in a very efficient transport system. In the lung capillaries, oxygen is taken up by the hemoglobin due to the effects of high oxygen tension and low CO_2 tension. The situa-tion is reversed in the tissues. There, CO_2 pressure is high and oxygen pressure is low, so oxygen dissociates from oxyhemoglobin and diffuses into the tissues. It is important to realize that the principal factor that determines the mode of hemoglobin dissocia-tion and rate of diffusion is the partial pressure (P_gas) of each particular gas. As is true for oxygen, the quantity of carbon dioxide that can physically dissolve in blood is quite small. Carbon dioxide can undergo the following re-action: CO_2 + H_2O \rightleftarrows H_2CO_3 (carbonic acid) This reaction would go quite slowly if it were not catalyzed by the enzyme carbonic anhydrase. The quan-tities of both dissolved carbon dioxide and carbonic acid are directly proportional to the partial pressure of CO_2. In this case, as in the case with oxygen, we are only concerned with the pressure of the gas in question. This is because in a mixture of gases, each one acts indepen-dently of the others, and exerts the same pressure it would if it were present alone. The actual quantity of carbonic acid in the blood is small, because, as we saw previously, it dissociates into H^+ and HCO_3^- ions. These ions are quite soluble in the blood. Thus, the addition of carbon dioxide to blood plasma results ultimately in the produc-tion of hydrogen and bicarbonate ions. Carbon dioxide is transported primarily as bicarbonate ions to the lungs, where it is excreted as carbon dioxide. Carbon dioxide can also react with proteins, particularly hemoglobin, to form carbamino compounds. CO_2 + Hb \rightleftarrows HbCO_2 (carbamino hemoglobin) Carbon dioxide diffuses from the tissues into the blood. In the blood, some (8%) of the CO_2 stays dissolved, and some (25%) reacts with hemoglobin to form HbCO_2. The largest fraction (67%) is converted to H^+ and HCO_3^-. This occurs primarily within the red cells because they contain large quantities of carbonic anhydrase, whereas the plasma does not. This dissociation into H^+ and HCO_3^- explains why tissues and capillaries, where CO_2 concentration is high, have a hydrogen ion concentration higher than that of arterial blood. This also explains the increase in H^+ concentration as metabolic rate increases. The CO_2 it-self passes from the tissues to the blood, and then to the lungs by diffusing from a region of high CO_2 tension to one of low CO_2 tension.

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Question:

A stone of mass 100 grams is whirled in a horizontal circle at the end of a cord 100 cm long. If the tension in the cord is 2.5 newtons, what is the speed of the stone?

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Solution:

First, we shall calculate the acceleration of the object, and from that we may calculate its veloci-ty. Firstly, F = mg 2.5 newtons = 100 gm × a Newtons have the units (kg - m)/(s^2), and 100 gm = 0.100 kg, so that 2.5[(kg - m)/(s^2)] = 0.100 kg × a a = 25 m/s^2 Alsoa = v^2/r for uniform circular motion, where a is the linear acceleration and v is the linear velocity of the object. Therefore, 25(m/s^2) = v^2/(1m) v = 5 (m/s) .

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Question:

A young scientist is watching a group of new-born ducklings. On the first day, he observes them to crouch in fear at the sight of falling leaves. He also sees them "practice" flying. Over the next few days he observes that they do not react to the falling leaves and can now fly expertly. He concludes that the ducklings have learned to ignore falling leaves and have also learned to fly. Why is he partly incorrect?

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Solution:

The scientist is incorrect because he uses the term "learning" to describe both processes. Learning is the adaptive change of behavior as a result of experience The animal modifies its behavior to respond to stimuli in ways that promote its survival and reproduction; that is, in ways that adapt it to its environment. One of the simplest forms of learning is called habituation. Habituation is the gradual decline in response to repeatedly presented stimuli not associated with reward or punishment. Young ducklings initially show fear and crouch when presented with any overhead flying objects, such as falling leaves. But the ducklings cease to respond to the leaves since experience has shown them that they are insignificant. (no associated reward or punishment) The duckling, however, is not "learning" to fly. One of the factors that complicates the study of learning is maturation. Maturation is the automatic disclosing of a behavioral process due to physiological development. The flight of birds is a prime example of maturation. As the scientist observed the flying attempts of the ducklings, he termed it "learning", thinking that the ducks flew better (change in behavior) because of this apparent "practice" (experience). But flying can be achieved solely by maturation, with no practice involved. This has been shown by experiments with pigeons. Pigeons were raised in narrow tubes that prevented them from moving their wings and practicing flying. However, when the pigeons were released at an age when they would have already been flying, they were able to fly as well as control pigeons raised under normal conditions. It is not practice that enables the flight of a young bird to improve; it is progressive maturation. The development of the bird's nervous and muscular systems enables it to improve its flying. The bird has not "learned" to fly since experience is unnecessary. Hence the young scientist is incorrect in this respect.

Question:

Large deficits of water can be only partly compensated for by renal conservation; drinking is the ultimate compensatory mechanism. What stimulates the subjective feeling of thirst which drives one to drink water?

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Solution:

The feeling of thirst is stimulated both by a low extracellular fluid volume and a high plasmaosmolarity. The production of ADH by the hypothalamus is also stimulated by these factors. The centers which mediate thirst are located in the hypothalamus very close to those areas which produce ADH. Should this area of the hypothalamus be damaged, water in-take would stop because the sensation of thirst would be impaired. Conversely, electricalexcitation of this area stimulates drinking. There has been much speculation that, because of the similarities between the stimuli for ADH secretion and thirst, the receptors which initiate the ADH controlling reflexes are identi-cal to those for thirst. These are the osmoreceptorsre-ferred to in the previous solution. In addition to hypothalamic pathways, there are also other pathways controlling thirst. For example, dryness of the mouth and throat causes profound thirst, which is relieved by moistening. There is also a learned control of thirst; the quantity of fluid drunk with each meal is, in large part, a learned response determined by past experience.

Question:

What is the de Broglie wavelength of a radium alpha particle (a helium nucleus,He^2+) having an energy of 4.8MeV(million electron volts)?

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Solution:

The de Broglie wavelength, \lambda, of a particle of mass, m, moving at speed, v, is given by \lambda = (h/mv), where h is Planck's constant (h = 6.6 × 10^-27 erg-sec). The mass of an alpha particle is 66 × 10^-24 g. The energy of an alpha particle is given by E = (1/2) mv^2 = 4.8 MeV . Since 1MeVis equal to 1.6 × 10^-6 erg, we have: (1/2) mv^2 = 4.8 × 1.6 × 10^-6 erg. We will employ this value of the energy in order to obtain the value ofmvwithout having to find the velocity of the particle. (mv)^2 = 2 [(1/2) mv^2] m = 2 (4.8 × 1.6 × 10^-6 erg) (6.6 × 10^-24 g) = 1.0 × 10^-28 erg-g = 1.0 × 10^-28 g^2 cm^2/sec^2, where we have used 1 erg = 1 g - cm^2/sec^2. Taking the square root of both sides gives: (mv)^2 = 1.0 × 10^-28 g^2 cm^2/sec^2 mv= 1.0 × 10^-14 g cm/sec. Substituting this value into the expression for the de Broglie wavelength gives: \lambda = (h/mv) = [(6.6 × 10^-27 erg-sec)/(1.0 × 10^-14 g-cm/sec)] = [(6.6 × 10^-27 g-cm^2/sec)/(1.0 x 10^-14 g-cm/sec)] = 6.6 × 10^-13 cm.

Question:

Find the wavelength (\lambda) of the first transition, from the first excited state to the ground state in the Lyman and Paschen series. These series show the lines of the emission spectra of H_2 gas. The Lyman series defines its lowest energy level n_1 as being equal to 1, the Paschen series defines n_1 as 3.

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Solution:

When an atom is in its ground state, its electrons are in the lowest energy state possible. When energy is added to the system, its electrons are promoted to a higher energy level. These electrons are said to be in the excited state. After a short time, the electrons fall back to lower energy levels and eventually to their ground states. The same amount of energy that was needed to excite the electron is released when the electron falls back to its ground state. The Paschen and Lyman series differ in the value of their lowest energy states. The Paschen series is the spectra of electrons falling from excited states to the third energy level (n_1 = 3) , whereas the Lyman series is the spectra that is exhibited when the electrons fall from excited states to the first energy level (n_1 = 1). This is illustrated in the accompanying figure. Wavelength and the energy levels of the transition are related in the Rydberg - Ritz equation, (1/\lambda) = R [(1/n_1 ^2) - (1/n_2 ^2)], where \lambda is the wavelength, R the Rydberg constant, and n_1 and n_2 are the energy levels of the transition. The electron falls from n_2 to n_1, thus, n_2 is greater than n_1. Solving for \lambda: Lyman series: In the Lyman series n_1 = 1. The first transition is from n_2 = 2 to n_1 = 1. One solved for \lambda by using the Rydberg - Ritz equation. (1/\lambda) = (109, 678 cm^-1) [1/(1)^2 - 1/(2)^2], (1/\lambda) = 82258 cm^-1 \lambda = [(1)/(82258 cm^-1] = 1.21 × 10^-5 cm. Wavelengths are usually expressed in angstroms. 1 \AA = 10^-8 cm. (1.21 × 10^-5 cm) [(1 \AA)/(10^-8 cm)] = 1210 \AA Paschen series: The first transition in the Paschen series is from n_2 = 4 to n_1 = 3 because the base energy level in this series is 3. Here again, one uses the Rydberg - Ritz equation to solve for \lambda. (1/\lambda) = 109678 cm^-1 [1/(3)^2 - 1/(4)^2] = 5332 cm^-1 \lambda = [(1) / (5332 cm^-1)] = 1.87 × 10^-4 cm. Converting to angstroms: 1.87 × 10^-4 cm ×[(1 \AA) / (10^-8 cm)] = 18700 \AA.

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Question:

In a hydrogen bomb, deuterium nuclei fuse to produce a helium nucleus. In the process, 5MeV of energy is re-leased per deuterium. If, 1/7000 of natural H is D, how many liters of gasoline (C_8H_18) are needed to be burned to produce the same amount of energy as derived from 1 liter of water. Assume the following: the heat of combustion of gasoline is 5100 KJ/mole the density of gasoline is .703 g/cm^3; leV = 1.60 × 10^-19 J.

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Solution:

The answer to this question involves several calculations. To determine the amount of gasoline which will yield an equal amount of energy upon combustion as one liter of water, first compute how much energy can be obtained from each. For gasoline, this entails calcula-tion of the number of moles per liter of the substance. Multiplication by the heat of combustion yields the total energy. For water, first determine, how many deuterium ions are present per liter and then, using conversion factors, multiply this by the energy re-leased per deuterium ion fused. To obtain the energy per liter of water, first find the number of deuterium ions. To obtain this, you need to first find the number of atoms of H per liter of water, since the density of water is defined as one and density = mass/volume, 1 liter H_2 O =1000 g H_2 O. The molecular weight of one mole of water is 18. Therefore, the number of moles of water in one liter is (1000) / (18) = 55.5 [(moles water) / (liter)].There are two moles of H atoms, per mole of water, since its molecular formula is H_2 O. Therefore, the number of moles of H atoms = [(55.5 moles water) / (liter)] × [(2 mole H atoms) / (moles H_2O)] = (111 mole H atoms) / (liter). However, in a mole of anything, there are 6.02 × 10^23 particles, (Avogadro's number). Therefore, the total number of H atoms per liter of water is [(111 mole H atoms) / (liter)] × 6.02 × 10^23 (atoms/mole) = 6.68 × 10^25 (H atoms/liter) since you have 1 deuterium atom per 7000H atoms. Therefore, the number of deuterium atoms per liter of water is 6.68 × 10^25 (H atoms/liter) × [(1 atom deuterium) / (7000 H atoms)] = 9.55 × 10^21 [D atoms/liter)] Since each deuterium atom yields 5MeV or 5 x 10^6 ev of energy, when it fuses to produce helium, the amount of energy per liter of water is 9.55 × 10^21 (D atoms/liter) × 5 × 10^6 (ev/D atom) = 47.75 × 10^27 (ev/liter). Since, 1.60 × 10^-19 J = one ev = 1.60 × 10^-22 kJ, 47.75 × 10^27 (ev/liter) = 47.75 × 10^27 (ev/liter) × 1.60 × 10^-22 kJ/ev = 7.64 × 10^6 kJ/liter of energy per liter of water. To compute the energy in one liter of gasoline, remember that the density of gasoline equals mass of gasoline divided by its volume. Therefore, the mass of one liter of gasoline is (density × volume) = .703 g/cm^3 × 1000 cm^3 = 703 g. Thus, there are 703 g/liter of gasoline. The molecular weight of gasoline is 142 grams per mole. Therefore, the number of moles of gasoline per liter is (703 g/l) / (142 g/mole) = 6.16 moles/liter for gasoline. The heat of combustion of gasoline is 5100 kJ/mole. Therefore, the energy per liter of gasoline is 6.16 moles/liter × 5100 kJ/mole = 31,400 kJ/liter = 3.14 × 10^4 kJ/liter. To find out how may liters of gasoline is equivalent to water in terms of energy, divide the energy of 1 liter H_2 O by the energy of 1 liter gasoline. For water, it is 7.64 × 10^6 kJ/l. For gasoline, it is 3.14 × 10^4 kJ/l. (7.64 × 10^6 kJ/lwater) / (3.14 × 10^4 kJ/l gasoline) = 243 [(liters gasoline) / (liter water)].

Question:

Hydrogen peroxide solution for hair bleaching is usually prepared by mixing 5.0 g of hydrogen peroxide (H_2O_2, molecular weight = 34 g/mole) per 100 ml of solution. What is themolarityof this solution?

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Solution:

Before employing the definition molarity= [(number of moles)/{volume (liters)}] we must convert 5.0 g of H_2O_2 to the corresponding number of moles. To do this, we use the formula moles = mass/molecular weight. Then, moles = (mass/molecular weight) = [5.0 g / (34 g/mole)]= 0.15 moles H_2O_2 Converting 100 ml to liters, 100 ml = 100 ml × (1 liter/1000 ml) = 0.10 liter. Themolarityis then molarity= [(number of mole) / {volume (liters)}] = (0.15 moles / 0.10 liters) = 1.5 moles/liter = 1.5 molar = 1.5 M.

Question:

Synthesize CH_2BrCH_2Br from CH_3CH_2OH.

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Solution:

You are given the alcohol, ethanol, and are asked to synthesize CH_2BrCH_2Br , 1,2 dibromoethane. The fact that it is a dibromo compound suggests that the bromine added across a double bond. Therefore, if you could convert the alcohol to an olefin, namely, ethylene, you could add Br_2 in the solvent CCI_4 , and via an addition reaction, obtain the desired product. To obtain ethylene, remember that the alcohol can be dehydrated to the corresponding olefin by the addition of acid. This well-documented mechanism is as follows: When you add the acid, you protonate the alcohol. Water is expelled and you obtain a carbonium ion, CH_3CH_2+, The carbonium expels a proton (H+), so that ethylene is obtained. Now that you have CH_2 = CH_2 , add Br_2 and CCI_4 and you obtain CH_2BrCH_2Br , your desired product. The Br_2 is added across the double bond in an addition reaction.

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Question:

What causes the characteristic sounds of a beating heart?

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Solution:

The heart produces two sounds per heartbeat which can be heard by using a stethoscope, which is an instrument which magnifies sounds and transmits them to the ear. The sound is sometimes de-scribed as "lub, dub, lub, dub..." The first sound, lub, is low- pitched and of long duration. It is associated with the closure of the atrio-ventricular valves (the tricuspid and bicuspid) at the beginning of ventricular systole. The second sound, dub, is higher-pitched, louder and shorter in duration. It is associated with the closure of the semilunar valves at the beginning of diastole. In the heart, the valves are cushioned by the blood, and it is difficult to understand why the valves create sound. It is not due to the slapping together of the "flaps" of the valves. The cause is believed to be the vibrations of the walls of the heart and the vessels around the heart. For example, contraction of the ventricles causes blood to flow against the A-V valves, clos-ing them. The elasticity of the valves causes the blood to surge backwards and bounce back toward the ventricles. This sets the ventricular walls into vibrations, which are carried to the chest wall (where it contacts the heart), creating the sound waves audible in a stethoscope. A very small portion of the first sound is also caused by the contraction of the heart muscle itself. As any muscle contracts, a small amount of noise can be heard as the internal filaments rearrange. The second heart sound is caused by the vibrations of the walls of the aorta, the pulmonary artery, and the ventricles. The closing of the semilunar valves causes the blood in the arteries to rever-berate between the arterial walls and valves. When these vibra-tions of the walls come into contact with the chest wall, an audible sound is created. When the heartbeat sounds normal, it is an indication that the valves are functioning properly. When the valves are damaged, a hissing or sloshing sound can be heard. This is caused by the blood leaking back through the damaged valve. This condition is known as a heart murmur. It is a common result of rheumatic fever, syphilis, and other diseases.

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Question:

200mlof oxygen is collected over water at 25\textdegreeC and 750 torr. If the oxygen thus obtained is dried at a constant temperature of 25\textdegreeC and 750torr, what volume will it occupy? What volume will be occupied by the water vapor removed from the oxygen if it is maintained at 25\textdegreeC and 750 torr? (The equilibrium vapor pressure of water at 25\textdegreeC is 28.3torr.)

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Solution:

This problem will be approached by determining the partial pressures of oxygen and of water vapor in the initial mixture and then using these partial press-ures to determine the separate volumes of each component. The partial pressures exerted by the oxygen and the water vapor must add up to the total pressure. Since the partial pressure of water vapor is equal to its equi-librium vapor pressure, we can determine the partial pressure of oxygen: total pressure = partial pressure of O_2 + partial press-ure of water 750torr= partial pressure of O_2 + 28.3torr or,partial pressure of O_2 = 750torr- 28.3torr = 721.7torr. By definition, this partial pressure is the press-ure the oxygen would exert if it filled the entire 200mlvolume. We must determine the volume this amount of oxygen would fill at 750torr. To do this we use the relationship P_1V_1 = P_2V_2, which is valid for constant temperature and constant number of moles of gas. Let P_1 = 721.7torr, V_1 = 200ml, and P_2 = 750torr. Then, V_2 = [P_1 / P_2] × V_1 = [721.7torr/ 750torr] × 200ml= 192.45ml. Hence, the dried oxygen would occupy a volume of 192.45ml. We follow a similar procedure in calculating the volume that the water vapor would occupy at 750torr. The vapor pressure of water, 28.3 torr, is the pressure the water vapor would exert if it filled the entire 200ml volume. Again, applying the relationship P_1V_1 = P_2V_2, remembering that the only reason we can do so is that we have constant temperature and constant number of moles of gas, we can calculate the volume that the water vapor occupies under a pressure of 750torr. Let P_1 = 28.3torr, V_1 = 200ml, and P_2 = 750torr. Then the volume at 750torris V_2 = [P_1 / P_2] × V_1 =[28.3torr/ 750torr] × 200ml= 7.55ml. As a check, we ascertain that the individual volumes at 750torradd up to the total volume at 750torr: Volume of O_2 + volume of water = 192.45ml+ 7.55ml = 200 ml.

Question:

For the following exothermic reaction at 25\textdegreeC, C_6 H_6(l) + 7(1/2) O_2(g) \rightarrow 6CO_2 (g) + 3H_2O(l), the change in energy, ∆E, is - 780 Kcal/mole. Find the change in enthalpy, ∆H.

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Solution:

The solution is based on the following equation for the enthalpy change ∆H = ∆E + ∆nRT, where ∆n is the number of moles of gaseous products minus the number of moles of gaseous reactants, R is the gas constant (1.987 cal/deg-mole), and T is the absolute tem-perature. Since there are 6 moles of gaseous products (6CO_2(g)) and 7(1/2) moles of gaseous reactants (7(1/2) O_2 (g)), ∆n = 6 - 7(1/2) = - 1.5. The absolute temperature is T = 25\textdegree + 273 = 298\textdegreeK. Hence, ∆H= ∆E + ∆nRT = - 780 Kcal + (- 1.5 mole) (1.987 cal/deg-mole) × (298\textdegreeK) = - 780 Kcal - 888 cal = - 780 Kcal - 0.888 Kcal \cong - 781 Kcal.

Question:

What is the function of the leaves? Describe the structures of a typical dicot leaf.

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Solution:

The leaf is a specialized nutritive organ of the plant. It's function is to carry out photosynthesis, a process requiring a continuous supply of carbon dioxide, water, and radiant energy. The leaf of a typical dicot consists of a stalk, called the petiole, attached to the stem, and a broad blade, which may be simple or compound. The petiole, like a stem in cross-section, contains vascular bundles. Within the blade, the vascular bundles fork repeatedly and form a network of veins. A cross-section of the leaf shows that it is composed of several types of cells. The outer cells, lining both the top and bottom surfaces of the leaf, make up the epidermis which secretes a protective, waterproof cutin covering. Distributed throughout the epidermal surface are many small, specialized pores, the stomata, each surrounded by two guard cells. There are many more stomata on the lower surface than the upper surface of the leaf in most species. The chloroplast- containing guard cells requlate the size of the stomata by changing their turgor pressure and thus their shape. The stomata allow oxygen to diffuse out and carbon dioxide to diffuse into the leaf. Most of the space between the upper and lower epidermal layers is filled with thin-walled cells, called mesophyll, Most of the space between the upper and lower epidermal layers is filled with thin-walled cells, called mesophyll, which are full of chloroplasts. The mesophyll layer near the upper epidermis is usually made of cylindrical palisade cells, closely packed together with their long axes perpendicular to the epidermal surface. The rest of the mesophyll cells are very loosely packed together, with large air spaces between them. These loosely - packed cells form the spongy layer of the leaf. The cells of the palisade layer, containing abundant chloroplasts, are chiefly responsible for the photosynthetic functioning of the leaf, while the air spaces between the mesophyll of the spongy layer hold moisture and gases. These air spaces are continuous with the stomata where exchange of gases takes place.

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Question:

Calculate (a) the root-mean-square speed of the molecules of nitrogen under standard conditions, and (b) the kinetic energy of translation of one of these molecules when it is moving with the most probable speed in amaxwelliandistri-bution.

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Solution:

(a) Each molecule of an ideal gas may have a velocity with rectangular componentsv_x,v_y,v_z. Hence, the square of the velocity of one molecule is v^2 = (v_x^2 + v_y^2 + v_z^2) If we calculate this for every gas molecule, add the re-sults, and divide by the number of molecules in the box, N, we find (^N\sum_i_=1 v_i^2)/N =v^2=v^2_x+v^2_y+v^2_z where the bar over a quantity indicates an average value. (The summation symbol \sum indicates that we calculate v^2 for each gas molecule and add the results.) Since no direc-tion of motion is referred for the gas molecule, v^2_x=v^2_y=v^2_z andv^2=3v^2_x we call \surd(v^2) the root mean square speed,v_rms. Now, the average kinetic energy of one molecule is related to the temperature T of the gas by (1/2)mv^2= (3/2)kT(1) where m is the mass of one gas molecule, and k is Boltzmann's constant. Then the kinetic energy of 1 mole of molecules is found by multiplying (1) by N_0, Avogadro's number, or (1/2)N_0mv^2_rms = (3/2)N_0kT The gas constant, R, is, however R = kN_0 and mN_0 = M the mass of 1 mole of gas. Hence (1/2)mv^2= (3/2)RTandv^2= 3RT/M If we consider a kg mole of the gas, we have v^2= 3 × 8.31 × 103(joule/\textdegree K\bulletmole K\bulletmole ) × 273\textdegreeK × (1 mole/28kgm) ) × 273\textdegreeK × (1 mole/28kgm) = 2.43 × 10^5 m^2/sec^2 = 2.43 × 10^5 m^2/sec^2 ; v_rms= \surd(v^2) = \surd(2.43 × 10^5 m^2/sec^2) = 492 m/sec. (b)E_k= (1/2)mv^2_m , wherev_mis the most probable speed. The mass of a nitrogen molecule can be calculated from its molecular weight and Avogadro's number: m= M/N_0 = 28 kg/kmole× [(1kmole)/(6.02 × 10^26 molecules)] = 4.64 × 10^-26 kg/molecule. v_m= \surd(8/3\pi)v_rms= .921v_rms HenceE_k= (1/2)mv^2_m = (1/2)m \bullet (.921v_rms)^2 = (1/2)(4.64 × 10^-26 kg)(.921)^2(2.43 × 10^5 m^2/s^2) = 3.76 × 10^-21 Joule.

Question:

The best vacuum that can be produced corresponds to a pressure of about 10^-10 dyne cm^-2 at 300\textdegree K. How many molecules remain in 1 cm^3?

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Solution:

We can use the ideal gas equation to calculate N, the number of molecules in the given volume: N =pV/kT = (10^-10 dyne/cm^2 × 1 cm^3)/(1.38 × 10^-16 dyne-cm/\textdegreeK × 300\textdegree K) \approx 2,500 There is still a large number of molecules left.

Question:

Suppose 100 ml of oxygen were collected over water in the laboratory at a pressure of 700torrand a temperature of 20\textdegreeC. What would the volume of the dry oxygen gas be at STP?

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Solution:

STP means Standard Temperature and Pressure, which is 0\textdegreeC and 760torr. In this problem oxygen is gathered over water, therefore, Dalton's law of partial pressure (each of the gases in a gaseous mixture behaves independently of the other gases and exerts its own pressure, the total pressure of the mixture being the sum of the partial pressures exerted by each gas present; that isP_total= P_1 + P_2 + P_3 \textbullet\textbullet\textbulletP_n) is used to calculate the original pressure of the oxygen. There is both water vapor and oxygen gas present. After you obtain the original pressure of the oxygen, you can use the combined gas law to calculate the final volume of the oxygen. The combined gas law stated that for a given mass of gas, the volume is inversely proportional to the pressure and directly proportional to the absolute temperature. It can be written as follows: [P_1V_1] T_1 = [P_2V_2] / T_2 where P_1 is the original pressure, V_1 is the original volume, T_1 is the original absolute temperature, P_2 is the final temperature, V_2 is the final volume, and T_2 is the final absolute temperature. You are given the temperature in \textdegreeC, so it must be converted to the absolute temperature by adding 273. P_total= 700torr= P_(O)2 + P_(H)2O P_(O)2= 700 - 17.5 = 682.5torr. (b) Converting the temperature to the absolute scale. T_1 = 20 + 273 = 293\textdegreeK T_2 =0 + 273 = 273\textdegreeK (c) Using the combined law [P_1V_1] / T_1 = [P_2V_2] / T_2 P_1 = 682.5torrP_2 = 760torr V_1 = 100 mlV_2 = ? T_1 = 293\textdegreeKT_2 = 273\textdegreeK [(682.5torr) (.100 ml)] / [293\textdegreeK ] = [(.760torr) V_2] / [273\textdegreeK] V_2 = [(682.5torr)(100 ml)(273\textdegreeK)] / [(760torr) (293\textdegreeK)] = 83.7 ml.

Question:

One of the most outstanding features of any assembler lan-guage is its close association with the main memory of a computer . Discuss the uses of explicit addresses and state the importance of checking the format of an instruction be-fore coding it.

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Solution:

The process of explicit addressing generally in-volves replacing a symbolin a program by its displacement, its length, and a substitute base register, not necessarily in this order. Going from left to right, the displacementindicates what portion, (how many bytes) of a symbol is to beignored in the calculations. The length tells how many of the remaining bytesare to be used in the calculations, and the substitute base register givesthe address of the symbol relative to the start of the program in main memory. The address of this symbol usually has to be loaded into a registerbefore it is used in an explicit address. The re-gister is referred to asa substitute base register because it is generally not the main base registerof the program. To illustrate this we will add two packed decimal fields, PKDAYS and PKHRS. We will assume PKDAYS to be 8 bytes long andPKHRS to be 6 bytes long. Our intentions are to add the last 3 bytes ofPKHRS to the last 4 bytes of PKDAYS: AK 3(4, 14), 2(3, 7) The above instruction accomplishes our task. Operand 1 is PKDAYS and operand 2 is PKHRS. In each operand, the number outside the parentheses is the displacement D, the one fol-lowing it is the length, L and the last is the substitute base register, B. Therefore, the general format is D( L,B). One should be especially careful when coding the displacement. Note that we are operating on the last 4 bytes of PKDAYS and on the last 3 bytes of PKHRS, but we have used a displacement of 3 and 2 respectively . This is so because the first byte of the symbol is not included in D. As you can see, explicit addressing enables us to op-erate on desired portions of a field. This method of cod-ing is especially useful when table handling is to be done. For example, let us place zeros in a 16 byte table . Each entry in the table will be 4 bytes long, therefore, our zeroing symbol , ZTAB, will be 4 bytes in length. We will name our table TABZERO. Example L 9, = F, 16' STORING LENGTH OF TABLE LA 6, TABZERO LOAD ADDRESS OF TABLE ZERO ZAP 0 (4, 6), ZTAB ZERO 4 BYTES OF TABLE S 9, = F '4' 4 BYTES JUST SET TO ZERO, SO ADJUST REG. 9 TO AMOUNT LEFT C 9, = F '0' SEE IF ZEROING IS COMPLETED BE OUT A 6, = F '4' GO TO NEXT 4 BYTES OF TABLE IF NOT FINISHED B ZERO CONTINUE ZEROING As you can see, through explicit addressing, we are able to operate on a table with great ease. The general operational rules for string handling and mathematics also apply to ex-plicit coding. Thus, we should be careful of such things as overflows. When coding explicit operands you must remember that different instructions have different formats, and so, what can be coded in some instructions cannot be coded in others. Each explicit value is critical to the resulting object code, so the formats must be checked to make sure each explicit value means the right thing.

Question:

Write a program segment in C to display the following message: Hi from C C is a unique and exciting language

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Solution:

To display the above message the program segment is as follows: main ( ) { printf ("Hi from C\textbackslashn"); printf ("C is a unique and exciting language"); } The \textbackslashn in the firstprintf( ) statement is an escape sequence that causes the output line to be advanced. If it is not in-cluded then the program segment becomes, main ( ) { printf ("Hi from C"); printf ("C is a unique and exciting language); } and will output both messages on the same line as follows: Hi from CC is a unique and exciting language

Question:

Determine ∆G\textdegree for the reaction 4 NH_3 (g) + 5O_2 (g) \rightarrow 4NO (g) + 6H_2 O (l) ∆G\textdegree_f of NH_3 (g) = - 4.0 Kcal/mole ∆G\textdegree_f of NO (g) = 20.7 Kcal/mole ∆G\textdegree_f of H_2 O (l) = - 56.7 Kcal/mole

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Solution:

The change in free energy (∆G\textdegree) may be found by subtracting the sum of free energies (∆G\textdegree_f) of the reactants from the free energies of the products. The free energy of formation of pure elements is always 0. When more than 1 mole of a compound is either reacted or formed, the ∆G\textdegree_f of that compound must be multiplied by thestoichiometriccoefficient for the specific compound, ∆G\textdegree = [4 moles × ∆G\textdegree_f of NO (g) + 6 moles × ∆G\textdegree_f of H_2 O (l)] - (4 moles × ∆G\textdegree_fof NH_3 (g) + 5 moles × ∆G\textdegree_f of O_2) G\textdegree = [4 moles × 20.7 kcal/mole + 6 moles × (-56.7 kcal/mole)] - (4 moles × (-4.0 kcal/mole) + 5 moles × 0 kcal/mole) = - 241.4 kcal.

Question:

How many moles are there in one atom?

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Solution:

A mole of atoms is defined as containing Avogadro's Number of atoms. Avogadro's number is 6.02 × 10^23. Therefore, the number of moles in one atom is equal to 1 atom divided by 6.02 × 10^23 atoms/mole. No. of moles = [(1 atom)/(6.02 × 10^23 atoms/mole)] = 1.66 × 10^-24 moles.

Question:

Explain the mechanism of the genetic determination of sex in man .

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Solution:

The sex chromosomes are an exception to the general rule that the members of a pair of chromosomes are identical in size and shape and carry allelic pairs. The sex chromosomes are not homologous chromosomes . In man, the cells of females contain two identicalsex chromosomes or X chromosomes. In males there is only one X chromosome and a smaller Y chromosome with which the X pairs during meiotic synapsis. Men have 22 pairs of ordinary chromosomes (autosomes), plus one X and one Y chromosome, and women have 22 pairs ofautosomesplus two X chromosomes. Thus it is the presence of the Y chromosome which determines that an individual will be male. Although the mechanism is quite complex, we know that the presence of the Y chromosome stimulates thegonadal medulla , or sex-organ forming portion of the egg; to develop into male gonads , or sex-organs. In theabscenceof the Y chromosome, and in the presence of two X chromo-somes, the medulla develops into female gametes . [Note, that a full complement of two X chromosomes are needed for normal female development.] In man, since the male has one X and one Y chromosome, two types of sperm, or male gametes, are produced during spermatogenesis (theprocess of sperm formation, which includes meiosis) One half of the sperm population contains an X chromosome and the other half contains a Y chromosome. Each egg, or female gamete, contains a single X chromosome . This is because a female has only X chromo-somes, and meiosis produces only gametes with X chromo-somes. Fertilization of the X-bearing egg by an X-bearing sperm results in an XX, or female offspring . The ferti-lization of an X-bearing egg byaY-bearing sperm results in an XY, or male offspring. Since there are approximately equal numbers of X- and Y- bearing sperm, the numbers of boys and girls born in a population are nearly equal.

Question:

A light emitting diode (LED) used to display a number in decimal form has the following 7-segment configuration. Segment a lights up when input a is at logic level 1 and, segment b lights up when input b is at logic level 1, etc. Design a decoder which will accept a 3 bit number and de-code it to a seven bit number to be accepted by the LED. The LED will then display the decimal form of the three bit number.

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Solution:

To design this decoder we can begin by setting up a table of the input possibilities. Let the 3 bit input number be represented by ABC, C being the least significant digit. INPUT EQUIVALENT DECIMAL SEGMENTS NEED TO LIGHT UP ABC NUMBER a b c d e f g 000 0 1 1 1 1 1 1 0 001 1 0 1 1 0 0 0 0 010 2 1 1 0 1 1 0 1 011 3 1 1 1 1 0 0 1 100 4 0 1 1 0 0 1 1 101 5 1 0 1 1 0 1 1 110 6 1 0 1 1 1 1 1 111 7 1 1 1 0 0 0 0 Note: 3 bits can give you a maximum of 8 possibilities. Next we find which numbers light up a given segment, in other words, which numbers are common to a given segment. From the table we find: a = \sum(0, 2, 3, 5, 6, 7) or a = \sum (000, 010, 011, 101, 110, 111) b = \sum(0, 1, 2, 3, 4, 7) or b = \sum (000, 001, 010, 011, 100, 111) c = \sum(0, 1, 2, 3, 4, 5, 6, 7) or c = \sum (000, 001, 011, 100, 101, 110, 111) d = \sum(0, 2, 3, 5, 6) or d = \sum (000, 010, 011, 101, 110) e = \sum(0, 2, 6) or e = \sum (000, 010, 010) f = \sum(0, 4, 5, 6) or f = \sum (000, 100, 101, 110) g = \sum(2, 3, 4, 5, 6) or g = \sum (010, 011, 100, 101, 110) Next we minimize the logic for each segment a through g using K-maps: Now the logic diagram of figure 1 is synthesized using the simplified segment equations a through g.

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Question:

As we have seen in the previous problem, character strings in Pascal have fixed length. However, we treat a string as having a varying length if we ignore all trailing blanks. Consider, for example, the following implementation: CONST STRSIZE = 80; TYPE STRING = RECORD WORD: PACKED ARRAY [1. .STRSIZE] of CHAR; length: 1..strsize END; where the field length represents the position of the last non trailing-blank character. For example, the string 'HELLObbb... 'haslength 5 and 75 trailing blanks, and the string 'GOODbbbBYEbbbb....' has length and 70 trailing blanks . Assuming the above defini-tions, write PROCEDURE SUBSTRING (S1:STRING;var S2:string;p,n: integer); which copies the first n characters, starting with posi-tion p in the string S1 into string S2. For example, if S1='ABCDEFGHIJbbbb.....', then after substring (S1, S2, 4,3 )S2 becomes 'DEFbbb....'.

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Solution:

PROCEDURE SUBSTRING (S1: string;varS2: string, p, n : integer); VAR I, J: integer; BEGIN IF (P+N-1)>strsizethen Write1n ('ERROR-SUBSTRING is out of RANGE') {checkto see if desired substring is within the first one} ELSE BEGIN J: = 1; {initial position of the second string} FOR I: = P to P + N - 1 DO {start copying with position p} BEGIN S2.W0RD[ J]: = S1.W0RD[I] J: = J +1; {increment position count for 2string} END;{all required characters copied, pad the rest of the string S2 with blanks) S2.length: = J - 1; FOR I:=J tostrsizeDO S2.W0RD[I]: '' END END;

Question:

An ideal gas is contained within a volume of 2ft^3, when the pressure is 137 atmosphere and the temperature is 27\textdegree C. What volume would be occupied by this gas if it were allowed to expand to atmospheric pressure at a temperature of 50\textdegree C?

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Solution:

We may use the ideal gas law to analyze the behavior of the ideal gas. PV =nRT where P is the pressure of a container (of volume V) of gas at an absolute temperature T. n is the number of moles of the gas and R is the universal gas constant. Since we are dealing with a fixed mass of gas, we may write PV/T =nR= constant. Alternatively, (P_1V_1)/T_1 = (P_2V_2)/T_2 where (V_1,P_1,T_1) AND (V_2,P_2,T_2) are the conditions which describe the behavior of the gas before and after it expands, respectively. Since T_1 = 273\textdegree + 27\textdegree = 300\textdegreeK and T^2 = 273\textdegree + 50\textdegree = 323\textdegreeK then V_2 = V_1 (P_1/P_2)(T_2/T_1) = [2 × (137/1) × (323/300)ft^3] = 295ft^3.

Question:

What pressure is required to compress 5 liters of gas at 1 atm. pressure to 1 liter at a constant temperature?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E02-0028.htm

Solution:

In solving this problem, one uses Boyle's Law: The volume of a given mass of gas at constant temperature varies inversely with the pressure. This means that, for a given gas, the pressure and the volume are proportional; at a constant temperature, and their product equals a constant. P × V = K where P is the pressure, V is the volume and K is a constant. From this one can propose the following equation P_1 V_1 = P_2 V_2 , where P_1 is the original pressure, V_1 is the original volume, P_2 is the new pressure and V_2 is the new volume. In this problem, one is asked to find the new pressure and is given the original pressure and volume and the new volume. P_1 V_1 = P_2 V_2P_1 = 1 atm. V_1 = 5 liters 1atm× 5 liters = P_2 × 1 literP_2 = ? [(1atm× 5 liters)/(1 liter)] = P_2V_2 = 1liter 5 atm. = P_2 .

Question:

A chemist decides to find the vapor pressure of water by the gas saturation method. 100 liters of N_2 gas is passed through 65.44 g of water. After passage of the gas, 63.13 g remained. The temperature of the H_2O (water) is 25\textdegreeC.Find the vapor pressure of water at this temperature.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E07-0253.htm

Solution:

In the gas saturation method, a dry,unreactivegas such as nitrogen or air is bubbled through a specific amount of liquid maintained at constant temperature. After the gas has been bubbled away, the loss in weight of the liquid is determined. The weight loss is the number of grams of liquid in the vapor state. There exists an equation that relates the volume, pressure, weight loss, and molecular weight of the liquid. P = (g / MV)RT, where P = vapor pressure, g = grams of vapor, M = molecular weight of liquid, R = universal gas constant [.0821 (liter- atm )] / (mole \textdegreeK), V = volume, and T = temperature in Kelvin (Celsius plus 273\textdegree) The pressure will be expressed in mm, so that you use the conversion factor of (760 mm) / (atm)g = (65.44) - (63.13) = 2.31atm grams or the weight of the liquid in the vapor state. The molecular weight of water = (18.02 g / mole). Thus, P = (gRT)(MV) = [(2.31 grams) (.0821) {(literatm) / (mole \textdegreeK)} (298\textdegreeK) (760 mm / 1atm)] / [(18.02 g / mole)(200 liters)] = 23.8 mm = vapor pressure of H_2O.

Question:

What is wrong in the following array declaration? int my _ array (200);

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G17-0427.htm

Solution:

Arrays in C are created based on the four data types (int, char, float or double). The basic format of an array declaration is as follows: [number _ of _ subscripts]; The items appearing above i.e. array's __ type and variable _ name and number _ of _ subscripts have to be provided to declare an ar-ray. < > brackets are used only for identity purpose, but they are restricted in C (used only in defining directories). If we check the above statement, int my _ array (200); we note that individual items are as follows: int : array's _ type my _ array: variable _ name 200: number _ of _ subscripts. But the standard notation of enclosing number _ of _ sub-scripts in square brackets i.e. [], instead of which ( ) has been used, is missing. Therefore the correct declaration is: int my _ array [200];

Question:

The world's total reserve of coal and natural gas is estimated to be equivalent to 3.0 × 10^19 Kcal. The world's present consumption of these fuels is 4.2 × 102 7ergs/year. At this rate of consumption, how long will the fuel reserves last?

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Solution:

This problem is an example of the conversion between calories and ergs (1 cal = 4.2 × 10^7 ergs). The total reserve is 3.0 × 10^19 Kcal = 3.0 × 10^19 Kcal × 10^3 cal/Kcal = 3.0 × 10^22 cal, or,3.0 × 10^22 cal × 4.2 × 10^7 ergs/cal = 1.26 × 10^30 ergs. Dividing the total reserve by the rate of consumption gives (1.26 × 10^30 ergs) / (4.2 × 10^27 ergs/year) = 300 years. Hence, at the present rate of consumption, the world's supply of gas and coal will last 300 years.

Question:

For the gas-phase decomposition of acetaldehyde, CH_3CHO\rightarrow CH_4 + CO, the second-order rate constant changes from 0.105 m^-1 sec^-1 at 759\textdegreeK to 0.343 m^-1 sec^-1 at 791\textdegreeK. Calculate the activation energy that this corresponds to. What rate constant would you predict for 836\textdegreeK?

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Solution:

The rate constant is related to the activation energy by the Arrhenius equation, which is stated k =Ae^-E/RT where k is the rate constant, A is a constant characteristic to the reaction, e is the base of natural logarithms, E is the activation energy, R is the gas constant (8.314 J/mole \textdegreeK) natural log: lnk=lnA- E/RT One is given k, R and T in the problem for two trials. For 759\textdegreeK:k = 0.105 m^-1 sec^-1 a)ln(0.105) =lnA - E/(8.314 J/mole \textdegreeK) (759\textdegreeK) For 791\textdegreeK:k = 0.343 m^-1 sec^-1 b)ln(0.343) =lnA - E/(8.314 J/mole \textdegreeK) (791\textdegreeK) Subtract equation b from equation a to solve for E. ln(0.105) =lnA- E/(8.314 J/mole\textdegreeK) (759\textdegreeK) -[ln(0.343) =lnA- E/(8.314 J/mole\textdegreeK)(791\textdegreeK)] C)ln(0.105) -ln(0..343) = - E/(8.314 J/mole\textdegreeK) (759\textdegreeK) + E/(8.314 J/mole\textdegreeK)(791\textdegreeK) -2.25 - (- 1.07) = - E/6310.33 J/mole + E/6576.37 J/mole (6310.33 J/mole)(6576.37 J/mole)(- 1.18) = E/6310.33 J/mole + E/6576.37 J/mole(6310.33J/mole)(6576.37 J/mole) - 4.8969 × 10^7 J^2/mole^2 = - 6576.37 E J/mole + 6310.33 E J/mole - 4.8969 × 10^7 J^2/mole^2 = - 2.6604 × 10^2 JE/mole E = (- 4.8969 × 10^7 J^2/mole^2 ) /(- 2.6604 × 10^2 J/mole ) E = 1.8407 × 10^5 J/mole One can solve for k at 836\textdegreeK by replacing the values of this third trial for the values found at 791\textdegreeK. ln(0.105) -lnk= -[(1.8407 × 10^5 J/mole) /{(8.314 J/mole\textdegreeK)(759\textdegreeK)}] + [(1.8407 × 10^5 J/mole) / {(8.314 J/mole\textdegreeK)(836\textdegreeK)}] - 2.25 -lnk= - 2.917 × 10^1 + 2.648 × 10^1 -lnk= - 4.4 × 10^-1 k = 1.55 m^-1 sec^-1.

Question:

Hydrogen peroxide, H_2 O_2, can be synthesized in two ways. The first method involves reduction of oxygen by hydrogen, H_2 (g) + O_2 (g) \rightarrow H_2 O_2 (l) . The second method involves oxidation of water: 2H_2 O (l) + O_2 (g) \rightarrow 2H_2 O_2 (l). Find the free energy of formation, ∆G\textdegree, for both processes and predict which process is more efficient for the com-mercial preparation of hydrogen peroxide.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E15-0529.htm

Solution:

The standard free energy change for a reaction is equal to the difference between the standard free energy of formation of the products and the standard free energy of formation of the reactants, ∆G\textdegree = ∆G\textdegree _Products- ∆G\textdegree _reactants . The reaction with the more negative free energy change will be the more efficient, since it proceeds spontaneously as written. We will require the following standard free energies of formation: ∆G\textdegree_(H)2 (O)2 (l) = - 27.2 Kcal/mole ∆G\textdegree_(H)2 O (l)= 56.7 Kcal/mole ∆G\textdegree_(H)2 (g)= ∆G\textdegree_(O)2 (g) = O Kcal/mole For the first process H_2 (g) + O_2 (g) \rightarrow H_2 O_2 (l), the standard free energy change is ∆G\textdegree = ∆G\textdegree _products - ∆G\textdegree _reactants = ∆G\textdegree(H)2 (O)2(l) - [∆G\textdegree(H)2 (g)+ ∆G\textdegree(O)2 (g)] = - 27.2 Kcal/mole - (O Kcal/mole + O Kcal/mole) = - 27.2 Kcal/mole For the second process, 2H_2 O (l) + O_2 (g) \rightarrow 2H_2 O_2 (l), the standard free energy change is the standard free energy change is ∆G\textdegree = ∆G\textdegree _products - ∆G\textdegree _reactants = 2∆G\textdegree(H)2 (O)2 (l)- [2∆G\textdegree(H)2 O (l)+ ∆G\textdegree_(O)2 (g)] = 2 (-27.2 kcal/mole) - [2 × (-56.7) kcal/mole + O kcal/mole] = 59.0 Kcal/2 moles H_2 O_2 produced. = 29.5 Kcal/mole. Since the first process proceeds spontaneously as written (negative ∆G\textdegree) and the second process requires energy to proceed as written (positive ∆G\textdegree), the first method for preparing H_2 O_2 is more efficient than the second.

Question:

A motion, starting from rest, has the acceleration - time A motion, starting from rest, has the acceleration - time graph shown in figure (a) Draw the v - t graph and calculate the net displacement.

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/Users/wenhuchen/Documents/Crawler/Physics/D03-0067.htm

Solution:

Between t = 0 and t = 2 sec, a = 2 m/sec^2. Thus \Deltav = a\Deltat = 4 Between t = 0 and t = 2 sec, a = 2 m/sec^2. Thus \Deltav = a\Deltat = 4 m/sec. Thus at t = 2, v = 4 m/sec. Between t = 2 and t = 6, a = 3 m/sec^2; thus \Deltav = 3 × (6 - 2) = 12 m/sec. At t = 6, 4 + 12 =, 16 m/sec and so forth. Having found the velocities at various times and plotting the points as in figure (b), we can connect them with straight lines, since, as we can see from the acceleration - time graph, all accelerations are constant (therefore velocity is a linear function of t). Since displacement equals the product of velocity and time, the net displacement can be found by calculat-ing the area under the v - t curve until t = 10 sec. In figure (c), we break the area under the v-t curve into triangles and trapezoid. The total area under the curve is equal to the sum of the areas of these figures: d= area = (1/2)(4 × 2) + (1/2)(4 + 16) × 4 + 16 × 2 + (1/2)(16 + 8) × 2 = 4 + 40 + 32 + 24 = 100 m

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Question:

Write a program to find the prime numbers between 1 and 50.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G16-0411.htm

Solution:

The classical algorithm for enumerating prime numbers is the sieve of Eratosthenes. Suppose that the prime numbers less than 10 are to be found . The prime numbers less than 10 can be found by writing down the numbers from 2 to 10: 2345678910 Then the lowest number is to be removed claiming that it is prime. The multiples of 2 are also to be removed. After the first step, 2 is a prime, and the sieve contains odd numbers only. 3579. After the second step, 3 is a prime, and only 5 and 7 remain in the sieve. The process terminates when the sieve is empty. Declare CONST maximum = 100000 VAR Sieve: PACKED ARRAY [2. .maximum] ofboolean; Initially, each component of sieve is to be set to 'true\textasteriskcentered indicating that alltbe numbers are present. As the numbers are removed, the corresponding components are set to 'false'. The program consists of two nested loops, one to find the lowest number still in the sieve, and the other to remove its multiples . The termination condition for the outer loop is that there are no numbers left in the sieve, and it can be expedited by maintaining a count of the numbers currently in the sieve. The program is given below. PROGRAM prime numbers (input, output); CONST first prime = 2; maximum = 100000; VAR sieve : PACKED ARRAY[first prime. .maximum] ofboolean; Left in, range, factor, multiple: 0..maximum; BEGIN read (range); FOR factor: = first primeTorange DO sieve [factor]:=true; Left in:=range - first prime +1; factor :=first prime - 1; REPEAT factor : =factor + 1; IF sieve [factor] THEN {factor isprime} BEGIN write1n (factor); multiple: = 1; WHILE factor \textasteriskcentered multiple \leq range DO BEGIN IF sieve [factor \textasteriskcentered multiple] THEN {remove multiple} BEGIN sieve[factor \textasteriskcentered multiple]:= false; leftin:= left in - 1 END; Multiple: = multiple + 1 END {while} END UNTIL left in = 0 END. {primenumbers} Input: 50 Output: 241 3 543 7 1147 13 17 19 23 29 31 37

Question:

A coil having resistance and inductance is connected in series with an ac ammeter across a 100 volt dc line. The meter reads 1.1 am-peres. The combination is then connected across a 110 volt ac 60 cycle line and the meter reads .55 ampere. What are the resistance, the impedance, the reactance, and the inductance of the coil?

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/Users/wenhuchen/Documents/Crawler/Physics/D23-0767.htm

Solution:

Note that with dc by ohm's law, I = (E/R)R = (E/I) R = (110v/1.1A) = 100 ohms (Resistance) On ac, however the reactance of the inductance must also be taken into account (it is zero for dc current). Ohm's law still applies with the impedance taking the place of resistance. Therefore (E/I) = Z = (110/.55) = 200 ohms (Impedance) ButẒ = \surd(R^2 + X^2) where X is the impedance of the inductance. = 100^2 + X^2 = (200\Omega)^2 therefore X^2 =(40000 - 10000)\Omega^2 = 30000 \Omega^2 AndX = \surd(30000\Omega^2) = 173\Omega (Reactance) X = 2\pinL where n is the number of times per second the current alternates. For dc current, n = 0, and the inductance has no effect on the circuit. (The effect of the inductance on the behavior of the circuit increases appreciably as the frequency of the ac current increases). Therefore L = (X/2\pin) = {173\Omega/[2\pi(60)s^-1]} = .459 henry (inductance)

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Question:

What is the function of the contractile vacuole?

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/Users/wenhuchen/Documents/Crawler/Biology/F11-0270.htm

Solution:

Unicellular and simplemulticellularorganisms lack special excretory structures for the eli-mination of nitrogenous wastes. In these organisms, wastes are simply excreted across the general cell membranes. Someprotozoansdo, however, have a special excretory organelle called the contractile vacuole. It appears that this organelle eliminates water from the cell but not nitrogenous wastes. These organelles are more common in fresh waterprotozoansthan in marine forms. This is because, in fresh water forms, the concentration of solutes is greater in the cytoplasm than in the sur-rounding water, and water passively flows into the cell through the cell membrane. The fresh water protozoan is said to live in hypotonic environment. This inflow of water would cause a fatal bloating if the water were not removed by the contractile vacuole. The vacuole swells and shrinks in a steady cycle, slowly ballooning as water collects in it, then rapidly contracting as it expels its contents, and then slowly ballooning again. The exact process of how the cell pumps water out of the vacuole is still unclear, but it is believed that the process is an energy consuming one.

Question:

What are van der Waals forces? What is their significance in biological systems?

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/Users/wenhuchen/Documents/Crawler/Biology/F01-0016.htm

Solution:

Van der Waals forces are the weak attractive forces that molecules of non-polar compounds have for one another. These are the forces that allow non-polar compounds to liquefy and/or solidify. These forces are based on the existence of momentary dipoles within molecules of non- polar compounds. A dipole is the separation of opposite charges (positive and negative). A non-polar compound's average distribution of charge is symmetrical, so there is no net dipole. But, electrons are not static, they are constantly moving about. Thus, at any instant in time a small dipole will probably exist. This momentary dipole will affect the distri-bution of charge in nearby non-polar molecules, inducing charges in them. This induction happens because the negative end of the temporary dipole will repel electrons and the positive end attracts electrons. Thus, the neighboring non-polar molecules will have oppositely oriented dipoles: These momentary, induced dipoles are constantly changing, short range forces. But, their net result is attraction between molecules. The attraction due to van der Waals forces steadily increases when two non-bonded atoms are brought closer together reaching its maximum when they are just touching. Every atom has a van der Waals radius. The atoms are said to be touching when the distance between their nuclei is equal to the sum of their van der Waals radii. If the two atoms are then forced closer together, van der Walls attraction is replaced by van der Waals repulsion (the repulsion of the positively charged nuclei). The atoms then try to restore the state in which the distance be-tween their two nuclei equals the sum of their van der Waals radii. Both attractive and repulsive van der Waals forces play important roles in many biological systems. It is these forces, acting between non- polar chains of phospholipids, which serve as the cement holding together the membranes of living cells.

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Question:

Show that if k is the radius of gyration of a body, then Show that if k is the radius of gyration of a body, then k = \surd(I/m) where I is the body's rotational inertia about a given axis.

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/Users/wenhuchen/Documents/Crawler/Physics/D04-0181.htm

Solution:

The radius of gyration of a rigid body is defined as the radial distance from a given axis at which the entire mass of the body can be concentrated, without altering the object's moment of inertia about the axis. The definition implies 2 equivalent configurations (see figure), either of which may be used to calculate the body's moment of inertia about axis Z. From the discussion above, the moment of inertia of the body of mass m about axis Z in figure (a) is equivalent to the moment of inertia, about Z; of a mass particle m at a radial distance k from Z. (See figure (b)). Hence mk^2 = I ork = \surd(I/m)

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Question:

What trends in the evolution of the life cycle are evident from the algae to flowering plants?

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Solution:

In the evolution from the algae to the flowering plants there is, first, a change from a domi-nating population that is mostly haploid to one that is almost entirely diploid. Inmulticellularfilamentous algae such aUlothrix, only one cell during the life cycle, namely the zygote, is diploid; all the rest are haploid. In mosses, the haploid phase is more conspicu-ous and long-lived than the diploid phase, but the latter has evolved a complex, multicellularplant body. The relative importance of the two phases is reversed in the ferns: the diploid phase is the obvious, larger plant, and the haploid gametophyte, though still an independent plant, is small and inconspicuous. Thegymnosperrasand angiosperms show progressive reductions of the haploid generation. In angiosperms, the male gametophyte consists only of three cells and the female gametophyte of eight cells. The evolutionary advantage of a long-lived, dominant sporophyteand a short-lived, inconspicu-ous gametophyte generation is that a diploid individual can survive despite the presence of a deleterious recessive gene. A haploid individual would be more susceptible to the effects of such genes, since there would be no way to counterbalance the lethality of such an allele. Besides the reduction of the gametophyte, there is a gradual reduction in the dependence of fertilization on the presence of moisture. Aquatic plants such as algae have motile sperm that swim to the eggs. Bryophytes and lower vascular plants, which develop on the ground near water, also require a moist habitat. The seed plants have replaced sperm motility by pollination. Pollen may be very light and carried by the wind over large distances. The angiosperms have further evolved showy floral parts and nectar glands to attract insects, birds and other animals which serve to carry pollen on their feet or fur. Pollination has thus two evolutionary advantages: it allows sperm transport and fertilization to occur without moisture, and it enables long distance dispersal of sperm cells. Greater embryonic protection is a third evolutionary trend in the life cycle of plants. The green algaeChlamydomonashas its zygote protected only within a thick wall. Bryophytes and lower vascular plants have their embryos developing within themulticellulararchegoniumwhich draws nutrients from the gametophyte. Gymnosperms and angiosperms have evolved a seed structure for main-taining and protecting the embryo independent of the gametophyte. The seed has contributed immensely to the success of the seed plants. The stored food nourishes the embryo and the tough outer coat protects it from heat, cold, desiccation, drying, and parasites. Seeds also provide an effective means for the dispersal of the species. They may be wind-borne, water-borne, or carried by animals. To secure even greater protection, the angi-osperms have evolved fruit which encloses one or more seeds. Fruits may either decay or be eaten by animals, liberating the seeds.

Question:

What are the characteristics of ecological succession?

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Solution:

Succession is a fairly orderly process of changes of communities in aregion. It involves replace-ment, in the course of time, of the dominant specieswithin a given area by other species. Communities succeed each otherin an orderly sequence in which each successive stage is thought to bedependent on the one which preceded it. Each community in the successionis called aseralstage. The first stage of succession is usually the colonization of barren spacesuch as a sandy beach or a rock, by simple pioneer species such asgrasses. The pioneer species are able to grow and breed rapidly. They haveadopted the strategy of finding and utilizing empty space. Then, graduallythe pioneers are replaced by more complex and bulky species untilfinally the community is characterized by climax species which representthe final stage of succession. Climax commu-nities are the most stableand will only if some outside agent such as new species or a geographicchange dis-places them. Most succession can be classified into eitherprimary or secondary successions. Primary successions proceed by pioneeringnew uninhabited sites. Secondary successions occur on disturbedland where a climax community that existed before had been destroyedby such occurrences as fire, severe wind storms, flooding and landslides. Although succession in different places and at different times is not identical(the species involved are often completely different) some ecologistshave nevertheless formulated generalizations that hold true for mostcases where bothautotrophsandheterotrophsare involved: 1) The species composition changes continuously during the succession, but the change is usually more rapid in the earlier stages than inthe later ones. 2) The total number of species represented increases initiallyand then becomes fairly stabilized in the older stages. This is particularlytrue of theheterotrophs, whose variety is usually much greater inthe later stages of the succession. 3) Both the total biomass in the ecosystemand the amount of nonliving organic matter increase during the successionuntil a more stable stage is reached. 4) The food webs becomemore complex, and the relations between species in them becomebetter defined. 5) Although the amount of new organic matter synthesizedby the producers remains approximately the same, except at thebeginning of suc-cession, the percentage utilized at the varioustrophic levelsrises. In summary, the trend of most successions is toward a more complexecosystem in which less energy is wasted and hence a greater biomass can be supported.

Question:

The normal boiling point of benzene is 80.10\textdegreeC. When 1 mole of a solute is dissolved in 1000 g of benzene, the boiling point of the resulting solution is 82.73\textdegreeC. When 1.2 g of elemental sulfur is dissolved in 50 g of benzene, the boiling point of the solution is 80.36\textdegreeC. What is the molecular weight of sulfur?

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Solution:

In order to determine the molecular weight of sulfur, we must determine how many moles of sulfur corres-pond to 1.2 g. When 1 mole of solute is added to 1000 g of benzene, the normal boiling point is raised by 2.63\textdegreeC, from 80.10\textdegreeC to 82.73\textdegreeC. This corresponds to a concentration of 1molal(1 mole solute / 1000 g solvent = 1molal). The ratio of sulfur to benzene in the sulfur-benzene solution is (1.2 g sulfur / 50g)benzene = 0.024. Thus, adding 1.2 g of sulfur to 50 g of benzene is equivalent to adding 24 g of sulfur to 1000 g of benzene (because 24 g sulfur / 1000 g benzene is also 0.024). The observed boiling point elevation in the sulfur- benzene solution is 0.26\textdegreeC (from 80.10\textdegreeC to 80.36\textdegreeC).This is the boiling point elevation one would obtain by adding 24 g of sulfur to 1000 g of benzene. But this is about one-tenth the rise one would obtain by adding 1 mole of sulfur (the rise would then be 2.63\textdegreeC) , hence 24 g of sulfur must correspond to about one-tenth of a mole. Thus, the apparent molecular weight of sulfur is (24 g / 0.1 mole) = 240 (g / mole).

Question:

The density of diamond is 3.5 gm/cm^3. What is the order of magnitude of the average velocity and average energy of a carbon atom in diamond at absolute zero?

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Solution:

Density is defined as \varphi = M/V where M is the mass in a volume V. Solving for M M =\varphiV The number of carbon atoms, n, in this mass is M divided by the atomic mass of carbon n = [M/(12amu)] = [\varphiV/(12amu)] Hence the volume occupied by 1 carbon atom is (V/n) = [(12amu)/\varphi] (V/n) = [[(12amu) (1.66 × 10^-27 kg/amu)}/(3.5 kg/m^3)] (V/n) = [(19.8 × 10^-27 m3)/3.5] (V/n) = 5.7 × 10^-27 m^3 Assuming that each carbon atom occupies a cubical volume, the length of one side of this cube, l will be such that l^3 = 5.7 × 10^-27 m^3 l = 1.8 × 10^-10 m The mass of a carbon atom, m, is m = (12amu)(1.66 × 10^-27 kg/amu) m \approx 2 × 10^-26 kg We may now use Heisenberg's uncertainty relationship to find the aver-age velocity of the carbon atom. Since the atom is assumed to be in a cubical volume of length l, the uncertainty in its position is \Deltax= l Hence, using the uncertainty relationship (where\Deltapis the uncertainty in momentumhand is Planck's constant divided by 2\pi) \Deltap\Deltax\geqh \Deltap\geq (h/l) = [(6.63 × 10^-34 J \textbullet s)/{(2\pi) (1.8 × 10^-10 m)}] \Deltap\geq[(6.63 × 10^-34 N \textbullet m \textbullet s)/(11.3 × 10^-10 m)] \Deltap\geq .586 × 10^-24 N \textbullet s \Deltap\geq 5.86 × 10^-25 N \textbullet s The minimum uncertainty in p is then \Deltap_min= 5.86 × 10^-25 N \textbullets But\Deltap_min=m\Deltav_min' since m is constant. Hence \Deltav_min= (\Deltap_min/m) = [(5.86 × 10^-25 N \textbullets)/(2 × 10^-26 kg)] \Deltav_min= 2.93 × 10^1 m/s \Deltav_min= 29.3 m/s This quantity gives us an approximate value for the velocity of the carbon atom. The average energy is then E = (1/2)mv^2 = (1/2) (2 × 10^-26 kg) (29.3 m/s)^2 = (10^-26 kg) (858.5 m^2/s^2) = 8.59 × 10^-24 Joules The zero point energy is the least energy a system can have and occurs at absolute zero. It is E_0 = (1/2)hf and f = (2E_0/h). Assuming that the magnitude of E_0 is the E calculated above, f = [{(2) (8.59 × 10^-24 Joules)} / (6.63 × 10^-34 Joules-sec)] f = [(17.18 × 10^-24) / (6.63 × 10^-34)] s^-1 f = 2.6 × 10^10 s^-1

Question:

What is the radius of the cyclotron orbit in a field of magnetic induction of 1weber/m^2 for an electron traveling with velocity 10^6 m/s in a direction normal to B^\ding{217}? (This radius is termed thegyroradius).

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Solution:

Before starting this problem, we must recognize what type of motion the electron is executing. Since the electron is a charged particle traveling perpendicular to a uniform magnetic field (a field having the same value over all space), the particle will travel in a circle. To find the exact radius of this circular motion, we relate the magnetic force acting on the electron to the electron's acceleration via Newton's Second Law, F^\ding{217} = ma^\ding{217}. The magnetic force for a particle of charge q traveling perpendicular to a magnetic field with velocity v^\ding{217} is F =qvB(1) Using Newton's Second Law, qvB = ma(2) Then because the motion is circular, we know that the acceleration the electron experiences is centri-petal (that is, it always points to the center of thecircle) and has the value (v^2 /R) , where v is the speed of the electron, and R is the radius of its orbit. Substituting this into equation (2), we find qvB = (mv^2 /R)(3) Solving for R, we obtain R = mv /qB(4) Now, the charge of an electron is q = 1.6 × 10^-19 Coulombs, and its mass is m = 9.11 x 10^-31 kilograms. Substituting this information and the values of v and B given in the statement of the problem into (4) R = [{(9.11 × 10^-31 kg) (10^6 m/s)}/{(1.6 × 10^-19 C) (1 w/m^2 )}] R = 5.7 × 10^-6 m

Question:

The pendulum in figure 1 consists of a homogeneous prismatic bar of mass m and length l, which hangs at rest on its frictionless suspension O. A spherical pellet of mass m travels with a horizontal velocity v_1 and strikes it in the middle. Immediately after the impact, the pellet moves with velocity v_2 to the right and the pendulum begins to rotate about O with initial angular velocity \omega. Solve for v_2 and \omega in terms of v_1 for the cases in which the collision is: 1.) completely elastic, and 2.) completely inelastic.

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Solution:

Since not only the normal force, but also the reaction at O, becomes large during the impact, the principle of impulsive forces is useless for the determination of v_2 and \omega. However, here we can apply the principle of impul-sive moments about O. The angular momentum about O of the system of two bodies before the impulse originates from the pellet alone. It is normal to the plane of motion and has the magnitude. H_(0)1 = m(l / 2)v_1 .(1) The total angular momentum immediately after the impulse has the same direction and the magnitude H_(0)2 = m(l / 2)v_2 + (ml^2 / 3)\omega.(2) The second term in equation (2) is the angular momen-tum of the bar, I\omega, where I is the moment of inertia of the bar about O, which is equal to ml^2 / 3. Since no external forces that have a static moment about O become large during the impulse, we have H(0)1= H(0)2or substituting the values from (1) and (2), m(l / 2)v_2 + (ml^2 / 3)\omega = m(l / 2)v_1 .(3)- If the impulse is elastic the kinetic energy of the system is conserved. Thus we have (m/2)^v2 _2 + (1/2) (ml^2 / 3)\omega^2 = (m/2)v^2 _1.(4) Hence, by (3) and (4) we have the following equations for the determination of v_2 and \omega: v_2 + (2 / 3)l\omega= v_1,v2_2 + (1 / 3)l^2\omega^2 = v2_1 . In addition, we have the requirement that (l / 2)\omega \geq v_2. The solution yields \omega = (12 / 7)(v_1 / l) and v_2 = \rule{1em}{1pt}(1 / 7) v_1. Thus, both bodies move after the impulse, and in such a way that the speed of the pellet relative to the pendulum if v^1_2 = (l / 2)\omega \rule{1em}{1pt} v_2 = v_1 to the left. The velocity of the pellet relative to the pendulum is reversed. If the pellet remains imbedded in the pendulum, the impulse is completely inelastic. Then, instead of (4) , we have the condition v_2 = (l / 2)\omega.(5) and, from (3) and (4) it follows that \omega = (6 / 7)(v_1 / l) and v_2 = (3 / 7)v_1.

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Question:

H_2(g) + 2AgCl (s) + 2H_2O \rightarrow 2Ag(s) + 2H_3O^+(Ag) + 2Cl^-(Ag). It has been determined that at 25\textdegreeC the standard free energy of forma-tion of AgCl(s) is -109.7 kJ/mole, that of H_2O(l) is -237.2 kJ / mole, and that of (H_3O^+ + Cl^-) (ag) is - 368.4 KJ / mole. Now assuming the reaction is run at 25\textdegreec and 1 atm of pressure in a cell in which the H_2(g) activity is unity and H_3O+(ag) and Cl^-(ag) activities are .01,what will be the cell voltage?

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Solution:

The answer to this question requires the use of the Nernst equation and the relationship between overall free energy change (∆G) and the cell voltage E. The Nernst equation implies that the voltage of a cell depends on the concentrations or activity of the species involved. It states E = E\textdegree - (RT / nF) 1n Q , where E is the potential for a reaction at nonstandard conditions, E\textdegree = the standard potential, n = number of faradays of electricity, F = the value of a faraday, (for conversion to kilo joules per mole is 96.49 kJ V^-1 equiv^-1 ), T = temperature in kelvin (celsius plus 273\textdegree), R = universal gas constant, (8.31 J mol^-1 deg^-1), and Q is the mass- action expression. The relationship between the overall free energy and cell voltage is ∆G\textdegree = -nFE , where ∆G is the free energy. If all the species are in their standard states, then you can say ∆G\textdegree = -nFE\textdegree . The sum of ∆G\textdegree must be the free energy contribution from each component. Therefore, ∆G\textdegree = 2∆G\textdegreeH3O+ Cl-^ -2∆G\textdegreeAgCl-2∆G\textdegree_H2O = 2(-368.4) - 2(-109.7) - 2(-237.2)kJ = -43.0 kJ per mole of reaction. E\textdegree = (- ∆G\textdegree) / nF = -[(-43.0k J) / {2(96.49kJ / v)}] = .223v. Since the activities of solid Ag, of solid AgCl and of H_2O are unity, they can be eliminated from the mass-action expression: {[H_3O^+]^2 [Cl^-]^2 } / [H_2] . Thus, E = E\textdegree -[RT / nF] ln({[H_3O^+]^2 [Cl^-]^2 } / [H_2]) . Therefore, E = .223 -[{(8.31)(298)(2.303)} / {2(96490)}] log [{(.0100)^2(.0100)^2} / (1.00)] E = .223 + .236 = .459v.

Question:

For the first order decomposition of azomethane at 600\textdegreeK, it takes30 minutes for the original concentration to decrease tohalf its value. After 60.0 minutes have elapsed, what percentageof the azomethane originally present remains?

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Solution:

The decomposition is first order which means rate = k [A], where k isa constant and [A] is the con-centration of azomethane. The half-life is independentof time, i.e., it is a constant and is defined as the time necessaryfor half the particular reactant present to decompose. Thus, if youknew the half-life of azomethane, you could calculate the percentage ofazomethane that remains after a certain period of time. The half-life of azomethaneis 30 minutes. Thus, after 30 minutes, you have half of the originalmaterial left. After 60 minutes half of this original material is left or 1/4 of the original material remains.

Question:

Distinguish between copulation, insemination and fertilization.

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Solution:

Copulation is the act of physical joining of two sex partners. In copulation, accessory sex organs play an important role and synchronized hormonal control is often extensively involved. Insemination is the actual process of depositing semen into the vagina. This occurs during copulation or sexual intercourse, but could also be accomplished by means of artificial injection. When copulation and insemination are complete, fertilization can occur. Fertilization is the union of the male and female gametes to produce the zygote. It should be noted that successful copulation and insemination do not guarantee that fertilization will occur. This may be due simply to the fact that the female has not ovulated and thus provided a mature ovum at the time of insemination. However, it is possible that an ovum may be present, and yet not be fertilized. For fertilization to occur, sperm must not only survive and reach the ovum, they must also succeed in penetrating it. In addition, if one or both partners are sterile, fertilization cannot take place. Sterility refers to any number of specific conditions which result in the failure to produce viable gametes. For example, sterile males may produce viable sperm, but in such low amounts that sufficient numbers do not survive in the female to insure fertilization. Sterile females may produce viable eggs but have a blockage preventing their passage from the ovaries to the oviducts or uterus. Some may produce sperm antibodies in their cervical secretions which resist sperm passage.

Question:

Calculate the kinetic energy of an automobile weighing 2000 pounds and moving at a speed of 50 miles per hour. Record your answer in (a) Joules and (b) ergs.

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Solution:

Kinetic energy is equal to 1/2 mV^2, where m is the mass of the object and V is its velocity. One is given that the mass of the car is 2000 lbs and that its velocity is 50 miles/hr. Solving for the kinetic energy in lbs miles^2/hr^2: kinetic energy = (1/2) (2000 lb) (50 mile/hr)^2 = (1/2) (2000 lb) (2500 mile^2/hr^2) = 2.5 × 10^6 lb mile^2/hr^2 (a) 1 erg = 1 g cm^2/sec^2. Thus, the units of the kinetic energy found must be converted. Pounds must be converted to grams, miles^2 to cm^2 and hours^2 to sec^2. This is done by the use of conversion factors. There are 454 g in 1 lb, thus to convert pounds to grams one must multiply the number pounds by 454 g/lb. There are 1.0 × 10^5 cm in 0.6214 mi, squaring these two quantities, one finds that there are 1.0 × 10^10 cm^2 in .386 mi^2. The conversion factor from mi^2 to cm^2 is then 1.0 × 10^10 cm^2/ .386 mi^2 . There are 3600 seconds in an hour, therefore there are (3600 sec)^2 in 1 hour^2 or 1.296 × 10^7 sec^2 per 1 hr^2. The appropriate conversion factor from hr^2 to sec^2 is (1 hr^2) / (1.296 × 10^7 sec^2 ) Solving for ergs: [2.5 × 10^6 (lb mi^2 / hr^2)] [454 (g/lb)] [(1.0 × 10^10 cm^2 ) / (.386 mi^2)] [(1 hr^2) / (1.296 × 10^7 sec^2 )] = 2.27 ×10^12 (g cm^2 / sec^2 ) = 2.27 × 10^12 ergs. (b) 1 Joule = 10^7 ergs, thus the number of Joules is equal to the number of ergs divided by 10^7. The conversion factor used when converting ergs to Joules is 1 Joule/10^7 ergs. no. of Joules = 2.27 × 10^12 ergs × [(1 Joule) / (10^7 ergs)] = 2.27 × 10^5 Joules.

Question:

A camera photographs a 120-W sodium lamp 100 m away, thefilm being exposed for 1/100s. The aperture of the camerahas a diameter of 4cm, How many photons enter the cameraduring the exposure? The wavelength of sodium lightis 5893 \AA, and all the energy of the lamp is assumed to beemitted as light.

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Solution:

The energy possessed by one photon of sodium light is E =hf, whereh is Planck's constant and f is the photon frequency. But c =\lambdaf where\lambda is the wavelength of light and c is the speed of light. Therefore, hf= h(c/\lambda) = 6.625 × 10^-34 J\textbullets × [(2.998 × 10^8 m\textbullets^-1)/(5893 × 10^-10 m)] = 3.37 × 10^-19 J. The lamp consumes 120 J of energy per second. Therefore the number of photonsemitted per second is n = [(120 J \textbullet s^-1)/(3.37 × 10^-19 J)] = 3.55 × 10^20 s^-1. These photons are emitted in all directions and 100 m from the lamp are distributedevenly over a spherical surface of that radius. The number N enteringthe aperture of the camera per second is thus n multiplied by the fractionof this surface occupied by the aperture. That is, N =n[\pir^2/4\piR^2] whereR is the distance from the lens to the lamp, and r is the aperture radius. Then N = 3.55 × 10^20 s^-1 × [{\pi × (0.02)^2m^2}/{4\pi × (100)^2m^2}] = 3.55 × 10^12 s-1. . Thus in 1/100 s the number of photons entering the camera is 3.55 × 10^10.

Question:

A chemistfinds that an unknown compound contains 50.05 % S and 49.95 % O by weight. Calculate its simplest formula.

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Solution:

To calculate a compound's simplest formula, you need the relative number of moles of atoms in the compound. The percentages by weight of the elements allow for this calculation. If you had 100 g of the unknown compound, it would consist of 50.05 g of S and 49.95 g of O. A mole is de-fined as weight in grams/atomic weight. Therefore, the number of moles of sulfur (s) = (50.05 g) / (32.06 g/mole)= 1.561 moles . The number of moles of oxygen = (49.66) / (15.999 g/mole) = 3.122 moles. You see that the ratio of S to O is 1.561 / 3.122 = 1 / 2 ; 1 ; 2. Therefore, the simplest formula can be expressed as SO_2.

Question:

Why are organ transplants generally unsuccessful?

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Solution:

Under normal circumstances, if a piece of skin, or some organ such as the heart or kidney, is trans-planted from one person to another, it initially appears to be accepted. But within two weeks, the graft (trans planted tissue) is invaded by white blood cells and the tissue dies. This graft rejection is an immunological reaction, also known as a primary set reaction. Second grafts from the same donor are rejected within a few days. This accelerated rejection is termed the second set reaction and can be explained on the basis of theclonalselection theory. Althoughhumoral(circulating) antibodies may be involved in the rejection of transplanted tissues and organs, the active agents are intact small lymphocytes.Immunoglobulinson the surface of these lymphocytes form a complex with the surface antigens on the foreign tissue cells. These lymphocytes specifically destroy these cells by secreting toxic substances or by stimulating increasedphagocyticactivity of macrophages. This immunological response is a reflection of the ability of the body's cells (self) to recognize and destroy anything that is foreign (unself). The rejection of tissue from another member of a different species is termed a heterograft reaction.Heterograftsare usually unsuccess-ful. When the donor is a member of the same species, the transplantation, called a homograft, is also usually un-successful. But when the donor and recipient are either identical twins or members of the same purebred strain, the transplantation is usually successful. These trans plants, calledisografts, are successful because the donor and recipient are genetically very similar. The genes that determine the antigens responsible for the trans-plantation immunological response are termed histocompati-bility genes. The greater the number ofhistocompatibility genes that the donor and recipient, have in common, the greater the probability that the transplant will be successful. That is whyautografts, or tissue transplants from one location to another on the same individual, are usually successful. The success ofhomograftsdepend on how well the donor's and recipient's tissues are matched, that is, how well the donor and recipient resemble each other genetically. Sometimes certain tissues of an individual's own body may be rejected, resulting in disease. The immuno-logical mechanism for distinguishing between self and non-self becomes disrupted and the body begins to destroy itself. These autoimmune diseases include rheumatoid arthritis, rheumatic fever and multiple sclerosis.

Question:

An electric kettle contains 2 liters of water which it heats from 20\textdegreeC to boiling point in 5 min. The supply voltage is 200 V and a kWh (kilowatt-hour) unit costs 2 cents. Calculate (a) the power consumed (assume that heat losses are negligible),(b) the cost of using the kettle under these conditions six times,(c) the re-sistance of the heating element, and (d) the current in the element.

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Solution:

The heat gained by the water in being raised to the boiling point is given by the expression H = mc(t_2 - t_1) where c is the specific heat of water (the amount of heat required to raise the temperature of the substance 1\textdegreeC), m is the mass of the water being heated and (t_2 - t_1) is the temperature difference before and after heating. (a) H = 2 × 10^3 cm^3 × 1 g \bullet cm^-3 × 4.18J \bullet g^-1 \bullet C deg^-1 × (100 - 20)C deg = 6.69 × 10^5 J . Since heat losses are neglected, the conservation of energy requires that the heat energy generated be equal to the electrical energy consumed by the kettle. Thus the electric energy E = 6.69 × 10^5 J. The power is the energy consumed per second, which is thus P = (H/\tau) = [(6.69 × 10^5 J)/(5 × 60 s)] = 2.23 x 10^3 J \bullet s^-1 = 2.23 kW.[for IW = 1 J \bullet s^-1 .] (b)The kettle uses 2.23 kW for 5 min each time the water is boiled. When it is used six times, 2.23 kW is used for 30 min = (1/2) hr. The cost is thus 2.23 kW × (1/2) hr × 2 cents \textbullet kWh^-1 = 2.23 cents. (c) The power P consumed is 2.23 kW and the supply voltage V is 200 V. But P = V^2/R, where R is the resistance of the kettle's heating element. R = (V^2 /P) = [(200^2 V^2 )/(2.23 × 10^3 W)] = 17.9\Omega. \Omega (d)But one may also write the power as P = IV, where I is the current through the heating element. I = (P/V) = [(2.23 × 10^3 W)/(200 V)] = 11.2 A .

Question:

Identify the minerals most essential to the body and give the function of each.

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Solution:

About fifteen minerals are known to be essen-tial in the human diet. A few of these are required in minute quantities and are referred to as trace elements. The essential minerals include sodium, chlorine, potassium, magnesium, phosphorus, calcium, iron, copper, manganese and iodine. Sodium chloride is a major salt in blood and other body fluids. Sodium and chloride ions play an important role in maintaining osmotic balance and acid-base balance in the body fluid. Secretions of the digestive tract such as the hydrochloric acid of the stomach and the pancreatic and intestinal juices, contain a large amount of sodium and chloride ions. Potassium and magnesium are required for muscle contraction and for the functioning of many enzymes. Calcium and phosphorus form the major components of bones and teeth, and are thus essential for maintenance of locomotion and support of the body. Phosphorus is also im-portant in the synthesis of DNA, RNA, NAD+ (NADH) , NADP+ (NADPH) and ATP. Moreover,phosphorylation(the addition of phosphate) of sugars and glycerol must take place before they can becatabolizedto produce biologically useful energy. Trace elements such as iodine, iron, copper and zinc generally function as metal or non-organic components of specific enzyme systems. Iodine is a constituent ofthyroxine, the hormone secreted by the thyroid gland. Hemoglobin and thecytochromescontain iron. Small amounts of copper are necessary in the diet to facilitate proper utilization of iron for normal growth and as a component of certain enzymes. Traces of manganese, molybdenum, zinc and cobalt are required for normal growth and as activators of certain enzymes. Fluorine, present in trace amounts in drinking water, is effective in preventing dental decay.

Question:

What are the functions of the xylem and the phloem?

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Solution:

In vascular plants, the vital function of the transportation of food, water and minerals is performed by the vascular system, composed of the xylem and the phloem. The xylem is chiefly concerned with the conduction of water and mineral salts from the roots to the above-ground portion of the plant, where they are used for photosynthesis and other metabolic purposes. The xylem also serves as a means of support in larger vascular plants. It contains several specialized cell types, the two major kinds being the tracheids and vessel elements. Both are elongated cells with thickened cellulose walls heavily impregnated with lignin; both contain no living protoplasm, and hence are dead cells. The tracheids (see Figure 1) have tapering ends and lateral pores, or pits, which al-low water to flow between adjacent cells. Although structurally similar, vessel elements are a more efficient system for conduction than tracheids. At maturity, the end of the vessels are broken down to form a perforation area, so that each individual cell is continuous with the cell above and the cell below. Vessel cells are thought to have evolved from tracheids, having increasing degrees of perforation with evolutionary advancement, (see Figure 2) Indeed, vessel elements are characteristic of the flowering plants and do not occur at all in most of the less advanced gymnosperms. Vessel elements have lateral pits but movement of materials is chiefly through their ends. Carbohydrates are manufactured primarily in the leaves of a plant. They are transported to the other parts of the plant by the phloem, which runs parallel to the xylem throughout the plant body. The phloem consists of elongated, tube-like cells, called sieve-tube elements, with specialized pores at each end (See Figure 3). Unlike the tracheids and vessels, mature sieve-tube elements are living and have a protoplasm, but no nuclei. The sieve-tube-elements are associated with small, narrow cells having dense contents and prominent nuclei, knowns as the companion cells. These cells probably function as the `nuclei' for the sieve - tube elements and make it possible for them to continue to function. Besides conduction, the phloem serves as a supporting tissue, due to the presence of strong fibers in the walls of its cells.

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Question:

A Martian performs an experiment to determine the Martian g with the local type of Atwood's machine (see figure). He hangs two equal weights of mass 0.02 slug over a frictionless pulley and adds a rider of mass 0.002 slug to one side. When the heavier side has descended 2 ft the rider is removed and the system travels 4 ft in the next 3.5s. What value does he obtain for g?

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Solution:

Initially, there are equal masses on both sides of the Atwood's machine (see figure). The apparatus is then in equilibrium. When we add the rider mass (m), the system is no longer in equilibrium, and each mass accelerates, as shown In the figure. After the heavier side drops 2 ft., the system Is moving with a velocity v^\ding{217}, and then we remove the rider. By Newton's First Law, the left side must continue to move with a velocity v\ding{217} since the system is no longer accelerating. As will be shown below, the acceleration of the system is related to the Martian value of g. If we know the velocity attained by the left hand mass at the instant the rider is removed, we can find a numerical value for g. To find the acceleration of the system, we apply Newton's Second Law to each separate mass shown in the figure. Therefore, taking the acceleration of the left hand mass as positive downward, we obtain (M + m)g - T = (M + m)a_1 where a_1 is the acceleration of the left hand mass. Similarly, taking the acceleration (a_2) of the right hand mass as positive upward, we find T - Mg = Ma_2 But a_1 = a_2, since the 2 masses are connected by a string, and, therefore accelerate as a unit, whence (M + m)g - T = a(M + m)(1) T - Mg = Ma(2) To find a, we eliminate T by adding (1) and (2) mg = (2M+m)a whencea = mg/2M + m(3) Since a is constant, we know that a is related to the position (y) and velocity (v) of the left hand mass by v^2 = v2_0 + 2a y - y_0 where y_0 and v_0 are the initial position and velocity of this mass. Since the left hand mass is initially at rest, v_0 =0. v^2 = 2a(y - y_0) Using (3)v^2 = [{(2mg) / (m + 2M)}] {y - y_0} Hence[{(m + 2 M)v^2} / {2m(y - y_0)}] = g and, to find g, we must know v. Note that y - y_0 is the distance traversed during acceleration. But, we observe that the left side must move with constant velocity after the rider is removed (m = 0) since, by (3), a = 0. Since this mass moves 4 ft in 3.5s after the rider is removed, and its velocity is constant, we may write v = (4 ft) / (3.5 s) = (8/7)ft/s Therefore,g = [{(.042 sl)(64/49 ft^2/s^2)} / {(2)(.002 sl)(2 ft)}] g = 6.86 ft/s^2

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Question:

If 2.02 g of hydrogen gas occupies 22.4 liters, calculate the density of hydrogen gas in grams per milliliter.

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Solution:

Density = (weight / volume) = g/ml. One is given the weight as 2.02 g but the volume is given in liters. Therefore, before calculating the density in g/ml,, one must convert liters to milliliters. 1 liter = 1,000 ml; therefore, 22.4 l = 22,400 ml. Solving for the density: density = (2.02 g / 22,400 ml) = 9.0 × 10 ^\Elzbar5g/ml.

Question:

A Rowland ring made of iron has a mean circumferential length of 50 cm and a cross-sectional area of 4 cm^2 . It is wound with 450 turns of wire which carry a current of 1.2 A. The relative permeability of iron under these conditions is 550. What is the magnetic flux through the ring? What would be the flux through the ring if a gap of 2 cm were to be cut in its length, assuming that the flux did not spread from the gap?

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/Users/wenhuchen/Documents/Crawler/Physics/D21-0734.htm

Solution:

Assume that the iron the Rowland ring is operated in the linear region. Then the magnetic induction through a Rowland ring (see fig. A) is given by the formula B = \muH Applying Ampere's circuital law about the circumference (of length l) of the Rowland ring, \int H^\ding{217} \bullet dl^\ding{217} = NI_C H^\ding{217} is constant along the circumference and is parallel to dl^\ding{217}, an infinitesimal element of length. N is the number of windings of current- carrying wire around the ring. The H \int dl = Hl = NI_C H = (NI_C /l) Hence B = \mu (NI_C /l) = K_m \mu_0 (NI_C /l) Where the relative permeability of iron K_m = \mu/\mu_0 . The flux through the ring is \varphi = \int B^\ding{217} \bullet dA^\ding{217} = BA = K_m \mu_0 A (NI_C /l)(1) where A is the cross-sectional area of the ring Since we are given A = 4 cm^2 , N = 450, l = 50 cm = 0.5 m, I_C = 1.2 A and K_m = 550, then \varphi = 550 × 4\pi × 10^-7 N \bullet A^-2 × (4 × 10^-4 m^2 ) × [(450 × 1.2 A)/(0.5 m)] = 2.99 × 10^-4 Wb. Equation (1) may be written as \varphi = [(NI_C )/(l/\muA)] = ( M/R_1 ) where M is the magnetomotive force, and R the reluctance of the ring. This is the magnetic analogue of Ohm's law, with \varphi the analogue of current, M the analogue of emf and R the analogue of resistance (see figure A). When a gap is cut in the ring (see fig. B), the flux may be obtained by use of the magnetic circuit relation \varphi = [M/(R_1 + R_2)], where R_1 and R_2 are the reluctance of ring and gap, respectively (see fig. B), Thus \varphi = [(NI_C )/[(l_1 /\mu_1 A_1 ) + (l_2 / \mu_2 A_2 )] l_1 is the length of the ring minus the gap (= 0.5 cm - .02m = 0.48 m), \mu_1 the permeability of iron ( = 550 \mu_0) , A_1 the cross sectional area of the ring, l_2 the length of the gap (= 0.02 m), \mu_2 = \mu_0 and A_2 = A_1 . Thus \varphi = (450 × 1.2 A) / [{0.48 m/(550 × 4\pi × 10^-7 N \bullet A^-2 × 4 × 10^-4 m^2)} + {0.02 m/(4\pi × 10^-7 N \bullet A^-2 × 4 × 10^-4 m^2)}] = [(450 × 1.2 × 4\pi × 10^-7 × 4 × 10^-4 )/[(0.48/550) + 0.02)] = 1.30 × 10^-5 Wb.

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Question:

Use the laws of Boolean algebra to simplify the following Boolean expression: [(A+B) + (A+C) + (A+D)] \bullet [A \bullet \textasciitilde B] \textbullet (1) + is the inclusive OR, \bullet is AND and N is NOT. Show the switching and gate representations of the resulting expression.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G03-0043.htm

Solution:

One suggested plan of attack is to eliminate redundancy in the expression. Since A appears four times, we try to simplify the terms using A. The law that allows each step is given on the right: [(A+B) + (A+C) + (A+D)] \textbullet [A \textbullet \backsimB] = [(B+A) + (A+C) + (A+D)] \textbullet [A \bullet \backsimB] Comm. of + = [B + (A+A) + C + (A+D) ] \bullet [A \bullet \backsimB] Assoc. of + = [B + A + C + (A+D) ] \textbullet [A \textbullet \textasciitildeB] Idempotent law = [B + C + A + (A+D) ] \bullet [A \textbullet \textasciitildeB] Comm. of + = [B + C + (A+A) + D] \textbullet [A \bullet \textasciitildeB] Assoc. of + = [B + C + A + D] \textbullet [A \textbullet \textasciitilde B] Idempotent law(1) Working on B and \textasciitilde B, (1)= [B + C + A + D] \textbullet [\textasciitilde B \textbullet A] Commutativity of \bullet = [(B + C + A + D) \textbullet \textasciitilde B] \bullet A Associativity of \textbullet = [(B \bullet \textasciitilde B) + (C \textbullet \textasciitilde B) + (A \bullet \backsim B) + (D \bullet \backsim B) \bullet A Distributive law of \bullet over + = [0 + (C \textbullet \textasciitilde B) + (A \textbullet \textasciitilde B) \textbullet (D \bullet \backsim B) \bullet A Law of complement = [(C \textbullet \textasciitilde B) + (A \textbullet \backsim B) + (D \textbullet \textasciitilde B)] \bullet A Identity of \textbullet = [(C + A + D) \textbullet \textasciitilde B) \textbullet A Dist. law of \textbullet over +(2) The switching representation of (2) would be as shown in Fig. 1 Note that the circuit conducts if and only if both A andare closed, regardless of the settings of C and D. The gate representation of (2) would be as shown in Fig. 2. The reduction of statement (1) to statement (2) was tedious but straightforward. It allows an eight-switch circuit to be replaced by a five-switch circuit, or a seven-gate circuit to be replaced by a three-gate circuit. However, the reduction method depended on observation and a "feel" for algebraic manipulation. Fortunately, minimization algorithms exist which do not depend on such subjective methods and can, in fact, be implemented on a computer.

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Question:

Write the micro-operations for the indirect cycle of the computer whose registers and control signals are specified as follows: M - memory PC - program counter MAR - memory address register MBR - memory buffer register OPR - op-code register F } - cycle identification registers R } I - indirect register M - memory The cycle code is shown in Figure 1. Cycle ID Registers FR Cyclecode Cycledescription 00 C_0 Fetch cycle 01 C_1 Indirect cycle 10 C_2 Execute cycle 11 C_3 Interrupt cycle Fig. 1 Each cycle consists of four steps T_0, T_1, T_2, T_3.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G05-0116.htm

Solution:

The difference between a direct memory reference instruction and an indirect memory reference in-struction is explained with the use of Figure 2. In a direct memory-reference instruction the address por-tion of the instruction format (bits 4 through 15) con-tains the address of the operand; hence, when the fetch cycle is completed, bits 4 through 15 of the MBR contain the address of the operand. The operand is not taken from memory until the execution cycle. In an indirect memory-reference location the address por-tion of the instruction format contains the location of the address of the operand; hence, after the fetch cycle the address of the operand must still be found. Thus, an extra cycle called the indirect cycle is used during an indirect memory reference instruction to fetch the ad-dress of the operand. The indirect cycle is designated by the control signal Instruction format : 013415 I OP-CODE ADDRESS Direct address instruction : c_1 (F = 0, R = 1).A step is specified when the control signal expression to the left of the colon is true (Logical 1). C_1 t_0 :MAR \leftarrow [MBR]_4-15(Step 1) Note: [] means "contents of" In the first step of the indirect cycle the address por-tion of the fetched instruction is transferred to the MAR; thus, the computer is preparing to fetch the address of the operand. c_1t_1 :MBR \leftarrow [M](Step 2) After the second step, the address of the operand has been fetched. .The only thing that remains to be done is to go to the execute cycle. c_1t_2 :(Step 3) c_1t_3 :F \leftarrow 1, R \leftarrow 0(Step 4) The execute cycle control signal cannot be set until the last step. If it were set on step three, control would go to the fourth step of the execute cycle rather than the first step of the execute cycle. (In other words, the computer would skip three steps in the execute cycle.)

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Question:

When a small drop of stearic acid is placed on the surface of water, the liquid spreads out on the water to form a single layer of molecules which are all standing on end, as shown in the figure. In a stearic acid molecule, one end is polar. This occurs in the 0-H bond, since the electron the two atoms share does not orbit an equal time around the individual atoms. The electron spends more time in orbit around the oxygen atom, causing it to be slightly negative and the hydrogen atom to be slightly positive. The charge of the two atoms as a whole remains neutral. Since water molecules (H_2 O) contain 0-H bonds, they are also polarized. The stearic acid and water molecules therefore attract each other causing the acid molecules to stand on end. Estimate the length of the stearic acid molecule if the volume of the drop was 1.56 × 10^-10 m^3 and the area of the stearic acid film is 6.25×10^-2 m^2. Also compute the approximate size of and atom, assuming that the carbon, oxygen, and hydrogen atoms are the same size. The chain is 20 atoms long.

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/Users/wenhuchen/Documents/Crawler/Physics/D31-0925.htm

Solution:

The volume V is related to the area, A, of the liquid and L, the length of the stearic acid molecule, as follows: V = LA The length of the molecule is, therefore, L = V/A = [(1.56 × 10^-10m^3)/(6.25 × 10^-9 m^2)] = 2.5 × 10^-9 m Since the chain is 20 atoms long, (see figure), each atom has a diameter d that is approximately d = L/20 = [(2.5 × 10^-9 m)/(2.0 × 10^1)] = 1.3 × 10^-10 m This method is a very simple way to estimate the length of linear molecules. The technique was suggested independently in 1890 by both W.C. Rontgen and Lord Rayleigh.

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Question:

Why are genetic ratios more reliable when there are large numbers of offspring? Discuss the above with a reference to one inherited character in human beings.

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/Users/wenhuchen/Documents/Crawler/Biology/F26-0693.htm

Solution:

Genetic ratios are often misunderstood. When we say that a certain type of cross yields a three-to-one ratio, this does not mean that for every four offspring, there will always be three of one type and one of another. A ratio is worked out on the basis of mathematical pro-bability and will be approached when large numbers are considered. In tossing a coin there is an equal probability of obtaining either a head or a tail, and the ratio of heads to tails would be 1:1. This is because each toss or event is independent of, and therefore not influenced by, the results of any preceding or subsequent tosses. The 50% chance of obtaining either side of the coin is the pro-bability within which each individual event operates, and does not change, regardless of the number of time the event occurs. However, when the event takes place a large number of times, the results do tend to average out to the expected probability, and the actual ratio approaches the anticipated ratio of 1/2. The same is true for certain genetic events. For instance, the cross between a heterozygous brown-eyed man (Bb) and a blue-eyed woman (bb) would be depicted as the following: p Each offspring produced has a 1/2 chance of being brown eyed and a 1/2 chance of being blue eyed. Thus, the expected ratios are; 1 Bb : 1 bb; or phenotypically, 1 brown : 1 blue. We might expect then, a one-to-one ratio in eye color among their offspring. But suppose their first child had blue eyes. This does not automatically mean that their second child must have brown eyes. That child has, like the first child, an equal probability of being either brown or blue-eyed; either outcome is equally probable regardless of what eye color the first child has. This is because the separation of alleles in gamete formation is, like a coin toss, a purely random event, not effected by preceding events. So, while we know that in this case the egg can only be carrying the gene for blue eyes, (because the female in this case is homozygous for blue eyes) . The gene carried by the sperm that will fertilize the egg has an equal chance of being brown or blue, and we cannot predict which it will be because of the randomness of gene segregation. Therefore, though we might expect a one-to-one ratio for brown-eyed and blue-eyed offspring, the actual ratio may deviate from expected entirely by chance. However - and this is the important point - if we were to tabulate the eye color of thousands of children from many families of parents having the same genotypic combinations as this couple, we would indeed find that close to one half of them will have blue eyes and the other half will have brown eyes. As was the case for the coin toss, the larger our sampling population, the closer our ratio approaches one-to-one, the expected ratio. Our actual ratio approaches the probable ratio, and our results become more reliable.

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Question:

Construct a flowchart describing the problem-solving procedure. Include a description of debugging the program.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G08-0185.htm

Solution:

The novice is apt to imagine that bugs crawl into the program only during the process of converting the written program to a form acceptable to the computer. In fact, errors can arise even during the phase when one is formulating a mathematical solution to the problem. Hence, the process of debugging a program should properly start by studying, once again, the algorithm for solving the problem and then proceed to checking the written program. Finally, the input medium (cards, C.R.T., etc.) should be checked for typographical errors. A typical flowchart of the problem-solving procedure might be as follows:

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Question:

In the simple case of the field due to a single point charge q, check the method for obtaining E from \varphi See the figure.

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/Users/wenhuchen/Documents/Crawler/Physics/D18-0585.htm

Solution:

Choose the points A and B to be on the same radius, in which case the component of E in the direction AB^\ding{217} is the total field. Let A be at a distance R from q and B at a distance (R + \sigma), where \sigma is very small. In general, work = \intF^\ding{217} \textbullet dr^\ding{217} where the integral is evaluated over the path of travel of the object which we are working on. For this problem, F^\ding{217} and dr^\ding{217} are parallel because we are moving q along a radius. Hence work =(\sigma + R)\int_R F dr Now, because \sigma is very small, F will be essentially constant along the path of travel, and we may replace it by its average value, F^\ding{217}. Therefore work \approxF\sigma(1) (This relation becomes more exact as \sigma gets smaller.) Also note that, by definition of the potential difference, between R and R + \sigma, we obtain work = q∆v Substituting this n (1). We find q∆V =F\sigma ButF= qE Hence ∆V =E̅̅̅\sigma orE = ∆V / \sigma(2) As \sigma \rightarrow0, we will obtain E, the exact value of the electric field intensity. The potential at A is by definition \varphi^A _e = q / R The potential at B is \varphi^B _e = q / R + \sigma Using equation (2) ,E= Using equation (2) ,E= [(q / R) \rule{1em}{1pt} {q / (R + \sigma)}] / \sigma = 1 / \sigma [{(q(R + \sigma)) / (R(R + \sigma))} \rule{1em}{1pt} {qR / ((R + \sigma)R)}] = 1/ \sigma [(qR+q\sigma-qR) / {R(R + \sigma)}] = q / [R(R +\sigma)] = q / [R^2{1 + (\sigma / R)}] Taking the limit as \sigma \rightarrow 0, we find E = q / R^2 for the value of E at a distance R from q.

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Question:

Working-Storage entries describing two heading lines are shown in fig. 2. Description of report items, including data- names and starting columns to line up with the headings, are given in the table of fig. 1. Use these descriptions of edited items to write a record description for TRANS-LINE.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G11-0265.htm

Solution:

In writing the record description of fig. 3, defined in WORKING STORAGE SECTION and the data table, the VALUE clause is used to specify the appearance of spaces. Observe that the solution is following the data table step by step. The editing characters used in the PICTURE of each data item are described in fig. 4. Data-Name Starting Column Description C-NUMB 9 6 digits; insert 0 between sets of two C-NAME 26 20 characters; insert blank bet. 12th. and 13th. C-CODE 55 1 digit C-AMT 64 6 digits, 2 to right of decimal point; all digits replaced with blanks if zero. C-NEW-BAL 79 6 digits, 2 to right of decimal point; comma in appropriate position, fixed dollar sign, sign if value negative, item all asterisks if zero. C-DUE 96 5 digits, 2 to right of decimal point; dollar sign just before first nonzero digit. Blank if zero. C-DATE 110 3 sets of two-digit numbers; insert blanks between sets. Fig. 1. 0 B Z \textbullet $ \textasteriskcentered Inserts a zero at this position Inserts a blank at this posi-tion Replaces leading zeros with blanks up to the amount ofZ's used. Inserts a decimal point in this position Inserts dollar sign at this position if just one, or at farthest right position for more than one, suppressing leading zeros. Replace leading zeros with as-terisks up to the amount of asterisks data PICTURE printed 5432 AB26 00265 00023 265 125 0026 00015 0003 0002 9099 XXB9(2) Z 9 (5) Z Z Z Z Z 9.99 $9(4) $$$$ $$9(5) $\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered \textasteriskcentered\textasteriskcentered 50432 AB 26 0265 23 2.65 $125 $26 $015 $\textasteriskcentered\textasteriskcentered\textasteriskcentered3 \textasteriskcentered\textasteriskcentered02 fig. 4

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Question:

Suppose someone told you that plants bend toward the light so that their leaves will be better exposed for photosynthesis. Tell how you could demonstrate by experiments that this may not be a scientific explanation of the growth toward the light.

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/Users/wenhuchen/Documents/Crawler/Biology/F09-0229.htm

Solution:

If plants bend toward the light in order to have their leaves exposed to sunlight, one would assume then, that the signal for stem-bending comes from the leaves. In other words, there will be a mechanism present in the leaves which allows them to receive the stimulus of sunlight and consequently direct the stem to elongate and bend in the direction of the light. We can test the validity of this hypothesis by defoliating the plant. If our hypothesis is correct, removing the leaves would also remove the receptors of the stimulus by light and the effectors of the signal to bend, and no bending of the stem should occur. We know, however, that as long as the apex is intact, a defoliated stem will bend toward the light. The coleoptile tip of the oat seedling, for instance, is a non-leaf-bearing structure which has been used in classical demonstrations of phototropism (the growth response to light). Since a stem will bend toward the sun even if it has no leaves, one cannot really say that plants bend toward the sun purposefully in order to better expose their leaves for photosynthesis. Our experiment has shown that the leaves themselves do not hold the controlling key to bending. We can advance a step further and test the role of the tip of the stem in phototropism. If the apex is involved in the bending of the stem toward light, covering the tip from the light or removing it totally should prevent phototropism. It has, in fact, been observed in experiments that both operations do prevent phototropism as well as inhibit growth. Replacing the tip on the decapitated stem actually brings about resumed growth and bending of the stem. Thus it is clear that the apical meristem of the stem rather than the leaves of the plant is responsible for the phenomenon of phototropism.

Question:

Write a subroutine in FORTRAN to compute the square root of the complex number w = u + iv (where i = \surd-1 and where u and v are any real numbers) according to the formula w1/2 = + \surdw = [{(u2 + v2)1^/2 + u} / 2]1^/2 \pm i [{(u2 - v2)1^/2 - u} / 2]1^/2 Note that the imaginary part of w1^/2 is positive if v > 0, but negative if v < 0.

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Solution:

It is convenient to begin the calculation by first computing the modulus of w, which is defined as │w│ = R = (u2 + v2)1^/2 . Once R is evaluated, it is simple to calculate the real part of w1^/2, denoted by the variable REAL, as REAL= ((R + u)/2)1^/2. Then one calculates the imaginary part of w1^/2, denoted by AMAG, as AMAG = ((R - U)/2)1/2 . If v < 0, one must then change the sign of AMAG from positive to negative. This completes the calculation of the real and imaginary parts of w. The logic of these calculations is present-ed in the flowchart. COMPUTER PROGRAM SOLUTION [Note all FORTRAN statements begin in column 7.] SUBROUTINE C0MSQ (U,V,REAL,AMAG) CSQRT IS A LIBRARY FUNCTION FOR TAKING THE CSQUARE ROOT OF A REAL NUMBER R = SQRT (U \textasteriskcentered U + V \textasteriskcentered V) REAL = SQRT ((R + U)/2.) AMAG = SQRT ((R - U)/2.) IF (V.LT.O.O) AM&G = -AMA CABOVE STATEMENT SWITCHES SIGN OF CIMAGINARY PART IF V IS LESS THAN ZERO RETURN END

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Question:

A high-powered rifle whose mass is 5 kg fires a 15-g bullet with a muzzle velocity of 3 × 10^4 cm/sec. What is the recoil velocity of the rifle?

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/Users/wenhuchen/Documents/Crawler/Physics/D06-0304.htm

Solution:

The momentum of the system after the gun has fired must equal the momentum before the gun went off. Originally, the momentum of the bullet and rifle is zero since they are at rest. Using the conservation of momentum equation: (m_1 + m_2)v_0 = m_1v_1 + m_2v_2 = 0 m_1v_1 = - m_2v2 v_1 = -[(m_2v_2)/(m_1)] = - [ {(15g) × (3 × 10^4_ cm/sec)}/ {5 ×10^3g } ] = - 90 cm/sec. This is a sizable recoil velocity and if the rifle is not held firmly against the shoulder, the shooter will receive a substantial "kick". However, if he does hold the rifle firmly against his shoulder, the shooter's body as a whole absorbs the momentum. That is, we must use for m_1 the mass of the rifle plus the mass of the shooter. If his mass is 100 kg, then the recoil veloci-ty (now of the rifle plus shooter) is v_1 = - [ {(15g) × (3 × 10^4cm/sec)} / {5 × 10^3g + 10^5g} ] \cong - 4.5 cm/sec This magnitude of recoil is quite tolerable.

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Question:

For the following voltaic cell, write the half-reactions, designating which is oxidation and which is reduction. Write the cell reaction and calculate the voltage of the cell made from standard electrodes. The cell is Co; Co^+2 ││Ni^+2 ; Ni .

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/Users/wenhuchen/Documents/Crawler/Chemistry/E16-0587.htm

Solution:

The cell reaction is the algebraic sum of the reactions that take place at the electrodes. Every cell has 2 electrodes an anode and a cathode. Oxidation, which is the loss of electrons, occurs at the anode. Reduction, which is the gain of electrons, takes place at the cathode. The cell is always written as solid; ion in solution ││ ion in solu-tion; solid (anode)(cathode) Oxidation and reduction are the half reactions that take place in the cell. For this cell, they are Co \rightarrow Co^+2 + 2e^- (oxidation at anode) Ni^+2 + 2e^- \rightarrow Ni (reduction at cathode) Sum: Ni^2+ + Co \rightarrow Ni + Co^+2 (Cell reaction). Since Co is losing electrons, it provides the oxidation reaction and Ni^+2, gaining these electrons, takes part in the reduction re-action. The voltage of a cell is the sum of the oxidation and reduction potentials in units of volts and is designated by E\textdegree (under standard conditions) . E\textdegree_cell = E\textdegreeoxidation+ E\textdegree_reduction. The voltages of half cell reactions are usually given as the reduction potentials. The oxidation potential is opposite in sign to the re-duction potential. The potentials can be obtained from a table of standard reduction potentials. Ni^+2 + 2e^- \rightarrow Ni , the potential is -.25v, E\textdegree_red = -.25v. For Co^+2 + 2e^- \rightarrow Co, the potential is -.277 v. Since Co \rightarrow Co+2+ 2e^- is the oxidation reaction, E\textdegree_ox equals the negative of -.277v, or E\textdegreeox= .277v. Substituting these values into the equation E\textdegree = E\textdegreeoxid+ E\textdegree_red , one obtains E\textdegree = +.277 + (-.25) = .027 v . Since E\textdegree is positive, the reaction proceeds spontaneously and can be used to supply current.

Question:

Classify the bonds in the following as ionic, ionic - covalent, covalent - ionic and covalent: I_2,LiCl,MgTe, Cl_2O, H_2S.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E17-0623.htm

Solution:

The bonds in these molecules can be classified by considering the electronegativities of the atoms in-volved . Theelectronegativityof any element is the tendency of that element to attract electrons. The greater theelectronegativity, the stronger the attraction. When 2 atoms join, the type of bond formed can be determined by calculating the difference inelectronegativityof the two atoms involved. When the difference is greater than the bond is classified as ionic, if less than 1.7, then the bond is covalent. An ionic bond involves the complete transfer of electrons from 1 atom to another, while in a covalent bond, the electrons are shared between like or similar atoms. Covalent-ionic indicates a mixed character which is more covalent in nature. Ionic - covalent bonds have more of an ionic character. Consider the cases in point. For I_2, there is no difference in electronegativity because the 2 atoms involved in the bond are the same, and hence contains a covalent bond with equal sharing of the electrons. InLiCl, theelectronagetivitiesare Li = 1.0 andCl= 3.0. (The electronegativity values are obtained from anelectronegativitytable). The electronegative difference is 2.0. Therefore, the bond is ionic. One proceeds in a similar manner for the remainder of the compounds. Compounds and Electronegatives Of the Elements Difference in Electronegatives Types of Bond Mg - Te 1.22.1 2.1 - 1.2 = 0.9 Covalent-ionic Cl_2 - O 3.03.5 3.5 - 3.0 = 0.5 Covalent-ionic H_2 - S 2.12.5 2.5 - 2.1 = 0.4 Covalent-ionic

Question:

Water flows at the rate of 300 ft^3/min through an inclined pipe as shown in the figure. At A, where the diameter is 12 in., the pressure is 15 lb/in^2. What is the pressure at B, where the diameter is 6.0 in. and the center of the pipe is 2.0 ft lower than at A?

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/Users/wenhuchen/Documents/Crawler/Physics/D11-0435.htm

Solution:

The mass of liquid entering the tube at point A, in a time ∆t, should be equal to the mass leaving it at point B in ∆t. Let the velocities and the densities at A and B be v_1,v_2 and \rho_1,\rho_2,respectively. In a time ∆t, the liquid entering at A moves a distancev_1∆t, hence the volume of liquid entering is Av_1∆t . Therefore, the mass of this fluid is \rho_1v_1A_1∆t. Similarly, the mass of liquid leaving at B is \rho_2v_2A_2∆t. Dividing both terms by ∆t, we get the continuity equation for the flow \rho_1v_1A_1 = \rho_2v_2A_2 . For all practical purposes, we can assume liquids to be incompressible, therefore the density of liquid during flow remains constant \rho1= \rho_2 and we have v_1A1= v_2A_2 For this problemA_1v1= [(300 ft^3/min)/(60 sec/min)] = 5.0 ft^3/sec Hencev_1 = (A_1v_1)/(A_1) = (A_1v_1)/(\pir^2_1) = (5.0 ft^3/sec)/(3.14 × {1/2 ft}^2) = 6.4 ft/sec v_2= [(A_1)/(A_2-)]v_1 = [\pi(1/2 ft)^2]/[\pi(1/4 ft)^2] v_1 = 4v_1 v_2= 25.6 ft/soc. The pressure P_1 at point A is P_1= (15 lb/in.^2)(144 in.^2/ft^2) = 2200 lb/ft^2 The weight density of water D = 62.4 lb/ft^3, and therefore \rho = 1.94slugs/ft^3. The Bernoulli equation for the pressures at A and B is P_1 - P_2 = \rhogh + (1/2) \rho(v^2_2 - v^2_1) where h is the difference between the heights of the centers of the cross- sections at the two ends of the pipe. The pressure at point B is P_2 = P_1 + \rhog(h_1 - h_2) + \rho/2 (v^2_1 - v^2_2) = 2200 lb/ft^2 + (62.4 lb/ft^3)(2.0 ft) + [(1.94 slug/ft^3)/2][(6.4)^2 - (25.6)^2] ft^2/sec^2 = 2200 lb/ft^2 + 125 lb/ft^2 - 596 lb/ft^2 = 1729 lb/ft^2 = 12 lb/in.^2

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Question:

Electrons of energies 10.20eV, 12.09eV, and 13.06eV colliding with a hydrogen atom, can cause radiation to be emitted from the latter. Calculate in each case the principal quantum number of the orbit to which the electron in the hydrogen atom is raised and the wavelength of the radiation emitted if it drops back to the ground state.

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/Users/wenhuchen/Documents/Crawler/Physics/D33-1009.htm

Solution:

Bohr theory relates the wavelength of the radiation emitted by a hydrogen atom to the principal quantum numbers of the energy levels involved by the equation (1/\lambda) = R [(1/n^2_1) - (1/n^2_2)] or f = (c/\lambda) =Rc[(1/n^2_1) - (1/n^2_2)], where c is the speed of light, f is the frequency of the radiation and R is theRydbergconstant. The energy of the quantum of radiation emitted is E =hf=Rch[(1/n^2_1) - (1/n^2_2)]. (Here, h isPlahcK'sconstant.) This is the energy emitted in the form of a quantum of radiation when the electron drops from the orbit characterized by the quantum number n_2 to that character-ized by the quantum number n_1 If the electron is raised from the orbit with quantum number n_1 to the orbit with quantum number n_2, it must absorb an equal amount of energy. Before being struck by an electron, a hydrogen atom will be in the ground state with n_1 = 1. It is, in any case, true that if n_1 had a value higher than 1, then E, the absorbed energy, could not be greater than 3.40eV, the limiting energy for n_1 =2 . Thus, if E has the values 10.20eV, 12.09 eV, and 13.06eV, the atom absorbing these energies from the incoming electrons, then E =Rch[(1/n^2_1) - (1/n^2_2)]. E = (1.097 × 10^7 m^-1) (2.997 × 10^8 m \textbullet s^-1) (6.625 × 10^-34 J \textbullet s) [(1/n^2_1) - (1/n^2_2)] E = 2.178 × 10^-18 [(1/n^2_1) - (1/n^2_2)] J Since the given energies are calculated in terms of electron volts, we transform 2.178 × 10^-18 J intoeVnoting that 1 J = 6.242 × 10^18eV whence E = (2.178 × 10^-18 J) (6.242 × 10^18eV/J) × [(1/n^2_1) - (1/n^2_2)] E = (13.60eV) [(1/n^2_1) - (1/n^2_2)] Assuming that n_1 =1, and using the given data, we obtain [1 - (1/n^2_2)] = (10.20/13.60) = (3/4)or n_2 = 2 [1 - (1/n^2_2)] = (12.09/13.60) = (8/9)or n_2 = 3 [1 - (1/n^2_2)] = (13.06/13.60) = (24/25)or n_2 = 5 We interpret these results to mean that an incident electron, of energy 10.20eV, will cause the hydrogen atom electron to jump from n = 1 to n = 2, and similarly for the other 2 cases. In dropping back to the ground state, the wavelength emitted in the three cases will be given from the original equation quoted. Thus (1/\lambda) = R [1 - (1/n^2_2)] (1/\lambda) = R [1 - (1/4)] = (3R)/4or \lambda = 4/(3R) (1/\lambda) = R [1 - (1/9)] = (8R)/9or \lambda = 9/(8R) (1/\lambda) = R [1 - (1/25)] = (24R)/25or \lambda = 25/(24R) Whence \lambda = [4/(3 × 1.097 × 10^7 m^-1)] = 121.57 × 10^-9 m \lambda = [9/(8 × 1.097 × 10^7 m^-1)] = 102.57 × 10^-9 m \lambda = [25/(24 × 1.097 × 10^7 m^-1)] = 94.97 × 10^-9 m

Question:

A + 2B + C = AB + BC. You collect the following data: Concentration(moles/liter) [A] [B] [C] -dA/dt (moles/liter - sec) 1 1.00 1.00 2.00 1.00 2 2.00 1.00 2.00 2.00 3 2.00 2.00 2.00 8.00 4 2.00 2.00 4.00 8.00 (a)Determine the experimental rate law expression. (b) Find the specific reaction rate constant.(c) Calculate the rate of reaction, if the [A], [B], and [c] are, respectively,1.0, 2.0, and 3.0: (d) Speculate on the rate-controlling step, i.e., the slow step.

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Solution:

First write a general rate law for the reaction, and then write a specific law based on the given data. The specific rate constant can be found from a given set of data. The rate with given concentrations is obtainable from the specific rate law. From this, a rate-controlling step can be guessed at. Proceed as follows: (a)To write the rate law, note that, in general, the rate = r = k[A]x[B]^y [C]^z, where the brackets indicate concentrations of A, B, and C, k = specific rate constant and the exponents (x, y, and z) are the orders ofthe re-action with respect to each reactant. The order of a chemical reaction is the number of particles needed to form the activated complex or transition state. To find the actual law, you must determine these exponents. To do this, consult the data. You are told the actual rate of the re-action, since you are given - dA/dt , which is an indication of rate. Thus, to find the order, see what happens when one of the concentrations is changed. In going from (1) to (2) , the concentration of A is doubled from 1.00 to 2.00 . Notice - dA/dt also doubles, from 1.00 to 2.00, which means the re-action is first order, x = 1, with respect to A. In going from (2) to (3), [B] goes from 1.00 to 2.00 (doubles). But notice, - dA/dt quadruples going from 2.00 to 8.00. Thus, the reaction is second order, y = 2, with respect to B. In going from (3) to (4) , [C] is doubled going from 2.00 to 4.00, while - dA/dt remains constant at 8.00. Thus, the reaction is zero order, z = 0, with respect to C, i.e., rate doesn't depend on [C]. You then have, r = k[A][B]^2[C]^0 orr = k[A][B]^2 for the experimental rate law expression. (b)To find k, substitute any of the four sets of data in the experimental rate law expression. rate = (-dA/dt) = k[A][B]^2 For example, let us take the data from line (1). You have 1.00 mole/liter- sec = k[1.00 mole/liter] [1.00 mole/liter]^2 , or k = 1.00. Thus, the value of the specific rate constant k is 1.00 liter^2/mole2- sec. (c)To find the rate of the reaction at the concen-trations [A] = 1, [B] = 2, and [c] =3, substitute these values into the rate law, r = k [A][B]^2, Thus, r = (1)[1][2]^2 = 4.0 moles/liter-sec. (d)You can use previous information gained in this problem to speculate about what the rate-controlling step must be. From part (a) you know the rate does not depend on [C]. Thus, C cannot be involved in the slow step. The rate of the reaction is influenced by [A] and [B], however. The rate-controlling step is probably the combination of B_2 and A,i.e., B_2 + Aslow\rightarrow AB_2.

Question:

Explain how the stack area of a computer's memory oper-ates.

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Solution:

The stack is similar to a plate dispenser in a cafeteria. Figure 1 shows a plate dispenser with yel-low, blue, and red plates. Notice that plates can only be put in or taken from the top of the stack. If someone wants the yellow plate, the red and then the blue plate must be removed before the yellow plate. Notice that the last plate in is always the first plate out. This "last in-first out" (LIFO) rule is characteristic of all computer stacks. Most computers have a specific section in memory which is used as a stack and a specific register called the stack pointer. The stack pointer (SP) contains the ad-dress of the first "empty\textquotedblright memory location in the stack. Figure 2 shows the stack with contents A, B, C, D and E at memory locations 25, 26, 27, 28 and 29 respectively. To remove A the POP instruction is used and the SP is incremented. With A removed the SP still contains the first empty position. To insert F a PUSH instruction is used;\textasteriskcentered the SP is decremented, and F goes into the stack at location 25. When using the stack, care must be taken to ensure that the stack is not in "over fill." If too many Push instructions are used, the stack will grow upward and interfere with a main program. The programmer must always keep track of what is at the top of the stack and how many items are in the stack, otherwise errors will be made. Also, care must be taken to see that not too many items are removed from a stack, since it is im-possible to "pop" an. "empty" stack.

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Question:

The electric field in the space between the plates of a discharge tube is 3.25 × 10^4 newtons / coul. What is the force of the electric field on a proton in this field? Compare this force with the weight of the proton, if the mass of the proton is 1.67 × 10^\rule{1em}{1pt}27 kg and its charge is 1.60 × 10^\rule{1em}{1pt}19 coul

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Solution:

E^\ding{217}, the electric field strength, is defined as the force per unit charge q, or E^\ding{217} = F^\ding{217} / q The magnitude of F^\ding{217} is F = Eq = 3.25 × 10^4 nt/coul× 1.60 × 10^\rule{1em}{1pt}19 coul = 5.20 × 10^\rule{1em}{1pt}15 nt The weight of the proton is given by the force of gravitational attraction on m. By Newton's Second Law, F_g = ma where a is the acceleration of m and F_g is the force of gravitational attraction on m (the weight of m). Be-cause the acceleration of m is due to the earth's gravity, we may write F_g = mg = (1.67 × 10^\rule{1em}{1pt}27 kg)(9.80 m/s^2) = 1.64 × 10^\rule{1em}{1pt}26 nt To compare the force on the proton in the electric field to the weight on a proton (also a force), we take their ratio. Hence F / W = (5.20 × 10^\rule{1em}{1pt}15 nt ) / (1.64 × 10^\rule{1em}{1pt}26 nt) = 3.17 × 1011 This emphasizes the fact that the force caused by an electric field may be much greater than that due to a gravitational field.

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Question:

The net dipole moment of water is 1.84 debyes and the bond angle is 104.45\textdegree. What moment can be assigned to each O-H bond?

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Solution:

To answer this problem, consider the definition of a dipole, how it is expressed quantitatively, and the geometry of a water molecule. A dipole consists of a positive and negative charge separated by some distance. Quantitatively, it is ex-pressed as dipole moment, which is the charge times the distance between the positive and negative centers. The units are debyes. The geometry of a water molecule is not linear. It is represented by FIG. a. The oxygen atom shares its electrons with the hydrogen atoms; covalent bonding. Because the oxygen atom is electronegative, which means it has a tendency to attract electrons, the shared electrons will spend more time on the oxygen atom than on the hydrogen atoms. Thus, a polar covalent bond exists. The oxygen tends to develop a negative charge because the electrons spend more time on the oxygen atom. Since a water molecule is neutral (overall), the charge on the hydrogen atoms become positively charged (Fig. b). Therefore, a dipole exists. The net dipole is represented by Fig. c. As given, the dipole moment is 1.84 D. One is asked to compute the moment of each OH bond. One can represent the net moment and the OH moment as in Fig. d. The net dipole moment is the longer line and bisects the bond angle of 104.45\textdegree. The shorter lines represent the components of the net dipole, which comes from each bond. This can be simplified and drawn as Fig. e. This is the representation for one OH bond. The net dipole must be divided by two (1.84/2 = 0.92), since the total number O-H bonds is 2. The bisected angle of 104.45\textdegree yields 52.22\textdegree. To find X, the unknown OH dipole, use trigonometry. The cos 52.22\textdegree is = adjacent leg / hypotenuse = .92 / X, since a right triangle is present, cos 52.22\textdegree = 0.6127. Thus, equating 0.6127 = 0.92 / X. Therefore, X = .92 / .6127 = 1.50 Debyes.

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Question:

What do each of the embryonic germ layers give rise to?

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Solution:

Ectoderm gives rise to the epidermis of the skin, including the skin glands, hair, nails, and enamel of teeth. In addition, the epithelial lining of the mouth, nasal cavity, sinuses, sense organs, and the anal canal are ectodermal in origin. Nervous tissue, including the brain, spinal cord, and nerves, are all derived from embryonic ectoderm. Mesoderm gives rise to muscle tissue, cartilage, bone, and the notochord which in man is replaced in the embryo by vertebrae. It also provides the foundation for the organs of circulation (bone marrow, blood, lymphoid tissue, blood ves-sels), excretion (kidneys, ureters) and reproduction (go-nads, genital ducts). The mesoderm produces by far the greatest amount of tissue in the vertebrate body. Endoderm gives rise to the epithelium of the digestive tract, the tonsils, the parathyroid and thymus glands, the larynx, trachea, lungs, the bladder and the urethra with its associated glands.

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Question:

Pentobarbital, sometimes prescribed for the treatment of chronic insomnia under the trade name Nembutal, has the following structure: It is toxic in doses of 0.5 g and lethal in doses of 1.5 g. Barbituric acid, denoted by H Bar, and which may be used in the synthesis of pentobarbital, has the following structure: H Bar dissociates according to the following equation: H Bar \rightleftarrows H^+ + Bar^-. It is a weak acid, having a pK_a of only 4.01. Approximately what percent of H Bar molecules are dissociated in a 0.10 N solution?

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Solution:

We will solve this problem by determining the concentrations of H^+ and Bar- formed from the dissociation of H Bar. If we let x denote the equilibrium concentration of H+, then, since one mole of Bar- is formed per mole of H^+ formed, the equilibrium concentration of Bar- is also x. Furthermore, the equilibrium concentration of H Bar is 0.10 - x. Then K_a = [H^+] [Bar-] / [HBar] = x^2/(0.10 - x). Note that we have neglected the contribution to [H^+] from the dissociation of water. Assuming that x is much smaller than 0.10 (a justifiable assumption, since H bar is so weak an acid), the 0.10 - x \allequal 0.10 and we have K_a = x^2/(0.10 - x) \allequal x^2 / 0.10 orx^2 = 0.10 × K_a 0 x = (0.10 × K_a)^1/2. The value of K_a is determined from the pK_a. By definition, pK_a = - log K_a. Hence K_a = 10^-pKa = 10^-4.01, and x = (0.10 x K_a)^1/2 = (0.10 ×10^-4.01)^1/2 = (10^-5.01)^1/2 = 10^-2.50 \allequal 3.2 × 10^-3. The equilibrium concentrations of H^+, Bar-, and H Bar are then [H^+] = x = 3.2 × 10^-3 M, [Bar-] = x = 3.2 ×10^-3 M, and [H Bar] = 0.10 - x = 0.10 - 3.2 ×10^-3 \allequal 0.10 M. The percent dissociation of H Bar is given by % dissociation = {[H^+] / [HBar]}× 100 % = {[Bar-] / [HBar]}x 100 % = [(3.2 × 10^-3 M) / (0.10 M)] × 100 % \allequal 3%.

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Question:

Refer to the accompanying figure and determine the number of unit particles (atoms) in a unit cell of each of the three type lattices. Unit cells in cubic lattices: (a) simple cubic lattice; (b) body- Unit cells in cubic lattices: (a) simple cubic lattice; (b) body- centered lattice; (c) face-centered lattice. centered lattice; (c) face-centered lattice.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E06-0219.htm

Solution:

The following rules may be used to determine the number of unit particles (atoms) associated with each type of cubic unit cell: 1) An atom at a corner contributes 1/8 of its volume to each of eight adjacent cubes. 2) An atom on an edge of a cube contributes 1/4 of its volume to each of 4 adjacent cubes. 3) An atom in the face of a cube contributes 1/2 of its volume to each of 2 adjacent cubes. 4) An atom completely within a cube contributes all of its volume to the unit cell. The number of unit particles in each of the three types of lattices can be found as follows: Simple cubic: In the simple cubic lattice all eight units are of the same variety - they are all at corners, thus rule 1 is used. Total atoms in cell = 8 corners × 1/8 atom/corner= 1 atom Therefore the area of unit cube is equivalent to 1 atom. Body-centered: In the body-centered lattice there are nine atoms contributing to the unit cube. There are 8 corners and one atom completely-within the cube, thus, rules 1 and 4 are used. Total atoms in cell = 8 corners × 1/8 atom/corner + 1 atom in center = 2 atoms. Face-centered: In the face-centered lattice there are 14 atoms contributing to the unit cube. There are 8 corners and 6 faces, thus, rules 1 and 3 will be used. Total atoms in a unit cell = 8 corners × 1/8 atom/corner + 6 faces × 1/2 atom/corner = 4 atoms.

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Question:

What is the energy equivalent in MeV of a particle whose mass is 1 amu?

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Solution:

The energy equivalent is given by the Einstein relation E = mc^2. In M.K.S. units, a mass of 1 amu is 1.66 × 10^-27 kg. Hence its equivalent energy is E = mc^2 = (1.66 × 10^-27 kg)(3 × 10^8 m/s)^2 = 1.49 × 10^-10J Since 1 eV = 1.6 × 10^-19 J, this energy can be expressed in eV E = (1.49 × 10^-10J)/(1.6 × 10^-19J/ev) =9.31 × 10^8 eV = 931 MeV Thus the energy equivalent of 1 amu is 931 MeV.

Question:

CH_3COC_4H_9; C_2H_5OH; CH_3NH_2; (C_2H_5)_2O; CH_3COOC_3H_7; CH_3COOH.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E20-0747.htm

Solution:

All of these compounds are derivatives of hydrocarbons, i.e. compounds that contain only hydrogen and carbon atoms. The derivative of a hydrocarbon is the substitution of elements or radicals for one or more hydrogen atoms. These derivatives may be classified into functional groups, which, to a great extent, determine the properties of the compound. This is the key to naming these compounds, identify the functional groups and name it accordingly. First one must describe these functional groups. Alcohols: They have the general formula R\rule{1em}{1pt}OH, where R is any alkyl group. Amines: They are derivatives of ammonia (NH_3), where one or more H's have been replaced by an alkyl group. Ethers: General formula is R\rule{1em}{1pt}O\rule{1em}{1pt}R', where R may be the same as R'. C_2H_5OH fits into the alcohol classification group. It is named ethanol or ethyl alcohol. CH_3NH_2 fits the description of an amine and is named methylamine. (C_2H_5)_2O can be written as H_5C_2\rule{1em}{1pt}O\rule{1em}{1pt}C_2H_5, which fits the description of an ether. It is named ethyl ether. CH_3COOC_3H_7 indicates that an ester is present, once it

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Question:

Given a solution of 1.00M HC_2H_3O_2 , what is the concentration of all solute species? What is the percentage of acid that dissociated? AssumeK_diss= 1.8 × 10^-5 .

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Solution:

Determine the equilibrium equation, write anequilibriamconstant expression, and substitute the concentrations of the species into the expression. By definition, an acid is a substance which donates protons (H^+). Since water is the only other species in the solution, it must act as a base; it receives protons. HC_2H_3O_2 + H_2O \rightleftarrows H_3O^+ + C_2H_3O_2^- . There is a constant calledK_dissociation_, which measures the extent of the HC_2H_3O_2 donation of protons. For the general reaction, HA + H_2O \rightarrow H_3O+ + A^- ,K_diss= {[A^-][H_3O^+]} / [HA] . For the reaction in this problem, you have {[C_2H_3O_2^-][H_3O^+]} / [HC_2H_3O_2] =K_diss= 1.8 × 10^-5 . You are asked to find these concentrations. Let x = the moles per liter of HC_2H_3O_2 that dissociated. From the chemical equation, you see that x must also be the concentration of H_3O^+ and C_2H_3O_2^- , since for each mole HC_2H_3O_2 of that dissociates, one mole of H_3O^+ and one mole of C_2H_3O_2^- is produced. With this in mind, you can represent HC_2H_3O_2 as 1-x, since you started with 1M of HC_2H_3O_2 and x moles per liter dissociate. You have left 1-x moles per liter. Now substituting, you obtain {[H_3O^+][ C_2H_3O_2^-]} / [HC_2H_3O_2] = 1.8 × 10^-5 = [(x \textbullet x) / (1 - x)] . Solving for x, you obtain x = .0042. Therefore, the concentrations of the species are [HC_2H_3O_2] = 1 - .0042 = .9958M [H_3O^+] = x = .0042M [C_2H_3O_2^-] = x = .0042M The percentage of acid that dissociated is, thus, (.0042/1) × 100 = .42% .

Question:

Write the Born-Haber cycle for the formation of crystalline sodium fluoride (Na^+F^-), starting with solid Na and gaseous F. Then, using the thermochemical data supplied below, de-termine its heat of formation: (1) Na(s)\ding{217} Na(g)∆H = +26.0 Kcal : sublimation (2) F_2 (g)\ding{217} 2F(g)∆H = +36.6 Kcal : dissociation (3) Na(g) \ding{217} Na^+(g) + e-∆H = +120.0 Kcal : ionization (4) F(g) + e^-\ding{217} F^-(g)∆H = -83.5 Kcal : electron addition (5) Na^+(g) + F^-(g)\ding{217} Na^+, F^-(s)∆H = -216.7 Kcal: lattice formation.

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Solution:

The Born-Haber cycle shows that the heat of the reaction is the sum of the ∆H's for the sequence of the reactions used in the formation of crystalline sodium fluoride. The Born-Haber cycle relates reactions which produce the gaseous atoms from solids, those that produce the gaseous ions, and those that indicate the union of these ions to form the crystalline product. Thus, one can write the Born-Haber cycle for this reaction as (The numbers above the arrows correspond to the numbered reactions in the problem.) One is given the energies, i.e., the thermodynamic data, for these processes. If one numbers the reactions in this problem (1) to (5), the reaction becomes (1) +(1/2)(2) + (3) + (4) + (5). The 1/2 value for (2) is derived from the fact that one mole of NaF is required, but, in (2), 2 moles of F(g) atoms are generated. Thus, take 1/2 this number. Therefore, heat of reaction is: ∆H= ∆H (1) + 1/2∆H (2) + ∆H(3) + ∆H(4) + ∆H(5) = 26 + 18.3 + 120 - 83.5 - 216.7 = - 135.9 Kcal / mole.

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Question:

For a uniform sphere with moments of inertia I_1 = I_2 = I_3, use the Euler equations to find the equation of motion of the sphere.

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Solution:

The Euler equation are I_1(d\omega_1/dt) + (I_3 - I_2)\omega_3\omega_2 = N_1 I_2(d\omega_2/dt) + (I_1 - I_3)\omega_1\omega_3 = N_2 I_3(d\omega_3/dt) + (I_2 - I_1)\omega_2\omega_1 = N_3 where the subscript 1 refers to the first principal axis of the sphere, N and \omega are the net external torque and angular velocity of the sphere. Noting that I_1 = I_2 = I_3, we obtain I_1(d\omega_1/dt) = N_1 I_2(d\omega_2/dt) = N_2 I_3(d\omega_3/dt) = N_3 Defining I_1 = I_2 = I_3 = I, we nay write (d\omega_1/dt) = (N_1/I), (d\omega_2/dt) = (N_2/I), (d\omega_3/dt) = (N_3/I)(1) In free motion N = 0, and (1) tells us that \omega = const. The result \omega = const is a special feature of the free rotating sphere.

Question:

Describe the mechanism that prevents food from entering the trachea or nasal pasages during swallowing.

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Solution:

Swallowing is a complex reflex initiated when the tongue forces a bolus of food into the rear of the mouth. Here, pressure-sensitive cells are stimulated to send impulses to the swallowing control center in the medulla, which coordinates the remainder of the process. Swallowing is an all-or-none response, and once initiated, it cannot be stopped. The swallowing reflex is a sequen-tial process in which several responses occur at timed in-tervals determined by nerve connections in the control center. As the bolus is pushed into the pharynx, the soft palate rises and lodges against the back wall of the pharynx, preventing the entrance of food into the nasal cavity. Breathing is temporarily stopped and the glottis, the opening between the vocal cords, is closed as the larynx, the upper region of the trachea, rises. The rear portion of the tongue pushes the bolus down the throat over the epiglottis and the epiglottis is simultaneously bent backwards over the glottis. It is primarily the closure of the glottis, not the folding of the epiglottis, that pre-vents food from entering the trachea. At the top of the esophagus is the hypopharyngeal sphincter. This sphincter is normally closed but opens as food reaches it. The sphincter immediately closes after the food passes it, preventing the food from reversing its direction. Peristalsis, a series of rhythmic wave-like muscular con-tractions, pushes the food through the esophagus into the opening of the stomach.

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Question:

A research worker isolated an unknown compound from one of his reaction products. This compound was a volatile liquid with a boiling point of 65\textdegreeC. A.134 g sample of the liquid was vaporized and the gas was collected. The temperature of the collecting chamber was kept at 98\textdegreeC in order that liquefaction would not occur. After all the sample was vaporized, the volume of the gas was measured as .0532 liters at 98\textdegreeC and 737 mm. Calculate the molecular weight.

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Solution:

We are dealing with the variables m = mass, V = volume, P = pressure, and T = temperature. Whenever this occurs, it indicates that one should employ the ideal gas equation, PV = nRT, where n = moles and R = universal gas constant. An ideal gas is one in which the gas molecules take up no space and do not attract one another. Real gases do not completely meet these conditions, but at values in the neighborhood of standard conditions ( 0\textdegreeC and 1 atm) , the difference in the real and ideal values is small enough such that the ideal gas law is accurate enough to carry out the calculations. One further bit of information is required. Since M.W. (Molecular Weight) does not appear anywhere in the equation, a relation must be found between M.W. and one of the 4 variables. The relation is n = moles = mass/M.W. Substituting this into the ideal gas equation we obtain PV = (m/M.W.) RT Solving for the molecular weight, we obtain: MW = (mRT/PV) m = mass of the sample = .134 g R = gas constant = .082 liter atm/mole - \textdegreeK T = 98\textdegreeC = 273 + 98 = 371\textdegreeK V = 0.0532 l P = 737 mm. For the formula, P must be given in atmospheres. To do this, we multiply 737 mm by the conversion factor of [(1 atm)/(760 mm)].Thus, 737 mm = 737 mm \bullet [(1 atm)/(760 mm)] = [(737)/(760)] atm. MW = [{(0.134) (0.082) (371)}/{(737/760) (0.0532)} = 79.018 g/mole.

Question:

The total speed of a projectile at its greatest height v_1, is \surd(6/7) of its total speed when it is at half its greatest height, v_2. Show that the angle of projection is 30\textdegree.

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Solution:

When a particle is projected as shown in the figure i the component of the velocity in the x- direction stays at all times the same, v_x = v-_0 cos \texttheta_0, since there is no acceleration in that direction, owing to the fact that there is no horizontal component of force acting on the projectile. In the y-direction, the upward velocity is initially v_0 sin \texttheta_0 and gradually decreases, due to the acceleration g acting downward. At its greatest height h, the upward velocity is reduced to zero. The kinematic relation for constant acceleration which does not involve time is used to find the greatest height of the trajectory. It is v^2_f = v^2_i + 2as(1) In this case v_f = 0, v_i, the initial velocity, is v_0 sin \texttheta_0, a = - g and s = h. Then 0 = (v_0 sin \texttheta_0)^2 - 2gh orh = [(v_0 sin \texttheta_0)^2]/2g. The total velocity at the highest point is thus the x-component only. That is, v_1 = v_0 cos \texttheta_0. At half the greatest height, h/2 = (v_0 sin \texttheta_0)^2/4g, the velocity in the y-direction, v_y, is obtained from the equation (1) with v_f = v_y, v_i = v_0 sin \texttheta_0, a = -g, and s = h/2. v^2_y2 = (v_0 sin \texttheta)^2 - 2g (h/2) = (v_0 sin \texttheta)^2 - (1/2) (v_0 sin \texttheta)^2 = (1/2) (v_0 sin \texttheta)^2.(2) In addition, there is also the ever-present x- component of the velocity v_0 cos \texttheta_0. Hence the total velocity at this point is obtained by the Pythagorean theorem, v^2_2 = v^2_x + v^2_y2 = (v_0 cos \texttheta_0)^2 + (1/2) (v_0 sin \texttheta_0)^2 = (v_0 cos \texttheta_0)^2 + (1/2)v^2_0(1 - cos^2 \texttheta_0) = (1/2)v^2_0 + (1/2) (v_0 cos \texttheta_0) 2.(3) Here we used the trigonometric identity sin^2 \texttheta + cos_2 \texttheta = 1. However, we are given that v_1 = \surd(6/7) v_2or(v^2_1)/(v^2_2) = 6/7 . Therefore,[(v_0 cos \texttheta_0)^2]/[(1/2)v^2_0 + (1/2)(v_0 cos \texttheta_0)^2] = 6/7 ; or7(v_0 cos \texttheta_0)^2 = 3v^2_0 + 3(v_0 cos \texttheta_0)^2, or4 cos2\texttheta_0 = 3. One can therefore say that cos \texttheta_0 = \surd3/2or\texttheta_0 = 30\textdegree.

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Question:

A saturated solution of CaF_2 contains .00168g of CaF_2 per 100g of water. Determine theK_sp.

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Solution:

When a salt is added to water, it dissociates into ions. Once a saturated solution is obtained, the addition of more salt re-sults in precipitation of the solid. In such a case, an equilibrium is established between the solid phase and the ions in solution. This equilibrium is expressed as the solubility product constant orK_sp. To determine theK_sp, you must set up an equation that expresses this equilibrium. You must also determine themolarityof the solution.Molarityrefers to the number of moles per liter of solution. The equilibrium reaction is CaF_2 (s) \rightleftarrows Ca^2+ (aq) + 2F-(aq). The products have their oxidation states expressed because they are ions in solution. TheK_spfor the reaction is defined as K_sp= [Ca^2+] [F-]^2 . Because this is a heterogeneous equilibrium, there is no need to in-clude CaF_2 (s) in the expression forK_sp. You include only those substances with variable molar concentrations. The fluoride ion con-centration is squared because 2 moles of fluoride ion are generated in the reaction. In general, then, the exponent of the concentration will equal the number of moles of that ion that is generated. Now that theK_spis defined, determine the molarityof the solution. Sincemolarityis defined as moles of solute per liter of solution, you have M =molarity= [(.00168g) / {78g/mole (CaF_2)}] / [100 ml / (1000ml/1)] Solving, you obtain M = .214 × 10-^2 M. You will notice that the 100 grams of water do not appear in this calculation. The density of water is 1g/ml. Since Density = Mass/Volume, 100g H_2 O has a volume of 100ml. Now that you know themolarity, go back to CaF_2 (s) = Ca^2+ (aq) + 2F-(aq). If CaF_2 (s) has M = .214 × 10-^2 , then by looking at the coefficients, you fina in solution 2.14 × 10-^3 M Ca^2+ and 4.28 × 10-^3 M F- . These results are obtained because for every mole/liter of CaF_2 in solution, 1 mole/liter of Ca^2+ and 2 moles/liter of F- is obtained, Therefore, recalling K_sp= [Ca^2+] [F-]^2 , you have K_sp= (.214 × 10-^2 ) (.428 × 10-^2 )^2 = 3.92 × 10-^8 .

Question:

Draw the structure of the isomer of trinitrobenzene that would be easiest to synthesize from benzene, nitric acid, and sulfuric acid.

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Solution:

The nitration of benzene proceeds quickly if a mixture of concentrated nitric and sulfuric acids is used. For each mole of nitric acid dissolved in sulfuric acid, four moles of particles are formed: 2H_2SO_4 + HNO_3 \rightarrow H_3O+ + 2HSO-_4 + NO_2+ The nitronium ion (NO_2+) is the reagent responsible for the nitration of aromatic compounds. When benzene derivatives undergo substitution, a "directing effect" on the position of the new substituent is observed. The first phase of the nitration reaction is the attack of the nitronium ion on the electrons of the benzene ring (seen in figure A). The loss of a hydrogen atom, or proton, to HSO-_4 (a proton acceptor) leads to the restoration of the original \pi - bonding system. The attack on the benzene ring is generally accomplished by an electron-deficient species such as a positive ion, or the positive end of a polar molecule. The electron withdraw-ing nitro group makes it more difficult for an electrophilic species to add onto the benzene ring; vigorous conditions are required for further nitration of nitrobenzene. The nitro group will "direct" the substitution of additional nitronium ions in a meta orientation. With this in mind, the structure of the isomer of trinitrobenzene is at once known. Each nitronium ion is di-rected to the meta position of the other ion. Thus, the structure is that of figure B and has the formula 1,3,5-trinitrobenzene.

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Question:

A wire of diameter 1 mm carrying a heavy current has a temperature of 1200\textdegreeK when equilibrium has been achieved. We assume that the heat lost from the wire is purely by radiation. The temperature of the sur-roundings is 300\textdegreeK, the resistivity of the wire at this temperature is 5 × 10^-8 \Omega \bullet m, and the temperature coefficient of resistance of the wire is 4 × 10^-3 per C deg. What is the magnitude of the current in the wire?

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Solution:

Since the head is being lost by radiation only, the energy lost per second by a 1 - m length of the wire (i.e. the intensity of radiation emitted by the wire)is given by Stefan's law W =A\sigma(T^4 - T^4 _0 ) , where A is the surface area of the length of wire and \sigma is Stefan's constant. The wire is assumed to radiate as a blackbody (i.e. it emits all of the radiation produced by the excited atoms of the wire). T_0 is the wire temperature and T is the ambient temperature. But this energy is supplied by the current flowing. Thus, if R is resistance of 1 m of the wire, then the energy supplied per second is Power = I^2 R = W =A\sigma(T^4 - T^4 _0 ). But R = (\rhol/A') where A' is the cross-sectional area, l is the length and \rho is the resistivity of the wire at temperature T. It is related to T by the following relation: \rho = \rho_0 [1 + \alpha (T - T_0)], where \rho_0 is the resistivity at 300\textdegreeK, and \alpha is the temperature coefficient of resistance. Hence R = [{\rho_0 [1 + \alpha (T - T_0)] l}/(A')] But, for 1 meter of wire, P = W and I^2 = [{A\sigma(T^4 - T^4 _0)} / ({\rho_0 [1 + \alpha (T - T_0)] l} / {A'})] = [{AA'\sigma(T^4 - T^4 _0)} / {\rho_0 [1 + \alpha (T - T_0)] l}] = [{1m×2\pi×(1/2)×10^-3m×\pi[(1/2)×10^-3 ]^2 m^2 ×5.67×10^-8 W\bulletm^-2 (K deg)^-4 × (1200^4 - 300^4 ) (K deg)} / {5×10^-8 \Omega\bulletm[1 + 4 ×10^-3 (K deg)^-1 (1200 - 300) K deg] × 1m}] = 1258 A^2 . \thereforeI = 35.5A.

Question:

Construct a tree which represents the Polish prefix expres-sion + XA - B12 \div C4. Remember that in Polish notation, all oper-ators have the same precedence; only their position is sig-nificant to the order of operations.

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Solution:

Trees are useful to display the way in which the computer translates the algebraic form of an expression (the way in which you generally enter various formulae) into Polish notation during the process of compiling a program. As a general rule, a tree will have operators (+, -, ×, \div , \uparrow) in its nonterminal nodes and operands (constants or variables) in its terminal nodes. The root node is the exception; it may be thought of as representing the middle operator of an expression. In Polish prefix notation, however, the middle operator actually comes at the beginning of the string. To begin constructing the tree, we number each element to correspond with the number of the node it will occupy. Hence, + × A - B 12 \div C 4 1 2 3 4 5 6 7 8 9 The first element (+) will be the root node. We see that the next element is (x). Remembering that trees are always constructed left to right, this element becomes the left subnode of the root. Since it is also an operator, we will neglect the right subnode for a moment in order to find the two operands which appear on either side of the (x). (See Figure 1.) The following figures should illustrate the growth of the tree in a clearer fashion: At this point, we need to find the operands which should appear on either side of the (-). B and 12, since they are the next two elements, will be those operators. Fig. 2 shows that the left side of the tree is complete: Finally, we notice that the next operator is (\div). The two operands, C and 4, follow directly, so we have the com-pleted tree in Figure 3.

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Question:

What is a resting potential? Describe the chemical mechanism responsible for the resting potential. How can a resting potential be detected?

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Solution:

There is a difference in electrical potential between the inside and the outside of all living cells. For example, the potential difference across the plasma membrane of the neuron is measured to be about 60 milli-volts, the inside being negative with respect to the outside. This potential difference is called the resting potential. The chemical basis for the resting potential is as follows (refer to Figure). By active transport, an energy- requiring process that transfers substances across the plasma membrane against their concentration gradients, the concentration of potassium (K^+) ions is kept higher inside the cell than outside. At the same time there is a lower concentration of sodium (Na^+) ions in the cell interior than the exterior. Moreover, in the resting state the permeability of the plasma membrane is different to K+ and Na^+ ions. The membrane is more permeable to K^+ ions than to Na^+ ions. Hence, K^+ ions can move across the membrane by simple diffusion to the outside more easily than Na^+ ions can move in to replace them. Because more positive charges (K^+) leave the inside of the cell than are replenished (by Na^+), there is a net negative charge on the inside and a net positive charge on the outside. An electrical potential is established across the membrane. This potential is the resting membrane potential. We can measure the resting potential by placing one electrode, insulated except at the tip, inside the cell and a second electrode on the outside surface and con-necting the two with a suitable recording device such as a sensitive galvanometer. The reading on the galvanometer should be approximately 60 millivolts if the cell tested is a neuron. (Different types of cells, such as skeletal and cardiac muscle cells, vary in their values of resting potential.) Note, however, that if both electrodes are placed on the outside surface of the cell, no potential dif-ference between them is registered because all points on the outside are at equal potential. The same is true if both electrodes are placed on the inside surface of the plasma membrane.

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Question:

Calculate the energy, in ergs and eV, of the photon which has just sufficient energy to disintegrate a deuteron. What are its frequency and wavelength?

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Solution:

We can find the minimum photon energy needed to disintegrate the deuteron by assuming that the products of the photon-deuteron collision (a proton and neutron) are at rest. Hence, the conserva-tion of total relativistic energy yields hv + M_Dc^2 = (m_p + m_n)c^2(1) where h is Planck's constant, v is the photon's frequency, and c is the speed of light. Note that the initial energy (left side of (1)) is that of the incident photon, plus the rest mass energy of the target deuteron. The final energy [right side of (1)] is the rest mass energy of the reaction products. hv = (m_p + m_n - m_D)c^2 Since m_p \approx m_n = 1.6747 × 10^-27 kg M_D = 3.3441 ×10^-27 kg hv = [(3.3494 - 3.3441) × 10^-27 kg] × (9 × 10^16 m^2/s^2) hv = 0.0053 × 9 × 10^-11 Joules hv = 4.77 × 10-1 3Joules(2) But1 Joule = 6.242 × 101 8eV hv = (4.77 × 10-1 3J) (6.242 × 101 8J/eV) hv = 2.977 × 10^6 eV From(2) , we obtain v = [(4.77 × 10-1 3J)/(6.63 × 10-3 4J\textbullets)] = 7.2 × 10^20 s^-1 Because the wavelength of the photon is we find \lambda = (c/v) we find \lambda = [(3 × 10^8 m/s) / (7.2 × 10^20 s^-1)] = 4.2 × 10^-13 m Note that the above analysis gives only a lower bound on hv. The reason for this can be seen by examining the figure. In order to conserve momentum, the react-ion products must be in motion, contrary to what we assumed in equation (1). Hence, in reality, hv is greater than the value we calculated, since some of the photon's energy goes into the kinetic energy of the reaction products.

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Question:

Describe the procedures involved in adding and subtracting hexadecimalnumbers.

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Solution:

The process of adding and subtracting hexadecimal (Hex) digits is almostidentical to that of adding and subtracting Arabic digits. Actually, thereare a few ways in which these calculations may be per-formed. The mostefficient one is the method used in the decimal cal-culations. The operationswill be done from right to left. In using this method, each Hex symbolwill be translated to a decimal digit before the calculations, then, uponcompletion of the calculations, each re-sulting decimal digit will be retranslatedto a corresponding Hex symbol. Addition: In decimal calculations, as we go from one column to the next, wecarry units of tens. For example: col2col1col\o 128 88 216 In Hex addition, instead of carrying unitsof ten, we carry units of sixteen. Example 1 col3col2col1col\o 6E88 +1D3 Forcol\o: Hex \rightarrow Decimal \rightarrow Decimal Sum \rightarrow Hex Sum 888 333 11\rightarrow B Note in this column the decimal sum is less than 16, and no units of sixteenare carried. Forcol1: Hex \rightarrow Decimal \rightarrow Decimal Sum \rightarrow Hex Sum 888 D1313 21\rightarrow 16 + 5 Here the decimal sum is greater than 16. After factoring out the units of sixteen(here only one), the remaining digit should be placed in this column. Forcol2: Hex \rightarrow Decimal \rightarrow Decimal Sum \rightarrow Hex Sum E1414 111 15 + 1(fromcol1) 16 \rightarrow 16 + \o One unit of sixteen can be factored out, leaving zero as the remaining digit forcol2. Forcol3: Hex\rightarrowDecimal\rightarrow Decimal Sum \rightarrow Hex Sum 666 6 + 1(fromcol2) 7 \rightarrow7 Thus, the final result is: 6 E 88 + 1 D 3 7 0 5 B Subtraction: In the subtraction procedure the units of sixteen are borrowed insteadof being carried. Example 2: col2col1col\o 35E -2B8 Forcol\o Hex \rightarrow Decimal \rightarrow Decimal Difference \rightarrow Hex Difference E1414 88- 8 6 \rightarrow 6 Forcol1: Hex \rightarrow Decimal \rightarrow Decimal Difference \rightarrow Hex Difference 55(5) + 16 (Borrowed unit fromcol2) B1111 10\rightarrowA The difference incol2 is zero since one unit was borrowed from 3. The finalresult is: 2+16 35E -2B8 A6

Question:

.0324 Faradays (F) liberated .651g of Calcium. What is the atomic weight of Calcium?

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Solution:

A faraday may be defined as the number of coulombs or charges of electricity that liberate one gram-equivalent weight of an element in solution. Note, the equivalent weight of a substance is the amount of substance which 1 mole of electrons can oxidize or reduce into a neutral species. Calcium has an equivalent weight of (1 / 2) its atomic weight, since its oxidation state is +2. In other words, one mole of electrons will neutralize only half of the calcium ions present, since each ion needs two electrons. Thus, if you can find the gram-equivalent weight of calcium, you can find its atomic weight by multiplying it by two. However, you can determine the number of gram-equivalents given the number of faradays of electricity used. You are told that .0324F deposits .651g. By definition, 1F deposits the gram-equivalent weight. You have, then, the following proportion: (.651g Ca) / (.0324 )F = (g-eq.wt. of Ca) / (1F) = 20.1. Hence, 20.1g is the gram-equivalent of calcium. Thus, the atomic weight is found to be 2 × 20.1 = 40.2.

Question:

On a given diet yielding 3000 Kcal per day, a 70 kg man excretes27.0g of urea daily. What percentage of his daily energy requirement is met by protein? Assume that gram of protein yields 4 Kcal and 0.16g of nitrogen as urea.

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Solution:

If 1.0 g of protein yields 0.16 g of urea, and 27.0 g of urea is excreted daily, then the man eats 27. 0 g/0.16 g or 168.75 gof protein a day. Each gram of protein contributes 4 Kcal towards the energy re-quirement. Therefore, 168.75 gof protein produce 4 × 168.75or 675 Kcal.675 Kcal is (675/3000) × 100 or 22.5% of the daily energy requirement.

Question:

Given the function f(x,y,z) below, write f(x,y,z) as a product ofmaxterms. f(x,y,z) = (z +x) (y +z) (x + y + z) (x+y)

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Solution:

If a switching function is specified in Product of Sum form, then it may be expanded to canonical PS form by repeatedly using the theorem given below. THEOREM: (a + b) (a +b) = a Note: in the original function f(x,y,z), the first, second, and the fourth terms can be changed using the theorem defined above. These changes will be; (z +x) = (x+ y + z) (x+y+ z) (y +z) = (x + y +z) (x+ y +z) (x+y) = (x+y+ z) (x+y+z) then the function f(x,y,z) becomes; f(x,y,z) = (x+ y + z) (x+y+z) (x + y +z) \bullet (x+ y +z) (x+y+ z) (x+y+z) \bullet (x + y + z) this is the desired final form of the function, now it can be translated into MAXTERM code by assigning 1's to false variables, and 0's to true variables. This procedure of coding MAXTERMS is illustrated below using the function f(x,y,z); f(x,y,z) = (x+ y + z) (x+y+z) (x + y +z) 100111001 (x+ y +z) (x+y+ z) (x+y+z) 101110111 (x + y + z) 000 In decimal arithmetic each code group represents numbers such as 4,7,1,5,6,0, with the redundant 7 being used only once, and the solution can be written as; f(x,y,z) = M_4 M_7 M_1 M_5 M_6 M_0 or f(x,y,z) = \prod M(0,1,4,5,6,7)

Question:

Using Vein der Waal's equation, calculate the pressure exerted by 1 mole of carbon dioxide at 0\textdegreeC in a volume of (a) 1.00 liter,(b) 0.05 liter.

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Solution:

Real gases are more compressible than ideal gases because their molecules attract each other. The intermolecular attraction is provided for by adding to the observed pressure a term of n^2 a/V^2 in the ideal gas law (PV = nRT) where n is the number of moles, V is the volume, and a is a van der Waal's constant. At very high pressures, real gases occupy larger volumes than ideal gases; this effect is provided for by subtracting an excluded volume nb from the observed volume V to give the actual volume in which the molecules move (b is a constant). Thus, van der Waal's full equation is: [p + (n^2 a/V^2 )] (V - nb) = nRT To solve this problem one must 1) convert the tem-perature from units of \textdegreeC to \textdegreeK by adding 273\textdegree to the \textdegreeC temperature. This is done because T is expressed in the absolute. (P = pressure and R = gas constant.) 2)Find the values for the constants (a) and (b) (these can be found in any reference text). For CO_2; a = 3.592 liter^2 atm/mole^2; b = 0.04267 liter/mole. 3)Sub-stitute the known values into the van der Waal's equation and solve for pressure. Known values: n = 1R = 0.082 [(liter-atm)/(mole-\textdegreeK)]T = 273\textdegreeK (a)V = 1.00 liter.Substituting, [P + {(1)^2 (3.592)}/(1.00)^2 ] [1.00 - (1) (0.04267)] = (1) (0.082) (273) P = [{(0.082) ( 273\textdegreeK)}/(1.00 - 0.04267)] - [(3.592)/(1.00)^2 ] = 23.38 - 3.592 = 19.79 atm (b)V = 0.05 liter.Substitute [P + {(1)^2 (3.592)}/(0.05)^2 ] [0.05 - (1) 0.04267] = (1) (0.082) (273 K) P = [{(0.082) ( 273\textdegreeK)}/(0.05 - 0.04267)] - [(3.592)/(0.05)^2 ] = 3054 - 1437 = 1617 atm.

Question:

Memory cell, Registers, Decoders, Multiplexers.

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Solution:

The memory cell is the basic unit of memory. It can store 1 bit of information (i.e. it can have value "1" in it or value "0" in it). The cell may be either a core memory or a flip-flop. Registers: A group of flip-flops constitutes a Register. A 4 bit Register is made up of 4 flip-flops and holds 4 bits of information. Registers can be used in various ways; for example, as counters, shift registers, Incre-ment/Decrement counters with parallel load, etc. These are temporary storage devices used in the internal storage of intermediate data in the CPU. Decoders: Decoders are devices by which we can divide n number of input lines to it into 2^n output lines; see Figure 1. The decoder is used in conjunction with a truth table, Figure 2. From the truth table we can find the decoder's output for any combination of the inputs; enable and bits A_1, A_2, A_3. INPUTS OUTPUT Enable A_1 A_2 A_3 C1 C2 C3 C4 C5 C6 C7 C8 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 0 0 0 1 0 0 1 0 1 0 0 0 0 0 0 1 0 1 0 0 0 1 0 0 0 0 0 1 0 1 1 0 0 0 1 0 0 0 0 1 1 0 0 0 0 0 0 1 0 0 0 1 1 0 1 0 0 0 0 0 1 0 0 1 1 1 0 0 0 0 0 0 0 1 0 1 1 1 1 0 0 0 0 0 0 0 1 Fig. 2. Truth Table for 3 line to 8 line decoder Multiplexer: This is just the reverse of the Decoder. Given 2^n input lines there will be only one output line, see Figure 3. Here, which input lines proceed to output depends on the bit combination of the n select lines. The truth table for the 4 line to 1 line multiplexer, shown in Figure 3, is given in Figure 4. INPUTS OUTPUT S_0 S_1 X 0 0 A 0 1 B 1 0 C 1 1 D Fig. 4

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Question:

In the change from unicellular organisms to multi-cellular organisms, a number of different events occur. Describe these changes as they are evidenced by the motile-colony series of green algae.

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Solution:

The motile-colony series, or the volvocine series, of Chlorophyta shows a gradual progression from the unicellular condition of Chlamydomonas to elaborate colonial organization. Gonium is an example of the simplest colonial stage. The cells of a Gonium colony are morphologically similar to the unicellular Chlamydomonas. Each cell has a single large cup-shaped chloroplast, containing a pyrenoid which functions in starch synthesis, and an eye spot or stigma which functions in light detection. There are two flagella at the anterior end of each cell, and near the base of the flagella two small contractile vacuoles, which dis-charge rhythmically, expelling excess water from the cell. The Gonium cells are embedded in a mucilaginous matrix. There is coordination of activity between the cells, since the flagella of all the cells beat together, enabling the colony to swim as a unit. Pandorina is at a more advanced stage of develop-ment than Gonium. The colony has anterior and posterior halves, and there is regional differentiation. The colony swims in an oriented fashion, with a definite anterior half. The stigmas are larger in the anterior cells. In addition, the vegetative cells of the colony are so dependent on one another that they cannot live apart from the colony, and the colony cannot survive if broken up. The genus Pleodorina is further advanced with a considerable division of labor among the cells. The anterior cells are purely vegetative, and the posterior cells are larger and function in both sexual and asexual reproduction. Each Gonium colony consists of 4, 16 or 32 cells and the Pandorina colony contains 8, 16, or 32 cells, and the Pleodorina colony has 32 to 128 cells. Volvox is the most advanced of the volvocine series. The colony consists of 500 to 50,000 cells. All the anterior cells are exclusively vegetative, and the cells in the posterior half of the colony are much larger than the vegetative cells, and are specialized for asexual reproduction. This is the production of male gametes, or of non-motile egg cells. The changes represented by the volvocine series are as follows: (1) increasing size of colonies, (2) increasing coordination of activity among the cells, (3) increasing interdependence among the cells, so that vegetative cells cannot survive apart from the colony, and (4) increasing division of labor, particularly between vegetative and reproductive cells.

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Question:

The two adrenal glands lie very close to the kidneys. Each adrenal gland in mammals is actually a double gland, composed of an inner corelike medulla and an outer cortex. Each of these is functionally unrelated. Outline the function of the adrenal medulla.

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Solution:

The adrenal medulla secretes two hormones, adrenalin (epinephrine) and noradrenalin (norepinephrine, NE), whose functions are very similar but not identical. The adrenal medulla is derived embryologically from neural tis-sue. It has been likened to an overgrown sympathetic gan-glion whose cell bodies do not send out nerve fibers, but release their active substances directly into the blood, thereby fulfilling the criteria for an endocrine gland. In controlling epinephrine secretion, the adrenal medulla behaves just like any sympathetic ganglion, and is dependent upon stimulation by sympathetic preganglionic fibers, as shown below: Epinephrine promotes several responses, all of which are helpful in coping with emergencies: the blood pressure rises, the heart rate increases, the glucose content of the blood rises because of glycogen breakdown, the spleen contracts and squeezes out a reserve supply of blood, the clotting time of blood is decreased, the pupils dilate, the blood flow to skeletal muscle increases, the blood supply to intestinal smooth muscle decreases, and hairs become erect. These adrenal functions, which mobilize the resources of the body in emergencies, have been called the fight-or-flight response. Norepinephrine stimulates reactions similar to those produced by epineph-rine, but is less effective in the conversion of glycogen into glucose. The significance of the adrenal medulla may seem questionable since the complete removal of the gland causes few noticeable changes in the animal; the animal can still exhibit the fight-or-flight response. This occurs because the sympathetic nervous system compliments the adrenal medulla in stimulating the fight-or-flight response, and the absence of the hormonal control will be compensated for by the nervous system.

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Question:

The resistance of a copper wire 2500 cm long and 0.90 cm in diameter is 0.67 ohm at 20\textdegreeC. What is the resistivity of copper at this temperature?

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Solution:

The resistivity of a conductor is directly proportional to the cross-sectional area A and its resistance, and inversely proportional to its length l. Therefore, knowing the resistance, we have \rho = R (A/l) = [(0.67 ohm)/(2500 cm)] [{\pi (0.090 cm)^2 } / (4)] = 1.7 × 10^-6 ohm-cm Resistivity is a characteristic of a material as a whole, rather than of a particular piece of it. For a given temperature, it is constant for the particular medium, as opposed to resistance which depends on the dimensions of the piece.

Question:

A horizontal shelf moves vertically with simple harmonic motion, the period of which is 1 s and the amplitude of which is 30 cm. A light particle is laid on the shelf when it is at its lowest position. Determine the point at which the particle leaves the shelf and the height to which it rises from that position, g being taken as \pi^2 m\bullets^-2.

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Solution:

The period of a simple harmonic motion is given by the expression T = 1/f = 2\pi/2\pif = 2\pi/\omega Therefore, the angular frequency in this case is \omega = 2\pi/T = 2\pi/l s = 2\pi rad\bullets^-1. The only forces acting on the particle are its weight mg^\ding{217} downward and the normal force N\ding{217}exerted by the shelf upward. (See the figure.) At \ding{217} any time, according to Newton's second law, N - mg = ma, where a is the upward acceleration of shelf and particle. If a = -g, (i.e. the shelf accelerates downward with magnitude g) N becomes zero and, if a becomes more negative, the shelf is retarded at a greater rate than the particle; therefore the particle moves away from the shelf. Since the shelf undergoes simple harmonic motion, its displacement may be described by x = A cos (\omegat + \delta), where A is the amplitude of the oscillation, and \delta is a phase factor dependent on the initial conditions (i.e. where the shelf is at t = 0). The acceleration a = d^2x/dt^2 = -\omega^2 A cos (\omegat + \delta) = -\omega^2x. Hence, the displacement x at which the particle leaves the shelf is given by -g = -\omega^2xor x = g/\omega^2 = (\pi^2 m\bullets^-2)/(4\pir2rad^2\bullets^-2) = 1/4 m. (for g = \pi^2 m\bullets^-2 and \omega = 2\pi rad\bulletsec^-1). The particle thus leaves the shelf when it is 1/4 m above the mean position. At that point the common velocity of shelf and particle is given by the formula relating to velocity to displacement for simple harmonic motion: v = \pm \omega\surd(A^2 - x^2) Since the shelf is rising, v is positive and v = 2\pi rad\bulletsec^-1 ×\surd[(0.3m)^2 - (0.25m)^2] = 2\pi\surd(0.0275) m\bullets^-1. We now have a new problem concerning a particle thrown upward from a platform with an initial speed v. If the platform level is taken as the reference level for measuring potential energy, then the law of conservation of energy requires that at each moment of time that the particle is in motion, the sum of its kinetic and potential energies must remain constant. At the time the particle leaves the platform, its potential energy is zero and E_T = PE + KE = 0 + 1/2 mv^2. At its maximum height h, v = 0 and E_T = PE + KE = mgh + 0 \therefore 1/2 mv^2 = mghor h = v^2/2g = (4\pi^2 ×0.0275 m^2\bullets^-2)/(2 ×\pi^2 m\bullets^-2) = 0.055 m = 5.5 cm.

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Question:

Two springs are joined and connected to a mass m which rides on a frictionless surface (see figure (a)). What is the frequency v of the oscillation that will result if the mass is displaced a small distance x?

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Solution:

We must show that the configuration of figure (a), is equivalent to that of mass m, connected to a single spring having force constant k'. In figure (b), a mass m is shown connected to such a spring. A segment of the spring is divided into two sections of length x_1 and x_2. The tension in these two sections must be equal, as must be true of the tension in any other arbitrary sections into which the spring may be divided. If this were not the case, the spring would buckle. In general, when the spring is stretched or compressed by a force F, the changes in length ∆x_1 and ∆x_2 of sections x_1 and x_2 will not be equal since the magnitudes of such distortions will be proportional to the lengths of the segments involved. We can see this by symmetry. If a spring is stretched to twice its usual length, for example, each arbitrary section must stretch to twice its usual length. If we choose the sections to be of unequal length, the amount that each section stretches will be different. Thus: F = -k_1 ∆x_1 = -k_2 ∆x_2 since we know that the force that the spring exerts on the mass equals the tension within each section. The force constants of sections x_1 and x_2 are represent-ed by k_1 and k_2. It is not unreasonable to assume that each arbitrary section of the spring can be considered to have a dif-ferent force constant as long as: F= -k_1 ∆x_1 = -k_2 ∆x_2 = -k_3 ∆x_3 = . . . -k_n ∆x_n = -k'x where x = ∆x_1 + ∆x_2 + ∆x_3 + . . . + ∆x_n is the total displacement of the mass. Thus we see that in the case of the two connected springs above, the tension in each must be equal: F = -k_1 ∆x_1 = -k_2 ∆x_2 = -k'x = -k'(∆x_1 + ∆x_2) ∆x_1 = -(F/k_1) ∆x_2 =-(F/k_2) Thus: k' = -[F/(∆x_1 + ∆x_2)] = -[F/{-(F/k_1) - (F/k_1)}] = [1/{(1/k_1) + (1/k_2)}] = 1/[(k_1 +k_2)/(k_1k_2)] = (k_1k_2)/(k_1 +k_2) Since the frequency of an oscillator having force constant k' is: v = [1/(2\pi)]\surd(k'/m) then:v = [1/(2\pi)]\surd[(k_1k_2)/{(k_1 + k_2)m}]

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Question:

The Eiffel Tower is 984 feet high. Express this height in meters, in kilometers, in centimeters, and in millimeters.

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Solution:

A meter is equivalent to 39.370 inches. In this problem, the height of the tower in feet must be converted to inches and then the inches can be converted to meters. There are 12 inches in 1 foot. Therefore, feet can be converted to inches by using the factor 12 inches/1 foot. 984 feet × 12 inches/1 foot = 118 × 10^2 inches. Once the height is found in inches, this can be converted to meters-by the factor 1 meter/39.370 inches. 11808 inches × 1 meter/39.370 inches = 300 m. Therefore, the height in meters is 300 m. There are 1,000 meters in one kilometer. Meters can be converted to kilometers by using the factor 1 km/1000 m. 300 m × 1 km/1000 m = .300 km. As such, there are .300 kilometers in 300m. There are 100 centimeters in 1 meter, thus meters can be converted to centimeters by multiplying by the factor 100 cm/1 m. 300 m × 100 cm/1 m = 300 × 102cm. There are 30,000 centimeters in 300 m. There are 1,000 millimeters in 1 meter; therefore, meters can be converted to millimeters by the factor 1000 mm./1 m. 300 m × 1,000 mm/1 m = 300 × 10^3 mm. There are 300,000 millimeters in 300 meters.

Question:

Draw the inter-connections between the control section (CPU) and the memory section of the computer and state the purpose of each inter-connection. What takes place when the CPU transfers data to or from memory?

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Solution:

Figure 1 shows the inter-connections between the CPU and memory section of the computer. The address bus transfers the address of the location to be read from, or written into, the memory section The read/write line designates whether data is to be read from memory or written into memory. The output data bus is used to transmit data to be written into memory. The input data bus is used to transmit data read from memory. Many times a single data bus is used for more than one purpose. However, more control lines are needed to designate the purpose of the bus. Such multifunction buses are said to be "time shared" or "multiplexed." The CPU transfers data to the memory section as follows: 1.The CPU sends the address to the memory section via the address bus. 2.The CPU sends the data to the memory section via the output data bus. 3.The CPU signals a Write operation to the memory sec-tion via the Read/Write line. The CPU transfers data from memory as follows: 1.The CPU sends the address to the memory section via the address bus. 2.The CPU signals a Read operation to the memory sec-tion via the Read/Write line. 3.The CPU waits, then the data is sent out to the CPU via the input data bus.

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Question:

A bullet weighing 4 g is fired at a speed of 600 m/s into a ballistic pendulum of weight 1 kg and thickness 25 cm. The bullet passes through the pendulum and emerges with a speed of 100 m/s. Calculate the constant retarding force acting on the bullet in its passage through the block, and the height to which the pendulum rises.

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Solution:

The ballistic pendulum is used to measure bullet speeds. The three stages of the motion of the system are as shown in the diagram. In the first stage, the bullet of mass m approaches with velocity V^\ding{217} the ballistic pendulum of mass M. In the second, the bullet, having passed through the pendulum, is moving off with velocity v_1^\ding{217}, leaving the pendulum just starting to move with velocity v_2^\ding{217} Since momentum must be conserved, mV^\ding{217} = mv_1^\ding{217} + Mv_2^\ding{217}. \thereforev_2 = [mg(V - v_1)] / [Mg] = (4 × 10-3Kg (600 - 100) m/s)] / (1 kg) = 2 m/s. In the third stage, the pendulum, which has acquired kinetic energy (1/2) Mv2_2, swings through a certain angle such that the bob loses all its kinetic energy but gains an equivalent quantity of potential energy in rising a height h, in accordance with the principle of conservation of energy. But = Mgh = (1/2)MV^2 _2.Therefore h = \surd(V22/2g) = 0.45 m = 45 cm. The loss of kinetic energy as the bullet passes through the pendulum is (1/2) mV^2 - (1/2) mv2_1 - (1/2) Mv22 = [(4 × 10^-3 kg) / (2)] (600^2 - 100^2) m^2/s2 - [(1kg) / 2] × 4 m^2/s^2 = 698 J. By the principle of conservation of energy, this quan-tity of energy must represent the work done against the retarding force, F^\ding{217}, as the bullet pushes its way through the pendulum. Thus, W = F^\ding{217} \bullet s^\ding{217} = F × 0.25 m = 698 J or F = 4 m^-1 × 698 J = 2792 N.

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Question:

The space between the plates of a parallel-plate capacitor is filled with dielectric of coefficient 2.5 and strength \textasciigrave5 ×10^6V \bullet m^\rule{1em}{1pt}1. The plates are 2mm apart. What is the maximum voltage which can be applied between the plates? What area of plates will give a capacitance of 10^\rule{1em}{1pt}3\muF, and at maximum voltage what are the free and bound charges per unit area of the plate and dielectric surface?

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Solution:

The dielectric strength of the dielectric is the largest electric field which it can withstand before becoming a conductor. We must relate the voltage between the capacitor plates to E_max (dielectric strength). This can be done by realizing that voltage differences are defined V_b \rule{1em}{1pt} V_a = \rule{1em}{1pt}^b \int_a E^\ding{217} \bullet dl^\ding{217}(1) where V_b,V_a refer to the potentials at z = b and z = a, respectively (see figure), and E^\ding{217} is the electric field. The integral in (1) is a line integral, and it is to be evaluated over an arbitrary path connect-ing a and b, dl^\ding{217} being a small element of path length. Looking at the figure, we see that E^\ding{217} is composed of contributions from two sources-the conducting plates (E^\ding{217}_c)and the dielectric material (E^\ding{217}_D). Note that the latter is in opposition to E^\ding{217}_C . Since E^\ding{217} is uniform in direction and magnitude for a parallel-plate capacitor,(1) becomes V_b \rule{1em}{1pt} V_a = \rule{1em}{1pt}E^\ding{217} \bulletb\int_a dl^\ding{217}(2) The easiest path to evaluate this line integral along is straight line path from z = a to z = b. In this case. dl^\ding{217} = dl k˄ Since E^\ding{217} = \rule{1em}{1pt}E k˄ (2) becomes V_b \rule{1em}{1pt} V_a = \rule{1em}{1pt}(\rule{1em}{1pt}Ek˄) \bullet ^b\int_a (dl k˄) V_b \rule{1em}{1pt} V_a = E(b-a) But b \rule{1em}{1pt} a = d (see figure) and we define Vb\rule{1em}{1pt} V_a = ∆V as the potential difference between the plates. Therefore, ∆V = Ed(3) The maximum voltage which can be applied between the plates is ∆V_max = E_max d = [5 × 106(V / m)](2 × 10^\rule{1em}{1pt}3 m) ∆V_max = 10^4V To answer the second part of the problem, we must relate the capacitance of the capacitor to its geometry. We do this by beginning with the definition of capacitance (C) C = Q / ∆V(4) where Q is the charge on 1 capacitor plate. If the free charge density on the plate of cross-sectional area A is \sigma_f, (see figure) then Q = \sigma_fA(5) Using (5) and (3) in (4) C = (\sigma_fA) / Ed(6) Since we don't know E in this portion of the problem, we must eliminate it from (6), This may be done by using Gauss' Law for a dielectric \oint_s E^\ding{217}. ds^\ding{217} = (Q / \epsilon)net(7) where \epsilon is a constant, andQ_net is the total free charge enclosed by the surface s over which the integral is evaluated. Evaluating (7) over the Gaussian surface indicated in the figure, \oint_a E^\ding{217} \bullet ds^\ding{217} = \int_(s)T E^\ding{217}(s_T)\bullet ds^\ding{217} + \int_(S)B E^\ding{217}(s_E)\bulletda^\ding{217}= Q_net / \epsilon(8) ^\ding{217} where S_T and S_B are the top and bottom of the rectangular surface. Now, ds^\ding{217} is a vector element of surface area. For S_T, ds^\ding{217} = ds k˄ and for S_B, ds^\ding{217} = \rule{1em}{1pt}ds k˄. Then, using the fact that E^\ding{217} = \rule{1em}{1pt}Ek˄ everywhere, (8) becomes \oint_S E^\ding{217} \bullet ds^\ding{217} = \int_(s)T {\rule{1em}{1pt}E^\ding{217}(s_T)k˄}\bullet(ds^\ding{217}k˄) + \int_(S)B {\rule{1em}{1pt}E(s_B)k˄}\bullet(-ds k˄) = Q_net / \epsilon \oint_s E^\ding{217} \bullet ds^\ding{217} = \rule{1em}{1pt}\int_(s)T E(s_T)(ds) + \int E(s_B)(ds) = Q_net / \epsilon Now, E(S_T) is the value of E at S_T, which lies in the top conduct-ing plate. By definition of a conductor, then,E(S_T) = 0, Therefore \oint_s E^\ding{217} \bullet ds^\ding{217} = \int_(s)B E(s_B)(ds) = Q_net / \epsilon(9) But E(s_B) = E, the electric field between the capacitor plates. This is constant, and (9) becomes E \int_(s)B ds =EA = Q_net / \epsilon since the area of s_B is A. Hence E = Q_net / \epsilonA. Now, Q_net includes the free charge found within the Gaussian surface. Therefore (see figure) Q_net / \epsilon = \sigma_fA and, E = \sigma_f / \epsilon(10) Using (10) in (6) C = [(\sigma_fA) / {(\sigma_f / \epsilon)d}] = \epsilonA / d But K, the dielectric constant) is defined by K = \epsilon / \epsilon_0 Hence,\epsilon = \epsilon_0Kand C = k\epsilon_0A / d(11) Solving for A A = Cd / K\epsilon_0(12) Substituting the given data in (12) A ={(10^\rule{1em}{1pt}3 \muF)(2 × 10^\rule{1em}{1pt}3m)} / {(2.5)(8.85 × 10^\rule{1em}{1pt}12 c^2/N \bullet m^2)} A = .09m^2 The free charge density can be found using (10) \sigma_f = \epsilonE = \epsilon_0KE or \sigma_f = (5 × 10^6 V/m)(2.5)( 8.85 × 10^\rule{1em}{1pt}12 C^2 / N \bullet m^2) \sigma_f = 110.6 × 10^\rule{1em}{1pt}6 C / m^2 From the figure, note that the net field E^\ding{217} is the superposition of the field due to the bound charges of the dielectric (E^\ding{217}_D) and the field due to the free charge on the capacitor plates (E^\ding{217}_C). Then E^\ding{217}= E^\ding{217}_C + E^\ding{217}_D(13) ^\ding{217} But by (10) and the figure, E^\ding{217} = \rule{1em}{1pt} (\sigma_f / \epsilon)(k˄) . Since the bound charge accumulates only on the surface of the dielectric, we may consider the dielectric slab to be equivalent to a parallel plate capacitor with surface density \sigma_b and an air dielectric. E^\ding{217}_D = (\sigma_b / \epsilon_0 )(k˄) Similarly, E^\ding{217}_C = \rule{1em}{1pt} (\sigma_f / \epsilon_0 )(k˄) Hence, from (13) \rule{1em}{1pt}(\sigma_f / \epsilon) (k˄) = \rule{1em}{1pt}(\sigma_f / \epsilon_0 )(k˄) + (\sigma_b / \epsilon_0 )(k˄) or \rule{1em}{1pt}\sigma_f = \rule{1em}{1pt}\sigma_f(\epsilon / \epsilon_0) + \sigma_b (\epsilon / \epsilon_0) \sigma_f [(\epsilon / \epsilon_0) \rule{1em}{1pt} 1] = \sigma_b(\epsilon / \epsilon_0) Since K = \epsilon / \epsilon_0 \sigma_b = \sigma_f [(k \rule{1em}{1pt} 1) / (k)] = [(1.5) / (2.5)] \sigma_f = [3 / 5] \sigmaf Noting that \sigma_f = 110.6 × 10\rule{1em}{1pt}6c/m^2 \sigma_b = (331.8 / 5)×10\rule{1em}{1pt}6c/m^2 \sigmab=66.4 ×10\rule{1em}{1pt}5c/m^2

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Question:

A woman who is Rh-negative marries a man who is Rh- positive. What possible problems could arise if the couple wants more than one child?

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Solution:

Besides the A and B agglutinogens, there is another type of blood antigen, designated the Rh factor. There are at least eight different types of Rh antigens which may be present on human erythrocytes. People are classified into two groups with respect to the Rh antigen: Rh-positive individuals have an Rh antigen, while Rh- negative individuals have such a weakly antigenic factor that it can be disregarded. The A\rule{1em}{1pt}B\rule{1em}{1pt}O blood type system is an example of a rare case in which the body normally synthesizes antibodies in large amounts against an antigen to which it has not yet been exposed. The Rh system follows the rule rather than the exception. An individual's plasma will not con-tain an antibodies against the Rh antigen unless it has been sensitized to the antigen by previous exposure. Thus, while Rh-positive individuals have the Rh antigen, Rh-negative individuals have neither an effective antigen nor an antibody against the antigen. If Rh-positive blood is given to an Rh negative individual, the Rh-negative individual will gradually synthesize antibodies against the Rh antigen introduced. Agglutination of the cells occurs, plugging up the small blood vessels, inactivating the erythrocytes as oxygen carriers, and eventually leading to death of the recipient. In our particular problem, the couple's first child might inherit the Rh-negative characteristic. There is thus no particular problem since the mother is also Rh-negative. But if the fetus is Rh-positive, trouble may occur. If some blood from the Rh-positive fetus enters the maternal circulation, its antigens stimulate the mother's plasma cells to synthesize Rh antibodies. Normally, the mother cannot produce antibodies rapidly enough and in large enough amounts to affect the first child before birth. But if the mother, who is Rh-sensitized by the first pregnancy, has subsequent pregnancies, she will have produced sufficient antibodies which can diffuse into the blood of an Rh-positive fetus and react with its erythrocytes. This agglutination of the blood of the fetus with the antibodies produced in its mother is a disease known as erythroblastosis fetalis. Only about 3% of the second or later Rh-positive fetuses born to Rh-negative mothers show signs of erythroblastosis fetalis. The blood of the mother may not diffuse into the fetal circulation, thus preventing agglutination. In addition, only one of the types of Rh antigens is responsible for erythroblastosis; other forms will not cause any trouble. Also, the longer the time interval between pregnancies, the less intense is the anti-Rh immunal response in the mother. One method of preventing erythroblastosis is to in-ject Rh antibodies into a Rh-negative mother immediately after the birth of her first Rh positive infant. The antibodies kill any fetal red blood cells that have en tered her circulation and thus prevent stimulation of maternal antibody production. The injected antibodies eventually are destroyed arid the mother is now unsensi-tized and able to bear a second Rh-positive infant.

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Question:

Discuss various types of computer systems in terms of the level of expense, sophistication, and capabilities to carry out intended commercial and scientific data processing tasks.

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Solution:

Traditionally, the computer systems from lowest to highest level of capabilities have been classified as: Microcom-puters, Minicomputers, and Mainframes. It used to be that micros such as INTEL 8085 had 8-bit registers, minis such as PDP-11 had 16-bit registers, and mainframes such as IBM 370 had 32-bit registers. Note that the amount of addressable Random Access Memory (RAM) is dependent upon the register length. Accordingly, micros had small, minis had larger, and mainframes had much larger memories. Although the above classification still serves a useful purpose for quick sizing and comparisons, the rapid developments in VLSI technologies for fourth generation computers have blurred these distinctions. There have also been many grades and shades in between this family of computers. Generally, the mi-crocomputers have caught up with mainframes in terms of process-ing speeds and memory capacities, but the mainframes still re-tain the edge in terms of secondary storage, communication, and I/O peripherals. As of this time (1989-90), a plausible (by no means complete or precise) classification is as follows: Types and sub-types Description and example Home computers 8-16 bits registers and memory words. - Simple Commodore VIC-20 and 64 (64K RAM) -Complex Commodore AMIGA with sound and graphics Personal computers 16-32 bits registers and memory words. -Low-end IBM PC/XT, AppleIIe, 0p6 AT&T 7300, etc.m -High-end IBM PC/AT, Apple. SE, and com-patibles. -Super high-end IBM PS/2 80, Apple MACIIx, AT&T 6386, and NeXT. Personal Supercomputers Either employs 64-bit vector type processors, or a number of 16/32-bit high-end processors. A personal computer that employs INTEL 860 with 64-bit vector register would be an example of the former. IntelSugarCubethat uses 8 16-bit processors that are connected according to hypercube topology is an example the of latter. Work Stations These are generally UNIX machines that can support multiple users and have large memories and special constructions for various types of tasks such as high-resolution graph-ics symbolic processing for artifi-cial intelligence applications. The typical examples are Sun 3 and Sun series, Texas Instrument Explorer II, andSymbolicsseries. Minicomputers DEC PDP-11 and HP series of comput-ers that use 16-bit registers are typical examples. Superminicomputers DEC VAX 11 and 85XX series that use 32-bit registers, and AT&T 3B20 machines are typical examples. Mainframes IBM 370 series (formerly), and 4000 series recently are typical ex-amples . Supercomputers Cray 1, 2, X-MP, and Y-MP that use 64-bit vector CPUs are examples. Parallel Computers These are multiple processor systems that use 8 to 1024 processors that are connected according to a regular topology and cooperatively solve a large-scale problem with parallelism.

Question:

A slender homogeneous bar, shown in Figure 1, is 4 ft long and weighs 1.2 lb. A spring is attached to it as shown, 3 ft from the lower end which is hinged. When the bar is in a vertical position the spring is compressed 0.12 ft. The bar is released from rest. What should the spring con- stant be if the bar comes to rest when it just reaches the horizontal position, Figure 2?

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Solution:

The system is conservative since only spring and gravity forces are acting, and if the hinge is assumed to be frictionless. We therefore use the principle of conser-vation of mechanical energy, i.e., (KE)_1 + (PE)_1=(KE)_2 + (PE)_2 (Bar in vertical position 1)(Bar in horizontal position, 2.) Further, since the bar is at rest in both positions, the above equation reduces to (PE)_1 = (PE)_2(1) The potential energy of the system is the sum of that of the spring and the gravitational potential energy. Potential energy of a spring is (1/2) K\sigma^2 where K is the spring con-stant and \sigma its extension or compression. In position 1, \sigma_1 = 0.12 ft. Thus the free length of the spring is 1.2 + 0.12 = 1.32 ft. In position 2, the length of the spring is (4.2^2 + 3^2)^1/2 = 5.16 ft. Thus \sigma_2 = 5.16 \rule{1em}{1pt} 1.32 = 3.84 ft. The gravitational potential energy of a mass m is mgh where h is the height of its center of gravity above the datum. Let us assume the datum to be the horizontal plane passing through the hinge at O. Thus the total potential energies in the two positions are: (PE)_1 = (1/2)K (0.12)^2 + (1.2) (2) = 0.0072 K + 2.4 ft-lb (PE)_2 = (1/2)K (3.84)^2 + (1.2) (0) = 7.378 K ft-lb. Substituting these values into equation (1) yields 0.0072 K + 2.4 = 7.378 K orK = 0.3256 lb / ft. Note that this is not the equilibrium position. The bar will oscillate. The bottom position will be an amplitude extreme, where the bar is momentarily at rest. To consider the equilibrium position we need to use force equilibrium.

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Question:

What are spinal nerves and what are their functions?

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Solution:

Spinal nerves belong to the peripheral nervous system (PNS). They arise as pairs at regular intervals from the spinal cord, branch, and run to various parts of the body to innervate them. In human beings, there are 31 symmetrical pairs of spinal nerves. The size of each spinal nerve is related to the size of the body area it innervates. All the spinal nerves are mixed nerves, in the sense that all have both motor and sensory components in roughly equal amounts. They contain fibers of the somatic nervous system, which are both afferent (conducting impulses towards the nervous system) and efferent (conducting impulses to effector organs). They also contain fibers of the auto-nomic nervous system, which are uniquely efferent and are separable into parasympathetic and sympathetic nervous systems. Because of this nature, the spinal nerves function to convey messages from the external environment to the central nervous system, and from the central nervous system to various effectors of the body. In other words, the spinal nerves, as part of the peripheral nervous system, serve as a link between the central nervous system and the effector organs. Somatic fibers of the spinal nerves innervate skeletal muscles of the body, and are under voluntary regulations. We can bend our arm or leg at will. The auto-nomic nerves innervate the smooth and cardiac muscles and glands of the body, and cannot be voluntarily controlled. We cannot speed up our stomach contractions or heart beat at will. The autonomic nerve fibers leave the spinal cord and run for a certain distance with the somatic fibers in the spinal nerves. Then the two types of fibers diverge and run to their respective body areas which they innervate.

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Question:

The lens system of a certain portrait camera may be taken as equivalent to a thin converging lens of focal length 10.0 in. How far behind the lens should the film be located to receive the image of a person seated 50.0 in. from the lens? How large will the image be in comparison with the object?

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Solution:

The general equation for thin lenses states 1/p + 1/q = 1/f where p is the distance of the object, in this case the person, in front of the lens, and q is the distance between the image formed and the lens. The focal length is repre-sented by f. All distances are measured with respect to the optical center of the lens. This center is defined is the point through which light can pass without being bent. Since the lens is a converging one, it is convex. This means that it is thicker at the center than at the edges. Substitute values in the above equation noting that q is positive when it is on the opposite side of the lens in relation to the object. 1/q = 1/10 - 1/50 = (50 - 10) / {(10)(50)} = 4/50 q = 50/4 = 12.5 in. To find the magnification M ], the equation M = q/p can be used Substituting, M = 12.5 in. / 50.0 in. = 0.250 . The image will be one-fourth as large as the object.

Question:

Trace the sequence of events between nerve action potential and contraction and relaxation of a muscle fiber.

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/Users/wenhuchen/Documents/Crawler/Biology/F19-0466.htm

Solution:

The cell membranes of muscle fibers are excitable membranes capable of generating and propagating action potentials by mechanisms very similar to those found in nerve cells. An action potential in the muscle cell membrane results from nervous stimulation and provides the signal for the initiation of contractile activity within the muscle cell by a mechanism known as excitation-contraction coupling. An action potential is initiated and propagated in a motor axon which innervates the muscle fiber. This potential is the result of synaptic events on the cell body and dendrites of the motor neuron in the central nervous system, (see Figure 1) The action potential in the motor axon causes the release of the neurotransmitter known as acetylcholine from the axon terminals into the synaptic space between the nerve and muscle, (see Figure 2) This junction between nerve and skeletal muscles is known as the neuromuscular junction. The nerve cells which form junctions with skeletal muscles are known as motor neurons, and the cell bodies of these neurons are located in the brain and spinal cord. Once released, acetylcholine binds to receptor sites on the muscle membrane which lie directly under the terminal portion of the axon. This region of the muscle membrane is known as the motor end plate. Acetylcholine increases the permeability of the motor end plate to sodium and potassium ions, causing a depolarization of the end plate called an end-plate potential (EPP). The EPP depolarizes the muscle membrane to its threshold potential, generating a muscle action potential which is propagated over the surface of the muscle membrane. The molecules of acetylcholine re-leased from the motor-neuron have a lifetime of only about 5 milliseconds before they are destroyed by the acetylcholin-esterase, an enzyme on the muscle cell membrane near the receptors for acetylcholine. Once acetylcholine is de-stroyed, the muscle membrane permeability to sodium and po-tassium ions returns to its initial state, and the depolar-ized end plate returns to its resting potential. The muscle action potential depolarizes the T tubules at the A-I junction of the sarcomeres. This depolarization leads influx of calcium from the extracellular fluid and to the release of calcium ions from the terminal cisternae of the sarcoplasmic reticulum surrounding the myofibrils. Cal-cium ions bind to troponin in the thin actin myofilaments. This calcium-troponin complex causes tropomyosin to shift its position, releasing the inhibition that prevented actin from combining with myosin. Actin then combines with myosin. This binding activates the myosin ATPase, which splits ATP, releasing energy which is used to produce the movement of the cross bridge of the myosin molecule. ATP then binds to the myosin bridge, breaking the bond between the actin and myosin, thus allowing the cross bridge to dissociate from actin. As long as the concent-ration of calcium ions is high enough to counteract the inhibitory action of the troponin-tropomyosin system, cycles of cross-bridge contraction and relaxation will continue. The concentration of calcium ions falls as they are moved back into the sarcoplasmic reticulum by an energy- requiring process which splits ATP. Removal of calcium ions restores the inhibitory action of troponin- tropomyosin, and in the presence of ATP, actin and myosin remain in the dissociated, relaxed state.

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Question:

Draw and explain a system dynamics diagram of a market model.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G18-0467.htm

Solution:

System dynamics analyzes the forces operating in a system in order to determine their influence on the stability or growth of the system. Here, stability is determined by the rates of changes of different variables. Consider the following system dynamics diagram of market model. This diagram depicts a modified exponential growth model in which a variable increases in value at first at an increas-ing rate and then later at a decreasing rate until it reaches an upper limit. This corresponds to a typical marketing situation. At first, as the product is intro-duced into the market, sales will increase rapidly. Later as the saturation point is approached, the increase in sales will gradually taper off. Next, consider the various symbols used in the System Dynamics model. The number of products sold can be viewed as a level in a reservoir of products. Levels are represented by boxes. Rates are represented by boxes with valves attached. The constants of the system are indicated by the symbol \cyrchar\CYROTLD. The differential equation corresponding to the system diagram is: ẋ = k(X - x)(4) with initial condition x(o) = 0 The solution is x = X(1 - e^-kt) which is known as a modified exponential curve. A computer simulation model would solve (4) for various values of k, the sales rate constant, and X, the market limit.

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Question:

80 standard characters,and E-MESSAGE: 52 standard characters.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G11-0268.htm

Solution:

The DATA DIVISION is shown in fig. 1. The statement BLOCK CONTAINS 5 RECORDS in the INDEXED file description means that the record BAS-REC will be stored in the indexed file in blocks of 5. This allows more efficient searching for the unique keys requested in the program. The computer handles all the details of record storage and retrieval.

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Question:

What is the difference betweenelectronegativityand electron affinity? Use specific examples.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E17-0625.htm

Solution:

Electron affinity represents the energy released when an electron is added to an isolated neutral gaseous atom.Electronegativityis the tendency of an element to attract electrons, not in the gaseous state, but from a nucleus joined to it by a covalent bond. The electron affinity is a precise quantitative term while theelectro-negativityis a more qualitative concept.

Question:

Two hollow spherical shells are mounted concentrically, but are insulated from one another. The inner shell has a charge Q and the outer shell is grounded. What is the electric intensity and potential in the space between them? When the outer shell is not grounded, why does a charge outside the system experience a force when the inner shell is charged?

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/Users/wenhuchen/Documents/Crawler/Physics/D18-0614.htm

Solution:

In the region between the shells, because of the symmetry of the arrangement, the electric in-tensity E^\ding{217} must have the same magnitude at all points a distance r from the common center, and must thus be everywhere radial in direction. Hence, applying Gauss's law to a region bounded by a spherical surface of radius r (see the figure), we have \int E^\ding{217} \bullet ds^\ding{217} = Q / \epsilon_0(1) where Q is the total charge enclosed by the Gaussian surface, and dS is an element of surface area of the Gaussian surface. E^\ding{217} is parallel to dS^\ding{217}, thus (1) becomes E \intds = 4\pir^2E = Q / \epsilon_0orE = Q / 4\pi\epsilonr^2 . But E = \rule{1em}{1pt} (dV / dr) by definition of E as a potential gradient. \therefore dV = \rule{1em}{1pt}(Q / 4\pi\epsilon_0r^2)drorV =(Q / 4\pi\epsilon_0r) + C, where C is a constant of integration. But at the outer spherical shell of radiusr = b, V = 0. \therefore 0 = (Q / 4\pi\epsilon_0b) + CorC = \rule{1em}{1pt}(Q / 4\pi\epsilon_0b). \thereforeV = Q / 4\pi\epsilon_0 [(1 / r) \rule{1em}{1pt} (1 / b)]. Lines of force come from the inner shell and all end on the outer shell. The same number of lines of force end on the outer shell as start on the inner shell; hence the charge induced on the inside of the outer shell is equal and opposite to that on the inner shell. Thus a charge \rule{1em}{1pt}Q is induced on the inside of the outer shell. But this shell was initially un-charged. Hence a charge + Q must be left on the out-side of the outer shell. If the outer shell is grounded, electrons flow from the earth to neutralize this positive charge. In the absence of grounding, the positive charge remains on the outside of the shell and produces a field of force around the system which affects any other charge in the vicinity.

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Question:

What cell components typically found in aeucaryoticcell are missing from a bacterial (procaryotic) cell?

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/Users/wenhuchen/Documents/Crawler/Biology/F05-0118.htm

Solution:

Cells are classified as eukaryotic or prokaryotic. The former means "truenucleus" while the latter means "before the nucleus". Prokaryotes include all bacteria including thecyanobacteria(blue-green algae), whereas eukaryotes include theprotistan, fungal, animal and plant cells. The most obvious difference between bacteria andeucaryotesis the absence of the nuclear membrane in bacteria. The DNA of the bacterial cell is generally confined to a specific area but this region is not enclosed by a membrane. In addition, proteins, such as thehistonesofeucaryotic cells , are not found in association with bacterial DNA. Another distinctive feature is that during replication of the bacterial nuclear region, the mitotic spindle apparatus is not seen. Bacterial cells also lack Golgiapparati, mitochondria, and endoplasmic reticulum. This means that theribosomesfound in the cytoplasmic region are free, i.e. they are not bound to endoplasmic reticulum . Membranous systems are not completely absent in bacteria. The plasma membrane folds inward at various points to formmesosomes. These membranes may be involved with the origin of other intra-cellular structures , and the compartmentalization and integration of biochemical systems . For example, although the electron transport system is located in the plasma membrane of bacteria, most of the respiratory enzymes are located in themesosomes. The plasma membrane andmesosomesare the bacterial counterpart of theeucaryoticcells' mitochondria, serving to compartmentalize the respiratory enzymes. Those bacteria that contain chlorophyll do not contain plastids to "house" the chlorophyll. Instead, the chlorophyll is associated with membranous vesicles arising from mesosomes .

Question:

Discuss some properties and functions of (a) carbohydrates (b) lipids, (c) proteins and (d) nucleic acids.

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Solution:

(a) Carbohydrates are made up of carbon, oxygen and hydrogen, and have the general formula (CH_2O)_n. Carbohydrates can be classified as monosaccharides, disaccharides, oligosaccharides or polysac-charides. The monosaccharides ("simple sugars") are further categorized according to the number of carbons in the molecule. Trioses contain 3 carbons: pentoses contain 5 carbons (i.e. ribose, dexyribose): and hexoses contain 6 carbons (i.e. glucose, fructose, galactose). The hexoses, which exist as straight chains or rings, are important building blocks for disaccharides and the more complex carbohydrates. Disaccharides, important in nutrition, are chemical combinations of two monosaccharides: Lactose = glucose + galactose Sucrose (table sugar) = glucose + fructose Maltose = glucose + glucose As can be seen by the double arrows, the reverse of this reaction, hydrolysis, is also possible. Oligosaccharide means "few sugars" and is arbitrarily defined as com-pounds which upon hydrolysis yield 2 - 10 monosaccharides. Polysaccharides are complex carbohydrates made up of many monosaccharides bonded by glycosidic linkages. These long chains are formed by dehydration synthesis. They also can be broke down into monosaccharide units by hydrolysis. There are many complex polysaccharides that are of great biological significance. Their primary functions include both storage and structural properties. Examples of these are: starches (principal storage product of animals), and cellulose (major supporting material in plants). These are all polymers of glucose. (b) Lipids are also composed principally of carbon, hydrogen and oxygen. However, they can also contain other elements, particularly phosphorus and nitrogen, and typically contain a much smaller proportion of oxygen than do carbohydrates. Lipids are insoluble in water. There are many known lipids, the most common include the fats, the phospholipids and the steroids. Fats are composed of two different types of compounds: glycerol (an alcohol) and fatty acids (organic compounds with a carboxyl group . Each molecule of fat contains one glycerol molecule and three fatty acids joined together by dehydration reactions. There are basically two groups of fats: the saturated (those fats which have the maximum possible number of hydrogen atoms and therefore have no carbon to carbon double or triple bonds) and the unsaturated (those which have at least one carbon to carbon double or triple bond). Unsaturated fats are an important part of our diet due to their storage capabilities. Phospho-lipids are composed of glycerol, fatty acids, phoshoric acid and usually a nitrogenous compound. They are important components of many cellular membranes. Steroids are classified as lipids because their solubility characteristics are similar to those of fats and phospholipids - they are all insoluble in water, but soluble in ether. However, steroids differ structurally from fats and phospholipids - they are not based on the bonding together of fatty acids and an alcohol. Instead, they are composed of four interlocking rings of carbon atoms with various side groups attached to the rings: They are very important biologically. Steriods include the sex hormones, various other hormones and some vitamins. In addition they are also important structural elements in living cells, especially plants. c) Proteins are much more complex than either carbohydrates or lipids. They are made up of the four essential elements: carbon, hydrogen, oxygen and nitrogen bonded together to form compounds called amino acids. Some amino acids contain sulfur. The amino acid is represented as: where R is a chain which can bevery simple (as in the amino acid glycine where R=H) or it can be very complex (as in tryptophan where R contains two ring structures). Proteins are long and complex polymers of varying combinations of the twenty amino acids which are formed by condensation reactions between the -COO^- and -NH_3^+ groups of the amino acid building blocks. The bonds formed by these reactions are called peptide bonds. A dipeptide is a molecule with two amino acids joined together by one peptide bond. A tripeptide refers to three linked amino acids. Oligopeptides, which includes tripeptides, is the term for a short chain of amino acids. Polypeptides are the polymer of amino acids and may contain 1000 amino acids. Finally, a protein is one or more polypeptide chains coiled or folded into complex three- dimensional configurations. Often a metal ion or organic molecule is an integral part of the protein structure. Proteins are found in every part of the cell and are an integral part in both the structure and function of living things. They play an important part in many of the chemical reactions that occur within cells because the enzymes that catalyze these reactions are proteins themselves. Proteins also function as the structural and binding materials of organisms. Hair, fingernails, muscle, cartilage, tendons and ligaments are all structures which contain large amounts of proteins. d) Nucleic acids, as their name implies, are found primarily in the nucleus. There are two types of nucleic acid: deoxyribonucleic acid (DNA) and ribonucleic acid (RNA). DNA and RNA molecules are very long chains composed of repeating subunits called nucleotides. A nucleotide is composed of any one of the following five nitrogenous bases: adenine, guanine, cytosine, thymine (only in DNA) or uracil (only in RNA), a five carbon sugar (ribose in RNA, deoxyribose in DNA) and a phosphate group. Both DNA and RNA are composed of many nucleotides linked together and are called poly-nucleotides or nucleic acids. In a polynucleotide, any two nucleotides are linked together by a dehydration re-action between the phosphate group of one nucleotide and the sugar group of another. The bases are attached to the sugars. Nucleic acids primarily function in heredity and governing the synthesis of many different kinds of proteins and other substances present in organisms. Chromosomes and genes are predominantly composed of DNA. Some DNA is also found in the mitochondria and the chloroplasts. Large quanitites of RNA are present in the nucleoli, the cytoplasm, and the ribosomes of most cells.

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Question:

What are the chemical and physical processes involved in transmission at the synapse?

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Solution:

The nervous system is composed of discrete units, the neurons, yet it behaves like a continuous system of transmission of impulses. For this to occur, there have to be functional connections between neurons. These con-nections are known as synapses. A synapse is an anatomical-ly specialized junction between two neurons, lying adjacent to each other, where the electric activity in one neuron (the presynaptic neuron) influences the excitability of the second (the postsynaptic neuron). At the synapse, the electric impulse is transformed into a chemical form of transmission. Chemical transmission at the synapse involves the processes of neurosecretion and chemoreception. The arrival of a nerve impulse at the axon terminal stimulates the release of a specific chemical substance, which has been synthesized in the cell body and stored in the tip of the axon, into the narrow synaptic space between the ad-jacent neurons. This process constitutes neurosecretion. The chemical secreted, known as neurotransmitter, can cause local depolarization of the membrane of the postsynaptic region and thus transmit the excitation to the adjacent neuron. Chemoreception is the process in which the neurotransmitter becomes attached to specific molecular sites on the membrane of the dendrite (postsynaptic region), producing a change in the properties of the cell membrane so that a new impulse is established. The chemical transmitter, say acetylcholine, passes from the presynaptic axon to the postsynaptic dendrite by simple diffusion across the narrow space, called the synaptic cleft, separating the two neurons involved in the synapse. The synaptic clefts have been measured under the electron microscope to be about 200 \AA in width. Diffusion is rapid enough to account for the speed of trans-mission observed at the synapse. After the neurotransmitter has exerted its effect on the postsynaptic membrane, it is promptly destroyed by an enzyme called cholinesterase. This destruction is of critical importance. If the acetyl-choline were not destroyed, it would continue its stimulatory action indefinitely and all control would be lost.

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Question:

What is a cone? Describe its structure.

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/Users/wenhuchen/Documents/Crawler/Biology/F08-0193.htm

Solution:

Cones are the typical reproductive structures of the gymnosperms. A cone or strobilus is a spiral aggregation of modified leaves called cone scales (see figure problem 8\rule{1em}{1pt}3). These cone scales are thesporophylls, and each scale bears on its surface the sporangia. In most gymnosperms, themicrosporangiaandmegasporangiaare borne on separate cones. In the conifers, both male and female cones are borne on the same plants. It would thus appear that self-fertilization occurs. However, the male cones are borne on the lower branches of the conifers, while the female cones are borne on the higher branches. Thus, self-fertilization is essentially prevented. The ovulate, or female cones, are larger. In these cones, the megasporangiaare enclosed by enveloping integuments to form structures called ovules. Each scale bears two ovules; each ovule is composed of the outer integuments and the innernucellus (megasporangium) and has a small opening at one end known as the micropyle, through which the pollen grains will enter. It is from the tissue of thenucellus, that the megaspore mother cell will be produced. It is of interest to note that the megaspore, which gives rise to the megagametophyte, is never released from themegasporangium. It remains embedded in the sporangium, which is enclosed to form the ovule. It is the ovule, which when mature, constitutes the naked seed of the gymnosperms. Male or staminate cones are smaller and bear themicrosporangia on their scales. The number ofmicro-sporangiaproduced by each microsporophyll varies among gymnosperms; in conifers, two is the usual number, while cycads have manymicrosporangiascattered on the lower surface of each scale. Each microsporangium will produce the microspore mother cell, which will give rise to numerousmicrogametesor pollen grains.

Question:

Given ∆H = 2800 kJ for the production of one mole of C_6H_12O_6 from CO_2 and H_2O, calculate what wavelength of light would be needed per molecule of CO_2 to achieve this photosynthesis in a one-step process.

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Solution:

To solve this problem: (1) set up the balanced equation for the production of 1 mole of C_6H_12O_6from CO_2 and H_2O. (2) From thestoichiometryof this equation determine the number of moles of CO_2 necessary to make 1 mole of C_6H_12O_6. (3) Calculate the number of molecules of CO_2 necess-ary to make 1 mole of C_6H_12O_6 using Avogadro's number (6.022 × 10^23 molecules / mole). (4) calculate the energy necessary per molecule of CO_2 by dividing ∆H by the total number of CO_2 molecules. (5) Use the expression E =hѵwhere E is the energy of light, h is Planck's constant, and ѵ is the frequency of light. Set the energy per molecule of CO_2 equal tohѵand solve for the frequency, ѵ. (6) Use the expression c =\lambdaѵ, (where c is the speed of light, ѵ is the frequency of light (from step 5) and \lambda is the wavelength of light) to find \lambda. To begin, the balanced equation for the production of 1 mole of C_6H_12O_6 is 6CO_2 + 6H_2O \rightarrow C_6H_12O_6 + 6O_2 Note that this process is photosynthesis. From thestoichiometryof this equation, 1 mole of C_6H_12O_6 is produced for every 6molesof CO_2 consumed. And, therefore, the number of molecules of CO_2 necessary to make 1 mole of C_6H_12O_6 is (6 moles CO_2) (6.02 × 10^23 molecules/mole) = 3.61 × 10^24 molecules. The energy necessary per molecule of CO_2 is (2,800,000 J) / (3.61 × 10^24 molecules CO_2) = 7.75 × 10-^19 J/molecule. Steps 5 and 6 are combined to give the expression E = h (c/\lambda) where h = 6.63 × 10-^34 J-s, and c = 3.00 × 10^10 cm/s. Setting this equal to 7.75 × 10-^19 J and solving for \lambda gives \lambda = (hc/ E) = {(6.63 × 10-^34 J-s) (3.00 × 10^10 cm/s)} / {7.75 × 10-^19 J} = 2.57 × 10-^5 cm = 2.57 × 10-^5 cm × [(10^8 \AA) / cm] = 2570 \AA.

Question:

BothHClandNaOHare strong electrolytes. What is the pH of the solution formed by adding 40 ml of 0.10 MNaOHto 10 ml of 0.45 MHCl?

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Solution:

We will solve this problem by considering the number of moles of H_3O^+ and of OH^- formed by the complete dissociation ofHClandNaOH, respectively. If we assume thatHClandNaOHdissociate completely, then the concentration of H_3O^+ in theHClsolution is equal to the initial concentration ofHCl, or [H_3O^+] = 0.45 M, and the concentration of OH^- in theNaOHsolution is equal to the initial concentration ofNaOH, or [OH^-] = 0.10 M. Since moles = concentration × volume, the number of moles of H_3O^+ in 10 ml of the acid solution is moles H_3O^+ = [H_3O^+] × volume = 0.45 M × 10 ml = 0.45 M × 0.01 liter = .0045 mole = 4.5 × 10^-3 mole. Similarly, the number of moles of OH^- in 40 ml of the basic solution is moles OH^- = [OH^-] × volume = 0.10 M × 40 ml = 0.10 M × 0.04 liter = 0.004 mole = 4 × 10^-3 mole H_3O^+ and OH^- neutralize each other according to the reaction H_3O^+ + OH^- \rightarrow 2H_2O. Ignoring the dissociation of water, we can assume that this reaction is complete. Hence, the 4 × 10^-3 moles of OH^- will be neutralized by 4 × 10^-3 moles of H_3O^+, leaving 4.5 ×10^-3 - 4.0 ×10^-3 = 0.5 ×10^-3 = 5 ×10^-4 mole of H_3O^+ remaining. Thus, when the two solutions are mixed, no OH^- remains and 5 ×10^-4 mole of H_3O^+ remains. Since the final volume of the solution is 40 ml + 10 ml = 50 ml = 0.05 liter = 5 ×10^-2 liter, the concentration of H_3O^+ is [H_3O^+] = 5 ×10^-4 mole/5 × 10^-2 liter = 10^-2 mole/liter. The pH is defined as pH = - log [H_3O^+], hence pH = - log [H_3O^+] = - log (10^-2) = - (- 2) = 2.

Question:

If, in tracing evolutionary relationships, anatomic evidence pointedone way and biochemical evidence the other, which doyou think would be the more reliable? Why?

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Solution:

It has been stressed again and again that evolutioncan notoccur withouta change in the genotype. Biochemical properties are controlled to agreater extent by genes than are anatomic ones. Anatomic characteristicsare more susceptible to modification by the external environmentin which the organism possessing them lives. Plants of the samespecies growing in two quite different habitats may demonstrate strikinglydifferent characteris-tics; this is because the two groups of plants areexposed to different environmental forces and achieve a different developmentalpotential even though the genes present in them are very similar. When we compare two organisms or groups of organisms in seekingan evolutionary relationship, we are essentially looking for a geneticrelationship between them. Since biochemical characteristics are moregreatly controlled by genes than are anatomic characteristics, a biochemicalsimilarity between the two organisms should be more reliable thanan anatomic similarity in indicating a genetic linkage and thus an evolutionaryrelationship. Therefore, biochemical evidence is more useful inthe tracing of evolutionary relationships between organisms.

Question:

If the year 1975 (year 1) is started with a 2 year old truck, what decision should be made at the start of this year and the start of the next four years in order to maximize the total return.

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Solution:

This is a problem in dynamic optimization. A series of decisions must be made over time so as to maxi-mize some chosen quantity. Each decision by itself may be non-optimal; the optimality of the sequence of decisions is what matters. As an example, suppose X wishes to travel from city A to city F and has the choice of routes ABCF and ADEF. It is possible for AB to cost less than AD and yet ADEF may be cheaper overall (DE is less than BC and EF is less than CF). Going back to the initial problem, let 1)r_i (t) = revenue in period i from a truck that was made in year (i - t) and is t years old at the start of period i. For example r_3(3) = 8. 2)ui(t) = upkeep in period i on a truck that was made in year (i - t) and is t years old at the start of period i. For example, a truck made in 1975 requires an up-keep of 2 in 1978, i.e., u_5 (3) = 2. 3)c_i(t) = cost to replace a truck that was made in year (i - t) and is t years old at the start of period i. 4)IT = age of the given truck = 2. 5)fi(t) = optimal return for periods i, i+1,...., 5 when the i-th period is started with a truck that is t years old. 6)X_i (t) = decision to make at the start of period i that will yield f_i(t). The only two possible decisions are to keep the old truck or to purchase a new one. Suppose the current truck was 2 years old at the start of 1975 (year 1), and the estimates in table I are available. Suppose we decide to keep the current truck until 1979. Then the total revenue is 10 + 8 + 8 + 6 + 4 = 36. Total upkeep is3 + 3 + 4 + 4 + 5 = 19. Thus total return is 17. On the other hand, suppose we decide to replace the 1973 truck with a truck made in 1975 at the start of 1975. Then, total revenue is 14 + 16 + 16 + 14 + 12 = 72 and the upkeep costs are TABLE I:Truck made... 1973 (year minus 1) Age Revenue Upkeep Replacement 2 10 3 25 3 8 3 26 4 8 4 27 5 6 4 28 6 4 5 29 Truck made in 1975 (year 1) Age Revenue Upkeep Replacement 0 14 1 20 1 16 1 22 2 16 2 24 3 14 2 25 4 12 3 26 Truck made in 1976 (year 2) Age Revenue Upkeep Replacement 0 16 1 20 1 14 1 22 2 14 2 22 3 12 2 25 Truck made in 1977 (year 3) Age Revenue Upkeep Replacement 0 18 1 20 1 16 1 22 2 16 2 24 Truck made in 1978 (year 4) Age Revenue Upkeep Replacement 0 18 1 21 1 16 1 22 Truck made in 1979 (year 5) Age Revenue Upkeep Replacement 0 20 1 21 1 + 1 + 2 + 2 + 3 = 9 Replacement costs are 25 and hence total return is 38. Of course we could keep the truck for a year and then sell it. The total return in this case works out to 37. Clearly there is a large number of possible choices. The functional equation for period i is fi(t) = max│purchase:ri(0) - u_i (0) + Ci(t)_ + fi+1(1)│ │keep:ri(t) - u_i (t) + fi+1(t + 1)│ for i = 1,2,...,5 and t = 1,2,....(i - 1), (i+ IT - 1) We now apply the functional equation working back-wards. That is, assume that the fifth year is reached with a truck that is 1, 2, 3, 4, or 6 years old. Thus, for i = 5 and t = 1 f5(1) = max│Purchase: r5(0) - u_5 (0) + C5(1)+ f6(1) │ │keep:r_5 (1) - u_5 (1) + f6+(2)│ Note that f_6 (j) = 0 for all j. Then f5(1) = max│20 - 1 - 24 + 0│ │16 - 1 + 0│ = 15 Thus, we keep the 1978 truck in 1979, i.e., x_5 (1) is keep. At t = 2 f5(2) = max│Purchase: r5(0) - u_5 (0) + C5(2)│ │keep:r5(2) - u_5 (2)│ = max│20 - 1 - 24│ │16 - 2│ = 14 x_5(2) = keep t = 3 f5(3) = max│Purchase: 20 - 1 - 25│ │Keep:12 - 2│ = 10 x5(3) = keep t= 4 f5(4) = max│Purchase:20 - 1 - 26│ │Keep:12 - 3│ = 9 x5(4) = keep t = 6 f5(6) = max│Purchase:20 - 1 - 29│ │Keep:4 - 5│ = -1 X5(6) = keep. Now, assume that the start of the fourth period is reached with a 1, 2, 3, or 5 year old truck. The optimal decision to make in order to maximize the return from the last two periods is given by the functional equation: f4(t) = max│P: r_4(0) - u_4(0) - C_4(t) + f_5(1)│ │K: r_4(t) - u_4(t) + f_5(t + 1)│ t = 1,2,3,4,5 Thus, f_4 (1) = max│P: 18 - 1 - 22 + 15│ │K: 16 - 1 + 14│ = 29 x_4(1) = Keep f_4 (2) = max│P: 18 - 1 - 24 + 15│ │K: 14 - 2 + 10│ = 22 x_4(2) = Keep f_4 (3) = max│P: 18 - 1 - 25 + 15│ │K: 14 - 2 + 9│ = 21 x_4(3) = Keep f_4 (5) = max│P: 18 - 1 - 28 + 15│ │K: 6 - 4 + (-1)│ = 4 x_4(5) = purchase Moving back to the start of the third period and consider-ing the optimal policy over the last three periods when the third period is started with a 1, 2 or 4 year old truck, we have f_3(t) = max│P: r_3(0) - u_3(0) - C_3(t) + f_4 (1)│ │K: r_3(t) - u_3(t) + f_4(t + 1)│ t = 1,2,4 f_3(1) = max│P: 18 - 1 - 22 + 29│ │K: 14 - 1 + 22│ = 35 x_3(1) = Keep f_3(2) = max│P: 18 - 1 - 24 + 29│ │K: 16 - 2 + 21│ = 35 x_3(2) = Keep f_3(4) = max│P: 18 - 1 - 27 + 29│ │K: 8 - 4 + 4│ = 19 x_3(4) = purchase Now move backwards from period 2 with t = 1, 3 f_2(1) = max│P: 16 - 1 - 22 + 35│ │K: 16 - 1 + 35│ = 50 x_2(1) = Keep f_2(3) = max│P: 16 - 1 - 26 + 35│ │K: 8 - 3 + 19│ = 24 x_2(3) = purchase or keep Finally, period 1 is reached where we have a 2-year old truck. To arrive at a decision whether to keep it or replace it in order to maximize the total return over the next five periods, we have f_1(2) = max│P: 14 - 1 - 25 + 50│ │K: 10 - 3 + 24│ = 38 x_1(2) = purchase Thus, we purchase a truck at the start of 1975. We keep this truck in 1976, 1977, 1978 and 1979 in order to maxi-mize total returns over the next four years.

Question:

An automobile moves with a constant speed of 50 mi/hr around a track of 1 mi diameter. What is the angular velocity and the period of the motion?

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/Users/wenhuchen/Documents/Crawler/Physics/D09-0358.htm

Solution:

For circular motion, the angular velocity \omega, the radius r, and the linear velocity v obey the re-lation: v\ding{217}= \omega\ding{217}× r\ding{217}as shown in the figure. \ding{217} \ding{217} \ding{217} Since \omega and r are perpendicular to each other, this reduces to v = \omegar. \omega = v/r = (50 mi/hr)/(0.5 mi) = 100 rad/hr The period \cyrchar\cyrt is the time duration of one complete cycle of motion around the circular path. In linear motion, x = vt. This equation can be applied to circular motion with linear velocity v replaced by \omega and linear distance x replaced by \texttheta expressed in radians. In one cycle of motion, the automobile travels 2\pi radians. Therefore\omegat = \textthetat = \texttheta/\omega\cyrchar\cyrt = 2\pi/\omega \cyrchar\cyrt = 2\pi/\omega= 2\pi/(100 rad/hr) = 0.063 hr = 3.8 min.

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Question:

Discuss the current theory of the evolution ofautotrophs fromprimitiveheterotrophs.

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/Users/wenhuchen/Documents/Crawler/Biology/F27-0712.htm

Solution:

The first organisms on earth were almost certainlyheterotrophs, which, in an atmosphere lacking free oxygen, presumably obtained energy bythe fermentation of certain organic substances in the sea. Because of theirdependence upon the external environment for food, they could surviveonly as long as the supply of organic molecules in the sea lasted. Because food in the sea was necessarily limited, before the supply was exhausted, some of the then existingheterotrophsmust have evolved the mechanismsto becomeautotrophs. They would then be able to make theirown organic molecules either by chemosynthesis or photosynthesis. An explanation of how anautotrophmay have evolved from one of theprimitiveheterotrophswas offered by N.H. Horowitz in 1945. Horowitz postulatedthat the evolution was a result of successive gene mutations in certainheterotrophswhich resulted in the enzymes needed to synthesize complexsubstances from simple ones. According to this hypothesis, the primitiveheterotrophswere able to survive only if a certain specific organic substancex, limited in supply, was present in the surrounding sea. If a mutationoccurred for a new enzyme enabling theheterotrophto synthesizesubstance x from some other more abundantly available organicform, the strain ofheterotrophswith this mutation would be able to survivewhen the supply of x in the sea was exhausted. A second mutation thatestablished an enzyme catalyzing another reaction by which x could bemade from inorganic substances would again have survival value. Finally, when all the organic molecules in the sea were exhausted, a mutationfor the synthesis of x from some inorganic substances, present in largequantities, marked an important evolutionary advantage. In this way, successive mutations over time led to the production of astrain of organisms able to synthesize all of its metabolic requirements fromsimple inorganic compounds. Such organisms, called theautotrophs, areexemplified by the many photosynthetic algae and chemosynthetic bacteriathat are believed to be among the earliest inhabitants of the earth.

Question:

Compute the speed of sound in the steel rails of a rail road track. The weight density of steel is 490 lb/ft^3, and Young's modulus for steel is 29 x 10^6 lb/in^2.

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/Users/wenhuchen/Documents/Crawler/Physics/D26-0828.htm

Solution:

For an elastic medium, the speed of longitudinal waves is given by v = \surd(Y/\rho) = \surd[(Yg)/D] Where Y is Young's modulus, \rho is the density of the medium and \rho = D/g where D is the weight density of the medium. v = \surd[{(29 × 10^6 lb/n^2) (32 ft/s^2) }/{ 490 lb/ft^3 }] In order to keep all length dimensions consistent, we must change 29 × 10^6 lb/in^2 to lb/ft^2, Hence, 29 × 10^6 lb/in^2 = [(29 × 10^6 1b) / (1/44 ft^2)] 29 × 10^6 lb/in^2 = 29 × 144 x 10^6 lb/ft^2 Therefore, v = \surd[{(29 × 10^6 144 lb/ft^2) (32 ft/sec^2)}/{ 490 lb/ft^3 }] = 1.6 × 10^4 ft/sec

Question:

Write a Basic program to simulate the game ofAwari.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G23-0568.htm

Solution:

Awariis an African game played with seven sticks and 36 beans, laid out as shown: Y 6 5 4 3 2 1 Y's Home Away From Home 000 000 000 000 000 000 X's Home 000 000 000 000 000 000 1 2 3 4 5 6 X In the given diagram X and Y are the two players. Suppose X has the first move. A move is made by taking all the beans from any non-empty pit on X's side and sowing them in a counterclockwise direction, one in each pit. A turn consists of one or two moves. If the last bean of X's move is sown in his own home he can take another move. Also, if the lastbeansown in a move lands in an empty pit and the pit opposite is not empty, then all the beans in that pit (together with the last bean sown) are captured and moved to the player's home. When either side is empty, the game is finished and the player with the most beans is the winner. Consider the following sample part game: Y 333333 00 333333 X X starts sowing from 4 333333 01 333344 Y starts sowing from 2: 344403 01 333344 X starts from 1 344403 01 044144 (He now captures the 4 stones on Y (3) and his X (4) to obtain 344003 06 044044 Y plays from 4: 040003 56 044044 X starts from 3 to obtain 040003 57 040155 Since he finished at home he can play again. He plays from 5: 040114 58 040006 and so on... The program follows: 10PRINT "AWARI\textquotedblright 15DATA 0 20DIM B (13), G (13), F (50): READ N 30E = 0 40FOR I = 0 TO 12 50B (I) = 3 60NEXT I 70C = 0 80F (N) =0 90B(13) = 0 100B (6) = 0 110GOSUB 540 120PRINT "YOUR MOVE"; 130GOSUB 310 140IF E = 0 THEN 210 145IF M = H THEN GOSUB 300 146IF E = 0 THEN 210 150PRINT "MY MOVE IS"; 160GOSUB 820 170IF E = 0 THEN 210 180IF M = H THEN PRINT " , " 190GOSUB 820 200IF E > 0 THEN 110 210PRINT "GAME OVER" 220D = B (6) - B (13) 230IF D < 0 THEN PRINT "I WIN BY"; -D; "POINTS" 240GOTO 30 250N = N + 1 260IF D = 0 THEN PRINT "DRAWN GAME" 270GOTO 30 280PRINT "YOU WIN BY"; D ;"POINTS" 290GOTO 30 300PRINT "AGAIN"; 310INPUT M 320IF M < 7 THEN IF M > 0 THEN M = M - 1 330GOTO 360 340PRINT "ILLEGAL MOVE" 350GOTO 300 360IF B (M) = 0 THEN 340 370H = 6 380GOSUB 400 390GOTO 540 400K = M 410GOSUB 680 420E = 0 430IF K > 6 THEN K = K - 7 440C = C + 1 450IF C < 9 THEN F (N) = F (N)\textasteriskcentered6 + K 460FOR I = 0 TO 5 470IF B (I) <> 0 THEN 500 480NEXT I 490RETURN 500FOR I = 7 TO 12 510IF B(I) <> 0 THEN E = 1 520RETURN 530GOTO 480 540FOR I = 12 TO 7 STEP - 1 550GOSUB 650 560NEXT I 570PRINT: I = 13 580GOSUB 650 590PRINT ""; 600PRINT B (6) 610PRINT ""; 620FOR I = 0 TO 5: GOSUB 650 630NEXT I 640RETURN 650IF B (I) < 10 THEN PRINT ""; 660PRINT B (I); 670RETURN 680P = B (M) 690B (M) =0 700FOR P = P TO 1 STEP - 1 710M = M + 1 720IF M > 13 THEN M = M - 14 730B (M) = B (M) + 1 740NEXT P 750IF B (M) = 1 THEN IF M <> 6 THEN IF M <> 13 THEN IF B (12 - M) <> 0 THEN 780 770RETURN 780B (H) = B(H) + B(12 - M) + 1 790B (M) = 0 800B (12 - M) = 0 810RETURN 820D = - 99 830H = 13 840FOR I = 0 TO 13 850G (I) = B (I) 860NEXT I 870FOR J = 7 TO 12 880IF B(J) = 0 THEN 1110 885G = 0: M = J: GOSUB 680 886FOR I = 0 TO 5: IF B (I) = 0 THEN 970 890L = B (I) + I 900R = 0 910IF L > 13 THEN L = L - 14 920R = 1 930GOTO 910 940IF B (L) = 0 THEN IF L <> 6 THEN IF L <> 13 THEN R = B(12 - L) + R 960IF R > Q THEN Q = R 970NEXT I 980Q = B(13) - B (6) - Q 990IF C > 8 THEN 1050 1000K = J 1010IF K > 6 THEN K = K - 7 1020FOR I = 0 TO N - 1 1030IF F (N)\textasteriskcentered6 + K = INT (F(I)/6 \uparrow (7 - C) +.1) THEN Q = Q - 2 1040NEXT I 1050FOR I = 0 TO 13 1060B (I) = G (I) 1070NEXT I 1080IF Q > = D THEN A = J 1090D = Q 1100NEXT J 1110M = A 1120PRINT CHR$ (42 + M); 1130GOTO 400 1140FOR I = 0 TO N - 1 1150PRINT B (I) 1160NEXT I 1170END

Question:

Use the uncertainty principle to estimate the kinetic energy of a neutron in a nucleus of radius R = 2 × 10^-13 cm.

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/Users/wenhuchen/Documents/Crawler/Physics/D33-0985.htm

Solution:

We write the uncertainty principle in the form \Deltap\Deltax= h Then \Deltax\cong 2R for the neutron can be found anywhere in the nucleus. The\Deltaxgiven expresses this lack of certainty as to where in the nucleus it can be found. As a result of the uncertainty principle, the neutron can acquire a momentum P =\Deltap = (h/2R) The kinetic energy,T = (1/2) mv^2 = [1/2{(mv)^2/m}] = (p^2/2m) = [h^2/(8mR^2)] T = [{(6.63 × 10^-27 erg-sec)^2}/{8(1.67 × 10^-24 gm)(2 × 10^-13 cm)^2}] = 8.22 × 10^-5 [(erg^2-sec^2)/(gm-cm^2)] = 8.22 × 10^-5 [(erg^2) / {(gm-cm/sec^2) -cm}] = 8.22 ×10^-5 (erg^2/dyne-cm) = 8.22 ×10^-5 erg

Question:

What are the roles of coenzyme A in metabolism ?

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/Users/wenhuchen/Documents/Crawler/Biology/F03-0074.htm

Solution:

Coenzyme A is a large, complex organic molecule involved in a number of key reactions in the cell. The functions of coenzyme A are several. In glycolysis, glucose is converted to pyruvate. To enter the Krebs acid cycle (also known as the tricarboxylic acid, or TCA cycle and the citric acid cycle) , pyruvate must be con-verted to another form; such a conversion first involves decarboxylation of pyruvate (loss of carbon dioxide), then a reaction with coenzyme A, and finally, a dehydrogenation re-action, to yield acetyl coenzyme A. The reactions are rep-resented as follows: The acetate group is bonded to the coenzyme A molecule via a high energy bond. The energy of this bond in acetyl coenzyme A is utilized to transfer the acetyl group to oxaloacetic acid, the first reaction of the TCA cycle. Coenzyme A is also involved in a reaction farther along in the cycle. An intermediate (derived from \alpha- Ketoglutarate) combines with coenzyme A to yield succinyl coenzyme A. The bond connecting coenzyme A to succinate is also of high energy. The energy in this bond is converted to an energy-rich phosphate bond,\simP, through the addition of an inorganic \sim phosphate. The phosphate group is subsequently transfered to GDP to produce GTP and free succinate. The reactions are diagrammed below: Coenzyme A is also utilized in fatty acid oxidation. The metabolism of a fatty acid initially involves its activation by ATP and coenzyme A to fatty acyl coenzyme A. Through a repeating series of reactions, fatty acyl coenyzme A is converted to acetyl coenzyme A, which, in turn, goes into the TCA cycle As a general rule, coenzyme A is involved in reactions which utilize its high-energy bond in order to more effectively facilitate metabolism. The sequence of reactions, wherein Acyl CoA is shortened by two carbon atoms, repeats itself until the Acyl CoA is shortened to Acetyl CoA.

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Question:

If I_2 and Br_2 are added to a solution containing I^- and Br^-, what reaction will occur if the concentration of each species is 1m? Some half-reactions and their standard reduction potentials. Increasing strength as oxidixing agents Half - reaction Increasing strength as reducing agents Standard reduction potential, V 2e^- + F_2(g) \rightarrow 2F^- +2.87 2e^- + Cl_2(g) \rightarrow 2Cl^- +1.36 4e^- + 4H_3O^+ +O_2(g) \rightarrow 6H_2O +1.23 2e^- + Br_2(g) \rightarrow 2Br^- +1.09 e^- + Ag^+ \rightarrow Ag(s) +0.80 2e^- + I_2 \rightarrow 2I^- +.054 2e^- + Cu^2+ \rightarrow Cu(s) +0.34 2e^- + 2H_3O^+ \rightarrow H_2(g) + 2H_2O Zero 2e^- + Fe^2+ \rightarrow Fe(s) -0.44 2e^- + Zn^2+ \rightarrow Zn(s) -0.76 3e^- + Al^3+ \rightarrow Al(s) -1.66 2e^- + Mg^2+ \rightarrow Mg(s) -2.37 e^- + Na^+ \rightarrow Na(s) -2.71 e^- + Li^+ \rightarrow Li(s) -3.05

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/Users/wenhuchen/Documents/Crawler/Chemistry/E16-0585.htm

Solution:

When I_2 and Br_2 are added to the solution, a dynamic equilibrium is attained between the non-charged element and its ion form. These reactions, redox reactions, are written as: 2e^- + I_2 \rightarrow 2I^-+ 0.54V 2e^- + Br_2 \rightarrow 2Br^-+ 1.09V The voltages listed indicate the electric potential between two electrodes. In other words, it is a measurement of the work done in moving a unit charge (an electron) from one electrode to the other. For any reaction to occur, the half reactions must be written such that the overall vol-tage is positive. It is written positively to indicate that the reaction tends to go spontaneously. Any redox reaction for which the overall potential is positive has the tendency to take place as written. Therefore one must pick the larger voltage in the half reaction to be positive and the other negative. Thus, 2e^- + Br_2 \rightarrow 2Br^-+ 1.09V 2I^- + I_2+2e^- -0.54V 2I^- + Br_2 \rightarrow I_2 + 2Br^- The half-reactions are written so that when added together, the electrons cancel out. Because the reaction, 2I^- + Br_2 \rightarrow I_2 + 2Br^- , has a positive voltage, it will proceed as written.

Question:

A girl is a hemophiliac, a) What are the possible genotypes and phenotypes of her parents?b) Assuming that her mother is normal, what were this girl's chances of being born with the disease?c) Several cases of hemophilia in girls have been reported within a small region in England where there is much close intermarriage. Explain this high frequency of hemophilia in girls.

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/Users/wenhuchen/Documents/Crawler/Biology/F25-0669.htm

Solution:

a) Hemophilia is a sex-linked trait and its gene is carried on the X chromosome. It is also a recessive trait, and so will not be expressed in the presence of a normal X chromosome. It will be expressed, however, in the presence of a normal Y chromosome, because that chromosome does not carry any genes that would mask its expression. This means that in order for a boy to be a hemophiliac, he need only have one parent who carries the gene. A girl, however, would have to have both parents carry the gene, since she receives one X chromosome from each parent. Let ^+X represent the chromosome carrying the gene for hemophilia, and X and Y represent the normal sex chromosomes. The father of a hemophiliac girl, in order to be able to give his daughter the gene for hemophilia, must be ^+X Y, and therefore a hemophiliac himself. The mother could be either ^+X^+X, (a hemophiliac), or ^+XX, (a carrier), in order to pass the hemophilia gene to her daughter. b) If the mother is normal, her genotype must be ^+XX. Let us look at the cross that produced the hemop-hiliac girl. F^1 + X Y + X + X^+X + XY X + XX XY The offspring will be as follows: 1/4 female carrier (X^+X) 1/2 normal1/4 normal male (^+X^+X) 1/4 hemophiliac male (^+X Y) 1/2 hemop-hiliac1/4 hemophiliac female (^+X^+X) 1/2 hemop-hiliac Since half the possible offspring are hemophiliac, the girl had a 50% chance of being born with the disease. c) While within the family above, a female has the same chance as a male of being a hemophiliac, this is not the case in the population as a whole. Because of the fact that it requires that a father be hemophiliac and a mother be at least a carrier in order to produce a hemo-philiac girl, there is usually only a rare case now and then of a girl being born with hemophilia. The hemophilia gene has a lower frequency than the normal gene in the population, and the chances that both parents will have this gene are very low. However, within the particular region in question, there is much intermarriage. If a gene is carried by a family member, its frequency among other family members will be much higher than in the population as a whole. If an individual carries a particular gene, in this case the gene for hemophilia, and marries within the family, there is a good chance that their spouse will also carry the gene. There is thus an increased probability that it will be carried by offspring in the homozygous state, in this case resulting in hemophiliac girls.

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Question:

In a double slit interference experiment the distance between the slits is 0.05 cm and the screen is 2 meters from the slits. The light is yellow light from a sodium lamp and it has a wavelength of 5.89 × 10^-5 cm. What is the distance between the fringes?

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Solution:

To find the distance between fringes in the double slit experiment, we must first derive the formulas for the location of the maxima and minima of the fringe pattern. Let us examine this experiment in more detail. Light is incident on the 2 slits from the left. (See figure) MP and AP represent 2 rays of light, one from each slit, arriving at P. Typically, L >> d, and we may con-sider MP to be equal to BP. Assuming that the light rays emerging from the slits are in phase, the two light rays arriving at P will be out of phase because light from A must travel the extra distance AB when compared with light from M. If this path difference (AB = d sin \textphi ) is equal to an even number of half wavelengths, P will be a maximum point. If AB equals an odd number of half wavelengths, P will be a minimum point. Hence, For a maximum sin \textphi = (2n) \lambda/2d(n = 0, 1, 2,...) For a minimum sin \textphi = [(2n + 1)] \lambda/2d(n = 0, 1, 2,...) Therefore, the angular location of adjacent maxima on the screen (say the nth and (n + 1) th maxima,) is sin (\textphi_n+1) = [{2(n + 1)\lambda} / {2d}] = [(n + 1)\lambda/d](1) sin (\textphi_n) = [{(2n)\lambda} / {2d}] = n\lambda/d But, if \textphi is small, sin (\textphi_n+1) \approx tan (\textphi_n+1) sin (\textphi_n) \approx tan (\textphi_n). Hence, using (1) and the figure, (Y_n+1)/L = [{(n + 1)\lambda}/d] Y_n/L = n\lambda/dhence, Y_n+1 - Y_n = [{(n + 1)\lambdaL}/d] - (n\lambdaL)/d Y_n+1 - Y_n = \lambdaL/d. This is the screen separation of 2 adjacent maxima. If\lambda = 5.89 × 10^-5 cm L = 200 cm d = 0.05 cm. Y_n+1 - Y_n = [{(5.89 ×10-5× 200)cm^2} / {0.05 cm}] = .233 cm.

Question:

A rectangular coll 30 cm long and 10 cm wide is mounte in a uniform field of flux density 8.0 × 10^-4 nt/amp -m. There is a current of 20 amp in the coil, which has 15 turns. When the plane of the coil makes an angle of 40\textdegree with the direction of the field, what is the torque tending to rotate the coil?

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/Users/wenhuchen/Documents/Crawler/Physics/D21-0731.htm

Solution:

The torque on a circuit in a field of magnetic induction, B^\ding{217}, is T^\ding{217} = \mu^\ding{217} × B^\ding{217}(1) where \mu^\ding{217} is the magnetic moment of the circuit. (This is the property of the circuit which causes the torque to be exerted.) The magnitude of the magnetic moment is \mu = NIA(2) where N is the number of turns in the circuit, I is the current in the circuit, and A is the area it encloses. The direction of \mu^\ding{217} is given by the right hand rule: wrap the fingers of Your right hand around the circuit in the direction of the current, and the direction in which your thumb points will then be the sense of \mu^\ding{217}. Since we ^\ding{217} only want the magnitude of T^\ding{217}, we write T = \muB sin \texttheta(3) where T, \mu, B are the magnitudes of T^\ding{217}, \mu^\ding{217}, B^\ding{217}, and \texttheta is the angle between the directions of \mu^\ding{217} and B^\ding{217} (see figure). Sub-stituting (2) into (3), we obtain T = NiAB sin \texttheta(4) However, the data is given in terms of flux density, not in terms of B. But flux density is actually equal to B because Flux density = (\cyrchar\CYRF/A) = (BA/A) = B where A is the area enclosed by the circuit, and \cyrchar\CYRF is the flux cutting through the circuit. We still cannot proceed yet, because we do not have \texttheta. The question gives us the angle between the plane of the coil and the direction of B^\ding{217}. (In the figure this is \alpha.) The angle we need, \texttheta, is 90\textdegree - a = 50\textdegree. Inserting the given data in (4), we find T = [8 × 10^-4 (nt/A \bullet m)] (15) (20A) (.3m) (.1m) (.77) T = .0055 nt \bullet m

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Question:

1) Two linearly polarized waves are in phase, but have different amplitudes. At x = 0 E^\ding{217}_1 = \^{\i} A_1 cos 2\piѵt + \^{\j} B_1cos 2\piѵt, E^\ding{217}_2 = \^{\i} A_2 cos 2\piѵt + \^{\j} B_2cos 2\piѵt. Show that E^\ding{217}_T = E^\ding{217}_1 + E^\ding{217}_2 is also linearly polarized, and find its polarization direction. 2) Two circularly polarized waves (a right and a left) can be added to form a linearly polarized wave: At x = 0, E^\ding{217}_1 = \^{\i} E_o cos 2\piѵt + \^{\j} Eosin 2\piѵt, E^\ding{217}_2 = \^{\i} E_o cos (2\piѵt + \alpha) - \^{\j} E_osin (2\piѵt + \alpha) . Show that E^\ding{217}_T is linearly polarized, and find its polariza-tion direction. 3) An elliptically polarized wave is written (at z = 0) as E^\ding{217}_T = \^{\i} A_1 sin 2\piѵt + \^{\j} Bcos 2\piѵt, Show that this can be decomposed into a linearly and a circularly polarized wave.

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/Users/wenhuchen/Documents/Crawler/Physics/D37-1093.htm

Solution:

1) E_1 = (\^{\i}A_1 + \^{\j}B_1) cos 2\piѵt, E_2 = (\^{\i}A_2 + \^{\j}B_2) cos 2\piѵt. Now, add the components of E_1 and E_2 together to find the expression for E_T. Then, E_T = {\^{\i}(A_1 + A_2) + \^{\j}(B_1 + B_2)} cos 2\piѵt. The point here is that the quantity in { } just specifies a vector, and the time variation is the same for all parts of it. So the vector lies along a line and changes size, but not direction. Therefore, E^\ding{217}_T is linearly polarized. In the case of circular polarization, this is not true, since the x and y components vary differently in time, one being large when the other is small. 2) Left: E_1 = \^{\i}E_o cos 2\piѵt + \^{\j}E_o sin 2\pivt, Right: E_2 = \^{\i}E_o cos (2\piѵt + \alpha) - \^{\j}E_o sin (2\pivt + \alpha). E_T = \^{\i}E_0{cos 2\piѵt + cos (2\piѵt + \alpha)} + \^{\j}E_0{ sin (2\piѵt - sin(2\piѵt + \alpha)} cos(2\piѵt + \alpha) = cos [2\piѵt + (\alpha / 2) + (\alpha / 2)] Using one of the double angle formulas, we find cos (2\piѵt + \alpha) = cos[2\piѵt + (\alpha / 2)] cos (\alpha/2) - sin [2\pivt + (\alpha/2)] sin (\alpha/2) = cos [2\piѵt + (\alpha / 2) cos (\alpha / 2) - [sin 2\piѵtcos(\alpha / 2) + cos 2\piѵt sin(\alpha / 2)] sin(\alpha / 2) where we have made use of the formula sin (x + y) = sin x cos y + cos x sin y. Then, cos (2\piѵt + \alpha) = cos [2\piѵt + (\alpha / 2)] cos (\alpha / 2) - sin 2\piѵtcos (\alpha / 2) sin(\alpha / 2) - cos 2\piѵtsin^2 (\alpha / 2). Therefore, cos 2\piѵt + cos(2\piѵt + \alpha) = cos 2\piѵt + cos[2\piѵt +(\alpha / 2)] cos (\alpha / 2) - sin 2\piѵtcos (\alpha / 2) sin(\alpha/ 2) - cos 2\piѵt sin^2 (\alpha / 2). Substituting 1 - cos^2(\alpha / 2) for sin^2(\alpha / 2) in the preceding expres-sion yields the following result: cos(2\piѵt) + cos(2\piѵt + \alpha) = cos(2\piѵt) + cos[2\piѵt + (\alpha / 2)] cos(\alpha / 2) -sin(2\piѵt) cos(\alpha / 2) sin(\alpha / 2) + cos(2\piѵt) cos^2(\alpha / 2) - cos(2\piѵt). The cos(2\piѵt) terms cancel one another, and so, cos (2\piѵt) + cos (2\piѵt + \alpha) = cos [2\piѵt + (\alpha / 2)] cos(\alpha / 2) + cos(\alpha / 2)[cos(2\piѵt)cos (\alpha / 2) - sin(2\piѵt) sin(\alpha / 2)]. By applying the double angle formula cos (x + y) = cos x cos y -sin x sin y, we have the result that cos (2\piѵt) + cos (2\piѵt + \alpha) = 2 cos [2\piѵt + (\alpha / 2)] cos(\alpha / 2) Similarly, sin (2\piѵt) - sin (2\piѵt + \alpha) = 2 sin(\alpha / 2) cos [2\piѵt + (\alpha / 2)]. Therefore, E_T = 2\^{\i}E_0 cos [2\piѵt + (\alpha / 2)] cos(\alpha / 2) + 2\^{\j}E_0 sin (\alpha / 2) cos[2\piѵt + (\alpha / 2)] = 2E_0 [(\^{\i}cos (\alpha / 2) + \^{\j} sin(\alpha / 2)] cos [2\piѵt + (\alpha / 2)] . Since the two components are in phase, we can add them in a vector diagrams:ET is a vector of magnitude 2A cos { 2\piѵt + (\varphi_1 / 2) } at an angle of \varphi_(1/2) to the y axis. Notice that we can see some easy limiting cases in the first two lines: If \alpha = 0 the y components cancel out, while the x components add to E_T = \^{\i}2E_0 cos (2\piѵt )(\alpha = 2\pi just multiplies everything by -1). If \alpha = (\pi / 2), we get E_T =(2E_0) [\^{\i} cos (\pi / 4)+ \^{\j}sin (\pi / 4)] cos [2\piѵt + (\alpha / 4)] = \surd(2E_0)_ (\^{\i} + \^{\j}) (cos 2\piѵt cos (\alpha / 4)-sin 2\piѵt sin(\alpha / 4)] = E_0(\^{\i} + \^{\j}) (cos 2\piѵt - sin 2\piѵt) linearly polarized at 45 degrees 3) E = \^{\i}A sin 2\piѵt + \^{\j}B cos 2\piѵt. First take out a circularly polarized part : E_o = \^{\i}A sin 2\piѵt + \^{\j}A cos 2\piѵt . What is left is E_lin =\^{\j}(B - A) cos 2\piѵt , a linearly polarized wave. To show that the original wave is elliptically polar-ized, draw the vector at various times, as shown in the figure.

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Question:

The figure shows a box being dragged along a horizontal surface by a constant force P making a constant angle \texttheta with the direction of motion. The other forces on the box are its weight W^\ding{217} = mg^\ding{217}, the nor-mal upward force N^\ding{217} exerted by the surface, and the friction force f^\ding{217}. What is the work done by each force when the box moves a distance s along the surface to the right?

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Solution:

The component of P^\ding{217} in the direction of motion is P cos \texttheta. The work of the force P^\ding{217} is by definition W_p = \int P^\ding{217} \textbullet ds^\ding{217} = P cos \texttheta \int ds = P(cos \texttheta)(s) for P^\ding{217} is a constant force, ds^\ding{217} is a vector in the direction of horizontal motion. The forces w^\ding{217} and N^\ding{217} are both at right angles to the displacement. Hence W_w = \intw^f \textbullet ds^\ding{217} = ws cos 90\textdegree = 0 (W^\ding{217} is constant. Therefore it may be taken out of the integral). W_N = \intw^\ding{217} \textbullet ds^\ding{217} = Ns cos 90\textdegree = 0 (N^\ding{217} is constant. Therefore it was taken out of the integral). The friction force f^\ding{217} is opposite to the displacement, so the work of the friction force is W_F = \intf^\ding{217} \textbullet ds^\ding{217} = fs cos 180\textdegree = -fs Since work is a scalar quantity, the total work W of all forces on the body is the algebraic (not the vector) sum of the individual works. W = W_p + W_w + W_N + W_f = (P cos \texttheta) \bullet s + 0 + 0 - f \bullet s = (P cos \texttheta - f)s. But (P cos \texttheta - f) is the resultant force on the body. The sum of the forces in the vertical direction, acting on the body is zero, for the object moves only in the horizontal direction. Hence the total work of all forces is equal to the work of the resultant force. Suppose that w = 100 lb, P = 50 lb, f = 15 lb, \texttheta = 37\textdegree , and s = 20 ft. Then W_p = (P cos \texttheta) \textbullet s = 50 × 0.8 × 20 = 800 ft.lb, W_f= - fs = -15 × 20 = -300 ft.lb, W= W_p + W_w + W_N + W_f = 800 ft\bulletlb + 0 + 0 - 300 ft\bulletlb = 500 ft\bulletlb

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Question:

The quadrilateral ABCD is a square of side 1 ft which can rotate about the fixed point 0, which is the midpoint of the diagonals. Forces 2, 3, 2, and 1 lb act along sides AB, BC, CD, and DA, respectively. Find the magnitude and line of action of a single force which would produce the same effect as these four forces.

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Solution:

At first sight this may not appear to be a problem in equilibrium, but the easiest method of solution is obtained when it is changed into one. Add a fifth force E\ding{217}, the force necessary to produce equilibrium. This force is just sufficient to negate the translational and rotational effects of the resultant force. E\ding{217} (see diagram) acts at the angle \texttheta to AB, at a distance of x from A. The resultant required in the problem must be equal and opposite to E\ding{217}, since the single force equivalent to the four given forces must, with E\ding{217}, produce equilibrium. Resolve the five forces parallel to AB and at right angles to AB. Since the forces are in equilibrium, 2 lb - 2 lb - E cos \texttheta = 0 and1 lb - 3 lb + E sin \texttheta= 0. E cos \texttheta = 0andE sin \texttheta = 2 lb. \texttheta = 90\textdegreeandE = 2 lb. Taking moments about 0 and using the condition for equilibrium, one obtains -2lb × (1/2)ft - 3lb × (1/2)ft - 2lb × (1/2)ft - lb × (1/2)ft + E[x - (1/2)ft] = 0 E[x - (1/2)ft] = 4 ft \textbullet lb x - (1/2)ft = (4ft \bullet lb)/(2lb) = 2 ftorx = 2(1/2)ft. Thus, the equilibrant has a magnitude of 2 lb and acts at right angles to AB in a direction away from 0, along a line passing through a point at a distance of 2(1/2) ft from A. The resultant required has thus the same magnitude and position but acts at right angles to AB toward 0.

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Question:

What must be the width a of a rectangular guide such that the energy of electromagnetic radiation whose free-space wavelength is 3.0cm travels down the guide (a) at 95% of the speed of light? (b) At 50% of the speed of light?

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Solution:

The group speed is given by: v_gr v_gr = 0.95c = c\surd = 0.95c = c\surd [1 - (\lambda / 2a)^2] . Solving for a yields a = 4.8cm; repeating forv_gr= 0.50c yields a = 1.7cm. If \lambda = 2a, thenv_gr= 0 and energy cannot travel down the guide. For the radiation considered in this example \lambda = 3.0cm, so that the guide must have a width a of at least (1/2) × 3.0cm = 1.5 cm if it is to transmit this wave.The guide whose width we calculated in (a) above can transmit radiations whose free-space wavelength is 2 × 4.8cm = 9.6 cm or less.

Question:

How do higher plants get rid of their waste products?

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Solution:

Since green plants undergo both photosyn-thesis and respiration, the products of one process may become the raw materials of the other and vice versa. Thus oxygen, the product of photosynthesis, is utilized in cellular respiration while the products of respiration, carbon dioxide and water, are used in photosynthesis. These products may also diffuse out through the pores of the leaves, depending on the dominance of either process at a particular time. For example, during the night, when respiration predominates, carbon dioxide and water vapor escape from the pores of the leaf surfaces. The amount of nitrogenous wastes in plants is small compared to that in animals, and can be eliminated by diffusion either as ammonia gas through the pores of the leaves or as nitrogen-containing salts through the membrane of the root cells into the soil. Other wastes such as oxalic acid accumulate in the cells of the leaves and are eliminated from the plant when the leaves are shed.

Question:

The solubility products of Fe(OH)_2 and Fe(OH)_3 are 10-^17 and 10-^38 , respectively. If the concentrations of Fe^2+ and Fe^3+ are each 10-^5 M, at what pH will each hydroxide just begin to precipitate?

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Solution:

To solve this problem, set up the solubility product equa-tions for Fe(OH)_2 and Fe(OH)_3 . Once this is done, substitute the values of Fe^2+ and Fe^3+ concentrations and solve for the OH- ion concentrations. Since pH +pOH= 14, then pH = 14 -pOH, wherepOH= -log[OH-]. One has, therefore, [Fe^2+] [OH-]^2 = 10-^17 and [Fe^3+] [OH-]^3= 10-^33 . For precipitation to occur in these solutions [Fe^2+] [OH-]^2 \geq 10-^17 and [Fe^3+] [OH-]^3 \geq 10-^33 . When the product of the concentrations is equal to zero the concentrations are just sufficient for preci-pitation to occur. Solving for the pH at these concentrations: For Fe(OH-)_2 , [OH-]^2 =(10-^17 ) / [Fe^2+] = (10-^17 ) / (10-^5 ) = 10-^12, [OH-]^2 =(10-^17 ) / [Fe^2+] = (10-^17 ) / (10-^5 ) = 10-^12, so that [OH-] = 10-^6 therefore,pOH= 6 and thus pH = 8 . For Fe(OH)_3 , [OH-]^3 = (10-^38 ) / [Fe^2+] = (10-^38 ) / ((10-^5 ) = 10-^33 , so that [OH-] = 10-^11 . Consequently, pOH= 11andpH = 3 .

Question:

Find a general formula to give information about the number of passes through the loop in both the best case and the worst case for the selection sort. Express in terms of N elements.

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Solution:

Take the worst case first. For the integers 5, 4, 3, 2, 1, the selection sort will take the following ac-tions: 1. Find the smallest entry and interchange it with the element in the first position. 1, 4, 3, 2, 5 2. Sort the remaining N - 1 elements in positions 2 through N of the array 1, 2, 3, 4, 5 If you refer back to the program, notice that on the first pass through the outer loop, the inner loop is executed for K from 2 to N; on the second pass, from 3 to N, and so on. The total number of passes through the inner loop is (N - 1) + (N - 2) + (N - 3) + ... 3 + 2 + 1, which comes to N (N - 1) / 2 passes through the inner loop. Unfortunately, the same number of passes is required in the best case to check the list. The best way to understand the formula above is to write the series in the order given and in reverse: (N - 1) + (N - 2) + (N - 3) + ... 3 + 2 + 1 1 + 2 + 3 + ... + (N - 3) + (N - 2) + (N - 1) Adding term by term, we get twice the sum of the series, which is N + N + N + ... + N + N + N Since there are N-l terms in the above series, twice the sum of the series may be written as N (N - 1). Dividing by 2 gives us the sum of the series, or N (N -1) / 2.

Question:

Explain how a "Skinner box" can be used in operant conditioning.

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Solution:

In operant conditioning, a particular behavior is rewarded or punishedwhen it occurs, thereby increasing or decreasing the probability thatit will occur again. The Skinner box (named after B. F. Skinner, a behavioralpsychologist) is a principal means of studying this type of learning. The typical Skinner box is a small cage with a bar and food dish. When a rat or other small animal is placed in the box, its exploratorybehavior might cause it to press the bar accidentally. If the rat isrewarded by being given a food pellet, it is more probable that it will pressthe bar again. This is called positive reinforce-ment. It is critical that thereward be given within a second or so after the response in order for conditioningto be affective. The reward is given whenever the rat presses thebar. The rat eventually learns to associate bar-pressing with food. Repetition produces a degree of learning beyond which there is no improvement. One can eliminate the bar pressing behavior by eliminating the reinforcement. This is called extinction, which is the decay of the original learnedbehavior in the absence of the reward.Therateof extinction dependson the extent and schedule of positive reinforcement. In-creased repetitionmakes a response more resistant to extinction. Also, reinforcementat temporal intervals (giving a reward for every fourth bar press) is more difficult to extinguish than continuous reinforcement (rewardfor every bar press). It is also possible to condition the rat not to pressthe bar through the use of punishment. By electrically shocking the ratcon-currently with bar pressing behavior, the rat associates the two and learnsnot to press the bar. Operant conditioning is sometimes called trial-and- error learning whenit occurs in nature. Young chicks will initially peck at any small object; eventually they peck only at edible objects. When the chick pecks atinedible objects, it gets no reward, so this behavior is extinguished. It learnsto peck only at edible objects since they act as a positive reinforcement.

Question:

Why isn't a biome map of the earth a true representationof thevegetation formations found on land?

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Solution:

The naturalist travelers of the eighteenth and nineteenth centuries sawstrikingly different kinds of vegetations parceled out into formations. They drew maps showing the extent of each formation. However, drawing amap involves drawing boundaries. Where there are real, distinct boundarieson the ground this is an easy task. Such boundaries include thosebetween sea and land, along deserts or beside mountain ranges. But often there are no real boundaries on the ground. In fact, most of the formationboundaries of the earth are not distinct. Real vegetation types usuallygrade one into another so that it is impossible to tell where one formationends and another begins. The apparent distinctness of the vegetation formations, when viewedfrom afar, is largely an optical illusion as the eye picks out bands whereindividual species are concentrated. A vegetation map-maker could plotthe position of a tree line well enough, if he had a few reports from placeswith distinct tree lines which he could plot and link up. However, thegradual transitions, such as between the, temperate woods and tropicalrain forests would be much more difficult to plot. So mapmakers drawboundaries using what seems to be the middle of the transitions. When the map is finished, the area is parceled out by blocks of formations ofplants. A map like this shows the vegetation of the earth to be more neatly set into compartments than it really is.

Question:

Chlorine is found in nature in two isotopic forms, one of atomic mass 35amuand one of atomic mass 37amu. The average atomic mass of chlorine is 35.453amu. What is the percent with which each of these isotopes occurs in nature?

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Solution:

This is a mathematical problem, the solution to which centers on defining the average value properly. Consider a set of N observations or measurements, M_1, M_2, .... , M_N, Let the probability that the observations M_1 is madebep_1, the probability that the observation M_2 is madebeP_2, and so on. Then, the average A is defined as: A = p_1M_1 + p_2M_2 + .... +p_NM_N. Since, by definition, the sum of the probabilities must be one, then: 1 = p_1 + p_2 + ... + P_N. For this problem there are two observations, hence N = 2. Let the first observation be M_1 = 35amuand the second observation be M_2 - 37 amu. The average A = 35.453amu. Hence, A= p_1M_1 + p_2M_2 35.453amu= p_1 × 35amu+ p2× 37amu where1= p_1 + p_2. One must determine p_1 and p_2 (the probabilities of occurrence of isotope 35 and 37, respectively). From the second of these relationships, 1 = p_1 + p_2 one obtains p_2 = 1 - p_1. Then A= p_1M_1 + p_2M_2 = p_1M_1 + (1 - p_1)M_2 = p_1M_1 + M_2 - p_1M_2 = M_2 + (M_1 - M_2)p_1 or, p_1 = (A - M_2)/(M_1 - M_2) = (35.453amu- 37amu)/(35amu- 37amu) = 0.2265 Then, p_2 = 1 - p_1 = 1 - 0.7735 = 0.2265. Thus, the isotope of mass 35amuoccurs with probability of 0.7735 or 0.7735 × 100 % = 77.35 % and the isotope of mass 37amuoccurs with a probability of 0.2265 or 0.2265 × 100 % = 22.65 %. In reality, the two isotopes do not have integral atomic masses, and the percent occurrences calculated above are not exactly correct.

Question:

What is the critical angle of incidence for a ray of light passing from glass into water. Assume n_glass = 1. 50 and n_water =1.33

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Solution:

When light passes from a more optically dense to a less optically dense medium, there is an angle of incidence i, called the critical angle, at which the light ray (ray c of the diagram) will be refracted parallel to the interface of the media. The angle of refraction r will be 90\textdegree in this case. Therefore, by Snell's Law, n_1 sin i = n_2 sin r. Also critical angle of incidence means sin r = 1. \therefore sin i = (n_2/n_1) = (1.33/1.50) = .887 \therefore i_c = 62\textdegree (approx).

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Question:

Explain the four important evolutionary events that occurred inthe Paleozoic Era.

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Solution:

The Paleozoic Era was marked by four important evolutionary events. First, in the Ordovician Period, the ancestral backbone-bearing animalsor vertebrates appeared. These early vertebrates were the ostracoderms, small, jawless, armored, fresh-water, bottom-dwelling fisheswithout fins. Theostracodermsthen gave rise in the later periods to agreat variety of finned fish. In the period following the Ordovician, the Silurian, two events of great biologic importance marked the second and thirdmajor evolutionary advancement of the Paleozoic Era. The first land plantsappeared, signifying the beginning of the invasion of the land by plants. These plants most closely resembled the ferns. At about the same time, the first air-breathing animals arose. These were thearachnids- landdwellers resembling to some extent modern scorpions. Finally, in the Carboniferous Period, the first reptiles, called the stem reptiles, appeared. This was an important evolutionary event because the stem reptiles were thecommon primal ancestors of a great variety of animals that still exist today. The modern reptiles, birds, and mammals are all derivatives of the stem reptiles.

Question:

What are tropisms?Where in a plant cantropistic responses occur?

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Solution:

Most animals perceive external stimuli via specialized sense organs and respond with the aid of an elaborate nervous system. Plants have neither sense organs nor nervous systems, but react to stimuli by means of tropisms. A tropism can be defined as a growth movement effected by an actively growing plant in response to a stimulus coming from a given direction. This results in the differential growth or elongation of the plant toward or away from the stimulus. Tropisms are named after the kind of stimulus eliciting them. Phototropism is a response to light; geotropism, a response to gravity; chemotropism, a response to a certain chemical; and thigmotropism, a response to contact or touch. The mechanisms of tropisms are now under much investigation. At this point, they are believed to involve certain plant hormones which are produced primarily inmeristemicregions (regions showing rapid cell division) of the plant. In appropriately low concentrations, these hormones stimulate growth and differentiation. These plant hormones are along the phloem and exert their effects on parts somewhat removed from the site of production. But because they can stimulate only those cells that are capable of rapid division,tropisticresponses can occur only in those parts of the plant which are actively growing or elongating, such as the apical part of the stem and the tip of the root.

Question:

What volume of hydrogen at STP is produced as sulfuric acid acts on 120 g. of metallic calcium. Equation for the re-action is Ca + H_2 SO_4 \rightarrow CaSO_4 + H_2

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Solution:

This problem may be solved by using either the mole method or the proportion method. Mole method: In using the mole method, one looks at this equation and sees that for every mole of calcium acted on by H_2 SO_4 one mole of hydrogen is produced. This means that to find how muchhyrdogenis produced, one must first find out how much calcium is present. There will be the same number of moles of hydrogen produced as there are calcium reacted. After one knows how many moles of hydrogen are produced, one can calculate the volume. To calculate the number of moles of calcium present, one must divide the amount present by the molecular weight (molecular weight of Ca = 40). number of moles = [(number of grams present) / (molecularweigth)] number of moles of Ca = [(120 g)/(40 g/mole)] = 3.00 moles Therefore, 3.00molesof hydrogen gas are produced. At STP (Standard Temperature and Pressure), the volume of one mole of any gas occupies 22.4 liters. Thus, when 3 moles of gas are generated, as in this problem it occupies 3 × 22.4 liters = 67.2 liters. Proportion method: In the proportion method, the molecular weights (multiplied by the proper coefficients) are placed below the formula in the equation and the amounts of substances (given and unknown) are placed above. In this case, because one is trying to find the volume of hydrogen and not its weight, the volume of one mole will be placed below the equation as shown. 120 gX( Vol.) Ca+H2 SO_4\rightarrowCaSO_4+H2 40.0g22.4 liters [(120 g) / (40.0 g)] = [(X) / (22.4 liters)] X = unknown volume of H_2 produced. Solving for X: X = [{(22.4 liters) (120 q)} / {40.0}] = 67.2 liters.

Question:

What is a hermaphrodite? How and why is self-fertilization in hermaphrodites prevented?

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Solution:

Any organism that has both male and female gonads is said to be hermaphroditic. The majority of flowering plants are hermaphroditic in that they bear both the pistil (the female sex organ) and the stamen (the male sex organ).Hermaphroditismis rarer in animals than in plants. It occurs chiefly in sessile forms, notably the sponges and some mollusks. It is also found in the earthworms, the parasitic tapeworms and flukes. Hermaph-roditism due to abnormal embryonic sexual differentiation resulting in individuals born with both a testis and an ovary has also been documented in man. Some hermaphroditic animals, the parasitic tapeworms, for example, are capable of self-fertilization. Most hermaphrodites, however, do not reproduce by self-ferti-lization; instead, two animals, such as two earthworms, copulate and each inseminates the other. Self-fertiliza-tion is prevented by a variety of methods. In some animal species, self- fertilization is prevented by the development of the testes and ovaries at different times. In others, such as the oysters, the different gonads produce gametes at different times, so that self-fertilization cannot occur. The advantage accompanying cross-fertilization is that genetic diversity is enhanced. Self-fertilization preserves the same genetic composition and with mutation occurring at a slow rate genetic variation is limited. Hermaphrodites undergoing cross-fertilization produce offspring showing the same genetic diversity as those resulted from individuals having a distinct sex. Thus, by ensuring gene exchange between individuals, cross--fertilization enhances genetic variation in the population. Only when a population carries a sufficient degree of genetic variation, can physical and environmental forces act to ultimately improve the characteristics of the individuals. And, as we shall see in a later chapter, the survival andpropagativeability of a species depends to a large extent on its genetic diversity.

Question:

Three forces acting on a particle and keeping it in equilibrium must be coplanar and concurrent. Show that the vectors representing the forces, when added in order, form a closed triangle; and further show that the magnitude of any force divided by the sine of the angle between the lines of action of the other two is a constant quantity.

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Solution:

Let the three forces be P\ding{217}, Q\ding{217}, and S\ding{217}, at angles \alpha, \beta, and \Upsilon to one another as shown in figure(a). In order that the three forces shall be in equilibrium, the resultant R\ding{217} of P\ding{217} and Q\ding{217} must be equal and opposite to S\ding{217}. The vectors P\ding{217}, Q\ding{217}, and S\ding{217} are concurrent and, since the vector R\ding{217} is in the same plane as P\ding{217} and Q\ding{217}, they are coplanar. But the resultant of P\ding{217} and Q\ding{217} is obtained by vector addition, as in figure (b). That is, R\ding{217} is the third side of the triangle formed by placing the tail of Q\ding{217} at the head of P\ding{217}. The force S\ding{217} is equal and opposite to R\ding{217} and thus will occupy the same space as R\ding{217}, the third side of the triangle, but will be opposite in di-rection to R\ding{217}. Thus P\ding{217} + Q\ding{217} + S\ding{217} taken in order, form a closed triangle and their sum is of necessity zero. Applying the law of sines to the triangle of figure (b)P/(sin \texttheta) = Q/(sin \varphi) = S/(sin \Psi) \thereforeP/[sin(180 - \alpha)] =Q/[sin(180 - \beta)] = S/[sin(180 - \Upsilon)] \thereforeP/[sin \alpha] =Q/[sin \beta] = S/[sin \Upsilon] = const.

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Question:

It can be shown that the set of operations S = {+, \bullet, \textasciitilde} is functionally complete. That is, every Boolean function can be represent-ed by a form f(x_1,....,x_n)in variables x_1,...,x_n and operations + ,\bullet, \textasciitilde . Equivalently, the set of gates S^I = {OR, AND, NOT} is function-ally complete. Show that: (a) S_1 = {+, \backsim} (b) S_0 = {\bullet, \backsim} (c) S3= {\uparrow} where x_1\uparrowx_2 = \backsim(x_1 \bullet x_2)(1) (d) S4= {\downarrow} where x_1\downarrowx_2 = \backsim(x_1 + x_2)(2) Are functionally complete.

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Solution:

(a) It is sufficient to show that we can construct the \textbullet operation from S_1, since we know that S = { +, \textbullet , \backsim } is functionally complete. DeMorgan's law suggests itself here: \textasciitilde(A \bullet B) = \textasciitilde A + \textasciitilde B9B Applying the \textasciitilde operator to both sides and using the involution law for the left side, we get: A \textbullet B = \textasciitilde [\textasciitilde (A \bullet B)] = \backsim [\backsim A + \backsim B] Hence, S_1 = [+, \backsim ] is functionally complete. The gate representation of A \textbullet B = \textasciitilde [ \backsim A + \textasciitilde B] is shown in figure 1 This means that every gating function can be implemented by a network of OR and NOT gates, by substituting the network of fig. 1 for any AND gate in the original OR-AND-NOT network. (b) It is sufficient to show that we can construct the + operation from S_2. But DeMorgan's law states that: \textasciitilde(A + B) = \textasciitilde A \textbullet \textasciitilde B Applying the \textasciitilde operator to both sides and using the involution law for the left side, we get: A + B = \textasciitilde[\backsim(A + B) ] = \textasciitilde(\textasciitilde A \textbullet \backsim B] Hence, S_2 = {\bullet, \backsim } is functionally complete. The gate representation of A + B = \textasciitilde[\textasciitilde A \bullet \textasciitilde B] is shown in fig. 2 . This means that every gating function can be implemented by a network of AND and NOT gates. (c) The \uparrow operator is called the Shaeffer stroke function and its associated gate is a NAND(for NOT-AND) gate. It is sufficient to show that we can construct S_2 = { \textbullet , \backsim} from the stroke function alone, since it has already been shown that S_2. is functionally complete. The idempotent law for \textbullet allows substitution of x_1 \textbullet x_1 for x_1 so that: \backsim x_1 = \backsim (x_1 \bullet x_1) But from the definition of the stroke function, equation (1), we have: \backsim (x_1 \bullet x_1) = x_1 \uparrow x_1 Therefore, \textasciitilde x_1 = x_1 \uparrow x_1 \bullet(3) To obtain \textbullet from the stroke function, recall that the involution law allows us to write: x_1 \bullet x_2 = \backsim [\backsim(x_1 \bullet x_2)] But from the definition of the stroke function, equation (1), we have: \textasciitilde[\textasciitilde(x_1 \textbullet x_2)] = \textasciitilde[x1\uparrow x_2] Using the result obtained equation (3), we can write: \backsim[x_1 \uparrow x_2] = (x_1 \uparrow x_2) \uparrow (x_1 \uparrow x_2) Therefore,x_1- \bullet x_2 = (x_1 \uparrow x_2) \uparrow (x_1 \uparrowx_2)(4) and S_3 is functionally complete. The gate representation of equation (3) is shown in fig. 3 (3) is shown in fig. 3 . The gate representation of equation (4) is shown in fig. 4 . This means that every gating function can be implemented by a network of NAND gates only. (d)The \downarrow operator is called the NOR (from NOT-OR) function and its associated gate is a NOR gate. It is sufficient to show that we can construct the S_1 = { + , \textasciitilde} from the NOR function alone. Using the idempotent law for + and the definition of the NOR function, equation (2), we can write: \textasciitildex1= \textasciitilde(x_1 +x_1) = x_1 \downarrow x_1(5) To obtain + from the NOR function, we apply the involution law and use the definition of the NOR function and equation (5): x_1 + x_2 = \textasciitilde[\textasciitilde(x_1 + x_2)] = \textasciitilde[x_1 \downarrow x_2] = [x_1 \downarrow x_2] \downarrow[x_1 x_2](6) Hence, S_4 = {\downarrow} is functionally complete. The gate representation of equation (5) is shown in fig. 5 The gate representation of equation (6) is shown in fig. 6 Every gating function can be implemented by a network of NOR gates only: Though, for simplicity, only two variables were used in showing the functional completeness of S_1 - S_4, the results can be shown to be true for any number of variables by a simple induction argument.

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Question:

Compare cardiac muscle to skeletal and smooth muscle.

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Solution:

Cardiac muscle is the tissue of which the heart is composed. Cardiac muscle shows some characteristics of both skeletal and smooth muscle. Like skeletal muscle, it is stria-ted; it has myofibrils composed of thickand thinmyofilaments, and contains numerous nuclei per cell. The slidingfilament mech-anism of contraction is found in cardiac muscle. Cardiac muscle resembles smooth muscle in that it is innervated by the autonomicnervous system. The cells of cardiac muscle are very tightly compressedagainst each other and are so intricatelyinterdigitatedthat previouslyno junctions were thought to exist between cells. They do, however, exist, and are visible under the light micro-scope as dark-colored discs, calledintercalcalateddiscs. It is believed that these discs may help totransfer the electrical impulses generated by the S-A node between musclecells due to their low resistance to the flow of current. The metabolism of cardiac muscle is designed for endurance rather thanspeed or strength. A continuous supply of oxygen and ATP must be providedin order for the heart muscle to main-tain its contractile machinery. Cardiac cells deprived of oxygen for as little as 30 seconds ceaseto contract, and heart failure ensues.

Question:

The pineal gland is a lobe on the upper portion of the forebrain, and has long intrigued investigators by its glandular appearance. Is this lobe an endocrine gland? If yes, what is its function(s)?

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Solution:

It was only recently demonstrated that the pineal gland has an endocrine function. In some primitive vertebrates, the pineal responds to light by generating nerve impulses and by secreting a hormone called melatonin. Melatonin lightens the skin by concentrating the pigment granules in melanophores, which are specialized pigment- containing cells. In frogs, the action of melatonin is opposed by melanocyte-stimulating hormone (MSH). MSH is secreted by the pituitary gland, and serves to disperse the pigment granules, causing a darkening of the skin. In mammals, the pineal secretes melatonin but does not generate impulses because light-sensitive cells are not present. Although the pineal does secrete melatonin in mammals, a problem arises when we realize that mammals have no melanophores (the sites of the hormonal action). Thus, the function of the pineal and its hormone in mammals has been a source of puzzlement for a long time. Recently, a theory has been proposed that the pineal gland functions as a neuroendocrine transducer. It was theorized that the pineal receives neural input from the visual receptors, and transduces this data into chemical messages (hormones). Apparently, impulses from the eyes reach the pineal via sympathetic nervous pathways from the cervical ganglia. The pineal responds by secreting more or less melatonin. In mammals, melatonin no longer affects skin pigmentation, but appears to have become a regulator of the anterior pituitary. Melatonin seems to in-hibit reproductive activities of rodents and birds by inhib-iting the release of gonadotrophic hormones (LH and FSH) . The diurnal rhythm of the light/dark cycles seems to mediate this function of melatonin. Thus, in these animals, repro-duction coincides with the optimal season, as light of day is interpreted by the brain via stimulation of the photore-ceptors. The function of the pineal gland in man remains highly speculative, though it may play a role in circadian rhythms.

Question:

Two hikers set off in an eastward direction. Hiker 1 travels 3 km while hiker 2 travels 6 times the distance covered by hiker 1. What is the displacement of hiker 2?

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Solution:

From the information given the displacement vector is directed east. The magnitude of the displace-ment vector for hiker 2 is 6 times the magnitude of the displacement vector for hiker 1. Therefore, its magnitude is 6 × (3 km) = 18 km

Question:

Plot the graph of a sine curve with point at 15\textdegree intervals.

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Solution:

The TAB function is useful here. By specifying the appropriate coordinates, it causes the printer to place an X in the correct column. We use column 40 to correspond to the X-axis so that there will be room to plot negative values. In the program, C = \pi/180. It converts from degrees to radians. The PRINT statement, however, represents the points in 150 steps, as given by the FOR ... NEXT construct. The program is given below: 1\O REM A PLOT OF THE SINE CURVE 2\O LET C = 3.14159/18\O. \O 3\O FOR A = \O TO 36\O STEP 15 4\O LET R = C\textasteriskcenteredA 5\O PRINT A; "DEG"; TAB(4\O. \O + 25\textasteriskcenteredSIN(R)); "X" 6\O NEXT A 7\O END -10+1 \O DEG X X X X X X X X X X X X X X X X X X X X X X X X X 15 DEG 3\O DEG 45 DEG 6\O DEG 75 DEG 9\O DEG 1\O5 DEG 12\O DEG 135 DEG 15\O DEG 165 DEG 18\O DEG 195 DEG 21\O DEG 225 DEG 24\O DEG 255 DEG 27\O DEG 285 DEG 3\O\O DEG 315 DEG 33\O DEG 345 DEG 36\O DEG

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Question:

Discuss the structure of the chloroplast and the possible relation of its structure to the process of photosynthesis. What is a quantosome?

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Solution:

Chloroplasts are small organelles found in plant cells that serve as the sites of photosynthesis. There are some 20 to 100 chloroplasts in a plant cell; these can grow and divide to form daughter chloroplasts. A chloroplast, like a mitochondrion, has a double - layered outer membrane. Lying within each chloroplast are many smaller stacks called grana (see figures). Under the electron microscope, the grana are seen as part of an elaborate system of membranes organized in parallel pairs. The pairs of membranes are joined at their ends to form a closed disk or thylakoid. The grana themselves are stacks of these thylakoids. Each granum is composed struc-turally of layers of protein molecules alternating with layers of chlorophyll, carotenes and other pigments, and special types of lipids containing galactose or sulfur but only one fatty acid. These surface - active lipids are believed to be absorbed between the layers and serve in stabilizing the lamellae composed of the alternate layers of protein and pigments. This lamellar structure of the grana is important in permitting the transfer of energy captured from the sun from one molecule to the adjacent one during the light phase of the photosynthetic process. This is evidenced by the fact that chlorophyll extracted from the chloroplast and isolated in a test tube can no longer carry out any photosynthetic reaction owing to the destruc-tion of the lamellar structure. The semifluid matrix within the chloroplast and surrounding the grana is called the stroma. The stroma contains the enzymes responsible for the dark reactions of photosynthesis. Electron microscopy has revealed the presence of repeating unit structures on the surface of the lamellae within the chloroplast. These structures, given the name quantosomes, are found to each contain some 230 molecules of chlorophyll. Botanists now believe that quantosomes are the functional photosynthetic units.

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Question:

What average force is necessary to stop a bullet of mass 20 gm and speed 250 m/sec as it penetrates wood to a distance of 12 cm?

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Solution:

As it travels through the block, the bullet experiences an average force, F_avg^\ding{217}, which retards its motion. By the work-energy theorem, the work done by the net force on an object equals the change in kinetic energy of the object. Hence \int F^\ding{217} \textbullet ds^\ding{217} = (1/2)mv^2_f - (1/2)mv^2_0 But we only know F^\ding{217} as an average value. Hence \int F^\ding{217} \textbullet ds^\ding{217} \approx F_avg^\ding{217} \textbullet ∆s^\ding{217} = (1/2)mv^2_f - (1/2)mv^2_0 By definition F_avg^\ding{217} \textbullet ∆s^\ding{217} = \vert F_avg^\ding{217}\vert\vert∆s^\ding{217}\vert cos \texttheta = (1/2)mv^2_f - (1/2)mv^2_0- where \texttheta is the angle between F_avg^\ding{217} and∆s, 180\textdegree in this problem. Whence -\vert F_avg^\ding{217}\vert\vert∆s^\ding{217}\vert = (1/2)mv^2_f - (1/2)mv^2_0- \vert F_avg^\ding{217}\vert = [(1/2)mv^2_o - (1/2)mv^2_f] / [\vert∆s^\ding{217}\vert] Hence,\vert F_avg^\ding{217}\vert = [(1/2)(0.2 kg)(250m/s)^2 - 0] / [.12 m] = 5.2 × 10^3 nt This force is nearly 30,000 times the weight of the bullet. The initial kinetic energy, (1/2) mv^2 = 620 joules, is largely wasted in heat and in work done in deforming the bullet.

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Question:

Develop a FORTRAN program to useLagrangian interpolationto evaluate F(x) = x3 at x = 3, given the table below: i 1 2 3 4 x_i 1 2 4 7 f_i 1 8 64 343 f_i \equiv f(x_i)

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Solution:

The Lagrange polynomials of degree m are defined as P_j(x) =A_j^m^+1\prod(k=1)k\not =j(x -x_k) whereA_j= ^m+1\prod(k=1)k\not =j[1/(x_j-x_k)] Here m = 3 and we approximate f(x) by 4\sum_i_=1f(x_i)P_i (x). The evaluations of P_j(x) are shown below, for x = 3. P1(x) \equiv P(x,1) = {(x-2)(x-4)(x-7)} / {(x1-x2)(x1-x3)(x1-x4)} = {(x-2)(x-4)(x-7)}/ (-18) P2(x) \equiv P(x,2) = {(x-1)(x-4)(x-7)} / {(x2-x1)(x2-x3)(x2-x4)} = {(x-1)(x-4)(x-7)} / 10 P3(x) \equiv P(x,3) = {(x-1)(x-2)(x-7)} / {(x3-x1)(x3-x2)(x3-x4)} = {(x-1)(x-2)(x-7)} / 18 P4(x) = P(x,4) = {(x-1)(x-2)(x-4)} / {(x4-x1)(x4-x2)(x4-x3)} = {(x-1)(x-2)(x-4)} / 90 ThusF(3) \approx 4\sum_i_=1 f(x_i) P(3,I) = 1(-2/9) + 8(4/5) + 64(4/9) + 343(-2/90) = 26.9999980 which is very close to 33 = 27. The program and sample output are given below: CPROGRAM 3.3 CPROGRAM FOR LAGRANGIAN INTERPOLATION CUNEVENLY SPACED PIVOTAL POINTS DIMENSIONX(50), P(50), F(50) PUNCH 994 READ 999, N, (X(I), F(I), I = 1,N) 1READ 998, XO DO 10 J = 1,N P(J) = 1 DO 10 I = 1,N IF (I - J) 9, 10, 9 9P(J) = P(J) \textasteriskcentered (XO - X(I))/(X(J) - X(I)) 10CONTINUE FO = O DO 20 I = 1,N 20FO = FO +P(I) \textasteriskcentered F(I) PUNCH 997 PUNCH 996, (I,X(I),F(I), P(I), I = 1,N) PUNCH 995, XO, FO GO TO 1 994FORMAT (//, 19X, 24H RESULTS FROM PROGRAM 3.3) 995FORMAT (19X, 6H AT X = F5.2, 21H THE VALUE OF F(X) IS F12.7) 996FORMAT (I10, 3F14.7) 997FORMAT (//, 9X, 1H1, 8X,4HX(I), 10X, 4HF(I), 9X, 6HP(X,I)) 998FORMAT (8F10.0) 999FORMAT (I5/8F10.0) END [SAMPLE]RESULTSFROMPROGRAM 3.3 X(I) F(I) P(X,I) 1 1.0000000 1.0000000 -0.2222222 2 2.0000000 8.0000000 0.8000000 3 4.0000000 64.0000000 0.4444444 4 7.0000000 343.0000000 -0.0222222 AT X = 3.00 THE VALUE OF F(X) IS 26.9999980

Question:

What are the mole fractions of solute and solvent in a solution prepared by dissolving 98 g H_2SO_4 (M.W. 98) in 162 g H_20 (M.W. 18)?

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Solution:

The mole fraction of solute is defined as the moles of solute divided by the sum of the number of moles of the solute and the number of moles of the solvent. Mole fraction of solute = [(moles of solute)/( moles of solute + moles of solvent)] The mole fraction of the solvent is defined similarly, Mole fraction of solvent = [(moles of solvent)/(moles of solute + moles of solvent)] Here, the solute is H_2SO_4 and the solvent is H_2O. One is given the amount of H_2SO_4and H_2O in grams. Therefore, these quantities must be converted to moles. This can be done by dividing the number of grams available by the molecular weight. No. of moles = [(no. of grams)/MW] No. of moles of H_2SO_4 = [98 g/(98 g/mole)] = 1 mole No. of moles of H_2O = [162 g/(18 g/mole)] = 9 mole Now that the number of moles of both solvent and solute are known, the mole fraction can be found. Mole fraction of H_2SO_4 = [(1mole)/(1 mole + 9 moles)] = 0.1 Mole fraction of H_2O = [(9mole)/(1 mole + 9 moles)] = 0.9.

Question:

What would be the surface tension of a liquid (density 0.876 g / cm^3) which rises 2.0 cm in a capillary where water at 20\textdegreeC rises 4.5 cm?

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Solution:

Surface tension refers specifically to the force within a liquid that acts parallel to the surface and tends to stretch the surface out. The equation that relates surface tension to the density of a liquid is \Elzpgamma = (1 / 2 rh\rhog where \Elzpgamma is the surface tension, r is the radius of the tube, h is the height the liquid rises, \rho is the density of the liquid and g is the downward acceleration of gravity. This equation can be rewritten rg = (2\Elzpgamma) / (h\rho) Because r and g are constant for a given capillary tube (2\Elzpgamma H_2O)/(h_H(2)O \rho_H(2)O) = (2\Elzpgamma liq)/(h_liq \rho_liq) One can find \Elzpgamma_H(2)O and \rho_H(2)O from standard tables. \Elzpgamma_H(2)O = (72.62 dyn / cm), \rho_H(2)O = (1 g / cm^3). Solving for liq: [2(72.62 dyn / cm) / (4.5 cm)(1g / cm^3)] = (2\Elzpgamma_liq) / [(2.0cm)(0.876 g / cm^3)] \Elzpgamma_liq = [2(72.62 dyn / cm)(2.0cm)(0.876 g / cm^3)] / [2(4.5 cm)(1g / cm^3)] \Elzpgamma_liq = 28.27 (dyn / cm)

Question:

A 0.10 M solution ofNaOHis prepared. What species of ions are present atequlibrium, and what will be their equilibrium concentrations?

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Solution:

Two processes are occurring simultaneously, the dissociation of NaOH, and theautoionizationof H_2O. NaOHdissociates according to the equation NaOH\rightleftarrows Na^+ + OH^-. For every mole ofNaOHthat dissociates, one mole of Na^+ and one mole of OH^- are produced. The initial concentration ofNaOHis 0.10 M. Thus, if we assume thatNaOHdis-sociates completely[Na^+] = 0.10 M and [OH^-] = 0.10 M. Waterautoionizesaccording to the equation H_2O + H_2O \rightleftarrows H_3O^+ + OH^-. The water constant for this process is K_W = 10^-14 moles^2/liter^2 = [H_3O+][OH^-]. Hence, [H_3O^+] = [(10^-14 moles^2/liter^2)/(OH^-)] Since [OH^-] was determined to be 0.10 M = 10^-1 M = 10^-1 moles/liter, by substitution, we obtain [H_3O^+] = [(10^-14 moles^2/liter^2)/(OH^-)] = [(10^-14 moles^2/liter^2)/ (10^-1 moles/liter)] = 10^-13 moles/liter = 10^-13 M. Hence, at equilibrium, H_3O^+, OH^-, and Na^+ are present in the concentrations [H_3O^+] = 10^-13 M, [OH^-] = 0.10 M, [Na^+] = 0.10 M.

Question:

A snowball at - 15\textdegreeC is thrown at a large tree trunk. If we disregard all frictional heat loss, at what speed must the snowball travel so as to melt on impact? The specific heat of fusion of ice is 79.7 cal/gm.

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Solution:

To solve this problem, one must know that the translational energy or kinetic energy of the snowball is completely converted into heat energy. The kinetic energy is given by the equation k. e. = 1/2 mv2, where m = mass and v = velocity. The total heat energy for this particular problem is Q =m∆Tc+m∆H_fus_ , where m = mass, ∆T = change in temperature, c = heat capacity (for H_2O, 0.5 cal/\textdegreeC gm) , and ∆H_fus= 79.9 cal/gm. These two equations are set equal to each other. The kinetic energy must be large enough so that the snowball can increase its temperature to 0\textdegreeC and then be completely converted to liquid water at 0\textdegreeC. Thus, m∆Tc+m∆H_fus= (1/2) mv^2 ∆Tc+ ∆H_fus= (1/2) v^2 (15\textdegreeC) (0.5 cal/\textdegreeC gm) + 79.7 cal/gm= (1/2) v^2 v^2 = 174 cal/gm = (174 cal/mg) (4.18 J/cal) = 727 J/gm = 727 kgm^2 / s^2 / gm = 7.27 × 10^5 m^2/s^2 [Note: 1 kg = 1000 gm; 727 × 1000 m^2/s^2 = 7.27 × 10^5 m^2/s^2] Thus, v = \surd(7.27 × 10^5 m^2/s^2) = 853 m/s.

Question:

Differentiate betweenisolecithal,telolecithal, and centrolecithal eggs. Which organisms are characteristicof each ?

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Solution:

The eggs of different animals vary greatly in the amount and distribution of the yolk they contain. Some, such as those of many annelids , mollusks, and echinoderms, are small and contain only a little yolk . If the yolk is evenly distributed they are calledisolecithaleggs. Other eggs are larger and have a moderate supply of yolk. Of this sort are amphibian eggs and those of primitive fish. Still others, such as those of reptiles and birds, have a tremendous supply of yolk. Since, in these types , the yolk is concentrated toward one pole of the egg (known as the vegetal pole), these eggs are termedtelolecithal. Arthropod eggs usually have the yolk massed toward the center, and hence are called centrolecithal .

Question:

A skier is filmed by a motion-picture photographer who notices him traveling down a ski run. The skier travels 36 ft during the fourth second of the filming and 48 ft during the sixth second. What distance did he cover in the eight seconds of filming? Assume that the acceleration is uniform throughout.

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Solution:

The fact that the acceleration is uniform gives us a big advantage since, in this case, the instantaneous acceleration is equivalent to the aver-age acceleration: a\ding{217} =\Deltav\ding{217}/\Deltat= (v_f\ding{217} - v_0\ding{217})/(t_f- t_0) wherev_f\ding{217} and v_0\ding{217} are the velocities at timest_fand t_0 respectively. To solve for a we use the kinematic equation s = v_0t + (1/2)at^2 where s is the distance covered in time t. 36 = v_0(1) + (1/2)a(1) = v_0 + (1/2)a 48 =v_f(1) + (1/2)a(1) =v_f+ (1/2)a. where v_0 andv_fare the velocities at the beginning of the fourth and sixth seconds respectively, and both time intervals are one second long. (1/2)a = 36 - v_0 = 48 -v_f v_f= v_0 + 12 Since there is a two second interval between the times when the skier has velocities v_0 andv_f: a= (v_f- v_0)/\Deltat = {(v_0 + 12) - v_0}/2 = 12/2 = 6 ft/sec^2 Knowing the acceleration, we can now solve .for the skier's velocity v_0 at the beginning of the 4th second: 36 = v_0(1) + {(1/2)(6)} (1) v_0 = 36 - 3 = 33 ft/sec Now, we may solve for v_0', the velocity at the beginning of the filming v_0 = v_0' + at,v_0' = v_0 - at v_0' = 33 - (6) (3) = 15 ft/sec Thus the distance covered in the eight seconds of filming is: s= v_0' t + (1/2)at^2 = (15 ft/sec) (8 sec) + {(1/2)(6 ft/sec^2)} (8 sec)^2 = 312 ft.

Question:

A certain gas occupies a volume of 100 ml at a temperature of 20\textdegreeC. What will its volume be at 10\textdegreeC, if the pressure remains constant?

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Solution:

In a gaseous system, when the volume is changed by increasing the temperature, keeping the pressure constant, Charles' Law can be used to determine the new volume. Charles' Law states that, at a constant pressure, the volume of a given mass of gas is directly proportional to the absolute temperature. Charles' Law may also be written (V_1 /T_1 ) = (V_2 /T_2 ) where V_1 is the volume at the original temperature T_1 and V_2 is the volume at the new temperature T_2. To use Charles1 Law, the temperature must be ex-pressed on the absolute scale. The absolute temperature is calculated by adding 273 to the temperature in degrees Centigrade. In this problem, the centigrade temperatures are given and one must convert them to the absolute scale. T_1 = 20\textdegreeC + 273 = 293\textdegreeK T_2 = 10\textdegreeC + 273 = 283\textdegreeK Using Charles' Law, V_1 = 100 ml T_1 = 293\textdegreeK T_2 = 283\textdegreeK V_2 = ? (V_1 /T_1 ) = (V_2 /T_2 )V_2 = (V_1 T_2 /T_1 ) V_2 = [{(100 ml) (283\textdegreeK)}/(293\textdegreeK)] V_2 = 96.6 ml.

Question:

A mass of 5 kg hangs on a spring which has a spring constant of 2 × 10^3 new tons/m. (a) Calculate the minimum energy of the system, (b) By optical methods the position of the mass can be determined to within 10^-7 m. What is the uncertainty in the velocity of the mass?

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/Users/wenhuchen/Documents/Crawler/Physics/D33-0990.htm

Solution:

(a) The natural frequency of the vibrating mass is given by \omega = \surd(k/m) = \surd[(2 × 10^3nt/m)/5 kg] = 20 sec^-1 The zero point energy is the residual energy the vibrating particle has at absolute zero. It is given by E = (1/2)hf = (1/4\pi) × (6.63 × 10^-34 × 20)J \textbullet s \textbullet s^-1 = 1.05 × 10^-32 J (b) If we equate this to the maximum kinetic energy the vibrating particle has (when the elastic potential energy of the particle is zero) then E = (1/2) mv^2 v^2 = [(2 × 1.05 × 10^-32 J)/5 kg] v = 2.0 × 10^-17 m/sec If the uncertainty in the position is\Deltax= 10^-7 m, then, according to the uncertainty principle, \Deltax\Deltap= (h/2\pi) \Deltap= (h/2\pi\Deltax) where\Deltap=m\Deltavis the uncertainty in the momentum of the particle. Therefore the uncertainty in the velocity is \Deltav= (h/2\pim\Deltax) = [(6.63 × 10^-34 J \textbullet sec)/(2\pi × 5 × 10^-7 m \textbullet kg)] = 2.1 × 10^-28 m/sec This uncertainty is less than the zero point velocity. Suppose that we attempt to measure the velocity by measuring the time required for the mass to move through x = 2 × 10^-7 m. That is, t = (x/v) = [(2 × 10^-7 m)/(2 × 10^-17 sec)] = 10^10 sec(about 300 years) Such a measurement is liable to be interrupted before it is completed.

Question:

Calculate ∆G\textdegree andK_pat 25\textdegreeC for CO (g) + H_2 O (g) \rightarrow CO_2 (g) + H_2 (g), CO (g) + H_2 O (g) \rightarrow CO_2 (g) + H_2 (g), ∆G\textdegree_ffor H_2 (g), CO_2 (g), H_2 O (g), and CO (g) are 0, - 94, 2598, - 54.6357, and - 32.8079 Kcal/mole, respectively.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E15-0547.htm

Solution:

The Gibbs free energy of formation (∆G\textdegree_f) of a substance is the Gibbs free energy change, i.e., the amount of free energy available for work, for the re-action in which the substance in its standard state is formed from its elements in their standard states. By use of the following equation, ∆G\textdegree = ∆G\textdegree_f,products- ∆G\textdegree_f,_ reactants the Gibbs free-energy change for a reaction may be calculated from the Gibbs free energies of formation. The equilibrium constant for the reaction may then be calculated using the standard Gibbs free energy. Thus,∆G\textdegree = [∆G\textdegree_f, _CO2 + ∆G\textdegree_f, _H2] - [∆G\textdegree_f,_CO + ∆G\textdegree_f, _H2O] = (- 94.2598 + 0) - (- 32.8079 - 54.6357) = - 6.8162 Kcal/mole = - 6816.2 cal/mole. ∆G\textdegree = - RT InK_p, where R = gas constant, T = absolute temperature (Celsius plus 273\textdegree) and K_p = equilibrium constant, which is to be calculated. As such, one can, via substitution, write ∆G\textdegree = - 6816.2 cal/mole = - RT InK_p = - (1.987 cal\textdegreeK^-1 mole^-1) (298\textdegreeK) (2.303) logK_p logK_p= 4.997 K_p= [(P_H2 P_CO2)/(P_CO PH2 O)] = 9.931 × 10^4.

Question:

A ball is thrown upward with an initial speed of 80 ft/sec. How high does it go? What is its speed at the end of 3.0 sec? How high is it at that time?

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Solution:

Since both upward and downward quantities are involved, upward will be called positive. At the highest point the ball stops, and hence at that point v_1 = 0. The only force acting on the ball is the gravitational force which gives a constant acceleration of g =- 32 ft/sec^2. For constant acceleration and unidirectional motion, 2as = v^2_1 - v^2_0 2(-32 ft/sec^2)s_1 = 0 - (80 ft/sec)^2 s_1 = [-(80 ft/sec^2)^2]/[2(-32 ft/sec^2)] \approx 100 ft. s_1 is the highest point the ball reaches. To find the speed of the ball after 3 seconds, v_2 = v_0 + at = 80 ft/sec + (-32 ft/sec^2)(3.0 sec) = 80 ft/sec - 96 ft/sec = -16 ft/sec. After 3 seconds, the speed of the ball is v_2 = 16 ft/sec downward. The height of the ball after 3 seconds can be found from s_2 = v_0t + (1/2)at^2 s_2 = v_0t + (1/2)at^2 = (80 ft/sec)(3.0 sec) + (1/2)(-32 ft/sec^2)(3.0 sec)^2 = 240 ft - 144 ft = 96 ft. As a check, s_2 can also be found by using s_2 =vt= [(v_2 + v_0)/2]t = [(-16 ft/sec + 80 ft/sec)/2] × 3.0 sec = 96 ft wherevis the average velocity. Note that s is the magnitude of the displacement, not the total distance traveled. If the ball returns to the starting point or goes on past it, s will be zero or negative, respectively.

Question:

The accompanying figures show a cathode-ray tube (figure A) and the deflection of an electron in a region of magnetic field (figure B). If deflection in the tube is being produced by the magnetic field alone, predict the effect on the observed deflection by in-creasing (a) mass of particles, (b) velocity of the particles, (c) magnetic field, and (d) charge on the particles.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E18-0663.htm

Solution:

The magnetic field, charge, velocity, mass and the deflection of a particle are related by the following equation, Hev = [(mv^2 )/(r)] where H is the strength of the magnetic field, e the charge on the particles, r the radius of deflection, v the velocity of the particles, and m the mass of the particles. The derivation of this equation need not be considered in solving this problem. Rewriting the given equation and solving for e/m, the charge to mass ratio, (e/m) = (v/Hr). Since e/m is a constant if (a) the mass is increased, then the radius of deflection must also increase. This is seen more clearly by solving the given equation for r, r = [(vm)/(He)]. Here, if m increases (keeping v, H, and e constant), then r increases. Following this type of reasoning, one can predict: (b) the radius of deflection increases with increasing velocity; (c) the radius of deflection decreases with in-creasing magnetic field strength; and (d) the radius of deflection increases with increasing charge on the particles. Note that the radius of curvature is inversely proportional to the distance the particle is actually deflected from its original path.

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Question:

Show that in theBalmerseries that the frequencies of Balmer series that the frequencies of successive lines tend toward a limiting value ofcR/4, where successive lines tend toward a limiting value ofcR/4, where R is theRydbergconstant and c the speed of light. R is theRydbergconstant and c the speed of light.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E18-0700.htm

Solution:

Here, one must try to relate frequency ѵ with c and R. TheRydberg

Question:

Using the given d-t curve calculate the velocity- time curve.

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/Users/wenhuchen/Documents/Crawler/Physics/D03-0066.htm

Solution:

A velocity-time curve is found from a displacement-time curve by plotting the slope of the d - t curve versus time. Our task in this particular case is made easier by the fact that the velocity in each segment of the trip is constant. From A to Bv_AB= d_AB/t_AB = {(3 - 0)m}/{(2 - 0)sec} = 1.5 m/sec From D to Ev_DE= d_DE/t_DE = {(0 - 5)m}/{(14 - 12)sec} = -2.5 m/sec. The corresponding segments on the v - t curve are represented by horizontal lines. The rest of the curve is found similarly. Note that the area under the v - t curve at each time gives the displacement. That is, at t = 6 sec,area= (1.5 × 2 + 0.5 × 4)m = 5 m at t = 14 sec,area= (1.5 × 2 + 0.5 × 4 - 2.5 × 2)m = 0 m

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Question:

Write a FORTRAN subroutine to determine the largest element (in absolute value) in theithrow of N × N array called ARRAY.

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Solution:

Let the arguments of the subroutine be denoted by ARRAY, N, I, BIG, J. Here ARRAY stands for a square array whose dimensions are N × N, I stands for the row in which we wish to determine the maximum element, denoted by BIG, is to be found and J stands for the column in which BIG is located. The only trick to the algorithm is to implement a sequential comparison of each element of theithrow to the largest preceding value. The subroutine looks as follows: SUBROUTINE LARGE (ARRAY, N, I, BIG, J) DIMENSION ARRAY (N, N) BIG = ABS (ARRAY (1, 1)) J = 1 DO 9 K = 2, N IF (ABS (ARRAY (I, K)) .LT.BIG) GO TO 9 CSET BIG TO THE NEW VALUE BIG = ABS (ARRAY (I,K)) CUPDATE THE COLUMN OF THE NEW BIG ELEMENT. J = K 9CONTINUE RETURN END

Question:

The pressure of the nitrogen in a constant-volume gas thermometer is 78.0 cm at 0\textdegreeC. What is the temperature of a liquid in which the bulb of the thermometer is immersed when the pressure is seen to be 87.7 cm?

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/Users/wenhuchen/Documents/Crawler/Physics/D14-0521.htm

Solution:

In a thermometer there is one physical property (thermometric property) whose change is used to indicate a change of temperature. The thermometric property of the thermometer, in this case the pressure, is taken as being directly proportional to the Kelvin temperature. Therefore T = cp where c is a constant of proportionality. We can then state that for two temperatures on this scale, the following relationship holds: T_1/T_2 = p_1/p2 Let T_1 be 0\textdegreeC or 273\textdegreeK. Substituting the known values, we get T_2 = T_1p_2/p_1 = (273\textdegreeK) (87.7 cm)/(78.0 cm) = 307\textdegree K Reconverting to the Celsius scale, T_2 = 307\textdegreeK - 273\textdegreeK = 34\textdegreeC.

Question:

Water at 30\textdegreeC has a vapor pressure of 31.82 mm Hg. When 25.0 g of ethylene glycol is added to 1000 g of water, the vapor pressure is lowered to 31.59 mm Hg. Determine the molecular weight of ethylene glycol.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E07-0260.htm

Solution:

This problem is an application of raoult's law. Let p\textdegree denote the vapor pressure of pure water, p the vapor pressure of the ethylene glycol - water solution, n_W the number of moles of water in the solution, and n_e the number of moles of ethylene glycol. Then from Raoult's law, p = p\textdegree (n_W) / (n_e +n_w). Multiplying by minus one and adding p\textdegree to both sides of this equation gives p\textdegree - p = p\textdegree[(n_W) / (n_e +n_w)] = p\textdegree[1 - {(n_W) / (n_e +n_w)}] = p\textdegree[{(n_e +n_w) / (n_e +n_w)} -{(n_W) / (n_e +n_w)}] or,p\textdegree - p = p\textdegree [n_e(n_e +n_w)] Since the number of moles is equal to the number of grams divided by the molecular weight, this equation becomes p\textdegree - p = p\textdegree(g_e / M_e) /[(g_e / M_e) + (g_w / M_w)] where g_e denotes the number of grams of ethylene glycol, g_w the number of grams of water, M_e the molecular weight of ethylene glycol (which we are trying to determine) , and M_w the molecular weight of water. For this problem, p\textdegree = 31.82 mm, p = 31.59 mm, g_e = 25.0 g. g_w = 1000g and the molecular weight of water, M_w = 18. Substituting into the last equation, we obtain 31.82 mm - 31.59 mm = 31.82 mm [(25.0g/M_e)/(1000g /18) + (25.0 g/M_e)] Or,0.23 mm = 31.82 mm [(25.0 g / M_e) / (1000 g / 18) + (25.0 g / M_e)]. Solving for M_e, (0.23 mm) / (31.82 mm)[(1000 g / 18) + (25.0 g / M_e)] = (25.0 g / M_e) [(0.23 mm) / (31.82 mm)] × [(1000 g / 18) = (25.0 g / M_e)] = [(0.23 mm) / (31.82 mm)] × (25.0 g/M_e) = (1 / M_e) × 25.0 g[1 - {(0.23 mm)/(31.82 mm)}], Or, M_e =[(31.82 mm) / (0.23 mm)] × (1000 g / 18) × 25.0 g[1 -[(0.23 mm) / (31.82 mm)] = 62. Thus, the molecular weight of ethylene glycol is 62.

Question:

Calculate the minimum photon energy (in electron volts) required to create a real electron-positron pair. Express all numbers to four signi-ficant figures.

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Solution:

The rest mass-energy E of an electron or positron is E = m_ec^2 =(9.110 × 10^-31kg) (2.998 × 10^8m/s)^2 = 8.188 × 10^-14 J If the created pair of particles possess no kinetic energy, then by the conservation of energy, the photon energy E_\Upsilon must be equal to the sum of the rest mass-energies of the two particles: E \Upsilon = 2E = 1.638 × 10^-13J Since 1 eV = 1.602 × 10^-19 J, the required energy may be written E_\Upsilon = (1.638 × 10^-13\Delta) / (1.602 × 10^-19\Delta/ eV) = 1.022 × 10^6 eV This energy is the minimum or threshold energy required for creation of the electron-positron pair. Photons with this such energy are from the x-ray or gamma-ray region of the electromagnetic spectrum. Note however, that although we have satisfied the conservation of energy in our analysis, we have not satisfied the conservation ofmomen- tum. The photon has an initial momentum, but the electron and positron have no final momentum, since we have required them to be at rest. Hence, the energy of the incoming photon must be greater than 1.022 × 10^6eV, since, in any practical experiment, some agent must carry away momentum (and energy) at the end of the reaction. This is why pair production cannot occur in a vacuum.

Question:

Describe what is known as the shunting-yard algorithm for converting infix (or algebraic) notation into the Polish postfix notation. Enumerate the five rules of priority that must be applied to the input string in order to explain the algorithm.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G07-0136.htm

Solution:

The metaphor of the shunting-yard is an appropriate one for this algorithm. We can think of a string of variables and operators as the cars of a train. The shunting-yard is the place where cars are added, removed, and rearranged. In programming terminology, the shunting-yard can be represented by a stack. This stack can be likened to a rail-road track siding, as shown in Figure 1. As the string is moved from right to left, each symbol is evaluated. A symbol will be placed either in the stack or directly into the output string, contingent upon the fol-lowing rules of priority: 1) Operands at the front of the input string are moved directly to the output string. 2) A left parenthesis at the front of the input string is pushed onto the stack. 3) A right parenthesis at the front of the input string causes all operators to be popped up from the stack into the output string until a left parenthesis (if necessary) appears. The parentheses are then discarded from the output string. 4) Operators at the front of the input string cause operators in the stack to be popped up and into the output string until the top item in the stack is one of lower priority than the operator in the front of the input string. Priority is assigned to items according to the following table: ItemsPriority (low + , -high \textasteriskcentered , /higher If the stack is empty, or if the item at the top of the stack is one of lower priority, an operator at the front of the input string is pushed onto the stack. 5) When the input string becomes empty, all items in the stack are popped up and into the output string sequen-tially. Let us now "walk through" this algorithm, showing the movement of the items in the string and stating the rules to be applied at each step.

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Question:

How can glycogen (a polymer of glucose) be converted into fats (triacylglycerals) in the body?

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/Users/wenhuchen/Documents/Crawler/Biology/F03-0093.htm

Solution:

A fat, or triacylglycerol molecule, consists of glycerol bonded to three fatty acids. Therefore, in order for glycogen to be converted into fats, it must first be broken down and converted into glycerol or an activated form of glycerol and fatty acids. Glycogen is broken down by the enzyme glycogen phosphorylase, to give glucose 1-phosphate (G1P). G1P is next converted to glucose 6-phosphate (G6P) by the enzyme phosphoglucomutase. G6P is then isomerized to fructose 6-phosphate by an isomerase enzyme. Further reaction of fructose 6-phosphate result in the formation of two triose phosphates: dihydroxy- acetone phosphate and glyceraldehyde 3-phosphate. At this point, the conversion branches. Dihydroxyacetone-P (where P indicates phosphate) is reduced to glycerol 3-P which then gets hydrolyzed by a phosphatase to glycerol. Glyceraldehyde 3-P is converted to pyruvate by a series of reactions. Pyruvate now undergoes oxidative decarboxylation in the mitochondria and is activated by Coenzyme A to become acetyl-CoA. The latter enters the citric acid cycle where it combines with oxaloacetate to form citrate. Citrate, unlike acetyl-CoA, can pass through the mitochond-rial membrane to the cytoplasm, where it is reconverted to acetyl-CoA. Acetyl- CoA is then converted in the cytoplasm into malonyl-CoA, and finally, via a complex sequence of reactions, into fatty acids. Three fatty acids then react with one glycerol molecule to form a triacylglycerol. This sequence of events, discussed only briefly above, is summarized below: Note that pyruvate can only be converted to acetyl-CoA in the Note that pyruvate can only be converted to acetyl-CoA in the mitochondria. Fatty acids, however, are synthesized in the cytoplasm. Since acetyl-CoA cannot directly leave the mitochondria, it must first be converted to citrate, which can leave the mitochondria, and then be reconverted to acetyl-CoA in the cytoplasm. Although the enzymes for each reaction step are not indicated, it is important to realize that each step is catalyzed by an enzyme or complex of enzymes.

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Question:

Compare the cost of operating 3 lamps in series and in Compare the cost of operating 3 lamps in series and in parallel on a 115 volt circuit, if each lamp has a resistance of 100ohms. ohms.

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Solution:

We calculate the cost of operation for each configuration (see figures) by calculating the net power expended by each circuit. Assuming both circuits run for the same time, the energy used by each can then be found. The circuit using less energy is the more economical. For the series circuit, the same current I flows through each lamp. Each one then uses power P = I^2 R where R is the lamp resistance. Since the resistance of each lamp is the same, the net power expended in the series circuit is P_net = 3I^2 R(1) Looking at the equivalent circuit in figure (a), we realize that the net resistance of the series con-figuration is R_net = 3R By Ohm's Law, the current in this circuit is I = (V/R_net ) = (V/3R)(2) Using (2) in (1) P_net = [(3V^2 )/(9R^2 )] R = (V^2 / 3R) Using the given data, the series configuration uses power P_net = [(115 V)^2 /(300 \Omega)] = 44.1 Watts For the parallel connection, each lamp has the same voltage V applied across it. Each one uses power P ' = (V^2 /R) where R is the lamp resistance. Since all the resist-ances are equal, the net power expended by the parallel circuit is P'_net = (3V^2 /R) = [{3 (115 V)^2 } / (100 \Omega)] = 396.75 Watts If both circuits operate for a time \cyrchar\cyrt, the energies used are E_series = P_net \cyrchar\cyrt = (44.1 Watts) \cyrchar\cyrt E_parallel = P'_net \cyrchar\cyrt = (396.75 Watts) \cyrchar\cyrt Hence, the series combination is cheaper to run.

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Question:

Write a computer program in APL to analyze an electrical circuit using the AND and OR operators of Boolean algebra. Assume four switches in the circuit; two are connected in parallel and two are connected in series.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G14-0377.htm

Solution:

The circuit in fig. 1 is equivalent to the truth function Y (A,B,C,D) = (AVB) V (C\LambdaD). The complete truth table for Y is: ABAVBCDC\LambdaBY = AVBVC\LambdaD 0000000 0000100 0001000 0001111 0110001 0110101 0111001 0111111 1010001 1010101 1011001 1011111 1110001 1110101 1111001 1111111 The truth-table is equivalent to the circuit. For example, if switches A,B,C are off and D is on, the light will not shine. Similarly, if propositions A,B,C are false (=0) and proposition D is true (=1), the compound proposition (A or B) or (C and D) is false. \nablaBOOLE [1]'ENTER TRUTH VALUES FOR A,B,C AND D' [2]L \leftarrow [] [3]TEST \leftarrow (L[1] V L[2]) V L[3] \Lambda L[4] [4]'THE VALUE OF Y IS:'; TEST [5]\nabla The header of the program is composed of \nabla (del) and BOOLE. The del operator causes the system to change to programming mode while BOOLE is the name of the program. As an example, suppose the values 0,1,1,0, are entered for A, B, C and D respectively. Then the value of Y would be 1. The program may be modified to handle other types of circuits. (FIG. 2) This circuit is equivalent to the truth function Y (A,B,C,D) = (AVB)V(CVD). To test this function replace [3] by [3] TEST \leftarrow (L[1]VL[2])V (L[3]VL[4]) The circuit, (FIG. 3) represents the function (AVB)\Lambda(CAD). The test for this function is [3] TEST \leftarrow (L[1]VL[2]) \Lambda L [3] \LambdaL [4].

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Question:

A man is using a uniform ladder of weight W = 75 lb, one end of which is leaning against a smooth vertical wall, the other end resting on the sidewalk. It is prevented from slipping by rubber suction pads rigid-ly attached to the feet of the ladder and stuck firmly to the concrete. If the man of weight w = 150 lb is standing symmetrically three-quarters of the way up the ladder, and if the normal force N\ding{217}exerted by \ding{217} the wall on each leg of the ladder is 43.3 lb, what is the force exerted on the ladder by each suction pad?

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Solution:

The ladder is uniform and thus its weight acts as its center. The man is symmetrically placed on the ladder. Hence, by symmetry, the normal forces exerted by the wall on the two legs of the ladder are equal, as are the forces exerted by the two suction pads. Let the force exerted by either suction pad on the ladder be resolved into component forces F_x and F_y along the sidewalk and normal to it, respectively. The complete force system acting on the ladder is as shown in the figure, the man exerting a force equal to his weight on the ladder. The ladder, of course, exerts an equal and opposite force on him, since he is in equilibrium. The whole system is in equilibrium. It follows from Newtons Second Law that 2F_x = 2N, F_x = N = 43.3 lb and2F_y = W + w,F_y = (W + w)/2 = (75 lb + 150lb)/2 = 112.5 lb The total force exerted by each suction pad on the ladder thus has magnitude F= \surd(F_x^2 + F_y^2) = \surd[(43.3 lb)^2 + (112.5 lb)^2] = 120.6 lb and it acts at an angle \texttheta to the horizontal, where tan \texttheta = (F_y/F_x) = (112.5 lb)/(43.3 lb) = 2.60,\texttheta = 69\textdegree.

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Question:

Give examples, in pseudocode or with flowcharts, of the following control constructs: a) DO-WHILE b) D0-F0R c) IF-THEN-ELSE

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G08-0183.htm

Solution:

Control constructs are program segments which interrupt the sequencing of the entire program to perform some specific task. These constructs may consist of decision instructions, loops, or a combination of both. If rendered faithfully, these constructs will make your program efficient and readable. The DO-WHILE construct requires the continued execution of a sequence of statements, as long as a given condition is satisfied. A pseudocode description looks something like this: do while (X) tasks A - Z end do while X is some logical expression. The loop is repeated until X is no longer satisfied. Program control drops out of the loop and on to the next statement. A flowchart of the situation is as shown in Fig. 1 The DO-F0R construct differs from the DO-WHILE construct in that it specifies explicitly the number of tines the loop should be executed. When the desired number of loops has been executed, control passes out of the loop. Here are pseudocode and flowchart descriptions of the DO-FOR: (see Fig. 2) do for I=1 through N tasks A - Z end do for When translated into a high-level language, the loop, will contain some statement that declares the value of N. Exit from the loop is contingent upon N executions. It should be noted that there are two distinct types of DO-FOR loops. The PL/1 implementation of a DO-FOR loop is as shown in Fig. 1(a). If N = 0, the body of the loop (tasks A - Z) will not be performed. The FORTRAN implementation of a DO-FOR loop is described by the flowchart see Fig. 2(b). Note that the body of the loop (tasks A - Z) will always be perforated at least once in a FORTRAN DO-loop, even if N = 0. An unwary student might look at the flowchart given for a DO-FOR loop and think that it describes FORTRAN'S DO-loop, but actually it does not. Note also that DO FOR constructs may be nested inside each other, such as in the following pseudocode: do for I = 1 through N do for J = 1 through M tasks A - Z end do for J end do for I Tasks A - Z will be performed M × N times. Notice two important conventions in this example. First, in both the DO FOR and DO WHILE constructs, only the first and last statements are left-justified; all other statements within the loop are indented. This is purely a stylistic convention, but it serves to clarify the difference between the loop's control statements and its task-performing, or assignment, statements. Second, realize that you may not terminate a nested loop outside the bounds of the "nest". In the example above, because the DO-FOR J loop is the nested loop, you must terminate it before term-inating the DO-FOR I loop. A violation of this rule will cause serious logical errors. The IF-THEN-ELSE construct is a decision-rendering construct. If the given condition is satisfied, then a certain sequence of statements is executed. If the condition is not satisfied, a different sequence is followed. In pseudocode we have If (x) then tasks A - M else tasks N - Z end if then else with x as the given condition. Notice again that the first and last statements are not indented, while the others are. See flowchart Fig. 3. The true condition corresponds to the THEN part of the construct, The true condition corresponds to the THEN part of the construct, while the false condition is interpreted as the ELSE part.

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Question:

A flywheel of mass 12 kg and radius of gyration 20 cm is mounted on a light horizontal axle of radius 5 cm which rotates on frictionless bearings. A string wound round the axle has attached to its free end a hanging mass of 4 kg, and the system is allowed to start from rest. If the string leaves the axle after the mass has descended 3 m, what torque must be applied to the fly-wheel to bring it to rest in 5 revs?

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/Users/wenhuchen/Documents/Crawler/Physics/D05-0249.htm

Solution:

Consider the hanging mass. It has two forces acting on it, its weight, mg^\ding{217}, downward and the tension in the string, T^\ding{217} upward. Since the mass descends with acceleration a^\ding{217}, by Newton's second law mg - T = ma But, if the flywheel rotates with angular acceleration \alpha, then the tangential acceleration a = r\alpha,and thusmg - T = mr\alpha(1) The tension T^\ding{217} acts at distance r from the axis of the flywheel of mass M and radius of gyration k. Therefore the torque is \Gamma = Tr. Using the rigid body analogue of Newton's second law, where torque takes the place of force, moment of inertia, I, takes the place of m, and angular acceleration \alpha takes the place of linear acceleration, \Gamma = Tr = l\alpha. Since I = Mk^2, Tr = Mk^2\alpha. Also, upon multiplication of both sides of equation (1) by r, mgr - Tr = mra = mr^2\alpha. \thereforemgr = \alpha(mr^2 + Mk^2). \alpha= mgr/(mr^2 + Mk^2) = (4 kg × 9.8 m\bullet^-2 × 0.05 m) / (4 kg × (0.05)^2m^2 + 12 kg × (0.2)^2 m^2) = [(1.96) / (0.49)]s^-2 = 4 rad\bullets^-2. The flywheel starts from rest and accelerates as long as the string is exerting a couple on it. In that time the mass descends 3 m. But in 1 rev, the string unwraps a length equal to its circumference, 2\pir, and the mass descends by this distance. Thus the angular distance the flywheel turns during the period of acceleration is 3m / (2\pi × 0.05m/rev) = (3/0.05) × (1rev/2\pi) × (2 \pirad/1rev) = (3/0.5) rad = 60 rad The angular speed when the string leaves the axle can be found by using the rigid body analogue of the kinematic equations for constant acceleration. If \omega is the analogue of linear velocity, v, \alpha the analogue of linear acceleration, a, and \texttheta the analogue of linear displacement, s, then the kinematic equation not involving the time variable is \omega^2 = \omega^2_0 + 2\alpha\texttheta. The initial angular velocity \omega_0 = 0 and \omega^2 = 2 × 4 rad\bullets^-2 × 60 rad or\omega= 4 × \surd30 rad\bullets^-1= 21.9 rad\bullets^-1. If a torque \Gamma' is now applied to the wheel, it produces a constant deceleration \alpha', and since the flywheel is being brought to rest, \omega_final = 0 or 0^2 = \omega^2 + 2\alpha' × 10\pi rad. \therefore\alpha' = (-\omega^2)/(20\pi rad) = - [(480 rad^2\bullets^-2)/(20\pi rad)] = - (24/\pi) rad\bullets^-2.. But\Gamma' = l\alpha' = Mk^2\alpha' = - 12 kg × 0.04 m^2 ×(24/\pi) rad\bullets^-2 = - 3.67 N\bulletm. Thus a torque of 3.67 N\bulletm applied against the direction of rotation is necessary. Note that the result is more easily obtained from a consideration of energy of the mass-flywheel system. The mass m descends a height h. In so doing it loses potential energy, which reappears in the form of kinetic energy of the mass and of the flywheel. Thus, if the bottom of the fall is taken as the reference level, then mgh = (1/2)mv^2 + (1/2)I\omega^2 = (1/2)mr^2\omega^2 + (1/2)Mk^2\omega^2 \therefore\omega^2 = (2mgh)/(mr^2 + Mk^2) = 480 rad^2\bullets^-2. In the final stage, the work done by the retarding torque in the 5 rev = 10\pi radians must equal the kinetic energy possessed by the flywheel before the couple is applied. Thus - 10\pi\Gamma = (1/2)I\omega^2 = (1/2)Mk^2\omega^2. \therefore\Gamma = - [(Mk^2\omega^2)/(20 \pi)] = - 3.67 N\bulletm.

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Question:

A mass m = 2 kg is hung on a spring of constant k = 3.92 × 10^3N/m. The natural angular frequency of the system is \omega_0 = 44.3 sec^-1. Now we will force the system to vibrate at different frequencies by an alternating force F = 2 cos \omegat newtons That is, the maximum force is two newtons. A static force of two newtons would cause a deflection of about 1/2 mm to the system. What will be the amplitude of vibration for \omega = 15 sec^-1, and 60 sec^-1?

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Solution:

This problem is a case of harmonic motion with a periodic force applied to a vibrating mass. The forces acting on the mass m are the given applied force F = 2 cos \omegat, and the restoring force F_R = -ky due to the spring. y is the vertical displacement of the mass from its rest position. The negative sign indicates that the restoring force acts in such a direction as to oppose a change in vertical displacement. By Newton's second law, -ky + 2 cos \omegat = m[(d^2y)/(dt^2)] or[(d^2y)/(dt^2)] + (k/m)y = 2/m cos \omegat(1) The following function is a solution to the above differential equation, y(t) = A cos \omegat(2) where A is the amplitude of vibration. This can be verified by sub-stitution into equation (1): -\omega^2 A cos \omegat + (k/m)A cos \omegat = 2/m cos \omegat Dividing out the common factor, and solving for A, we find A = (2/m)/[(k/m) - \omega^2] Let k/m = \omega^2 _0 (where \omega_0 is the natural frequency of vibration) then A = (2/m)/(\omega^2 _0 - \omega^2) Since m = 2kg, we can calculate values of \vertA\vert as shown in the table. \omega \vertA\vert 15 sec^-1 5.8 × 10^-4m 44 3.8 × 10^-3 60 6.1 × 10^-4 When the frequency \omega of the applied force is near the natural fre-quency \omega_0 the amplitude becomes much larger (see figure (b)). We can also calculate the average total energy of the system. The total energy H of the system at any moment is H = (1/2) mv^2 + (1/2)ky^2(3) where the first term is the kinetic energy of the object and the second term is the elastic potential energy of the object. Upon differentiating equation (2), substituting it into equation (3), we obtain H = (1/2) mA^2\omega^2 sin^2\omegat + (1/2) kA^2 cos^2\omegat. But k = \omega^2 _0m (by definition of \omega2_0). Therefore H = (1/2) mA^2\omega^2 sin^2\omegat + (1/2) mA^2\omega^2 _0 cos^3\omegat The average value of cos^2\omegat over 1 period is, by definition = 1/2\pi^2\pi\int_0 cos^2\omegat d(\omegat) = 1/2 Similarly, = 1/2 Hence = 1/4 mA^2\omega^2 + 1/4 mA^2\omega^2 _0 or = 1/4 mA^2(\omega^2 +\omega2_0) which shows that the energy of the system becomes much larger as \omega \ding{217} \omega_0. \omega \vertA\vert 15 sec^-1 3.7 × 10^-4 44 2.8 60 1.0 × 10^-3 As the frequency of the applied force nears the resonance frequency it is possible to transfer considerably greater energy to the system.

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Question:

A cell possesses two electrodes. Both half-cells are .01M MnO^-_4 ion. One cell is .01MH_3O^+ ion, while the other has a H_3O^+ concentration of.10M. The electrode reaction for the reduction half-cell may be written: MnO^-_4 + 4H^+ + 3e^- \rightarrow MnO_2 + 2H_2O . The oxidation half-cell is the reverse of this reaction. 1) Write the net equation for the spontaneous cell process taking place; 2) Find ∆E for the reaction; (3) Find the value of the equilibrium constant.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E16-0593.htm

Solution:

1) The net reaction, in such a situation, is the sum of the balanced half-reactions, i.e., oxidation reaction plus reduction reaction. Since, you are given both reactions, add the equations together to find the overall reaction. You have, with concentrations included, ox:MnO_2 + 2H_2O \rightarrow MnO_4^- (.01M) + 4H^+ (.01M) + 3e^- red:MnO^-_4 (.01M) + 4H (.10M) + 3e^- \rightarrow MnO_2 + H_2O Net Reaction: 4H^+(.10M) \rightarrow 4H^+(.01M) . Notice: all species cancelled out, except H^+ (actually H_3O^+ ). This is the net equation for the spontaneous reaction taking place. 2) To find ∆E for the reaction, use the Nernst equation, which states ∆E = ∆E\textdegree -(.059 / n)log K for a temperature at 25\textdegreec, where ∆E\textdegree = potential for cells under other than standard conditions, ∆E\textdegree = standard cell potential, n = number of electrons transferred and K = equilibrium concentration expression, n = 3, since from either the oxidation or reduction reaction, 3 electrons are being transferred; ∆E\textdegree = 0, since ∆E\textdegree = E\textdegree_prod - E\textdegree_reactants and both the product and reactant are the same species. K is defined as the ratio of products to reactants, each raised to the power of their coefficients in the net equation. Substituting these values into the Nernst equation, you have ∆E = 0 -(.059 / 3)log [(.01)^4 / (.10)^4 ] = 0 - .0197(log 10^-4) = (-.0197)(-4) = 0.079 volt. 3) To find the value of the equilibrium constant, note that there exists a relationship between ∆E\textdegree and the constant at 25\textdegreec. Namely, ∆E\textdegree =(.059 / n)log K. From part 2, you found ∆E\textdegree = 0 . Thus, K = unity (one), since this is the only value that permits log K = 0, which, then, allows ∆E\textdegree = 0, as it must.

Question:

Determine the approximate solubility ofAgClin 0.10MNaCl solution.K_spforAgCl= 1.1 × 10-^10 .

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/Users/wenhuchen/Documents/Crawler/Chemistry/E11-0391.htm

Solution:

The approximate solubility ofAgClin this solution is equal to the concentration of Ag+ in the solution. This is because the concentration of Ag+ is equal to the amount ofAgClionized. The ionization ofAgClcan be stated AgCl\rightleftarrows Ag+ +Cl+ . This means that one Ag+ ion is formed each time one molecule ofAgClis ionized. The concentration of Ag+ can be obtained by using the solubility product constant (K_sp).+This constant is equal to the pro-duct of the concentration of Ag+ ([Ag+]) times the concentration ofCl-([Cl-]) . K_sp= [Ag+] [Cl-] = 1.1 × 10-^10 . The concentration of theCl- ions is also affected by theCl- ions from the NaCIsolution. BecauseNaCIis a salt, it is 100% ionized in solution NaCI\rightleftarrows Na+ +Cl- . There is oneCl- ion formed for everyNaCIionized. The original concentration ofNaCIis 0.10 molar, and after ionization, the solu-tion is 0.10 molar inCl-ions. The Na+ ions can be disregarded because here we are looking at the solubility ofAgCl, which does not contain any Na+ ions. One can assume that the total concentration ofCl- ions in solution will be approximately 0.10, since the amount con-tributed by theAgClionization is small, and can be neglected. The [Ag+] can now be found. [Ag+] [Cl-] = 1.1× 10-^10 [Ag+ ] (0.10) = 1.1 × 10-^10 [Ag+] = 1.1 × 10-^9 M .

Question:

Why is genetic drift less important in human evolution at present than it was twenty thousand or more years ago?

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/Users/wenhuchen/Documents/Crawler/Biology/F29-0757.htm

Solution:

The communities that man lived in about twenty thousand years ago all had something in common - they all were very small villages. The total world population then was low and people living in these small social groups were relatively distant from people in other places. With no means of transportation and the danger of traveling alone, people tended to stay close to their shelters. As a result, most breeding took place within individual social groups with each probably ranging up to about 100 members. In other words, these people were probably genetically similar for in inbreeding, no genes could be exchanged with other populations that existed. As an illustration, we can consider the western European Neanderthal people who were cut off from the rest of the world population by geographic factors. They were no longer able to exchange genes and consequently their further evolution was independent. Thus, they were sheltered from different genes present in the world's gene pool. These people became more unlike the rest of the world population and more like each other. Because of their reduced variability, when faced with an environmen-tal crisis such as sudden cold or disease, they were not able to adapt successfully and faced extinction. Let us take a look at the genetic effects involved with small inbreeding populations. Evolutionary change is not usually automatic, it occurs only when something disturbs the genetic equilibrium. A condition necessary for a popu-lation to be in equilibrium is that it must be large enough to make it highly unlikely that chance alone could signi-ficantly alter gene frequencies. Any breeding population with more than 10,000 members of reproductive age is probably not significantly affected by random change. This is the condition that exists in the world today. But gene frequencies in small isolated populations of less than 100 reproductive members are highly susceptible to random fluctuations, which can easily lead to loss of an allele from the gene pool and the fixation of another even if the former allele is an adaptively superior one. This is the phenomenon called ge-netic drift. Genetic drift explains why small populations tend to have a high degree ofhomozygosity, while large popu-lations tend to be more variable and diverse.

Question:

A water tank standing on the floor has two small holes vertically above one another punched in one side. The holes are 3.6 cm and 10 cm above the floor. How high does water stand in the tank when the jets from the holes hit the floor at the same point?

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Solution:

For any hole in a tank, the velocity of discharge of the liquid is given by Torricelli's theorem. To derive it, consider a hole a distance d below the surface of the liquid in the tank (see figure 1). Using Bernoulli's equation, P_1 + (1/2) \rhov^2_1 + \rhogy_1 = p_2 + (1/2) \rhov^2_2 + \rhogy_2 take point 1 to be at the hole and point 2 at the surface of the liquid. The pressure at each point is the atmospheric pressure p_a since both are open to the atmosphere. If the hole is small, the level of the liquid in the tank falls slowly and its velocity v_2 can be assumed to be zero. Using the bottom of the tank as the reference level, P_a + (1/2) pv^2_1 + pgy_1 = p_a + 0 + pgy_2 orv^2_1 = 2g(y_2 - y_1) = 2g d Therefore, the velocity of discharge from a hole in a tank is given by v = \surd2gd In this problem, the velocities of efflux are horizontal from both holes. Using Torricelli's theorem, one gets for the upper hole (see figure 2), v_1 = \surd[2g(h - h_1)], and for the lower one, v_2 = \surd[2g(h - h_2)]. Water from the upper hole has a horizontal velocity v_1 and no initial vertical velocity u. In time t_1 applying the formula (y - y_0) = ut_1 + 1/2 gt^21to the vertical motion , one obtains h_1 = 0 + 1/2 gt^2_1 or t_1 = \surd(2h_1/g). In that time the horizontal distance gone is v_1jt_1, which is the distance from the tank at which the jet strikes the floor. Similarly, the distance at which the jet from the lower holes strikes the floor is v_2 t_2, where t_2 = \surd[2h_2/g]. But these distances are equal, and thus v_1t_1 = v_2t_2 or (v_1t_1)^2 = (v_2t_2)^2. Then 2a(h - h_-1) × (2h_1/g) = 2a(h - h_-2) × (2h_2/g) (h - h_-1) × h_1 = (h - h_-2) × h_2 hh_1 - h^2_1 = hh_2 - h^2_2 hh_1 - hh_2 = h^2_1 - h^2_2 h = (h^2_1 - h^2_2)/(h_1 - h_2) = h_1 + h_2 .\therefore h = 13.6 cm.

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Question:

The motor driving a grindstone is switched off when the latter has a rotational speed of 240 rev\bulletmin^-1 . After 10 s the speed is 180 rev\bulletmin^-1. If the angular retardation remains constant, how many additional revolutions does it make before coming to rest?

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/Users/wenhuchen/Documents/Crawler/Physics/D04-0186.htm

Solution:

The initial speed\omega_ois 240 rev\bulletmin^-1 and the later speed \omega is 180 rev\bulletmin^-1. Thus since the angular acceleration, \alpha, is constant, we may write \omega = \omega_0 +\alphat. Here, t is the time it takes for the grindstone's angular velocity to go from \omega_0 to \omega. Solving for \alpha, \alpha = (\omega - \omega_0)/t = (180 rev\bulletmin^-1 - 240 rev\bulletmin^-1)/(10 s) Noting that 1 min^-1= 1/60 s^-1, we find \alpha = - (60 rev\bullets^-1)/(60 × 10s) = -.1 rev\bullets^-2 Considering the subsequent slowing-down period, the final speed is zero and the grindstone traverses an angular distance \texttheta. Hence, using the equation \omega^2 = \omega^2_0 + 2\alpha\texttheta with \omega = 0 and \omega_0 = 180 rev\bulletmin^-1 = 3 rev\bullets^-1,we find 0 = 9 rev^2 \bullets^-2 + 2(-.1 rev\bullets^-2)\texttheta or\texttheta = (9 rev)/(.2) = 45 rev.

Question:

Using positive logic, show that the diode-transistor circuit of fig. 1 represents a three-input NAND gate.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G04-0066.htm

Solution:

The circuit is broken down into two "sub--circuits"; fig. 2(a) and (b). In sub-circuit a, fig. 2(a), if any of the inputs are low(or), its corresponding Diode will conduct, pulling the output X_4 Low. If all the inputs are high (\geq 5V), none of the diodes will conduct, leaving the output X_4 high (5V). The truth table of figure 3 illustrates sub-circuit a's action, that of an AND gate. InputsOutput X_1 X_2 X_3 X_4 L L L L L L H L L H L L L H H L H L L L H L H L H H L L H H H H Fig.3 In sub-circuit b, figure 2(b), the transistor acts as a switch, conducting when X_4 is high ( >.7V), and open when X_4 is low (OV). When X_4 is high the transistor is conduct-ing and the output Y is pulled low. When X_4 is low the transistor is open and Y floats to 5V (high). Thus, sub--circuit b acts as an inverter and the whole circuit, whose truth table is shown in fig. 4, acts as a NAND gate. INPUTS X_4 Y(OUTPUT) L L L L H L L H L H L H L L H L H H L H H L L L H H L H L H H H L L H H H H H L

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Question:

By photographic and optical means, an image can be enlarged almost indefinitely. However, most light microscopes offer magnifications of only 1000 to 1500 times the actual size, while electron microscopes offer magnifications of 100,000 times or more. Explain.

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/Users/wenhuchen/Documents/Crawler/Biology/F01-0003.htm

Solution:

All microscopes are characterized by limits of resolution. Resolution refers to the clarity of the image. Objects lying close to one another can not be distinguished (resolved) as separate objects if the distance between them is less than one half the wavelength of the light being used. The average wave-length of visible light is 550 nanometers (or 5500 \AA). Thus, for light microscopes, objects can be distinguished only if they lie farther apart than about 275 nanometers. Objects closer together than 275 nm are not resolved and appear to be one object. Increasing the size of image, or the magnification, will not give meaningful information unless resolution is also increased. Increasing magnification without increasing resolution results in a larger image that is still blurred. Electron microscopes offer resolution of details separated by .1 to .5 nanometers. Electrons, rather than light, are the radiation used in electron micros-copes. Recall that electrons have a wave property in addition to a particle property and may be regarded as a radiation of extremely short wavelength. Since the wavelength of an electron in motion is so much shorter than the wavelength of light, resolution is more than a thousandfold better. Structures such as the plasma membrane, endoplasmic reticulum,ribosomes, microtubules and microfilaments were not visible until the advent of the electron microscope. These structures are all less than 275 nm in width. Theplagmamembrane has a thickness of 7.5 to 10 nm (or 75 to 100 \AA). The ribosome is 15 to 25 nm in diameter (or 150 to 250 \AA) . Microtubules are 20 to 30 nm in diameter, and microfilaments range from 5 to 10 nm. Electron microscopy has also made possible the visualization of the nuclear envelope and the internal membranes of mitochondria and chloroplasts.

Question:

Two true-breeding varieties of unicorns differ with respect to three traits. The first has a straight horn, a green coat, and a straight mane. The other variety has a twisted horn, a blue coat, and a curly mane. When a male of either variety is crossed with a female of the other, all the progeny have straight horns, green coats, and straight manes. These hybrid offspring are then crossed to unicorns with twisted horns, blue coats, and curly manes. This cross produces offspring in the following numbers: straight horn, green coat, straight mane855 twisted horn, blue coat, curly mane855 twisted horn, green coat, straight mane95 straight horn, blue coat, curly mane95 straight horn, blue coat, straight mane50 twisted horn, green coat, curly mane50 50 2000 Construct a linkage map for the three genes involved in this cross.

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Solution:

This is a complicated problem on linkage. Go over the previous problems on recombination. Some new concepts will be introduced here. This cross involves three traits (that is, a three-point cross) each governed by a gene. To begin, we have to determine the dominance or recessiveness of the phenotypes. This can be obtained from the F_1 since all the F_1 descendants have straight horn, green coat, and straight mane, the gene for straight horn (H) is dominant over twisted horn (h); green coat (C) is dominant over blue coat (c); straight mane (M) over curly mane (m) . Since the parents are true-breeding, one of them isand the other; The actual sequence of the genes has yet to be determined. In the cross between two homozygous parents, Note that all the dominant genes are on one chromosome while all the recessive ones are on the other. These are the parental typesHCMand hcm . All other combinations of genes are recombinants. To calculate the recombination frequency we have to know the number of recombinants. There are several ways to do this; one of them begins by translating the F_2 phenotypes to genotypes keeping the F_2 progeny in the same order we get: 1 HCM 855 2 hcm 855 3 hCM 95 4 Hcm 95 5 HcM 50 6 hCm 50 Consider a cross-over between H and C. The number of recombinants is 95 3 + 95 4 + 50 5 + 50 6 = 290. recombinants is 95 3 + 95 4 + 50 5 + 50 6 = 290. The RF between H and C = (290) / (2000) = 0.145. The RF between H and C = (290) / (2000) = 0.145. An RF of 0.145 × 100 = 14.5 map units, meaning H and C are 14.5 An RF of 0.145 × 100 = 14.5 map units, meaning H and C are 14.5 map units apart. map units apart. Consider a recombination between C and M . By similar reasoning, Consider a recombination between C and M . By similar reasoning, the number of recombinants = 50 5 + 50 6 = 100 the number of recombinants = 50 5 + 50 6 = 100 RF = (100) / (2000) RF = (100) / (2000) That is, a map distance of 5 units separates C and M. Finally, consider a cross-over between H and M. The number of recombinants = 95 3 + 95 4 = 190. RF = (190)/(2000) = .095 RF = (190)/(2000) = .095 That is, H and M are 9.5 map units apart. Now that we have calculated a map distance for any two of the three genes, we are ready to plot the linkage map. As our reference points we use H and M, which are 9.5 units apart. We know that H and C are 14.5 units apart. But C can be to the right or left of H. To determine which side of C is on, we look at the distance between C and M , which is 5 units. This means C must be on the right side of H. If C were 14.5 units to the left of H, then C and M would be separated by 24 units.' This is not the case. Therefore, the correct linkage map for the three genes is

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Question:

Estimate the upper limit to the frequency of "sound" waves in ordinary matter.

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Solution:

The relation v = v/\lambda ,(1) where ѵ, v, and \lambda are the frequency, velocity and wave-length of the sound wave, respectively, makes it clear that the higher the frequency the shorter is the wavelength. We are therefore asking what is the shortest wavelength sound can have. The answer is that, when the wavelength becomes shorter than the distance apart of the atoms, the concept of a wave breaks down because there is nothing in between the atoms to oscillate (see figure). The atoms are closest together in solids and highly compressed gases. However, when two atoms are about 10^-8 cm apart they repel one another very strongly and it is difficult to force them any nearer. Consequently, in no material are the atoms or molecules found to be closer together than about 10^-8 cm. This is therefore the order of magnitude of the shortest wavelength of sound. The highest value found for the velocity of sound is about 10^6 cm/sec. Using equation (1). Highest frequency = (Largest wave velocity) / (Shortest wavelength) = (10^6 / 10^-8) approximately = 10^14 sec^-1 approximately. This is much larger than the highest frequency yet produced experimentally, which is 2.5 x 10^10 sec^-1. The reader should note the character of these arguments, which are not precise calculations, but rather intelligent guesses. We have made it very plausible that the highest frequency is not very much bigger than 10^14 sec^-1. It is conceivable that, by applying a very high pressure to the right gas or solid, we could force its atoms a little closer than 10^-8 cm and achieve a velocity of sound a little larger than 10^6 cm/sec. In this way we might perhaps realize a frequency of, say, 2 × 10^14 sec^-1. We should be very surprised, though, if we ever pushed the frequency up to 10^15 sec^-1. In fact we know that there is a limit to the pro-cedure, because there is reason to believe that, when the pressure on a substance becomes extremely large, its atoms break up into separate electrons and nuclei.

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Question:

The reaction A + 2B \rightarrow products was studied kinetically and the following data was obtained. Experiment [A] (moles/liter) [B] (moles/liter) Rate(mole/liter - sec) 1 2.0 × 10^-3 4.0 × 10^-3 5.7 × 10^-7 2 2.0 × 10^-3 8.0 × 10^-3 11.4 × 10^-7 3 4.0 × 10^-3 4.0 × 10^-3 22.8 × 10^-7 Determine the rate law for this reaction.

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Solution:

The rate of reaction is equal to some rate constant, k, multiplied by the concentrations of A and B raised to the appropriate powers. That is, rate = k [A]^m [B]^n where the exponents m and n are to be determined. The presence of the coefficient 2 in the reaction equation A + 2B \rightarrow products in no way affects the form of the rate equation. Comparing experiments 2 and 1 we see that doubling [B] while holding [A] constant doubles the rate of reaction (2 × 5.7 × 10^-7 = 11.4 × 10^-7) . Hence, the rate is directly proportional to [B] and n = 1. Comparing experiments 3 and 1 we see that doubling [A] while holding [B] constant quadruples the rate (4 × 5.7 × 10^-7 = 22.8 × 10^-7). Thus, the rate has a square dependence on [A] (2^2 = 4), that is, it is second order in [A],and m = 2. Substituting m = 2 and n = 1 into the rate expression gives rate = k [A]^2 [B]^1, or rate = k [A]^2 [B] .

Question:

Iron may crystallize in the face-centered cubic system. If the radius of an Fe atom is 1.26 \AA, (a) Determine the length of the unit cell, (b) Calculate the density of Fe if its atomic weight is 55.85.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E06-0223.htm

Solution:

Picture the face of the cube with an iron atom at each, corner and one in the center of each face as shown in the figure. The diagonal BC will be equal to the length L 4 radii of Fe atoms. Hence, BC = 4 × 1.26 \AA = 5.04 \AA Using the Pythagorean Theorem, one knows that the squares of the two sides of a right triangle added together equal the square of the hypotenuse. Thus, AB 2 +AC^2 =BC^2 Because the sides of a square are equal -AB=AC. From this it can be seen that 2AB2=BC^2.ABis defined as the length of the unit cell. Using 2AB^2 =BC^2, it can be seen that AB^2 =BC^2 / 2andAB = \surd[BC^2 / 2] = BC / \surd2 Since we have already found BC, this equation can be solved. (BC = 5.04 \AA.) AB =BC/ \surd2 =5.04 \AA / 1.42 = 3.55 \AA The length of the unit cell is 3.55 \AA. (b) The density is defined as the mass in grams of a unit cell in one unit volume. In this problem we are given the mass of one mole of Fe and we have calculated the length of one side of the unit cube. To find the density one must calculate the number of atoms in one unit cell, the weight of one atom of Fe and the volume of one unit cube first. The following rules may be used to determine the number of unit particles (atoms) associated with the face-centimeter lattice: 1) An atom at a corner contributes 1/8 of its volume to each of 8 adjacent cubes. 2) An atom in the face of a cube contributes 1/2 of its volume to each of 2 adjacent cubes. In this lattice we have 8 corner atoms and 6 face atoms. To calculate the total number of atoms contribut-ing to the unit cube one can use the above rules, Total number of atoms = 8 corners × 1/8 atom/corner + 6 faces × 1/2 atom/face = 4 atoms. The unit cube therefore contains the equivalent of 4 atoms. To find the weight of 1 atom one divides the molecular weight by Avogadro's Number (6.02 × 10^23), because there are Avogadro's Number of atoms contained in a mole of atoms. weight of 1 atom = (55.85 g) / (6.02 × 10^23) = 9.28 × 10^-23 g Since there are 4 atoms in this unit cube this figure is multiplied by 4. weight of 4 atoms = [4×(55.85 g)] / [6.02 × 10^23] = weight of unit cube The volume of the cube is found by cubing the length of a side of the square because the volume is equal to the length × width × height of a rectangular solid. In a cube all three of these quantities are equal. The volume of the cube = (3.55 × 10^-8 cm)^3. (1 \AA = 10^-8 cm) . We now have the quantities necessary to find the density, cc = cubic centimeters = cm^3. density= weight of unit cube / volume of unit cube in cc = {[4×(55.85 g)] / [6.02 × 10^23]} / (3.55 × 10^-8 cm)^3 = 8.30 g/cc The density of Fe is therefore 8.30 grams per cubic centimeter.

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Question:

Two possible metabolic fates of \alpha-D-glucose are given below. Compare the standard free energy changes at 25\textdegreeC for these processes. Which process represents the more efficient utilization of \alpha-D-glucose? (a) \alpha-D-glucose (1 M) \rightarrow 2 ethanol (1 M) + 2CO_2 (g) (b) 6O2(g) + \alpha-D-glucose (1 M) 6CO_2(g) + 6 H_2O (l) ∆G\textdegree_fin Kcal/mole are CO_2(g) = - 94.26, H_2O (l) = - 56.69, ethanol (1 M) = - 43.39, \alpha-D-glucose = - 219.22.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E28-0909.htm

Solution:

For the standard free energy changes, ∆G_reaction=\sum ∆G_formationof products-\sum∆G_formationof reactants The free energy of any element in its standard state is taken as zero. Thus, for reaction (a), ∆G\textdegree_f, reaction (a) is: 2∆G\textdegree_f ,CO_2 + 2∆G\textdegree_f , ethanol -∆G\textdegree_f, \alpha-D-glucose. Substituting the known values into this equation yields 2(- 94.26) + 2(- 43.39) - (- 219.22) = - 56.08 Kcal/mole For reaction (b), ∆G\textdegree_f, reaction (b) is 6∆G\textdegree_f, H_2O + 6∆G\textdegree_f, CO_2 -∆G\textdegree_f, \alpha-D-glucose-∆G\textdegree_f, O_2 Substituting in the known values yields 6 (- 56.69) + 6(- 94.26) - (- 219.22) - (0) = - 685.5 Kcal/mole Reaction (b) is more efficient since its standard free energy change of formation is more negative or, in other words, more exothermic and releases more energy per mole of \alpha-D-glucose.

Question:

N_2O_5 \rightarrow 2NO_2 + (1/2)O_2 . The rate expression = {-d[N_2O_5]} / dt = k [N_2O_5]. The following mechanism has been proposed:N_2O_5k\rightleftarrowsNO_2 + NO_3 NO_2 + NO_3k_1\rightarrow NO_2 + O_2 + NO NO + NO_3k_2\rightarrow^2NO_2 Show the rate of O_2 formation is directly proportional to [N_2O_5].

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Solution:

To relate O_2 formation with the concentration of N_2O_5 investigate the mechanism of this reaction. In the mechanism, oxygen (O_2) is formed only from the second reaction, the rate being {d[O_2]} /dt= k_1 [NO_2] [NO_3]. To relate {d[O_2]} /dtto [N_2O_5], express [NO_2] [NO_3] in terms of [N_2O_5]. This can be done from the first reaction, which is at equlibrium. The equilibrium constant of this reaction, k, measures the ratio of products to reactants, each raised to the power of its coefficient in the chemical equation. Thus, k_eq = {[NO_2][NO_3]} / [N_2O_5]or[NO_2][NO_3] = k_eq [N_2O_5] Recall, {d[O_2]} /dt= k_1 [NO_2] [NO_3]. Thus [NO_2] [NO_3] can be replaced byk_eq[N_2O_5] , so that {d[O_2]} /dt= k_1k_eq[N_2O_5], which means O_2 formation is directly proportional to [N_2O_5] .

Question:

Discuss the excretory system of planaria.

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/Users/wenhuchen/Documents/Crawler/Biology/F11-0283.htm

Solution:

The excretory system of planaria involves a network of tubules running the length of the body on each side. These highly-branching tubules open to the body surface through a number of tiny pores. Side branches of the tubules end in specialized cells called flame cells (see Figure), also referred to as protonephridia. Each flame cell consists of a hollow, bulb--shaped cavity containing a tuft of long, beating cilia. It is very probable that flame cells function primarily in the regulation of water balance. The presence of better developed flame cells in freshwater species lends support to the osmoregulatory function of these cells. Primarily, water, and some waste materials, move from the tissues into the flame cells. The constant undulating movement of the cilia creates a current that moves the collected liquid through the excretory tubules to the nephridiopores, through which it leaves the body. The motion of the cilia resembles a flickering flame, hence this type of excretory system is often called a flame-cell system. Most metabolic wastes move from the body tissues into the gastrovascular cavity, and from there they are eliminated to the outside through the mouth. Nitrogenous wastes are excreted in the form of ammonia via diffusion across the general body surface to the external aquatic environment.

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Question:

You have just received a Basic Assembler Language programyou ran. As you turn to the execution page to check theprintout, you find that every fourth line of the program reads, for example, PSW =... 00004F17B4C4. What is the PSW and what information does it provide? In addition, what informationis provided in the three lines following the PSW statement?

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G09-0210.htm

Solution:

PSW is a mnemonic that stands for Program Status Word. The PSW has a length of 64 bits (8 bytes), and in general, contains the value of the location counter, system protection information, and program interrupt status. The system protection information protects the operating system (data needed by the computer) from errors that could be - detrimental to the system. This feature is an imperative one, since quite often programmers unknowingly make critical system errors. The interrupt status indicates the condi-tion of the Central Processing Unit (CPU) at the time of the interruption. When an interruption occurs, the proces-sor (CPU) places specific interrupt information in the PSW. In a simplified manner, we can say that when an interrupt occurs, the PSW stores the address of the next instruction to be executed; we may say the PSW 'looks-ahead'. The ad-dress of this instruction is indicated by the location counter portion of PSW. Thus, it is this part of the PSW that the beginning assembler language programmer should concern himself with. When an interrupt occurs, starting with the third byte of the 8-bytes PSW, the contents of the PSW are printed on the execution page of the program . It is printed in hexa-decimal notation. To find the instruction about to be ex-ecuted when the interruptoccured, the programmer need only concern himself with the rightmost four digits of the PSW. Using these digits, he can find the instruction in the body of the program. As this is the instruction that was to be executed when the interruption occured , it is the instruc-tion immediately preceding it that caused the problem . By closely inspecting this instruction, the cause of the in-terrupt can be found. The three lines following the PSW statement display the contents of the system's registers at the time of inter-ruption. The first two lines contain the 16 general purpose registers; this is indicated by the GPR (GPR 0-7; GPR 8-F) statement at the beginning of the line. The third line holds the Floating Point Registers (FPR). It should be noted that each time an interrupt occurs, the system prints the contents of the PSW. The program will be terminated if the number of allowed program inter-ruptions is exceeded.

Question:

What is the resistance of an electric toaster if it takes a current of 5 amperes when connected to a 120 - volt circuit?

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/Users/wenhuchen/Documents/Crawler/Physics/D20-0644.htm

Solution:

This is an application of Ohm's law. Since we wish to find the resistance, we use R = (E/I) . E = 120 volts,I = 5 amperes Therefore R = [(120 volts)/(5 amperes)] = 24 ohms

Question:

A bar with a cross-sectional area of 6 square centimeters and a permeability of 1000 units is placed in a magnetic field of intensity 5 oersteds. What is the total number of lines of magnetic flux in the iron?

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/Users/wenhuchen/Documents/Crawler/Physics/D21-0700.htm

Solution:

Let B = flux density \mu = magnetic permeability H = magnetic field intensity B = \muH and total flux \cyrchar\CYRF = B x area Since H = 5 oersteds, B = 1000 × 5 oersteds = 5000 gauss Then the flux \cyrchar\CYRF = a 5000 gauss × 6cm^2 = 30,000 lines of flux.

Question:

750 ml of gas at 300torrpressure and 50\textdegreeC is heated until the volume of gas is 2000 ml at a pressure of 700torr. What is the final temperature of the gas?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E02-0040.htm

Solution:

Here, one is given a gaseous system involving pressure, volume and temperature, where two of the variables are changed in going from the original system to the final system. This indicates that the combined gas law should be used. It can be stated: For a given mass of gas, the volume is inversely proportional to the pressure and directly proportional to the absolute temperature. This gas law can also be stated (P_1 V_1 )/T_1= (P_2 V_2 )/T_2 where P_1 is the original pressure, V_1 is the original volume, T_1 is the original absolute temperature, P_2 is the final pressure, V_2 is the final volume, and T_2 is the final absolute temperature. In this problem, you are given the original pressure, volume and temperature and the final pressure and volume. You are asked to find the final temperature. The temperature in \textdegreeC must be converted to the absolute scale before using the combined law. This can be done by adding 273 to the temperature in \textdegreeC. Converting the temperature: T_1 = 50 + 273 = 323\textdegreeK Using the combined law: (P_1 V_1 )/T_1= (P_2 V_2 )/T_2 P_1 = 300torrP_2 = 700torr V_1 = 750 mlV_2 = 2000 ml. T_1 = 323\textdegreeKT_2 = ? [{(300torr) (750 ml)}/(323\textdegreeK)] = [{(700torr) (2000 ml)}/(T_2 )] T_2 = [{(700torr) (2000 ml) (323\textdegreeK)}/{(300torr) (750 ml)}] T_2 = 2010\textdegreeK Convert T_2 to centigrade by subtracting 273 from it. 2010 - 273 = 1737\textdegreeC.

Question:

When 2.00 lb of brass at 2120F is dropped into 5.00 lb of water at 35 .00F the resulting temperature is 41.2 F. Find the specific heat of the brass. (Neglect the effect of the container.)

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/Users/wenhuchen/Documents/Crawler/Physics/D13-0486.htm

Solution:

The quantity of heatQ_wadded to the water is Q_w= m_wc_w∆T_w where m_w andc_ware respectively the mass and specific heat of the water, and ∆T_wis the increase in temperature of the water. Similarly, the heat QBlost by the brass to the water is Q_B = m_B c_B∆T_B , ∆T_B being the decrease in the temperature of the brass. The heat will flow from the brass to the water until they are at the same temperature. Since the total energy of the system is conserved, the heat leaving the brass equals the heat entering the water, m_Bc_B∆T_B = m_wc_w∆T_w (2.00 lb)(c_B)(2120F - 41.20F) = (5.00 lb)(1.00 Btu/lbF0)(41.20F - 35.00F) c_B= 0.091 Btu/lbF0

Question:

The combination of electric and magnetic fields used in J.J. Thomson's experiment can be used to measure the speed of the electrons. This measurement is possible if the electric and magnetic fields are arranged so that they produce forces acting in opposite directions. The strength of the electric and magnetic fields are then ad-justed so that the resultant force is zero and the beam isundeflected, (a) Show in this case that the electron speed v is given by, v = (E/B) Where E and B are theelecticand magnetic field strengths, respectively, (b) Compute the velocity of an electron beam that isundeflectedin passing through electric and magnetic fields of 3.7 × 10^4 N/C and 0.23 Weber /m^2 respectively.

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/Users/wenhuchen/Documents/Crawler/Physics/D21-0719.htm

Solution:

Thomson measured the ratio of the charge q of the electron to its mass m. Electrons were emitted from a hot filament and accelerated by an applied potential difference in a direction perpendicular to an electric field E and a magnetic field B. E and B are also at right angles so that they accelerate the electron along the same direction. The resultant force F on the electron is, F =qE+qv× B For the electron not to be deflected, this force must be equal to zero. The above equation reduces to, qE =qvB Since v and B are perpendicular. Solving this equation for v, v = E/B (b)Since the values of the electric and magnetic fields are 3.7 × 10_4 N/C and 2.3 × 10^1 Weber/m^2 respectively, v = (E/B) = [(3.7 × 10^4 N/C)/(2.3 × 10^-1 Weber /m^2 )] Since, 1 (Weber/m^2 ) = 1 [N/(A \bullet m)] v = 1.6 × 10^5 m/s The speed of the electrons is 1.6 × 10^5 m/s. Electric and magnetic fields arranged in this manner are used as a ve-locity selector for charged particles (ions) in several types of apparatus.

Question:

Show that a particle subjected to two simple harmonic vibrations of the same frequency, at right angles and out of phase, traces an elliptical path which de-generates to two coincident straight lines if the phase difference is \pi. Indicate the relevance of this to a half-wave plate.

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Solution:

Let the vibrations be taking place along the x\rule{1em}{1pt} and y-axes with a phase difference of \textphi between them. Then if x = a sin \omegat, y = b sin (\omegat + \textphi). Since sin (\alpha + \beta) = sin \alpha cos \beta + cos \alpha sin \beta (y/b) = sin \omegat cos \textphi + cos \omegat sin \textphi = (x/a) cos \textphi + \surd{1 - (x^2/a^2)} \textbullet sin \textphi Here we have used the fact that cos \omegat = \surd(1 - sin^2 \omegat) (y^2/b^2) + (x^2/a^2) cos^2 \textphi \rule{1em}{1pt} (2xy/ab) cos \textphi = [1 - (x^2/a^2)] sin^2 \textphi (x^2/a^2) (cos^2 \textphi + sin^2 \textphi) + (y^2/b^2) - (2xy/ab) cos \textphi = sin^2 \textphi or(x^2/a^2) + (y^2/b^2) - (2xy/ab) cos \textphi = sin^2 \textphi This is the general equation of an ellipse where the major and minor axes do not coincide with the x- and y-axes. Thus the particle always has x- and y- coordinates such that the point they define lies on an ellipse. The particle thus follows an elliptical path. If = \textphi = \pi/2, 3\pi/2, 5\pi/2, ..., the equation of the path reduces to (x^2/a^2) + (y^2/b^2) = 1, which is an ellipse with the major and minor axes coincident with the coordinate axes. When \textphi = \pi, the equation of the path becomes (x^2/a^2) + (y^2/b^2) + (2xy/ab) = 0, that is [(x/a) + (y/b)]^2 = 0. This is the equation of two coincident straight lines x/a = \rule{1em}{1pt} y/b, inclined to the negative x-axis at an angle tan^\rule{1em}{1pt}1 (b/a). (See figure.) In the case of a half-wave plate, plane-polarized light striking the plate is split up in two components, O and E, plane-polarized at right angles to one another and initially in phase. These pass through the plate at different speeds and the thickness is such that on emergence the two beams are out of phase by \pi. Any particle affected by the two components will thus be affected by two simple harmonic vibrations at right angles, out of phase by \pi. As can be seen from the above analysis, the particle would trace a straight- line path. This means that the two components are equivalent to a single vibration at an angle tan^\rule{1em}{1pt}1 (b/a) to the slower component, b/a being the ratio of the amplitudes of the components of the incident light on entering the plate. If the plane- polarized light is striking the plate at an angle of 45\textdegree to the two transmission directions, then it is resolved into two equal components so that b = a. The emerging light is thus plane-polarized in a direction making an angle of \rule{1em}{1pt} 45\textdegree with each of the principal directions in the plate.

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Question:

Suppose Stan is on the earth and Mavis is flying past the earth in a spaceship with a speed v = (3/4) c, as shown in the figure. Another spaceship is approaching the earth in the opposite direction and Stan measures its velocity of approach v_s = (3/4)c. The negative sign denotes the fact that the spaceship is moving to the left along the x-axis. What is the speed of the other spaceship as measured by Mavis?

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Solution:

The given quantities are v =(3/4)c and v_s = {─(3/4)c In the nonrelativistic case (i.e., where the speeds of the objects are much less than the speed of light, c) the speed of the other spaceship relative to Mavis is, by the law of addition of velocities, equal to the sum of the velocity of the spaceship relative to Stan and the velocity of Stan relative to Mavis. Or v_m =v_s + (─ v) . The negative sign here indicates the velocity of Stan re-lative to Mavis. It is equal but opposite to the velocity of Mavis relative to Stan (v). In the relativistic case, (where, as in this case, the velocities of the objects are comparable to the speed of light ) , the result above must be modified according to the special theory of relativity ,as follows v_m = {v_s + (─v)} / (1 + [{vs(─ v)} / c^2]) = (vs─ v ) / [1 ─ (vv_s / c^2)] = [{─(3/4)c} ─ (3/4)] / [1─ [{(3/4)c} {─(3/4)c}] / c^2] = {─ (3/2)c} / {1 + (9/16)} = ─ (24/25)c . Thus Mavis measures a speed less than the speed of light.

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Question:

Write a single APL statement that gives values for the current, according to Ohm's Law, where I = E/R. I is current, E is voltage, and R is resistance. You can assume that E and R are vectors of four elements each. If E \leftarrow 4 8 9 12 and R \leftarrow 2 4 3 4, find I.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G14-0365.htm

Solution:

Since E and R are vectors of equal length, one can write: I \leftarrow E \div R. This will divide the first elements of each vector, then the second, and so on. The result is a four element vector I. E is given by E \leftarrow 4 8 9 12 andR is given by R \leftarrow 2 4 3 4, therefore I \leftarrow E \div r execution will be performed as follows: I \leftarrow 4 \div 28 \div 49 \div 312 \div 4 I \leftarrow2233

Question:

Brass weights are used in weighing an aluminum cylinder whose approximate mass is 89 gm. What error is introduced if the buoyant effect of air (\rho = 0.0013 gm/cm^3) is neglected? \rho_brass = 8.9 gm/cm^3\rhoAl = 2.7 gm/cm^3

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Solution:

If the objects on the pans occupy different volumes, then the volume, and therefore the weight of the air is different on the two objects and must be accounted for. The additional weight on the object with less volume is just equal to the weight of the additional volume of air. V = m/\rho V_B = (89 gm)/(8.9 gm/cm^3) = 10 cm^3 V_Al = (89 gm)/(2.7 gm/cm^3) = 33 cm^3 The difference in volume V of air displaced on the two pans of the balance is V = V_Al - V_B = 33 cm^3 - 10 cm^3 = 23 cm^3 Hence, the mass error m = V\rho = 23 cm^3 × 0.0013 gm/cm^3 = 0.030 gm The error introduced is only a small fraction of the total mass, but in many experiments where accuracy is important an error of 0.030 gm in 89 gm is too great to allow.

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Question:

An atom emits a photon of green light \lambda = 5200 \AA in \tau = 2 × 10^-10 sec, Estimate the spread of wavelengths in the photon.

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Solution:

The Heisenberg uncertainty relationship for energy and time is \DeltaE\Deltat\geqh(1) where \DeltaE is the uncertainty in the energy of the system we are observ-ing, and\Deltatis the time interval over which we observe the system. Also,his Planck's constant divided by 2\pi . The energy of a photon is given by E =hv(2) where v is the frequency of the photon. But c =v\lambda(3) where c and \lambda are the speed and wavelength of light, respectively. From (3) and (2) E = (hc/\lambda) Taking the derivative of E with respect to \lambda, (dE/d\lambda) = (-hc/\lambda^2)(4) Hence \DeltaE \approx (-hc/\lambda^2)\Delta\lambda(5) Using (5) in (1) (-hc/\lambda^2)\Delta\lambda\Deltat\geqh \Delta\lambda\geq (h\lambda^2/-hc\Deltat) \Delta\lambda\geq (\lambda^2/-2\pic\Deltat)(6) Therefore, using the given data, \Delta\lambda\geq [{(5200 \AA)^2}/{(-2) (3.14) (3 × 10^8 m/s) (2 × 10^-10 s)}] Since 1 \AA = 10^-10 m \Delta\lambda\geq - [{(5.2 × 10^-7 m)^2}/(37.68 × 10^-2 m)] \Delta\lambda\geq - .717 × 10^-12 m \Delta\lambda\geq - 7.17 × 10^-13 m(7) Hence, the spread in wavelength is at least 7.17 × 10^-13 m. (We may neglect the minus sign in (7), because, looking at (6). it only indicates that as\Deltatincreases,\Delta\lambdadecreases).

Question:

One curie is defined as the amount of material, which is equivalent to one gram of radium, that gives 3.7 × 10^10 nuclear disintegrations per second. The half-life of ^210At is 8.3 hr. How many grains of astatine (At) would equal one curie?

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Solution:

The half life is defined as the time it takes for one half of the amount of a radioactive substance to disintegrate. To find the number of grams of ^210At that would be equal to 1 curie, one should first find the number of disintegrations of ^210At per second. One can then solve for the number of moles necessary and from that solve for the number of grams. Solving: (1) determining the half-life in seconds. This unit is used because curies are measured in disintegrations per second. t(1/2)= 8.3 hr × 60 min/hr × 60 sec/min = 29880 sec (2) Solving for the number of disintegrations of ^210At in 1 curie. Let x = the total number of nuclei of ^210At in 1 curie, x/2 the number of nuclei left after 1 half-life, 29880 sec. The following ratio can be set up: [(^210At) number of nuclei after 1 half life] / [half-life] = 3.7 × 10^10dis/sec. One can solve this ratio for x, the original number of nuclei present. [(x/2) / 29880 sec] = 3.7 x lo10dis/sec [x/2] = 29880 sec × 3.7 × 10^10dis/sec x = 2 × 29880 sec × 3.7 × 10^10dis/sec = 2.21 × 10^15 dis. Thus, 2.21 × 10^15disof ^210At are equal to 1 curie. (3) Solving for the number of moles. There are 6.02 × 10^23 nuclei per mole. Therefore, the number of moles present is equal to the number of disintegrations divided by 6.02 × 10^23 . no. of moles= [(2.21 × 10^15 nuclei) / (6.02 × 10^23 nuclei / mole) = 3.67 ×10^-9 moles (4) One solves for the number of grams necessary by multiplying the number of moles by the molecular weight of ^210At (MW = 210) no. of grams= 210 g/mole × 3.67 × 10^-9 moles = 7.71 × 10^-7 g. This would be the number of grams necessary at t = t_(1/2).

Question:

Given the reaction A + B \rightleftarrows C + D, find the equilibrium constant for this reaction if .7 moles of C are formed when 1 mole of A and 1 mole of B are initially present. For this same equilibrium, find the equilibrium composition when 1 mole each of A, B, C, and D are initial-ly present.

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Solution:

In general, an equilibrium constant measures the ratio of the concentrations of products to reactants, each raised to the power of their respective coefficients in the chemical equation. Thus, for this reaction, the equilibrium constant, K, is equal to {[C] [D] } / {[A] [B]} . You are asked to find K in the first part of this problem. Therefore, you must evaluate this expression. To do this, usestoichiometry. From the equation for the chemical reaction, you see that all the quantities are in equimolaramounts; 1:1:1:1 . It is given that .7 moles of C is produced. This means, therefore, that .7 mole of D must also be pro-duced. If the equimolarquantities are to be maintained, then, .7 mole of each A and B must have been consumed. If you started 1 with 1 mole of each, then .3 mole must be left. These mole amounts are the concentra-tions if you assume they are in a given amount of volume. Thus, sub-stituting these values in K = {[C] [D]} / {[A] [B]} K = {[C] [D]} / {[A] [B]} , you obtain K = (.7)^2 / (.3)^2 = 5.44. Therefore, you have calculated the equilibrium constant of this reaction to be 5.44. The second part of this problem asks you to use this same equi-librium for a reaction that starts with 1 mole each of A, B, C, and D. To find its composition, determine the concentration of each species at equilibrium. This can be found from K = {[C] [D]} / {[A] [B]} = 5.44 . To find the composition, assume X moles/Iiter(concentration) react. Thus, at equilibrium, both C and D have a final concentration of 1 + x, since you started with 1 mole/liter and had x moles/liter of each produced. This means, therefore, that A's and B's initial concentration must be reduced by x moles/liter to 1 - x. Substituting, 5.44 = (1 + x)^2 / (1 - x)^2 . Solving, x = .40 moles/liter . Thus, the concentration of both A and B = 1 -.4 = .6M and of both C and D = 1 + .4 = 1.4M .

Question:

A gyroscope consists of a uniform circular disk of mass M = 1 kg and radius R = 0.2 m. The disk spins with an an-gular velocity\omega =400 sec^\rule{1em}{1pt}1. The gyroscope precesses with the axes making an angle of 30\textdegree with the horizontal. The gyroscopewheel is attached to its axis at a point a dis-tance l = 0.3 m from the pivot which supports the whole structure. What is the precessional angular velocity?

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/Users/wenhuchen/Documents/Crawler/Physics/D37-1073.htm

Solution:

First, establish the general formula for regular precession. Refer to the figure, \vert L \vert = I\omegafor a symmetric object spinning around its symmetry axis. The torque \tau from the weight is mgl sin \alpha this is directed perpendicular to the plane containing L^\ding{217} and the vertical axis, ac-cording to the rule for a vector cross product. In a time ∆t the torque effects a change ∆L = \tau∆t = mgl sin \alpha∆t in the angular momentum, also perpendicular to L^\ding{217}. Thus, after ∆t, the angular momentum will have changed its direction; the horizontal component will have moved by an angle ∆\texttheta = \Omega∆t. (\Omega is the precessional angular velocity.) There is no change in the magnitude of L^\ding{217} since its increment ∆L^\ding{217} is perpendicular. Comparing the arc length to radius in the circle described by the tip of L^\ding{217}, the result is ∆\texttheta = \textthetaL / (L sin \alpha); L sin \alpha is the radius of this circle. Thus, \Omega = ∆\texttheta /∆t = (1 / L sin \alpha) (∆L / ∆t). Substituting the above expressions for L and ∆L gives \Omega = 1 / (I\omegasin \alpha)(mgl sin \alpha∆t / ∆t) = (mgl) / (I\omega). For the uniform disk we have I = (1 / 2)mR^2 , so \Omega = 2gl / (R^2\omega). Use g = 9.8 m / s^2, R = 0.2 m, \omega = 400 s^\rule{1em}{1pt}1, l = 0.3 m. All these are mks units. Then, \Omega = 2 \textbullet 9.8 \textbullet 0.3 / [(0.2)^2 \textbullet 400] \textasciitilde 0.368 s^\rule{1em}{1pt}1. Note that the precessional angular velocity does not depend on the mass of the disk nor on the angle that its axes makes with the horizontal.

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Question:

Write an APL program to build 10 rows of Pascal's triangle.

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Solution:

Each row ofpascal'striangle is a set of binomial coefficients of (x + y)^n-1, n > 0. Another way of representing these coefficients is (^N_K) for N = 1,2,....., and for K = 0, 1, 2,..., such that (^N_K) = [n!] / [K!(N - K)!] or,= [N(N-1) .... (N-K+1)] / [K!] Our strategy rests on the fact that each number in the triangle is the sum of two numbers right above it. For ex-ample, let us take rows 1 to 5. ROWS N = 11K = 0 N = 211K = 1 N = 3121K = 2 N = 41331K = 3 N = 514641K = 4 To get from row 4 to row 5, we can do the following: ROW 41331 +1331 14641 By taking a row, moving it one place to the right, and then adding it to itself, we can generate the next row. In APL, the symbol used for catenation is the comma. Catenation means "chaining" elements together. This can be done by the statement P \leftarrow (0,P) + (P,0) . This statement will do the following: 01331 +13310 A zerocatenatedwith string P in (0,P) part, moves P one place to the right, while the zero in the second part, (P,0), just adds itself to the string P. Then, the addition is performed, and the new row is created. The program is given below: \nablaPASCAL [1]P \leftarrow 1 [2]P [3]\rightarrow 2 × N \geq\rho\Rho\leftarrow (0,P) + (P,0) \nabla Line [3] contains the symbol \rho, which is called the re-shaping operator. When you enter the value for N (in this case, N = 10), \rho determines the number of times P \leftarrow (0,P) + (P,0) is to be executed. When ten branches have been execut-ed, the comparison N \geq \rho will be false or, in APL, it will have the value zero. This zero will then be multiplied by 2 telling the program to branch to instruction zero, which is the way of indicating the termination of the program.

Question:

What is the physiological purpose of shivering?

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Solution:

Shivering is a means by which the body maintainsitsnormal temperature when the ambient tem-perature is cold. Recall that heat is produced by virtually all chemical reactions within the body. One way by which the basal level of heat production can be increased is to increase the rate of skeletal muscle contractions. The first muscle changes in response to cold are a gradual and general increase in skeletal muscle tone . This soon leads to shivering, which consists of oscillatory, rhythmic muscle tremors occurring at the rate of about 10 to 20 per second. This intensive muscle activity rapidly uses up ATP, thus stimulating more cellular respiration and more heat production. These contractions are so effective that body heat pro-duction may increase several fold within seconds . Because no external work is performed, all the energy liberated by the metabolic machinery becomes internal body heat. Thus shivering is an adaptation to cold. Shivering tends to ruffle the body hair in most animals, creating dead air space that serves as insulation. In man, shivering causes the erection of body hair, but his sparse coat of hair is insufficient to act as an insulatory mechanism.

Question:

Hydras, like most cnidarians, are carnivorous and feed mainly on small crustaceans. Describe the method of ingestion and digestion in these organisms.

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Solution:

Should a small crustacean make contact with the tentacles of the hydra, the nematocysts, which are lodged in the tentacle walls, would discharge. The nematocysts will entangle the prey and paralyze it. The captured organism is pulled toward the mouth by the tentacles. The mouth then opens to receive the prey. The retraction and bending inward of the tentacles and the opening of the mouth is a reflex response. Mucus secretions in the mouth aid in swallowing. Eventually the prey arrives at thegastrovascularcavity.Secretorycells lining the cavity release enzymes that begin the digestion of protein. Eventually, the tissues of the prey animal are reduced to a soupy broth. Flagel-lated cells lining the cavity continually beat their flagella, thus aiding in mixing the digesting food with enzyme molecules. Subsequent to this extracellular phase of digestion comes intracellular digestion. Amoeba-like cells lining the cavity extend pseudopodia which engulf small fragments of what is left of the prey animal. Continued digestion of proteins and the digestion of fats occur within food vacuoles in these amoeba-like cells lining thegastrodermis, Products of digestion are circulated by cellular diffusion. Glycogen and fat are the chief storage products of excess food. Undigested materials are ejected from the mouth upon contraction of the body.

Question:

Write a program to determine the prime factorizations of integers greater than 1.

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Solution:

According to the Fundamental Theorem of Arithmetic any integer can be expressed as the product of primes. Some examples 108 = 2 × 2 × 3 × 3 × 3 = 22 × 33 390 = 2 × 3 × 5 × 13 -210 = -(2 × 3 × 5 × 7) The algorithm for computing the prime factors of a positive integer greater than 1 is based on the following facts: 1) Let p be a prime \leq \surdn. If n is divisible by no such p, then n is a prime. 2) Let p be a divisor of n. Then q = n/p is an integer, and any other prime divisor of n divides q. 3) n has at most one prime factor greater than \surdn . Using these facts we can construct the following computer program. The program assumes that a table of primes for the range 1-1000 was developed earlier and can be obtained from the secondary memory files. Thus, the program can find prime factorizations for all numbers less than (1000)2 = 1,000,000. Execution terminates when the considered integer N is 1 or less. The program looks as follows: PRIME FACTORIZATION CRESERVE MEMORY FOR PRIME DIVISORS (NPR), FOR PRIME FACTORS (NFAC) , AND FOR THE NUMBER OF TIMES EACH PRIME FACTOR OCCURS (NBR). DIMENSION NPR(168), NFAC(8), NBR(8). CREAD IN THE PRIMES LESS THAN 1000. C(THERE ARE 168 OF THEM) READ (5,5)(NPR(I), I = 1, 168) 5FORMAT (20I4) 10READ (5,15) N 15FORMAT (I7) IF (N - 2) 85,75,20 CCOPY N INTO INT FOR TESTING PURPOSES. 20INT = N RN = INT NRT = SQRT(RN) CSTART A COUNTER FOR THE NUMBER OF DIFFERENT PRIMES THAT DIVIDE N. J = 0 DO 45 I = 1, 168 CNF COUNTS THE OCCURRENCES OF EACH PRIME FACTOR NF = 0 CBRANCH OUT OF THE DO LOOP IF CTHE CURRENT PRIME IS GREATER CTHAN THE SQUARE ROOT OF INT. IF (NPR(I) - NRT) 25,25,50 25KINT = INT/NPR(I) IF (INT - KINT\textasteriskcenteredNPR(I)) 35,30,35 CIF NPR(I) DIVIDES INT, STORE THE CQUOTIENT KINT IN INT. THUS, THIS CQUOTIENT IS NOW THE DIVIDEND FOR FURTHER TESTING. 30INT = KINT NF = NF + 1 GO TO 25 35IF (NF) 45,45,40 CINCREMENT J BY 1 IF NF IS POSITIVE CAND IF NPR(I) DOES NOT DIVIDE INT. CTHEN STORE THE PRIME NPR(I), AS THE CJ-TH MEMBER OF NFAC AND STORE CTHE MULTIPLICITY NF IN NBR(J). 40J = J + 1 NFAC(J) = NPR(I) NBR(J) = NF RN = INT NRT = SQRT(RN) 45CONTINUE CN IS PRIME IF AND ONLY IF J = 0 50IF (J) 75,75,55 55IF (INT-1) 65,65,60 60J = J + 1 NFAC (J) = INT NBR(J) = 1 CPRINT PRIME FACTORIZATION 65WRITE (6,70) N, (NFAC(I) ,NBR(I), I = 1,J) 70FORMAT (1HO, I7, 3H = , 8(I5, 1H( ,I2, 1H))) GO TO 10 75WRITE (6,80) N 80FORMAT (1HO, I7, 9H IS PRIME) GO TO 10 85STOP END

Question:

In contrasting a food chain and a food web, which one do you think actually operates in a real community?

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Solution:

Food is the common word used to describe the various nutrients that all living heterotrophic organisms must ingest in order to obtain energy to sus-tain their life processes. Autotrophic photosynthetic organisms, such as the green plants, can manufacture their own food from simple inorganic molecules with the energy from sunlight. Life on earth ultimately depends on food energy which in turn is dependent on the sun. The radiant energy from the sun that reaches the photosynthetic green plants is responsible for the transformation of basic raw materials such as water, carbon dioxide, nitrogenous compounds and minerals into the development of the plants themselves. Plants store nutrients (starches and sugars) within them and in turn are eaten by heterotrophic organisms such as animals. A food chain is most commonly a sequence of organisms that are related to each other as prey and predators. One species is eaten by another, which is eaten in turn by a third, and so on. Each species forms a step or link in one or more food chains. Every food chain begins with the autotrophic organisms (mainly green plants) that serve as the producers of the community. Any organism that does not produce its own food and must therefore depend on another for nutrients is a consumer. Every food chain ends with decomposers, the organisms of decay, which are usually bacteria and fungi that degrade complex organic materials to simple substances which are reusable by the producers. The links between the producers and the decomposers are variable because the producers may die and be acted upon directly by the decomposers, or the producers may be eaten by primary consumers, called the herbivores. The herbivores consume the green plants and in turn may be either acted upon directly by the decomposers or fed upon by the secondary consumers, the carnivores (animal eaters). Some food chains consist of tertiary consumers, or secondary carnivores, which eat the secondary consumers and sometimes also the primary consumers. In almost all ecosystems there are top carnivores: one or more large, specialized animal species that prey on organisms on the lower steps, but are not ordinarily consumed by predators themselves. The larger whales enjoy this status, as do lions, wolves, and man. The decomposers can feed on any dead organism in the food chain. In addition, there are parasite chains in which small or-ganisms live on or within larger ones from where they obtain food. The successive levels in the food chains of a community are referred to as trophic levels. Thus all the producers together constitute the first (or lowest) trophic level and the primary consumers (herbivores) constitute the second trophic level. The herbivore-eating carnivores constitute the third trophic level, and so on. Energy-flow in an ecosystem is actually more complicated than the flow of nutrients that simplified food chains imply. In most real communities, there are many different possible food chains that are tied together by cross linkages. Any one animal usually eats a variety of food and thus may be part of several food chains that intersect. Also, food chains starting from a common plant source may radiate outward as the plant food is eaten by different herbivores, and as the latter are eaten by different carnivores, and so on. Thus, there is actually a complex food web formed as the food chains first radiate outward from the plants and then come together at the top carnivore level of the web (see Figure). For example, the grasshoppers of the grasslands of Canada eat grass and are eaten by robins. Robins eat many kinds of insects as well as grasshoppers. The great horned owl feeds upon the robin as well as other mammals such as the prairie vole. But the prairie vole is also eaten by the coyote. Hence, even though organisms in a community can be classified into the various trophic levels according to the nature of their prey, it is more characteristic to assign organisms to a food web than to a food chain in a real community.

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Question:

Explain briefly the background and applications of the COBOL programming language.

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Solution:

On May 28 and 29, 1959, a meeting was called in the Pentagon for the purpose of considering the desirability and feasibility of establishing a common language for pro-gramming of electronic computers in business data processing. As a result, in a short period of time, a new computer lan-guage COBOL (Common Business Oriented Language) was devel-oped. Since that time COBOL has definitely become the most popular and useful language inside the business community. The basic language element in COBOL is the "word." As in the English language, COBOL contains many types of words with which meaningful thoughts are formed. This makes the language intelligible for many people who do not have much programming experience and only use the computer occasion-ally for solving problems. Every COBOL program is divided into four parts: 1.The Identification Division - identifies the program. 2.The Environment Division - specifies the input-output devices to be used. 3.The Data Division - describes the data to be used in the program. 4.The Procedure Division - includes the instructions that the system will follow in solving the problem. Such a division provides greater simplicity of understanding and analysis of a COBOL program. These are only a few of the factors that account for the constantly growing popularity of the COBOL language in the business world.

Question:

A forced oscillator is a system whose behavior can be described by a second-order linear differential equation of the form: \"{y} + A_1ẏ + A_2y (t) = E(t)(1) where A_1, A_2 are positive constants and E(t) is an external forcing input. An automobile suspension system, with the road as a vertical forcing input, is a forced oscillator, for example, as shown in Figure #1. Another example is an RLC circuit connected in series with an electro-motive force generator E(t), as shown in Figure #2. Given the initial conditions y(0) = y0and ẏ(0) = z_0 , write a FORTRAN program that uses the modified Euler method to simulate this system from t = 0 to t = t_f if: Case 1:E(t) = h where h is constant Case 2:E(t) is a pulse of height h and width (t_2 - t_1) . Case 3:E(t) is a sinusoid of amplitude A, period 2\pi/\omega \omega \omega and phase angle \varphi . Case 4:E(t) is a pulse train with height h, width W, period pW and beginning at time t = t_1 .

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G24-0589.htm

Solution:

If equation (1) is rearranged to the form \"{y} = E(t) - A_1ẏ - A_2y It can be seen to be of the form d^2y/dx^2 = 8_3(x,y,y') , so one can use the Euler 3 subroutine developed earlier In this chapter. Case 1:This Is the CONST FUNCTION also developed earlier in the chapter . The parameters need to be read in only once: they are read in the main program and kept in a COMMON block. REAL T/0.0/,TFIN,N,ACCUR,Y0,Z0,A1,A2,H COMMONA1,A2,H READ,N,TFIN,ACCUR,Y0,Z0,A1,A2,H CALL EULER3(N,TFIN,T,Y0,Z0,ACCUR) STOP END FUNCTION G3(T,V,W) REAL G3,T,V,W,A1,A2,H COMMON A1,A2,H G3 = H - A1\textasteriskcenteredW - A2\textasteriskcenteredV RETURN END Case 2:This is the PULSE FUNCTION developed earlier. REAL T/0.0/,TFIN,N,ACCUR,Y0,Z0,A1,A2,H,T1,T2 COMMON A1,A2,H,T1,T2 READ, N, TFIN, ACC UR, YO, ZO ,A1 ,A2, H, T1, T2 CALL EULER3 (N,TFIN,T,YO,ZO,ACCUR) STOP END FUNCTION G3(T,V,W) REAL G3,T,V,W,A1,A2,H,T1,T2 COMMON A1,A2,H,T1,T2 G3 = -A1\textasteriskcenteredW - A2\textasteriskcenteredV IF(T.GE.T1.AND.T.LT.T2)G = H - A1\textasteriskcenteredW - A2\textasteriskcenteredV RETURN END Case 3:This is the SINE FUNCTION developed earlier. The variable OMEGA is used for\omegaand PHI is used for \varphi . \omega REAL T/0.0/, TFIN,N,ACCUR,YO, ZO,A1,A2 ,A,OMEGA,PHI COMMON A1,A2,A,0MEGA,PHI READ N, TF IN, ACC UR, YO, ZO, A1, A2, A, OMEGA, PHI CALL EULER3(N,TFIN,T,YO,ZO,ACCUR) STOP END FUNCTION G3(T,V,W) REAL G3,T,V,W,A1,A2,A,OMEGA,PHI COMMON A1,A2,OMEGA,PHI G3 = A\textasteriskcenteredSIN(OMEGA\textasteriskcenteredT + PHI) -A1\textasteriskcenteredW - A2\textasteriskcenteredV RETURN END Case 4:This is the TRAIN FUNCTION developed earlier. The parameters for the G3 FUNCTION subprogram (A1,A2) are kept in the COMMON block named G, while the parameters for the TRAIN FUNCTION subprogram (H,T1,W,P) are kept in the COMMON block named TRANE. The subprogram calling sequence is as follows: The main program calls the EULER3 subroutine,EULER3 calls the G3 FUNCTION subprogram, and G3 calls the TRAIN FUNCTION subprogram. CMAIN PROGRAM REAL T/0.0/, TFIN, N, ACCUR, YO, 20, A1, A2, H, Tl, W, P) COMMON /G/A1,A2/TRANE/H,T1,W,P READ,N,TFIN,ACCUR, Y0,Z0,A1,A2,H,T1,W,P CALLEULER3(N,TFIN,T,YO,ZO,ACCUR) STOP END FUNCTION G3(T,V,U) REAL G3,T,V,U,A1 ,A2,TRAIN COMMON /G/A1,A2 G3 = TRAIN(T) -A1\textasteriskcenteredU - A2\textasteriskcenteredV RETURN END FUNCTION TRAIN(T) REAL TRAIN,T,H,T1,W,P COMMON/TRANE/H, T1, W, P TIME = T IF(T.GE. (T1 + P\textasteriskcenteredW)) TIME =AMOD((T - T1)/(P\textasteriskcenteredW))\textasteriskcenteredP\textasteriskcenteredW + T1 IF((TIME.LT.T1).0R. (TIME.GE. (T1 + W)))TRAIN = 0. IF((TIME.GE.T1) .AND. (TIME.LT. (T1 + W)))TRAIN = H RETURN END

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Question:

1.3625 g of an unknown metal X reacts with oxygen to form 1.4158 g of the oxide X_2O. What is the atomic mass of X?

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Solution:

This problem will be solved by calculating the mass and, from this, the number of moles of oxygen in the compound. The corresponding number of moles of metal is twice this amount. The molecular weight is the number of moles of metal divided by the weight of metal that was used. The mass of oxygen in the compound is mass O = mass oxide - mass X = 1.4158 g - 1.3625 g = 0.0533 g. The corresponding number of moles is moles O = mass O \div atomic mass O = 0.0533 g \div 15.9994 g/mole = 0.0033 mole . Since, in the oxide, there are two moles of X per mole of O, the number of moles of X present is 2 × 0.0033 mole = 0.0066 mole X. Hence, 1.3625 g of X corresponds to 0.0066 mole X. The atomic mass of X is then: atomic mass X = (mass X) / (moles X) = 1.3625g / 0.0066 mole =206.4394 g/mole.

Question:

Themolalfreezing point constant for a certain liquid is 0.500\textdegreeC. 26.4 g of a solute dissolved in 250 g of this liquid yields a solution which has a freezing point 0.125\textdegree below that of the pure liquid. Calculate the molecular weight of this solute.

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Solution:

A mole of a substance in 1000 g of H_2O gives a definite and known lowering of the freezing point. By determining the freezing point of a solution of known con-centration, one can calculate the molecular weight of the dissolved substance. A general formula may be developed for this kind of calculation: molecular weight = (grams solute) / (kg of solvent) × (k_f) / (∆T_f) wherek_fis the freezing point constant and ∆T_fis change in the freezing point. Here, 26.4 g of a solute is dissolved in 250 g of liquid. One can find the number of grams of solute using the following ratio: (Note: there are 1000 g in 1 kg.) Let X = number of g of solute in 1000 g (26.4 g) / (250 g) = (X / 1000 g) X = [(1000 g × 26.4 g)] / (250g) = 105.6 g One can now solve for the molecular weight molecular weight= (105.6 g / 1kg) × (0.500\textdegreeC / 0.125\textdegree) = 422.4 g

Question:

A 3200-lb car is slowed down uniformly from 60 mph to 15 mph along a level road by a force of 1100 lb. How far does it travel while being slowed down?

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Solution:

A diagram should first be drawn so that our sign conventions are consistent (see diagram). we are given the change in velocity of the car in the positive x direction, the force acting against it in the negative x direction, and the weight of the car. We can calculate the deceleration of the car, and from this we can calculate the time the car is decelerated by its change in velocity. First, we must find the mass of the car, given its weight. W = mg 3200 lb = m 32 ft/s^2 m = 100 lb-s^2/ft[slugs] From this we can calculate the deceleration of the car. Remember that the force acts in the negative × direction. F = ma -1100 lb = 100 lb-s^2 / fta a = -11 ft/s^2 Assuming constant deceleration: a = (∆v/∆t) where ∆v = 15 mph - 60 mph = - 45 mph. A useful conversion factor to remember is 60 mph = 88 ft/s, so that - 45 mph = -66 ft/s. Hence, -11 ft/s^2 = -[(66 ft/s)/∆t] ∆t = 6s Distance is given by the formula 1/2 at^2 + v_it + d_i = d where v_i is initial velocity, and d_i is the initial position (which we will here set equal to 0). Therefore d = 1/2 (-11 ft/s^2)(6s)^2 + 88 ft/s^2 (6s) + 0 d = 330 ft.

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Question:

Suppose that the first half of the distance between two points is covered at a speed v_1 =10mi/hr and, that during the 10 second half, the speed is v2= 40 mi/hr. What is the average 2 speed for the entire trip?

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Solution:

The average speed is the total distance traveled divided by the total traveling time. The average speed is not v= {(10 mi/hr) + (40 mi/hr)} / 2 = 25 mi/hr. Let 2x be the total distance traveled and let t_1 and t_2 de-note the times necessary for the two parts of the trip. Then, v= (2x)/(t_1 + t_2) Since only the velocities are known, the average velocity must be expressed in terms of these variables. In order to eliminate unknown variables, we see that t_1 = x/v_1; t2= x/v_2. t_1 + t_2 = x/v_1 + x/v_2 = [x(v_1 + v_2)]/(v_1v_2) Therefore,v= 2x/[{x(v_1 + v_2)}/(v_1v_2)] = (2v_1 + v_2)/(v_1 + v_2) = [2(10 mi/hr)(40 mi/hr)]/(10 mi/hr + 40 mi/hr) = (800/50)mi/hr = 16mi/hr.

Question:

What is the evidence for the theory that theneurilemma sheathplays a role in the regeneration of severed nerves?

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Solution:

When an axon is separated from its cell body by a cut, it soon degenerates . However, the part of the axon still attached to theperikaryon can re-generate. A healthyperikaryonis important in regene-ration. As long as the cell body of the neuron has not been injured, it is capable of making a new axon. Regeneration begins within a few days following the severing of the nerve. Axon regeneration is believed to involve theneurilemma, a cellular sheath composed of Schwann cells which envelopes the axon. The role the neurilemmasheath plays in regeneration is to provide a channel for the axon to grow back to its former position. What happens is that the growing axon enters the old sheath tube and proceeds along it to its final destination in the central nervous system or periphery. In some experiments, theneurilemmasheath is removed and replaced by a conduit, for example, a section of blood vessel or extremely fine plastic tube. The severed axon is able to regenerate normally within the substituted conduit. This result shows that the sheath is not an absolute requirement for the regeneration of an axon in vitro, since a plastic sheath serves the same function as well.

Question:

Insurance companies have worked out probability tables to estimate the chances a person has of living until a certain age. A study of just over a million people was made, and in 1941, a mortality table was published. It listed the number of persons still alive at. age 0,1,2,3, ..., 99. LetL(X) = number of persons living at age X then,P(X) = probability that a person of age X will live for one more year = [L(X+1)] / [L(X)] Develop a program in BASIC that allows the user to input the year of birth, the present age, and some year after the year of birth. The output should list the age, the year, and the chances of that person living until that year.

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Solution:

To make the problem more meaningful, let us present a con-crete example. Suppose we want to calculate the chances of a 35-year old born in 1945 living until the year 2000. Adding the person's age to hisbirthdate gives us the year 1980. This person must live for 20 more years to reach 2000. His chances of making it are [L(35+20)] / [L(35)] = [L(55)] / [L(35)] = P(X) If we look at the data tables given in the program, we find that L(55) = 754191 and L(35) = 906554. So the answer is [754191/ 906554] = .8319316 or, approximately 0.83. The strategy for our computer program is fairly easy. The program reads in the values, figures out the probability, and returns the answer. To terminate the procedure, input a negative value for the year of birth. We list the data (as it is stored in array L) in a DATA statement at the end of the program. The first entry is L(0), the total number of persons included in the study. 1\O DIM M(99,4) 15 FOR J = 0 TO 99 2\O READ M (J,0) 25 IF J = 0 THEN 30 28 LET M(J-1,1) = M(J-1,0) - M(J,0) 3\O NEXT J 32 LET M(99,1) = M(99,0) 33 REM INPUT YEAR OF BIRTH, AGE, AND EXPECTED YEAR 35 PRINT "INPUT YEAR OF BIRTH, AGE, AND EXPECTED YEAR'' 4\O INPUT B,A,X 45 IF B < 0 THEN 99 48 REM N IS THE PRESENT YEAR; M IS THE 49 REM YEARS BETWEEN PRESENT AND EXPECTED. 5\O LET N = B + A 52 LET M = X - N 55 LET P = L (A+M) / L(A) 58 PRINT 6\O PRINT "CHANCES OF LIVING UNTIL YEAR"; X; "ARE"; P 65 GO TO 35 7\O DATA 1023102, 1000000, 994230, 990114, 986767 72 DATA 983817, 981102, 978541, 976124, 973869 74 DATA 971804, 969890, 968038, 966179, 964266 76 DATA 962270, 960201, 958098, 955942, 953743 78 DATA 951483, 949171, 946789, 944337, 941806 8\O DATA 939197, 936492, 933692, 930788, 927763 82 DATA 924609, 921317, 917880, 914282, 910515 84 DATA 906554, 902393, 898007, 893382, 888504 86 DATA 883342, 877883, 872098, 865967, 859464 87 DATA 852554, 845214, 837413, 829114, 820292 88 DATA 810900, 800910, 790282, 778981, 766967 89 DATA 754191, 740631, 726241, 710990, 694843 9\O DATA 677771, 659749, 640761, 620782, 599824 91 DATA 577882, 554975, 531133, 506403, 480850 92 DATA 454548, 427593, 400112, 372240, 344136 93 DATA 315982, 287973, 260322, 233251, 206989 94 DATA 181765, 157799, 135297, 114440, 93378 95 DATA 78221, 63036, 49838, 38593, 29215 96 DATA 21577, 15514, 10833, 7327, 4787 97 DATA 3011, 1818, 1005, 454, 125 99 END

Question:

Two bodies having masses m_1 = 30 gm and m_2 = 40 gm are attached to the ends of a string of negligible mass and suspended from a light frictionless pulley as shown in the diagram. Find the accelerations of the bodies and the tension in the string.

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Solution:

In order to find the acceleration of the masses, we use Newton's Second Law (F = ma) to relate the external forces applied to each mass to the accelera-tion of each mass. Consider the body of mass m_1. Two external forces act on it, the weight m_1g downward and the upward pull T of the string. The resultant force on this body is T -m_1g upward. Using the Second Law T - m_1g = m_1a(1) where the negative sign implies that T and m_1g are in opposite directions, and a is the upward acceleration of this body. Now consider the body of mass m_2. The forces acting on this body are its weight m_2g downward and the tension T upward. The resultant force is m_2g - T downward and m_2g - T = m_2a(2) where a is the downward acceleration of this body. Since the two bodies move together, the accelerations are equal in magnitude but they are opposite in di-rection. To find the acceleration of the string and two masses, we add (1) and (2) and solve for a: m_2g - m_1g = m_1a + m_2a = (m_1 + m_2)a ora= (m_2g - m_1g) / (m_1 + m_2) = {(m_2 - m_1)g} / {m_1 + m_2} [{(.004 kg - .03 kg)(9.8 m/s^2)} / {.04 kg + .03 kg}] = [{(0.1 kg)(9.8 m/s^2)} / {.07 kg}] = (.098 Nt) / (.07 kg) = 1.4 m/s^2(3) We can substitute this value for a in either of the original equations {(1) or (2)} to obtain the value for T, the string tension. Substituting (3) in (1): T - m_1g = m_1a T = m_1g + m_1a = m_1(g + a) = .03 kg (9.8 m/s^2 + 1.4 m/s^2) = .03 kg (11.2 m/s^2) = .336 Nt.

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Question:

Name the two basic classes of electronic computers. Give the basic similarities and differences between them.

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Solution:

The two main classes of electronic computers are analog and digital, although analog computers are seldom used these days. The analog computers use continuous electrical quan-tities such as voltages, currents, capacitances, etc. to repre-sent the continuous real-world quantities such as temperature, pressure, mass, and so on. For example, 1 + 1 = 2 may be repre-sented as 1 volt + 1 volt = 2 volts using appropriate circuitry. On the other hand, a more complicated mechanical system shown in Figure 1-2a may be represented by the electrical circuit shown in Figure 1 - 2b. Figure 1-2. Representation of A Problem by Analog Computer. The general equation of oscillation for the mechanical system is given by: {MD + F + (1 / KD)} X = 0, where D is called the differential operator, and X is the posi-tion of the weight at any moment. After the switch is closed, the equation of the electrical oscillation will take the form: {LD + R (1 / CD)} I = 0, where D is same as before, and I is the current at any moment. Note that these two equations are identical, except for some of the letters used. Therefore, the behavior of the two systems will be identical. Although analog computers are good in solving problems that involve continuous quantities, the representation is not always easy and the accuracy is in doubt. For example, an analog com-puter calculates the interest on a million-dollar loan, the result could be off by at least a few dollars. The accountant who must reckon every penny may not be satisfied with that kind of result. The digital computer deals with numbers represented inter-nally as a sequence of binary digits (or bits). The architecture and operation of these systems are detailed throughout this chapter. Although a digital computer cannot directly represent and solve the above problem whose nature is continuous, it can be programmed to solve the problem through a mathematical proce-dure. Moreover, the results will be as accurate as the math-ematical formulation allows. In the case of interest calcula-tions on a million-dollar loan, the result will be accurate down to the pennies, except for possible round-off error that is really the property of mathematical solution procedure and not the fault of the digital computer. The digital computers have advantages of programmability with reliable accuracy, since exactly the same result will be obtained the next time the program is executed. The analog com-puter may sometimes have advantage in solving continuous type problems in a direct way to produce results that are not precise but acceptable for all practical purposes. The similarity be-tween the two classes of computers is the fact that both are electromechanical machines. The difference is that the digital computers, sometimes at the expense of extra storage and fre-quency bandwidths (e.g. more complex electronic signals), are more reliable and accurate.

Question:

Explain briefly the background and applications of APL programming .

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Solution:

APL (A Programming Language) is an extremely pow-erful, simple to learn, easy to use computer language, which lends itself to many applications . Widely recognized as a scientific language, APL is equally useful in business and in education. APL is one of the best-suited languages for special applications, such as computer aided instruction (CAI), operations research, and simulation. Being a versatile language, it can be used in either execution or definition modes. In execution mode the lan-guage enables you to operate the terminal as though it were a desk calculator. In definition mode you are able to enter, store, execute, and debug programs. APL is efficient. When compared with other languages: it takes the least amount of programming effort to accomplish a given job. When in execution mode, APL has immediate diagnostics. The system reports errors as soon as they are made and gives their type and location . All these features make APL a strong tool in solving a wide range of complicated problems.

Question:

Calculate the number of liters in one cubic meter.

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Solution:

There are 1,000 milliliters (ml) or cubic centimeters (cc) in one liter. Thus, if one wishes to convert one cubic meter to liters, the cubic meter mustbe converted to cubic centimeters. 1 meter =100 centimeters (1 meter)^3= (100 centimeters)^3 = 1,000,000 centimeters^3 = 1 × 10^6 cubic centimeters Cubic centimeters can be converted to liters by multiplying the number of cubiccentimeters by the factor 1 liter/1,000 cubic centimeters. 1 × 10^6 cubic centimeters × 1 liter/1,000 cubic centimeters = 1,000 liters. There are 1,000 liters in one cubic meter.

Question:

Find the logarithm of 3^2.

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Solution:

Recall that log_b x^y = y log_b x; thus log_10 3^2 = 2 log_10 3 Referring to a table of common logarithms we find: log_10 3 = .4771; hence, = 2 (.4771) = .9542 . Thus, log_10 3^2 = .9542.

Question:

Discuss the basic features of the molluscan body. How do they differ in the several classes of mollusks ?

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Solution:

The phylum Mollusca is the second largest in the animal kingdom. Snails, clams, oysters, slugs, squids, and octopuses are among the best known molluscs. The adult body plan is remarkably different from that of any other group of invertebrates. The soft body consists of three principal parts: (1) a large ventral muscular foot which can be extruded from the shell (if a shell is present) and functions in locomotion; (2) a visceral mass above the foot, containing most of the organs of the body; and (3) a mantle, a heavy fold of tissue that covers the visceral mass. In most species, the mantle contains glands that secrete a shell. The mantle often projects over the edges of the foot and overhangs the sides of the visceral mass, thus enclosing a mantle cavity, in which gills frequently lie. The chitons, members of the class Amphineura, have an ovoid bilaterally symmetrical body with an anterior mouth and posterior anus. They have paired excretory organs, nerves, gonads, and gills. The shell consists of eight dorsal plates. Class Gastropoda is a large class, containing the snails, slugs and their relatives. Most gastropods have a coiled shell; however, in some species, coiling is minimal, and in others, the shell has been lost. The body plan of the adult gastropods is not symmetrical. During develop-ment, two rotations of the body occur, so that the anus comes to lie dorsal to the head in the anterior part of the body. The organs on one side of the body atrophy, so that the adult has one heart, one kidney, one gonad, and one gill. This embryonic twisting, called torsion, occurs in all gastropods, including those with a flat shell, slugs and other species without a shell Class Pelecypoda or Bivalvia contains the bivalves. These mollusks have two hinged shells. Scallops, clams, mussels, and oysters are well known bivalves. These animals have well-developed muscles for opening and clos-ing the shells. These animals also have a muscular siphon for the intake and output of water. The nautilus squid, and octopus are members of the class cephalopoda. Cephalopod means head-foot, and in these molluscs the foot is fused with the head. In squids and octopuses, the foot is divided into ten or eight tentacles, and the shell is greatly reduced or absent. These molluscs have a well-developed nervous system with eyes similar to vertebrate eyes. The mantle is thick and muscular. Giant squids are the largest living invertebrates; they have attained lengths of 55 feet, and weights of two tons.

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Question:

Find the net dipole moment inDebyesfor each of the following situations: (a) One + 1 and one - 1 ion se-parated by 2 × 10^-8 cm, and (b) one +2 and one - 2 ion separated by 2 × 10^-8 cm. One charge = 4.80 × 10^-10esuand 1 Debye = 1 × 10^-18esu-cm.

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Solution:

A dipole consists of a positive and a negative charge separated by a distance. Quantitatively, a dipole is described by its dipole moment, which is equal to the charge times the distance between the positive and negative centers. For both situations, substitute the values into this quantitative expression: net dipole moment = charge × distance (a) Net dipole moment = (4.80 × 10^-10esu) (2 × 10-8cm) = 9.6 × 10^-18esu-cm. 1 Debye= 1 × 10^-18esu-cm. [(9.6 × 10^-18esu-cm) / (1 × 10^-18esu-cm / Debye)] = 9.6Debyes (b) Net dipole moment = (2) (4.80 × 10^-10) (2 × 10^-8) = 19.2 × 10^-18esu-cm Converting toDebyes [(19.2 × 10^-10esu-cm) / (1 × 10^-18esu-cm / Debye)] = 19.2 Debye.

Question:

Calculate ∆G\textdegree for the reaction N_2 (g) + 3H_2 (g) = 2NH_3 (g) at 400\textdegreeC = 673\textdegreeK using K_C . The value ofK_pfor this reaction is 1.64 × 10^-4. Interpret the different values of ∆G\textdegree calculated fromK_pand K_C.

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Solution:

Given the equilibrium reaction: (i)aA_(g) +bB_(g) + . . . =cC_(g) +dD_(g) + . . . there are two constants that measure the equilibrium: K_C relates the concentrations. (ii)K_C = {[C]^c [D]^d . . .}/{[A]^a [B]^b . . .} K_p relates the pressures (iii)K_p= [(Pc_C P^d _D . . .)/(Pa_AP^b_B. . .)] P^b _B To derive K_C fromK_psubstitute for each pressure P =cRT, where R is the gas constant; T = absolute temperature; and c is the number of moles of a gas per liter. For the reaction, N_2 (g) + 3H_2 (g) = 2NH_3 (g), K_p = [(P^2 _NH3)/(P_N2 P^3 _H2)] = [{C^2 _NH3 (RT)^2 }/{C_N2 C^3 _H2 (RT)^4} = [(K_C)/(RT)^2], whereK_C = [(C^2 _NH3)/(C_N2 C^3 _H2)] = (RT)^2K_p.Thus, K_C = [{0.0824 (liter-atm/\textdegreeK-mole)} (673\textdegreeK K_C = [{0.0824 (liter-atm/\textdegreeK-mole)} (673\textdegreeK )]^2 (1.64 × 10^-4 ) = 0.500 and∆G\textdegree_C = - RT InK_p = - (1.987 cal\textdegreeK^-1 mole^-1) (673\textdegreeK) (2.303) log (0.500) = 927 cal/mole. ∆G\textdegree_p= - RT InK_p = - (1.987 cal\textdegreeK^-1 mole^-1) (673\textdegreeK) (2.303) log (1.64 × 10^-4) = 11, 657 cal/mole. ∆G\textdegree_C is the change in Gibbs free energy when 1 mole of N_2 in the ideal gas state at the concentration of 1 mole per liter and 3 moles of H_2 in the ideal gas state at 1 mole per liter react to form 2 moles of NH_3 in the ideal gas state at 1 mole per liter. ∆G\textdegree_pis the change in Gibbs free energy when 1 mole of N_2 in the ideal gas state at 1atm, and 3 moles of H_2 in the ideal gas state at 1atm react to form 2 moles of NH_3 in the ideal gas state at 1 atm.

Question:

When lithium is bombarded by 10-MeV deuterons, neutrons are observed to emerge at right angles to the direction of the Incident beam. Calculate the energy of these neutrons and the energy and angle of recoil of the associated beryllium atom. Relevant masses in amu are: _0n^1 = 1.00893, _3Li^7 = 7.01784, _1H^2 = 2.01472, and _4Be^8 = 8.00776

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Solution:

Since the deuteron's kinetic energy is much smaller than its rest mass energy (\approx1876 MeV), the collision is non-relativistic. The equation governing the reaction is _3Li^7 + _1H^2 =_0n1+ _4Be^8 . The sum of the masses of the initial atoms is 9.03256 amu, and of the final atoms 9.01669 amu. The difference in mass, 0.01587 amu, appears, by Einstein's mass-energy relation, as E = (.01587 amu)c^2 amount of kinetic energy. Hence, E = (.01587 amu) (1.66 × 10^-27kg) (9 × 10^16 m^2/s^2) E = 14.77 MeV Since total energy (that is, rest mass energy plus kinetic energy) is conserved in the reaction, the final products have a combined kinetic energy of (10 + 14.77)MeV = 24.77 MeV . When we remember that the momentum of a particle p is related to its kinetic energy E by the relation p = \surd(2mE), then the principle of conservation of momentum, applied In the Initial direction and at right angles to it, gives the following equations, In self-explanatory notation: (see figure) \surd(2m_DE_D)_ = \surd(2m_BE_B) cos \texttheta(1) and\surd(2m_NE_N)_ = \surd(2m_BE_B) sin \texttheta(2) 2m_DE_D + 2m_NEN= 2m_BE_B(3) It was also shown above that E_N + E_B = 24.77 MeV(4) Solving (4) for EN EN= 24.77 MeV - EB(5) Using (5) in (3) 2m_DE_D + 2m_N( 24.77 MeV) -2m_N EB= 2m_BE_B Solving for E_B E_B (m_B + m_N) = m_DE_D+ m_N( 24.77 MeV) orEB= {m_DE_D+ m_N( 24.77 MeV)} / (m_B + m_N) whence E_B = {(2.01472 amu)(10MeV) + (1.00893amu) ( 24.77 MeV)} / (8.00776 amu + 1.00893 amu) E_B = 45.13840 amu .MeV / 9.01669 amu = 5.01 MeV Using this in (5) E_n = 24.77 MeV - 5.01 MeV = = 19.76 MeV Further, we obtain for the angle of recoil (using (1) and (4) tan \texttheta = \surd{(m_NE_N)_ / (m_DE_C)} = \surd[{(1.00893amu) (19.76 MeV)} / {(2.01472 amu) (10MeV)}] = \surd(.9895) = .9948 Hence ,\texttheta \approx 44050' .

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Question:

A copper coil has a resistance of 100 ohms at 20\textdegreeC. What is its resistance at 50\textdegreeC if the temperature coefficient of resistivity for copper is 0.0038 ohm per ohm per degree centigrade?

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Solution:

We are given the resistance of a wire at one temperature, and are asked to calculate its resistance at another temperature given the temperature coefficient of resistivity. The new resistance is given by the old re-sistance plus a factor proportional to the old resistance, the temperature change, and the constant coefficient for the substance. R2 0= 100 ohms,t = 50\textdegreeC,\alpha = 0.0038 ohm/ohm/\textdegreeC. UsingR_t= R2 0+ R2 0\alpha(t - 20\textdegreeC), we have R_t = 100 ohms + 100 ohms × 0.0038 ohm/ohm/\textdegreeC (50\textdegreeC - 20\textdegreeC) = 100 ohms + 11.4 ohms = 111 ohms, approximately.

Question:

Ethanol boils at 78.5\textdegreeC. If 10 g of sucrose (C_12H_22O_11) is dissolved in 150 g of ethanol, at what temperature will the solution boil? Assume k_b = (1.20\textdegreeC / M) for the alcohol.

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Solution:

When a non-volatile solute, such as sucrose, is dissolved in a solvent, such as ethanol, it will raise the boiling point of the solvent. The boiling point elevation can be found by using the equation ∆T_b =k_b\eta, where ∆T_b is the boiling point elevation, kbis the elevation constant, and \eta is themolalityof the solution. You want to determine ∆T_b and are given K_b. Thus, you need to determine themolalityof the solution.Molalityis defined as the number of moles of solute per 1 kg of solvent, i.e. (moles solute / 1 kg solvent). Solving formolarityof sucrose: (MW of C_12H_22O_11 = 342.3). Moles of solute= (grams solute) / (molecular wt. solute) = 10g / [(342.3 (g / mole)] = 2.921 × 10^-2 moles. You have 150 g of solvent, ethanol. Butmolalityis per 1 kg, so that you must multiply 150 g by (1 kg / 1000 g). Molality = (.02921mole) / [(150 / 1000 kg)] = .195 M Thus, the elevation of the boiling point is ∆T_b =K_b\eta= (1.20\textdegreeC / M)(.195 M) = .23\textdegreeC. Thus, the boiling point of the solution is 78.5\textdegreeC + .23\textdegreeC = 78.73\textdegreeC.

Question:

What wavelength of light is needed to excite an electron in a 0.2 nanometer (1 nm = 10^-9 m) box from the ground state to the second excited state? What wavelength of light is emitted when the same electron falls from the second excited state to the first excited state?

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Solution:

The solution to this problem requires the application of two equations, one to determine the energies of the various states, from which may be obtained the energy of a transition between states, and one to relate energy to wavelength. The first equation is the energy of a particle-in-a-box, E_n = [(n^2 h^2) / (8mL^2)] where n is the principal quantum number (n = 1, 2, 3, ...), h is Planck's constant (6.626 × 10^-27 erg-sec = 6.626× 10^-34 J-sec), m is the mass of the particle, and L is the length of the box. For an electron in a 0.2 nm = 0.2 × 10^-9 m = 2 x 10^-10 m box, E_n = (n^2 h^2) / (8mL^2)] = [{n^2 (6.626 × 10^-34 J-sec)^2} / {8 × 9.11 × 10^-31 kg × (2 × 10^-10 m)^2}] = 1.506 × 10^-18 j × n^2. The second equation relates the energy difference, ∆E, between two states and the wavelength \lambda: ∆E =hc/\lambda, or \lambda =hc/∆E, where c is the speed of light. In the following table, we calculate the energies corresponding to the ground state (n = 1), the first ex-cited state (n = 2) , and the second excited state (n = 3). n E_n = n^2 × 1.506 × 10^-18 J 1E_1 = (1)^2 × 1.506 × 10^-18 J = 1.506 × 10^-18 J 2E_2 = (2)^2 × 1.506 × 10^-18 J = 6.024 × 10^-18 J 3E_3 = (3)^2 × 1.506 × 10^-18 J = 1.355 × 10^-17 J In going from the ground state to the second excited state, the electron must absorb a quantum of light of energy ∆E = E_3 - E_1 = 1.355 × 10^-17 J - 1.506 × 10^-18 J = 1.204 × 10^-17 J. The corresponding wavelength is \lambda = [(hc) / (∆E)] = (6.626 × 10^-34 J × 3 × 10^8 m/sec) / (1.204 × 10^-17 J)] = 16.510 × 10^-9 m = 16.510 nm. In falling from the second excited state to the first excited state the electron will emit a quantum of light of energy ∆E = E_3 - E_2 = 1.355 × 10^-17 J - 6.024 × 10^-18 J = 7.526 × 10^-18 J. This corresponds to a wavelength of \lambda = [(hc) / (∆E)] = (6.626 × 10^-34 J × 3 × 10^8 m/sec) / (7.526 × 10^-18 J)] = 26.412 × 10^-9 m = 26.412 nm.

Question:

AssumingpD= - log [D_3O^+] in analogy to pH, what ispDof pure D_2O? K = 2 × 10^-15.

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Solution:

This question can be solved with the knowledge that D_2O undergoes dissociation in a similar manner to H_2O. The only difference is that D_2O is composed of deuterium instead of hydrogen. Therefore, one can write: D_2O + D_2O \rightleftarrows D_3O^+ + OD^- K = [D_3O^+] [OD^-] = 2 × 10^-15. [D_2O] is a constant and does not appear in the above equi-librium expression . From the reaction, it becomes apparent that the concentrations of D_3O^+ and OD^- are equal. There-fore, both concentrations can be represented by x, x \textbullet x = 2 × 10^-15. Solving for x, x = 4.47 × 10^-8. This, then, represents the concentration of D_3O^+ (and OD^-). pD = - log [D_3O^+] = - log 4.47 × 10^-8 = 7.35.

Question:

Differentiate between a placenta and an umbilical cord.

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Solution:

As the human embryo grows, the region on the ventral side from which the folds of the amnion, the yolk sac, and the allantois grow, becomes relatively smaller, and the edges of the amniotic folds come together to form a tube which encloses the other membranes. This tube is called the umbilical cord and contains, in addition to the yolk sac and allantois, the large umbilical blood vessels through which the fetus obtains nourishment from the wall of the uterus. The umbilical cord, about 1 cm in diameter and about 70 cm long at birth, is composed chiefly of a peculiar jelly- like material found nowhere else. The placenta is the area where a portion of the chorion and a portion of the uterine wall join. The placenta is thus a structure of double origin, partly embryonic (chorion) and partly maternal (uterus). It functions to exchange nutrients, wastes, and gases between the mother and the fetus. In mammals such as the human, the placenta is a disk about 7 inches in diameter and about 1 to 1(1/2) inches thick. Not all mammals have such disklike placentas. For example, cows have numerous isolated clusters of fetal-maternal associations, other animals have fingerlike projections of the chorion that provide a very diffuse type of association. In cats and dogs the placenta forms a ring around the fetus.

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Question:

You have 100-proof (50 percent alcohol by volume) bonded Scotch whisky. Calculate itsmolarity, mole fraction, and molality. If the temperature were to drop to - 10\textdegreeC, could you still drink the Scotch? Assume density = .79 g/ml for ethyl alcohol (C_2H_5OH) andK_f= 1.86\textdegreeC/m for water.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E08-0293.htm

Solution:

To findmolarity, mole fraction, andmolality, you need to know how many moles of alcohol are present in the whiskey. You can obtain this from a calculation using the density. To find whether you can still drink the whisky at - 10\textdegreeC, requires you to know its freezing point, since you cannot drink something if it is frozen. You can deter mine the freezing point of the Scotch by determining the freezing point depression of the water containing the alcohol. This is found from\DeltaT_f=K_fm, where\DeltaT_f= the depression,K_f= freezing point depression constant, and m =molality. You now proceed as follows: You are told that 100 proof means 50 percent alcohol by volume. This means that 1 liter of components possess 500 cm^3 of ethyl alcohol (a liter = 1000 cm^3). Density = mass/volume. You know the density and volume, so that mass of alcohol = density × volume = .79(500) = 395 g. Molecular wt. of the alcohol = 46.07 g. Thus, you have 395/46.07 = 8.57 moles of alcohol. Hence, molarity= [(moles of solute)/(liters of solution)] = 8.57 / 1 = 8.57 M. Molarity= [(moles of solute)/(kilograms of solvent)]. To find the number of grams of solvent, you need to use the density again. In 1 liter of Scotch Whisky, you have 500 cm^3 of water. Density of water = 1 g/ cm^3. Thus, mass of water = 500 cm^3 (1 g/ cm^3) = 500 grams. Thus, Molality= [(8.57) / {(500/1000)}] = 17.14 m You divided 500 by 1000 sincemolalityis moles per 1000 g of solvent. Mole fraction: (MW of H_2O =18.) moles of H_2O = 500 g/18.0 g/mole. Mole fraction of ethyl alcohol = [(moles ethyl alcohol)/(moles ethyl alcohol + moles H_2O)] = [(8.57) / {8.57 + (500/18)}] = .2357. Now, you answer whether the scotch can be consumed at - 10\textdegreeC. You already calculated themolality. Thus,\DeltaT_f=k_fm = (1.86)(17.14) = 31.88\textdegreeC. Thus, the freezing point of whisky is 0\textdegreeC - 31.88\textdegreeC = - 31.88\textdegreeC. The temperature is only - 10\textdegreeC, however. This means whisky is still a liquid and can be consumed.

Question:

A positively charged ring of radius R lies in the Y-Z plane. Compute the electric intensity at point P on the axis of the ring.

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/Users/wenhuchen/Documents/Crawler/Physics/D17-0569.htm

Solution:

To calculate the electric field intensity at a point P from a continuous distribution, we calculate the electric field intensity due to one arbitrary element of charge, dq, at the point P, and then integrate this over the entire region under consideration. The electric field intensity, due to a differential charge element dq is given by dE^\ding{217} = (dq r^\ding{217})/(4\pi\epsilon_0 r^3)(1) where dE^\ding{217} is the field intensity due to dq, dq is an element of charge, r^\ding{217} is a vector from dq to P, and r is the magnitude of this vector. In this problem, however, we can simplify the integral in (1). Note from the figure that the Z com-ponents of the field intensities due to charge elements dq on opposite sides of the ring will cancel because they are of equal magnitude, but opposite in direction. Hence, the net electric field intensity will point along the x-axis and will be due to the x component of dE^\ding{217}, which is of magni-tude (dE) cos \alpha. Now we may neglect the vector nature of the integrand in (1), because the problem has been reduced to a calculation in one dimension. Therefore, E_net = \intdE cos \alpha. From (1), we obtain E_net = \int(dE cos \alpha)/(4\pi\epsilon_0r^2). Because cos \alpha, 4\pi\epsilon_0, and r are constants, we may write E_net = (1/4\pi\epsilon_0) [(cos \alpha)/r^2 ]\int dq(2) Integrating (2), we find E_net = (q cos \alpha)/(4\pi\epsilon_0 r^2)(3) where q is the total charge on the ring. Note that this field intensity is directed along the x-axis, and varies inversely with r^2. At the center of the ring, \alpha = 90\textdegree, cos \alpha = 0, and E = 0, as is evident by symmetry. At distances large compared to R, the angle \alpha is small, cos \alpha is approximately 1, and s^2(see figure) is nearly equal to r^2. Hence at large distances E = [1/(4\pi\epsilon0)] (\alpha/s^2) In other words, at large distances the ring can be con-sidered a point charge.

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Question:

Devise a BASIC program to calculate the first N numbers of the FIBONACCI series.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G10-0240.htm

Solution:

The FIBONACCI series is defined recursively by F(1) = 1, F(2) = 1, and for I greater or equal to 3, F(I) = F(I - 1) + F(I - 2). It is convenient to tabulate the functional values in an array dimensioned to be, say, 50. DIM (dimension) statement is used for this purpose. BASIC automatically provides for 10 places in its memory cells for any one dimensional array (11 places in those systems, which accept 0 subscript). If one desires N \not = 10 places to be reserved, he can use the statement of the form: DIM V(N) where V can be any valid variable. Thus the program will handle N up to 50. 1\O DIM F(5\O) 2\O INPUT N 3\O LET F(1) = 1 4\O LET F(2) = 1 45 PRINT F(1), F(2), 5\O FOR 1=3 TO N 6\O LET F(I) - F(I - 1) + F(I - 2) 7\O PRINT F(I), 8\O NEXT I 1\O\O END

Question:

You have been given an unordered table to search for a par-ticular entry. When this entry is found you are to replace it with an updated piece of data.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G09-0205.htm

Solution:

Since the table is not ordered it would be best to use a linear search in which all the elements of the table are inspected. To do this problem we will assume that the table has 100 entries and each entry is 14 bytes long. Program: RDCDREADCARDDATA, CRDEOF LA4, SYMTBLSTORE ADDRESS OF TABLE SEARCHCLC0(14, 4), SYMBOLCOMPARE ENTRY WITH ITEM TO BE SEARCHED FOR (SYMBOL) BESYMFOUNDIF EQUAL WE'VE FOUND SYMBOL C4, LAST'LAST' HOLDS THE ADDRESS OF THE LAST ENTRY IN THE TABLE BENOTFOUNDHERE, IF EQUAL AND THE SYM-BOL IS NOT FOUND, THEN IT IS NOT IN THE TABLE A4, = F '14'GO TO NEXT ENTRY IF SYMBOL IS NOT FOUND BSEARCHCONTINUE TO SEARCH NOTFOUND \textbullet \textbullet \textbullet \textbullet \bullet \bullet SYMFOUND MVC0 (14 , 4),REPLACE ENTRY WITH \textbulletNUDATAUPDATED DATA \textbullet CRDE OF- - - - - \textasteriskcenteredDATA DEFINITION FOR INPUT AREAS SYMTBLDS100CL14ALLOCATE SPACE FOR A HUN-DRED 14-BYTE ENTRIES. SYMBOLDSCL14HOLDS SYMBOL BEING CHECKED. LASTDCA (FINDAT)STORES ADDRESS OF FINAL DATA IN TABLE. \textasteriskcenteredDATA DEFINITION FOR INPUT AREA DATADC\varphiCL80 \varphi \varphi NUDATADSCL14 DSCL66 Comments: The 'NOTFOUND' segment of this program would usually involve printing a message which indicates the data was not in the table. This is always a good procedure, since when we do any table searching there is the possiblility that the given data won't be found. For the 'LAST DC A (FINDAT)' instruction, the 'A' indicates that the address of the symbol in the parentheses should be stored in the location represented by the symbol in the label field. In other words, this instruction tells the computer to store the address of 'FINDAT' in the location denoted by LAST. In coding the table, in our program, we would attach the label 'LAST' to the final piece of data in the table. On reaching the label 'CRDEOF', program execution would have generally been completed, and as such, we would restore the contents of the registers stored at the begin-ning of our program.

Question:

Small amounts of cyanide can kill a person. Why is cyanide such a deadly poison?

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/Users/wenhuchen/Documents/Crawler/Biology/F03-0085.htm

Solution:

Cyanide acts by inhibiting the final reaction of the respiratory chain, shutting it off and preventing the energy extraction process from taking place. Normally, reduced cytochrome a, the final element in the respiratory chain, would be reoxidized by oxygen, but since cyanide inhibits this from occuring, the whole chain is shut off as the intermediate acceptors are all converted to the reduced condition. Consequently, there is a severe decrease in the amount of ATP that can be produced fox life processes, leading to the death of the victim. Hydrogen sulfide and carbon monoxide have similar effects at this site. Other poisons block either of the first two sites. Looking at the electron transport chain it may not be obvious how the inhibition of the reoxidation of reduced cytochrome a can shut off the entire respiratory chain. Cyanide blocks ATP formation of the third site of ATP synthesis in the chain. However, this does not mean that ATP production still goes on as usual at the first and second site. Some ATP may be produced immedi-ately after cyanide enters the electron transport chain, but eventually production stops. ATP production ceases because of the inability of the intermediate acceptors to oxidize NADH. Inhibition by cyanide generates an accumulation of reduced cytochrome a by preventing its oxidation. Since cytochrome a is in a reduced state, it cannot oxidize its neighbor, reduced cytochrome c. This leads to an accumulation of reduced cytochrome c and then c_1, which, being unable to oxidize cytochrome b, prevents the formation of ATP at the second site. Of more importance is the "backing up" of reduced states towards the beginning of the chain leading even-tually to the reduction of all intermediate acceptors. These reduced intermediates cannot accept electrons; ATP formation in the respiratory chain requires inter-mediates to receive electrons which are liberated going from a reduced state to an oxidized state. Therefore, ATP formation cannot proceed normally and, in fact, finally stops if any step in the electron transport chain is blocked.

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Question:

John is injured and badly needs a blood transfusion. He is type B. His girlfriend Jane volunteers, but she has type O blood. The physician, in view of the urgency of the situation, uses Jane's blood anyway. A few months later, Jane needs a transfusion. John volunteers but the physician turns him down. Explain.

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/Users/wenhuchen/Documents/Crawler/Biology/F14-0370.htm

Solution:

Before we explain this particular problem, it would be worthwhile to discuss the A\rule{1em}{1pt}B\rule{1em}{1pt}O blood type system. There are four types of blood determined geneti-cally: A, B, AB and O. The basis of these blood types is the presence or absence of certain antigens, calledagglutinogens, on the surface of the erythrocytes. The erythrocytes of type A blood carry agglutinogen A; those of type B blood carryagglutinogenB. Type O blood has neither of theseagglutinogens, while type AB has both A and B agglutinogens . Theseagglutinogensreact with certain antibodies, called agglutinins, that may be present in the plasma. An agglutinin-agglutinogen reaction causes the cells to adhere to each other. This clumping, or agglutination, would then block the small blood vessels in the body, causing death. Agglutination is simply the clumping of erythrocytes by agglutinins. The blood clotting mechanism, or co-agulation is not involved. A Type A person has the A antigen (agglutinogen) and the anti-B antibody (agglutinin) . A Type B person has the B antigen and anti-A antibody. A Type O person has no antigens but both anti-bodies. Conversely, a Type AB person has bothanitgensbut no antibodies. Basically, a person does not have the agglutinin in his plasma which would clump his own red blood cells. A summary of the antibodies and anti-gens that each blood type contains is outlined in the following table: Blood Type Agglutinogens (antigens) Agglutinins (antibodies) A A anti-B B B anti-A AB A and B none O none anti-A and anti-B Blood typing is a critical factor in blood trans-fusions. Since the agglutinins in the plasma of one type will react with theagglutinogenson the erythrocytes of the other types, transfusions are usually between people with the same blood type. However, one may use another blood type provided that the transfusion is small and the plasma of the recipient and the erythrocytes of the donor are compatible (does not cause a reaction) . This type of transfusion was performed in the given case. Type O blood from Jane was given to John, a blood-type B patient. The anti-A agglutinin in John's plasma finds noagglutinogensto attack on Jane's erythrocytes. We can usually ignore theagglunitins, specifically anti-B, in Jane's plasma, which would normally react with theagglutinogenB on John's erythrocytes. This is because Jane's plasma is sufficiently diluted during transfusion so that the con-centration of agglutinins is too low to cause appreciable agglutination. Jane's blood, type O, can be given to any-one since her erythrocytes contain noagglutinogens. Type O blood is therefore called the universal donor. Although type O blood can be donated to any type, it can only accept type O since it has the agglutinins, anti-A and anti-B, which react with theagglutinogensof every other blood type. In contrast, type AB blood can receive from any type since it has no agglutinins in its plasma to react with the donor's erythrocytes. Type AB blood is therefore called the universal recipient. But it can only donate to other AB patients since itsagglutinogens, A and B, react with the plasma of every other type. The reason John's offer was turned down is that his blood, type B, would agglutinate with Jane's blood, type O. The agglutinins in Jape's plasma, specifically anti-B, would react with the Bagglutinogenon John's erythrocytes and cause agglutination.

Question:

Three coordinate systems S, S', and S" have a common x-axis. With respect to S, S' moves in the direction of the x-axis with a speed v, and S" accelerates along the x-axis with acceleration a. At time t = 0, the origins of all three coordinate systems coincide and S" has zero velocity with respect to S. At that instant/ a man starts out running from the origin along the x-axis and an observer in S measures his speed as constant and of magnitude u(u > v). How do observers in S' and S" describe his motion, using Galilean relativity?

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/Users/wenhuchen/Documents/Crawler/Physics/D04-0217.htm

Solution:

The observer in S sees the man running along the x-axis with constant speed u. The motion is thus observed as one in which the x displacement increases linearly with time. On a displacement-time diagram, the motion appears as the dot-dashed straight line at an angle \texttheta to the t-axis, where tan \texttheta = x/t = u; and on a velocity -time diagram it appears as the dot-dashed straight line parallel to the t-axis. (see figs. (a) and (b)) . To an observer in s', the man has only the speed (u - v), the relative speed between the two. But the man is still seen as moving with constant speed and his motion is shown in the diagrams by the dashed straight lines. The angle \o is such that tan \o= u - v. An observer in S\textquotedblright, who is accelerating along the x-axis with acceleration a relative to S and therefore to the running man, considers himself at rest and therefore attributes to the runner an acceleration of -a. The runner appears to start off with velocity u but to decelerate gradually to rest and then go backward. His vel-ocity at any time is seen to be, using the kinematics equations for constant acceleration, V = u - at, and on the velocity-time diagram this is represented by the full line at an anglevto the t-axis, where tanv= -a. Also his x-displacement relative to the origin of s" is given by the constant acceleration kinematics equation x = ut - (1/2)at^2, which is represented by a parabola in the displacement- time diagram, tangent to the dot-dashed line at the origin and having its highest point at the time t where the corresponding velocity--time graph cuts the t-axis.

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Question:

Find the ionic radius of I-, given that the unit-cell edge length of Lil is measured to be .6240 nm. Assume that the large negative ions are in actual contact with the diagonal.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E06-0227.htm

Solution:

To solve this problem, determine the length of the face diagonal in Lil space lattice, which is the pattern of the arrangement of atoms or molecules in a crystal. Because Lil has the same form of the cubic unit cell as NaCl, its space lattice will possess 3 I- ions on the face diagonal. Thus .if one knows the length of the face diagonal, one can divide by 4 to deduce the radius of I-. The face diagonal can be found (from the Pythagorean Theorem) by multiplying (\surd2) times (cubic- edge length) . Thus, the face diagonal = (\surd2) (.6240) = .8823 nm. However, face diagonal = 4I- radii. Thus, I- radius = .8823 / 4 = .2206 nm.

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Question:

Using the pedigree below, fill in the genotype of each individual. Assume the trait is recessive and an individual who marries into the family and does not exhibit a trait does not carry the recessive gene for it,

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/Users/wenhuchen/Documents/Crawler/Biology/F26-0686.htm

Solution:

For simplicity and conciseness, we shall use the following system to refer to any one of the members in the pedigree: Let the allele for left-handedness be a and the allele for right-handedness be A. Consider P_1 (that is, the first member of the Parental generation). Since she expresses the trait, her genotype is aa. To determine the genotype of P_2 , we look at the progeny, F_1 (first generation). From the diagram, we see that F_1,2 is left- handed; hence his genetic make-up is aa. This means that he must have received a copy of a from each of his parents. Therefore we know that P_2 carries a copy of the recessive gene. But since P_2 is phenotypically normal, his genotype must be Aa. F_1,1 and F_1,3 are right-handed. Yet they must have received an a allele from P, since she has only a alleles to transmit. Therefore their genotype of F_1,1 and F_1,3 is Aa. Consider the cross between P_3 and P_4. We know P_3 is aa (because he is left-handed) and P_4, can be AA or Aa (since she is left handed). Since all the offspring are right-handed, it seems probable that P_4 is AA; however, we cannot be entirely certain of this, since both genotypes AA and A_2 are compatible with the phenotypes of the offspring. We know, however, that F_1,4, F_1,5, and F_1,6 must all have the genotype Aa since they carry an a allele donated by P_3. In the F_2 generation, we see that F_2,2 is left- handed, and therefore her genotype is aa. This is compatible with what we have determined to be the genotypes of the parents (F_1,3 and F_1,4) namely Aa. F_2,1 and F_2,3 are both right-handed, and are either AA or Aa. Since both parents are Aa, either genotype is possible, though Aa is more probable.

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Question:

500 g of alcohol at 75\textdegree C are poured into 500 g of water at 30\textdegree C in a 300 g glass container (c_glass = .14). The mixture displays a temperature of 46\textdegree C. What is the specific heat of alcohol?

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/Users/wenhuchen/Documents/Crawler/Physics/D13-0479.htm

Solution:

When the alcohol is poured into the glass- water system, the former loses heat energy, and its temperature drops. The latter gains heat energy and its temperature rises. Hence, by the principle of conservation of energy heat loss by alcohol = heat gained by H_20 + heat gained by glass(1) In general, if a sample of mass m, composed of a substance of specific heat c, is exposed to a temperature change ∆T, it will lose or gain heat energy Q = m c ∆T. Using this fact, we may write heat gained by H_20 = m(H)2 0c(H)2 0∆T(H)2 0 = (500 g)(1 cal/g\textdegreeC) (46\textdegreeC - 30\textdegreeC) = 8000 cal heat gained by glass = m_g c_g ∆T_g = (300 g) (.14 cal/g\textdegreeC) (46\textdegreeC - 30\textdegreeC) = 672 cal heat lost by alcohol = m_a c_a ∆T_a = (500 g)(c_a) (75\textdegreeC - 46\textdegreeC) = (14500 cal) c_a (Note that the heat gained by the alcohol is negative. Hence, the heat lost by the alcohol is positive. This is the reason why ∆T_a > 0). Using these facts in (1) (14500 g \textdegreeC)c_a = 8000 cal + 672 cal = 8672 cal orc_a = (8672 / 14500) (cal / g0C) = .598 (cal / g0C)

Question:

When ultraviolet light of frequency 1.3 × 10^15 sec^-1 is shined on a metal, photoelectrons are ejected with a maximum energy of 1.8 electron volts. Calculate the work function of the metal in ergs and electron volts. What is the threshold frequency of this metal?

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/Users/wenhuchen/Documents/Crawler/Physics/D33-0963.htm

Solution:

This problem is an example of the photo- electric effect. We want to relate the work function of a metal to the energy of the incident radiation. (The work function is the amount of energy needed to release an electron from the attraction of the rest of the metal, and thereby pull it out of the metal.) Let us examine the reaction occurring. A photon of energyhu(the ultraviolet radiation) collides with an electron of the metal, and, in the process, gives up all of its energy. Ifhuis large enough, the electron will travel through the metal and be released with kinetic energy (1/2) mv^2, where m and v are the electron mass and velocity, respectively. In leaving the metal, the electron loses energy due to collisions with atoms of the metal. The electron also loses energy, equal in amount to the work function, in order to overcome the attraction of the metal and escape. If we examine only those electrons which lie close to the surface of the metal, we observe that they do not go through any collisions and hence can-not lose energy by this means. These surface electrons lose less energy than the interior electrons and there-fore will have a maximum amount of kinetic energy when they leave the metal. Using the principle of conservation of energy, we obtain hu= (1/2) m v^2_max + \textphi(1) where \textphi is the work function of the metal. (Note that this relation is true only for surface electrons.) Solving (1) for \textphi, we obtain \textphi =hu- (1/2) mv^2_max(2) The energy of a photon of the ultraviolet light is hu= 6.625 × 10^-27 × 1.3 × 10^15 erg = 8.61 × 10^-12 erg Since 1 electron volt = 1.60 × 10^-12 erg, the maximum kinetic energy of the photoelectrons is (1/2) mv^2_max = 1.8 X 1.6 X 10^-12 erg = 2.88 X 10^-12 erg From equation 2, the work function is \textphi =hu- (1/2) mv^2_max = 8.61 X 10^-12 - 2.88 X 10^-12 erg = 5.73 X 10^-12 erg Since 1 erg = .625 × 10^12eV \textphi = (5.73 × 10^-12 erg) [.625 × 10^12 (eV/erg)] \textphi = 3.58eV The threshold frequency is the smallest frequency of indecent radiation for which the electrons will be released from the metal with no kinetic energy. Setting (1/2) mv^2_max = 0 in equation (1) we find that the threshold frequency, u_0, is u_0 = (\textphi/h)(3) Using the value of \textphi calculated in the first part of the problem, u_0 = [(5.73 X 10^-12 erg)/(6.625 X 10^-27erg\textbulletsec)] oru_0 = 8.65 X 10^14 sec^-1 This frequency is in the ultraviolet, and just beyond the blue end of the visible spectrum.

Question:

Suppose there are three typists in a typing pool. Each typist can type an average of 6 letters/hr. If letters arrive to be typed at the rate of 15 letters/hr., a) What fraction of the time are all three typists busy? b) What is the average number of letters waiting to be typed? c) What is the average time a letter spends in the system (waiting and being typed)? d) What is the probability of a letter taking longer than 20 min. waiting to be typed and being typed? e) Suppose each individual typist receives letters at the average rate of 5/hr. Assume each typist can type at the average rate of 6 letters/hr., what is the average time a letter would spend in the system (waiting and being typed)?

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G18-0456.htm

Solution:

a) The following information is given \lambda = 15 letters/hr. \mu = 6 letters/hr. S = 3 Thus, we want P(n \geq 3). To obtain this, we need Po, the probability of no customers in the system given by the equation: Substituting the given values into the above equation we have the following: Therefore, P(n \geq S) = probability of the arrived letter wait-ing for service = probability of at least S customers in the system P(n \geq 3)= {(15/3)3(0.044944)} / [6{1 - (15/18)}] = 0.70225 b) Lq=average # of customers in queue = {(\lambda/\mu)S+1Po} / [s. s! {1 - (\lambda / \mus)2}] = {(15/6)4(0.044944)} / [(3) (6) {1-(15/18)^2}] = 3.51124 c) We first need the average number of letters in the system. L = L_q + (\lambda/\mu) = 3.51124 + (15/6) = 6.01124 W = average time a customer spends in a system = (L/\lambda) = (6.01124) / 5 = 0.40075 hr. \approx 24 min. d) The probability of a letter taking longer than 20 min. waiting to be typed and being typed e) Each queue and server can be treated as a separate single queue, single-server system so that W = 1 / (\mu - \lambda) = 1 / (6 - 5) = 1 hr. \textasteriskcentered\textasteriskcentered QUEUING PROGRAM \textasteriskcentered\textasteriskcentered \textasteriskcentered\textasteriskcentered INFINITE SOURCE, INFINITE QUEUE, MULTIPLE SERVERS \textasteriskcentered\textasteriskcentered WE ASSUME ARRIVALS FORM A SINGLE QUEUE INFINITE SOURCE AND INFINITE QUEUE FIFO QUEUE DISCIPLINE THERE ARE MULTIPLE SERVERS WITH EXPONENTIAL SERVICE TIME DEPARTURES FROM SYSTEM OCCUR COMPLETELY AT RANDOM REAL\textasteriskcentered4 TITLE (20) , LAMBDA, MU, N, L, LQ, NFACT 5READ(5, 10, END = 2000) TITLE 10FORMAT (20A4) WRITE (6, 11) TITLE 11FORMAT ('1', 20A4, //) READ (5, 20) LAMBDA MU 20FORMAT (3F10.0) READ (5, 20) N, S, T IF (LAMBDA. GE. S\textasteriskcenteredMU) GO TO 700 C\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered C\textasteriskcenteredCALCULATE PZERO\textasteriskcentered C\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered NS=S SUM = 0.0 DO 205 NK = 1, NS SK = NK-1 CALL FACT (SK, NFACT) SUM = SUM+ (1/NFACT) \textasteriskcentered (LAMBDA/MU) \textasteriskcentered\textasteriskcenteredSK 205CONTINUE CALL FACT(S, SFACT) PZERO = 1/ (SUM+ (1/ (SFACT\textasteriskcentered (1-LAMBDA/ (MU\textasteriskcenteredS) ) )) \textasteriskcentered (LAMBDA/MU) \textasteriskcentered\textasteriskcenteredS) C\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered C\textasteriskcenteredCALCULATE PN\textasteriskcentered C\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered IF(N.GE.S) GO TO 210 CALL FACT (N, NFACT) PN = (1/NFACT)\textasteriskcentered((LAMBDA/MU)\textasteriskcentered\textasteriskcenteredN) \textasteriskcenteredPZERO GO TO 215 210CALL FACT(S, SFACT) PN = (1/ (SFACT\textasteriskcenteredS\textasteriskcentered\textasteriskcentered (N-S) ) ) \textasteriskcentered ((LAMBDA/MU) \textasteriskcentered\textasteriskcenteredN) \textasteriskcenteredPZERO C\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered C\textasteriskcenteredCALCULATE PS, PT, L, LQ, W, WQ\textasteriskcentered C\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered 215CALL FACT(S, SFACT) PS = ( ( (LAMBDA/MU) \textasteriskcentered\textasteriskcenteredS) \textasteriskcenteredPZERO) / (SFACT\textasteriskcentered(1-LAMBDA/ (MU\textasteriskcenteredS)) ) PTN = ((LAMBDA/MU) \textasteriskcentered\textasteriskcenteredS) \textasteriskcenteredPZERO\textasteriskcentered (1-EXP (-MU\textasteriskcenteredT\textasteriskcentered (S- 1-LAMBDA/ MU))) PTD = SFACT\textasteriskcentered (1-LAMBDA/MU\textasteriskcenteredS)) \textasteriskcentered (S- 1-LAMBDA/MU) PT = EXP(-MU\textasteriskcenteredT)\textasteriskcentered(1+PTN/PTD) LQ = (((LAMBDA/MU) \textasteriskcentered\textasteriskcentered (S+1)) \textasteriskcenteredPZERO) /(S\textasteriskcenteredSFACT\textasteriskcentered (1-LAMBDA / (MU\textasteriskcenteredS))\textasteriskcentered\textasteriskcentered2) L = LQ+LAMBDA/MU W = L/LAMBDA WQ = LQ/LAMBDA C\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered C\textasteriskcenteredPRINT RESULTS\textasteriskcentered C\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered WRITE (6,100) LAMBDA, MU, N, S, T, L, LQ, W, WQ, PT, PN, PZERO, PS 100FORMAT (' LAMBDA = ', F10.5/' MU = ', F10.5/' N = ', F10.5/ ' S \textasteriskcentered = ', F10.5/'T = ' , F10.5/' L = ', F10.5/1 LQ = ', F10.5 \textasteriskcentered/' W = ' , F10.5/' WQ = ', F10.5/' PT = ', F10.5/' PN = '\textasteriskcentered, F10.5/'PZERO = ', F10.5/'PS= ', F10.5) GO TO 5 700WRITE(6, 705) 705FORMAT('QUEUING SYSTEM NOT VALID BECAUSE LAMBDA \leq S\textasteriskcenteredMU') GO TO 5 2000STOP END SUBROUTINE FACT (P, PROD) NUM = P PROD = 1. IF (NUM.EQ.0) GO TO 20 DO 10 K = 1, NUM 10PROD = PROD\textasteriskcenteredK 20RETURN END /DATA SINGLE QUEUE - MULTIPLE SERVER MODEL 15.6. 1.3..3333 SINGLE QUEUE - MULTIPLE SERVER MODEL LAMBDA= 15.00000 MU= 6.00000 N= 1.00000 S= 3.00000 T= 0.33330 L= 6.01124 LQ= 3.51124 W= 0.40075 WQ= 0.23408 PT= 0.46198 PN= 0.11236 PZERO= 0.04494 PS= 0.70225

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Question:

Convert the following logical expressions to reverse Polish string notation: (a) (Z \downarrow ((((X) \odot Y) / (X \odot (Z))) > ((Y) \equiv X))) (b) (X / Y) / ((X / X) / (Y / Y))

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G07-0135.htm

Solution:

Polish notation is a way of representing logical and algebraic expressions without using parentheses, unless parentheses represent some variables or operators in the particular expression. If you think about it, the purpose of parentheses is to override the natural hierarchy of the operators. Polish notation does away with that notion. All operators have the same precedence: their position in the string determines their order of evaluation. With reverse Polish notation, also called Polish post-fix, the operator is written after its two operands. For example, the usual logical equation A \oplus B would be written AB \oplus in postfix notation. It is possible to use trees to evaluate Polish strings, but we will not do so here. a) As we can see in this example, the abundance of parentheses makes this expression look formidable. We will proceed step-by-step, enclosing in brackets each portion of the expression that has been converted. Step 1. First, we will take care of the negations. We use the symbol \rceil for this operator. (Z \downarrow ((([X\rceil] \odot Y) / (X \odot [Z\rceil])) > ([Y\rceil] \equiv X))) (Z \downarrow ((([X\rceil] \odot Y) / (X \odot [Z\rceil])) > ([Y\rceil] \equiv X))) Step 2. We consider next the innermost partial expressions and convert them. (Z \downarrow (([X\rceilY \odot] / (X Z\rceil \odot]) > ([Y\rceilX\equiv])) Step 3. Now we include the NAND operator / to complete the innermost partial expression. (Z \downarrow ([X\rceilY \odot XZ\rceil \odot / ] > [Y\rceilX\equiv])) Step 4. We evaluate the operator >, inserting it at the end of the string. (Z \downarrow [X\rceilY \odot XZ\rceil \odot / Y\rceilX \equiv >]) (Z \downarrow [X\rceilY \odot XZ\rceil \odot / Y\rceilX \equiv >]) Step 5. Finally, we add the NOR operator \downarrow to the end of the string. The final reverse Polish string is ZX\rceilY \odot XZ\rceil \odot / Y\rceilX\equiv> ZX\rceilY \odot XZ\rceil \odot / Y\rceilX\equiv> \downarrow b) This problem uses solely the NAND operator/and several parentheses. We will proceed as before, evaluating the innermost parentheses first. Step 1. The two inner expressions on the right side get translated and bracketed first. (X / Y) / ([XX / ] / [YY / ]) (X / Y) / ([XX / ] / [YY / ]) Step 2. Next we merge the two bracketed expressions together over the NAND operator. (X / Y) / [XX / YY / / ] (X / Y) / [XX / YY / / ] Step 3. Now we convert the right side. [XY / ] / [XX / YY / / ] [XY / ] / [XX / YY / / ] Step 4. Finally, the entire expression, translated into a reverse Polish string looks like this. XY / XX / YY / / /

Question:

A copper bar laid perpendicularly across a pair of parallel metal tracks running north-south, one meter apart, is slid in a northerly direction. At one end, the tracks are connected through a galvanom-eter. At the given location the vertical component of the earth's magnetic field is .55 oersted. Is there any reason to suppose that the galvanometer will read a deflection as the bar is moved at the rate of 150 cm/sec? If so, what is it, which way does the current point, how much electromotive force is developed in the bar between the rails, and which end of the bar is at the higher potential?

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/Users/wenhuchen/Documents/Crawler/Physics/D22-0751.htm

Solution:

Note that as the bar moves along the track, as indicated in the diagram, it cuts the vertical component of the earth's field, which points downward, or into the paper. Hence, an electromotive force must be generated between the points A and B, and a current must be established in the' circuit, to be detected by the galvano-meter. This results from Faraday's Law because the magnetic flux is changing. The direction of this current follows from Lenz' Law, which states that the induced current will flow in a direction such as to set up an induced flux opposite in direction to the change in flux. The current points from B to A in the bar. This means that the point A is at a higher potential than point B. To determine the magnitude of the electromotive force Induced, use Faraday's law E = - (d\varphi/dt) where the negative sign indicates that the EMF is Induced in such a direction as to oppose the change in flux that created it. The flux \varphi through the loop formed by the bar and the galvanometer is \varphi = \int B^\ding{217} \bullet dA^\ding{217} = BLx for B is constant in the loop and it is parallel to the area vector (i.e. , \texttheta = 0 o ) . Then E = - BL (dx/dt) Sincev = (dx/dt) E = - BLv where B here is 5.5 × 10^-5 weber/meter^2 , L = 1.0 m, and v = 1.5 (m/sec) E = - {5.5 × 10^-5 (w/m^2)} (1.0 m) {1.5 (m/sec)} = -8.25 × 10^-5 volt .

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Question:

A water pipe in which water pressure is 80 lb/in.^2 springs a leak, the size of which is .01 sq in. in area. How much force is required of a patch to "stop" the leak?

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/Users/wenhuchen/Documents/Crawler/Physics/D10-0395.htm

Solution:

This problem is quickly recognized as an application of the definition of fluid pressure. Commencing with the definition of pressure as the ratio of the force f on an area a, and the area a, we find p = f/a It follows that:f = pa Substituting values: f = 80 lb/in^2 × .01 in^2 = .8 lb.

Question:

(ClO^-_3) + Fe^2+ \rightarrow (Cl^-) + Fe^3+

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/Users/wenhuchen/Documents/Crawler/Chemistry/E16-0567.htm

Solution:

Reactions in which electrons are transferred from one atom to another are known as oxidation-reduction re-actions or as redox reactions. To balance this type of re-action , you want to conserve charge and matter, i.e., one side of the equation must not have an excess of charge or matter. To perform this balancing, you need to (1)Balance charge by adding H^+ (in acid) or OH^- (in base) . (2) Balance oxygen by adding water. (3) Balance atoms (of hydrogen) by adding hydrogen to appropriate side. These three rules will balance the redox equation. These rules apply to balancing only the half-reactions. The overall reaction; the sum of these, will be balanced by their addition. Proceed as follows: Fe^2+ goes to Fe^3+. It lost an electron thus, it's the oxidation half reaction. To balance charge, add H^+. Thus, Fe^2+ + H^+ \rightarrow Fe^3+. To conserve mass, add an H atom to right. You obtain Fe^2+ + H^+ \rightarrow Fe^3+ + (1 / 2)H_2. The reduction must be ClO^-_3 \rightarrow Cl^-. The charges are already balanced. To balance the 3 oxygen atoms on left side, add 3 water molecules on right side. You obtain ClO^-_3 \rightarrow Cl-+ 3H_2O. To balance hydrogens, add 6 hydrogens (or 3H_2molecules) to left side. 3H_2 + ClO^-_3 \rightarrow Cl^- + 3H_2O. In summary, you have oxid:Fe^2+ + H^+ \rightarrow Fe^3+ + (1 / 2)H_2 red:3H_2 + ClO^-_3 \rightarrow Cl^- + 3H_2O. So that no free H's appear in the overall equation, multiply the oxidation reaction by six. You obtain 6Fe^2+ + 6H^+ \rightarrow 6Fe^3+ + 3H_2 Thus,oxid:6Fe^2+ + 6H^+ \rightarrow 6Fe^3+ + 3H_2 red:3H_2 + ClO^-_3 \rightarrow Cl^- + 3H_2O overall:6Fe^2+ + ClO^-_3 + 6H^+ \rightarrow 6Fe^3+ + Cl^- + 3H_2O Notice: The H atoms dropped out. This is the balanced equation for charge and mass are equal on both sides of the equation. H^+ comes from the acid; recall, it's in an acidic solution.

Question:

A chemist found the value of the specific rate constant for the decomposition of nitrous oxide, 2NO + N_2 + O_2, at two separate temperatures: k, = .14 liter/mole-sec at 970\textdegreeK, k_2 =3.7 liter/mole-sec at 1085\textdegreeK. (a) Calculate the activation energy, Ea, for the reaction, (b) Calculate the A factor in the Arrhenius equation, (c) Calculate the specific rate constant at 800\textdegreeK.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E13-0472.htm

Solution:

This problem deals with the effect of tem-perature on the kinetics of a reaction. (a) E_a is the activation energy, the difference between the heat content of (a) E_a is the activation energy, the difference between the heat content of the active molecules and that of the inert molecules. To find the value of E_a for the decomposition of nitrous oxide, use the Arrhenius equation: Ea= [(-2.303) / {(1/T_2) (1/T_1)}] log (k_2 / k_1) where T = temperature in degrees Kelvin (Celsius plus 273\textdegree) , k = specificrate constant and R = universal gas constant. Substitute known values to obtain (T_1 = 970\textdegree, T 2 = 1085\textdegree) E_a = [{(-2.303)(1.987)(1/1000)} / {(1/1085) (1/970)}] log (3.7 / .14) = 59.2k cal/mole [Note: 1/1000 is a conversion factor of 1 k cal/1000 cal.] (b) The Arrhenius equation can also be written in the from l_nk=l_nA- (E_A / RT) , where A is called the frequency factor. To find A, first solve the equation in term of A . l_nA=l_nk+(E_A / RT) or A = log k + (E_a /2.303 RT). l_nA =l_nk+(E_A / RT) or A = log k + (E_a /2.303 RT). Substitute for k and T and calculate log A = log 3.7 + [{(59.2) (1000cal / 1 k cal) } / {(2.303)(1.987)(1085)}] = 0.57 + 11.9 = 12.5. Since antilog 12.5 = A, A = 3 × 10^12 liters/mole-sec. (Notice: k2and T2 Since antilog 12.5 = A, A = 3 × 10^12 liters/mole-sec. (Notice: k 2 and T 2 valueswere used in the calculations but k_1 and T_1 could have been used instead.) (c) To find k at 800 solve the Arrhenius equation for k log k = log A - (E_a / 2.303RT) . Substitute in known values for A, Ea, R, and T to obtain: log (3 × 10^12) - [{(59.2)(1000cal / 1k cal)} / {2.303 (1.987)( 800 )}] = 12.5 - 16.2 = - 3.7 k. = antilog (- 3.7) = 2 × 10^-4 liters / mole-sec.

Question:

The mean proper lifetime (T) of \pi ^+ mesons is 2.5 × 10^-8 s. In a beam of \pi^+ mesons of speed 0.99c, what is the average distance a meson travels before it decays? What would this value be if the relativtistic time dilation did not exist?

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/Users/wenhuchen/Documents/Crawler/Physics/D32-0945.htm

Solution:

The proper lifetime, T, of a particle is the lifetime of the particle as measured by an observer traveling with the particle. In a coordinate system S' moving with the mesons the average lifetime of the mesons is 2.5 × 10^-8 s. A laboratory observer makes his measurements in a systemS fixed in the laboratory, with respect to which S' is moving at a speed of 0.99 c. According to the theory of special relativity, an event occurring in space-time will have coordinates (x, y, z, t) re-lative to S, and coordinates(x', y', z', t') relative to S'. If the relative velocity between the 2 frames is v, these 2 sets of coordinates are related by the Lorentz Transformation x = (x' +vt') / \surd{1 - (v/c)^2}(1a) y = y'(1b) z = z'(1c) t = {t' + (vx' / c^2)} / \surd{1 - (v/c)^2}(1d) where c is the speed of light. Now, note that the equations in (1) relate the co-ordinates of one event as recorded in 2 different reference frames. We need 2 events to describe the life-time of the meson: its "birth" and its "death". If we travel with the meson, we record the following coordi-nates for these 2 events: Birth:(x_1', y_1', z_1' , t_1') = (0, 0, 0, 0) Death: (x_2' , y_2', z_2' , t_2') = (0, 0, 0, T) Here, x_2' = 0 because, if we travel with the meson, we do not see it move. It appears stationary. The coordi-nates of these 2 events in S are found from equations using the above data: x_1 = {(0) + v(0)} / \surd{1 - (v/c)^2} = 0 y_1 = 0 z_1 = 0 t1= {(0) + (v (0) / c^2)} / \surd{1 - (v/c)^2} = 0 and x_2 =vT/ \surd{1 - (v/c)^2} = 0 y_2 = 0 z_2 = 0 t_2= T / \surd{1 - (v/c)^2} = 0 Relative to S, the meson travels a distance x_2 - x1=vT/ \surd{1 - (v/c)^2} in a time t2- t1= T / \surd{1 - (v/c)^2} = 0 Hence, t2- t1is the lifetime of the particle relative to S. (Notice that t2- t1> T) Using the given data x_2 - x_1 = (.99c) (2.5 × 10-8s) / \surd{1 - (.99)^2} x_2 - x_1 = {(2.5 × 10-8s) (.99c) (3 × 10^8 m/s)} / .141 x_2 - x1= 52.7 m. If time dilation did not exist, the distance traveled would be d_0, where d_0 = 2.5 × 10-8s × 0.99c × 3 × 10^8 m/s^-1 = 7.43 m.

Question:

A transverse sinusoidal wave is generated at one end of a long horizontal string by a bar which moves the end up and down through a distance of 0.50cm. The motion is continuous and is repeated regularly 120 times per second. a) If the string has a linear density of 0.25(kg / m) and is kept under tension of 90 N, find the speed, amplitude, fre-quency, and wavelength of the wave motion. b) Assuming the wave moves in the +x-direction and that, at t = 0, the end of the string described by x = 0 is in its equilibrium position y = 0, write the equation of the wave. c) As this wave passes along the string, each particle of the string moves up and down at right angles to the direc-tion of the wave motion. Find the velocity and acceleration of a particle 2.0 ft. from the end.

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/Users/wenhuchen/Documents/Crawler/Physics/D37-1091.htm

Solution:

(a) The end moves 0.25cm away from the equilibrium position, first above it, then below it; therefore, the ampli-tudey_mis 0.25cm. The entire motion is repeated 120 times each second so that the frequency is 120 vibrations per second, or 120 Hz. The wave speed is given by v = \surd(F / \mu). But F = 90 N and \mu = 0.25(kg / m), so that v = \surd[(90N) / (0.25 kg/m)] = 19(m/s). The wavelength is given by \lambda = v/\lambda, so that The wavelength is given by \lambda = v/\lambda, so that \lambda = [(19 m/s) / (120vib/s)] = 16cm. (b) The general expression for a transverse sinusoidal wave moving in the +x-direction is y =y_msin(kx-\cyrchar\cyromegat- \varphi). ). Requiring that y = 0 for the conditions x = 0 and t = 0yeilds 0 =y_msin(- \varphi) ) which means that the phase constant \varphimay be taken to be zero. Hence may be taken to be zero. Hence for this wave y=y_msin(kx-\cyrchar\cyromegat), and with the values just found, y_m= 0.25cm, \lambda = 16cmork = (2\pi / \lambda) = (2\pi / 16cm) = 0.39cm^-1, v = 19 m/s = 1900 cm/s or \cyrchar\cyromega =vk= (1900 cm/s) (0.39cm^-1) = 740s^-1 = 740Hz, the equation for the wave is y = 0.25 sin(0.39x - 740t) where x and y are in centimeters and t is in seconds. c) The general form of this wave is y =y_msin(kx-\cyrchar\cyromegat) =y_msink (x -vt). The v in this equation is the constant horizontal velocity of thewavetrain. It is not the velocity of a particle in the string through which this wave moves; this particle velocity is neither horizontal nor constant. Each particle moves vertically, that is, in the y-direction. In order to deter-mine the particle velocity designated by the symbol u, con-centrate on a particle at a particular position x-that is, x is now a constant in this equation-and see how the particle displacement y changes with time. With x constant u = (\partialy / \partialt) = -y_m\cyrchar\cyromegacos(kx-\cyrchar\cyromegat), in which the partial derivative (\partialy / \partialt) shows that although in general y is a function of both x and t, assume that x remains constant so that t becomes the only variable. The accelera-tion a of the particle at this (constant) value of x is a = (\partial^2y / \partialt^2) = (\partialu / \partialt) = -y_m\cyrchar\cyromega^2 sin(kx-\cyrchar\cyromegat) = -\cyrchar\cyromega^2y. For a particle at x = 62cm with y_m= 0.25cm, k = 0.39cm^-1, \cyrchar\cyromega = 740 s^-1, u = -y_mvcos(kx-\cyrchar\cyromegat) oru = -0.25(740)cos[(0.39) (62) - (740)t] = -185cos(24 - 740t) anda = -\cyrchar\cyromega^2y ora = -(740)^20.25 sin[(0.39) (62)-(740)t] = -13.7 ×10^4sin(24 - 740t) where t is expressed in seconds u in (cm/s) and a in cm/s^2 .

Question:

How does thetypedefstatement increase portability?

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Solution:

In C new data types can be created usingtypedefstatement. Once a type is defined, it can be used in the same manner as the standard C types. The main advantage of usingtypedefstatement is for porta-bility of code, i.e. code which can be run without change on a variety of hardware. Some machines define the typeintto contain 16 bits while other machines provide 32 bits. If we create a type called integer based on the word-size of the machine, then we can de-fine all our integer variables in terms of this type. This is done asfollows: typedeflongintinteger; main ( ) { integer x; integer y; /\textasteriskcenteredstatements\textasteriskcentered/ } Two things are to be noted: 1)typedefstatement is defined outside the main. 2) Qualifier long tells the compiler to use two words (32 bits) for storage. Now if we want the same program to run on 16-bit machine instead of modifying variables x, y, and z, all we have to do is modify thetypedef i.e. typedeflongintinteger; changeto typedefintinteger; All the following integer type statements in the program will be changed automatically.

Question:

What are the functions of (a) the vascular cambium, (b)stomata, (c) heartwood, (d) lenticels (e) abscission layer and (f)cutin?

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Solution:

The vascular cambium is ameristemicregion found in the stem. Sometimes it is found in the roots of higher plants. It exists as a continuous ring of cells extending between the cluster of phloem cells and xylem cells. The cambium produces these two types of cells during periods of active mitotic cell division. The ones facing the center of the stem differentiate into xylem cells; and the ones facing the periphery become phloem cells. The stomata are small specialized pores scattered over the epidermal surfaces of leaves and some stems. These pores are found mainly on the lower surface of the leaves. The stomata are delineated by a pair of cells called the guard cells. The guard cells, by changing their shape (due to changes in theirturgorpressure), regulate the size of the stomatal aperture. An open stoma permits the escape of water and the exchange of gases. The heartwood is the inner layer of the xylem in a woody plant. They are composed of dead, thick-walled xylem cells which have lost the ability to conduct water and minerals. Instead it has become tough supporting fibers. The heartwood increases the strength of the stem, and accommodates the increasing load of foliage as the tree grows. Lenticels are masses of cells which rupture the epidermis and form swellings . These masses of cells are formed from cells of the cork cambium which divide repeatedly and rupture the epidermis lying above. Lenticels represent a continuation of the inner plant tissues with the external environment, and permits a direct diffusion of gases into and out of the stem or twig. Such direct passages are necessary because the cork cambium forms a complete sheath around the vascular bundles and effecti-vely obstructs the ventilation of the vascular bundles. Anabcissionlayer is a sheet of thin-walled cells, loosely joined together , which extends across the base of the petiole. The formation of the abscission layer separates and loosens the point of attachment of the petiole to the stem. After the abscission layer is formed, the petiole is held on only by the epidermis and the fragile vascular bundles, so that wind or other mechanical disturbances will cause the leaf to fall off. This is the mechanism that accounts for the fall of leaves during autumn. Cutin is a waxy organic substance secreted by the epidermal cells of the stems and leaves, but not the roots. Its waterproof property retards evaporative water loss to the atmosphere.

Question:

Two thirds of the atoms in a molecule of water (H_2O) are hydrogen. What percentage of the weight of a water molecule is the weight of the two hydrogen atoms? The atomic weight of hydrogen is 1.008 g/mole and of oxygen is 16.00 g/mole.

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Solution:

The most direct way to solve this composition problem is to consider the total weight of one molecule of water: mass H_2O = 2 mass_H + mass_O. = 2(1.008 g/mole) + 16.00 g/mole = 18.016 g/mole. The mass of two hydrogen atoms is 2 × 1.008 = 2.016 g/mole. Hence, the percentage mass of hydrogen in water is: % mass = (2 mass_H-) / (mass H_2O) ×100% = [(2.016 g/mole) / (18.016 g/mole)] × 100% = 11.19

Question:

What are the lower vascular plants? Why are they considered more advanced than the bryophytes but less advanced than the seed plants?

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Solution:

The vascular plants (phylumTracheophyta) are subdivided into five groups . These are thePsilophyta(fork ferns),Lycophyta(club mosses, quill worts)Sphenophyta(horsetails or the "scouring rushes"),Pterophyte (ferns) andSpermophyta(the seed plants gymnosperms, and angiosperms ). Vascular plants, in contrast to the algae, fungi, and bryophytes , posses a vascular system that serves for support and for the conduction of water, mineral salts, and foods. The lower vascular plants refer to all the non-seed bearing classes, namely all classes except the gymnosperms and angiosperms. Most of the lower vascular plants are land plants. As mentioned before , they contain xylem and phloem, a system missing in the bryophytes . They have also evolved a more complicated anatomy than the bryophytes - more efficient roots, more elaborate stem and leaves, more complicated embryonic structure and a larger body size. Thesporophyte of the lower vascular plants has become the independent, dominant generation , and corres-pondingly, the gametophyte has become reduced. These characteristics of the lower vascular plants place them higher on the evolutionary ladder than the bryophytes. Structurally,most lowervascular plants resemble the seed plants far more than the bryophytes in their roots, stems and leaves. However, they are not that much better adapted for a life on land than the bryophytes . Because their sperm retain thebiflagellatedstructure necessitating a film of moisture to be active, and because the young sporophyte develops directly from the zygote without passing through any stage where it is protected by a seed, these plants are restricted to moderately moist habitats for active growth and re-production. Most of the seed plants, on the other hand, have evolvednonflagellatedsperm, a mechanism of pol-lination for sperm transport andgameticunion, and a seed structure within which the embryonicsporophyteis protected and nourished . These three features make the seed plants extremely successful inhabitants of the land. In addition, while the lower vascular plants have a reduced gametophyte, the seed plants have reduced it even further in size and have simplified its structure to the point where it is completely dependent on thesporophyte. The seed plants have also evolved heterospory, that is, the production of two types of spores. The lower vascular plants arehomosporous, and produce only one type of spore . The seed plants are therefore most advanced in the evolutionary trend toward greater embryonic protection, flexible and efficient ways of fertilization , the production of two types of spores, and the reduction of the haploid generation. While the lower vascular plants, then, are more advanced than the bryophytes, they trail the seed plants in evolutionary development .

Question:

(a) 2^3 \textbullet 2^2(b) a^3 \bullet a^5(c) x^6 \textbullet x^4

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Solution:

If a is any number and n is any positive integer, the product of the n factors a\bulleta\bulleta ... a is denoted by a^n \bullet a is called the base and n is called the exponent. Also, for base a and exponents m and n, m and n being positive integers, we have the law: a^m \bullet a^n = a^m+n. Therefore, (a) 2^3 \bullet 2^2= (2 \bullet 2 \bullet 2) (2 \bullet 2) = 8 \bullet 4 = 32 or2^3 \bullet 2^2= 2^3+2 = 2^5 = 32 (b) a^3 \bullet a^5= (a \bullet a \bullet a) (a \bullet a \bullet a \bullet a \bullet a) = (a \bullet a \bullet a \bullet a \bullet a \bullet a \bullet a \bullet a) = a^8 ora^3 \bullet a^5= a^3+5 = a^8 (c) a^6 \bullet x^4 = x^6+4 = x^10.

Question:

A mixture of gaseous oxygen and nitrogen is stored at atmospheric pressure in a 3.7 l iron container maintained at constant temperature. After all the oxygen has re-acted with the iron walls of the container to form solid iron oxide of negligible volume, the pressure is measured at 450 torr. Determine the final volume of nitrogen and the initial and final partial pressures of nitrogen and oxygen.

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Solution:

The partial pressure of each component, which is independent of any other component in a gaseous mixture, at a defined volume, is equal to the pressure each com-ponent would exert if it were the only gas in that volume. The total pressure of a gaseous mixture is the sum of the partial pressures of the components. These two definitions are sufficient to solve this problem. Once all the oxygen has reacted, no oxygen is present in the gaseous phase, so that the final partial pressure of oxygen is zero. By the second definition, final total pressure = final partial pressure of N_2 + final partial pressure of O_2 450 torr = final partial pressure of N_2 + O torr or,final partial pressure of N_2 = 450 torr. But since, after all the oxygen has reacted, only nitrogen fills the entire volume, the final volume of nitrogen is 3.7 l, By the first definition, the initial partial pressure of nitrogen is the same as the final partial pressure of nitrogen, 450 torr. We now again employ the second definition to determine the last re-maining unknown quantity, the initial partial pressure of oxygen. Proceeding, we obtain initial total pressure = initial partial pressure of O_2 + initial partial pressure of N_2 1 atm = initial partial pressure of O_2 + 450 torr or,initial partial pressure of O_2 = 1 atm - 450 torr = 760 torr - 450 torr = 310 torr. Summarizing these results in tabular form we have: InitialFinal partial pressure of O_2310 torr0 torr partial pressure of N_2450 torr450 torr total pressure760 torr450 torr

Question:

A deuterium atom moving with kinetic energy of 0.81× 10^-13 J collides with a similar atom at rest. A nuclear inelastic reaction takes place and a neutron is observed to be emitted at right angles to the original direction of motion. Determine its kinetic energy, given that the other product of the reaction is an atom of the light isotope of helium and that the rest masses of a neutron, a deuterium atom, and a light helium atom are 1.6747, 3.3441, and 5.0076, respectively, in units of 10^-27 kg.

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Solution:

Before doing this problem, we must see if it is necessary to use relativistic considerations. The total relativistic energy of a particle is the sum of its rest mass energy and its relativistic kinetic energy, or E = m_0c^2 + T where m_0 is the particle's rest mass. Hence, T = E - m_0c^2 ButE = m_0c^2 / \surd{1 - (v2/ c^2)} where v is the particle's velocity. Therefore, T = m_0c^2 [ \surd{1 / (v2/ c^2)} - 1] Solving for v/c, T / m_0c^2 + 1 =1 / \surd{1 - (v2/ c^2)} 1 -(v2/ c^2) = [m_0c2/ (T + m_0c^2)]2 v/c =\surd {1 - [m_0c2/ (T + m_0c^2)]2} For the incident deuterium atom, v/c = \surd {1 - [(3.3441 × 10^-27 kg × 9 × 1016m^2 / s2)/ (.81 × 10^-13J +(3.3441 × 10^-27 kg × 9 × 1016m^2 / s^2 ))]2} or ,v/c =5.382 × 10^-4 Since this is very small, the problem is extremely non- relativistic. We now proceed with a classical analysis of the collision. Looking at the figure, we may write the following 2 equations from the principle of conservation of momentum. m_Dv = m_Hv_2 cos \texttheta(1) m_nv_1 = m_Hv_2 sin \texttheta(2) Since the collision is inelastic, we must account for the fact that energy, Q, is gained or lost in the reaction. Hence, the principle of conservation of energy yields. 1/2 (m_Dv^2) + Q = 1/2 (m_n v^2_1) + 1/2 (m_H v^2_2)(3) If Q > 0, energy is released in the reaction, and is absorbed as kinetic energy by the reaction products. Now we solve for v_1 in terms of Q. First, square (1) and (2), then add [m^2_Dv^2= m_H^2v^2_2 cos^2 \texttheta] + [m^2_nv^2_1= m_H^2v^2_2 sin^2 \texttheta] m^2_Dv^2 + m^2_nv^2_1 = m_H^2v^2_2 orm_Hv^2_2 = ( m^2_Dv2+ m^2_nv^2_1) / m_H(4) From (3)m_Hv^2_2 = 2Q +m_Dv2- m_nv^21(5) Combining (4) and (5) 2Q +m_Dv2- m_nv^2_1 = m_Hv^2_2 = ( m^2_Dv^2 + m^2_nv^2_1) / m_H or2QmH+ m_Dm_Hv^2 - m_nm_Hv^2_1 = m_D^2v^2 - m^2_nv^2_1 (m^2 n + m_nm_H ) v^2_1 = 2Qm_H + (m_DmH- m^2_D) v^2 Finally ,v^2_1 = {2Qm_H + m_D(mH- m_D) v^2} / {m_n(mn+ m_H)} and(1/2) m_n v^2_1 = {Qm_H + (1/2) m_D v^2 (mH- m_D) } / (mn+ m_H)(6) This is the neutron kinetic energy. Now, note that in (6), we have an unknown variable, namely, Q. But, we can obtain Q by noting that the only mechanism which can be responsible for Q is a difference in mass between the products and reactants of the collision. This follows from Einstein's mass-energy relation, since a mass difference \Deltam is equivalent to an energy E = (\Deltam)c^2(7) Now, the mass entering the reaction is the sum of the masses of the incident and target deuterons, or 2m_D(8) The sum of the masses of the products is m_n + m_H Hence,\Deltam = m_n + m_H - 2m_D \Deltam = [1.6747 + 5.0076 - 2(3.3441)] × 10^-27 kg \Deltam = 59 ×10^-31 kg Now,Q = E = (59 × 10^-31kg) (9 × 10^16 m^2/s^2) Q = 5.31 × 10^-13 J(9) Inserting (9) and the given data in (6), we obtain (1/2) m_nv^2_1 = [(5.31 × 10^-13 J) (5.0076 × 10^-27kg) + (.81 × 10^-11 J) (5.0076 - 3.3441) × 10^-27kg] / [(1.6747 + 5.0076) × 10^-27kg] = [(26.5904 × 10^-40 + 1.3474 ×10-40) / (6.6823 × 10^-27 )] J = 4.1809 × 10^-13J

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Question:

A toroidal coil has 3000 turns. The inner and outer diameters are 22 cm and 26 cm, respectively. Calculate the flux density inside the coil when there is a current of 5.0 amp.

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/Users/wenhuchen/Documents/Crawler/Physics/D21-0705.htm

Solution:

A toroid is a wire wound in a helix and bent into the shape of a doughnut with a current i running through it. The magnetic field B^\ding{217} forms concentric circles inside the toroid. Let r be the mean radius of the toroid. Apply Ampere's law around the circular path of radius r, \oint B^\ding{217} \bullet dl^\ding{217} = \mu_0I where the symbol \oint indicates that the integral is taken over a closed path and I is the total current enclosed by the path of integration. We get (B) (2\pir) = \mu_0 N i Therefore,B = [(\mu_0 N i)/(2\pir)] For the toroid given, mean diameter = [(22 cm + cm)/2] = 24 cm = 0.24 m mean radius = r = (0.24 m/2) = 0.12 m The magnetic field is B = [{( 4 \pi × 10^-7 weber/arap-m) (3000) (5.0 amp)}/{(2\pi) (0.12 m)}] = 0.025 weber/m^2 Note that the magnetic field is not constant over the cross section of The toroid, but is inversely proportion-al to the radius. The magnetic field outside the toroid is zero since around any closed path encircling the toroid the net current enclosed is zero, due to equal amounts of current travelling in opposite directions.

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Question:

What type of flip-flop does the diode-transistor circuit of fig. 1 represent?

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Solution:

The circuit of fig. 1 is broken down into two identical "sub-circuits" in fig. 2(a) and (b). In sub--circuit (a) if either or both of the input diodes are con-nected to OV (200), current flows through R2. This makes V_be = OV and the transistor does not conduct, leaving its output Y_1 high (5V). If both diodes are connected to 25V (high), no current flows through R2. This makes V_be = 5V and the transistor conducts, pulling V_1 low (OV). Thus, each sub-circuit is a NAND gate and the output of each gate is connected to the input of the opposite gate. The logic diagram of fig. 1 is shown in fig. 3. The truth table for fig. 3 is shown in Fig. 4. InputsPresent StateNext State X_1X_2 Y_1Y_2 Y_1Y_2 00 01 not used 00 10 not used 01 01 10 01 10 10 10 01 01 10 10 01 11 01 01 11 10 10 Fig.4 This truth table is similar to that of an S-R flip-flop except in this case X_1 =S; X_2 =R. Hence, this is anS-Rflip-flop.

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Question:

Define antibiotics, where are they derived from? What is meant by broad-spectrum antibiotics?

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Solution:

Compounds that can inhibit and/or destroy bacteria may be defined as antibiotics. These substances may be bacteriostatic (growth-inhibiting) or bactericidal (bacteria killing) in their action. Currently, antibiotics are derived from microorganisms and are, in effect, metabolic products of these organisms. Common classes include the penicillins and the tetracyclines. Tetracyclines have been isolated from "streptomyces" bacteria strains and all have a basic structure of four fused rings. These compounds are also called broad- spectrum antibiotics because of their effectiveness against a wide variety of microorganisms. Aureomycin, the first member of the tetracycline family, has its structure written above:

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Question:

A chemist dissolves BaSO_4 , in pure water at 25\textdegree C. If its K_sp= 1 × 10-^10 , what is the solubility of the barium sulfate in the water?

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Solution:

The solubility of a compound is defined as the limiting concentration of the compound in a solution before precipitation occurs. To find the solubility of the barium sulfate, you need to know the concentration of its ions in solution. BaSO_4 will dissociate into ions because it is a salt. There will be an equilibrium between these ions and the BaSO_4 . The equilibrium can be measured in terms of a constant,K_sp, called the solubility constant. TheK_spis ex-pressed in terms of the concentrations of the ions. As such, to answer the question, you want to represent thisK_sp. For this reaction, the equation is BaSO_4\rightleftarrowsBa++ + SO-_4 K_sp= [Ba^2+ ] [ SO4^2-] = 1 × 10-^10 . Let x = [Ba^2+ ] . Thus, x = [SO4^2-], also, since both ions will be formed in equimolaramounts. Therefore, x \bullet x = 1 × 10-^10 . Solving, x = 1 × 10-^5 M = [Ba^2+ ] = [SO_4 ^2- ] .

Question:

A steel rod of length L = 20 cm and cross-sectional area A = 3 cm^2 is heated at one end to T_1 = 300\textdegreek while the other end rests in ice. Assuming that heat transmission occurs exclusively through the rod (without losses from the walls), calculate the mass m of the ice melting in time ∆t = 10 min. The thermal conductivity of steel is k = 0.16 cal deg^-1 sec^-1cm^-1.

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Solution:

Thermal conductivity k as given above, indicates the amount of heat transferred per second per square centi-meter per degree centigrade through 1 cm length of steel. The rate of heat transfer through the steel rod is given by heat/sec = ∆Q/∆t = k × [A × (T_1 - T_2)/L] where T_2 is the temperature of the colder end of the rod. Hence, ∆Q/∆t = [(0.16 cal K\textdegree^-1 sec^-1 cm^-1) × (3 cm^2) × (300 - 273)K\textdegree] / [20 cm] = 0.648 cal/sec In time ∆t = 10 min., the amount of heat transferred by the rod to the ice is ∆Q= (0.648 cal/sec) × ∆t = (0.648 cal/sec) × (10min×60sec/min) = 388.8 cal. The heat of fusion of ice is 80 cal/gr, which means in order to melt 1 gr. of ice, 80 cal. must be added to the ice. Therefore, 388.8 calories will melt m = (∆Q cal)/(80 cal/gr) = (388.8)/(80 gr) = 4.86 gr. of ice.

Question:

How much energy is emitted by Avogadro number of atoms, if they each emit a light wave of 400 nm wavelength?

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Solution:

Energy is related to wavelength in the following equation, E =hc/\lambda, where E is energy, h Planck's constant (6.626 × 10^-34 Js) , c the speed of light (3.0 × 10^8 m/sec) and \lambda the wavelength. Using this equation, one calculates the energy emitted by each atom. To find the total amount of energy produced byAvogrado'snumber of atoms, multiply the amount of energy one atom emits by 6.02 × 10^23. Solving for E:1 nm = 10^-9 m. E = (hc/\lambda) = [(6.626 × 10^-34 Js) (3.0 × 10^8 m/s)] / (400 × 10^-9 m) = 4.97 × 10^-19 J/atom Thus, the total amount of energy emitted is E = (4.97 × 10^-19 J/atom) (6.02 × 10^23 atoms/mole) = 2.98 × 10^5 J.

Question:

A rod of negligible mass with length 2 meters has a small 2kg mass mounted on each end. (a) Calculate the moment of inertia of this rod about an axis perpen-dicular to the rod and through its center. (b) Find its angular momentum if the rod rotates about this axis with an angular velocity of 10 radians per second.

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Solution:

(a) Rotational inertia I of a system is defined as the sum of the products of the masses m of the particles in the system and the squares of their respective distances r from the rotational axis. Then I = I = \sum_i m_i r^2 _i For the system shown in the figure, we have I = (2kg)(1m)^2 + (2kg)(1m)^2 = 4 kg-m^2. (b) The angular momentum is defined as the product of the rotational inertia and the angular velocity \cyrchar\cyromega. Therefore the angular momentum L in this problem is L = Iw = (4 kg-m^2)(10 rad/sec) = 40 kg-m^2/sec . Note that angular momentum is analogous to linear momentum which equals the product of inertial mass m and linear velocity v.

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Question:

It was found that a magnesium oxide contained .833 g of oxygen and 1.266 g of magnesium. Calculate the gram- equivalent weight of magnesium.

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Solution:

You begin this problem by establishing what is meant by the term gram-equivalent weight. Gram-equivalent weight may be defined as the number of grams of an element that will involve a gain or loss of N electrons, i.e., the Avogadro's number (6.02 × 10^23) of electrons, when the element enters into chemical combination with another element. In this problem, you know that oxygen is present. Oxygen has an oxidation state of - 2. In a reaction, one mole of oxygen will gain two moles of electrons. The molecular weight of oxygen is 16 g/mole. Its equivalent weight becomes 16/2 = 8.00 g, since 2 moles of electrons will be gained and, by definition, equivalent weight is the amount of a substance that will gain or lose one mole of electrons. The gram-equivalent weight of magnesium is that amount of the element that combines with 8.00 g of oxygen. You are told that 1.266 g Mg combines with .833 g O_2. If you let x = grams of Mg that will combine with 8.00 of oxygen, you can set up the following proportion: (1.266 g/Mg) / (.833 g O_2) = (x g Mg) / (8.00 g O_2) Solving for x, you obtain x = 12.16 g of Mg, which is its gram-equivalent weight.

Question:

If the atomic weight of oxygen was 50, what would its equivalentweight be?

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Solution:

The relationship of the atomic weight and the equivalent weight is (Atomic weight)/ (Equivalent weight) = valence number The valence number is a measure of the number of atoms of hydrogenthat will combine with one atom of the element.Two hydrogen atomscombine with one oxygen. The valence number of oxygen is therefore2. One can now solve for the equivalent weight of oxygen when theatomic weight is taken as 50. 50 / (equivalent weight) = 2 equivalent weight = 50/2 = 25.

Question:

Draw a labelled phase diagram for a substance Z which has the following properties? normal boiling point = 220\textdegreeC, normal freezing point 80\textdegreeC, and triple point 60\textdegreeC and .20 atm. Predict the freezing and boiling, if the pressure were .80 atm?

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Solution:

To draw this diagram you want to understand all the terms involved. The relation between solid, liquid, and gaseous states as a function of the given temperature and pressure can be summarized on a graph known as a phase diagram. From the given experimental observations, you can draw the diagram. The lines which separate the states in a diagram represent an equilibrium between the phases. The intersection of the three lines is called the triple point, where all three phases are in equilibrium with each other. By normal boiling and melting points you mean those readings taken at 1 atm. Thus, the phase diagram can be written as shown in the accompanying figure. From the diagram you see that if the atm was .80, the b.p. and f.p. would drop, respectively, to 2150C and85\textdegreeC.

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Question:

Starting from rest at the top, a small sphere rolls without slipping off a large fixed sphere. At what point will the small sphere leave the surface of the big sphere? (See figure).

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/Users/wenhuchen/Documents/Crawler/Physics/D04-0207.htm

Solution:

The small sphere will leave the surface of the large sphere when the normal force (N^\ding{217}) of the latter on the former is zero, for this means that contact has ceased. Applying Newton's Second Law to the tangential and radial components of motion of the small sphere (see figure), we obtain F_radtal = Ma_radial = Mg cos \varphi - N F_tangential = Ma_tangential = mg sin \varphi where a_radial is positive in the direction of BA^\ding{217}, and a_tangential is positive in the direction of motion of the sphere. Furthermore, a_radial = v^2/(R + r), since the small sphere is traveling along a circular arc. (Here, v is the latter's speed.) Then (Mv^2)/(R + r) = Mg cos \varphi - N orN = Mg cos \varphi - [(Mv^2)/(R + r)] We require N = 0, or Mg cos \varphi = [(Mv^2)/(R + r)] Thereforecos \varphi = v^2/[(R + r) g](1) We are not yet finished, since we don't know v in (1). In order to find it, we may use the principle of conservation of energy to relate the energy of the small sphere at points O and S. Since the sphere starts from rest at O, it has only potential energy. Measuring poten-tial energy from A, the energy at O is E = Mg(R + r)(2) At S, the sphere has potential and kinetic energy equal in amount to the energy at O. Hence E = (1/2) Mv^2 + (1/2) Iw^2 + Mg(R + r) cos \varphi(3) where the first and second terms are the translational and rotational kinetic energies of the sphere respectively. Equating (2) and (3) (1/2) Mv^2 + (1/2) Iw^2 + Mg(R + r) cos \varphi = Mg(R + r)(4) Since the small sphere rolls without slipping v = wr, whence, using (4) (1/2) Mv^2 + (1/2) I(v^2/r^2) + Mg(R + r)(1 - cos \varphi) The moment of inertia of a sphere about its center is I = (2/5) Mr^2 Then(1/2) Mv^2 + (1/2)(2/5)(Mr^2)(v^2/r^2) + Mg(R + r)(1 - cos \varphi) (1/2) Mv^2 + (1/5) Mv^2 = Mg(R + r)(1 - cos \varphi) (7/10) v^2 = g(R + r)(1 - cos \varphi) v^2 = (10/7) g(R + r)(1 - cos \varphi)(5) Utilizing (5) in (1), cos \varphi = [10g (R + r)(1 - cos \varphi)]/[7 (R + r) g] cos \varphi = (10/7) (1 - cos \varphi) Solving for \varphi (10/7) cos \varphi + cos \varphi = 10/7 cos \varphi (17/7) = (10/7) cos \varphi = (10/17) and\varphi = cos^-1 (10/17) \approx 54\textdegree

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Question:

When 10.0 g of silicon dust, Si, is exploded with 100.0 g of oxygen, O_2, forming silicon dioxide, SiO_2, how many grams of O_2 remain uncombined? The reaction equation is Si + O_2 \rightarrow SiO_2 .

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/Users/wenhuchen/Documents/Crawler/Chemistry/E05-0192.htm

Solution:

From the equation, it can be seen that Si and react in a 1 : 1 ratio. This means that 1 mole of Si will react with 1 mole of O_2. To determine the amount of left unreached after the reaction is performed, calculate the number of moles of O_2 and Si present and then subtract the number of moles of Si from the number of moles of O_2. (MW of Si = 28, MW of O_2 = 32.) number of moles = [(number of grams) / (MW)] number of moles of O_2 = (100.0 g) / (32 g/mole) = 3.12 moles number of moles of Si = [(10.0 g) / (28 g/mole) = 0.357 moles number of moles of excess O_2 = 3.12 - 0.357 = 2.763 moles weight of excess O_2 = number of moles × MW = 2.763 moles × 32 g/mole = 88.5 g.

Question:

What are the advantages and disadvantages of an exoskeleton? How have some of the disadvantages been overcome by themolluscsand by the arthropods?

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/Users/wenhuchen/Documents/Crawler/Biology/F12-0304.htm

Solution:

An exoskeleton serves several functions: (1) it provides support for the soft body to counter-balance the force of gravity; (2) it serves as a point of attachment for muscles; (3) it protects the body against desiccation; and (4) it serves to protect the animal from predators. The disadvantages of an exo-skeleton are that movement is somewhat restricted and that growth within the exoskeleton is limited. Different species ofmolluscshave evolved different mechanisms for overcoming these limitations. Mostmolluscslead very sluggish lives; many are sedentary or burrow in sand or wood. The fast moving squids have a greatly reduced exoskeleton enclosed within the mantle. Octopuses, also fast moving, have lost the exoskeleton entirely. Movement of bivalves" is made possible by muscles attached to two hinged shells. Growth of mostmolluscsproceeds by additions to the shell, creating a larger exoskeleton. In gastropods, growth proceeds spirally, resulting in a spiral shell. Nautilus, a cephalopod, builds a new chamber as it grows, and lives within the latest and largest chamber of the shell. The animal secretes a gas into the other chambers, and is thus able to float. The arthropods have a segmented body. The rigid exoskeleton is thinner in certain regions, such as the leg joints and between body segments. Muscle attachments at these points offer arthropods a fairly wide range of movement. Most arthropods walk, jump, or crawl. Some insects are capable of flying and many aquatic arthropods swim actively. Growth in arthropods occurs by a series of molts. When the growing animal reaches a certain size, the exoskeleton is shed and a new, larger exo-skeleton is formed. During the process, the arthropod is left "naked", and is thus temporarily vulnerable.

Question:

A braking device in a rifle is designed to reduce recoil. It consists of a piston attached to the barrel. The piston moves in a cylinder filled with oil. When the shot is fired the barrel and the piston recoil at an initial velocity V_0 . The oil under pressure passes through tiny holes in the piston causing the piston and the barrel to decelerate at a rate proportional to their velocity a = \rule{1em}{1pt}kv. Express (a) v in terms of t, (b) x in terms of t, (c) v in terms of x. Draw the corresponding motion curves.

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/Users/wenhuchen/Documents/Crawler/Physics/D37-1063.htm

Solution:

(a) The basic definition of acceleration is a = dv / dt .(a) Substituting the expression given for the accelera-tion yields \rule{1em}{1pt}kv = dv / dt .(b) Rearranging to separate variables yields \rule{1em}{1pt}k dt = dv / v .(c) Now it is possible to integrate equation (c) to get \rule{1em}{1pt}k^t\int_0dt = \rule{1em}{1pt}k^t\int_0dt = ^v\int_(v)0 (dv / v) \rule{1em}{1pt}kt = In (v / v_0) .(d) If antilogs of equation (d) are taken, we arrive at the desired form v = v0e^\rule{1em}{1pt}kt(e) This curve is shown in Figure 2. (b) Again, starting with a basic definition, this time for v, we have v = dx / dt(f) Substitution of the expression (e) into (f) yields v_0 e^\rule{1em}{1pt}kt = dx / dt, v_0 e^\rule{1em}{1pt}ktdt = dx.(g) Integrating (g) v_0^t\int_0 e^\rule{1em}{1pt}ktdt =^x\int_0 dx, \rule{1em}{1pt}v_0 / ke^\rule{1em}{1pt}kt│t_0 = x, \rule{1em}{1pt}v_0 / k (e^\rule{1em}{1pt}kt \rule{1em}{1pt} e^0) = x, x = \rule{1em}{1pt}v_0 / k (1 \rule{1em}{1pt} e^\rule{1em}{1pt}kt).(h) This curve is shown in Figure 3. (c) Part c may be done in two ways: Using the chain rule for differentials, equation (a) may be rewritten a =dv / dt = (dx / dt)( dv / dx) = v(dv / dx).(i) Substituting a = \rule{1em}{1pt}kv in equation (i), we get \rule{1em}{1pt}kv = v(dv / dx) ordv / dx = \rule{1em}{1pt}k, dv = \rule{1em}{1pt} k dx. Integrating to solve yields v \rule{1em}{1pt} v_0 = \rule{1em}{1pt} kx. orv = v_0 \rule{1em}{1pt} kx. As an alternate method, equations (e) and (h) can be combined. From (e) we know e^\rule{1em}{1pt}kt = v / v_0 . Substituting for the exponential in (h) yields x = v_0 / k [(1 \rule{1em}{1pt} (v / v_0)] orkx = v_0 \rule{1em}{1pt} v orv = v_0 \rule{1em}{1pt} kx. This curve is illustrated in Figure 4.

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Question:

If blood is carefully removed from a vessel and placed on a smooth plastic surface, will it clot? Explain why or why not.

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/Users/wenhuchen/Documents/Crawler/Biology/F14-0347.htm

Solution:

Vertebrates have developed a mechanism for preventing the accidental loss of blood. Whenever a blood vessel is ruptured, one of the soluble plasma proteins, fibrinogen, is enzymatically converted into an insoluble protein, fibrin, which forms a semisolid clot. Many people think that blood clots when it becomes exposed to the air or when it stops flowing. However, if one were to carefully remove blood from a vessel- without allowing it to contact the damaged part of the vessel, and then place this blood on a smooth plastic dish or one lined with paraffin, it would not clot. However, if this blood were allowed to touch any damaged tissues or were placed on glass or some other relatively rough surface, the blood would clot. Either the damaged tissues or the blood itself must release some chemical which initiates the clotting mechanism. Actually, it is both damaged tissues and disintegrated platelets in the plasma that release substances responsible for the clotting reaction. Platelets are very small disc-shaped bodies found in mammalian blood and formed in the bone marrow from large cells called megakaryocytes. Platelets seem to disintegrate more readily upon contacting glass surfaces than on plastic surfaces. Platelets are also called thrombocytes. When a blood vessel is cut, the damaged tissues release a lipoprotein, called thromboplastin, which initiates the clotting mechanism. Calcium ions and certain protein factors in the plasma must be present in order for thrombo-plastin to be effective. Thromboplastin interacts with Ca^+2 and these proteins to produce prothrombinase, the enzyme that catalyzes the second step in the clotting mechanism. Prothrombinase can also be made by the interaction of a substance released from the disintegrated platelets (platelet factor # 3) and other factors in the plasma, including Ca^2+ and proteins. The prothrombinase made from either the tissues or the platelets catalyses the conversion of prothrombin, a plasma globulin, into thrombin. Finally, thrombin enzymatically converts fibrinogen into fibrin, an insoluble protein. Fibrin forms long fibers which mesh and trap red cells, white cells, and platelets, forming the clot. Usually the clot forms within 5 minutes of the rupturing of the vessel. The clot then begins to contract and squeeze out most of the plasma from itself within an hour. This process, called clot retraction, serves to increase the strength of the clot, and also pulls the vessel walls adhering to the clot closer together. The extruded plasma is now called serum, since all the fibrinogen and most other clotting factors have been removed. Because it lacks these constituents, serum cannot clot.

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Question:

Discuss the sequence of reactions that constitute the "dark reactions" of photosynthesis. What are the products of these reactions?

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/Users/wenhuchen/Documents/Crawler/Biology/F09-0218.htm

Solution:

The dark reactions of photosynthesis, in which carbohydrates are synthesized, occur in a cyclic sequence of three phases - the carboxylative, reductive, and regenerative phases. These dark reactions do not require the presence of sunlight. In the carboxylative phase, a five - carbon sugar, ribulose- 5-phosphate, is phosphorylated by ATP to yield ribulose diphosphate (see Figure). Ribulose diphosphate is then carboxylated (that is, CO_2 is added), presumably yielding a six - carbon intermediate, which is split immedi-ately by the addition of water to give two molecules of phosphogiyceric acid. The three - carbon phosphogiyceric acid is reduced by NADPH in an enzymatic reaction in the reductive phase, utilizing energy from ATP. The product of this reduction is phosphoglyceraldehyde, another triose. Two molecules of phosphoglyceraldehyde can then condense in the regenerative phase to form one hexose molecule, fructose diphosphate. This is subsequently converted to a glucose-6-phosphate molecule and finally transformed into starch. Ribulose - 5 - phosphate, the cycle - initiating reactant, can be regenerated from fructose diphosphate via certain reactions of the pentose phosphate pathway. Some scientists tend to believe that the ribulose molecule is regenerated from phosphoglyceraldehyde by a complicated series of reactions. However, the cyclicity of the dark reactions is a well established and universally accepted fact. The materials consumed in the production of one hexose molecule are one molecule each of CO_2 and H_2O three ATP molecules, and four H atoms (from two molecules of NADPH_2). Ribulose - 5 - phosphate is not consumed but gene-rated at some point in the cycle. The glucose - 6 - phosphate produced can either be polymerized into starch and stored, or broken down to yield energy for work.

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Question:

A certain solution has pH 3.89 at 0\textdegreeC. FindpOHand [OH-].

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/Users/wenhuchen/Documents/Crawler/Chemistry/E10-0340.htm

Solution:

pH is a measure of the [H^+] andpOHis a measure of [OH^-]. Their product gives K_W, the ionization constant of water: [H^+][OH^-] = K_W. pH andpOHare related by the equation, pOH+ pH =pK_W. At 0\textdegreeC,pK_W= 14.94. Therefore: pOH=pK_W- pH = 14.94 - 3.89 = 11.05. To find [OH^-] use the equation, pOH= - log [OH^-] 11.05 = - log [OH^-] [OH^-] = 10^-11.05 = 10^-12+.95 = (10^-12)(10^.95) Find the antilog of 10^.95 . It is 8.9 which gives [OH^-] = 8.9 × 10-12.

Question:

A cube of copper metal has a mass of 1.46 × 10^-1 kg. If the length of each edge of the cube is 2.5 x 10^-2 m and a copper atom has a mass of 1.06 x 10^-25 kg, determine the number of atoms present in the sample and then estimate the size of a copper atom. Although the actual crystal structure is more complex, assume a simple cubic lattice.

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/Users/wenhuchen/Documents/Crawler/Physics/D31-0913.htm

Solution:

The number of atoms, N, is found in the follow-ing manner: Mass of N atoms = (Mass of 1 atom) (number of atoms, N) 1.46 × 10^-1 kg = [1.06 × 10^-25 (kg/atom)] (N atoms) N = [{1.46 × 10^-1 kg} / {1.06 × 10^-25 (kg/atom)}] = 1.38 × 10^24 atoms. Let us assume that each copper atom may be represented by a sphere whose diameter is d. If the atoms are touching one another, there will be the same number, say n, along each edge of the cube. The total number of atoms N is re-lated to n by the volume relationship n^3 = N = 1.38 × 10^24. The number of atoms along one edge is then approximately n= 1.1 × 10^8 atoms This number is approximate due to the inaccuracy of the initial assumption that each atom has the shape of a sphere. Also, Length of an edge of the cube = (diameter, d, of one copper atom) (number, n, of atoms along an edge) Thennd= 2.5 x 10^-2 m. Therefore, the diameter of a cop-per atom is approximately d = [(2.5 x 10^-2 m) / (1.1 x 10^8)] = 2.27 x 10^-10 This value agrees reasonably well with the value of 2.56 × 10^-10 m obtained using other methods.

Question:

Write a BASIC program which converts inches to feet and inches.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G10-0230.htm

Solution:

Use the INT function 1\O READ I 2\O LET F = INT(I/12) 3\O LET I1 = I - F\textasteriskcentered12 4\O PRINT I "INCHES = "; F;"FEET"; I1; "INCHES" 45 GO TO 1\O 5\O DATA 9, 86, 47, 37, 947, 48\O 6\O END

Question:

In hunting a submarine, two ships A and B, separated by 3000 ft, find a sonar echo on the line between them but 30\textdegree from vertical for A and 60\textdegree for B. However, 200 ft below the surface, the water temperature changes suddenly so that the velocity of sound in the depths is 0.9 that in the shallows, (a) For what depth will the depth charges be set, if the ship captains do not know of this? (b) What is the actual depth of the submarine? (c) What will the ship captains think the submarine's horizontal distance from ship A is? (d) What is the submarine's actual horizontal distance from ship A?

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/Users/wenhuchen/Documents/Crawler/Physics/D27-0868.htm

Solution:

Figure (a) shows the actual paths of the sonar waves represented by solid lines, and their apparent paths as broken lines. The submarine will appear to the ship captains to be at point 0' when in reality it is at point 0, as shown. AC + CB = 3000 (AC/CS) = tan 30 (CB/CS) = tan 60 AC + CB = CS (tan 30 + tan 60) CS = [(AC + CB)/(tan 30 + tan 60)] = [(3000 ft)/(0.577 + 1.732)] = 1310 ft since tan 30 = 0.577 tan 60 = 1.732 (b) In actual fact the sound waves are refracted at a depth of 200 ft. By definition, the refraction is such that: (sin i/sin r) = (v_s/v_d) = (v_s/0.9 v_s) = (1/0.9) where i is the angle that the path the sonar wave takes in the shallow water makes with an imaginary line normal to the interface between shallow and deep water. The letter r stands for a similar angle for the path in deep water, and v_s and v_d represent the velocities of sound in the shallow and deep water, respectively. If i = 30\textdegree, then r = 26.7\textdegree and if i = 60\textdegree, r = 51.3\textdegree We see from figure (c) that: AD + DC' + C'E + EB = 3000 ft AD = 200 tan 30 EB = 200 tan 60 (FH/HS) = tan 26.7(HG/HS) = tan 51.3 We can see from the diagram that: DC' = FH C'E = HG Thus, from the first equation in the above group: AD + FH + HG + EB = 3000 ft. 200 tan 30 + HS(tan 26.7 + tan 51.3) + 200 tan 60 = 3000 HS = [(3000 - 115 - 346) / (0.503 + 1.248)] = 1450 ft The actual depth of the submarine is 200 + 1450 = 1650 ft. The horizontal distance of the submarine from ship A will appear to be AC in figure (b): AC = 1310 tan 30 = 755 ft The submarine's actual horizontal distance is AC' in figure (c): AC' = 200 tan 30 + 1450 tan 26.7 = 115 + 730 = 845 ft Thus without knowledge of the water temperature the depth charges will be shallow by 340 ft and off in a horizontal direction by 90 ft. That is, the ex-plosion will be 350 ft away from the submarine, a near miss rather than a direct hit.

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Question:

The circulatory system of insects does not function in gas exchange. What is its function? Describe the circulatory and respiratory systems in insects.

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/Users/wenhuchen/Documents/Crawler/Biology/F12-0312.htm

Solution:

Insects, which have high metabolic rates, need oxygen in large amounts. However, insects do not rely on the blood to supply oxygen to their tissues. This function is fulfilled by the tracheal system. The blood serves only to deliver nutrients and remove wastes. The insect heart is a muscular dorsal tube, usually located within the first nine abdominal segments. The heart lies within a pericardial sinus. The pericardial sinus is not derived from the coelom, but is instead a part of the hemocoel. It is separated by connective tissue from the perivisceral sinus which is the hemocoel surrounding the other internal structures. Usually, the only vessel besides the heart is an anterior aorta. Blood flow is normally posterior to anterior in the heart and anterior to posterior in the perivisceral sinus. Blood from the perivisceral sinus drains into the pericardial sinus. The heart is pierced by a series of openings or ostia, which are regulated by valves, so that blood only flows in one direction. When the heart contracts, the ostia close and blood is pumped forward. When the heart relaxes, the ostia open and blood from the pericardial sinus is drawn into the heart through the ostia. After leaving the heart and aorta, the blood fills the spaces between the internal organs, bathing them directly. The rate of blood flow is regulated by the motion of the muscles of the body wall or the gut. A respiratory system delivers oxygen directly to the tissues in the insect. A pair of openings called spiracles is present on the first seven or eight abdominal segments and on the last one or two thoracic segments. Usually, the spiracle is provided with a valve for closing and with a filtering apparatus(composed of bristles) to prevent entrance of dust and parasites. The organization of the internal tracheal system is quite variable but usually a pair of longitudinal trunks with cross connections is found. Larger tracheae are supported by thickened rings of cuticle, called taenidia. The tracheae are widened in various places to form internal air sacs. The air sacs have no taenidia and are sensitive to ventilation pressures (see below). The tracheae branch to form smaller and smaller subdivisions, the smallest be-ing the tracheoles. The smallest tracheoles are in direct contact with the tissues and are filled with fluid at their tips. This is where gas exchange takes place. Within the tracheae, gas transport is brought about by diffusion, ventilation pressures, or both. Ventilation pressure gradients result from body movements. Body move-ments causing compression of the air sacs and certain elastic tracheae force air out; those causing expansion of the body wall result in air rushing into the tracheal system. In some insects, the opening and closing of spiracles is coordinated with body movements. Grasshoppers, for example, draw air into the first four pairs of spiracles as the abdomen expands, and expel air through the last six pairs of spiracles as the abdomen contracts.

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Question:

Assuming the same declarations as in Problem 14, write PROCEDURECONCAT(S1,S2:string;varS3:string) which concatenates strings S1 and S2 and puts the result in S3.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G16-0408.htm

Solution:

To concatenate strings S1 and S2 means to create a new string which consists of S1 and S2 appended to it. In the following program, the lengths of the two strings are first added to see if the resulting string could fit in the array "housing" it, which can hold only 80 characters. If this is the case , S3 is first set to S1, and then, starting with position S1.length+1, the significant characters of S2 are copied into S3. PROCEDUREC0NCAT(SI,S2:string:VAR S3:string); Var I : integer; BEGIN IF( S1.length + S2.length) >strsizeTHEN write1n ('concatenation impossible-strings too long') ELSE BEGIN FORI: =1 TO S1.length DO S3\bulletWord [I]: =SI\bulletWord[I]; FORI: =1 TO S2\bulletlength DO S3 \bullet Word [S1 \textbullet length + I] := S2\bulletWord [I] ; END END;

Question:

What Fahrenheit temperature corresponds to \rule{1em}{1pt}40\textdegree Centigrade?

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/Users/wenhuchen/Documents/Crawler/Physics/D12-0440.htm

Solution:

\rule{1em}{1pt}40\textdegree C is 40 Centigrade degrees below the freezing point of water. Now 40\textdegreeC = 9/5(40) = 72\textdegreeF since 1 Centigrade degree is 9/5 of 1 Fahrenheit degree. But 72\textdegreeF below the freezing point, which is 32\textdegreeF, is 40\textdegreeF below 0\textdegreeF (72 \rule{1em}{1pt} 32 = 40). Thus \rule{1em}{1pt}40\textdegreeC = \rule{1em}{1pt}40\textdegreeF. This could have been obtained directly as follows. The formula for converting temperature in degrees Centigrade to degrees Fahrenheit is \textdegreeF = (9/5) \textdegreeC + 32 If \textdegreeC = \rule{1em}{1pt}40\textdegree then \textdegreeC = (9/5)(\rule{1em}{1pt}40) + 32 = \rule{1em}{1pt}40\textdegree

Question:

What is the wavelength of the sound wave emitted by a standard 440 cycles per second tuning fork?

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/Users/wenhuchen/Documents/Crawler/Physics/D26-0830.htm

Solution:

Noting that the velocity (v), frequency (n), and wavelength (\lambda) of sound are related by v =n\lambda, and assuming the velocity of sound to be 34,000 cm/sec, or approx. 1100 ft/sec, we find \lambda = v/n = [1100 ft / 440] = 2.5 ft (approx.)

Question:

Consider the array of three charges shown in the diagram. Find the force on charge 1 caused by the other two charges. Calculate the field at the position of 1 due to the other two charges.

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/Users/wenhuchen/Documents/Crawler/Physics/D18-0584.htm

Solution:

The force at 1 will be the vector sum of forces caused by charges 2 and 3. The magnitudes of these forces are F_12 = (K_E) [(2q)q / L^2] = K_E [(2q)^2 / (L)^2 ] F_13 = K_E [(2q)q / (d)^2 ] F_13 = K_E [(2q)q / (d)^2 ] But d^2 = 2L^2(figure 1). Therefore But d^2 = 2L^2(figure 1). Therefore F_13 = [(K_Eq^2) / (2L)^2 ] F_13 = [(K_Eq^2) / (2L)^2 ] F_12 is an attractive force, and F_13 repulsive. Their x- and y-components are F_12 is an attractive force, and F_13 repulsive. Their x- and y-components are (figure 3). (figure 3). (F_12)_x = 0 (F_12)_x = 0 (F_12)_y = \rule{1em}{1pt}(K_E)[(2q)2/ (L)^2 ] (F_12)_y = \rule{1em}{1pt}(K_E)[(2q)2/ (L)^2 ] (F_12)_y = \rule{1em}{1pt}(K_E)[(2q)2/ (L)^2 ] (F_13)_x = \rule{1em}{1pt}(K_E)[(q)2/ (2L)^2 ] cos 45\textdegree (F_13)_x = \rule{1em}{1pt}(K_E)[(q)2/ (2L)^2 ] cos 45\textdegree (F_13)_x = \rule{1em}{1pt}(K_E)[(q)2/ (2L)^2 ] cos 45\textdegree (F_13)_y = (K_E)[(q)2/ (2L)^2 ] sin 45\textdegree (F_13)_y = (K_E)[(q)2/ (2L)^2 ] sin 45\textdegree Thus the resultant force F at 1 is Thus the resultant force F at 1 is F_x = (F_12)_x + (F_13)_x = \rule{1em}{1pt}(K_E)[(q)^2 / (2L)^2](1 / \surd2) for cos 45\textdegree = 1 / \surd2 F_x = (F_12)_x + (F_13)_x = \rule{1em}{1pt}(K_E)[(q)^2 / (2L)^2](1 / \surd2) for cos 45\textdegree = 1 / \surd2 F_y = (F_12)_y + (F_13)_y = \rule{1em}{1pt}(K_E)[(2q)^2 / (L)^2 ]+ (K_E)[(q)2/ (2L)^2](1 / \surd2) for sin F_y = (F_12)_y + (F_13)_y = \rule{1em}{1pt}(K_E)[(2q)^2 / (L)^2 ]+ (K_E)[(q)2/ (2L)^2](1 / \surd2) for sin 45\textdegree = 1 / \surd2 45\textdegree = 1 / \surd2 The magnitude of F is The magnitude of F is F = \surd( F^2_x + F^2_y) = 1.59 K_E (q2/ L^2) F = \surd( F^2_x + F^2_y) = 1.59 K_E (q2/ L^2) If q = e and L = 1 \AA = 10^\rule{1em}{1pt}10 m, then If q = e and L = 1 \AA = 10^\rule{1em}{1pt}10 m, then F = 1.59 × [9× 109{(nt\rule{1em}{1pt}m^2) / (coul)2}] × [{(1.6 × 10^\rule{1em}{1pt}19 coul)^2 } F = 1.59 × [9× 109{(nt\rule{1em}{1pt}m^2) / (coul)2}] × [{(1.6 × 10^\rule{1em}{1pt}19 coul)^2 } /(10^\rule{1em}{1pt}10m)^2 ] /(10^\rule{1em}{1pt}10m)^2 ] = 2.24 × 10^\rule{1em}{1pt}8N = 2.24 × 10^\rule{1em}{1pt}8N The electric field E at position 1 is given by the force at 1 divided by the The electric field E at position 1 is given by the force at 1 divided by the net charge at 1, 2q: net charge at 1, 2q: E = F / 2q =1.59 KE(q / 2L^2) E = F / 2q =1.59 KE(q / 2L^2) = (2.24 × 10^\rule{1em}{1pt}8N) / [2(1.6 × 10^\rule{1em}{1pt}19 coul)] = (2.24 × 10^\rule{1em}{1pt}8N) / [2(1.6 × 10^\rule{1em}{1pt}19 coul)] = 2.1 × 10^11 (N / coul) = 2.1 × 10^11 (N / coul)

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Question:

Describe briefly the characteristics of the ontogeny of a humanbeing from a zygote to a mature fetus. What is the significanceof this?

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/Users/wenhuchen/Documents/Crawler/Biology/F28-0746.htm

Solution:

A human being, like any .other organism, starts out as a fertilized egg, or zygote. The fertilized egg may be compared to the single-celled flagellatezygote of some ancient organism. The zygote then under-goes cellulardivisions and gives rise to the blastula. The blastula may be comparedto some spherical multicellular embryonic form of a primitive species. The blastula next develops into a gastrula with two layers of cells, thenan embryo with three layers of cells. The early human embryo resemblesa fish embryo, with gill slits, pairs of aortic arches, a fishlike heartwith a single atrium and ventricle, a primitive fish kidney, and a tail thatcan wag. Later in its development/ the human embryo resembles a reptilianembryo, with closed gill slits, fused vertebra, a new kidney, and anatrium par-titioned into left and right chambers. Still later, the human embryodevelops a mammalian, four-chambered heart, and a newer kidney. The seven-month old human embryo, covered with hair, resemblesa baby ape more than an adult human. The ontogeny of the human embryo reveals to some extent the evolutionaryhistory of man. The one-celled, zygotic stage may reflect the earlyperiod of life on earth. Some unicellular flagellate organism is believedto be the ultimate origin of all living things. The fish-like embryonicstage may parallel the age of the fish, theretipilianstage the ageof the reptiles, the mammalian stage the age of the mammals, and the apelikestage the period of the early primates. The sequence of all these embryonicstages parallels the chronological development of the ages as wellas the evolution of animal forms. Thus, one can trace man's evolution backfrom some more primitive, apelike primate, through the forms of someancient mammal reptile, amphibian, fish, primitive ver-tebrate, invertebrate, andmulticellularstructure, to an ultimate one-celled ancestor. The significance of this is that ontogeny recapitulates phylogeny.

Question:

Under ordinary conditions of temperature and pressure, nitrogen gas exists as diatomic molecules (N_2). As a result of its electronic configuration, N_2 is very inert and requires extreme conditions before it will react with any other species. What is the electronic ground state configuration of a nitrogen atom (atomic number =7)?

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Solution:

The solution to this problem involves the application of the Pauli Exclusion Principle and of Hund's Rule. Since the atomic number of nitrogen is 7 and it is a neutral atom, there are 7 electrons to place in atomic orbitals. The two orbitals of lowest energy are the is and the 2s. These are filled according to the Pauli Exclusion Principle - 2 electrons of unpaired spin in each orbital. The orbitals with the lowest energy of those remaining are the three 2p orbitals, 2p_x, 2p_y, 2p_z. Hund's rule dictates that the remaining three electrons enter these orbitals singly and with parallel spin. The ground state electronic distribution for nitrogen can therefore be pictured as shown in the accompanying Figure.

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Question:

The volume of the bulb of a mercury thermometer at 0\textdegreeC is V_0, and the cross section of the capillary is A_0. The coefficient of linear expansion of the glass \alpha_G is per C\textdegree, and the coefficient of cubical expansion of mercury is \beta_M per C\textdegree. If the mercury just fills the bulb at 0\textdegreeC, what is the length of the mercury column in the capillary at a temperature of t\textdegreeC?

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Solution:

An exaggerated view of the expansion is shown in figure (b). Figure (a) represents the initial situation. When exposed to a temperature change \DeltaT, the cross section of the capillary, the volume of the bulb, and the volume occupied by the mercury all change (see figures). The final volume occupied by the mercury is V_Hg = V_0 (1 + \beta_M \DeltaT)(1) where \beta_M is the coefficient of volume expansion of mercury. The new cross section of the capillary will be A = A_0 (1 + 2\alpha_G \DeltaT)(2) where \alpha_G is the coefficient of linear expansion of glass. Similarly, the new volume of the bulb is V_g = V_0 (1 + 3 \alpha_G \DeltaT)(3) (Note that, initially, the volume of Hg = the volume of the bulb). Now, the volume of mercury outside the bulb in figure (b) will be V_Hg - V_g = V_0 (1 + \beta_M \DeltaT - 1 - 3\alpha_G\DeltaT) V_Hg - V_g = V_0 (\beta_M - 3\alpha_G)\DeltaT(4) If the length of the mercury column in figure (b) is h, then V_Hg - V_g = hA = hA_0 (1 + 2 \alpha_G \DeltaT)(5) where we have used (2). Equating (5) and (4) hA_0 (1 + 2\alpha_G \DeltaT) = V_0 (\beta_M - 3\alpha_G)\DeltaT orh = (V_0/A_0) [(\beta_M - 3\alpha_G)/(1 + 2\alpha_G \DeltaT)] \DeltaT Since\DeltaT = t\textdegreeC - 0\textdegreeC h = (V_0/A_0) [(\beta_M - 3\alpha_G)/(1 + 2\alpha_Gt)]t This is the length of the mercury column in the capillary at temperature t.

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Question:

Suppose a salesman gets $6 commission on each sale as longas the sale is under $150. However, if the sale is over $150 he gets a bonus of 2% of the amount over the $150. Write a PL/I program that will take as data the values of the amountof sale and print the commission the salesman gets.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G18-0447.htm

Solution:

Formally, we can express the commission C as follows. If the amountA is less than $150, then C = $6. Otherwise, if A>150, then C = 6 + 2 x (A-150)/100 /\textasteriskcenteredCOMMISSION CALCULATION\textasteriskcentered/ L2\O:GET LIST (A); L3\O:IF A< = 15\O THEN GO TO L6\O; L4\O:C = 6 + 2\textasteriskcentered(A-15\O)/1\O\O; L5\O:GO TO L7\O; L6\O:C = 6; L7\O:PUT EDIT (A, C) (F (7, 2) F (7, 2)); L8\O:PUT SKIP; L9\O:GO TO L2\O;

Question:

Define the term "debugging."

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Solution:

The program, even after being tested and accepted as valid, may still have faults that will manifest themselves for certain input sequences. Such hidden faults that are not readily detectable are called "bugs." Debugging a program means the modification of programs, throughout its lifetime , in order to fix the faults that produce unacceptable or unexpected results . Thus, as opposed to "unit testing" that simply makes the program "acceptable" for the specified input-output sequences, the de-bugging process involves the entire life cycle of a computer program.

Question:

The moon revolves about the earth in a circle (very nearly) of radius R = 239,000 mi or 12.6 x 10^8 ft, and requires 27.3 days or 23.4 × 10^5 sec to make a complete revolution, (a) What is the acceleration of the moon toward the earth? (b) If the gravitational force exerted on a body by the earth is inversely proportional to the square of the distance from the earth's center, the acceleration produced by this force should vary in the same way. Therefore, if the acceleration of the moon is caused by the grav-itational attraction of the earth, the ratio of the moon's acceleration to that of a falling body at the earth's surface should equal the ratio of the square of the earth's radius (3950 mi or 2.09 × 10^6 ft) to the square of the radius of the moon's orbit. Is this true?

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Solution:

(a) The velocity of the moon is v = (distance)/(time) = (circumference)/(time for one orbit) = (2\pi R)/(T) = [(2\pi × 12.6 × 10^8 ft)/(23.4 × 10^5 sec)] = 3360 ft/sec . Its radial acceleration is therefore a = v^2/R = [(3360 ft/sec)^2]/[12.6 × 10^8 ft] = 0.00896 ft/sec^2 = 8.96 × 10^-3 ft/sec^2 (b) The ratio of the moon's acceleration to the acceleration of a falling body at the earth's surface is: a/g = (8.96 × 10^-3 ft/sec^2)/(32.2 ft/sec^2) = 2.78 × 10^-4 The ratio of the square of the earth's radius to the square of the moon's orbit is: (2.09 × 10^6 ft)^2/(12.6 ×10^8 ft)^2 = 2.75 × 10^-4 The agreement is very close, although not exact because we have used average values.

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Question:

How would you explain the secondary response to antigens on the basis of the clonal selection theory?

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/Users/wenhuchen/Documents/Crawler/Biology/F14-0366.htm

Solution:

When an antigen is first injected into an animal, only a small number of lymphocytes become trans-formed into plasma cells. The amount of antibodies produced rises slowly to a low peak and then decreases. This is termed the primary response. If a second injec-tion of the same antigen occurs some weeks later, a much more vigorous response is induced. Not only is the amount of antibodies much greater than in the primary response, but their production is much more rapid (see Fig.). Further injections will result in the maximum amount of antibody that can be produced. Primary and secondary responses of antibody formation can be explained on the basis of the clonal selection theory. When an antigen first comes in contact with its corresponding lymphocyte, it stimulates the lymphocyte to divide and differentiate into plasma cells (or lymphoblasts). The antibodies Of these cells interact with the antigen to form complexes which are engulfed and destroyed by macrophages. Some days after the primary response, these plasma cells (or lymphoblasts) eventually die. However, the antigen caused production of many more copies of the originally stimulated small lymphocytes. It is thought that these lymphocytes, called memory cells, arise from division of the originally stimulated cells without concomitant dif-ferentiation into plasma cells (lymphoblasts). This immunological memory explains the process in which an animal "remembers" previous exposure to an antigen. When the animal is exposed to the antigen the second time, many more lymphocytes (specific for the antigen) are stimulated, resulting in a greater and more rapid production of anti-bodies. This explains the characteristics of the secon-dary response.

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Question:

Mostmarine vertebrates have body fluids with osmotic marine vertebrates have body fluids with osmotic pressure lower than that of their saline environment. How do these organismsosmoregulatein face of aprepetualthreat of dehydration?

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Solution:

To survive their hostile saline environment, marine organisms show a variety of adaptations. The concentration of body fluids in marine fish is about one-third that of the sea water. Osmotic pressure, being higher in sea water, tends to drive water out of the body fluid of the fish.One site of severe water loss is the gill, which is exposed directly to the surrounding water for gaseous diffusion. To compensate for this loss, the fish must take in large amounts of water. How-ever, the only source of water available is sea water; if consumed, the sea water would cause further water loss from the body cells because of its higher osmotic pressure. Marine fish, however, overcome this problem. In the gills, excess salt from the consumed sea water is actively transported out of the blood and passed back into the sea. Marine birds and reptiles also face the same problem of losing water. Sea gulls and penguins take in much sea water along with the fish which they scoop from the sea. To remove the excess salt, there are glands in the bird's head which can secrete saltwater having double the osmolarityof sea water. Ducts from these glands lead into the nasal cavity and the saltwater drips out from the tip of the bill. The giant sea turtles, a marine reptile, has similar glands but the ducts open near the eyes. Tears from the eyes of turtles are not an emotional response, in contrast to popular belief. Rather, 'crying' is a physiological mechanism to get rid of excess salt.

Question:

Define a buffering system. What significance does it have in the living cell?

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/Users/wenhuchen/Documents/Crawler/Biology/F01-0027.htm

Solution:

A buffering system is one that will prevent significant changes in pH, upon addition of excess hydrogen or hydroxide ion to the system. Buffering systems are of great importance in the maintenance of the cell. For example, the enzymes within a cell have an optimal pH range, and outside this range, enzymatic activity will be sharply reduced. If the pH becomes too extreme, the enzymes and proteins within the cell may be denatured, which would cause cellular activity to drop to zero and the cell would die. Hence a buffering system is essential to the existence of the cell. The average pH of a cell is 7.2, which is slightly on the basic side. A buffer will prevent significant pH changes upon variation of the hydrogen ion concentration by abstracting or releasing a proton. Relatively weak diprotic acids (Ex: H_2CO_3, carbonic acid) are good buffers in living systems. When carbonic acid undergoes its first dissociation reaction, it forms a proton and a bicarbonate ion: H_2CO_3 \rightleftarrows H^+ + HCO_3^-. The pH of the system is due to the concentration of H^+ formed by the dissociation of H_2CO_3. Upon addition of H^+, the bicarbonate ion will become protonated so that the total H^+ concentration of the medium remains about the same. Similarly, upon addition of ^-OH, the bicarbonate ion releases a hydrogen to form water with the ^-OH. This maintains the H^+ concentration and hence keeps the pH constant. These reactions are illustrated below:

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Question:

The sodium D-lines, two very distinct lines of yellow light emitted by heated sodium metal, are often used in the calibration of spectrometers. The wavelength of one of these lines is 5890 \AA. How much energy does an electron emit (or absorb) in undergoing the electronic transition associated with this line?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E18-0689.htm

Solution:

To solve this problem we will combine two relationships: ѵ = c/\lambda, where ѵ is the frequency, \lambda the wavelength, and c the speed of light; and Planck's re-lationship, E = hѵ, where E is the energy and h is Planck's constant, 6.62 × 10^-27 erg-sec. Then, E = hѵ = hc/\lambda. Since 1 \AA = 10^-8 cm, 5890 \AA = 5890 \AA × 10^-8 cm/\AA = 5.890 × 10^-5 cm. The energy associated with the 5890 \AA sodium D-line is then E = (hc/\lambda) = [(6.62 × 10^-27 erg-sec × 3.0 × 10^10 cm/sec) / (5.89 × 10^-5 cm)] = 3.37 × 10^-12 erg.

Question:

Distinguish betweenpenetranceexpressivity.

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Solution:

A recessive gene produces a given trait when it is present in the homozygous state. A dominant gene produces its effect in both the homozygous and heterozygous states. Geneticists, however, have found that many genes do not always produce their phenotypes when they should. Genes that always produce the expected phenotype in individuals who carry the gene in an expressible combi-nation are said to have completepenetrance. If only 70 percent of such individuals express the characterphenotypically, then the gene is said to have 70 percent penetrance .Penetranceis thus defined as the percentage of individuals in a population who carry a gene in the correct combination for its expression (homozygous for recessive, homozygous or heterozygous for dominant) and who express the genephenotypically. Some genes that are expressed may show wide variations in the appearance of the character. Fruit flies homozygous for the recessive gene producing shortening of the Wings exhibit variations in the degree of shortening. Expres-sivity is defined as the degree of effect or the extent to which a gene expresses itself in different individuals. If it exhibits the expected trait fully, then the gene is said to be completely expressed. If the expected trait is not expressed fully, the gene shows incomplete expres-sivity. The difference between the two terms-penetranceand expressivity, lies in the fact that the former is a function of the gene at the population level, while the latter varies on an individual level. Thus a given gene having a certainpenetrancewithin a population may have varying expressivity in individuals of that population who express it. Bothpenetranceand expressivity are functions of theinterractionof a given genotype with the environment. Changes in environmental conditions can change both thepenetranceand expressivity of a gene. For example, a given gene may code for an enzyme required for the synthesis of a given metabolite in bacteria. If that metabo-lite is provided in the organisms nutrient environment, the organism might not produce the enzymes needed for its synthesis, and then the gene will not be expressed. If however, the nutrient is depleted from the media, the organism will begin to manufacture the enzyme, and thus express the gene. In humans, it is thought that allergy is caused by a single dominant gene; the different types of allergies are due to the varying expressivity of the gene, as a result of the interaction of the gene with both the environment and a given individual's genetic and physical make up.

Question:

A nail of mass M is driven into a board against a constant resistive force F by a hammer of mass m which is allowed to fall freely at each stroke through a height h. The hammer does not rebound after striking the nail. Find the distance the nail is driven in at each blow. Show that the total energy expended in raising the hammer during the operation of driving the nail fully in to a depth d is independent of the value of h, and can be decreased by making the hammer more massive.

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/Users/wenhuchen/Documents/Crawler/Physics/D06-0329.htm

Solution:

The hammer falls through a height h, losing potential energy and gaining kinetic energy. By the principle of conservation of energy, mgh = 1/2 mv^2,and that the hammer strikes the nail with a velocity v, where v = \surd(2gh). An inelastic collision takes place and momentum is conserved, so thatmv= (m + M)V. Therefore the kinetic energy of hammer and nail together, after the impact, is (1/2) (m + M)V^2 = (1/2) [m^2/(m + M)] v^2. When the hammer hits the nail, we assume it loses all its kinetic energy. This energy is converted into the work necessary to drive the nail a distance x against the re-sistive force F. If the hammer drives the nail this distance x at each stroke, then we have from the work- energy theorem, W = F^\ding{217} \bullet x^\ding{217} = -Fx0 - (1/2) [m^2/(m + M)] v^2 since the final kinetic energy is zero and F opposes the nail's displacement. Solving for x and substituting v = \surd(2gh), we have x = (m^2v^2) / [2F(m + M)] = [ghm^2] / [F(m + M)]. If it requires n strokes to drive the nail fully home a distance d, then d =nx= [nghm^2] / [F(m + M)].(1) But the energy E supplied to the system is that energy required to raise the hammer n times through a height h. Thus E =nmgh In equation (1), solve fornmgh. Then E = [Fd(m + M)] / m =Fd+Fd(M/m). Thus the energy does not depend on h and will be decreased if m is made larger.

Question:

Parturition is the separation of the fetus and its membranes from the mother's body at the time of birth. Describe this process.

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Solution:

As the period of gestation ends, a hormonal substance termed relaxin appears in the bloodstream. This substance seems to be a product of the corpusluteumof pregnancy and of the placenta and causes the cervix of the uterus to become readily dilatable.Oxytocin, one of the hormones released from the posterior pituitary, is an ex-tremely potent uterine-muscle stimulant.Oxytocinisreflexlyreleased into the bloodstream as a result of afferent input to the hypothalamus from receptors in the uterus, par-ticularly in the cervix.Oxytocinstimulates the uterine muscles to contract with greater force and frequency during labor. Relaxin andoxytocinare two important hormones in-volved in parturition. At the time of birth, rhythmic contractions of the smooth muscle of the uterus begin. Soon they are accompanied by reflex and voluntary activity of skeletal muscle. The result is that internal pressure forces the fetus, normally head first, into the cervix of the uterus. The fetal membranes then bulge through the cervix and burst. Amniotic fluid is discharged. The child then enters the vagina and is born, a process lasting from twenty minutes to one hour. Some fifteen minutes after birth, the uterus again goes into rhythmic contraction, and an "afterbirth" consisting of the fetal membranes is expelled. Surprisingly enough, there is only a minor amount of bleeding. Within the next few days, the lining of the uterus is restored and the uterus returns to nearly its original dimensions.

Question:

In certain abnormal conditions, the stomach does not secretehydrochloric acid. What effects might this haveon thedigestive process?

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/Users/wenhuchen/Documents/Crawler/Biology/F17-0429.htm

Solution:

The highHClconcentration in the stomach serves several importantfunctions. Without the acid, the gastric enzymes would show no activity. For example, pepsin, an important protein-digesting enzyme, wouldremain inactive. Pepsin exists in the form of a precursor compound, calledpepsinogen, which has noproteolyticactivity in the absence of acidity. Thepepsinogenwill remain in the stomach as is, untilHClis secretedand acti-vates it. Thus the absence ofHClwould lead to some lossof protein digestion. This loss, however, can be made up by the intestinalproteases. HClis also important for its direct effect on the conformation of proteins. The high acidity ofHClde-natures proteins.Denaturationdoes notmean cleavage of a molecular bond, but rather an unwinding of the molecule. Proteins are normally tightly packed, wovenchains, thatare difficultfor enzymes to attack. The acid alters the protein to facilitate enzymeaction. Another function of stomach acid is to kill ingested bacteria. Many ofthe bacteria ingested are harmless, but some, if allowed to grow, could haveserious effects.HClis also important to pancreatic secretion. Pancreatic juice secretion is controlled, in part, by the hormonesecretin. TheHClsecreted by the stomach glands passes through the pyloricsphincter, along with the food, into the duodenum. Some of the acidstimulates special receptor cells in the intestinal lining. These cells, in turn, secretesecretininto the blood. When thesecretinreaches the pancreas, a pancreatic juice rich in bicarbonate travels down the pan-creaticduct into the intestine. If noHClis present, there will be less thanone quarter of the secretion, and less one quarter of the normal amountof digestion.

Question:

In acommensalisticrelationship between two species living symbioticallywhat are the consequences of their interactions?

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Solution:

Commensalism is a relationship between two species in which one speciesbenefits while the other receives neither benefit nor harm. The advantagederived by thecommensalspecies from its association with the hostfrequently involves shelter, support, transport, food or a combination ofthese. For example, in tropical forests, numerous small plants called epiphytesusually grow on the branches of the larger trees or in the forks oftheir trunks. They use the host trees only as a base of attachment and donot obtain nourishment from them. A similar type of commensalism is theuse of trees as nesting places by birds. Such relationships do not produceany apparent harm to the hosts. Commensalism is widespread throughout the animal kingdom, and isespecially common among the marine in-vertebrates. The host organismsare typically slow- moving or sessile and live in shells or burrows, which can be readily shared by the smallercommensalspecies. One example is a certain species of fish that regularly lives in association withsea anemones, deriving pro-tection and shelter from them and sometimessharing some of their food. These fish swim freely among the tenta-clesof the anemones even though these tentacles quickly paralyze otherspecies of fish that touch them. The anemones feed on fish, yet the particularspecies that live ascommensalswith them can actually enter thegastrovascularcavity of their host, emerging later with no ill effects. This implies a certain amount of physiological and behavioral adaptation betweenmembers making up acommensalrelationship. Still another example is a small tropical fish (Fierasfer) that lives in therectum of a particular species of sea cucumber. The fish periodically emergesto feed and then returns by first poking its host's rectal opening withits snout and then quickly turning so that it is drawn tail first into the rectal chamber.

Question:

A chemist reacts ferric sulfate with barium chloride and obtains barium sulfate and ferric chloride. He writes the following balanced equation to express this reaction: Fe_2 (SO_4)_3 + 3BaCl_2 \rightarrow 3BaSO_4\downarrow + 2FeCl_3 (A)How much BaC1_2 should be used to react with 10 grams of Fe_2 (SO_4)_3? (B) How much Fe_2 (SO_4)_3 will be necessary to produce 100 g of BaSO_4? (C) From a mixture of 50 g of Fe_2 (SO_4)3 and 100 g of BaC1_2, how much FeC1_3 can be produced?

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Solution:

To answer these questions, you must understand the mole concept, and how it is used to calculate the amount of material required in a chemical reaction. A mole is defined as the number of grams of a sub-stance divided by its molecular weight. In other words, a mole = [(amount in grams of substance) / (molecular weight of substance)] In the given equation: Fe_2 (SO_4)_3 + 3BaCl_2 3BaSO_4\downarrow + 2FeCl_3, the numbers before each compound are termed coefficients (the coefficient of Fe_2 (SO_4)_3 is one, and by convention is not written). The equation is balanced. This means that all of the elements are equal in number on both sides of the equation. For example, we have a total of 12 oxygen atoms on the left and 12 on the right. Before doing any problem involving a chemical equation, we must always balance it. All the atoms must be accounted for. The coefficients serve this purpose. After balancing, they also tell you the relative mole amounts that will react. In other words, in this reaction 1 mole of Fe_2 (SO_4)_3 reacts with 3 moles of BaCl_2. For the reaction to occur, we must have a mole ratio of 1 : 3 between Fe_2 (SO_4)_3 and BaCl_2. If such a condition exists, then the equation tells us that 3 moles of BaSO_4 will be produced per mole of Fe_2 (SO_4)_3. In addition, 2molesof FeCl_3 will also be produced. There-fore, the coefficients tell the relative number of moles of each substance that must be either present or produced. With this information, we can now answer the questions. (A)We have 10 grams of Fe_2 (SO_4)_3 and want the number of grams of BaCl_2 that are required to react with it. The molecular weight of Fe_2 (SO_4)_3 is 399.88 g/mole. This number is obtained by adding up all the atomic weights of the individual atoms in the formula. Recalling the definition of a mole, we have 10/399.88molesof Fe_2 (SO_4)_3. The equation tells us that 3 moles of BaCl_2 react with one mole of Fe_2 (SO_4)_3. Therefore, we must have 3 times the number of moles of Fe_2 (SO_4)_3or 3 × [(10)/(399.88)] = [(30)/(399.88)]. The number of BaCl_2 required is, therefore, [(30)/(399.88)]. The molecular weight of BaCl_2 is 208.24 g/mole. Recall, mole = number of grams/mole-cular weight. The number of grams of BaCl_2 required is, therefore, [(30) / (399.88)] moles × 208.24 g/mole = 15.62 grams (B) We want to produce 100 grams of BaSO_4. The molecular weight of BaSO_4 is 233.40 g/mole. Therefore, we want to produce 100/233.40molesof BaSO_4. How much Fe_2 (SO_4)_3 should be used? Again, we must go back to the equation and look at the coefficients to determine the mole requirements. We see that 3 moles of BaSO_4 are pro-duced for every mole of Fe_2 (SO_4)_3. Therefore, the required number of moles of Fe_2 (SO_4)_3 is 1/3 the number of moles of BaSO_4 or (1/3) × [(100) / (233.40)] = [(100)/(700.20)]. The molecular weight of Fe_2 (SO_4)_3 is 399.88 g/mole. Consequently, the number of grams required of Fe_2 (SO_4)_3 is [(100) / (700.20) mole × 399.88 g/mole = 57.11 grams (c) In this question, we are working with a mixture of two substances to produce a third. The first thing is to compute the moles: Moles Fe_2 (SO_4)_3= [(50.00 g) / (399.88)] = .125 moles Moles BaCl_2= [(100 g) / (208.24 g/mole)] = .48moles Because we are dealing with two substances to yield one, we must also consider their mole ratios as well as with the product FeC1_3. The equation calls for three moles of BaC1_2 to react with one mole of Fe_2 (SO_4)_3 We have .125 moles of Fe_2 (SO_4)_3. 3 times this amount is .375moles. However, you have .48 moles of BaCl_2. This means, there-fore, that we have .48 - .375 or .105. moles of BaCl_2 that will not react. In other words, BaCl_2 is in excess. We always take the limiting reagent, which is Fe_2 (SO_4)_3. The amount of FeCl_3 present will reflect the number of moles of Fe_2 (SO_4)_3, and not BaCl_2. Because 2 moles of FeCl_3 will be produced for every mole of Fe_2 (SO_4)_3, .125 × 2 = .25 moles of FeCl_3 will be produced. The molecular weight of FeC1_3 is 162.20. There-fore, the maximum weight of FeCl_3 produced is .25(162.20) = 40.56 grams.

Question:

How many liters ofphosphine(PH_3) gas at STP could be made from 30 g of calcium by use of the following sequence of reactions: 3Ca + 2P\rightarrowCa_3 P_2 Ca_3 P_2 + 6HCl\rightarrow2PH3 + 3CaCl_2 (Molecular weights: Ca = 40, PH_3 = 34.)

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Solution:

From thestoichiometryof these two equations 3 moles of Ca will yield 2 moles of PH_3 gas. Thus, if one knows the number of moles of Ca, then one can determine the number of moles of PH_3 produced. The number of moles of Ca given is its weight divided by its molecular weight. Therefore, number of moles Ca = [(30 g Ca) / (40 g/mole)] = 0.75 moles From 0.75 moles of Ca, one produces 2/3 as much or 2/3(0.75) = 0.50 moles of PH_3 gas. At STP, one mole of any gas occupies 22.4 liters, hence, the total volume of PH_3 produced at STP = (22.4 l/mole) (0.50 moles PH_3) = 11.2 l.

Question:

Outline briefly the events occurring in each stage of meiosis. Illustrate your discussion with diagrams if necessary.

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/Users/wenhuchen/Documents/Crawler/Biology/F25-0643.htm

Solution:

Meiosis is the process by which diploid organisms (having two sets of chromosomes) produce haploid gametes (having only one set of chromosomes). When two gametes fuse in fertilization, the zygote formed will thus have the full diploid chromosomal complement. Meiosis consists of two cell divisions, the first (Meiosis I) called a reduction division, and the second (Meiosis II) a mitotic type division. Interphase I:This phase is similar to mitotic interphase The cell appears inactive in reference to cell division; but it is during interphase that chromosome duplication occurs. Prophase I:The chromosomes become thicker and more visible. While they are still long thin threads, an attractive force (as yet not identified) causes homologous chromosomes to come together in pairs, a process known as synapsis. This is the stage during which cross-over between homologous chromosomes will occur. After synapsis, the chromosomes continue to shorten and thicken; their double nature becomes visible, so that each homologous pair appears as a bundle of four chromatids called a tetrad. Each tetrad is composed of two doubled homologous chromosomes. The number of tetrads is thus equal to the haploid number. The cent-romeres of homologous chromosomes are connected, and there are thus two centromeres for the four chromatids. While these events are occurring, the centrioles migrate to opposite poles, the spindle begins to form between them, and the nucleolus and nuclear membrane dissolve. The tetrads move to the equatorial plane of the spindle. Metaphase I:Migration to the equatorial plane is complete, and the nuclear membrane and nucleolus have dissolved. Anaphase I:At this point the homologoues chromosomes that had paired in prophase separate and move to opposite poles of the cell. Each is still composed of two identical daughter chromatids joined at the centromere. Thus the number of chromosome types in each resultant cell is reduced to the haploid number. Telophase I:Cytoplasmic division occurs as in mitosis. Meiosis I concludes and meiosis II begins. There is no definable interphase between the two series of divisions. The chromosomes do not separate or duplicate, nor do they form chromatin threads. Prophase II:The centrioles that had migrated to each pole of the parental cell, now incorporated in each haploid daughter cell, divide, and a new spindle forms in each cell. The chromosomes move to the equator. Metaphase II:The chromosomes are lined up at the equator of the new spindle, which is at right angles to the old spindle. Anaphase II:The centromeres divide and the daughter chromatids, now chromosomes, separate and move to opposite poles. Telophase II:Cytoplasmic division occurs. The chromo-somes gradually return to the dispersed form and a nuclear membrane forms. The two meiotic divisions yield four cells, each carrying only one member of each homologous pair of chromosomes. These cells are for this reason called haploid cells.

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Question:

Consider the following variant ofNim. A pile of N objects is given. Two players take turns removing 1, 2 or 3 objects fromthe pile. The person who has to take the last object, loses. Write a Basic program which enables you to play this gamewith a computer.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G23-0566.htm

Solution:

Let us consider 2 players: you and the computer. The computer losesif it takes the last object. The question now is: How many objects mustyou leave to the computer at its penultimate move to ensure your win? The answer is:1 + the maximum number of objects always possible toremove in one move by each you and the computer. This number is 4, becauseif the computer removes the minimum possible number of objects (i.e., 1), and you - the maximum (i.e., 3) - that will give you the desired number. So, by leaving the compu-ter with 5 objects at its penultimate move, you ensure your win (providing you do not make a mistake). In fact, ifyou leave the computer with 1 + 4N (where N = 0,1,2,3...) objects before everymove, you guarantee yourself a win by removing (4 - A) objects, where"A" is the number of objects removed by the computer. Using this strategyand providing that you move first, you are guaranteed to win everytime, except when there are 4N objects in a pile. In that case, by usingthe same strategy, the computer can beat you. 10PRINT "31 OBJECTS GAME" 20PRINT "LET'S FLIP A COIN TO SEE WHO GOES FIRST." 25PRINT "IF IT COMES UP HEADS, I WIN THE TOSS." 30N = 31 40Q = INT (2\textasteriskcenteredRND (5)) 50IF Q = 1 THEN 80 60PRINT "TAILS! YOU GO FIRST" 70GOTO 140 80PRINT "HEADS! I GO FIRST" 90PRINT "I TAKE TWO OBJECTS" 100N = N - 2 110PRINT "THE NUMBER OF OBJECTS IS NOW" N 120PRINT "YOUR TURN. YOU MAY TAKE 1, 2 OR 1303 OBJECTS." 140PRINT "HOW MANY DO YOU WISH TO REMOVE?", 150INPUT K 160IF K > 3 THEN 360 170IF K < = 0 THEN 360 180N = N - K 190PRINT "THERE ARE NOW" ;N; "OBJECTS REMAINING." 200IF N = 4 THEN 260 210IF N = 3 THEN 280 220IF N = 2 THEN 300 230IF N < = 1 THEN 400 240Z = 4 - K 250GOTO 320 260Z = 3 270GOTO 320 280Z = 2 290GOTO 320 300Z = 1 320PRINT "MY TURN. I REMOVE" Z "OBJECTS" 330N = N - Z 340IF N < = 1 THEN 380 350GOTO 110 360PRINT "IMPOSSIBLE. HOW MANY; 1, 2 OR 3" 370GOTO 150 380PRINT "YOU LOST" 390GOTO 410 400PRINT "YOU WERE LUCKY, YOU WON" 410STOP 420END

Question:

Describe the life cycle of a typical fern plant.

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/Users/wenhuchen/Documents/Crawler/Biology/F07-0184.htm

Solution:

The fern plant, like other vascular plants, has a life cycle in which the sporophyte generation is dominant. The spprophyte generation is the relatively large, leafy green plant we recognize as the fern plant growing in both the tropics and temperate regions. The sporophyte consists of a horizontal stem, or rhizome, lying at or just under the surface of the soil and bearing fibrous roots. From the rhizome grows several leaves or fronds, usually finely divided into pinnae. On the under surfaces of certain pinnae develop small, brown sori. A sorus is a cluster of sporangia or spore cases in which haploid spores are formed. Most modern ferns are homosporous, that is, all their spores are alike. When mature and proper conditions, the spores are shed. After germination, each spore develops into a gametophyte which is typically tiny, thin, and often heartshaped. Small and obscure as it is, the fern gametophyte, also called a prothallus, is an independent photosynthetic organism. It grows in moist, shady places, especially on decaying logs and on moist soil and rocks. A number of rhizoids grow from the gametophyte into the substrate, anchoring it and absorbing water and salts. The male and female sex organs (antheridia and archegonia) develop on the under-surface of the game-tophyte. Each archegonium, usually located near the notch of the heart-shaped prothallus, contains a single egg. The antheridia, located at the other end of the gametophyte, each develop a number of flagellated sperm, which are ovoid in shape and have many flagellae on a spiral band at their anterior end. The sperm, released after a rain and attracted by a chemical substance released by the archegonium, swim through a film of moisture to the egg. Although the fern plant is usually monoecious (that is, it possesses both the male and female sex organs), the sperm of one plant usually fertilizes the egg of another, thus accomplishing cross--fertilization. The zygote begins to develop within the archegonium into a sporophyte embryo. At first the sporophyte develops as a parasite on the gametophyte, but it soon acquires its own roots, stems, and photosynthetic leaves and becomes an independent sporophyte, thus completing the life cycle.

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Question:

What hormone, if any, acts antagonistically to the parathormone?

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/Users/wenhuchen/Documents/Crawler/Biology/F21-0546.htm

Solution:

In 1961 a hormone called calcitonin was discovered, and was found to be directly antagonistic to parathormone, the hormone secreted by the parathyroid glands. Calcitonin, also called thyrocalcitonin, is secreted by cells within the thyroid gland which surround but are completely distinct from the thyroxine secreting cells. It is the function of calcitonin to lower the plasma calcium concentration. This is achieved primarily by the deposition of calcium-phosphate in the bones. Parathormone has an exactly opposite effect in calcium- phosphate metabolism. The secretion of calcitonin is regulated by the calcium content of the blood supplying the thyroid gland. When blood calcium rises above a certain level, there is an increase in calcitonin secretion in order to restore the normal concentration of calcium. In lower invertebrates calcitonin is produced by separate glands, which in mammals become incorporated into the thyroid during embryonic development.

Question:

What is the flux density B^\ding{217} in the region between the plates of a circular capacitor that is being charged?

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/Users/wenhuchen/Documents/Crawler/Physics/D19-0638.htm

Solution:

Figure 2 represents a plane midway between the plates of the capacitor in Fig. 1. During the process of charging, an effective current will pass across the plates as a result of charge transfer between the plates over a finite period of time. The electric field inside the capacitance will increase in time from zero to a maximum value, and give rise to the displacement current density J^\ding{217}D, which is given by Maxwell's equations density J^\ding{217}D, which is given by Maxwell's equations as \mu_0\epsilon_0 =\partialE^\ding{217} / \partialt = J^\ding{217}D The magnetic field due to J\ding{217}Dcan be obtained from Stoke' s theorem \ding{217} \int_C B^\ding{217} \bullet dr^\ding{217} = \mu0\int_S J^\ding{217}D\bullet dA^\ding{217}(1) where B^\ding{217} is the magnetic induction For the first integral, let the circle of Fig. 2 be the contour C about which we evaluate the left side of (1). The integral involving J^\ding{217}_D is then evaluated about the area S enclosed by C. Because of the circular symmetry of the charge distribution on the plates, i.e., the uni-formity of the J^\ding{217}_D lines, the induced magnetic field B^\ding{217} will be circular. Then, we find using (1), that at every radius r 2\pir B_r = \mu0ID,r,Br =(\mu_0 / 2\pi)[(I_D, r) / r)] where I_D, r is the displacement current passing through the indicated surface. If r is less than the radius of the plates, then I_D, ris smaller than the conduction current I_C, because not all the electric field lines go through S. In the idealized case in which fringing fields are neglected, and r equals the radius of the plates, the displacement current equals the conduction current and B_r = (\mu_0 / 2\pi)( I_C, / r) which is the same as the field at a distance r^\ding{217} from an infinitely long straight conductor.

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Question:

Suppose the velocity of the particle in the diagram is given by the equationv\ding{217} = [m + nt^2] where m = 10 cm/s and n = 2 cm/s^3, (a) Find the change in velocity of the particle in the time interval between t_1 = 2 sec and t_2 = 5 sec. (b) Find the average accelera-tion in this time interval. (c) Find the instantaneous acceleration at t_1 = 2 sec.

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/Users/wenhuchen/Documents/Crawler/Physics/D03-0069.htm

Solution:

(a) At time t_1 = 2 sec v_1\ding{217}= [10(cm/s) + {(2cm)/(s^3)} (2s)^2] v_1\ding{217}= 18(cm/s) (a) Particle moving on the x-axis. (b) Velocity time graph of the motion. The average acceleration between t_1, and t_2 equals the slope of the chord pq. The instantaneous acceleration at p equals the slope of the tangent at p. whereis a unit vector in the positive x direction. At time t_2 = 5 sec v_2\ding{217} = [10(cm/s) + (2cm/s^3)(5s)^2 ] v_2\ding{217} = 60cm/s The change in velocity is therefore v_2\ding{217} - v_1\ding{217} = (60cm/s)- (18cm/s)= (42cm/s) (b) The average acceleration is defined as a_avg\ding{217} = (v_2\ding{217} - v_1\ding{217})/(t_2 - t_1) where v_2\ding{217} and v_1\ding{217} are the velocities at t_2 and t_1, respec-tively. Hence, in the given interval a_avg\ding{217} = [{(60cm/s) - (18cm/s)}]/(5s - 2s) = 14(cm/s^2) This corresponds to the slope of the chord pq in the diagram. (c) The instantaneous acceleration is a\ding{217} = dv\ding{217}/dt = 2nt At t_1 = 2 sec, a= (2)(2cm/s^3)(2s) = 8 cm/s^2 This corresponds to the slope of the tangent at point P in the figure.

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Question:

We consider a particle with charge q which moves along the x axis with a high momentum p_0 and enters a region of length L in which there is a transverse electric field \epsilon\^{y}. Find the angle through which the particle is deflected by the electric field.

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/Users/wenhuchen/Documents/Crawler/Physics/D19-0631.htm

Solution:

Since we are given .the momentum and not the mass of the particle, we use a momentum-energy approach. We first note that we are given the electric field \epsilon. We can translate this into the force exerted on the particle by multiplying the field by the charge on the particle (q\epsilon) . But, force is the derivative of momentum. Since the field is only present in the y direction, the resulting force is exerted solely in the y direction. Thus, d(p)x/ dt = 0 ;d(p)y/ dt = q\epsilon, Integrating, we have p_x = p(o)x, py= q\epsilont + p_(o)y where p(o)xis the initial x momentum, and p_(o)y is the initial y momentum. Since the particle initially has no y velocity p(o)y= 0 andpx= p(o)xp_y = q\epsilont(1) We want to find the velocity v. If we can find the energy E, then we can find the velocity from the momentum by using the relation v = pc2/ E. The energy is given by E2= M^2c4+ p^2c^2 + M^2c4+ (p^2_(0)x + q^2 \epsilon^2 t^2)c2 from (1). But ^2E_0 = M^2c4+^2p_0 c^2and, then , E2= (M^2c4+ p^2_(0)x c^2) + (q\epsilontc)2 Since p_0 = p_(0)x E2= E20+ (q\epsilontc)2(2) where E_0 is the intial energy. Therefore from (1),(2) and the velocity- momentum relation we have v_x = p_(x)c^2 / [E20+ (q\epsilontc)^2]^1/2 =p_0(x)c^2 / [E^2_0 + (q\epsilontc)^2]^1/2 v_y = p_(y)c^2 / [E20+ (q\epsilontc)^2]^1/2 =q\epsilontc^2 / [E^2_0 + (q\epsilontc)^2]^1/2 Note that v_x decreases as t increases. At a time t, the angle \texttheta^X the trajectory makes with the x axis is given by tan \Phi (t) = v_y / v_x = q\epsilontc^2 / p_0 c^2 =q\epsilont / p_0 .

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Question:

The gold leaves of the machine in the figure periodically diverge and then collapse. Describe how this perpetual- motion machine operates and explain why it is not a perpetual-motion machine.

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/Users/wenhuchen/Documents/Crawler/Physics/D34-1017.htm

Solution:

Radium atoms are radioactive and have a half- life of 1,622 years. For radium ^226 _88 Ra \ding{217}^222 _86 Rn +4_2 He(1) The alpha particle (the ^42He nucleus) with 2 quanta of positive charge, is ejected with a kinetic energy of 4.77 MeV and escapes from the surface of the metal. Therefore, if a piece of radium is placed on an electroscope stem, the electroscope becomes negatively charged. This is due to the fact that only a part of the radium atom is emitted as an alpha particle. The num-ber of electrons around the radium nucleus remains the same. Thus, for each alpha particle that escapes 2 quanta of ne-gative charge are left on the electroscope mount and gold leaves. As radium atoms decay, the leaves slowly diverge due to the mutual repulsion of the negative charge on the two foil strips. At the instant they touch the grounded metal surface of the container, the negative charge mi-grates from the leaves. The gravitational forces cause the leaves to collapse to their original position. The leaves immediately begin to diverge again, repeating the cycle over and over until there are no more radium nuclei present. Since the half-life of radium is 1,622 years, the machine would "run" for many years without any apparent slowing down. From today's perspective, though, we see that the energy source is derived from the nuclear energy released in the nuclear reaction described in (1). This energy is converted to mechanical motion of the gold leaves. There-fore, the device is not a perpetual-motion machine.

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Question:

Explain what is meant by primary, secondary, and tertiary treatment of sewage.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E24-0855.htm

Solution:

Sewage treatment is carried out in three stages known as primary, secondary, and tertiary. Each uses a different process to eliminate different substances from the sewage. Primary treatment uses mechanical means to separate out solid objects such as sticks and rags. These solid objects are removed by means of a coarse screen. The liquid that passes through this screen flows into settling tanks, where insoluble material settles and forms sludge. This removes approximately one third of the pollutants. Whereas primary treatment is mechanical, secondary treatment is biological. The liquid or effluent from the primary facility filters slowly through a bed of rock. Bacteria present consume most of the organic material. In many plants, oxygen is bubbled in to accelerate the growth of bacteria and, as such, the decomposition of the organic waste. This process removes about 90 % of the biodegradable materials in the water. The last type of treatment, tertiary, is the most expensive. It is used to remove organic chemicals, nutrients, and excessive salts. To accomplish this, the treatment may include such processes as chemical coagulation, distillation and reverse osmosis.

Question:

A conductor of capacitance 10^-2 \muF to the ground is insulated from the ground by a silica plate 2.5 mm thick and 5 cm^2 in cross-sectional area. What is the minimum resistivity of the silica if the rate of decrease of potential is to be no greater than 0.1% per min of its instantaneous value?

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/Users/wenhuchen/Documents/Crawler/Physics/D20-0695.htm

Solution:

In this problem, the capacitor is initially charged to its maximum value, and then discharges to ground through the silica plate (see figure (a)). The latter may be considered to be an effective resistance (see figure (b)). Before solving the problem, we must know the voltage across the capacitor as a function of time. Kirchoff's Voltage Law states that the net voltage around a closed circuit loop is zero. Hence, V_cb + V_ba = 0(1) Starting at c, and traversing the loop in the given direction (see figure (b)), we note that c is at a higher potential than b. Hence, V_c > V_borV_b - V_c = V_cb < 0. But, by definition of capacitance C = [(\midQ\mid)/(V_bc )] orV_bc = (\midQ\mid/C) where \vertQ\mid is the absolute value of the charge on 1 plate of the capacitor. Therefore, V_cb = -V_bc = (-Q/C)(2) whereQ > 0.(2a) Furthermore,V_b > V_aorV_a - V_b = V_ab < 0.By Ohm's Law, V_ab = -iR(3) Using (3) and (2) in (1), (-Q/C) - iR = 0(4) But i = [(dQ)/(dt)], where Q is the net charge passing a point of the circuit per unit time. Equation (4) then becomes (1/C) Q + R [(dQ)/(dt)] = 0 orQ + RC [(dQ)/(dt)] = 0(5) Solving (5) (dQ/Q) = -(1/RC) dt \int (dQ/Q) = - (1/RC) \int dt ln Q = - (1/RC) t + F(6) where F is a constant. Taking the exponential of both sides of (6) Q = e- t/RC + F= e^F e^-t/RC Let A = e^F .Then Q(t) = Ae^- t/RC . We find A by noting that at t = 0, Q = Q_0 , the maximum charge C has. Then Q(0) = A = Q_0 whence Q(t) = Q_0 e- t/RC(7) The voltage across C is then, by definition of C V = (\midQ\mid/C) = (Q_0 /C) e- t/RC Note that Q_0 > 0 because Q(t) > 0. [See eq. (2a)]. Defining the initial voltage across C as V_0 = (Q_0 /C) we find V(t) = V_0 e^- t/RC(8) The time rate of change of (8) is [{dV(t)} / dt] = - (1/RC) V_0 e- t/RC= - (1/RC) V(t) But, the question states that [{dV(t)} / dt] \leq (.001 min^-1 ) V(t) or- (1/RC) \leq .001 min^-1 Hence, solving for R - 1 \leq (.001 min^-1 ) RC orR \geq [(-1)/{(.001 min^-1 ) (c)}](9) Now, R is the resistance of the silica, which may be written as R = (\rhol/A)(10) where \rho is silica's resistivity, and l and A are the thickness and cross-sectional area of the plate. Using (9) in (8), and solving for \rho \rho \geq - [(A)/{(.001 min^-1 ) (l) (c)}] Using the given value \rho \geq [(5 × 10^-4 m^2 )/{(.001/605) (2.5 × 10^-3 ) (10^-2 \muF)}] Since 1\muF = 10^-6 F \rho \geq [{(5 × 10^-4 m^2 )(605)} / {(10^-3)(2.5 × 10^-3 m)(10^-8 F)}] \rho \geq 1.2 × 10^12 \Omega \bullet m We have used the fact that1F = 1 C/V. Then \rho_min = 1.2 × 10^12 m \bullet \Omega

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Question:

One of the two most important uses of ammonia is as a reagent in the first step of theOsfwaldprocess, a synthetic route for the production of nitric acid. This first step proceeds according to the equation 4NH_3(g) + 50_2(g) \rightleftarrows 4N0(g) + 6H_20(g) . What is the expression for the equilibrium constant of this reaction?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E09-0304.htm

Solution:

This problem is an exercise in writing the equilibrium constant of a reaction. In general, for a reaction in which re-actants A,B,C,... go to products W,X,Y,... according to the equation aA+bB+cC+ ... \rightleftarrowswW+xX+yY+ ... , wherea,b,c,w,x,y,... are thestoichiometriccoefficients, the equilibrium constant is given by K = { [W]^w [X]^x [Y]^y ...} / { [A]^a [B]^b [C]^c ...} Hence, for the reaction 4NH_3 + 5 O_2 \rightleftarrows 4N0 + 6H_2O , the equilibrium constant is given by K = {[NO]^4 [H_2O]^6 } / { [NH_3]^4 [O_2]^5}.

Question:

What is the fractional increase of mass for a 600─mi/hr jetliner?

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/Users/wenhuchen/Documents/Crawler/Physics/D32-0939.htm

Solution:

Fractional increase of mass is defined as change in mass divided by the original mass or \Deltam/m_0, The equation for the variation of mass with velocity is m = m_0 / (\surd1 ─ \beta^2) ,\beta^2 = v^2 / c^2 Therefore the change in mass is \Deltam = m ─ m0[ {1 / (\surd1 ─ \beta^2)} ─ 1] And \Deltam / m_0 =[ {1 / (\surd1 ─ \beta^2)} ─ 1] For velocities much less than light, \beta = v/c is very small and{1 / (\surd1 ─ \beta^2)} can be approximated by 1 + \beta2/ 2. Since 600 mi/hr is small compared to c, we can say that for this problem the fractional mass is \Deltam/ m_0 \allequal (1 + \beta2/ 2) ─ 1 = \beta^2 / 2 V = 600 mi/hr \allequal 2.7 × 104cm / sec \beta = v / c = ( 2.7 × 104cm / sec) /(3 × 1010cm/sec) \allequal 10^─6 Therefore , \Deltam/ m_0 \allequal 1/2 \beta^2 \allequal .5 × 10─12 so that the mass is increased by only a trivial amount .

Question:

An ancient Egyptian mummy is found. A count of the radiationemitted from the C^14 in the body is only one eighth thatfound in persons who have died recently. About how old isthe mummy? (The half-life of C^14 is 5568 years.)

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/Users/wenhuchen/Documents/Crawler/Biology/F28-0733.htm

Solution:

Certain radioactive elements are transformed spontaneously into otherelements at rates which are gradual and essentially independent of thetemperatures and pressures to which the elements have been subjected. In addition, unstable isotopes of certain elements also undergo gradualtransformation to more stable forms. An example of this is the decayof radioactive isotope carbon-14 into the element nitrogen. This processof transformation is termed radioactive decay. It is established that carbon-14 has a half-life of 5568 years. A half- lifeis defined as the period of time (in units of seconds, minutes, hours, yearsand so forth) that it takes for one half of a sample to undergo radioactivedecay. Since one half of the original sample remains after one half-life, one fourth will remain after two half lives, and one eight after threehalf lives. Therefore, the fact that the mummy contains one- eighth the amountof C^14 of that in persons who have died recently indicates to us thatthree half lives have passed since the mummification. During this period, 7/8 of the total C -14 has decayed into the more stable element, nitrogen. The mummy is hence 3 × 5568 or 16,704 years old.

Question:

What are the differences between the growth patterns of dicotsthat use their cotyledons as an absorption organ and those that use them for both storage and absorption?

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/Users/wenhuchen/Documents/Crawler/Biology/F08-0203.htm

Solution:

In thedicots, the cotyledons can either serve as only an absorption organ or as both an absorption and storage organ. When the cotyledons function only in absorption, the embryo remains relatively small and is surrounded by endosperm. The cotyledons of these plants absorb the stored food of the endosperm. After the seed germinates, and the endosperm has been depleted, these cotyledons develop into leaf-like photosynthetic organs. In otherdicots, such as beans and peas, the cotyledons function in storage as well as absorption of nutrients. The embryo grows until all of the endosperm is absorbed by the cotyledons. Subsequently the cotyle-dons undertake the function of food reserve, resulting in their appearance as enlarged, thickened structures, which are not photosynthetic in function. In thesedicots, the endosperm is usually completely absorbed before the seed germinates. The cotyledons remain as a food supply until the seedling is capable of photosynthesis, at which point they shrivel and fall off.

Question:

Write a FORTRAN program to simulate the daily activity at a gasstation. Assuming that one customer is served every six minutes, what is the average waiting time per customer?

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G24-0571.htm

Solution:

This is a problem in discrete simulation. A random number generatoris used in the program to model the arrival of customers. Since thegas station attendant cannot know if the stream of customers will be continuous, randomness of arrivals is assumed. To calculate the average waiting time, the program should compute the waitingtime of each individual car, add the times, and divide that sum by thetotal number of cars. The first arrival incurs no waiting time, but all subsequentarrivals must wait six minutes for each car in front of them. To obtainan average the service time remaining for each car, the total waitingtime for all the cars, and the number of cars that are lined up must befound. For each simulated minute, the program must reduce the service time SERV by one and check to see if another car has arrived.If so, the waiting timefor that last car is given by SERV. WAIT - the total waiting time, is increasedby adding SERV to it, while the number of cars, CARTOT, is Increased by one, and SERV is increased by six minutes. The parameter N is the number of minutes to be simulated (N cannot contain more than 6 digits). Another assumption made is that customers arrive each minute with a probabilityof 0.1. To simplify the problem, it is assumed that there is only onegas pump at this station. Note, that several values of N can be entered in the DATA section. The programprocedure will be done for each of those time intervals. In order toend the program, enter N \leq 0 as the last value. CGASOLINE LINE SIMULATION GASOLINE LINE SIMULATION INTEGER WAIT, SERV, CARTOT, TIME 5READ (5,1)N 1 FORMATE (I6) CDO WHILE N GREATER THAN ZERO IF (N. LE. 0) GO TO 99 SERV = 0 WAIT = 0 CARTOT = 0 DO 10 TIME = 1,N SERV = MAX (SERV - 1,0) 222 CALL RAND (X) IF (X.GT.0.1) GO TO 22 CARTOT = CARTOT + 1 WAIT = WAIT + SERV SERV = SERV + 6 10CONTINUE AVWAIT = FLOAT (WAIT)/FLOAT (CARTOT) WRITE (6,101) CARTOT,AVWAIT 101FORMAT (IX,'AVERAGE WAIT FOR EACH OF THE', 16,'CUSTOMERS IS', F6.2,'MINUTES.') GO TO 5 CEND DO-WHILE 99STOP END

Question:

Outline the development of the reproductive organs in the human embryo.

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/Users/wenhuchen/Documents/Crawler/Biology/F22-0583.htm

Solution:

The reproductive system develops relatively late in the history of the embryo. In fact, its full development is not complete until after puberty. More-over, the reproductive system remains remarkably labile and is subject to the influence of the sex hormones. At first, male and female embryos cannot be distinguished. This stage is known as the indifferent stage of develop-ment; the primordial gonads have the potential to become either ovaries or testes. If the gonad is to different-iate into an ovary, the outer layer of cells, or cortex, progresses, and the inner core, or medulla, regresses (see Figure 1). If it is to be a testis, the opposite process occurs. In addition, the indifferent embryo has two sets of primordial cells that will give rise to the reproductive ducts. The indifferent germ glands, or gonads, first appear as thickenings of the coelomic epithelium on the side of the embryonic kidney, termed the mesonephros (see Figure 2) and contain the germ cells. There has been much dispute as to the origin of the germ cells themselves. The bulk of the evidence indicates that they are not derived from the initial germinal epithelium but migrate into it via the blood from some other part of the embryo. Soon each gonad protrudes into the body cavity as a longi-tudinal ridge, called the germinal ridge. Cord-like clumps of cells, called the medullary cords, from the germinal epithelium push into the gonad. In the male, the medullary cords push into the core of the gonad and become the seminiferous tubules. Their walls consist of immature germ cells (the spermatogonia) and accompanying nurse or Sertoli cells which serve a nourishing function. Ultimately, the spermatogonia will transform into mature sperm. The seminiferous tubules become connected to the rate testes, a system of thin tubules that develop from the dorsal part of the gonad, and form connections with the adjoining tubules of the mesonephros, (which later become the epididymis). The epididymis is attached to the Diagram showing development of gonads in higher vertebrates. epithelium of the ridge and located partly in the adjacent mesenchyme. (future seminiferous tubules). cortex contains germ cells. vas deferens, whose embryonic origin is a duct in the mesonephros called the Wolffian duct, which opens to the outside. The development of the male system proceeds under the stimulation of testosterone secreted by the interstitial cells of the developing testes. In female embryos, the Wolffian ducts degenerate when the mesonephros stops functioning as an excretory organ. The mesonephros is replaced by the metanephros, which remains as the functioning kidney all throughout life. New formations called Mullerian ducts develop lateral to the Wolffian ducts. These open at the anterior end by a funnel-shaped ostium into the coelomic cavity, and at the posterior end to the outside. The anterior portions become the Fallopian tubes. Farther back, the Mullerian ducts fuse and give rise to the uterus. The posterior parts of the Mullerian ducts unite to produce a single vagina. In females, the mesonephric and Wolffian ducts degenerate while the Mullerian ducts proceed to develop. In the male, the testis appears to secrete, in addition to testosterone, a second hormone, that inhibits the development of the Mullerian ducts so that no oviduct or uterus appears. The external genitals of the male and female are derived from the same embryonic source, and at the in-different stage they cannot be distinguished (see Figure 3). In mammals the genital opening of the embryo is flanked on both sides by elongated thickenings, the genital folds. A genital tubercle (phallus) is located on the ventral end where the genital folds meet. Prom this sexually indifferent condition, under the stimulation of the sex hormones, development proceeds toward the male or female condition. In the male, the phallus grows large and elongates to form the penis. A groove on its ventral surface is bounded by the genital folds. These close in beneath and form a tube, extending from the urethra to the tip of the penis. In the female, the phallus is smaller and is known as the clitoris. The genital folds do not fuse but remain as the labia minora. An outer fold bordering each labia minor develops into the labia majora, which is comparable to the skin of the scrotum.

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Question:

Name an organism which produces spores in its life cycle and explain how this aids in its proliferation.

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Solution:

While spores are primarily used as a means of reproduction in plants, there is one class of protozoans (the Sporozoa) which produces spores. These spores are highly resistant forms of the organism which can withstand extreme environmental conditions and are specialized for asexual reproduction. The sporozoanPlasmodiumis a spore-forming protozoan that causes the disease malaria.Plasmodiumspores are produced initially by sexual reproduction in the body of a mosquito (see figure 1) which has sucked the blood of an individual infected with the organism. These spores (sporo-zoites) may now be injected into another individual's bloodstream when the mosquito bites the new host. After invading the red blood cells, the spores reproduce asexually, producing many spores (merozoites) within each invaded cell. This causes the cell to burst, releasing the spores. Each is able to invade another red blood cell. It is thus through repeated sporulation that the organism proliferates in the bloodstream of the infected individual. Some merozoites develop into male and female gametocytes. It is these gameto-cytes which a mosquito sucks up when she bites. In the mos-quito the gametocytes develop into gametes. By sexual repro-duction, the gametes fuse into zygotes which become sporozoites and the cycle repeats.

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Question:

A metal sphere of radius 5 cm has an initial charge of 10^-6 coul. Another metal sphere of radius 15 cm has an initial charge of 10^-5 coul. If the two spheres touch each other what charge will remain on each?

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Solution:

First we find the field of the charged sphere by using Gauss's law, \int E^\ding{217} \bullet dS^\ding{217} = q/\epsilon_0 Here, q is the total charge enclosed by the Gaussian surface and the integral is a surface integral. Consider a spherical Gaussian sur-face of radius R centered at the center of the sphere. The magnitude of E^\ding{217} is the same at all points on the Gaussian surface by symmetry considerations, E^\ding{217} and dA^\ding{217} are in the same directions, as well (i.e., both point radially outward). Gauss' law then reduced to E \int dA = E 4\piR^2 = q/\epsilon_0 For 4\piR^2 is the surface area of the sphere. Thus, For 4\piR^2 is the surface area of the sphere. Thus, E = (1/4\pi\epsilon_0) (q/R^2) The field external to the sphere is as if all the charge were con-centrated at the center; the field internal to the sphere must be zero. Since the sphere is a conductor. Similarly the potential ex-ternal to the sphere will be as if all the charge were concentrated at the center. The potential inside the sphere must be constant since E, the potential gradient, [i.e.,E = \rule{1em}{1pt}(dV/dr)], is zero (see figure). The potential, then, at any point on the sphere or external to the sphere, is, by definition V(R) - V(\infty) = \rule{1em}{1pt} ^R\int_\infty E dR V(R) - V(\infty) = \rule{1em}{1pt} ^R\int_\infty E dR = \rule{1em}{1pt} (q/4\pi\epsilon_0) ^R\int_\infty dR/R^2 = \rule{1em}{1pt} (q/4\pi\epsilon_0) ^R\int_\infty dR/R^2 = (1/4\pi\epsilon_0)(q/R) ^R\vert_\infty = (1/4\pi\epsilon_0)(q/R) ^R\vert_\infty But V(\infty) = 0, V(R) = (1/4\pi\epsilon_0) q/R. If we set K_E = 1/4\pi\epsilon_0, V(R) = K_Eq/R. When the two spheres are touched, their potentials must become equal. This is accomplished by a movement of charge from one sphere to the other, until the potentials equalize. After touching, V_1 = V_2 or K_E (q_1/R_1)= K_E(q_2/R_2) where q_1 and q_2 are the charged on spheres 1 and 2 respect-ively. Thus q_1/q_2 = R_1/R_2 = 15/5 = 3 q_1 = 3q2 But the total charge is q_1 + q_2 = 10^-5 coul +10^06 coul = 10^-5 (1 + 0.1) coul(1) = 1.1 × 10^-5 coul Also,3q_2 = q_1, and using (1) 4q_2 = 1.1 × 10^-5 coul q2= 2.75 × 10^-6 coul q_1 = 8.25 × 10^-6 coul Initially we had q_1 = 10^-5 coul and q_2 = 10^-5 coul.

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Question:

Suppose that two particles are traveling opposite to each other with velocity v'_x = \pm 0.9 c as observed in the S' system. What is the velocity of one particle with respect to the other, that is as measured by the other?

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/Users/wenhuchen/Documents/Crawler/Physics/D32-0948.htm

Solution:

In the reference frame S', the two particles move with velocities - v'(1)x= v'_(2)x = 0.9 c (see Fig. 1). If we observe from the reference frame S in which the first particle is at rest, then S' will appear to move with velocity + 0.9 c with respect to this particle, i.e. S' has velocity v = + 0.9 c relative to S (see Fig. 2). The velocity of the second particle in S, as observed by the first particle, will be given by the formula for the relativistic addition of v'_(2)x and v: v_(2)x =(v'(2)x+ v ) / {1 + (v'_(2)xv / c^2)} = 1.8c / {1 + (0.9)^2 } = 0.994c , (i.e. we transformed the velocity of the second particle from S' to S). Notice that the relative velocity of the two particles is less than c. If a photon is travelling at velocity + c in S', and S' is traveling relative to S at velocity + c, the photon as viewed from S is traveling only at velocity + c, and not at + 2c. If v'_s is the velocity of the photon in S' and v the relative velocity of S' with respect to S, then in S we have v_s = (v's+ v ) / {1 + (v'_sv / c^2)} = (c + c) / {1 + (c2/ c^2)} = c , for the photon. The fact of an ultimate speed is a consequence of the structure of the velocity-addition equations which we have derived from the Lorentz transformation. Note further that there is no frame in which a photon (light quantum) is at rest.

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Question:

Derive the mirror formula for rays incident on a mirror of radius of curvature R, if the rays make a small angle with the mirror's axis. mirror

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Solution:

Our proof will be totally geometrical. (See figure.) Ray OD is incident on mirror DG, at an angle s with the normal to the mirror CD. (C is the center of curvature of the arc forming the mirror.) It is re-flected at an angle r, and intersects the mirror axis at point I. By the law of reflection, angle i is equal to angle r. Hence s = r(1) Furthermore, \beta = (DG/CG)(2) But CG is the mirror radius R, whence \beta = (DG/R) Alsotan \alpha = (DE/OE) andtan \Upsilon = (DE/IE)(3) However, if \alpha, \beta and \Upsilon are small angles (that is, ray OD is close to the mirror axis) we may write tan \alpha \approx \alpha(4) tan \Upsilon \approx \Upsilon Furthermore, DE \approx DG(5) Using (5) and (4) in (3) \alpha \approx (DG/OE) \Upsilon \approx (DG/IE)(6) We also note that, since \Upsilon is an exterior angle of triangle ICD \Upsilon = \beta + r(7) Similarly,\beta = \alpha + s(8) Using (1) in (7) and (8) \Upsilon = \beta + s(9) \beta = \alpha + s(10) Eliminating s in (9) and (10), \Upsilon - \beta = \beta - \alpha or2\beta - \Upsilon - \alpha = O(11) Substituting (6) and (2) in (11) [(2DG)/R] - (DG/IE) - (DG/OE) = O or(1/IE) + (1/OE) = (2/R)(12) However, if \alpha, \beta and \Upsilon are small, IE \approx IG = i OE \approx OG = O(13) where i and O are the distances of the image from the mirror (image distance) and the object from the mirror (object distance). Hence, using (13) in (12) we have the mirror formula (1/i) + (1/O) = (2/R)

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Question:

Besides their actions, the sympathetic and parasympathetic systems also differ in the neurotransmitter they release. Explain.

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Solution:

Nerves of the parasympathetic system secrete a neurotransmitter called acetylcholine. For this reason they are usually referred to as cholinergic neurons. Acetylcholine is also the transmitter at the neuromuscular junction. Nerves of the sympathetic system release noradrenaline, also called norepinephrine, and are thus noradrenergic. Acetylcholine is a strong base, containing a choline moiety. [-CH_2CH_2-N_+-(CH_3)_3] In exists as a cation (positive ion) at physiological pH (about 7.4). Because of its ability to attach to a membrane and create a reversible change in the membrane's permeability to different ions, acetylcholine released by the presynaptic neuron acts to bring about depola-rization and generate an impulse in the postsynaptic neuron. The molecular structure of acetylcholine is: Noradrenaline has a molecular structure containing a ring moiety: The closely related compound adrenaline (epinephrine) - is a hormone released by the adrenal glands, and although its action is similar to that of noradrenaline, it is not a neurotransmitter. The adrenal medulla (inner part of the adrenal gland) acts like a modified post- ganglionic sympa-thetic "neuron": it is derived embryologically from neural tissue; it is innervated by a sympathetic preganglionic fi-ber; and it has become specialized to secrete noradrenaline and adrenaline and thus prolongs sympathetic activation.

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Question:

What are the three laws of thermodynamics? Discuss their biological significance.

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/Users/wenhuchen/Documents/Crawler/Biology/F01-0007.htm

Solution:

The first law of thermodynamics states that energy can be converted from one form into another, but it cannot be created or destroyed. In other words, the energy of a closed system is constant. Thus, the first law is simply a statement of the law of conservation of energy. The second law of thermodynamics states that the total entropy (a measure of the disorder or randomness of a system) of the universe is increasing. This is characterized by a decrease in the free energy, which is the energy available to do work. Thus, any spontaneous change that occurs (chemical, physical, or biological) will tend to increase the entropy of the universe. The third law of thermodynamics refers to a com-pletely ordered system, particularly, a perfect crystal. It states that a perfect crystal at absolute zero (O Kelvin) would have perfect order, and therefore its entropy would be zero. These three laws affect the biological as well as the chemical and physical world. Living cells do their work by using the energy stored in chemical bonds. The first law of thermodynamics states that every chemical bond in a given molecule contains an amount of energy equal to the amount that was necessary to link the atoms together. Thus, living cells are both transducers that turn other forms of energy into chemical bond energy and liberators that free this energy by utilizing the chemical bond energy to do work. Considering that a living organism is a storehouse of potential chemical energy due to the many millions of atoms bonded together in each cell, it might appear that the same energy could be passed continuously from organism to organism with no required extracellular energy source. However, deeper consideration shows this to be false. The second law of thermodynamics tells us that every energy transformation results in a reduction in the usable or free energy of the system. Consequently, there is a steady increase in the amount of energy that is unavailable to do work (an increase in entropy). In addition, energy is constantly being passed from living organisms to non-living matter (e.g., when you write you expend energy to move the pencil, when it is cold out your body loses heat to warm the air, etc.). The system of living organisms thus cannot be a static energy system, and must be replenished by energy derived from the non-living world. The second law of thermodynamics is also helpful in explaining the loss of energy from the system at each successive trophic level in a food pyramid. In the following food pyramid, The energy at the producer level is greater than the energy at the consumer I level which is greater than the energy of the consumer II level. Every energy transformation between the members of the successive levels involves the loss of usable energy, and this loss serves to increase the entropy. Thus, this unavoidable loss causes the total amount of energy at each trophic level to be lower than at the preceding level .

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Question:

How does energy release in combustion differ fromenergy releasein aerobic respiration?

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/Users/wenhuchen/Documents/Crawler/Biology/F03-0080.htm

Solution:

Since glucose serves as the primary energy source, we will considerthe oxidation of glucose. Com-bustion refers to the rapid oxidation ofa substance with the liberation of heat. When a mole of glucose is burnedoutside of a living cell, approximately 690 kilocalories are released asheat. This great amount of heat would destroy a cell if released in one burst. Instead, the cell oxidizes glucose by a series of reactions (in aerobic respiration), each catalyzed by a specific enzyme. The re-lease of energy issequential, and in small packets thus preventing damage to the cell. During oxidativephosphor-ylation, the energy is transformed and stored in theform of ATP. The complete degradation of one mole of glucose yields 36 ATPs.Since each high energyphosphate bond (\simP) holds about 7 kilocaloriesof energy, 252 (36 × 7) kilocal-ories are biologically useful, whilethe other 438 (690 252) kilocalories are released as heat. Aerobic respira-tion, although only 36.5% efficient, is able to harness the energy fromthe oxidation of glucose to provide an en-ergy source for later needs, withoutresulting in trauma to the organism.

Question:

(a)Ag_2 O \rightarrow Ag + O_2 ; (b)Zn +HCl+ ZnCl_2 + H_2 ; (c)NaOH + H_2 SO_4 \rightarrow Na_2 SO_4 + H_2 O.

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Solution:

When balancing chemical equations, one must make sure that there are the same number of atoms of each element on both the left and right side of the arrow. For example, H_2 + O_2 \rightarrow H_2 O is not a balanced equation because there are 2 O' s on the left side and only one on the right. 2H_2 + O_2 \rightarrow 2H_2 O is the balanced equation for water be-cause there are the same number of H and 0 atoms on each side of the equation. (a)Ag_2 O \rightarrow Ag + O_2 is not a balanced equation because there are 2 Ag on the left and only one on the right, and because there is only one O on the left and two O on the right. To balance this equation one must first multiply the left side by 2 to have 2 O' s on each side. 2Ag_2 O \rightarrow Ag + O_2 There are now 4 Ag on the left and only one on the right, thus the Ag, on the right must be multiplied by 4. 2Ag_2 O \rightarrow 4Ag + O_2 The equation is now balanced. (b)Zn + HC1 \rightarrow ZnCl_2 + H_2 In this equation, there are 2 H and 2Clon the right and only one of each on the left, therefore, the equation can be balanced by multiplying the HC1 on the left by 2. Zn + 2HCl \rightarrow ZnCl_2 + H_2 Because there are the same number of Zn, C1, and H on both sides of the equation, it is balanced. (c)NaOH + H_2 SO_4 \rightarrow Na_2 SO_4, + H_2 O Here, there are 1 Na, five O, 3 H and 1 S on the left and 2 Na, 1 S, five O, and 2 H on the right. To balance this equation, one can first adjust the Na by multiplying the NaOH by 2. 2NaOH + H_2 SO_4 \rightarrow Na_2 SO_4 + H_2 O There are now 2 Na, six O, 4 H, and 1 S on the left and Na, five O, 2 H, and 1 S on the right. Because there are two more H and one more O on the left than on the right, you can balance this equation by multiplying the H_2 O by 2. 2NaOH + H_2 SO_4 Na_2 SO_4 + 2H_2 O The equation is now balanced.

Question:

A reacts with B in a one-step reaction to give C. The rate constant for the reaction is 2.0 × 10^-3 M^-1 sec^-1. If 0.50 mole of A and 0.30 mole of B are placed in a 0.50 liter-box, what is the initial rate of the reaction?

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Solution:

The equation for this reaction can be written: A + B \rightarrow C From this, one can write the rate law assuming that the reaction is first order in both A and B. When a re-action is first order in a particular reactant, it means that the rate is proportional to the concentration of the reactant. Thus, the rate law is written Rate = k[A] [B], where k is the rate constant and [ ] indicates concentration. In this problem, one is given k, [A] and [B]. Thus, Rate= 2.0 × 10^-3 M^-1 sec^-1 × (1 M) (.60M) = 1.20× 10^-3 M/sec k = 2.0 × 10^-3 M^-1 sec^-1 [A] = 0.50 mole / 0.5 liter = 1 M [B] = 0.30 mole / 0.5 liter = 0.60 M .

Question:

Show that the average distance, l, a molecule travels between collisions in a gas is related to the number density n and the molecular diameter d by l = [1/(n\pid^2)].

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Solution:

Consider a molecule moving through a region of stationary molecules. It can collide with those molecules whose centers are at a distance less than or equal to d from its center. If the speed of the molecule is v, it moves a distance vt in time t and collides with every molecule in the cylindrical volume vt\pid^2 (Fig,A). The molecules in this cylinder cannot avoid the incident molecule because the diameter of the cylinder is 2d, twice that of a molecule. If there are n molecules per unit volume, the number of molecules in the cylinder is N = nvt\pid^2. Actually, after each collision, the molecule changes its direction and the cylinder mentioned before makes zigzags as shown in Fig. B. The number of collisions is just the number of molecules in the cylinder, hence the average distance per collision is l \approx (vt)/(nvt\pid^2) = 1/(n\pid^2). l is called the mean free path.

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Question:

Insulin controls the glucose level in the blood. In 1969, Hodgkin announced the secondary structure of insulin, the first protein hormone to be resolved by X-ray diffraction, (a) Chemial analysis shows insulin is 2.15% zinc by weight. Calculate the minimum molecular weight, (b) Osmotic pressure studies show the molecular weight of insulin is about 6000. How many zinc atoms does the molecule contain?

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Solution:

The minimum molecular weight of insulin can be determined by assuming that there is one zinc atom per molecule. This means that in 1 mole of insulin there exists 1 mole of zinc. From the knowledge that zinc is 2.15% by weight of insulin, and that the MW of zinc is 65.3, one can determine the minimum MW of insulin by using, the following relationship. (.0215)(MW of insulin) = (MW of zinc) MW of insulin = (MW of Zn) / (.0215) = (65.3 g/mole) / (.0215) = 3037 (g/mole) (b) If the true molecular weight of insulin is about 6000, then two of the units weighing 3037 (g/mole) make up each molecule of insulin. Thus, there are 2 atoms of Zn per molecule.

Question:

Follow the flow chart and the group of items given to write a program called DATA-ALTERATION. You may omit the Configuration Section, but use the required division header. Use two independent data items, LAST-NAMB, which is fifteen characters long, and SOCIAL-SEC, which is nine characters long. FIRST-NAME STUDENT-RECORD INIT 1 char-acter REST-OF-NAME 8 characters STUDENT-NO. 9 char-acters STUDENT-NAME 12 characters S-INIT 1 char-acter In the above, note that FIRST-NAME, STUDENT-RECORD, INIT, REST-OF-NAME, STUDENT-NO., STUDENT-NAME, and S-INIT are variables used in the program to be written.

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Solution:

In following the flow chart, the START statement covers the IDENTIFICATION and ENVIRONMENT DIVISION. In reality a programmer who works for a company almost always uses the same IDENTIFICATION and ENVIRONMENT divisions, keeping in mind that changes such as DATE-WRITTEN and DATE- COMPILED are essential. In DATA-DIVISION the programmer is required to use two independent data items. In COBOL independent data items are identified by level number 77; therefore the requireditems may be written as: 77 LAST-NAME PIC X(15). 77 SOCIAL-SEC PIC X(9). Observe that the first item has 15, and the second item has 9 characters from the standard character set. Examining the group of items given in the question, it is clear that there are two group items each with their own subdivisions, namely, FIRST-NAME, and STUDENT-RECORD. In format, these group items get identified by level numbers 01 through 49, the first one being the highest level of the data item, and starting from Margin A. The rest of the level numbers are called secondary and start from margin B. Following these rules, the given group of items may be formatted as: AB 01FIRST-NAME. 02 INIT PIC X(1). 02 REST-OF-NAME PIC X(8). 01STUDENT-RECORD 02 STUDENT-NAME X(12) 02 STUDENT-NO PIC X(9). 02 S-INIT PIC X(1) Now having finished the IDENTIFICATION, ENVIRONMENT, and DATA DIVISIONS the flow chart may be followed to write the PROCEDURE DIVISION. Step by step examination of the given flow-chart will enable the programmer to write the required statements. There are two paragraph names given in the flow-chart; FIRST- PARA, and LAST-PARA. The first paragraph consists of the following opera-tions : 1) Display the sentence "ENTER LAST NAME" on console (screen or typewriter); after the name is entered by the user, ac-cept it from typewriter or screen. 2, 3) Do the same for FIRST NAME, and SOCIAL SECURITY Number. 4) Take these accepted items and move them to their appro-priate places in Student-RECORD 5) Print the STUDENT-RECORD on the standard output device. In the last paragraph the computer is told to display the sentence "THANK YOU THATS ALL" on screen or on the type-writer and stop the execution. Following these rules and explanations the complete program is given in fig. 1.

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Question:

The figure shows a battery and a resistor in series. What is the power P supplied by the battery and the power dissipated as heat?

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Solution:

The total resistance of the circuit is R_t = r + R = (5.5 + 0.5) \Omega = 6 \Omega Therefore the current in the circuit is, by Ohm's Law, i = (\epsilon/R_t) = (12 v/6 amp) = 2 amp. Point b is at a higher potential than point a, and V_ba = iR = 2 amp (5.5\Omega) = 11 volts or, by Kirchoff's Law V_ba = \epsilon \rule{1em}{1pt} ir = 12 \rule{1em}{1pt} (2 × 0.5) = 11 volts and V_ab = \rule{1em}{1pt} V_ba = \rule{1em}{1pt} 11 volts. In the lower rectangle, the direction of the current is from b to a, hence the power dissipated in R is P_R = V_ba × i = (2 amp) × (11 v) = + 22 watts. The energy supplied to R is converted to heat at a rate energy/time = i^2R = (2 amp)2 x 5.5 \Omega = 22 watts which equals the power input to R. The energy loss in the battery is due to the internal resistance r; P_r = i^2r = (2 amp)^2 × 0.5 \Omega = 2 watts. The rate of conversion of nonelectrical to electrical energy in the seat of emf is given by P_\epsilon = i\epsilon = (2 amp) × (12 v) = 24 watts. The total rate of energy dissipation in the resistors should add up to the rate at which energy is supplied by the emf, orP_\epsilon = P_R + P_r = (22 + 2) watts = 24 watts.

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Question:

Snails can usually be seen crawling around during the night and on dark, damp days. What happens to them on bright, sunny days and why?

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Solution:

The soft body of the snail is protected by a hard calcareous shell secreted by the mantle. The mantle is a fold of tissue that covers the visceral mass and projects laterally over the edges of the foot. During the daytime, the snail usually draws up inside the shell. In order to move and to feed, the foot and head must emerge from the shell. This usually occurs at night. In bright sunlight, the soft body of the snail would dry out. Slugs aremolluscssimilar to snails, but without hard shells. Slugs are generally found in damp forest environments, and remain in the moist earth during bright daylight. Mostmolluscsare aquatic animals, and hence are in no danger of drying out. Land snails are one of the few groups of fully terrestrial invertebrates. In most land snails, the gills have disappeared, but the mantle cavity has become highlyvascularizedand functions as a lung.

Question:

Describe the composition and structure of the plant cell wall.

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Solution:

Because plant cells must be able to withstand high osmotic pressure differences, they require rigid cell walls to prevent bursting. This rigidity is provided primarily by cellulose, the most abundant cell wall component in plants. Cellulose, a polysaccharide, is a long, linear, unbranched polymer of glucose molecules. In the cell wall cellulose molecules are organized in bundles of parallel chains to form fibrils. The fibrils are often arranged in criss-cross layers reinforced and held together by three other polymeric materials: pectin and hemicellulose (complex polysaccharides) and extensin (a complex glycoprotein). Lignin is an addi-tional polymeric substance found in the wood of trees. Together these substances provide a matrix capable of withstanding enormous stress. A layer known as the middle lamella lies between and is shared by adjacent cells. The lamella is composed primarily of pectin. By binding the cells together, it provides additional stiffness to the plant. The cell wall is usually interrupted at various locations by plasmodesmata. These, are tiny holes in the cell walls through which run protoplasmic connections between adjacent cells. This provides for intercellular exchange of such materials as water and dissolved sub- stances. Being extremely small, the plasmodesmata do not prevent the cell wall from exerting pressure on a swollen cell.

Question:

(a) If the x direction is normal to the surface of the earth and directed upward, the gravitational force is F_G = -Mgx˄, where g is the acceleration of gravity and has the approximate value 980 cm/sec^2. Calculate the work done by gravity when a mass of 100 gm falls through 10 cm. (b)If the particle in (a) was initially at rest, what is its kinetic energy and its velocity at the end of its 10-cm fall?

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/Users/wenhuchen/Documents/Crawler/Physics/D05-0264.htm

Solution:

(a) Work done by a force F^\ding{217} is calculated by \intF^\ding{217} \textbulletds^\ding{217}. Here F^\ding{217} =mgx˄ which is a constant, and since the object falls in a straight line, the work is given by W =mgx This is a scalar quantity since we are taking the dot product of F^\ding{217} \textbulletds^\ding{217}. Since F^\ding{217} and s^\ding{217} are parallel, we take the simple arithmetic product: W = (100 gm) (980 cm/s)(10 cm) = 980,000 ergs (b) The initial value K_A of the kinetic energy is zero; the terminal value K_B is equal to the work done by gravity on the particle, so that K_B = W \approx 10^6 ergs = (1/2) mv^2_B, whence v^2_B \approx 2(10^6 ergs)/(100 gm) \approx 2 × 10^4 cm^2/sec^2. Therefore,v_B\cong 1.41 × 10^2 cm/sec We may obtain the same result from the equations of motion. We have v =gtand h = (1/2) gt^2. (These equations are adaptations of the more complete equations of motion, d = (1/2) at^2 + v_0t + d_0 and v = at + v_0, to the initial conditions of our problem where v_0 = 0, d_0 is assumed to be 0, and a = g.) Eliminating t, we have v^2 = 2gh. Therefore, v^2 = 2(980 cm/s^2) (10 cm) v^2 \cong 2 × 10^4 cm^2/s^2 v \cong 1.41 × 10^2 cm/sec.

Question:

If the coefficient of sliding friction for steel on ice is 0.05, what force is required to keep a man weighing 150 pounds moving at constant speed along the ice?

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Solution:

To keep the man moving at constant velocity, we must oppose the force of friction tending to retard his motion with an equal but opposite force (see diagram). The force of friction is given by: F = \mu_kineticN By Newton's Third Law F_forward = F_friction Therefore F_forward = \mu_kineticN F_forward = (.05)(150 lb) = 7.5 lb.

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Question:

The standard E\textdegree for(1/2)F_2(g) + e^- \rightarrow F^-(aq), which is +2.87V, applies when the flouride ion concentration is 1M . What would the corresponding E be in 1M H_3O^+ , i.e., for the electrode reaction (1/2)F_2(g) + e^- + H_3O^+ \rightarrow HF(aq) +H_2O? (The K_diss of HF is 6.7 × 10^-4).

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Solution:

One can solve for E for the electrode reaction by using the Nernst equation. This is stated E = E\textdegree - (0.591 / n)logQ , where E is the actual electrode potential for the half-reaction, E\textdegree is the standard potential for the half-reaction, n is the number of electrons transferred in the reaction, Q is the dissociation con-stant. Thus, one must find the concentration of F^-, in order to eval-uate Q, before solving for E\textdegree . The dissociation constant is equal to. [H_3O^+][F^- ] / [HF] . One is given that the dissociation constant (K_diss)for 1M HF is 6.7 × 10^-4. Thus, K_diss = [H_3O^+][F^- ] / [HF] = 6.7 × 10^-4 = Q . Solving for the E of the electrode reaction: For the half-reaction (1/2)F_2(g) + e-\rightarrow F^-(aq), one electron is trans-ferred, thus n = 1. One is given that E\textdegree = 2.87V and one has de-termined Q. Thus, E = 2.87V -(0.0591 / 1) log (6.7 × 10^-4) E = 2.87V - 0.0591 × (-3.17) E = 2.87V + .188 = (+3.06V) .

Question:

Find the minimum radius for a planet of mean density 5500 kg \bullet m^-3 and temperature 400\textdegreeC which has retained oxygen in its atmosphere.

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Solution:

The escape velocity from a planet is given by the relation V = \surd(2GM/r) = \surd[2\surd{(G × (4/3)\pir^3\rho)/r}] = \surd[(8/3)G\pi\rhor^2](1) where r is the planet radius, M is its mass, and \rho is its density If most oxygen molecules have velocities greater than this, then, when they are traveling upward near the top of the atmosphere, they will escape into space and never re-turn. A slow loss of oxygen from the atmosphere will there-fore take place. In this case, however, we are told that the planet has retained its oxygen and we can assume that escape velocity from the planet is greater than thermsvelocity of the oxygen molecules. When the two are equated, the minimum radius for the planet results. We need thermsvelocity of oxygen molecules. This speed V is so defined that the internal energy U would be the same if all the atoms had this speed. For a gas consisting of N atoms, V is defined by U = N(1/2)mv^2(2) where m is the mass of one atom. If the gas is ideal and monatomic, then we also know that U = N(3/2)kT(3) where k is Boltzmann's constant and T is the temperature of the gas in \textdegreeK. Oxygen is neither ideal or monatomic but equation (3) is still a good approximation since the gas is not very dense and thereforeinteratomic forces can be ignored. Equating equations (2) and (3), we get mv^2 = 3kT. Multiplying both sides by Avogadro's number N_A (the number of molecules in one mole of the gas), N_Amv^2 = 3N_AkT(4) ButN_Amequals the mass, M', of one mole of the gas. Also, by definition N_Ak = R where R is the universal gas constant. Substituting these two expressions in equation (4) yields M'v^2 = 3RT. Solving for the velocity, we have v = \surd(3RT/M')(5) Set equations (1) and (5) equal to each other so as to find the minimum radius. Then \surd[(8/3)G\pi\rhor^2_min] = \surd(3RT/M'). where T is the absolute temperature in the atmosphere, and M' is the mass per mole of O_2. The temperature of the oxygen is 400\textdegreeC + 273\textdegree = 673\textdegreeK. Oxygen gas is diatomic, and its effective molecular mass is therefore twice the atomic mass of monatomic oxygen M' = (2 × 16)g \bullet mole^-1 = 32 × 10^-3 kg \bullet mole^-1 r_min= \surd[(9RT)/(8G\pi\rhoM')] = \surd[(9 × 8.315 J\bulletmole^-1\bulletK deg^-1 × 673 K deg)/(8 × 6.67 × 10^-11 N\bulletm^2\bulletkg-2 × \pi × 5500 kg\bulletm^-3 × 32 × 10^-3 kg\bulletmole^-1)] \thereforer_min= \surd(1.708 × 10^11 m^2) = 4.131 × 10^5m = 413.1 km.

Question:

Define top-down structured programming. Also define bottom-up structured programming. What importance does each of these terms have in relation to the computer implementation of an algorithm?

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Solution:

The top-down design technique is the offspring of the idea of refinement. This technique requires that the programmer define the algorithm coarsely, in plain English or with some sort of code. Next, a structured language or a flow chart should be used to explain the parts of the algorithm that need greater clarification. This process of writing more and more detailed descriptions continues until the algorithm is sufficiently documented. Thereafter, the translation of the algorithm into a standard computer language, such as COBOL or FORTRAN, becomes a simple mechanical procedure. This technique is especially useful when a team of programmers is working on a specific problem. If each programmer is assigned a small part of a large task, the parts can be combined easily by assembling the codes from each part. If the modular programming rule of one-entry- one- exit per module is followed as well, the sections of code become like interchangeable parts of a machine. Any other programmer can pick up your section of code and substitute it into a program that needs what you have written. Bottom-up design is just the opposite. It refers to the creation of the program in a highly detailed fashion before coding or documentation is begun. This is sometimes bad practice when writing programs because the programmer is prone to miss important details. Also, it is both time-saving and economical to iron out errors in the coding stage before typing out a program for computer execution. This idea of program- first, code-second may be used if one wants to debug a program. By looking at a program first, you might be able to find syntactical errors that cause problems. After this is done, then a look at the code allows you to debug any logical errors.

Question:

Profound changes in uterine morphology occur during the menstrual cycle and are completely attributable to the effects of estrogen and progesterone. Describe these changes.

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/Users/wenhuchen/Documents/Crawler/Biology/F21-0559.htm

Solution:

In the uterus estrogen stimulates growth of the smooth muscle layer called the myometrium, and the glandular epithelium called the endometrium, lining its inner surface. Progesterone acts on the endometrium and converts it to an actively secreting tissue. The glands become coiled and filled with glycogen; the blood vessels become more numerous; various enzymes accumulate in the glands and connective tissue of the lining. These changes are ideally suited to provide a favorable environment for implantation of a fertilized ovum. Progesterone also causes the mucus secreted by the cervix (a muscular ring of tissue at the mouth of the uterus which projects into the vagina) to become thick and sticky. This forms a "plug" which may constitute an important blockage against the entry of bacteria from the vagina. This is a protec-tive measure for the fetus should conception occur. As a result of the regression of the corpus luteum when no fertilization occurs, there is a fall in blood progesterone and estrogen levels. Consequently, the highly developed endometrial lining becomes deprived of its hormonal support. The most immediate result of this deprivation is the constriction of the uterine blood vessels, resulting in a decreased flow of oxygen and nutrients to the tissue. Subsequently, the lining disintegrates and gets sloughed off as the mentrual flow begins. This phase is called the menstrual phase and usually lasts for about 5 days. The flow ceases as the endometrium repairs itself, and then grows under the influence of the rising blood estrogen concentration. This phase is called the proli-ferative phase, and lasts for 9 days. This phase covers the time from menstruation up until ovulation. After ovulation and formation of the corpus luteum, progesterone acts synergistically with estrogen to induce formation of the glandular endometrium. This period, called the secretory phase (lasting about 14 days) is terminated by disintegration of the corpus luteum. Thus the cycle is completed and a new one begins. The phases of the menstrual cycle can be named either in terms of the ovarian or uterine events. The cycle is di-vided into follicular and luteal phases with respect to the ovary; proliferative and secretory phases are the terms re-ferring to the uterine state. It is important to realize that the uterine changes simply reflect the effects of vary-ing blood concentrations of estrogen and progesterone throughout the cycle.

Question:

When a Kundt's tube contains air, the distance between several nodes is 25 cm. When the air is pumped out and replaced by a gas, the distance between the same number of nodes is 35 cm. The velocity of sound in air is 340 m \bullet s^-1. What is the velocity of sound in the gas?

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/Users/wenhuchen/Documents/Crawler/Physics/D26-0838.htm

Solution:

A Kundt's tube (see the figure) is a glass tube with sawdust spread over its interior. A piston P is set into longitudinal vibration and the wall W moved until a standing wave pattern of nodes and antinodes is set up within the tube. The sawdust will accumulate at all the nodal points, for these are the points of the wave for which there is no vibration of the gas. By measuring the distance d between the nodes, we may find the velocity of sound of the gas in the tube. Let the number of nodes involved be 2n + 1. There are 2n intervals between these 2n + 1 nodes, and the total distance thus corresponds to 2n half-wavelengths, or to n full wavelengths. The frequency of the emitted sound is the same in both cases, for it depends only on the frequency of the vibra-ting membrane which is causing the standing wave within the Kundt's tube. Thus c_A = f\lambda_A and c_G = f\lambda_G. Since there are, in the case of the air filled tube, n wavelengths in 25 cm of space, then the number of centi-meters of space one wavelength will contain is found by the following proportion. (25 cm / n wavelength) = (x cm / 1 wavelength) = \lambda_A Similarly for the case of the gas filled tube, (35 cm / n wavelength) = (x cm / 1 wavelength) = \lambda_G Thus c_G / c_A = \lambda_G / \lambda_A = [(35/n) cm] / [(25/n) cm] = 7/5. \therefore c_G = (7/5) × 340 m \bullet s^-1 = 476 m \bullet s^-1. If the number of nodes involved is 2n, then there are 2n - 1 intervals between these 2 nodes. The total dis-tance thus corresponds to 2n - 1 half-wavelengths, or to n - 1/2 full wavelengths. The same analysis as before holds, and c_G / c_A = \lambda_G / \lambda_A = [{(35/n) - (1/2)}cm] / [{(25/n) - (1/2)}cm] = 7/5 The same result is obtained, as it must, if an odd number of nodes are counted as if an even number of nodes are counted.

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Question:

Calculate the pH of (a) a 0.5 M solution with respect to CH_3COOH and CH_3COONa; (b) the same solution after 0.1 mole HCl per liter has been added to it. Assume that the volume is unchanged. K_a = 1.75 × 10^-5.

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Solution:

Part (a) involves the determination of the concentration of acid that has dissociated. Since one is adding a strong acid, part (b) is a buffer type problem. For part (a), CH_3COOH is the acid and for each x moles of it that dissociates, x moles of H_3O^+ and x moles of CH_3COO^- are formed. This is seen more clearly below. CH_3COOH + H_2O \rightleftarrows H_3O^+ + CH_3COO^- After reaction 0.5 - xxx There is also CH_3COONa present in this solution. Its effect on the equilibrium must be also taken into account. Since it is a salt it will completely dissociate, according to the following equation, in this dilute solution. CH_3COONa \rightleftarrows CH_3COO^- + Na^+ It is 0.05 M, thus, in the solution there will be 0.05 M CH_3COO^- and Na^+ contributions from it. The con-centrations of the various species found in this solution are shown in the table below. CH_3COOH(0.05 - x)M H_3O^+x M CH_3COO^-(0.05 + x)M (contributions are from both the CH_3COOH and CH_3COONa equilibria.) Na^+0.05 M pH is defined as being equal to - log [H_3O^+].Thus, to solve for pH, one must first determine [H_3O^+]. This can be done by setting up the equation for the equilibrium of the reaction CH_3COOH + H_2O \rightleftarrows CH_3COO^- + H_3O+ K_a = {[H_3O^+] [CH_3COO^-]} / [CH_3COOH] = 1.75 × 10^-5. 1.75 = {(x) (0.05 + x)} / (0.05 - x) Since CH_3COOH is a weak acid, there will be little dissociation and x will be very small. Thus one can approximate 0.05 + x \cong 0.05 and 0.05 - x \cong 0.05. Solving: {x (0.5)} / (0.5) = 1.75 × 10^-5 x = [H_3O^+] = 1.75 × 10^-5 pH = - log [H_3O^+] = - log (1.75 × 10^-5) = 4.76. Part (b): A buffer solution contains an acid and its conjugate base, both are moderately weak. What this type of solution does is to prevent a large change in pH when more acid or base is added. In a nonbuffered solution, the pH will change drastically. If a strong acid is added to a buffered solution, it reacts with the base A^- to produce a weak acid. H_3O^+ + A^- \rightleftarrows HA + H_2O. When a strong base is added the same effect occurs. It reacts with the acid ro produce a weak base. OH^- + HA \rightleftarrows A^- + H_2O. The acid-base ratio is altered somewhat but not to any great extent, thus allowing the pH to stay relatively stable. However, if the number of moles of acid or base added is greater than the number of moles of base or acid present, then the pH will change, thereby destroying the buffer. It is given that 0.1 M HCl is added. Since HCl is a strong acid, and dissociates completely, the H_3O^+ and Cl^- formed each have an initial concentration of 0.1 M. How-ever, the H_3O^+ will react with the CH_3COO^- and drive the equilibrium to the left by Le Chatelier's principle and produce a weak acid. Therefore, the buffer has converted the acid ions so that they cannot change the pH. Le Chatelier's principle states that when a stress is brought to bear on a system at equilibrium, the system tends to change so as to relieve the stress. The stress is relieved by driving the reaction to the other side of the equilibrium sign to prevent excesses of ions from building up. The net effect can be seen below. CH_3COOH + H_2O \rightleftarrows H_3O^+ + CH_3COO^- Before:0.5 - xx0.5 + x After add of0.5 + 0.1 - xx0.5 - 0.1 + x 0.1 M H_3O^+(0.6 - x)(0.4 + x) At the new equilibrium, there will be x moles of H_3O^+ still in solution, which were formed from the redissociation of CH_3COOH. x moles of CH_3COO^- are also formed resulting in the new concentrations of CH_3COOH + H_2O \rightleftarrows H_3O^+ + CH_3COO^- 0.6 - xx0.4 + x Upon setting up the equilibrium constant equation, one obtains [H_3O^+]. K_a = {[H_3O^+] [CH_3COO^-]} / [CH_3COOH] = 1.75 × 10^-5 1.75 × 10^-5 = {(x)(0.4 + x)} / (0.6 - x) Using the assumption outlined above, 0.6 - x \cong 0.6, and 0.4 + x \cong 0.4. Solving: 1.75 × 10^-5 = {(x)(0.4)} / (0.6) x = 2.63 × 10^-5 = [H_3O^+] Since, pH = - log [H_3O^+], one finds by substitution, pH = - log 2.63 × 10^-5 = 4.58.

Question:

A flask containing H_2 at 0\textdegreeC was sealed off at a pressure of 1 atm and the gas was found to weigh, 4512 g. Calculate the number of moles and the number of molecules of H_2 present.

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Solution:

A mole is defined as 6.023 × 10^23 molecules of a substance. Hydrogen has a molecular weight of 1.008, therefore, 1 mole of H_2 weighs 2.016 grams. The number of moles present is therefore moles = (gram/M.W.) = [(.4512 g)/(2.016 g/mole)] = .2238 mole of H_2 To calculate the number of molecules, recall that 1 mole of any gas at STP has 6.023 × 10^23 molecules/mole. no. of molecules = 6.023 × 10^23 (molecules/mole) × no. of moles = 6.023 × 10^23 (molecules/mole) × 0.2238 moles = 1.348 × 10^23 molecules of H_2 .

Question:

Given 25 equally spaced data points x_i (i= 1,...,25) and 25 functionalvaluesy_i= f(x_i ) [arranged in ascending order ], writea program to evaluate^(X)25\int_(X)1 f(x)dxusing the trapezoidalrule.

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Solution:

According to the trapezoidal rule of integration, (X)25\int_(X)1 f (x)dx= Area \approx (h/2) (y1 + 2y2 + 2y3 + ...+ 2y24 + y25). We adopt the following computational strategy: accumulate 24\sum_i=2y_iin a DO loop, double the result, add it to (y1+ y25), and multiply by h =X(2) - X(1). The program (which assumes values of X and Y have alreadybeen read in) is given below. You should also remember that the error margin for the trapezoid rule is givenby E \leq (1/12 )( )( b - a)h2 \vertf"(\xi)\vertX_n\leq \xi \leq X1. b - a)h2 \vertf"(\xi)\vertX_n\leq \xi \leq X1. Knowing this, you can use a suitable N to calculate the integration to a closeapproximation. REAL H SUM = 0.0 DO 50 I = 2,24 50SUM = SUM +Y(I) H =X(2) - X(1) AREA = 0.5\textasteriskcenteredH\textasteriskcentered(Y(1) + 2.0\textasteriskcenteredSUM + Y(25))

Question:

Can complete equilibrium in a gene pool exist in real situations?

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Solution:

Of the four conditions necessary for the genetic equilibrium described by the Hardy-Weinberg Law, the first, large population size is met reasonably often; the second, absence of mutations, is never met; the third, no migration, is met sometimes; and the fourth, random reproduction, is rarely met in real situations. Therefore it follows that complete equilibrium in a gene pool is not expected. With regard to the first condition, many natural populations are large enough so that chance alone is not likely to cause any appreciable alteration in gene frequen-cies in their gene pools. Any breeding population with more than 10,000 members of breeding age is probably not significantly affected by random changes. The second condition for genetic equilibrium the absence of mutations, is never met in any population be-cause spontaneous mutations are always occurring. Most genes probably undergo mutation once in every 50,000 to 1,000,000 duplications, with the rate of mutation for different genes varying greatly. However, since the rate of spontaneous mutation is usually low, it is usually insignificant in altering the gene frequencies in a large population. The third condition for genetic equilibrium, implies that a gene pool cannot exchange its genes with the out-side. Immigration or emigration of individuals would change the gene frequencies in the gene pool. A high percentage of natural populations, however, experience some amount of migration. This factor, which enhances variation, tends to upset. Hardy Weinberg equilibrium. The fourth condition, random reproduction, refers not only to the indiscriminate selection of a mate but also to a host of other requirements that contribute to success in propagating the viable offspring. Such factors include the fertility of the mating pair, and the survival of the young to reproductive age. An organism's genotype actually influences its selection of a mate, the physical efficiency and frequency of its mating, its fertility, and so on. Thus entirely random reproduction in reality is not possible. We can conclude, therefore, that if any of the conditions of the Hardy-Weinberg Law are not met, then the gene pool of a population will not be in equilibrium and there will be an accompanying change in gene frequency for that population. Since it is virtually impossible to have a population existing in genetic equilibrium, even with animals under laboratory conditions, there must then be a continuous process of changing genetic constitutions in all populations. This is ultimately related to evolu-tion in that evolutionary change is not usually automatic, but occurs only when something disturbs the genetic equilibrium.

Question:

Discuss the responses of stomata to environmental factors.

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Solution:

A number of environmental factors affectstomatalopening and closing. Water loss is the most important among them. When theturgorof a leaf drops below a certain critical point because of rapid evapo-ration to the atmosphere, thestomatalaperture responds to this loss by becoming smaller or closed. In this situation, the need for water conservation overrides the need for the intake of carbon dioxide for photosynthesis. With the stomata closed, the leaf cells can use the carbon dioxide produced by their respiration, and thus be able to sustain a very low level of photosynthesis. Factors other than water loss that affect the stomata include light, carbon dioxide concentration, and temperature. It has been shown that when a turgid leaf which has been kept in darkness for a few hours is exposed to light, the stomata open. When the light is switched off, the stomata close. Light apparently plays a role in the opening and closing of the stomata. Light is now believed to initiate a series of enzymatic reactions which break down insoluble starch to many smaller, soluble glucose molecules in the guard cells of a stoma. As the solute concentration increases, water enters the guard cells by osmosis, increasing theirturgor, thereby opening the stomata. The role of carbon dioxide is similar to that of light, in the opening and closing of the stomata. It is found experimentally that a decrease in carbon dioxide increases the pH in a cell and an increase in pH facilitates the conversion of starch to glucose. This is explained by the fact that the enzymephosphorylaseis stimulated by the increased pH to convert starch to glucose-l-phosphate, which is in turn converted to glucose. As discussed earlier, water now enters the leaf cells,turgorpressure increases and as the guard cells bend, the stomata open. Temperature also fits into this scheme because of its influence on carbon dioxide concentration. An increase in temperature increases the rate of respiration, and so increases in turn the cellular carbon dioxide level. A decrease in temperature, on the other hand, retards respiration, and thus decreases the cellular carbon dioxide levels. It is clear that the stomata are responsive to a complex interplay of factors and are able to respond efficiently to different combinations of these factors.

Question:

Given the coordinates of three points (X_1 ,Y_1 ), (X_2 ,Y_2 ), and (X_3 ,Y_3), developea program which outputs a message regarding theircollinearity. (Three points are collinear if they all lie on the same line.) Include statements that will take into account round-off errors.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G12-0298.htm

Solution:

The solution to the problem rests on the fact that three points P_1 , P_2 , and P_3 are collinear if and only if the slopes of the segments P_1 P_2 and P_2 P_3 are equal. This implies (Y_2 -Y_1)/ (X_2 - X_1) = (Y3- Y_2)/ (X3- X_2) where (X_1,Y_1), (X_2 ,Y_2 ) and (X_3 ,Y_3 ) are the coordinates of three points. Instead of calculating slopes, we can test the expressions (Y2- Y_1) × (X_3 - X_2) and (Y_3 - Y_2) × (X_2 - Y_1) , for equality. If they are equal, the three points are collinear, and an appropriate message is output. This expression may also be used to establish the degree of round-off error. One way of doing this is with the statement ABS((Y2 - Y1){_\ast}(X3 - X2) - (Y3 - Y2){_\ast}(X2 - X1)) If this quantity is very small, say, less than .0001, we can accept the accuracy ofcollinearity. If this condition is not met, a message indicates that there is an unacceptable accuracy in the determination. We use a format-free READ statement in the program for simplicity's sake, a feature available on some machines. CCOLLINEARITY PROGRAM READ (5,{_\ast}) X1, Y1, X2, Y2, X3, Y3 ,ACCUR CARE THEY COLLINEAR? IF((Y2 - Y1){_\ast}(X3 - X2). NE. (Y3 - Y2){_\ast}(X2 - X1)) GO TO 30 WRITE (5,100) X1, Y1, X2, Y2, X3, Y3 100FORMAT (1X, 'THE PAIRS OF POINTS ', 3(1X,2(1X,F7.5)), 'ARE 1COLLINEAR') TEST = ABS((Y2 - Y1){_\ast}(X3 - X2) - (Y3 - Y2){_\ast}(X2 - X1)) IF (TEST. GE. ACCUR) GO TO 25 WRITE (5,101) ACCUR 101FORMAT(1X, 'WITH DESIRED ACCURACY TO ', F7.5) GO TO 99 25WRITE (5,101) ACCUR 101FORMAT (1X 'WITHOUT ACCURACY OF', F7.5) 30WRITE (5,102) 102FORMAT (1X,'POINTS ARE NOT COLL INEAR') 99STOP END

Question:

Give the approximate boiling points at sea level for the following: (a) 2molalHBr; (b) Suspension of 100 g of powdered glass in one liter of water; (c) 1.2 × 10^24 (sucrose molecules / liter); (d) 0.5molalBaCl_2.

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Solution:

The boiling point is acolligativeproperty and, therefore, depends upon the number of particles present in 1 kg of solvent. The boiling point constant for water is 0.52\textdegreeC. This means that for each mole of particles dis-solved in 1 kg of H_2O the boiling point will be elevated by 0.52\textdegreeC. boiling pt. elevation = 0.52\textdegreec × (molalityof solute). Themolalityis defined as the number of moles in one kg. of solvent. (a) 2molalHBr.HBris a strong acid and will thus ionize completely when diluted with H_2O to form H^+ and Br^- ions. HBr \rightleftarrows H^+ + Br^- Therefore, 2 particles will be formed by eachHBrthat ionizes. The effectivemolalityof the solution is then twice themolalityof theHBr. effectivemolality= 2 × 2molal= 4molal. One can now find the boiling point elevation, boiling pt. elevation = 0.52\textdegree × 4molal= 2.1\textdegree The normal boiling point of H2O is 100\textdegreeC the new boiling point is (100\textdegree + 2.1\textdegree or 102.1\textdegreeC). (b) Suspension of 100 g of powdered glass. These particles will not dissolve to their molecular components in H_2O, thus the boiling point of the water will not be changed. (c) 1.2 × 10^24 (sucrose molecules / liter). One liter of H_2O weighs 1 kg, therefore, 1.2 × 10^24 sucrose molecules are dissolved in 1 kg of H_2O. The number of moles present is equal to the number of molecules present divided by Avogadro's number, the number of particles in one mole. no. of moles = (1.2 × 10^24 molecules) / (6.02 × 10^23 (molecules /moles) = 2 moles Hence, there are 2 moles of sucrose in 1 kg of water, thus the solution is 2 molal . Sucrose does not ionize in water, therefore, the truemolalityis equal to the effectivemolality. Solving for the boiling point elevations: boiling pt. elevation = 0.52\textdegree × 2molal= 1.04\textdegree The boiling point of this solution is 100\textdegree + 1.04\textdegree or 101.04\textdegree. (d) 0.5molalBaCl_2. BaCl_2 is a strong electrolyte and will completely ionize in H_2O. BaCl_2\rightleftarrowsBa^++ + 2Cl^- 3 ions will be formed for each BaCl_2 present. The effectivemolality will therefore be 3 times themolalityof unionized BaCl_2. effectivemolality= 3 × 0.5molal= 1.5molal One can now solve for the boiling point elevation. boiling pt. elevation = 1.5molal× 0.52\textdegree = 0.78\textdegree The boiling point of H_2O, 100\textdegree, will be raised 0.78\textdegree to be 100.78\textdegreeC.

Question:

A string is wrapped around a cylinder of mass M and radius R (see figure (a)). The string is pulled vertically upward to prevent the center of mass from falling as the cylinder unwinds the string, (a) What is the tension in the string? (b) How much work has been done on the cylinder once it has reached an angular speed w? (c) What is the length of string unwound in this time?

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Solution:

We must first find the moment of inertia I of a cylinder of mass M and radius R. To start, we first calculate the moment of inertia of a disk of mass density \rho and radius R, whose thickness is negligible (see figure(b)). The differential mass element dm, shown as the dotted ring in the figure, has area 2\pir dr. Thus: dm = 2\pi \rhor dr Therefore: I = \int r^2 dm = \int r^2 (2\pi\rhor dr) = 2\pi\rho ^R\int_0 r dr = 2\pi\rho (r^4/4) ]^R_0 = (\pi\rhoR^4)/2 = (1/2)(\piR^2\rho)R^2 = (1/2) mR^2 We have used the fact that \piR^2\rho is the mass m of the disk, since \rho is the mass density, and \piR^2 is the total area. We can think of a cylinder as many disks squeezed together, and we know that the total moment of inertia about a common axis, of many objects, is the sum of the individual moments of inertia: I_t = I_1 + I_2 + I_3 + ..... In the case of a cylinder: I = (1/2) m_1R^2 + (1/2) m_2R^2 + (1/2) m_3R^2 + ... = (1/2) MR^2 where M = m_1 + m_2 + m_3 + ... (a) The tension T of the string must exactly balance the cylinder's weight if the cylinder is not allowed to fall. Thus: T = Mg (b) The amount of work done on the cylinder equals its gain in (rotational) kinetic energy. Since the initial angular velocity is assumed to be zero: W = KE_r = (1/2) I \omega^2 = (1/2) [(1/2) MR^2]\omega^2 1/4MR^2\omega^2 (c) To calculate the length of string unwound we place ourselves in the reference frame in which the string is at rest. In this frame, the cylinder rolls forward at a linear acceleration: a = R\alpha leaving a trail of string behind. Here \alpha is the angular acceleration of the rotating cylinder. From the laws of rotational dynamics: \cyrchar\cyrt = I\alpha\alpha = \cyrchar\cyrt/I where \cyrchar\cyrt = MgR is the torque that the string exerts about the cylinder's center of mass. The length of string, S, that the cylinder unwinds after time t equals the distance it travels in this reference frames: S= v_0t + (1/2) at^2 ,v_0 = initial velocity = 0 S= (1/2) at^2 = (1/2) \alphaRt^2 = (1/2) (\cyrchar\cyrt/I) Rt^2 = (1/2) [(MgR)/(1/2 MR^2)] Rt^2 = gt^2 (These are the kinematics equations for constant acceleration.) We know that the time t it takes for the cylinder's angular velocity to each w is given by the angular kinematics equations for constant \alpha: w= w_0 + \alphat,w_0 = initial velocity = 0 w= \alphat t= w/\alpha = [w / {MgR/(1/2) MR2} ] = (wR)/(2g) Thus:S= g [(wR)/(2g)]^2 = (w^2R^2)/(4g)

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Question:

A 0.240 g sample of a compound of oxygen and element X was found by analysis to contain 0.096 g of X and 0.144 g of oxygen, (a) Calculate the percentage compo-sition by weight, (b) Calculate from the above data, three possible atomic weights for X relative to oxygen (at. wt. = 16). (c) What additional information is needed to calculate the true atomic weight of X?

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Solution:

(a) The percent composition by weight of each element in the compound is found by dividing the weight of that element by the weight of the compound and then multiplying the quotient by 100. For O, there-fore, you obtain the following percentage: {(.144) / (0.240)} × 100 = 60 % O For X: {(.096) / (0.240)} × 100 = 40 % X (b) If you assume that the compound is XO, you can solve for the molecularweight of X by using the following ratio. Let x = MW of X. {(.144g O) / (16.0 g/mole O)} = {(.096 g) / (X g/mole x)} MW X = 10.7 If the compound is taken to be X_2O, there are twice as many X atoms present as O atoms. In solving for the molecular weight of X in this case, use the following ratio (.144) / (weight of 1 mole of O) = (.096) / (weight of 2 moles of x) Let x = MW of X .144g / 16.0 = 0.96g / x x = (16.0 × .096g) / .144g x = 5.35. Assume also that the formula for the compound might be XO_2. Then, the following ratio should be used: Let x = MW of X. .144g / (weight of 2 moles of O) = .096 / (weight of 1 moles of X) .144 / 32.0 = .096g / X X = (.096g × 32.0) / .144 g = 21.4 . (c) To solve for the true atomic weight of X, you must know the actual number of X atoms present per each O atom present.

Question:

What is the difference between a kinesis and a taxis?

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Solution:

Bothkinesesand taxes are types of animal orientation. Orientation is the process by which the animal directs theconsummatoryact at an appropriate object (the goal). It is involved in those forms of appetitive behavior in which the animal moves in a constant direction in search of a goal. Highly developed forms of orientation are involved in the migration and homing behavior of birds.Kinesesand taxes are two simpler forms of orientation. A kinesis is a form of orientation in which the animal does not necessarily direct its body toward the stimulus. The stimulus merely causes the animal to speed or slow down its movements, resulting in the eventual displacement of the animal toward or away from the stimulus. It is an undirected type of orientation. The movement of the woodlouse is a good example of a kinesis. A woodlouse is a crustacean that lives in damp environments beneath stones or pieces of wood. When the woodlouse finds itself dehydrating, it begins to move around. This increase in locomotoractivity may cause the woodlouse to chance upon a moist spot. Here, the dampness causes the woodlouse to slow down and it now establishes a new home. The woodlouse did not direct itself toward the moist spot, but simply encountered it by random movement. In contrast, a taxis is a type of orientation in which the animal directs its body toward or away from the stimulus. The type of stimulus is usually denoted by a prefix to the word taxis: a geotaxis is movement regulated by gravity; aphototaxisis movement guided by light; a chemotaxisis movement guided by chemical substances. Movement toward the stimulus is a positive taxis whereas movement away from the stimulus is a negative taxis. For example, the directed movement of cockroaches away from light is termed negativephototaxis. Another common example of a taxis is the directed movement of the bee flying to its hive or source of pollen. The bee maintains a straight course by moving at a constant angle to the sun, therefore exhibiting aphototaxis. The term taxis should be distinguished from tropism: Tropism indicates growth of an organism in a particular direction. The phototropism of plants is a typical tropism.

Question:

We know that if we drop an object of mass m while giving it a horizontal velocity component, the object will fall toward the surface of the Earth with the horizontal velocity remaining constant. With what velocity must an object be projected so that the curvature of its path is just equal to the curvature of the Earth?

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Solution:

In such a situation, the object would fall toward the Earth at the same rate that the surface of the Earth curves away from the instantaneous velocity vector; that is, the object would fall around the Earth. The height of the object above the surface of the Earth would therefore never decrease and the object would become a satellite of the Earth. Suppose that we start with the object at a distance h above the surface of the Earth. The radius of the Earth is R so that the radius of the desired circular path of the object is R + h. The centripetal force required to maintain the circular motion is F =ma_c= (mv^2)/r = (mv^2)/(R + h) The centripetal acceleration is furnished by gravity, so we can substitute g [=GM_e/(R + h)^2] for a_c where G is the gravitational constant and M_e is the mass of the Earth. Thus, m [v^2/(R + h)] = m [ GMe GMe /(R + h)^2] , /(R + h)^2] , hencev = \surd[(GMe)/(R + h)] hencev = \surd[(GMe)/(R + h)] The period of the motion is \cyrchar\cyrt = 2\pi/\omega = 2\pir/v forv = (length of one orbit)/(time to make one orbit) = (circumference)/(period)

Question:

Given that log_102 = 0.3010 and log_103 = 0.4771 find log_10\surd6.

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Solution:

\surd6 = 6^1/2, thus log_10\surd6 = log_106^1/2. Since log_bx^y = y log_bx, log_106^1/2 = (1/2) log_106. Therefore log_10\surd6 = (1/2) log_106. 6 = 3 \bullet 2, hence (1/2) log_106 = (1/2) log_10(3 \bullet 2). Recall log_10(a \bullet b) = log_10a + log_10b. Thus (1/2) log_10(3 \bullet 2) = (1/2) (log_103 + log_102). Replace our values for log_103 and log_102.

Question:

The maximum permissible dosage for scientific workers using \Upsilon- radiation is 6.25milliroentgensper hour. What is the safe working distance from a 20-Ci source of cobalt-60 which has a dose rate of 27 roentgens (R) per hour at a distance of lm? If the source is used in a lead container which reduces the \Upsilon-radiation emitted to 1%, how much closer can the scientists work?

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Solution:

The radiation from the source is emitted uniformly in all directions. Hence at any distance r the radiation is passing through an area 4\pir^2 . The intensity of radiation per unit volume thus de-creases as r increases according to an inverse-square-power law. The diminution of the intensity of the \Upsilon-radiation due to absorption in the air may be neglected in comparison with this inverse-square-law effect. Hence if r is the safe working distance, [I (r)] / [I(r_0 )] = (r^2 _0) / (r^2 ) where I(r) is the intensity at a distance r from the source, and I(r_0 ) is the value of the intensity atr_0. Hence r^2 = [I(r_0 )(r^2 _0)] / [I (r)] r = r_0 \surd[ {I(r_0 ) } / { I (r) } ] But we know that when r_0 = lm, l(r_0 ) is 27 roentgens/hr. Hence, if I = 6.25 milliroentgens/hr at r, r = (Im) \surd[(27 r/hr) / (6.25 ×10\Elzbar3r/hr)] or r = [\surd{(27 ×10^3 ) / (6.25 )} ] ×Im= 65.7 m. If the lead container cuts the radiation to 1% of its former value, then by the same arguments, the new safe working distance r_1 . is given by r_1 = r_0 \surd[ {I(r_0 ) } / { I (r) } ] Note that only (I r_0 ) is changed. I(r_0 ) is now 27 × 10\Elzbar2roentgens/hr, and r_1 = (Im)\surd{(.27 ×10^3 ) / (6.25 )} = 6.57 m \surd{(.27 ×10^3 ) / (6.25 )} = 6.57 m

Question:

Krypton crystallizes in the face-centered cubic system, with the edge of the unit cell 5.59 \AA. Calculate the density of solid krypton. Assume 1 \AA = 10^-8 cm.

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Solution:

To determine the density of krypton in g/cc, one can use the given information in the following way; 1) determine the number of atoms in one cube 2) determine the number of cubes in one cubic centimeter 3) determine the weight of one cube 4) determine the weight of one cubic centimeter (i.e. the density) . Determining the density: 1) In a face centered cube, there are 14 atoms contributing to the volume. 8 are corner atoms and 6 are face atoms. Corner atoms contribute 1/8 of their volume to the cube and face atoms contribute 1/2 of their volume. Thus, the number of atoms making up the cube can be found. no. of atoms = (1/8 atom/corner × 8 corners) + (1/2 atom/face × 6 faces) = 4 atoms. 2) 1 \AA = 10^-8 cm. Thus, the side of the cube is equal to 5.59 \AA or 5.59 × 10^-8 cm. The volume of a cube is equal to the length of the side cubed. volume of one cube = (5.59 × 10^-8 cm)^3 = 1.75 × 10^-22 cc The number of cubes in 1 cc is found by dividing 1 cc by the volume of one cube. no. of cubes in 1 cc = 1 cc / [1.75 × 10^-22 cc] = 5.71 × 10^21 3) The weight of one mole of krypton is 84. [6.02 × 10^23] atoms (Avogadro's number) of it must weigh 84 g. Therefore, the weight of one atom is found by dividing 84 g by 6.02 × 10^23 atoms. There are 4 atoms in one cube, thus the weight of one cube is equal to the weight of one atom times 4. weight of 1 atom = 84 g / [6.02 × 10^23 atoms] weight of 4 atoms = 4 atoms × {84 g / [6.02 × 10^23 atoms]} = 5.58 × 10^-22 g. 4) The density is now found by multiplying the weight of 1 cube by the number of cubes in 1 cc, density = no. of cubes in 1 cc × weight of 1 cube density = 5.71 × 10^21 /cc × 5.58 × 10^-22 g = 3.19 g/cc.

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Question:

A 10 kg mass resting on a table is attached to a steel wire 5 m longwhich hangs from a hook in the ceiling.If the table is removed, the wire is momentarily stretched and the mass commences to oscillate up and down. Calculate the frequency of the oscillation if the cross-sectional area of the wire is 1 sq mm and the value of Young's Modulus is taken to be 19 × 10^11 dynes/cm^2.

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Solution:

Young's modulus is defined as Y = [(F/A)/(∆x/x)] where∆x/x is the fractional change in length of the wire, A is its cross-section and F is the force normal to A. \therefore F =Y(∆x/x)A But F = ma (Newton's second law, where a = acceleration). \therefore a = F/m = [(Y/m)(A/x)]∆x(1) Since a is proportional to the extension of the wire from its equilibrium length (∆x), the motion is simple harmonic. For this type of motion , a = \omega^2 _0 x(2) where \omega_0 is the angular frequency of the oscillation. Comparing (2) and (1) \omega^2 _0 = YA/mx and w_0 = \surd(YA/mx) If f is the frequency of the oscillation \omega_0 = 2\pif =\surd(YA/mx) where f = 1/2\pi \surd(YA/mx) Using the data supplied f = 1/2\pi\surd[{(19 × 10^11 dynes/cm^2)(1 mm)^2}/(10 kg)(5 m)] Transforming all quantities toc.g.s. values f = 1/2\pi\surd[{(19 × 10^11 dynes/cm^2)(10\Elzbar2mm)^2}/{(10000 g)(500 cm)}] f\approx 10vib/s.

Question:

What is the centripetal force required to cause a 3200 pound car to go around a curve of 880-ft radius at a speed of 60 mph?

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Solution:

When a body travels in uniform circular motion, it experiences an acceleration towards the center of the circle. Since the object has a mass, a force towards the center of the circle is produced. In circular motion, the acceleration of the body is given by a = v^2/r where v is the linear velocity of the object and r is the radius of the circle. In our case (using 60 mph = 88 ft/sec): a= (88 ft/sec)^2/(880 ft) a= (88 ft/sec)^2/(880 ft) = 808 ft/sec^2 = 808 ft/sec^2 The mass of the car is m= Wt/a = (3200 lb)/(32 ft/sec^2) =100[(lb - sec^2)/(ft)] [slugs] AndF=ma F= 100[(lb - sec^2)/(ft)] × 8.8 ft F= 100[(lb - sec^2)/(ft)] × 8.8 ft /sec^2 = 880 lb.

Question:

An electron with velocity v = 10^9 cm/s enters a region of length L = 1 cm in which there exists a transverse deflecting field E = .1 statvolt/cm (see figure). What angle with the x axis does the electron make on leaving the deflecting region?

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Solution:

From the figure, we see that the angle made by v_f (the exit velocity) with the x axis is tan \Phi = v_(f)y / v_(f)x(1) where v_(f)y / v_(f)x are the y and x components of the-electron's final velocity. We note that since there is no acceleration in the x- direction, the x component of the electron's velocity must remain constant. Hence v_(f)x = v(0)x= v = 10^9 cm / s(2) We must now calculate v_(f)y. In order to find v_(f)y, we must apply Newton's Second Law (F = ma) to the vertical (y) component of motion of the electron. By definition of electric field intensity E^\ding{217} = F^\ding{217} / q where F^\ding{217} is the force acting on charge q. For the electron, q = -e where e is the electron charge. Therefore F^\ding{217} = -e E^\ding{217} Furthermore, from the figure E^\ding{217} = E \^{\j} where \^{\j} is a unit vector in the positive y direction. Then F^\ding{217} = (- e)(- E\^{\j}) = eE \^{\j} ButF^\ding{217} = ma^\ding{217} where m, a^\ding{217} are the mass and acceleration of the electron. Hence ma^\ding{217} = eE\^{\j} anda^\ding{217} = eE\^{\j} / m We note that E^\ding{217} is constant, and, therefore, a^\ding{217} is constant. We may then use the kinematics equations for constant acceleration to find v_(f)y . Also , a^\ding{217} is one dimensional and we may use a scalar relation for v_(f)y v_(f)y= v_(0)y+ at = v_(0)y+ (eE / m) t where t is the time it takes for the electron to travel the distance L, and v_(0)y = 0, as shown in the figure. Then v_(f)y= (eE / m) t But by definition of v_x , (and since v_x is constant) t = L / v_(0)x Thereforev_(f)y = (eE / m) (L / v_(0)x ) = eEL / mv_(0)x Substituting in (1) and using (2) tan \Phi = v_(f)y/ v(f)x= (eEL / m v_(0)x ) (1 / v_(0)x ) tan \Phi = eEL / mv^2_(0)x tan \Phi = [(4.8 × 10^\rule{1em}{1pt}10 esu) (10^\rule{1em}{1pt}1 statvolt / cm)(1cm)] / [(9.11 × 10\rule{1em}{1pt}28gm)(1018cm^2 / s^2) But1 statvolt = 1 esu, and tan \Phi = [(4.8 × 10^\rule{1em}{1pt}11) esu^2 ] / [(9.11 × 10\rule{1em}{1pt}10)gm\bullet cm^2 / s^2] Also,1 (esu^2 / cm) = 1 dyne Hence1 dyne \bullet cm = 1 esu^2 and1gm \bullet cm^2 / s2= 1 dyne \bullet cm Thereforetan \Phi = .0527 \Phi \approx 3\textdegree

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Question:

Design an experiment which would select for a nutritional mutant .

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Solution:

Organisms possess structures called genes which determine their characteristics . A gene is capable of changing, or mutating, to a different form so that it determines an altered characteristic. Any organism that has a specific mutation is called a mutant. Among the large variety of bacterial mutants are those which exhibit an increased tolerance to inhibitory agents such as antibiotics. There are also mutants that exhibit an altered ability to produce some end product, and mutants that are nutritionally deficient (unable to synthesize or utilize a particular nutrient.) These nutritional mutants are calledauxotrophsbecause they require some nutrient not required by the original cell type or prototroph. The first step in isolating a nutritional mutant is to increase the spontaneousmutation rate of the bacteria, which usually ranges from 10^\rule{1em}{1pt}6 to 10^\rule{1em}{1pt}10 bacterium per generation.(This means that only 1 bacterium in 1 million to 1 in 10 billion is likely to undergo a mutational change.) The mutationrate can be significantly increased by exposing a bacterial culture (e.g. Escherichia coli), to ultraviolet radiation or x-rays. A portion of the irradiatedE.coliculture is placed on the surface of a Petri dish, and spread over the surface to ensure the isolation of colonies. This Petri dish must contain a "complete" medium, such as nutrient agar, so that all bacteria will grow, including the nutritional mutants . Nutrient media contain all the essential nutrients needed for growth . After incubation of this "complete" medium plate, the exact position of the colonies on the plate is noted. A replica plating device is then gently pressed to the surface of the complete plate, raised, and then pressed to the surface of a "minimal" media plate. A replica plating device consists of a sterile velveteen (a cotton fabric woven like velvet) cloth. Cells from each colony adhere to the cloth in specific locations so that each colony remains isolated. These cells serve as aninoculumwhen the cloth is pressed to another plate. The replica plating devicetranfersthe exact pattern of colonies from plate to plate. The positioning of the cloth on the "minimal" agar plate must be identical to its positioning when originally pressed to the complete plate. Colony locations will then be comparable on each of the two plates, which are termed "replicas." The "minimal" medium consists only of glucose and inorganic salts, which are nutrients which normally permit the growth ofE.coli. From these basic nutrients , normalE. colican synthesize all required amino acids, vitamins, and other essential components. After incubation, colonies appear on the minimal plate at most of the positions corresponding to those on the complete plate. Those missing colonies on the minimal plate are assumed to be nutritional mutants, becausethey cannot grow on a glucose inorganic salts medium. These missingcolonies can be located on the complete media plate by comparing the location of colonies on the replicas. If they had not been irradiated , and if mutations did not occur, all the colonies would have been able to grow on this minimal medium. The colonies that did grow were not affected by the irradiation, and were non-mutants. To determine the exact nutritional deficiency of theauxotrophs, one can plate them on media which contain specific vitamins or amino acids in addition to glucose and inorganic salts. These compounds are normally synthesized byprototrophicE. coli. The mutation might have affected a gene which controls the formation of an enzyme which, in turn, regulates a stepin the biosynthesis of one of these nutrients. By plating the mutant on several plates with each plate containing an additional specific nutrient, wecould determine which nutrient the mutant cannot synthesize. The mutant will grow on that plate containing the specific nutrient which the mutant is unable to synthesize.

Question:

A car travels at the constant speed of 30 mph for 20 miles, at a speed of 40 mph for the next 20 miles, and then travels the final 20 miles at 50 mph. What was the average speed for the trip?

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Solution:

For situations in which the speed is variable, the rate at which distance d is traveled as a function of time, t, can be described by the average speed. The average speedvis equal to that constant speed which would be required for an object to travel the same distance d in the same time t. Therefore v= d/t. The total time the car travels is the sum of the times for each segment of the trip. t= t_1 + t_2 + t_3 = (d_1/v_1) + (d_2/v_2) + (d_3/v_3) t= [(20 mi)/(30 mph)] + [(20 mi)/(40 mph)] + [(20 mi)/(50 mph)] = (0.67 + 0.50 + 0.40)hr = 1.57 hr The total distance is d= d_1 + d_2 + d_3 = (20 + 20 + 20)mi = 60 mi Therefore, the average speed is v= d/t = (60 mi)/(1.57 hr) = 38.2 mph

Question:

Suppose a rain cloud has a water volume of 10^5 liters. If the cloud forms over a chemical factory which puts SO_3 into the air and each SO_3 molecule bonds with one out of 10^4 H_2O molecules (to form H_2SO_4), what is the molarity of the H_2SO_4 in the cloud?

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Solution:

To solve this problem, the number of moles of H_2SO_4 must be determined. This is done by applying the stoichiometric rules to the equation of this particular reaction: H_2O + SO_3 \rightarrow H_2SO_4 According to this equation for every mole of SO_3 consumed, one mole of H_2SO_4 is produced or, in other words, if the number of moles of SO_3 consumed is known, then the number of moles of H_2SO_3 produced is also known. The number of moles of SO_3 can be found since it is the same number as H_2O (from the stoichiometry of the equation) . The number of moles of H_2O consumed can be found by dividing its weight by its molecular weight (MW = 18.02) or number of moles of H_2O = [{(10^5 liter) (1000 g / liter)} / {18.02 g / mole}] = 5.5 × 10^6 moles Note: 1000 g/liter is the density of water (rain). However, the number of water molecules is needed and thus, number of H_2O molecules = (5.5 × 10^6 moles) (6.02 × 10^23 molecules/mole) = 3.3 × 10^30 molecules. The problem states that SO_3 molecules bond with one out of 10^4 H_2O molecules. From this, the number of SO_3 molecules is determined, {[3.3 × 10^30 H_2O molecules] / [10^4(H_2O molecules/SO_3 molecules)]} = 3.3 × 10^26 SO_3 molecules. The number of moles of SO_3 can now be obtained. number of moles of SO_3 = {[3.3 × 10^26 SO_3 molecules] / [6.02 × 10^23 (molecules / mole)]} = 5.5 × 10^2 moles or 5.5 × 10^2 moles of H_2SO_4 are formed. The molarity of the H_2SO_4 produced is thus (5.5 × 10^2 moles) / (10^5 liter) = 5.5 × 10^-3 M.

Question:

Prove that when light goes from one point to another via a plane mirror, the path chosen is the one which takes the least time.

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Solution:

Let the points be A and B, and let C be any general point on the mirror. Orient the diagram so that the x- and y-axes are as shown. Draw the normals to the mirror surface passing through A, B and C. Now in specular reflection, the reflected ray lies in the plane determined by the incident ray and the normal to the mirror at the point of reflection. Hence A, B and C must be in the same plane. The coordinates of the three points are A(x_1, 0); B(x_2, y_0); C(0, y). The length of the path ACB is, by the Pythagorean theorem, p = \surd(x^2_1 + y^2) + \surd[x^2_2 + (y_0 - y)^2] But the time of travel of light by this path, the velocity of light being, c, is t = p/c. For the path to be traveled in minimum time, we must have dt/dy = 0, where y is the variable which changes with path. Thus t = (1/c) [(x^2_1 + y^2)^1/2 + {x^2_2 + (y_0 - y)^2}^1/2] (dt/dy) = (1/c) [(1/2) (x^2_1 + y^2)^-1/2 (2y) + (1/2) { x^2_2 + (y_0 - y)^2}^-1/2 × 2(y_0 - y) (- 1)] (dt/dy) = (1/c) [{y / \surd(x^2_1 + y^2)} - {(y_0 - y) / \surd(x^2_2 + (y_0 - y)^2)}] To find the values of y which make t a minimum, we set dt/dy = 0. Hence, 0 = (1/c) = {y/\surd(x^2_1 + y^2)} - {(y_0 - y)/\surd[x^2_2 + (y_0 - y)^2]} But since c \not = \infty, the quantity in the braces must be zero. Therefore, [y/{\surd(x^2_1 + y^2)}] = (y_0 - y)/[\surd{x^2_2 + (y_0 - y)^2}](1) sin \texttheta_1 = [y/(\surdx^2_1 + y^2)]sin \texttheta_2 = (y_0 - y)/[\surd{x^2_2 + (y_0 - y)^2}] Hence, (1) becomes sin \texttheta_1 = sin \texttheta_2 and consulting the diagram we see that, \texttheta_1 = \texttheta_2 which is the law of reflection. Since light reflected specularly always satisfies this condition, the light ray follows the path which .takes the least time.

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Question:

A 12-V automobile battery supplies 20C of charge in 5 s to a stereo tape player. What is the electrical current flowing into the device and how much work is done by the battery.

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/Users/wenhuchen/Documents/Crawler/Physics/D20-0656.htm

Solution:

The charge supplied by the battery, q = 20 C, and the time interval, t = 5 s, are given. The current is, by definition, I = (q/t) = [(20C)/(5 s)] = 4 A The potential of the battery, V = 12 V, is also given. However, the potential difference between two points is the work required to move a unit charge from the point at lower potential to that of higher potential. Or V = (W/q ) ThenW = Vq = (12 V) (20 C) = 2.4 × 10^2 J The work done by the automobile battery is 240 J.

Question:

High levels of progesterone are essential for the maintenance of the uterine lining during pregnancy. But high levels of progesterone exerts a negative feedback action on LH secretion, with a consequent decrease in LH secretion and eventual atrophy of the corpus luteum. Atrophy of the corpus luteum cuts the progesterone supply resulting in menstruation. Clearly, these events cannot take place after conception, or the embryo would get sloughed off. When conception occurs, the corpus luteum is seen to last through most of pregnancy. How can this be explained?

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/Users/wenhuchen/Documents/Crawler/Biology/F21-0562.htm

Solution:

It has been shown that the portion of the placenta called the chorion secretes a hormone, called chorionic gonadotrophin, that is functionally very similar to LH. The function of this hormone is to take the place of LH in preserving the corpus luteum, LH production being inhibited by a high progesterone level. The corpus luteum is then able to secrete progesterone at high levels without being shut off. The amount of this chorionic hormone produced in a pregnant woman is very great. A pregnant woman produces so much of this hormone that most of it is excreted in the urine. Many commonly used tests for pregnancy are based on this fact. One test involves the injection of the patient's urine into a test animal such as a rabbit. If chorionic hormone is in the urine, development of a corpus luteum within 24 hours will take place in the rabbit, and pregnancy can be confirmed. Although the corpus luteum is essential during early pregnancy and is present during most of pregnancy, it has been shown that it is no longer necessary after about the first two months. It seems that the placenta begins to secrete progesterone (and estrogen) early in pregnancy, and once this secretion has reached a sufficiently high level the placenta itself can maintain the pregnancy in the absence of the corpus luteum.

Question:

The barred pattern of chicken feathers is inherited by a pair of sex-linked genes, B for barred and b for no bars. If a barred female is mated to a non-barred male, what will be the appearance of the progeny?

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/Users/wenhuchen/Documents/Crawler/Biology/F25-0668.htm

Solution:

This is an example of a sex-linked cross. Using the notations for the sex-linked traits: let X^B be the chromosome carrying the gene for the barred pattern, and X^b be the chromosome carrying the gene for the non-barred pattern. Since B is a dominant gene, a barred female could have one of the 2 genotypes: X^BX^B or X^BX^b. The only possible genotype for a non-barred male is X^bY. Therefore there are two possible crosses between a barred female and a non-barred male: Phenotypically, all the progeny of this cross have the barred feather pattern. F_1 X^b Y X^B X^BX^b X^BY X^b X^bX^b X^bY Phenotypically:X^B X^bisabarred female, X^b X^bisanon-barred female, X^B Xisabarred male,and X^b Xisanon-barred male.

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Question:

Write a FORTRAN program to evaluate the polynomial expression a1x^n-1 + a2x^n-2 + ... + a_n forgiven values of n, a1, a2,...,a_n using various values of x whichare read in. Let the program terminate when a zero valueof x is read in.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G21-0521.htm

Solution:

The most efficient polynomial evaluation procedure is based onthe nesting (...((a1x+ a2)x + a3)x + ... + a_n-1)x + a_n . We assume n \leq 25. The data deck consists of a value for n followed by the ncoefficientsa_ion one or more data cards. These are fol-lowed by successivedata cards, each with a value for x; the final card contains the zerovalue. This nesting procedure is known as Horner's method for evaluating poly-nomials. We present the program segment below. CPOLYNOMIAL EVALUATION DIMENSIONA(25) READ, N,(A(J), J = 1,N) 16READ, X IF (X.EQ.O.) STOP POLY =A(1) DO 12 I = 2,N 12POLY = PGLY\textasteriskcenteredX +A(1) GO TO 16 STOP END

Question:

A differential hoist is used to lift a load of 450 pounds. If the larger wheel of the hoist has 9 slots and the smaller wheel 8 slots and if the efficiency of the hoist is 33 (1/3)%, what is the effort required to raise the load?

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/Users/wenhuchen/Documents/Crawler/Physics/D05-0292.htm

Solution:

The IMA = 2 × 9 = 18. This is to since there are 9 strands supporting the load. Since efficiency = AMA/IMA , 0.33 (1/3) = AMA/18, whence AMA = 6. Since AMA = R/E , 6 =(450 pounds)/(E), whence E = 75 pounds.

Question:

A force of 0.20newtonacts on a mass of 100 grams. What is the acceleration?

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/Users/wenhuchen/Documents/Crawler/Physics/D04-0098.htm

Solution:

From Newton's Second Law we have F = ma a = F/m Also, 100 grams = 0.10 kg. Therefore a = (0.20nt)/(0.10 kg) = [(0.20 kg-m/g^2)/(0.10 kg)] = 2.0 m/sec^2.

Question:

A magnetic south pole of 15 units strength, when placed 10 cm away from another pole in air, experiences a repulsion of 300 dynes. What is the nature and strength of the second pole?

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/Users/wenhuchen/Documents/Crawler/Physics/D21-0721.htm

Solution:

Since the force is one of repulsion, the second pole is a like pole, i.e. , a south pole. Until the advent of quantum theory, the nature of the magnetic force of a magnet was not well understood. The concept of magnetic pole strength due to a magnetic pole was then de-veloped in a manner analogous to the concept of electric field strength and charge. The analogue to Coulomb's law for the electrostatic force between two charges is Coulomb's law for the magnetic force between two poles of field strength m and m' separated by a distance r. It is F = (1/\mu) (mm'/r^2 ) where \mu = 1 for the CGS system being used. We are given that F = 300 dynes, m is 15 poles, \mu = 1 for air, and r = 10 cm. Therefore, m = (\mur^2 F/m) = [{(1) (10 cm)^2 (300 dynes)}/(15 poles)] = 2000 poles Although this model of magnetic poles correctly predicts experimentally observed magnetic forces between poles, magnetic monopoles have never been observed.

Question:

Write a structural formula for each of the following molecular formulas. Name each structure that you draw, (a) CH_2O; (b) C_2H_6O_2; (c) C_2H_6O; (d) C_2H_4O_2; (e) C_2H_4O.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E20-0745.htm

Solution:

In order to do this problem, one must know how carbon, hydrogen, and oxygen bond. Carbon can share its 4 valence electrons to form 4 covalent bonds, hydrogen can share its 1 valence electron to form 1 covalent bond, and oxygen can share its 6 covalent electrons to form only 2 covalent bonds. Thus, with this knowledge in mind, one can construct several types of molecular structures. A few of these structures have the following functional groups: Alcohols contain the hydroxyl group, - OH, acids contain the carboxyl group, (represented as - COOH), aldehydes and ketones contain the carbonyl group, Structure (a), CH_2O, is an aldehyde, and has the name formaldehyde. Structure (b), C_2H_6O_2, is an alcohol and has the name 1,2 dihydroxyethane or ethylene glycol. Structure (c), C_2H_6O, is either an alcohol or an ether and is named ethanol or dimethyl ether, respectively. Structure (d) C_2H_4O_2, is an acid and has the name acetic acid. Structure (e), C_2H_4O, is an aldeyde and has the name acetaldehyde.

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Question:

In what distance can a 3000-lb automobile be stopped from a speed of 30 mi/hr (44 ft/sec) if the coefficient of friction between tires and roadway is 0.70?

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/Users/wenhuchen/Documents/Crawler/Physics/D04-0103.htm

Solution:

The retarding force furnished by the roadway can be no greater than F_s =\muN= (0.70)(3000 lb) = 2100 lb. Since the work done by this force is equal to the kinetic energy of the car, the stopping distance can be found from W = Fs = 1/2mv^2. We must divide the weight by the acceleration due to gravity, g, to obtain the mass m = W/g = (3000 lb) / (32 ft/sec^2) = 94 slugs s = (1/2mv^2)/F = [{94 slugs (44 ft/sec)^2}/{2 × 2100 lb}] = 43 ft.

Question:

An elevator is descending with an acceleration of 5 ft\bullets^-2. The shade on the ceiling light falls to the floor of the elevator, 9 ft below. At the instant that it falls, one of the passengers sees it, and realizes that the shade will hit his foot. How long does he have to get his foot out of the way?

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Solution:

At the instant that the light shade becomes detached, it has a downward speed of v_0, and it will drop freely under the action of gravity. Suppose that there is an observer outside the elevator. By the time the shade strikes the (moving) floor of the elevator, it will have been seen by the observer to have dropped a distance s in a time t. This distance is given by the free fall equation s = v_0t + (1/2)gt^2. In the same time t, the elevator floor will have traveled s - 9 ft (9 ft less than the shade, since the elevator floor doesn't have to traverse the length of the elevator) with the acceleration of 5 ft \textbullet s^-2, having started with the same downward speed v_0. \therefore s - 9 ft = v_0t + (1/2) × 5 ft\bullets^-2 × t^2. Subtract one equation from the other. Thus 9 ft = (1/2)(g - 5 ft\bullets^-2)t^2 = (1/2) × 27 ft\bullets^-2 × t^2. This equation could have been obtained more easily by considering the motion of the shade relative to the elevator in an accelerated frame of reference. For, relative to the elevator, the light shade starts off with zero velocity and has an acceleration of g - 5 ft\bullets^-2. Thus, applying the same equation of motion as before, we find that the shade drops 9 ft relative to the elevator in a time t, and 9 ft = 0 × t + (1/2)(g - 5 ft\bullets^-2)t^2. t = \surd[(2 × 9 ft)/(27 ft\bullets^-2)] = \surd(2 s^2/3) = 0.82 s. The passenger has therefore less than 1 s to get his foot out of the way, and must react rapidly.

Question:

The speed of a certaincompressionalwave in air at standard temperature and pressure is 330 m/sec. A point source of frequency 600/sec radiates energy uniformly in all directions at the rate of 5.00 watts What is the intensity of the wave at a distance of 20.0 m from the source? What is the amplitude of the wave there?

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/Users/wenhuchen/Documents/Crawler/Physics/D25-0803.htm

Solution:

At any concentric spherical surface the energy from a point source is spread over an area 4\pir^2. At a distance of r = 20.0 m, the intensity I is I = (E/tA) = (P/A) = (5.00 watts/[4\pi × (20.0 m)^2] = 0.99 × 10^\rule{1em}{1pt}3 watt/m^2 The rate of transfer of energy depends on the square of the wave amplitude and the square of the wave frequency for all types of waves, with I the average intensity and the density of air \rho = 1.29 gm/liter = 1.29 kg/m^3, we have, A^2 = I/(2\pi^2vpf^2) = [(0.99 × 10^\rule{1em}{1pt}3 watt/m^2)/{2\pi^2 (330 m/sec) (1.29 kg/m^3) (300/sec)^2}] = 1.32 × 10^\rule{1em}{1pt}12 m^2 A = 1.15 × 10^\rule{1em}{1pt}6 m = 1.15 × 10^\rule{1em}{1pt}4 cm

Question:

Using the data from the accompanying table find the enthalpy change for the combustion of a mole of C_2H_4 (g) to form CO_2 (g) and H_2O(g) at 298\textdegreeK and 1 atm standard state conditions.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E14-0497.htm

Solution:

When a chemical reaction occurs heat is usually absorbed or given off to the environment. This means that the enthalpy or heat content of the system will also change. If one knows the enthalpy change of the combustion of each reactant and product, one can find the change in enthalpy of a reaction. The reaction in this problem can be written as C_2H_4 (g) + 3O_2 (g) \rightarrow 2CO_2 (g) + 2H_2O(g) For the reaction, ∆H_combustion = \sum∆H_products - \sum∆H_reactants. Remembering that the enthalpy change for the formation of an element (i.e. O_2) in its standard state is zero, one then has, ∆H_combustion= 2∆H_CO2 + 2∆H_H2O -∆HC2 H4- 3∆H_O2 = 2(-394) + 2(-242) - (52) - 3(0) = -1324 KJ Enthalpies of formation in kilojoules per mole from the elements for various compounds at 298\textdegreek and 1 atm pressure Component ∆H C_2H_4 (g) + 52 CO_2 (g) - 394 H_2O (g) - 242 O_2 (g) 0

Question:

Calculate the equilibrium constant for the following reaction at 25\textdegreeC or 298\textdegreeK C _(graphite) + 2H_2 (g) \rightarrow CH_4 (g) ∆H\textdegree for this reaction is - 17, 889 cal.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E15-0539.htm

Solution:

At equilibrium the equilibrium constant of any reaction is independent of the amount of pure solid (or liquid) phase. To solve this problem (1) determine ∆S\textdegree, the entropy change or randomness of the system. (2) Determine ∆G\textdegree, the free energy or the energy available to do work, by using the formulas:∆G\textdegree = ∆H\textdegree - T∆S\textdegree, where ∆H\textdegree = the change in enthalpy or heat content, and T = absolute tem-perature (Celsius plus 273\textdegree) . (3) DetermineK_p, the equilibrium constant, by using ∆G\textdegree = - RT InK_p. The entropy change (from any entropy table) for this reaction is ∆S\textdegree of reaction = ∆S\textdegree _products - ∆S\textdegree _reactants = ∆S\textdegree_CH4 - 2∆S\textdegree_H2 - ∆S\textdegree_C = 44.50 - 2 (31.211) - 1.3609 = - 19.28 cal/k ∆G\textdegree = ∆H\textdegree - T∆S\textdegree = - 17, 889 - (298) (- 19.28) = - 12, 143 cal. The equilibrium constant is thus ∆G\textdegree = - RT InK_p= - (1.987 cal\textdegreeK^-1 mole^-1) (298\textdegreeK) × (2.303) logK_p - 12,143 cal = - 1364 logK_p logK_p= 8.90 K_p= 7.94 x 10^8 .

Question:

Write a FORTRAN program segment to compute values of the POISSON distribution given by P(X) = e^-mm^x/x! Assume (1) that X and m are integer variables; (2) that X and m are small enough that no term will exceed the size permitted for a real variable in your computer; and (3) that the function FACT (n) for computing n! is available.

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Solution:

Remember that the library functione^yis defined for real y, so we set y = FLOAT(m) to convert the integer m to a real variable when evaluating P. The rest of the Integer expressions, need not be FLOATED since the FORTRAN compiler allows mixing of modes in a product expression of real \textasteriskcentered integer; the result being converted to a real number. Bearing the above in mind, we can write out the program below: FUNCTION P(X,M) INTEGER X Y = FLOAT (M) P = EXP(-Y) \textasteriskcentered M \textasteriskcentered\textasteriskcentered X / FACT (X) RETURN END The factorial function, FACT (X), is defined in chapter 19.

Question:

One end of a horizontal wire is fixed and the other passes over a smooth pulley and has a heavy body attached to it. The frequency of the fundamental note emitted when the wire is plucked is 392 cycles\textbullets^\rule{1em}{1pt}1. When the body is totally immersed in water, the frequency drops to 343 cycles\textbullets^\rule{1em}{1pt}1. Calculate the density of the body.

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Solution:

Let the density of the body be \rho and its volume be V. The density of water is 1 g\textbulletcm^\rule{1em}{1pt}3 = \rho_0. In the first case, the weight of the body is balanced by the tension S^\ding{217} in the wire. In the second case, a third force, the buoyancy, enters into the calcula-tion. (See the figure.) The weight of the body is balanced partly by the new tension in the wire, S_0, and partly by the buoyancy, U^\ding{217}, acting on it according to Archimedes' principle. That is, U is equal to the weight of water displaced by the body. Since the volume of the displaced water is equal to the volume of the body, then U = (V\rho_0)g where V\rho_0 is the mass of the displaced water. Thus S = V \rhog where V\rho is the mass of the body and S_0 + U = S_0 + V \rho_0g = V \rhog, the buoyancy due to the air in the first case being ignored. The frequencies of the fundamental notes emitted in the two cases are f_1 = (1/2L) \surd(S/\mu)andf_0 = (1/2L) \surd(S_0/\mu) where \mu is the mass of the wire per unit length (the mass density). Thus (f^2_0)/(f^2_1) = (S_0/S) = [(V \rhog - V \rho_0g)/(V \rhog)] = [(\rho - \rho_0)/\rho] \therefore\rho = [(f^2_1 \rho_0)/(f^2_1 - f^2_0)] = [(392^2 s ^\rule{1em}{1pt}2 × 1 g\textbulletcm^\rule{1em}{1pt}3)/(392^2 - 343^2) s^\rule{1em}{1pt}2] = 4.27 g\textbulletcm^\rule{1em}{1pt}3.

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Question:

Use the Euler method to solve the equation of motion of a dampedhar-monic oscillator. Let the initial conditions be that att = 0, x(0) = 10 cm anddx/dt= v = 0.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G21-0531.htm

Solution:

The equation of motion is given by (md2x / dt2) = - ( cdx cdx /dt) -kx /dt) -kx where(-cdx/dt) is the damping force and -kxis the restoring force. Since Euler's method is based on first order differential equations, we rewrite the equationof motion as two first order equations: dx/dt= v(1) dv/dt= -(c/m) v - (k/m)x(2) According to Euler's method, the solution is given by dx\approx ∆x =x_new-x_old;dx=vdt) x_new=x_old+v_old∆t(1') v_new=v_old+ [(-c/m)v_old-(k/m)x_old)∆t(2') Here we have used the general result that ifdy/dx= f(x,y), then y_i+_1 =y_i+y'_i∆x, wherey'_i_-= f(x_i ,y_i). For the sake of concreteness we use the following data: m = k = 10, c = 2, x_0 = 10, v_0 =0, D \equiv ∆t = 0.1 andwe follow the motion for 2 seconds. 1\O\O REM DAMPED HARMONIC OSCILLATOR 1\O1 REM EULER'S METHOD 1\O5 PRINT "TIME\textquotedblright, "VELOCITY", "POSITION" 1\O6 PRINT 11\O READ M, 1 K, C, X\O, V\O, D 12\O LET X1 = X\O 121 LET V1 = V\O 13\O FOR T = \O TO 2 STEP D 14\O PRINT T, V1, X1 15\O LET X2 =X1 + V1 \textasteriskcenteredD 151 LET V2 = V1 + (-C\textasteriskcenteredV1/M - K\textasteriskcenteredX1/M)\textasteriskcenteredD 16\O LET X1 = X2 161 LET V1 =V2 162 NEXT T 8\O\O DATA 1\O,1\O, 2, 1\O, \O, 0.1 999 END Note that the exact solution of t = 2 gives x = -2.5807, compared with the Euler method result of -2.96507. More sophisticated numeri-cal methods (suchas the improved Euler method orRunge-Kuttatype methods) are necessaryfor better accuracy. Finally,v_newcan also be solved with the quadratic equation. If you rearrangethe equation of motion, you can obtain m(d2x / dt2) + c (dx/dt)kx= 0, whichis the quadratic form. This equation has a solution of the form x =e^\alphat.If we substitute x into the first equation and divide bye^\alphat,we get thesimpler form m\alpha2 +c\alpha+ k = 0. By using thediscrim-inantc2 - 4mk, we candetermine what type of motion is present: If c2 - 4mk is the motion is positive over-damped zero critically-damped negative oscillatory-damped

Question:

Consider an elementary inventory control system where goods are received at a rate R (units of good per time period) , where R is directly proportional to the number of goods on order, G(t), with proportionality constant 1/L, where L is the delay in ordering. Suppose the order rate is directly proportional to the difference between desired inventory and actual inventory, with proportionality con-stant 1/A where A is the time that would be required to correct the inventory if the order rate were constant. Write a FORTRAN program to simulate this system, using modified Euler's method (the method is fully explained, and the subroutine for it is given in the SIMULATION chap-ter) from time t=0 to t=t_f if I(t) is the inventory level at time t, I(o)=Io and G(o) = Go.

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Solution:

Note that this system's goal is maintaining the desired inventory. A first picture of this system might be as shown in Fig. 1. Where the solid arrows represent the flow of goods and the broken arrows represent the flow of information. If \.{I} is the time rate of change of I(t): \.{I} = R = G/L(1) If G is the net time rate of change of G(t), W the order rate, and D the desired inventory, then \.{G} = W - R \.{G} = (D-I)/A - G/L(2) Converting equations (1) and (2) into block diagram form details the relationships in this coupled negative feed-back system: The program applies the modified Euler method first to equation (1) and then to equation (2). The variable GOODS is used for G(t). I and GOODS are kept in a COMMON block so that the new computed value of I will be used in determining GOODS. A logical flag, ORDER, is used to determine which equation is being integrated: REAL T/0, 0/, TFIN, DT, REALN, ACCUR, I, GOODS, I\O, G\O, L, D, A INTEGER J, N LOGICAL ORDER/. FALSE. / COMMON ORDER, I, GOODS, L, D, A INPUT N, TFIN, ACCUR, I\O, G\O, L, D, A OUTPUT T, I\O, G\O REALN = N DT = TFIN/REALN I = I\O GOODS = G\O DO 1\O I = 1, N T = T + DT CALL MEULER (T, I, ACCUR, DT) ORDER = .TRUE. CALL MEULER (T, GOODS, ACCUR, DT) ORDER = .FALSE. OUTPUT T, I, GOODS 1\OCONTINUE STOP END FUNCTION G(W) REAL G, W, I, GOODS, D, L, A LOGICAL ORDER COMMON ORDER, I, GOODS, L, D, A G = GOODS/L IF (ORDER)G = (D - I)/A - W/L RETURN END

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Question:

In. the figure, suppose that the block weighs 20 lb., that the tension T can be increased to 8 lb. before the block starts to slide, and that a force of 4 lb. will keep the block moving at constant speed once it has been set in motion, (a) Find the coefficients of static and kinetic friction, (b) What is the frictional force if the block is at rest on the surface and a horizontal force of 5 lb. is exerted on it?

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/Users/wenhuchen/Documents/Crawler/Physics/D02-0049.htm

Solution:

a) There are 2 forces of friction which can act on a body. These are the forces of kinetic and static friction. If a body, such as that in the figure, is initially at rest and we begin pulling on it with a variable force T, the block will remain at rest. This means that no matter what force T we apply to the body, the frictional force f always balances it. However, at some value of T, the frictional force no longer balances it, and the block begins translating, We may describe this static frictional force by f_s \leq\mu_sN(1) where \mu_s is the coefficient of static friction and N is the normal force of the table on the block. (The equality holds when the block begins translating). Once the block begins translating, the static frictional force stops acting and the kinetic frictional force takes over. This force is f_k = \mu_kN(2) It is also found that f_k < fs max, and that once the block starts moving, we may reduce T and the block will still move. Applying the Second Law to the block of mass m N - mg = ma_y T - f = ma_x Here a_x and a_y are the x and y components of the block's acceleration. Because the block coesn't leave the surface of the table, a_y = 0. Hence N = mg(3) T - f = ma_x(4) The block is initially at rest, and just begins to slip when T = 8 lb. If we examine the block just before it moves, a_x = 0, and f is the maximum force of static friction. Then, using (1) T - fs max= 0 T= \mu_sN 8 lb= \mu_sN Using (3)\mu_s= (8 lb)/mg = (8 lb)/(20 lb) = .4 Once translation at constant velocity begins, T = 4 lb, a_x = 0 and f in (4) is f_k = \mu_kN. Hence T - f_k= 0 T= \mu_kN \mu_k= (4 lb)/(20 lb) = .20 b) Note that, if the block is initially at rest, a force of 8 lb is needed to start the motion of the block. Hence, if we pull the block with a force of 5 lb., a_x = 0 and, the force of static friction is acting. Then, from (4) T - f_s = 0 T = f_s = 5 lb.

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Question:

Find the change in volume of an aluminum sphere of 5.0-cm radius when it is heated from 0 to 300\textdegree C.

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/Users/wenhuchen/Documents/Crawler/Physics/D12-0454.htm

Solution:

In the case of one dimensional thermal ex-pansion, we may relate the change in length of a sample to the temperature change which it experiences by \DeltaL =\alphaL\DeltaT(1) where \alpha is the coefficient of linear thermal expansion. Dividing both sides of (1) by \DeltaT, and taking the limit as \DeltaT \rightarrow 0, we obtain the exact relation: (dL/dT) =\alphaL(2) We are specifically concerned with the change in the volume of a sphere due to a change in temperature, or (dV/dT)(3) where V is the volume of the sphere. Using the chain rule. (dV/dT) = (dV/dR) (dR/dT) assuming that V changes only as a result of a change in radius of the sphere (R). ( dV dV /dR) = (d/dR) [(4/3) \piR^3] = 4\piR^2 /dR) = (d/dR) [(4/3) \piR^3] = 4\piR^2 (dV/dT) = 4\piR^2 (dR/dT) Also, from (2), (dR/dT) =\alphaRhence, (dV/dT) = [(4\piR^3)/3] (3\alpha) = 3\alphaV(4) where V is the original volume of the sphere. If \DeltaT is small, we may write \DeltaV = 3\alphaV \DeltaT(5) Substituting the information provided into (5), we obtain: \DeltaV = 3(2.2 × 10^-5/C\textdegree) (4/3) (\pi) (5 × 10^-2 m)^3 (300\textdegreeC) \DeltaV = 10 cm^3

Question:

The most abundant element in sea water is chlorine (Cl, atomic weight = 35.5 g/mole), which is present as chloride ion, Cl^-, in a concentration of 19 g Cl^- per 1 kg of sea water. If the volume of the earth's oceans is 1.4 × 10^21 liters, how many moles of Cl^- are present in the oceans? Assume that the density of sea water is 1.0 g/cm3.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E04-0113.htm

Solution:

The total number of moles of Cl^-is equal to the total mass of Cl^- divided by its atomic weight. Hence we must find the total mass of Cl^-. This will be done by determining the total mass of the oceans and using this value in conjunction with the concentration of Cl^-to find the total mass of Cl^-. The volume of the oceans is 1.4 × 10^21 liters = 1.4 × 10^21 liters × 1000 cm^3/liter= 1.4 × 10^24 cm^3. Multi-plying this volume by the density of sea water gives the total mass of the oceans, or 1.4 × 10^24 cm3× 1.0 g/cm^3 = 1.4 × 10^24 g. Expressing this total mass as kilograms, we obtain 1.4 × 10^21g = 1.4 × 10^24 g × 10^-3 kg/g = 1.4 × 10^21 kg. For every kg of sea water, there are 19 g of Cl^-. Hence, for 1.4 × 10^21 kg there are 1.4 × 10^21 kg × 19 g Cl^-/kg of sea water = 2.7 × 10^22 g Cl^- . Therefore, the number of moles of Cl^- is massCl-= (massCl-) / (atomic weightCl-) = (2.7 × 10^22 g) / (35.5 g/mole) = 7.6 × 10^20 moles.

Question:

Two straight parallel wires each 90 cm long are 1.0 mm apart. There are currents of 5.0 amp in opposite directions in the wires. What are the magnitude and sense of the force between these currents?

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/Users/wenhuchen/Documents/Crawler/Physics/D21-0724.htm

Solution:

There are 2 steps in the solution of problems like these. First we find the magnetic field due to current #1 at any point on wire #2. (See figure). This field (B^\ding{217}_21) exerts a force on each moving charge that constitutes current #2. It is the sum of these forces which constitute the net force on wire #2 due to wire #1. (By Newton's Third Law, the force on wire #1 due to wire #2 is equal and opposite to the force on wire #2 due to wire #1). The force on an element dl_2^\ding{217} of wire #2, which is immersed in the field of wire #1 (B_21^\ding{217}) is, by definition, dF_21^\ding{217} = I_2 dl_2^\ding{217} × B_21^\ding{217} andF_21^\ding{217} = I_2\intdl_2^\ding{217} × B_21^\ding{217}(1) where the integral in (1) is evaluated over the length of wire #2 and dl_2 has the direction of I_2 . In order to calculate F_21^\ding{217} , we must find B_21^\ding{217}. The lines of magnetic induction of a long wire are circles centered on the axis of the wire. Hence, to calculate B_21^\ding{217} , we may use Ampere's Law, which is \oint B^\ding{217} \textbullet dl^\ding{217} = \mu_0 I_enc(2) In (2), the integral is evaluated over a closed path (this is the meaning of the circle on the integral sign) and I_enc is the net current enclosed by the closed path. In evaluating B_21^\ding{217}, we take the integral about a circular line of magnetic induction, of radius r, centered on wire #1.Therefore, 2\pirB_21 = \mu_0 I_1 since I_1 is enclosed by the path. Then B_21 = [\mu_0 I_1 /(2\pir)](3) The direction of this field is shown in the figure by the "X" symbol, which indicates that the field is perpen-dicular to the plane of the figure and directed into the figure. Now, we may write the magnitude of F_21^\ding{217} as F_21 = I_2\intdl_2 B_21 sin \O where \O is the angle between dl_2^\ding{217} and B_21^\ding{217} . Looking at the figure, \O is 90\textdegree and F_21 = I_1\intdl_2B_21 Using(3)F_21 = I_2\int[(\mu_0 I_1)/(2\pir)] dl_2 F_21 = [(\mu_0 I_1 I_2 l_2 )/(2\pir)] F_21 = [{4\pi × 10^-7 (Weber/amp.m)} (25 amp^2 ) (90 × 10^-2 m)}/{(2\pi) (10^-3 m)}] F_21 = 4.5 × 10^-3nt By Newton's Third Law, \vert F_21^\ding{217} \vert = \vert F_21^\ding{217} \vert

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Question:

A boy can throw a baseball horizontally with a speed of 20 m/sec. If he performs this feat in a convertible that is moving at 30 m/sec in a direction perpendicular to the direction in which he is throwing (see figure), what will be the actual speed and direction of motion of the base-ball?

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/Users/wenhuchen/Documents/Crawler/Physics/D01-0018.htm

Solution:

Since the baseball is originally travelling with the convertible, it has the speed of 30 m/sec in the direction the car is travelling. When the boy throws the ball perpendicular to the car's path, he imparts an additional velocity of 20 m/sec in that direction. The ball's velocity is then 30 m/sec in the direction the convertible is moving and 20 m/sec perpendicular to this movement. Its resultant velocity can be found through adding vectors as shown in the diagram. If the resultant velocity is R m/sec at an angle \texttheta to the direction in which the convertible is moving, then R^2 = (20)^2 + (30)^2 = 1300 R = \surd1300 = 36.06 m/sec Also, tan \texttheta = (20/30) = 0.666 From tables of tangents, \texttheta = 33.69\textdegree. Therefore, the ball has a speed of 36.06 m/sec in a direction at an angle of 33.69\textdegree to the direction in which the convertible is travelling.

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Question:

A surface probe on Mars transmitsradiowavesat a frequency of 6.0 × 10^5 sec-^1 to an earth station 8.0 × 10^7 km away. How long does it takeradiowavesto traverse this distance?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E30-0918.htm

Solution:

Sinceradiowavesare a form of electromagnetic radiation, they move at the speed of light and we can use the relationship time = distance / speed where the speed is that of light. We do not need to know the frequency. Since 1 km = 10^5 cm, 8.0 × 10^7 km = 8.0 × 10^7 km × 10^5 cm/km = 8.0 × 10^12 cm and we have time = distance / speed = (8.0 × 10^12 cm) / (3.0 × 10^10 cm/sec) = 2.7 × 10^2 sec, or4 (1/2) minutes.

Question:

Express 20\textdegreeC and - 5\textdegreeC on the Kelvin scale.

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/Users/wenhuchen/Documents/Crawler/Physics/D12-0443.htm

Solution:

The relationship between the Celsius and Kelvin scale is K\textdegree = 273\textdegree + C\textdegree Therefore 20\textdegreeC is T = 273\textdegree + 20\textdegree = 293\textdegreeK and \rule{1em}{1pt} 5\textdegreeC is T = 273\textdegree+ (\rule{1em}{1pt} 5\textdegree) = 268\textdegreeK

Question:

Ribonuclease has a partial specific volume of (0.707 cm^3 g^-1) and diffusion coefficient of (13.1 × 10^-7 cm^2 sec^-1) cor-rected to water at 20\textdegreeC. The density of water at 20\textdegreeC is 0.998 (g/cm^3). Calculate the molecular weight of ribonuclease using the Svedberg equation. Assume that the sedimentation coefficient for ribonuclease is 2.02 × 10^-13 sec.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E22-0819.htm

Solution:

The molecular weight M of a protein can be calculated from the sedimentation coefficient by means of the Svedberg equation, which is derived by equating the centrifugal force with the opposing frictional force, the condition existing when the rate of sedimentation is constant: M = [RT_s] / [D(1 - ѵ\rho)], where R is the gas constant in ergs per mole per degree (8.314 × 10^7 ergs/\textdegreeK mole), T the absolute temperature, s the sedimentation coefficient, ѵ ̅ the partial specific volume of the protein, \rho the density of the solvent, and D the diffusion coefficient. All of the values necessary to solve for M are given. 20\textdegreeC = 293\textdegreeK. Solving for M: M = [(8.314 × 10^7 ergs/\textdegreeK mole) (293\textdegreeK) (2.02 ×10^-13 sec)] / [(13.1 × 10^-7 cm^2 sec^-1) {1-(0.707 cm^3 g^-2)(0.998 g/cm^3)}] = 13,000 g/mole.

Question:

Which are the first two organs to appear in the human embryo, and how are they formed?

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/Users/wenhuchen/Documents/Crawler/Biology/F23-0602.htm

Solution:

The first two organs to appear in the embryo are the brain and spinal cord, which form by a process termed neurulation. During the third week of development, the ectoderm in front of the primitive streak develops a thickened plate of cells called the neural plate. At the center of this plate there appears a depression, known as the neural groove. At the same time, the outer edges of the plate rise in two longitudinal neural folds that meet at the anterior end. These appear, when viewed from above, like a horseshoe. These folds gradually come together at the top, forming a hollow neural tube. The cavity of the anterior part of this neural tube becomes the ventricles of the brain. At the same time, the cavity of the posterior part becomes the neural canal, extending the length of the spinal cord. The brain region is the first to form, and the long spinal cord develops slightly later. The anterior part of the neural tube, which gives rise to the brain, is much larger than the posterior part and continues to grow so rapidly that the head region comes to bend down at the anterior end of the embryo. By the fifth week of development, all the regions of the brain, i.e., the forebrain, midbrain, and hindbrain are established. A short time later, the outgrowths that will form the large cerebral hemispheres begin to grow. While the various motor nerves grow out of the brain and spinal cord, the. sensory nerves do not, having a separate origin. During the formation of the neural tube, bits of nervous tissue, known as neural crest cells, are left over on each side of the tube. These cells migrate downward from their original position and form the dorsal root ganglia of the spinal nerves and the postganglionic sympathetic neurons. From sensory cells in the dorsal root ganglia, dendrites grow out to the sense organs and axons grow in to the spinal cord. Other neural crest cells migrate and form the medullary cells of the adrenal glands, the neurilemma sheath cells of the peripheral neurons, and certain other structures.

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Question:

Let the acceleration of a non-inertial frame, a freely falling elevator, be a_O^\ding{217} = -g\^{y} where \^{y} is measured upward from the surface of the earth and g is the acceleration of gravity. Under these conditions, what is the net force acting on a mass M at rest in the elevator?

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/Users/wenhuchen/Documents/Crawler/Physics/D04-0214.htm

Solution:

Note that in this problem, we are doing an experiment in a non-inertial reference frame. It is very important to realize that Newton's Laws only hold in inertial frames. If we want to analyze this experiment from the point of view of an observer in a non inertial frame, we must use a modified form of Newton's Second Law, Ma"^\ding{217} = F^\ding{217} - F'^\ding{217}(1) where a"^\ding{217} is the acceleration of M as observed in the non-inertial frame (the elevator) ,F^\ding{217} is the sum of all real forces acting on M (i.e., tensions, gravity, etc.) and -F'^\ding{217} is the sum of all fictitious forces. These latter forces arise because we are doing our experiment in a non--inertial frame. (Fictitious forces include Carioles forces, centrifugal forces, etc.) In this particular problem, a"^\ding{217} = 0 because the mass M is at rest in the elevator. F^\ding{217}, the sum of the real forces acting on M, includes Mg, the weight of M, and N, the normal force exerted by the floor of the elevator on the mass. Hence, F = N - Mg(2) where the minus sign appears because N and Mg are in opposite directions. We may now substitute equation (2) into equation (1) and solve the latter for F'^\ding{217} : 0 = F - F' F' = F F' = N - Mg(3) Now, - F', the fictitious force, is defined as - Ma'^\ding{217}, where a'^\ding{217} is the acceleration of the non-inertial frame with respect to the earth (-g\^{y}in our case). Hence, - F' = Mg(4) Inserting this in (3), we find - Mg = N - Mg(5) N = 0(6) Hence, the floor exerts no force on M. If this is the case, then M must be floating inside the elevator, i.e., M is weightless.

Question:

What is reverse transcriptase and what is its function? How might it be used in cancer research?

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/Users/wenhuchen/Documents/Crawler/Biology/F24-0639.htm

Solution:

We have seen in protein synthesis that biological information flows from DNA to RNA to protein. This flow had become the rule until 1964, when Temin found that infection with certain RNA tumor viruses (can- cer- causing substances) is blocked by inhibitors of DNA synthesis and by inhibitors of DNA transcription. This suggested that DNA synthesis and transcription are re-quired for the multiplication of RNA tumor viruses. This would mean that the information carried by the RNA of the virus is first transferred to DNA, whereupon it is trans-cribed and translated, and consequently, that information flows in the reverse direction, that is, from RNA to DNA. Temin proposes that the RNA of these tumor viruses, in their replication, are able to form DNA. His hypothesis requires a new kind of enzyme - one that would catalyze the synthesis of DNA using RNA as a template. Such an enzyme was discovered by Temin and by Baltimore in 1970. This RNA - directed DNA polymerase, also known as re-verse transcriptase, has been found to be present in all RNA tumor viruses. An infecting RNA virus binds to and enters the host cell (the cell which the virus attacks). Once in the cytoplasm, the RNA genome is separated from its protein coat (see figure). Then, through an as yet unknown mechanism, the viral reverse transcriptase is used to form a DNA molecule using the viral RNA as a template. This DNA molecule is integrated into the host's chromoso-me, with a number of possible consequences. The viral DNA may now be duplicated along with the host's DNA, and its information thus propogated to all offspring of the infected cell. Its presence in the genome of the host may "transform" the cell and its offspring (cause them to become cancerous). In addition, the viral DNA may be used as a template for the synthesis of new viral RNA, and the virus thus multiplies itself and continues its infectious process, often without killing the cell. Because the viral genome is RNA, it cannot as such be integrated into the host's genome. Reverse transcriptase enables the virus to convert its genetic material to DNA, whereupon it is capable of inserting itself into the host chromosome. There are three reasons why reverse transcriptase is so important in cancer research. First, human leukemia sarcomas (different forms of cancer) have been shown to contain large RNA molecules similar to those of the tumor viruses that cause cancer in mice. Second, these human cancer cells contain particles with reverse transcriptase activity. Third, the DNA of some human cancer cells have virus-like sequences of nucleotides, sequences not found in the DNA of comparable normal cells. Since the action of RNA from tumor viruses depends on the presence of the enzyme reverse transcriptase, reseach into its chemical composition (amino acid' sequence) may bring better understanding of its function. The ultimate objective of such reseach would of course be the prevention of cancer, perhaps through the inhibition of reverse transcriptase activity.

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Question:

Metallic gold crystallizes in the face-centered cubic lattice. The length of the cubic unit cell, a, is 4.070 \AA. (a) What is the closest distance between gold atoms? (b) What is the density of gold?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E06-0224.htm

Solution:

An ordered array of atoms, ions, or molecules is called a lattice. Every lattice, is a three-dimensional stacking of identical building blocks called unit cells. The most symmetrical crystals have cubic lattices. There are three kinds of elementary cubic lattices: the simple cubic, the face- centered cubic, and the body-centered cubic. The length of the cube edge is designated by the symbol a. In a face-centered lattice, an atom would be located at each corner of the unit cell and at the center of each of the six faces. (a) In the gold crystal above, the closest distance from one corner atom to another corner atom is a, 4.070 \AA. The distance from any corner atom to an atom at the center of a face is one-half the diagonal of that face. Using the fact that the diagonal of any square is \surd2 × length of one side, the diagonal is then a\surd2. Half the diagonal is 1/2 a\surd2 or 1/2 (4.070 \AA) \surd2 = 2.878 \AA This distance is the closest distance between atoms, since this distance is shorter than the distance between adjacent corners. (b) The density is computed by first counting the number of gold atoms that can occupy one unit cell. This number is 1/8 times the number of occupied corners in the unit cell plus 1/2 times the number of occupied face- centers. Since a cube has 8 corners and 6 faces, the number is . 1/8 \textbullet 8 + 1/2 \textbullet 6 = 4 The volume of the cubic unit cell is a^3. The density if the mass of 4 gold atoms (4 atomic Wt / Avogadro's #) divided by the volume of the unit cell, a^3. Thus, the formula for the density of a gold crystal lattice is Density = {(4 atoms)(Atomic Wt. of gold)} / {a^3(Avogadro's number)} Substituting into the equation the Atomic Weight of gold, 197.0 g/mole, Avogadro's number, 6.023 × 10^23 atoms/mole, and converting 4.070 \AA to 4,070 × 10^-8 cm, then Density = {(4 atoms) (197.0 g/mole)} / {(4.070 × 10^-8 cm)^3 (6.023 × 10^23 atoms/mole)} = 19.4 g/cm^3.

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Question:

What is dormancy? What factors play a role in the termi-nation of dormancy and the germination of a seed?

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/Users/wenhuchen/Documents/Crawler/Biology/F08-0201.htm

Solution:

Dormancy is a special condition of rest which enables an embryo to survive long periods of un-favorable environmental conditions, such as water scarcity or low temperature. During this period of rest, the embryo ceases or limits its growth, and metabolizes at very low levels, although ordinary plant rest can be termi-nated and normal growth resumed by the onset of warmer temperatures or the availability of water. Germination or the resumption of normal growth by a dormant embryo requires certain, very precise combinations of environ-mental cues. This is of great survival importance to the plant in that it prevents the dormant seed from germinating in response to conditions such as a warm spell in winter, which, although apparently favorable, are only temporary. Should the seed germinate during such a period, the seedling would begin to grow, only to be killed by the ensuing cold and frost. Apparently the dormant seed contains certain endogenous inhibiting factors which must be overcome before germination can begin, and thus requires more complex triggers then just a favorable temperature or water supply. Some seeds are required to pass through a period of cold before they are able to germinate. This is seen in the seeds of almost all plants growing in areas with marked seasonal variations. Many seeds require drying before they can germinate. Such a requirement has adaptive significance in that it prevents germination within the moist fruit of the parent plant. Some seeds such as those of the lettuce, require a light exposure, while the germination of others is inhibited by light. Certain seeds cannot germinate, even in highly favorable conditions of water, light, oxygen, and temperature, until they have been abraded by factors such as soil action. This abrasion wears away the seed coat, allowing water and oxygen to enter the seed and signal the initia-tion of germination. The seeds of some desert species germinate only when sufficient rainfall has drained inhibitory chemicals from their seed coats. Most seeds "force" the embryo into dormancy by the fact that the seed coat is impermeable toundissolvedoxygen and the seed itself has a low water content. The seed is protected from freezing and desiccation by the seed coat and provided with nourishment by the endosperm. During germination, the embryo and endosperm absorb water and swell; then the seed coat ruptures, freeing the embryo and enabling it to resume development. It is important to note that most seeds do not need soil nutrients in order to germinate-they germinate equally well on moist paper as in soil. Dormancy plays a similar role in plant buds.

Question:

(a)Write a FORTRAN subroutine to find approximate solutionsto the initial-value problem: dy/dx = f(x,y), y(x_0) = y_0 using modified Euler's (predictor-corrector) method. (b)Write a FORTRAN program to find an approximate solution to the initial-value problem: dy/dx = g(y), y(x_0) = y_0 using modified Euler's method.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G24-0576.htm

Solution:

The modified Euler's method uses Euler's formula: y_n+1 = y_n + \Deltaxf(x ,y )(1) to obtain a predicted valuey_n+1,_p \cyrchar\cyro\pm y_i+1 and then iterates the following formula to obtain (k+1) corrected valuesyi+1,c: y^(k+1)_i+1,c = y_i + 1/2 [f(x_1 , y_1) + f(x_i+1,y^(k)_i+1)] (x_i+1 - x_i)(2) The iterative process is stopped as soon as y^(k+1)_i+1 = y^(k)_i+1to the required degree of accuracy. The following flowchart details the programming of this method : (a)The subroutine closely follows the flowchart, but the indices i and k have been omitted: SUBROUTINE EULER (N,XN,X,YCOR,ACCUR) REAL N CUSE A REAL VALUE FOR N TO AVOID MIXED CMODE ARITHMETIC DELTAX = (XN - X)/N 50Y = YCOR FOLD = F(X,Y) PRINT, X,Y X = X + DELTAX IF(X.GT.XN) STOP YPRED = Y + FOLD \textasteriskcentered DELTAX 100YCOR = Y + \O.5\textasteriskcentered (FOLD + F(X,YPRED))\textasteriskcenteredDELTAX YDIF = ABS(YCOR - YPRED) IF(YDIF.LE.ACCUR) GO TO 50 YPRED = YCOR GO TO 100 RETURN END (b)A subroutine to perform the loop that computes the corrected value, leaving 1/0 tasks to the main program will be used here. The advantage of this approach is the availability of each y(x). (An alternative approach would be to store the y's in an array, but for large N this might require a large storage area). The dis-advantage is the increased execution time required to pass arguments back and forth. The program looks as follows: CMAIN PROGRAM READ, N,XN,X,YCOR,ACCUR PRINT, X, XCOR REALN = FLOAT (N) DELTAX = (XN - X)/REALN DO 10 I = 1,N X = X + DELTAX CALL MEULER(X,YCOR,ACCUR,DELTAX) PRINT, X,YC0R 10CONTINUE STOP END SUBROUTINE MEULER(X,YCOR,ACCUR,DELTAX) Y = YCOR GOLD = G(Y) YPRED = Y + GOLD \textasteriskcentered DELTAX 100YCOR = Y + 0.5\textasteriskcentered(G0LD + G(YPRED))\textasteriskcenteredDELTAX YDIF a ABS(YCOR - YPRED) IF (YDIF.LE.ACCUR) GO TO 11 YPRED = YCOR GO TO 100 11RETURN END

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Question:

A solid cylinder 30 cm in diameter at the top of an incline 2.0 m high is released and rolls down the incline without loss of energy due to friction. Find the linear and angular speeds a the bottom.

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/Users/wenhuchen/Documents/Crawler/Physics/D04-0199.htm

Solution:

This problem can be solved using the conser-vation of energy principle. The cylinder initially at rest at the top of the incline has only gravitational (potential) energy. Taking the bottom of the incline as the zero level of the potential energy (see the figure above), we get E_P = mgh where m is the mass of the cylinder, g is the acceler-ation of gravity, and h = 2.0 m is its height above ground level. When the cylinder reaches the bottom of the incline, all of its energy will be kinetic: E_k = (1/2)mv^2 + (1//2) I\omega^2 where v is the cylinder's linear velocity, I its moment of inertia about the central axis, and \omega its angular momentum. (l = (1/2)mR^2 for cylinders, where R is the radius.) In the process of rolling down the incline the cylinder's potential energy turns to kinetic, the total change in each being equal to: ∆E_p = ∆E_k mgh= (1/2)mv^2 + (1/2) I\omega^2 = (1/2)mv^2 + (1/2)[(1/2) mR^2] [v/R]^2 = (1/2)mv^2 + (1/4)mv^2 = (3/4)mv^2 Using \omega = v/R Thus, gh = (3/4)v^2 v= [(2\surd3)/3](gh)^1/2 = 1.15 [(9.8 m/sec^2)(2.0 m)]^1/2 = 5.09 m/sec. Note that the linear speed does not depend upon the size or mass of the cylinder. To find \omega, we use the formula: w= v/R = (5.09 m/sec)/(0.15 m) = 34 rad/sec.

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Question:

Mesogleaand mesoderm are two terms used to describe the middle tissue layer of an organism. What distinction can be made between the two and in what organisms can each be found?

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/Users/wenhuchen/Documents/Crawler/Biology/F11-0281.htm

Solution:

Mesogleais the term used to describe the layer between the outer epidermis and the inner layer of cells lining thegastrovascularcavity. It is found in adult Coelenterates (Cnidarians) such as hydra,Obelia, and Portuguese man-of-war (Physalia). Themesogleaconsists of either a thin non-cellular membrane or thick, fibrous, jellylike,mucoidmaterial with or without cells. The term mesoderm is used to describe the middle of the three embryonic tissue layers first delineated during an early developmental stage of the embryo. This layer gives rise to the skeleton, circulatory system, muscles, excretory system, and most of the reproductive system. It is found in higher invertebrates, the insects, and all vertebrate groups. The mesoderm is always a cellular layer and always refers to an embryonic layer.Mesoglea, on the other hand, is usuallyacellularand merely denotes the middle layer of the body wall of an organism.

Question:

Predict the final hydrogenation products in each of the compounds in figure A.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E21-0769.htm

Solution:

Hydrogenation refers to the reduction of unsaturated hydrocarbons, by addition of the H_2 across the bond. Frequently a catalyst is used to increase the rate of the reaction. After complete hydrogenation, the product contains only single bonds. A triple bond requires two moles of hydrogen gas, while a double bond requires only 1 mole for complete hydrogenation. With this in mind, you have the reactions shown in figure B.

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Question:

A space vehicle ejects fuel at a velocity u relative to the vehicle. Its mass at some instant of time is m. Fuel is expelled at the constant rate ∆m/∆t . Set up and solve the equation of motion of the space vehicle, neglecting gravity.

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/Users/wenhuchen/Documents/Crawler/Physics/D11-0438.htm

Solution:

At any given time t, the momentum is mv. At time t + ∆t the mass of the rocket is m + ∆m where ∆m is negative since the total mass of the vehicle is de-creasing, while the velocity is v + ∆v and the momentum is (m + ∆m) (v + ∆v). However, after time ∆t, since ∆m in mass leaves the rocket, it introduces its own momentum. In the inertial frame of reference, the velocity of the fuel is v - u which is the velocity of the rocket minus the velocity of the fuel with respect to the rocket. The mass of the fuel ejected during time ∆t is ∆m, and so its momentum is ∆m(v - u). The law of conservation of momentum tells us that, in an isolated system the momentum at time t equals the momentum at t + ∆t. Therefore, mv = (m + ∆m) (v + ∆v) - ∆m (v - u) Simplifying, we have m∆v = -∆m (u + ∆v) If we let ∆v \rightarrow 0 to get instantaneous velocity and ∆m \rightarrow 0 we have mdv = - dm(u) dv = - dm(u/m) Integrating from v_0 to v (initial and instantane-ous velocities) and from M to m (where M is the initial mass and m is the instantaneous mass) we have: ^v\int_v0 dv = - ^m\int_M u(dm/m) Integrating, we have v = u In [1 + (m/M)] + v_0

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Question:

A flywheel has a mass of 30 kg and a radius of gyration of 2 m. If it turns at a rate of 2.4 rev/sec, what isits rotational kinetic energy?

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/Users/wenhuchen/Documents/Crawler/Physics/D05-0248.htm

Solution:

The rotational kinetic energy of a body about a given axis is T = (1/2) I \omega^2(1) where I and \omega are the rotational inertia and angular velocity of the body about the given axis. By definition of the radius of gyration, \rho, we may write m\rho^2 = I(2) where m is the body's mass. Using (2) in (1) T = (1/2) m \rho^2 \omega^2 Using the given data T = (1/2)(30 kg) (4 m^2) (2.4 rev/s)^2 T = 345.6 kg\bulletm^2\bulletrev^2/s^2 To put this answer in conventional energy units, note that 1 rev/s = 2\pirad/s whenceT = 1382.4\pi^2 Joules T = 13643.74 Joules

Question:

Find the mistakes in the following FORTRAN expressions.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G12-0273.htm

Solution:

a) In FORTRAN any constant or variable in an expression, unless it is the first constant or variable of that expression, must be preceded immediately by one of the following: a left parenthesis one of the operators +, -, {_\ast},/, or {_\ast}{_\ast} . Any constant or variable in an expression, unless it is the last constant or variable of that expression, must be followed by one of the following: a right parenthesis one of the operatorsor +, -, {_\ast},/, or {_\ast}{_\ast} . The number of opened and closed parenthesis must be equal. Therefore the correct expression is (A - (B - (C + D(4.7)))). b) For the same argument as was used in part (a), the correct expres-sions are (A/B) or (A)/(B). c) The Statement is incorrectly formed, and has no meaning. Except for the unary minus, each operator +, {_\ast}, /, or {_\ast}{_\ast} must have a term or factor both to its left and to its right, in order for an expression to be correctly formed.

Question:

A PL/I program uses the EDIT form of input. The declare statementof the program is as follows: DCL(A,B,C,D,E,F) FIXED(5,2) ; GCHAR(15)VAR, H CHAR(4) ; A data card used by the program has the following informationpunched on it (Note: The symbolbdenotes a blankspace). \downarrowcol.8\downarrowcol.34 AB-123 \textbullet 45b123 \textbullet 459123 \textbullet 456bb12345b18 \textbullet b 345912345B2GbCARS CARS CARS Show what will be stored in the memory locations correspond-ingto A, B, C, D, E, F, G and H when the followingstatement is executed: GETEDIT(A, B, C, D, E, F, G, H) [COL(10), F(7, 2), F(7, 2), X(2), F(4, 1), COL(34), F(5, 2), X(1), F(6, 3), F(1, 1), X(5), A(3), X(1), A(4)];

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G13-0324.htm

Solution:

As a result of the GET EDIT statement, the data card will be read in thefollowing manner: All entries on the data card up to col. 10 are ignored. Starting from col. 10, a numerical fieldF(7,2) is read off, for A. Thus, a value -123\textbullet45 is stored for A. This brings the computer to col. 17 ofthe card. Next, a numeric fieldF(7,2) for B is read off. Thus, a value 123\textbullet45 is stored for B. This brings the computer to col. 24 of the card. The nextinstruction is X(2), i.e., skip 2 columns. Hence the computer ignores the91, and goes to col. 26 of the card. Now, it must read a numeric field ofF(4,1) corresponding to C. Thus, a value 23\textbullet4 is read off, and the computercomes to col. 30 of the card. However, C is declared as FIXED(5,2). Therefore, the value 23\textbullet4 which is read from the card is storedas 023\textbullet40 in the memory location corresponding to C. The next 4 columns of the card are to be skipped be-cause the programmust jump to col. 34 of the card according to the next instruction. Starting from col. 34, a numeric field ofF(5,2) is to be read. Thus, 12345 are read off from the card and assigned to D. And, as D is specified aformat of F(5,2), therefore 12345 is stored for D as 123\textbullet45. The program comesto col. 39 Column 39 is skipped due toX(1), and the program reaches col. 40. The next instruction isF(6,3) corresponding to E. So, 1\textbullet8345 are read fromthe card and assigned to E. However, E is specified a format of F(6,3). The format specifications wants E to be of the form XX\textbulletXXX. In caseof a conflict between the format specification and the value read from thecard, the value read from the card gets pre-ference. Hence, E becomes 1\textbullet8345. Now, E is declared asFIXED(5,2). Thus, the memory al-located for E on the basis of the declare statement (DCL) expects E to be of the form XXX\textbulletXX. Hence, the value stored in memory for E is 001\textbullet83. Note that the extra two decimal places to the right of 3 are truncated. Next to be read is a fieldF(1,1) corresponding to F which is declaredas FIXED(5,2). Thus,F(1,1) reads off a single digit, 9, from the card. The value of F stored is 000\textbullet90. The next 5 card columns are skipped due toX(5). The next instructionspecifies A(3). This means a character field of 3 characters is tobe read off from the card cor-responding to G. Thus, B2G is stored for G. (2 is treated as a character). The next card column is to be skipped be- causeof X(1). Finally, a value CARS is read off for H, corresponding to thespecification A(4). The contents of the memory locations are shown below: Name of the memory location Content of the memory location A -123\textbullet45 B 123\textbullet45 C 023\textbullet40 D 123\textbullet45 E 011\textbullet83 F 000\textbullet90 G B2G H CARS

Question:

With reference to fluid pressures, explain the mechanism underlyingglomerularfiltration.

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/Users/wenhuchen/Documents/Crawler/Biology/F18-0445.htm

Solution:

The mean blood pressure in the large arteries of the body is approximately 100 mm Hg. However, as the blood passes through the arterioles connecting the renal artery to theglomeruli. the blood pressure decreases, so that the pressure in theglomerularcapillaries is usually only about half the mean arterial pressure or 50 mm Hg. This level is considerably higher than in other capillaries of the body, due to the fact that the arterioles leading to theglomeruliare wider than most other arterioles and therefore offer less resistance to flow. The force driving fluid out of theglomerulusand into Bowman's capsule is the pressure of the blood in theglomerularcapillaries. But this hydrostatic pressure favoring filtration is not completely unopposed. There is fluid within Bowman's cap-sule, resulting in a capsular hydrostatic pressure of about 10 mm Hg, which resists the flow into the capsule. The presence of protein in the plasma and its absence in Bowman's capsule, results in a second force opposing fil-tration. This difference in protein concentration between the capsule and the blood induces an osmotic flow of water into the blood. This osmotic pressure, also known as colloidal osmotic pressure, equals roughly 30 mm Hg. Summing the forces, there is aglomerularcapillary blood pressure of 50 mm Hg favoring filtration, a 10 mm Hg opposition to filtration due to fluid in Bowman's capsule, and a 30 mm Hg. pressure working against filtration due to a protein concentration difference between the blood and the capsule fluid, resulting in a netglomerularfiltration pressure of 10 mm Hg. This net pressure forces fluid from the blood into Bowman's capsule.

Question:

Liquid nitrogen is an excellent bath for keeping temperatures around 77\textdegreeK, its normal boiling point. What pressure would you need to maintain over the liquid nitrogen if you wanted to set the bath temperature at 85\textdegreeK? Heat of vaporization is about 5560 (J / mole).

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/Users/wenhuchen/Documents/Crawler/Chemistry/E07-0266.htm

Solution:

One uses the Clausius-Clapeyron equation to solve for the final pressure. This equation is written In (P_1)(P_2) = (∆H / R) [(1 / T_2) - (1 / T_1)] where P_1 is the initial pressure, P_2 is the final pressure at 85\textdegreeK, ∆H is the heat of vaporization, R is the gas constant, (8.314 J / mole-\textdegreeK), T_1 is the initial temperature and T_2 is the final temperature. The pressure of a liquid at its boiling point is equal to the atmospheric pressure, which is 1 atm. Solving for the final pressure: P_1 = 1 atmT_1 = 77\textdegreeK P_2 = ?T_2 = 85\textdegreeK ∆H = 5560(J / mole) R = 8.314 (J / mole \textdegreeK) 1n (1 / P_2) = [(5560 J / mole)] / [(8.314 J /mole)] [(1 / 85\textdegreeK) - (1 / 77\textdegreeK)] -1nP_2 = 6.687 × 10^2\textdegreeK (1.176 × 10-^2\textdegreeK) - (1.298) × (10^-2\textdegreeK) -1nP_2 = 6.687 × 10^2\textdegreeK (-1.22) × (10^-3 / \textdegreeK) -1nP2= -8.1587 × 10^-1 P_2 = 2.26 atm.

Question:

Show that the energy eigenvalues E_n, of a particle moving between impenetrable walls is given by the relationship E_n = (h^2/8mL^2)n^2 where h is Planck's constant, m the particle mass, L the distance between the walls, and n = 1, 2, 3, ...

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/Users/wenhuchen/Documents/Crawler/Physics/D33-1011.htm

Solution:

In order to find the energy eigenvalues, we must, in general, solve Schroedinger's Equation. Since our problem is one dimensional, this equation reduces to (-h^2/2m) (მ2\psi/მx^2) + V\psi = ih(მ\psi/მt)(1) where m is the mass of the particle we are studying i = \surd- 1,his Planck's constant over 2\pi, and V is the potential that m is subject to. \psi is the wave function which describes the motion of the particle and it is generally a function of x and t. In order to solve (1), we use the method of separation of variables, and express \psi as the product of 2 functions, one with only a time dependence, and the other with only a spatial depend-ence . Hence, \psi = \psi (x) \textphi (t)(2) Substituting (2) in (1) (-h^2/2m) (მ2\psi/მx^2) [\psi (x) \textphi (t)] + V \psi (x) \textphi (t) = ih(მ/მt) [\psi (x) \textphi (t)] (-h^2/2m) \textphi (t) {[d^2 \psi (x)]/dx^2} + V \textphi (t) \psi (x) = ih\psi (x) [{d \textphi (t)}/dt] Dividing both sides by \psi = \psi (x) \textphi (t) (-h^2/2m) [1/{\psi (x)}] [{d^2 \psi (x)}/dx^2] + V = ih[1/{\textphi (t)}] [{d \textphi (t)}/dt](3) Note that if v is a function of x alone, the left side of (3) is also a function of x alone, and the right side is a function only of t. The only way a function of x can equal a function of t is if both functions equal the same constant K. Hence, we separate (3) into (-h^2/2m) [1/{\psi (x)}] [{d^2 \psi (x)}/dx^2}] + V = K(4) ih[1/{\textphi (t)}] [{d \textphi (t)}/dt] = K(5) Remember that this is true only if V is time in-dependent. We first solve the differential equation in (5). Rewriting this equation [{d \textphi (t)}/dt] = (K/ih) \textphi (t) or[{d \textphi (t)}/{\textphi (t)}] = (K/ih) dt \int [{d \textphi (t)}/{\textphi (t)}] = (K/ih) \int dt In \textphi (t) = (K/ih) t + C where C is a constant. Taking the exponential of both sides \textphi (t) = e ^[(Kt/ih) + C]= e^c e^k ^t/ih(6) But e^c is a new constant. Letting e^c = A we may write (6) as \textphi (t) = A c^ Kt/ih= A e - ^i ^k ^t/h(7) Equation (7) may be written in terms of sines and cosines since e^-ix = cos x - i sin x Hence, \textphi (t) = A [cos Kt/h- i sin Kt/h] and \textphi (t) is an oscillatory function of time with fre-quency \omega_0 = K/h. The energy of particle is E = hv and\omega = (2\piE/h) = (E/h) But the only physical interpretation \omega_0 can have is that it is the particle's frequency. Hence \omega = \omega_0 or(E/h) = (E/h) andE = K.(8) Putting (8) in (5) [ih/{\textphi (t)}] [{d \textphi (t)}/dt] = E orih(d/dt) \textphi (t) = E \textphi (t)(9) Equation (9) is an eigenvalue equation, since an operator [(ihd)/dt] acts on a function [\textphi (t)] and gives back the function multiplied by a constant (E). The constant E, is called the eigenvalue of \textphi (t). Now that we have found K, we may rewrite (4) as (-h^2/2m) [1/{\psi (x)}] [{d^2 \psi (x)}/dx^2] + V = E or(-h^2/2m) [1/{\psi (x)}] [{d^2 \psi (x)}/dx^2] = E - V (+h^2/2m) [{d^2 \psi (x)}/dx^2] = (V - E) \psi (x)(10) The real problem is to solve (10) for the given V. A particle moving between impenetrable walls is a picturesque way of describing a particle in an infinite potential well. (See figure.) We may describe this V by \infty, x \leq 0 V =0, 0 \leq x \leq L \infty, x \geq L Hence, within the well, V = 0 and (10) becomes [{d^2 \psi (x)}/dx^2] + (2mE/h^2) \psi (x) = 0 This is the equation of simple harmonic motion, and has the general solution \psi (x) = B cos [\surd(2mE)/h] x + C sin [\surd(2mE)/h] x(11) In order to "taper" (11) to the problem at hand, note that \psi (0) = 0 \psi (L) = 0(12) since, if the potential is infinite at these points, we cannot expect to find the particle there. Using (12) in (11) \psi (0) = B = 0 \psi (L) = B cos [\surd(2mE)/h] L + C sin [\surd(2mE)/h] L = 0 HenceC sin [\surd(2mE)/h] L = 0 C cannot be 0, otherwise \psi (x) = 0 for all x, and this is a trivial solution. Therefore, it must be that sin [\surd(2mE)/h] L = 0 or[{L \surd(2mE)}/h] = n\pin = 1, 2, ...(13) Note that if n = 0 in (13), this implies that E = 0, always. Again, (11) would be a trivial solution for this case, since B = 0. Hence, E = [(n^2 \pi^2h^2)/(L^2 2m)] = [n^2 h^2/8mL]n = 1, 2, ... These are the energy eigenvalues of a particle in an infinite square well.

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Question:

A cylinder is rolled up an incline by a tape arranged as shown. What minimum force T is required if the angle \texttheta = 30\textdegree? The weight of the cylinder is 2 lb.

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/Users/wenhuchen/Documents/Crawler/Physics/D04-0205.htm

Solution:

The force, T, needed to pull the cylinder up the incline can be found by applying Newton's Second Law, F = ma, to the x and y directions of the cylinder's motion. Hence, using the figure, T - f - Mg sin \texttheta = ma_x(1) N - Mg cos \texttheta = ma_y(2) where f is the frictional force, N is the normal force of the plane, and a_x and a_y are the x and y directed accelerations of the cylinder. Since the cylinder does not leave the incline, a_y = 0, and, from (2) N = Mg cos \texttheta(3) Now, f is an unknown in (1) since we do not know the coefficient of rolling friction for the plane. In order to eliminate this unknown, we calculate the net torque, \cyrchar\cyrt^\ding{217}, on the cylinder. By definition, \cyrchar\cyrt^\ding{217} = r^\ding{217} × F^\ding{217} \cyrchar\cyrt= \vert\cyrchar\cyrt^\ding{217}\vert = rF sin \texttheta where \texttheta is the angle between r^\ding{217} and F^\ding{217}, and r^\ding{217} is a vector locating the point of application of F^\ding{217}. In our case, taking torques about the center of mass ButT=Rf + RT T=I\alpha where I and \alpha are the cylinder's moment of inertia and angular acceleration, respectively. Then I\alpha = fR + TR(4) Because the cylinder rotates without slipping v= wR(5) where v is the velocity of the cylinder's center of mass, and w is its angular velocity. Hence, differentiating (5) (dv/dt) = (dw/dt)R a = \alphaR Since v is directed along the x-axis, a is also, and a = a_x, whence a_x = \alphaR(6) Substituting (6) in (4) I(a_x/R) = (f + T)R f = (Ia_x/R^2 ) - T(7) Inserting (7) in (1) T - (Ia_x/R^2) + T - Mg sin\texttheta = ma_x 2T - Mg sin \texttheta = [m + (I/R^2)]a_x a_x = (2T - Mg sin \texttheta)/[m + (I/R^2)](8) The minimum applied force T will not accelerate the cylinder, but move it with constant velocity (a_x = 0). Hence, from (8) 2T= Mg sin \texttheta T= [(2 lb)(1/2)]/2 = 1/2 lb.

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Question:

What is the value of log 0.0148?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E33-0965.htm

Solution:

0.0148 = 1.48 × 10^\rule{1em}{1pt}2. The characteristic is the exponent of 10. Hence, the characteristic is \rule{1em}{1pt}2. The mantissa for 148 can be found in a table of logarithms. The mantissa is 0.1703. There-fore, log 0.0148 = \rule{1em}{1pt} 2 + 0.1703 = \rule{1em}{1pt} 2.0000 + 0.1703 = \rule{1em}{1pt}1.8297. Notice that the number 0.0148 is less than 1. Therefore, the value of log 0.0148 must be negative, as it was found to be.

Question:

A biologist deals with things on a microscopic level. To A biologist deals with things on a microscopic level. To describe cellular dimensions and the amount of materials present at the cellular level, units of an appropriately small size are needed. What are these units of measurements?

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/Users/wenhuchen/Documents/Crawler/Biology/F01-0001.htm

Solution:

In biology, the units of length commonly employed include the micron (abbreviated by \mu) and the \AAngstrom (abbreviated by \AA). A micron is equivalent to 10^-3 millimeters (mm). An \AAngstrom is equivalent to 10^-4 \mu or 10^-7 mm. Weights are expressed in milligrams (10^-3 grams) micrograms (10^-6 grams), and nanograms (10^-9 grams) . The unit of molecular weight employed is the dalton. A dalton is defined as the weight of a hydrogen atom. For example, one molecule of water (H_2O) weighs about 18 daltons. One dalton weighs 1.674 × 10^-24 grams.

Question:

A rigid rod whose own weight is negligible (see figure) is pivoted at point 0 and carries a body of weight w_1 at end A. Find the weight w_2 of a second body which must be attached at end B if the rod is to be in equilibrium, and find the force exerted on the rod by the pivot at 0.

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/Users/wenhuchen/Documents/Crawler/Physics/D02-0029.htm

Solution:

The question states that the rod is in equilibrium. In this case, the net force on the rod must be zero, and the net torque on the rod about the pivot must also be zero. The forces on the rod are T_1\ding{217} and T_2\ding{217}, the weights of masses 1 and 2, respectively, and P\ding{217}, the force of the pivot on the rod. Hence T_1 + T_2 - P = 0 orT_1 + T_2 = P(1) The torque about a point 0 is Շ\ding{217} = r × F where r\ding{217} is the vector from 0 locating the point of applica-tion of F\ding{217}. The net torque about 0 is T_2l_2 - T_1l_1 = 0 since the torque due to T_2 is opposite in direction to the torque due to T_1. Then T_2l_2 = T_1l_1(2) Substitution (2) in (1) T_1 + [(T_1l_1)/(l_2) = P P = T_1 [1+ (l_1-)/(l_2-)] But T_1 = w_1, and P = w_1 [1+ (l_1-)/(l_2-)] If l_2- = 3 ft, l_2= 4 ft and w_1 = 4 lb P= 4 lb[ 1+(3/4)]= 7 lb Furthermore, T_2 =[T_1l_1]/(l_2). Since T_2 = w_2, and T_1 = w_1 w_2 = (w_1l_1)/(l_2) = (4 lb)(3/4) = 3 lb

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Question:

A microscope has an objective lens of 10.0 mm focal length and an eyepiece of 25.0 mm focal length. What is the distance between the lenses, and what is the magnifica-tion if the object is in sharp focus when it is 10.5 mm from the objective?

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/Users/wenhuchen/Documents/Crawler/Physics/D28-0881.htm

Solution:

This is a compound microscope which uses converging lenses to produce large magnification. A short-focus lens, the objective, is placed near the object. It produces a real, inverted and magnified image. The eyepiece further magnifies this inverted image and produces a virtual image 250 mm from the eye-piece. This is the distance of most distinct vision. Considering the image produced by the objective, we use the optics equation 1/p + 1/q = 1/f p represents the distance of the object from the ob-jective lens, q the distance of image 1 from the lens, and f is the focal length. Substituting, (1/10.5mm) + 1/q = (1/10mm) q = 210 mm Therefore image 1 is 210 mm from the objective lens, and is real. The eyepiece magnifies image 1 and produces a second virtual image 250 mm from the lens so as to provide most distinct vision. Since image 2 is virtual, on the same side of the lens as image 1, the distance q' of image 2 from the eyepiece is negative. Using the optics equation again, 1/p'^ + 1/q'= 1/f ' 1/p'^ + 1/-250mm = 1/25.0mm p' = 22.7 mm Therefore the eyepiece is 22.7 mm from image 1. The distance between lenses is q + p' = 210 mm + 22.7 mm = 233 mm = 23.3 cm To find the magnification, first find the magnifica-tion produced by each lens using the equation M = q/p Magnification by objective: M_0 = (210 mm)/(10.5mm) = 20.0 Magnification by eyepiece: M_E = (-250mm)/(22.7mm) = -11.0 Total magnification: M = MeM0= -11.0 × 20.0 = -220

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Question:

Define and compare asexual reproduction with sexual reproduction in animals. Give examples.

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/Users/wenhuchen/Documents/Crawler/Biology/F22-0566.htm

Solution:

The fundamental difference between asexual reproduction and sexual reproduction is the number of parents involved in the production of offspring. In asexual reproduction only a single parent is needed. Offspring identical to the parent are produced when the parent splits, buds, or fragments. Asexual reproduction typically occurs in animals such as Euglena,Paramecium,Amoeba,Hydra, flatworms, and starfish. Two parents are usually involved in sexual repro-duction. A fertilized egg is produced through the union of specialized sex cells (egg and sperm) from each parent. Instead of having traits identical to a parent, the off-spring possess a variety of recombined traits inherited from both parents. In higher animals such as man, sexual reproduction is the most common form. It must be noted that in some cases, sexual repro-duction can take place with a single parent. In the hermaphroditic worms, such as the earthworm and the fluke, both male and female sex organs occur in the same individual. Whereas self-fertilization is avoided in the earthworm by certain adaptive mechanisms, it does take place in the parasitic fluke. This is an exception to the general observation that sexual repro-duction involves two parents.

Question:

The dew point (the temperature at which the water vapor in a given sample of air becomes saturated and condensation be-comes possible) of a mass of air at 15\textdegreeC is 10\textdegreeC. Calculate the relative humidity and the mass of 1 liter of the moist air if the barometric height is 76 cm of mercury. The gram- molecular weight of air is 28.9 g and the saturated aqueous vapor pressures at 10\textdegreeC and 15\textdegreeC are 9.2 mm of mercury and 12.8 mm of mercury, respectively.

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/Users/wenhuchen/Documents/Crawler/Physics/D14-0520.htm

Solution:

The relative humidity is defined as 100 × [(saturated vapor pressure at the dew point) / (saturated vapor pressure at the given temperature)] Therefore the relative humidity is RH = (9.2/12.8) × 100% = 71.9% Using the gas law, one can calculate the number of moles present independently for (a) the dry air and (b) the water vapor. Measuring the pressure p in atmospheres and the volume V in liters gives (a)n =pV/RT = (1atm× 1 liter)/(0.082litre-atm/mole - \textdegreeK × 288\textdegreeK) = 0.424 mole and (b)n' =p'V/RT = [(9.2 mm Hg/760 mm Hg/atm) × 1 liter]/[0.082 liter- atm/mole-\textdegreeK × 288\textdegreeK] = 0.00051 mole. The mass of moist air consists of dry air and water vapor. In one liter of moist air, there are 0.0424 moles of dry air and 0.00051 mole of water vapor, hence its mass is M = 0.0424 mole × 28.9 g/mole + 0.00051 mole × 18.0 g/mole = (1.2254 + 0.0092)g = 1.2346 g.

Question:

What is the resistance of a piece ofnichromewire 225 centimeters long with a cross-sectional area of 0.015 square centimeter?

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/Users/wenhuchen/Documents/Crawler/Physics/D20-0641.htm

Solution:

To solve this problem we use the relation R = \rho(L/A) Where R = resistanceL = wire length \rho = resistivityA = cross sectional area This basic relationship tells us that resistance is directly proportional to resistivity and length and in-versely proportional to cross sectional area. In the case of a wire this means that the resistance depends on the nature of the substance (which appears in the equation as the resistivity), that the resistance increases as the wire gets longer and decreases as the wire gets thicker. The resistivity (\rho) fornichromeis 100 × 10^-6 ohm-centimeter. The length is 225 centimeters, and the area is 0.015 square centimeter . ThenR = [(10^-4 ohm-cm × 225 cm)/(0.015 cm^2 )] = 1.5 ohms .

Question:

What properties of water make it an essential component of living matter?

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/Users/wenhuchen/Documents/Crawler/Biology/F01-0015.htm

Solution:

The chemistry of life on this planet is essentially the chemistry of water and carbon. Water is the most abundant molecule in the cell as well as on the earth. In fact, it makes up between 70 and 90 percent of the weight of most forms of life. Life began in the sea and the properties of water shape the chemistry of all living organisms. Life developed as a liquid phase phenomenon because reactions in solution are much more rapid than reactions between solids, and complex and highly structural molecules can behave in solution in a way that they cannot behave in a gas. Water is an excellent solvent for living systems. It can stay in the liquid stage throughout a very wide range of temperature variation. Almost all chemicals present in living matter are soluble in it. Water serves many functions in the living organism. It dissolves waste products of metabolism and assists in their removal from the cell and the organism. Water functions in heat regulation almost as an insulator would. It has a high heat capacity or specific heat in that it has a great capacity for absorbing heat with only minimal changes in its own temperature. This is because water molecules bond to one another by hydrogen bonds. Excess heat energy is dissipated by breaking these bonds, thus the living material is protected against sudden thermal changes. In addition, plants and animals utilize water loss to cool their bodies. When water changes from a liquid to a gas, it absorbs a great deal of heat. This enables the body to dissipate excess heat by the evapo-ration of water. In animals this process is sweating. Also, the good conductivity properties of water makes it possible for heat to be distributed evenly throughout the body tissues. Water serves as a lubricant, and is present in body fluids wherever one organ rubs against another, and in the joints where one bone moves on another. Water serves in the transport of nutrients and other materials within the organism. In plants, minerals dissolved in water are taken up by the roots and are transported up the stem to the leaves. Water is also very efficient in dissolving ionic salts and other polar compounds because of the polar physical properties of water molecules. The proper con-centration of these salts is necessary for life processes and it is important to keep them at extremely constant concentrations under normal conditions. These salts are important in maintaining osmotic relationships.

Question:

Describe locomotion in the amoeba.

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/Users/wenhuchen/Documents/Crawler/Biology/F11-0269.htm

Solution:

The locomotion of amoeba is considered to be the simplest type of animal locomotion. A moving amoeba sends out a projection, termed a pseudopodium. Following this, the organism advances as the inner, granular, gel--like endoplasm flows into the pseudopodium. Two or three pseudopodia may be formed simultaneously but ultimately one will become dominant for a time. As new pseudopodia are formed, the old ones withdraw into the general body region. In its locomotion the amoeba often changes its course in response to environmental stimulation, by forming a new dominant pseudopodium on the opposite side, thus moving in a very irregular fashion. Currently there is no fully complete explanation for the changes, both physical and chemical, which are involved in amoeboid movement. The theory accepted by zoologists at the present time is based on changes in the texture of the cytoplasm. As a result of some initial stimulus, ectoplasm, the outer clear, thin layer of the organism, undergoes a liquefaction and becomes endoplasm, which is gel-like. As a result of this change, internal pressure builds up and causes the endoplasm to flow out at this point, forming a pseudopodium. In the interior of the pseudopodium, the endoplasm flows forward along the line of progression; around the periphery, endoplasm is con-verted to ectoplasm, there by building up and extending the sides of the pseudopodium like a sleeve. At the posterior of the body, ectoplasm is assumed to be undergoing con-version to endoplasm. During this entire process, energy consumption is known to have taken place.

Question:

Verify that the fusion of four protons releases approximately 25 MeV for the proton-proton fusion cycle .

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/Users/wenhuchen/Documents/Crawler/Physics/D36-1061.htm

Solution:

The fusion of two hydrogen atoms produces one deuterium atom ^1_1H + ^1_1H^\ding{217} ^2_1H + e^+ + v_e .(1) The fusion of deuterium with hydrogen produces the helium atom ^1_1H + ^2_1H^\ding{217} ^3_2He + \Upsilon.(2) Finally, the fusion of helium with helium gives ^3_2He + ^3_2He^\ding{217} ^4_3He + ^1_1H + ^1_1H .(3) The chain of events (1), (2) and (3) is known as the proton-proton cycle, which is one of the schemes offered to describe the conversion of hydrogen into helium in stars. If we multiply equations (1) and (2) by two and sum the resulting equations and (3), we get (2^1_1H + 2^1_1H \ding{217} 2^2_1H +2e^+ + 2v_e) (2^1_1H + 2^2_1H \ding{217} 2^3_2He + 2\Upsilon) (3_2He + ^3_2He \ding{217} ^4_2He + ^1_1H + ^1_1H.) 6^1_1H + 2^2_1H+ 2^3_2He \ding{217}2^1_1H + 2^2_1H+ 2^3_2He + ^4_2He + 2e^+ + 2ve+ 2\Upsilon or4^1_1H \ding{217}^4_2He +2e^+ + 2ve+ 2\Upsilon Before the reaction the total nuclear mass is: 4 × (Atomic mass of hydrogen) = 4 × 1.0073 amu = 4.0292amu After reaction the total mass is Atomic mass of ^4He + 2 × (mass of positron) = (4.0015 + 2 × 0.0011) amu = 4.0026 amu The energy released in the fusion reaction is equivalent to the difference in mass: 4.0292 - 4.0026 = 0.0266 amu = {931 (MeV/amu)} (0.0266amu) = 25MeV.

Question:

The primary of a transformer, consisting of 20 turns, is connected to a varying voltage source which a volt-meter indicates to have a value of 110 volts. If the secondary circuit has 1000 turns and a resistance of 20,000 ohms, what voltage would a meter read if con-nected across it? What is the current in the primary if 100% efficiency is assumed?

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/Users/wenhuchen/Documents/Crawler/Physics/D23-0776.htm

Solution:

The voltmeter reads the root mean square (rms) value of the actual voltage. That is, if the actual instantaneous voltage is given by e = E_max sin \omegat where \omega is the angular frequency of the voltage variation, then E_rms = E = \surd[(1/2\pi) ^2\pi\int_0 E^2_max sin^2 \omegatd(\omegat)] If E_p and E_s are the rms values of the voltages on the primary (i.e. the side where the voltage source is located) and on the secondary sides, respectively, of the transformer, then according to the principle of the transformer (E_s/E_p) = (N_s/N_p) \therefore E_s = (N_s/N_p)E_p = (1000/20) (110) = 5500 volts The efficiency of a transformer is defined as E_ff = (Power out/Power in) × 100 If 100% efficiency is assumed, the power output is equal to the power input, where P = El. By Ohm's Law, I_s = (E_s/R_s) = (5500/20000) = .275 ampere \therefore P = E_s I_s = E_p I_p \therefore I_p = [(E_s I_s)/E_p] = (5500/110) (.275) = 50 (.275) = 13.75 amperes.

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Question:

What would be the period of rotation of the earth about its axis if its rotation speed increased until an object at the equator became weightless?

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/Users/wenhuchen/Documents/Crawler/Physics/D04-0146.htm

Solution:

The two forces acting on a body at the equator are the force exerted on it due to the gravitational attraction of the earth, mg\ding{217}_0, where g\ding{217}_0^ is the free-fall acceleration at the equator and acts toward the center, and the normal force exerted by the surface of the earth on the body, N\ding{217}. This latter force acts upward. On a non--rotating earth, or at the poles (which by definition do not rotate), these forces are equal since a body would be in equilibrium. At the equator, where a body does ex-perience rotation and therefore a centripetal accelera-tion, the forces are unequal so that their resultant provides the centripetal force necessary to keep the body traveling in a circle. Therefore, Newton's Second Law yields: mg_0 - N = mv^2/R where v is the speed of the body and R is the radius of the earth. But the distance traveled in one period of rotation, T, is 2\piR. Therefore T = 2\piR/v, and v^2 = 4\pi^2R^2/T^2. Substituting this into the first expression, we get N = mg_0 - mv^2/R = mg_0 - (4\pi^2mR/T^2 ) = m[g_0 - (4\pi^2R/T^2)] = mg, where g is the acceleration as measured at the earth's surface, and N is a measure of the apparent weight of the body, which is thus less than the gravitational force exerted on the body by the earth. If the speed of revolution of the earth increases, the body becomes weightless when the normal force exerted on it by the surface be-comes zero. Thus weightlessness occurs when N = 0 = m[g_0 - (4\pi^2R/T^2)] g_0 = 4\pir^2R/T^2 or when the period of rotation is T= 2\pi\surd(R/g_0) = 2\pi\surd[(4 × 10^3mi × 5280 ft/mi)/(32.4 ft/s^2)] = (2\pi/3600 s/hr) × \surd[(4 × 5280 × 10^3)/32.4]s = 1.41 hr.

Question:

What exactly do the terms `'primitive" and "advanced" mean inevolution? Are "generalized" organisms necessarilymore advancedthan "specialized" ones?

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/Users/wenhuchen/Documents/Crawler/Biology/F28-0727.htm

Solution:

"Primitive" and "advanced" are two contras-ting terms. "Primitive" meansolder, more like the ancestral condition. "Advanced" means newer, lesslike the ancestral form. "Primitive" and "advanced" are just two terms referringto relative sequence in evolutionary time, and do not imply that oneis necessarily superior to or more complex than the other. The terms "specialized" and "generalized" refer to the relationships oforganisms, or particular characteristics of organisms, to their environment. "Specialized" organisms or structures are adapted to a special, usually rather narrow, mode of life. "Generalized" organisms or structuresare broadly adapted to a greater variety of environments and waysof life. The terms "specialized" and "generalized" are sometimes confusedwith "advanced" and "primitive." Generalized characteristics are likelyto be more primitive, and specialized ones more advanced, but this isnot an absolute rule.

Question:

The gene for PKU is found in the heterozygous state in about one person in fifty. About how frequently would you expect babies to be born with PKU? Assume here that the homozygous state for PKU is either lethal or severely restricts the affected person's ability to produce offspring.

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/Users/wenhuchen/Documents/Crawler/Biology/F26-0692.htm

Solution:

The trait for PKU disease is recessive, and thus the genotype of babies born with the disease must be homozygous recessive. This genotype could be produced through the mating of two phenotypically normal people who are heterozygous for the PKU gene. The chance that two such heterozygotes will marry can be determined by using simple probability procedures. We know that the probability of two separate events occurring together is the product of their separate probabilities. If the heterozygous genotype occurs in one person out of fifty, then the chance that any two people who marry will both be carriers is 1/50 × 1/50 or 1/2,500 and only one marriage out of 2,500 will be capable of producing children with PKU. We must now determine the probability within such a marriage that a child will be born with PKU. Let us use the small letter p to represent the recessive PKU allele, and the capital letter P, to represent the dominant, normal allele. The cross between two Pp individuals is illustrated below: P The offspring are obtained in the ratio: 1/4 PP : 1/2 Pp : 1/4 pp We see that only one child in four will be homozygous recessive for the PKU allele, and consequently have the disease. Now, if only one of every four offspring produced has the PKU disease, and only one of every 2,500 marriages has the genetic makeup to produce a PKU child, then the frequency of children born with PKU would be [1/(2,000)] × [1/4] or 1/(10,000) Therefore, one out of every 10,000 children born will have PKU disease.

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Question:

When a block of wood of mass 1 kg is held in a vise, a bullet of mass 10 g fired into it penetrates to a depth of 10 cm. If the block is now suspended so that it can move freely and a second bullet is fired into it, to what depth will the bullet penetrate? (The retarding force in both cases is assumed constant.)

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/Users/wenhuchen/Documents/Crawler/Physics/D06-0331.htm

Solution:

The wood exerts a constant retarding force F^\ding{217} on the bullet. The work this force does on the bullet is equal to its change in kinetic energy in accordance with the work-energy theorem. This work is given by W = F^\ding{217}\bullet S^\ding{217}(1) where s^\ding{217} is the distance the bullet penetrates the block. Since the force acts in a direction opposite to the path of motion of the bullet, equation (1) reduces to W = - Fs In the first case, all of the bullet's kinetic energy is converted to heat due to the work done on it by the re-tarding force. If the distance the bullet penetrates the block is d_1, then the work done on it is W_1 = - Fd_1 = (1/2)mv2_1 - (1/2)mv^2 = 0 - (1/2) mv^2 = - (1/2) mv^2 Fd_1 = (1/2)mv^2(2) where v is the bullet's initial velocity and v_1 is the bullet's final velocity, which equals zero since the bullet comes to rest in the block. In the second case, the bullet's final velocity, v_2,is not zero. Even after it stops inside the block, it has a speed since the block is now free to move and the bullet moves with it. Since there are no external forces acting on the system, the principle of conservation of momentum can be applied. This is an inelastic collision where the block and bullet move together with the same final velocity, v_2.If the bullet's mass is represented by m and that of the block by M, then we have mv= (m + M)v_2(3) where v has the same significance as before. Further, the work done by the retarding force in stopping the bullet over a distance d_2, must equal the total change in kinetic energy of the system. W_2 = - Fd_2 = (1/2) (m + M)v^2 _2 - (1/2) mv^2(4) Solving for v_2 in equation (3) and substituting into equation (4) - Fd_2 = (1/2) (m + M) [(mv) / (m + M)]^2 - (1/2) mv^2 Fd_2 = (1/2) mv^2 - (1/2) [m2/ (m + M)] v^2(5) Dividing equation (5) by (2), We obtain Fd2/ Fd_1 = [(1/2) mv^2 - (1/2) {m^2/(m + M)} v^2] / [ (1/2) mv^2] (d2/ d_1) = [m - {m^2/(m + M)}] / m = [{(m^2 +mM- m^2)/(m + M)} / m] = M/(m + M) Substituting the known values, we have for the distance d_2 that the second bullet penetrates d_2 = (m d_1)/(m + M) = [(1kg)(0.1 m)] / [(0.01 kg + 1kg)] = (0.1)/(1.01) m = 0.099m = 9.9cm

Question:

Calculate the characteristicvibrationalfrequency of the lowest vibration state for H_2. Assume k = 5.1newtons/cm and h = 6.63 × 10^-34 J-sec.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E18-0702.htm

Solution:

Picture a diatomic molecule, such as H_2, as two balls connected by a spring. The spring represents the chemical bond. When the two balls (atoms) are pulled apart the string acts as a restoring force to bring them back together to the equilibrium separation. When they are com-pressed, the string tends to force them apart. Such a system, in which the restoring force is proportional to the amount of distortion, is called a harmonic oscillator. Thevibrationalfrequency of the spring going back and forth is given by the formula ѵ = (1/2) \pi \surd(K/M_eff) where ѵ = characteristicvibrationalfrequency of the particular bond, k = force constant andM_eff= effective mass for the vibrating atoms: M_eff= [(M_A M_B) / (M_A + M_B)] If the masses of the two atoms are equal,M_effbecomes (M^2 /2M) = (1/2)M. To find the characteristic vibration frequency, use this formula. To find theM_efffind the mass of one hydrogen atom. The number of grams in one mole (molecular weight) is one. There are 6.02 × 10^23 atoms of H in one mole. Thus, the mass of one atom is [(1 g/mole) / (6.02 × 10^23 /mole)] = 1.66 × 10^-24 g.Thus, M_eff= [(1.66 × 10^-24 g) / (2)] = 8.31 × 10^-25 g. Solving for ѵ : k = 5.1 [(newtons) / (cm)] = 5.1 [(kg M) / (cm \bullet sec^2)] ѵ = (1/2\pi) \surd[{(5.1 (kg M/cm sec^2) × 1000 (g/kg) × 100 (cm/m)} /(8.31 × 10^-25 g)] = 1.25 × 10^14 sec^-1. Note: 1000 g/kg and 100 cm/m are conversion factors used to obtain the correct units.

Question:

Consider a molecule suspended in a liquid in the test chamber of an ultracentrifuge. Suppose that the molecule lies 10 cm from the axis of rotation and that the ultracentrifuge rotates at 1000 revolutions per second (60,000 rpm). What is the magnitude of the acceleration associated with the circular motion?

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/Users/wenhuchen/Documents/Crawler/Physics/D04-0162.htm

Solution:

The angular velocity of the molecule is \omega = 2\pin where n is the number of revolutions per second that the molecule is executing. Or \omega = 2\pi × 1 × 10^3 (rev/sec) \cong 6 × 10^3 (rad/sec) The linear velocity is v = \omegar \cong [6 × 10^3 (rad/sec) ](10 cm) \cong 6 × 10^4 cm/sec The magnitude of the acceleration associated with circular motion is a = v^2/r = \omega^2r since the particle is going around in a circle. a \cong [6 × 10^3 (rad/sec)]^2 (10 cm) \cong 4 × 10^8(cm/sec^2) Now the acceleration g due to gravity is only 980 cm/s^2 at the surface of the earth , so that the ratio of the rotational acceleration to the gravitational acceleration is a/g \approx (4 × 10^8)/(10^3) \approx 4 × 10^5

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Question:

A capillary tube of length 50 cm is closed at both ends. It contains dry air at each end separated by a mercury column 10 cm long. With the tube horizontal, the air columns are both 20 cm long, but with the tube vertical the columns are 15 cm and 25 cm long. What is the pressure in the capillary tube when it is horizontal?

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/Users/wenhuchen/Documents/Crawler/Physics/D10-0412.htm

Solution:

When the mercury column is vertical, the pressure on the gas in the two parts is as shown in the diagram, where p_1 is the pressure at the foot of the mercury column and p_2 the pressure at the top. But the difference in pressure at two levels in a vertical column of liquid is known from the laws of hydrostatic pressure. Thus measuring distance from the top of the mercury column, p_2 = p_2 andp_1 = \rhog(h) + p_2 whencep_2 - p_1 = - \rhogh Here, \rho is the density of mercury. Applying Boyle's law to section C of the gas when the capillary is in its 2 positions, we obtain p_1Al_1 = p_0Al_0 Similarly, for section B p_2Al_2 = p_0Al_0 where p_0 and l_0 refer to conditions when the tube is horizontal. Thus p_1 = p_0 (l_0/l_1)andp_2 = p_0 (l_0/l_2). \thereforep_1 - p_2 = \rhogh = p_0l_0 [(1/l_1) - (1/l_2)] = \thereforep_1 - p_2 = \rhogh = p_0l_0 [(1/l_1) - (1/l_2)] = p_0l_0 [(l_2 - l_1)/(l_1l_2)] \thereforep_0 = [\rhogh l_1l_2]/[l_0(l_2 - l_1)] \therefore(p_0 / \rhog) = [(l_1l_2 ) / { l0(l2-l_1 )}] h = [ {15 cm × 25 cm} / {20 cm(25 - 15)cm} ] × 10 cm = 18.75 cm of mercury.

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Question:

Write a Basic program that allows a player to play Tic-Tac-Toe with a computer. Tic-Tac-Toe with a computer.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G23-0558.htm

Solution:

The game board is numbered

Question:

Write apseudocodedprogram to input a sequence of numbers(representing kilowatt-hours, KWH) and output electricbills according to the following schedule: Each of the first 100 KWHs costs 10 cents Each of the next 100 costs 6 cents Each of the next 200 costs 8 cents Each KWH over 400 costs 10 cents. The program should terminate when a negative number is inputted. Also, how could you improve this programto handlethe situation in which the rate schedule changed rapidly?

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G18-0448.htm

Solution:

The program logic is straightforward. A DO-WHILE construct is set up, so that each entry of kilowatt-hours is processed separately until a negativevalue is encoun-tered . realKWH, COST inputKWH (1) through KWH (N) dowhile KWH > 0 if0 \leq KWH < 100 thenCOST\leftarrow10.0\textasteriskcenteredKWH if100 \leq KWH < 200 thenCOST\leftarrow(10. 0\textasteriskcentered100.0) + (6. 0\textasteriskcentered (KWH-100.0)) if200 \leq WH \leq 400 thenCOST\leftarrow(10.0\textasteriskcentered100.0) + (6.0\textasteriskcentered100.0) + (8.0\textasteriskcentered(KWH-200.0)) elseCOST\leftarrow(10.0\textasteriskcentered100.0) + (6.0\textasteriskcentered100.0) + (8.0\textasteriskcentered200.0) + (10.0\textasteriskcentered(KWH-400.0)) outputKWH, COST enddo while endprogram If the rates of the electric bills climb in price often, the program wouldhave to be rewritten just as often. To avoid this nuisance, we could writesubprograms to do the job. After each IF-THEN construct, a statementcalling a particular subprogram with a certain billing schedule couldbe inserted. In this way, the main program is left untouched, and you wouldneed to rewrite only a few lines of code in a subprogram. This is veryoften the case in programming for business applications. An alternative way of handling the situation of rapid rate schedule changeswould be to insert 2 READ statements in the program, one for inputtingthe KWH segments, and another for inputting the price of each segment.

Question:

Write a program which counts the number of times each letter is used in input text.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G15-0387.htm

Solution:

1&TRTM = 1 2CHAR = LEN(1) .CH 3LETTERS = 'ABCDEFGHIJKLMNOPQRSTUVWXYZ' 4COUNT = TABLE (30) 5READOUTPUT = INPUT:F(DISPLAY) 6TEXT = OUTPUT 7NEXTTEXT CHAR =:F(READ) 8COUNT = COUNT + 1:(NEXT) 9DISPLAYOUTPUT = 10LOOPLETTERS CHAR =: F (END) 11OUTPUT = NE(COUNT) CH 'OCCURS' +COUNT 'TIMES'(LOOP) 12END The first statement includes a keyword '&TRIM'. Keywords in SNOBGL IV allow access to several parameters and switches internal to the SNOBOL IV system. Keywords begin with ampersand (&) followed by certain identifiers. For example, keyword &TRIM controls the handling of trailing blanks on input of data. If the value of &TRIM is nonzero, trailing blanks are deleted. Other keywords will be explained upon occurrence in the programs. The second statement, CHAR = LEN (1). CH, includes two important features. First, LEN (1), is an example of primitive (Built-in) functions. This function returns a pattern that matches any string of the length specified by the integer in parentheses. INPUT, PUNCH, and OUTPUT are some other primitive functions. The second important feature of the second statement is that it is an example of conditional value assignment (in this case .CH). The conditional value assignment operator (.) is separated from its operands by blanks. In the state-ment CHAR = LEN (1) . CH, variable CHAR is assigned the very first found string of length one, and, in case of success, assigns the same string to CH, which is used as an array subscript later in the program. The fourth statement, COUNT = TABLE (30), introduces the use of tables. Tables of variables can be created by using the primitive function TABLE, for example: T = TABLE ( ) In a very similar way arrays of variables are constructed by using the primitive function ARRAY, for example: A = ARRAY ( ) The arguments of an array, written in parentheses, describe the number of dimensions, the bounds of each dimension, and the initial value of each variable in the array. Thus A = ARRAY (10,1.0) creates and assigns to Aaone-dimensional array of 10 variables, each initialized to the real value 1.0. A table is similar to a one-dimensional array. However, instead of referencing an element with an integer, any data object can be used. In this program, letters of the alphabet are used as subscripts for table elements. The fifth statement of the program introduces the most important feature of the flow of control - a transfer to alabelledstatement. It is specified in the GOTO field which appears at the end of a statement and is separated from the rest of the statement by a colon. Two types of transfers can be specified in the GOTO field: conditional and unconditional, A conditional transfer consists of letters F or S corresponding to failure or success respectively, followed by a label enclosed within parentheses. Thus OUTPUT = INPUT: F(DISPLAY) makes the computerprint outwhatever appears on the data card. If, however, there is no data in the input file, i.e., an end of file is encountered, the transfer is made to the statement labeled display. An unconditional transfer is indicated by the absence of an F or S before the enclosing parentheses. For example, consider state-ment #8 of this program: COUNT = COUNT + 1: (NEXT) Each time this statement is executed, the transfer is made to the statementlabelledNEXT. Statement #11 uses one of the primitive functions of a specific group called NUMERICAL PREDICATES. The main ones include: LT - less than... LE - less than or equal to... EQ - equal to... NE - not equal to... GE - greater or equal to... GT - greater than... Statement OUTPUT = NE (COUNT).... prints out whatever is following the parentheses if the value of COUNT is not equal to zero. Finally, a statement that is longer than one line can be continued onto successive lines by starting the continuation lines with a period or plus sign. This program employs a plus sign for continua-tion of statement #11. Also note, that the statements in this program are numbered only for easy reference. All labels in SNOBCL IV start with a letter, as was mentioned earlier. If a statement does not start with a label, it must start with at least one blank. The general flow of this program is as follows: statement 1 - deletes trailing blanks; statement 2 - indicates that CHAR and CH will be assigned the first one-character string matched; statement 3 - assigns a string of the 26 letters of the alphabet to variable LETTERS; statement 4 - sets a 30-element table and assigns it to COUNT; statement 5 - prints out the first card of the input text; statement 6 - assigns it to variable TEXT; statement 7 - tests the text for presence of the first one-character string (which is the first letter of the first word in the text), deletes them all one at a time, adding 1 to COUNT in statement 8 each time around; statement 8 - goes back to statement 7 every time after adding 1 to COUNT; statement 7 - repeats the loop until all the appearances of each letter are counted, and the table COUNT is set up. Then COUNT fails when no more characters are left; the transfer is made back to statement #5; statement 5 - repeats the whole procedure for every following input card, adding missing members (if there are any) to the table COUNT. Finally it fails when there is no input left and transfers the execution to statement #9; statement 9 - prints a blank line; statement 10 - tests string LETTERS for the presence of a one- character string, finds 26 of them, assigns the first one (i.e., A) to CHAR and deletes it from LETTERS; statement 11 - since CH takes the same value as CHAR, looks in the table for C0UNT which represents the number of times letter A appeared in the text. If finds it, outputs the letter A and the number of its occurrences. In any case it goes back to statement #10; statement 10 - repeats the same procedure for every letter. It fails when no more letters are left and finishes the program.

Question:

Differentiate clearly between diffusion, dialysis and osmosis.

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/Users/wenhuchen/Documents/Crawler/Biology/F02-0058.htm

Solution:

Diffusion is the general term for the net movement of the particles ofa substance from a region where the substance is at a high concentrationto regions where the substance is at a low concentration. The particles are in constant random motion with their speed being directly relatedto their size and the temperature. When the movements of all the particlesare considered jointly, there is a net movement away from the regionof high concentration towards regions of low concentration. This resultsin the particles of a given substance distri-buting themselves with relativelyuniform density or con-centration within any available space. Diffusion tends to be faster in gases than in liquids, and much slower in solids. The movement or diffusion of water or solvent molecules through a semipermeablemembrane is called osmosis. The osmotic pressure of a solutionis a measure of the tendency of the solvent to enter the solu-tion byosmosis. The more concentrated a solution, the more water will tend to moveinto it, and the higher is its osmotic pressure. The diffusion of a dissolved substance through asemipermeable membraneis termed dialysis. Dialysis is the movement of the solute, while osmosisis the movement of the solvent through asemipermeable membrane. Dialysis and osmosis are just two special forms of diffusion.

Question:

Find the pH of a solution in which [H^+] = 6.38 × 10^-6 mole/liter.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E10-0342.htm

Solution:

pH indicates the acidity orbasicityof a solution. It runs on a scale of1 to 14. 1 is most acidic, 7isneutral, and 14 is most basic.pHis a measureof [H^+]. The higher the [H^+], the lower the pH, and the stronger theacid. To calculatepH usethe equation: pH= - log [H^+] = - log [6.38 × 10^-6] = - [(- 6) + (log 6.38)] = 6 + (- .805) pH= 5.195.

Question:

Find the pH of 0.15 M H_2SO_4, solution, assuming K_2 =1.26 × 10^-2.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E12-0425.htm

Solution:

Begin this problem by noting that pH = - log [H^+]. Thus, to find pH, calculate the [H^+] in 0.15 M H_2SO_4. H_2SO_4 is an acid and dissociates into H^+ (= H_3O^+) ions. The general reaction for acid dissociation can be written as HA + H_2O \rightleftarrows H_3O^+ + A^-. H_2SO_4 undergoes this reaction, as H_2SO_4 + H_2O \rightleftarrows H_3O^+ + HSO_4^-. But note, HSO_4^- is also an acid, it still possesses a hydrogen that can dissociate. Thus, one also has HSO_4^- + H_2O \rightleftarrows H_3O^+ + SO_4^-. In other words, there exist two dissociation reactions. As given, the dissociation constant for the second reaction, K_2, measures the ratio of the concentrations of products to reactants. It is from this expression that [H^+] can be determined. H_2SO_4 is a very strong acid in its first dissociation reaction. This means it is completely ionized; i.e. 100% of the H^+ comes off (the first H^+ only). Thus, if there are 0.15 M of H_2SO_4, then, there are also 0.15 = [H^+] for the first dissociation. [H^+] = [HSO_4^-], as can be seen from the first dissociation reaction and are formed inequimolaramounts. Also, [HSO_4^-] = 0.15 (initially). For the second ionization, one is given K_2 = 1.26 × 10^-2, indicating dissociation is not complete. From the prior explanation of what K_2 indicates, one can write K_2 = 1.26 × 10^-2 = {[H_3O^+] [SO_4-]} / [HSO_4^-]. Note: Water is not included since it is assumed to be a constant. From the first dissociation, it is known that [H_3O^+] = 0.15. If x moles/liter of H^+ is produced in the second dissociation, the total [H_3O^+] = 0.15 + x. [SO_4 =] = [H_3O^+] from the second dissociation, since here they are formed in equimolar amounts. Thus, [SO_4-] = x. If one started with [HSO_4^-] = 0.15M (from the first dissociation) and x moles/liter dissociate in the second, one has [HSO_4^-] = 0.15 - x left. Substituting these values, 1.26 × 10^-2 = {[H_3O^+] [SO_4-]} / [HSO_4^-] = {(0.15 + x) (x)} / (0.15 - x) Solving for x (using the quadratic formula), x = .011. Thus, [H^+] = 0.15 + 0.011 = 0.161. pH = - log [H^+], so that pH = - log [.161] = 0.79.

Question:

You have the reaction HCrO^-4+ 3Fe^2+ + 7H^+ \rightarrow Cr3++ 3Fe^3 + 4H_2O with the rate law expression:rate = k [HCrO^-_4] [Fe^2+] [H^+].Explain why the rate of reaction is not proportional to the number of each species reacting according to the stoichiometryof the balanced equation.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E13-0476.htm

Solution:

Stoichiometrypredicts only the amount of products expected; it cannot predict when they will be produced. The rate of production is dependent only on the number of molecules in the activated complex and their concentrations in the system. A reaction can occur only when the species come in contact with each other. Thus, the concentration reflects the probability of a collision among the species.Stoichiometrycannot provide any in-formation in these areas, and, as such, the rate of the reaction is not necessarily reflective of thestoichio-metryof the balanced equation.

Question:

A 3.60 x 10^4 -kg rocket rises vertically from rest. It ejects gas at an exhaust velocity of 1800 m/sec at a mass rate of 580 kg/sec for 40 sec before the fuel is expended. Determine the upward acceleration of the rocket at times t = 0,20, and 40 sec.

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/Users/wenhuchen/Documents/Crawler/Physics/D06-0333.htm

Solution:

Suppose a rocket is travelling with a velocity v^\ding{217} relative to a stationary coordinate system (S), and emits fuel at velocity u^\ding{217} with respect to the rocket (S') (see figure). The exhaust velocity with respect to S, w^\ding{217}, is the exhaust velocity with respect to S' plus the velocity of S' with respect to S or w^\ding{217} = u^\ding{217} + v^\ding{217}. Now that we know all the velocities with respect to a stationary frame, we may use the law of conservation of momentum in this frame to find the velocity of the rocket after a mass \Deltam has been emitted. Att = 0, no fuel has been emitted and the initial momentum of the fuel-rocket system is P_0^\ding{217} = Mv^\ding{217}. Att = \Deltat, a mass \DeltaM (where \DeltaM > 0) of fuel has been emitted and travels with velocity w^\ding{217}. The rocket now has mass M - \DeltaM and travels with a velocity v^\ding{217} + \Deltav^\ding{217}. The momentum is then P_f^\ding{217} = (M - \DeltaM)(v^\ding{217} + \Deltav^\ding{217}) + \DeltaM(w^\ding{217}) orP_f^\ding{217} = (M - \DeltaM)(v^\ding{217} + \Deltav^\ding{217}) + \DeltaM(u^\ding{217} + v^\ding{217}) Hence, the change in momentum is \DeltaP^\ding{217} = P_f^\ding{217} - P_0^\ding{217} = (M - \DeltaM) (v^\ding{217} + \Deltav^\ding{217}) + \DeltaM(u^\ding{217} + v^\ding{217}) - Mv^\ding{217} = Mv^\ding{217} - \DeltaMv^\ding{217} + M\Deltav^\ding{217} - \DeltaM\Deltav^\ding{217} + u^\ding{217}\DeltaM + v^\ding{217}\DeltaM - Mv^\ding{217} = M\Deltav^\ding{217} + u^\ding{217}\DeltaM - \DeltaM\Deltav^\ding{217} Then\DeltaP^\ding{217}/\Deltat = M (\Deltav^\ding{217}/\Deltat) + u^\ding{217} (\DeltaM/\Deltat) - (\DeltaM\Deltav^\ding{217})/(\Deltat) Taking the limit as \Deltat\rightarrow0 dP^\ding{217}/dt = lim_\Deltat\rightarrow0 \DeltaP^\ding{217}/\Deltat = M (dv^\ding{217}/dt) + u^\ding{217} (dM/dt) because \DeltaM∆v^\ding{217}/\Deltat\rightarrow0 as \Deltat\rightarrow0, whence dP^\ding{217}/dt = M (dv^\ding{217}/dt) + u^\ding{217} (dM/dt) But, for a constant mass system, which our's is, (see figure) the time rate of change of momentum of the system equals the net external force on the system and M (dv^\ding{217}/dt) + u^\ding{217} (dM/dt) = F^\ding{217}_ext(1) In this equation, dv^\ding{217}/dt is the rocket's acceleration, M is the instantaneous mass of the rocket, F^\ding{217}_ext is the netforce on the fuel-rocket system, and dM/dt is the rate of change of the rocket's mass. Note that in (1), dM/dt > 0 due to our derivation. Hence, we may replace dM/dt by- dM/dt if we redefine dM/dt to be less than zero. Then M (dv^\ding{217}/dt) - u^\ding{217} (dM/dt) = F^\ding{217}ext orM (dv^\ding{217}/dt) = orM (dv^\ding{217}/dt) = F^\ding{217}_ext + u^\ding{217} (dM/dt)(2) We define u^\ding{217} (dM/dt) as the rocket's thrust. Solving for (dv^\ding{217}/dt) dv^\ding{217}/dt = [(F^\ding{217}_ext)/M] + (u^\ding{217}/M) (dM/dt)(3) For our problem, F^\ding{217}_ext = - Mg\^{J}, where \^{J} is a unit vector in the positive y direction. Furthermore, if the rocket is propelled straight up, u^\ding{217} = - u\^{J}. Hence, (dv^\ding{217}/dt) = -g\^{J} - (u/M) \^{J} (dM/dt) = -\^{J} [g + (u/m) (dM/dt)](4) Note that since M is a function of time, dv^\ding{217}/dt will also be time dependent. Att = 0, M = 3.60 × 10^4 kg anddv^\ding{217}/dt = -\^{J} [9.8m/S^2 + {(1800 m/s)/(3.6 × 10^4 kg)} {(- 580 kg)/S}] dv^\ding{217}/dt = -\^{J} (9.8m/s^2 - 29m/s^2) dv^\ding{217}/dt = 19.2 m/s^2 \^{J} Att = 20 secs, M = 3.6 × 10^4 kg - (580 kg/s)(20s) = 3.6 × 10^3 kg - 11.6 × 10^3 kg = 24.4 × 10^3 kg anddv^\ding{217}/dt = -\^{J} [9.8 m/S^2 + {(1800 m/s)/(2.44 x 10^4)} {- (580 kg/S)} ] dv^\ding{217}/dt= -\^{J} (9.8m/s^2 - 42.79m/s^2) = 32.99 m/s2\^{J} Att = 40 secs, M = 3.6 × 10^4 kg - (580 kg/s) (40s) = 3.6 × 10^4 kg - 2.32 × 10^4 kg = 1.28 × 10^4 kg anddv^\ding{217}/dt = -\^{J} [9.8 m/S^2 + {(1800 m/s)/(1.28 x 104kg)} {- (580 kg/S)}] dv^\ding{217}/dt = - \^{J} (9.8 m/S^2 - 81.56 m/s^2) dv^\ding{217}/dt = 71.76 m/s^2 \^{J} In this example we have neglected air friction and the variation of g^\ding{217} with altitude.

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